8.1 PROBLEM KNOWN: Flowrate and temperature of water in fully developed flow through a tube of prescribed diameter. FIND: Maximum velocity and pressure gradient. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Isothermal flow. PROPERTIES: Table A-6, Water (300K): = 998 kg/m , = 855 10 ANALYSIS: From Eq. 8.6, Re D = & 4m 4 0.01 kg/s = = 596. D ( 0.025m ) 855 10-6 kg m/s
3 -6
Ns/m .
2
Hence the flow is laminar and the velocity profile is given by Eq. 8.15, u ( r) = 2 1 - ( r/ro )2 . um
The maximum velocity is therefore at r = 0, the centerline, where u ( 0 ) = 2 u m. From Eq. 8.5 um = hence u ( 0 ) = 0.041 m/s. Combining Eqs. 8.16 and 8.19, the pressure gradient is dp 64 u 2 m =- dx Re D 2D
2 3 dp 64 998 kg/m ( 0.020 m/s) =- = -0.86 kg/m 2 s 2 dx 596 2 0.025 m
& m D2 / 4
=
4 0.01 kg/s 998 kg/m3 ( 0.025m )2
= 0.020 m/s,
dp = -0.86N/m 2 m = -0.86 10 -5 bar/m. dx
<
PROBLEM 8.2 KNOWN: Temperature and mean velocity of water flow through a cast iron pipe of prescribed length and diameter. FIND: Pressure drop. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed flow, (3) Constant properties. PROPERTIES: Table A-6, Water (300K): = 997 kg/m , = 855 10 ANALYSIS: From Eq. 8.22, the pressure drop is p = f With Re D = u mD 997 kg/m3 0.2 m/s 0.15 m = = 3.50 10 4 -6 N s/m2 855 10
-4 3 -6
Ns/m .
2
u2 m L. 2D
the flow is turbulent and with e = 2.6 10 -3 10 and f 0.027. Hence,
m for cast iron (see Fig. 8.3), it follows that e/D = 1.73
997 kg/m 3 ( 0.2 m/s) 2 p = 0.027 ( 600m ) 2 0.15 m p = 2154 kg/s 2 m = 2154 N/m2 p = 0.0215 bar.
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COMMENTS: For the prescribed geometry, L/D = (600/0.15) = 4000 >> (xfd,h/D)turb 10, and the assumption of fully developed flow throughout the pipe is justified.
PROBLEM 8.3
KNOWN: Temperature and velocity of water flow in a pipe of prescribed dimensions. FIND: Pressure drop and pump power requirement for (a) a smooth pipe, (b) a cast iron pipe with a clean surface, and (c) smooth pipe for a range of mean velocities 0.05 to 1.5 m/s. SCHEMATIC:
ASSUMPTIONS: (1) Steady, fully developed flow. PROPERTIES: Table A.6, Water (300 K): = 997 kg/m3, = 855 10-6 Ns/m2, = / = 8.576 10-7 m2/s. ANALYSIS: From Eq. 8.22a and 8.22b, the pressure drop and pump power requirement are u2 mL P = pV = p D2 4 u m p = f
2D
(
)
(1,2)
The friction factor, f, may be determined from Figure 8.3 for different relative roughness, e/D, surfaces or from Eq. 8.21 for the smooth condition, 3000 ReD 5 106, -2 f = 0.790 ln ( ReD ) - 1.64 (3)
(
)
where the Reynolds number is
u D 1m s 0.25 m ReD = m = = 2.915 105 -7 m 2 s 8.576 10
(a) Smooth surface: from Eqs. (3), (1) and (2),
(4)
f = 0.790 ln 2.915 105 - 1.64
3 2
( ) = 0.01451 p = 0.01451 (997 kg m 1m s 2 0.25 m )1000 m = 2.89 10 kg s m = 0.289 bar P = 2.89 104 N m 2 ( 0.252 m 2 4 )1m s = 1418 N m s = 1.42 kW )
-2
2 4 2
(
< < <
(b) Cast iron clean surface: with e = 260 m, the relative roughness is e/D = 260 10-6 m/0.25 m = 1.04 10-3. From Figure 8.3 with ReD = 2.92 105, find f = 0.021. Hence, p = 0.419 bar P = 2.06 kW
(c) Smooth surface: Using IHT with the expressions of part (a), the pressure drop and pump power requirement as a function of mean velocity, um, for the range 0.05 um 1.5 m/s are computed and plotted below. Continued...
PROBLEM 8.3 (Cont.)
8 4 deltap (bar) or P (kW)
1 0.6 0.2 0.08 0.04
0.01 0 0.5 1 1.5
Mean velocity, um (m/s) Pressure drop, deltap (bar) Pump power, P (kW)
The pressure drop is a strong function of the mean velocity. So is the pump power since it is proportional to both p and the mean velocity. COMMENTS: (1) Note that L/D = 4000 >> (xfg,h/D) 10 for turbulent flow and the assumption of fully developed conditions is justified. (2) Surface fouling results in increased surface roughness and increases operating costs through increasing pump power requirements. (3) The IHT Workspace used to generate the graphical results follows.
// Pressure drop: deltap = f * rho * um^2 * L / ( 2 * D ) deltap_bar = deltap / 1.00e5 Power = deltap * ( pi * D^2 / 4 ) * um Power_kW = Power / 1000 // Reynolds number and friction factor: ReD = um * D / nu f = (0.790 * ln (ReD) - 1.64 ) ^ (-2) // Eq (1); Eq 8.22a // Conversion, Pa to bar units // Eq (2); Eq 8.22b // Useful for scaling graphical result
// Eq (3) // Eq (4); Eq 8.21, smooth surface condition
// Properties Tool - Water: // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); x=0 // Quality (0=sat liquid or 1=sat vapor) rho = rho_Tx("Water",Tm,x) // Density, kg/m^3 nu = nu_Tx("Water",Tm,x) // Kinematic viscosity, m^2/s // Assigned variables: um = 1 Tm = 300 D = 0.25 L = 1000
// Mean velocity, m/s // Mean temperature, K // Tube diameter, m // Tube length, m
PROBLEM 8.4 KNOWN: Temperature and mass flow rate of various liquids moving through a tube of prescribed diameter. FIND: Mean velocity and hydrodynamic and thermal entry lengths. SCHEMATIC:
ASSUMPTIONS: Constant properties. PROPERTIES: (T = 300K) Liquid Engine oil Mercury Water Table A-5 A-5 A-6 (kg/m3) 884 13,529 1000 (Ns/m2) (m2/s)
-6
Pr
0.486 550 10 6400 -2 -6 0.152 10 0.113 10 0.0248 -3 -6 0.855 10 0.855 10 5.83
ANALYSIS: The mean velocity is given by & m 0.03 kg/s 61.1 kg/s m 2 um = = = . Ac ( 0.025m )2 / 4 The hydrodynamic and thermal entry lengths depend on ReD , & 4m 4 0.03 kg/s 1.53 kg/s m Re D = . = = D ( 0.025m ) Hence, even for water ( = 0.855 10 8.3 and 8.23 it follows that xfd,h = 0.05 D Re D =
-3
Ns/m ), ReD < 2300 and the flow is laminar. From Eqs.
2
1.91 10-3 kg/s
(1.9110-3 kg/s ) Pr . xfd,t = 0.05 D ReD Pr =
Hence: Liquid Oil Mercury Water um(m/s) 0.069 0.0045 0.061 x fd,h(m) 0.0039 1.257 2.234 x fd,t(m) 25.2 0.031 13.02
COMMENTS: Note the effect of viscosity on the hydrodynamic entry length and the effect of Pr on the thermal entry length.
PROBLEM 8.5
KNOWN: Number, diameter and length of tubes and flow rate for an engine oil cooler. FIND: Pressure drop and pump power (a) for flow rate of 24 kg/s and (b) as a function of flow rate for the range 10 m 30 kg/s. SCHEMATIC:
ASSUMPTIONS: (1) Fully developed flow throughout the tubes. PROPERTIES: Table A.5, Engine oil (300 K): = 884 kg/m3, = 0.486 kg/sm. ANALYSIS: (a) Considering flow through a single tube, find 4 ( 24 kg s ) 4m = = 251.5 Re D = D 25 ( 0.010 m ) 0.486 kg s m Hence, the flow is laminar and from Equation 8.19, 64 64 f= = = 0.2545 . Re D 251.5 With m1 ( 25 / 25 ) kg s ( 4 ) um = = = 13.8 m s 2 2 3 884 kg m ( 0.010 m ) D 4
(1)
(2)
(
) (
)
(3)
Equation 8.22a yields
p = f
um
2D
2
(884 kg m )(13.8 m s ) L = 0.2545
3
2
2 ( 0.010 m )
2.5 m = 5.38 10 N m = 53.8 bar
6
2
(4)
< <
The pump power requirement from Equation 8.23b, m 24 kg s 5 (5) = 1.459 10 Nm/s = 146 W. P = p V = p = 5.38 106 N m 2 884 kg m3 (b) Using IHT with the expressions of part (a), the pressure drop and pump power requirement as a function of flow rate, m , for the range 10 m 30 kg/s are computed and plotted below.
70 250
60 Pumping power (kW) P res s u re dro p (b ar)
200
50
150
40
100
30
50
20 10 20 Flo w ra te (kg /s ) 30
0 10 20 Flow rate (kg/s) 30
Continued...
PROBLEM 8.5 (Cont.)
In the plot above, note that the pressure drop is linear with the flow rate since, from Eqs. (2), the friction factor is inversely dependent upon mean velocity. The pump power, however, is quadratic with the flow rate. COMMENTS: (1) If there is a hydrodynamic entry region, the average friction factor for the entire tube length would exceed the fully developed value, thereby increasing p and P. (2) The IHT Workspace used to generate the graphical results follows.
/* Results: base case, part (a) P_kW ReD deltap_bar mdot 145.9 251.5 53.75 24 */ f 0.2545 mu 0.486 rho 884.1 um 13.83 D 0.01 N 25
// Reynolds number and friction factor ReD = 4 * mdot1 / (pi * D * mu) // Reynolds number, Eq (1) f = 64 / ReD // Friction factor, laminar flow, Eq. 8.19, Eq. (2) // Average velocity and flow rate mdot1 = rho * Ac * um // Flow rate, kg/s; single tube mdot = mdot1 * N // Total flow rate, kg/s; N tubes Ac = pi * D^2 / 4 // Tube cross-sectional area, m^2 // Pressure drop and power deltap = f * rho * um^2 * L / (2 * D) deltap_bar = deltap * 1e-5 P = deltap * mdot / rho P_kW = P / 1000 // Input variables D = 0.01 mdot = 24 L = 2.5 N = 25 Tm = 300
// Pressure drop, N/m^2 // Pressure drop, bar // Power, W // Power, kW
// Diameter, m // Total flow rate, kg/s // Tube length, m // Number of tubes // Mean temperature of oil, K
// Engine Oil property functions : From Table A.5 rho = rho_T("Engine Oil",Tm) // Density, kg/m^3 mu = mu_T("Engine Oil",Tm) // Viscosity, Ns/m^2
PROBLEM 8.6
KNOWN: The x-momentum equation for fully developed laminar flow in a parallel-plate channel
dP d 2u = constant = 2 dx dy
FIND: Following the same approach as for the circular tube in Section 8.1: (a) Show that the velocity profile, u(y), is parabolic of the form
3 y2 u ( y ) = um 1 - 2 ( a / 2 )2
where um is the mean velocity expressed as
um =
a 2 dP - 12 dx
and -dp/dx = p/L where p is the pressure drop across the channel of length L; (b) Write the expression defining the friction factor, f, using the hydraulic diameter as the characteristic length, Dh; What is the hydraulic diameter for the parallel-plate channel? (c) The friction factor is estimated from the expression f = C ReDh where C depends upon the flow cross-section as shown in Table 8.1; What is the coefficient C for the parallel-plate channel (b/a ) ? (d) Calculate the mean air velocity and the Reynolds number for air at atmospheric pressure and 300 K in a parallel-plate channel 2 with separation of 5 mm and length of 100 mm subjected to a pressure drop of P = 3.75 N/m ; Is the assumption of fully developed flow reasonable for this application? If not, what effect does this have on the estimate for um? SCHEMATIC:
ASSUMPTIONS: (1) Fully developed laminar flow, (2) Parallel-plate channel, a << b. PROPERTIES: Table A-4, Air (300 K, 1 atm): = 184.6 10 Ns/m , = 15.89 10 m /s. ANALYSIS: (a) The x-momentum equation for fully developed laminar flow is
-7 2 -6 2
d 2u dp = = constant dy2 dx
Since the longitudinal pressure gradient is constant, separate variables and integrate twice,
(1)
d du 1 dp = dy dy dx u= 1 dp 2 y + C1 y + C2 2 dx
du 1 dp = y + C1 dy dx
Continued .....
PROBLEM 8.6 (Cont.)
The integration constants are determined from the boundary conditions,
du =0 dy y = 0
to find
u (a / 2 ) = 0
C1 = 0
giving
C2 = -
1 dp 2 (a / 2 ) 2 dx
u (y) = -
(a / 2 )2 dp 1 -
2 dx
2 (a / 2 ) y2
a/2
(2)
The mean velocity is
um =
2 a 0
a/2
u (y)
2 2 (a / 2 ) dp y - dy = -
a
2
dx
y3 / 3
(a / 2 )
2
0
(3)
a 2 dp um = - 12 dx 3 y2 u m 1 - 2 ( a / 2 )2
Substituting Eq. (3) for dp/dx into Eq. (2) find the velocity distribution in terms of the mean velocity
u (y) =
<
(4)
(b) The friction factor follows from its definition, Eq. 8.16,
f=
- ( dp / dx ) Dh
u2 / 2 m
4 Ac 4 (a b ) = = 2a P 2 (a + b )
(5)
where the hydraulic diameter for the channel using Eq. 8.67 is
Dh =
since a << b.
<
(6)
(c) Substituting for the pressure gradient, Eq. (3), and rearranging, find using Eq. (6),
f=
um
Dh
2 a 2 /12 u m / 2
=
96 96 = u m D h / Re Dh
<
(7)
where the Reynolds number is
ReDh = u m Dh /
Continued .....
(8)
PROBLEM 8.6 (Cont.)
This result is in agreement with Table 8.1 for the cross-section with b/a where C = 96.
<
(0.005 m )2
(d) For the conditions shown in the schematic, with air properties evaluated at 300 K, using Eqs. (3) and (8), find
3.75 N / m 2 = 1.06 m / s um = 12 184.6 10-7 N s / m 2 0.100 m ReD = 1.06 m / s 2 0.005 m 15.89 10-6 m 2 / s
h
= 667
The flow is laminar as Re D < 2300, and from Eq. 8.3, the entry length is
x fd,h = 0.05 ReDh D h am x fd,h = 2 0.005 m 0.05 667 = 0.334 m = 334 mm
We conclude that the flow is not fully developed, and the friction factor in the entry region will be higher than for fully developed conditions. Hence, for the same pressure drop, the mean velocity will be less than our estimate.
PROBLEM 8.7 KNOWN: Mean velocity and temperature of oil, water and mercury flowing through a tube of prescribed diameter. FIND: Corresponding hydrodynamic and thermal entry lengths. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties. PROPERTIES: (Tm = 300K) Liquid Engine Oil Mercury Water ANALYSIS: With Re D = It follows that ReD xfd,h = 0.05 Red Hence: Liquid Oil Mercury Water x fd,h(m) 2.84 10 1.39 0.183
-4 3 2
Table A-5 A-5 A-6
(kg/m ) 884 13,529 997
(Ns/m ) 0.486 -2 0.152 10 -3 0.855 10 6400
Pr 0.0248 5.83
u mD = 5 10 -3 m/s 0.025m =1.25 10 -4 m 2 / s
Oil 0.227
Mercury 1113
Water 146
Hence for each fluid, the flow is laminar and from Eqs. 8.3 and 8.23, xfd,t = 0.5 D ReD Pr.
x fd,t(m) 1.82 0.0345 1.06
<
COMMENTS: Note the effect of viscosity on the hydrodynamic entry length and the effect of Prandtl number on the thermal entry length.
PROBLEM 8.8
KNOWN: Velocity and temperature profiles for laminar flow in a tube of radius ro = 10 mm. FIND: Mean (or bulk) temperature, Tm, at this axial position. SCHEMATIC:
ASSUMPTIONS: (1) Laminar incompressible flow, (2) Constant properties. ANALYSIS: The prescribed velocity and temperature profiles, (m/s and K, respectively) are u(r) = 0.1 [1-(r/ro)2] T(r) = 344.8 + 75.0 (r/ro)2 - 18.8 (r/ro)4 (1,2)
For incompressible flow with constant cv in a circular tube, from Eq. 8.27, the mean temperature and um, the mean velocity, from Eq. 8.8 are, respectively,
Tm =
2
um o
o 0 u ( r ) T ( r ) r dr r2
r
um =
2 ro 0
2 ro
u ( r ) r dr
(3,4)
Substituting the velocity profile, Eq. (1), into Eq. (4) and integrating, find um = 2 2 1 1 2 2 1 4 ro 0.1 1 - ( r ro ) ( r ro ) d ( r ro ) = 2 0.1 ( r ro ) - ( r ro ) = 0.05 m / s 2 0 4 0 2 ro
1
Substituting the profiles and um into Eq. (3), find Tm =
Tm = 4 1 2 r2 0.1 1 - ( r ro ) 2 o 0 (0.05 m s ) ro
2
{
}{344.8 + 75.0 (r r ) - 18.8 (r r ) } (r r ) d (r r )
o 2 o 4 o o
3 5 7 o o o o
0
1
{344.8 (r r ) + 75.0 (r r ) - 18.8 (r r ) - 344.8 (r r ) + 75.0 (r r ) - 18.8 (r r ) }d (r r )
3 5 o o o
Tm = 4 {[172.40 + 18.75 - 3.13] - [86.20 + 12.50 - 2.35]} = 367 K
<
The velocity and temperature profiles appear as shown below. Do the values of um and Tm found above compare with their respective profiles as you thought? Is the fluid being heated or cooled?
0.1 0.08 0.06 0.04 0.02 0 0 0.2 0.4 0.6 0.8 1 Temperature, T(r) (K) Velocity, u(r) (m/s) 440 420 400 380 360 340 0 0.2 0.4 0.6 0.8 1
Radial coordinate, r/ro
Radial coordinate, r/ro
PROBLEM 8.9
KNOWN: Velocity and temperature profiles for laminar flow in a parallel plate channel. FIND: Mean velocity, um, and mean (or bulk) temperature, Tm, at this axial position. Plot the velocity and temperature distributions. Comment on whether values of um and Tm appear reasonable. SCHEMATIC:
ASSUMPTIONS: (1) Laminar incompressible flow, (2) Constant properties. ANALYSIS: The prescribed velocity and temperature profiles (m/s and C, respectively) are 2 u ( y ) = 0.75 1 - ( y yo )
T ( y ) = 5.0 + 95.66 ( y yo ) - 47.83 ( y yo ) (1,2)
2 4
u ( y ) dAc Ac where the flow cross-sectional area is dAc = 1dy, and Ac = 2yo, 1 1 +y um = u ( y ) dy = u ( y )dy Ac Ac 2yo - yo m = Ac u m =
The mean velocity, um, follows from its definition, Eq. 8.7,
(3)
um =
+1 1 2 yo 0.75 1 - ( y yo ) d ( y yo ) -1 2yo
3 u m = 1 2 0.75 ( y yo ) - 1 3 ( y yo ) -1
{
}
+1
u m = 1 2 0.75 {[1 - 1 3] - [-1 + 1 3]} = 1 2 0.75 4 3 = 2 3 0.75 = 0.50 m s
The mean temperature, Tm, follows from its definition, Eq. 8.25, E t = mc v Tm m = Ac u m where
<
u ( y ) T ( y ) dA c Ac Hence, substituting velocity and temperature profiles, + yo 1 Tm = u ( y ) T ( y ) dy u m A c - yo
A c u mc vTm = c v
(4)
2 4 o ) - 47.83 ( y y o ) d ( y y o )
+1 -1
Tm =
0.75
1
(0.5 m s ) 2y o
o
yo
-1 {0.75 1 - ( y
+1
3 o
yo )
2
}{5.0 + 95.66 ( y y
3 o
}
Tm =
{5 ( y y ) + 31.89 ( y y ) - 9.57 ( y y ) - 1.67 ( y y ) + 19.13 ( y y ) - 6.83 ( y y ) } 0.5 2
5 5 7 o o o
Tm =
{[27.32 - 13.97] - [-27.32 - (-13.97 )]} = 20.0 C 0.5 2
0.75
<
Continued...
PROBLEM 8.9 (Cont.)
The velocity and temperature profiles along with the um and Tm values are plotted below.
Velocity, u(y) (m/s) 1 0.8 0.6 0.4 0.2 0 -1 -0.6 -0.2 0.2 0.6 1 Temperature, T(y) (C) 60 50 40 30 20 10 0 -1 -0.6 -0.2 0.2 0.6 1
Dimensionless coordinate, x/xo Velocity profile, u(y) Mean velocity, um = 0.5 m/s
Dimensionless coordinate, y/yo Temperature profile, T(y) Mean temperature, Tm = 20 C
For the velocity profile, the mean velocity is 2/3 that of the centerline velocity, um = 2u(0)/3. Note that the areas above and below the um line appear to be equal. Considering the temperature profile, we'd expect the mean temperature to be closer to the centerline temperature since the velocity profile weights the integral toward the centerline. COMMENTS: The integrations required to obtain um and Tm, Eqs. (3) and (4), could also be performed using the intrinsic function INTEGRAL (y,x) in the IHT Workspace.
PROBLEM 8.10 KNOWN: Flow rate, inlet temperature and pressure, and outlet pressure of water flowing through a pipe with a prescribed surface heat rate. FIND: (a) Outlet temperature, (b) Outlet temperature assuming negligible flow work changes. SCHEMATIC:
ASSUMPTIONS: (1) Negligible kinetic and potential energy changes, (2) Constant properties, (3) Incompressible liquid. PROPERTIES: Table A-6, Water (T = 300K): = 997 kg/m , cp = cv = 4179 J/kgK. ANALYSIS: (a) Accounting for the flow work effect, Eq. 8.35 may be integrated from inlet to outlet to obtain & q conv = m c v Tm,o - Tm,i + ( pv )o - ( pv )i Hence, q 1 Tm,o = Tm,i + conv + ( pi - po ) & mc v cv
3
(
)
(100 - 2 ) bar 105 N/m2 /bar 105 W Tm,o = 25o C + + 2 kg/s 4179 J/kg K 997 kg/m 3 4179 J/kg K
Tm,o = 25o C+ 12o C + 2.4o C Tm,o = 39.4o C. (b) Neglecting the flow work effect, it follows from Eq. 8.37 that, q Tm,o = Tm,i + conv = 25o C + 12oC & mc p Tm,o = 37 o C.
(
)
<
<
COMMENTS: Even for the large pressure drop of this problem, flow work effects make a small contribution to heating the water. The effects may justifiably be neglected in most practical problems.
PROBLEM 8.11
KNOWN: Internal flow with prescribed wall heat flux as a function of distance. FIND: (a) Beginning with a properly defined differential control volume, the temperature distribution, Tm(x), (b) Outlet temperature, Tm,o, (c) Sketch Tm(x), and Ts(x) for fully developed and developing flow conditions, and (d) Value of uniform wall flux q s (instead of q T = ax) providing same outlet temperature TT s as found in part (a); sketch Tm(x) and Ts(x) for this heating condition. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Incompressible flow. PROPERTIES: Table A.6, Water (300 K): cp = 4.179 kJ/kgK. ANALYSIS: (a) Applying energy conservation to the control volume above,
dq conv = mcp dTm
where Tm(x) is the mean temperature at any cross-section and dqconv = q dx . Hence,
(1)
ax = mcp a
dTm . dx
(2)
Separating and integrating with proper limits gives x T (x ) xdx = mcp m dTm x =0 Tm,i
Tm ( x ) = Tm,i +
ax 2 2mcp
(3,4)
<
(b) To find the outlet temperature, let x = L, then
Tm ( L ) = Tm,o = Tm,i + aL2 2mcp .
Solving for Tm,o and substituting numerical values, find
(5)
Tm,o = 27 C +
20 W m 2 30 m 2
2 ( 450 kg h (3600s h )) 4179 J kg K
(
)
= 27 C + 17.2 C = 44.2 C .
<
(c) For linear wall heating, q s ax , the fluid temperature distribution along the length of the tube is T quadratic as prescribed by Eq. (4). From the convection rate equation,
q = h ( x ) D ( Ts ( x ) - Tm ( x )) s
(6)
For fully developed flow conditions, h(x) = h is a constant; hence, Ts(x) - Tm(x) increases linearly with x. For developing conditions, h(x) will decrease with increasing distance along the tube eventually achieving the fully developed value. Continued...
PROBLEM 8.11 (Cont.)
(d) For uniform wall heat flux heating, the overall energy balance on the tube yields
q = q DL = mcp Tm,o - Tm,i s
(
)
= 95.3 / D W m 2
Requiring that Tm,o = 44.2C from part (a), find
q = s
( 450 3600 ) kg
s 4179 J kg K ( 44.2 - 27 ) K
D 30 m
<
where D is the diameter (m) of the tube which, when specified, would permit determining the required heat flux, q . For uniform heating, Section 3.3.2, we know that Tm(x) will be linear with distance. Ts(x) s will also be linear for fully developed conditions and appear as shown below when the flow is developing.
COMMENTS: (1) Note that cp should be evaluated at Tm = (27 + 44)C/2 = 309 K. (2) Why did we show Ts(0) = Tm(0) for both types of history when the flow was developing? (3) Why must Tm(x) be linear with distance in the case of uniform wall flux heating?
PROBLEM 8.12 KNOWN: Internal flow with constant surface heat flux, q . s FIND: (a) Qualitative temperature distributions, T(x), under developing and fully-developed flow, (b) Exit mean temperature for both situations. SCHEMATIC:
ASSUMPTIONS: (a) Steady-state conditions, (b) Constant properties, (c) Incompressible flow. ANALYSIS: Based upon the analysis leading to Eq. 8.40, note for the case of constant surface heat flux conditions, dTm = constant. dx Hence, regardless of whether the hydrodynamic or thermal boundary layer is fully developed, it follows that Tm ( x ) Tm,2 is linear and will be the same for all flow conditions.
<
The surface heat flux can also be written, using Eq. 8.28, as q = h Ts( x ) - T m ( x ) . s Under fully-developed flow and thermal conditions, h = hfd is a constant. When flow is developing h > hfd. Hence, the temperature distributions appear as below.
<
PROBLEM 8.13 KNOWN: Geometry and coolant flow conditions associated with a nuclear fuel rod. Axial variation of heat generation within the rod. FIND: (a) Axial variation of local heat flux and total heat transfer rate, (b) Axial variation of mean coolant temperature, (c) Axial variation of rod surface temperature and location of maximum temperature. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant fluid properties, (3) Uniform surface convection coefficient, (4) Negligible axial conduction in rod and fluid, (5) Negligible kinetic energy, potential energy and flow work changes, (6) Outer surface is adiabatic. ANALYSIS: (a) Performing an energy balance for a control volume about the rod, & & & Ein - Eout + Eg = 0 or & - dq + Eg = 0
& - q ( D dx ) + qo sin ( x/L ) D2 / 4 dx=0
(
)
& q = q o ( D/4 ) sin ( x/L ) .
<
The total heat transfer rate is then q= q= q=
L 0
& q D dx = D2 / 4 qo
(
)
L 0
sin ( x/L ) dx
& D2 L x L D 2qo L & qo - cos (1 + 1) 0 = 4 L 4 D2L & q o. 2 (1)
<
(b) Performing an energy balance for a control volume about the coolant, & & m c p Tm + dq = m c p ( Tm + dTm ) = 0. Hence & m c p d Tm = dq = ( D dx ) q & d Tm D q o D x = sin . & dx m cp 4 L Continued .....
PROBLEM 8.13 (Cont.) Integrating,
x & D2 q o x Tm ( x ) - Tm,i = sin dx & 4 m cp 0 L
Tm ( x ) = Tm,i + (2)
& L D2 qo x 1 - cos L & 4 m cp
<
q = h ( Ts - T m ).
(c) From Newton's law of cooling,
Hence Ts = q + Tm h
& & qo D x LD2 qo x Ts = sin + Tm,i + 1 - cos L . & 4h L 4 m cp To determine the location of the maximum surface temperature, evaluate & & d Ts qo D x LD2 q o x =0= cos + sin & dx 4hL L 4 m cp L L or 1 x D x cos + sin = 0. & hL L m cp L Hence tan x= & m cp x =- L DhL L tan -1 & m cp - = xmax . D h L
<
<
COMMENTS: Note from Eq. (2) that Tm,o = Tm ( L) = Tm,i + & L D2q o & 2 m cp
which is equivalent to the result obtained by combining Eq. (1) and Eq. 8.37.
PROBLEM 8.14 KNOWN: Axial variation of surface heat flux for flow through a tube. FIND: Axial variation of fluid and surface temperatures. SCHEMATIC:
ASSUMPTIONS: (1) Convection coefficient is independent of x, (2) Negligible axial conduction and kinetic and potential energy changes, (3) Fluid is an ideal gas or a liquid for which d(pv) < < d(cv Tm). ANALYSIS: Since Equation 8.38 is applicable, dTm q P ( D ) q sin ( x/L ) s,m = s = & & dx m cp m cp Separating variables and integrating from x = 0
Tm,o Tm,i
dTm =
Dq x s,m x sin dx & 0 m cp L LDq x s,m cos & m cp L
x 0
Tm ( x ) -Tm,i = - Tm ( x ) = Tm,i +
LDq s,m (1 - cos x/L ) . & m cp
<
From Newton's law of cooling, Eq. 8.28, Ts ( x ) = ( qs / h ) + Tm ( x ) q LDq s,m x Ts ( x ) = s,m sin + Tm,i + (1 - cos x/L ) . & h L m cp
<
COMMENTS: For the prescribed surface condition, the flow is not fully developed. Hence, the assumption of constant h should be viewed as a first approximation.
PROBLEM 8.15 KNOWN: Surface heat flux for air flow through a rectangular channel. FIND: (a) Differential equation describing variation in air mean temperature, (b) Air outlet temperature for prescribed conditions. SCHEMATIC:
ASSUMPTIONS: (1) Negligible change in kinetic and potential energy of air, (2) No heat loss through bottom of channel, (3) Uniform heat flux at top of channel. PROPERTIES: Table A-4, Air (T 50C, 1 atm): cp = 1008 J/kgK. ANALYSIS: (a) For the differential control volume about the air, & & E in =E out & & m c p Tm + q (w dx ) = m c p ( Tm + d Tm ) o d Tm q w = o & dx m cp Separating and integrating between the limits of x = 0 and x, find q ( w x ) Tm ( x) = Tm,i + o & m cp q ( w L ) Tm,o = Tm,i + o . & m cp (b) Substituting numerical values, the air outlet temperature is
<
(700 W/m2 ) (13 )m 2 Tm,o = 40o C +
Tm,o = 60.8o C.
0.1 kg/s (1008 J/kg K )
<
COMMENTS: Due to increasing heat loss with increasing Tm, the net flux q will actually o decrease slightly with increasing x.
PROBLEM 8.16
KNOWN: Air inlet conditions and heat transfer coefficient for a circular tube of prescribed geometry. Surface heat flux. FIND: (a) Tube heat transfer rate, q, air outlet temperature, Tm,o, and surface inlet and outlet temperatures, Ts,i and Ts,o, for a uniform surface heat flux, q s . Air mean and surface temperature TT distributions. (b) Values of q, Tm,o, Ts,i and Ts,o for a linearly varying surface heat flux q s = 500x (m). TT Air mean and surface temperature distributions, (c) For each type of heating process (a & b), compute and plot the mean fluid and surface temperatures, Tm(x) and Ts(x), respectively, as a function of distance; What is effect of four-fold increase in convection coefficient, and (d) For each type of heating process, heat fluxes required to achieve an outlet temperature of Tm,o = 125C; Plot temperatures. SCHEMATIC:
ASSUMPTIONS: (1) Fully developed conditions in the tube, (2) Applicability of Eq. 8.36, (3) Heat transfer coefficient is the same for both heating conditions. PROPERTIES: Table A.4, Air (for an assumed value of Tm,o = 100C, Tm = (Tm,i + Tm,o)/2 = 60C = 333 K): cp = 1.008 kJ/kgK. ANALYSIS: (a) With constant heat flux, from Eq. 8.39, q = qs ( DL ) = 1000 W m 2 ( 0.05 m 3m ) = 471W . From the overall energy balance, Eq. 8.37,
(1)
q 471W Tm,o = Tm,i + = 20 C + = 113.5 C mcp 0.005 kg s 1008 J kg K
From the convection rate equation, it follows that
(2)
< < <
q 1000 W m 2 Ts,i = Tm,i + s = 20 C + = 60 C 2 K h 25 W m Ts,o = Tm,o + qs h = 113.5 C + 40 C = 153.5 C
(3)
From Eq. 8.40, (dTm/dx) is a constant, as is (dTs/dx) for constant h from Eq. 8.31. In the more realistic case for which h decreases with x in the entry region, (dTm/dx) is still constant but (dTs/dx) decreases with increasing x. See the plot below. (b) From Eq. 8.38, 2 dTm 500x ( D ) 500x W m ( 0.05 m ) = = = 15.6x K m . dx mcp 0.005 kg s 1008 J kg K (4) Continued...
PROBLEM 8.16 (Cont.)
Integrating from x = 0 to L it follows that 3 3 x2 Tm,o = Tm,i + 15.6 xdx = 20 C + 15.6 = 20 C + 70.2 C = 90.2C. 0 2 0 The heat rate is 3 3 x2 q = q dAs = 500 ( 0.05 m ) xdx = 78.5 = 353 W s 0 2 0 From Eq. 8.28 it then follows that x 2 500 Ts = Tm + qs h = Tm,i + 15.6 x = 20 C + 7.8x 2 + 20x +
(5)
< <
(6)
2
25
Hence, at the inlet (x = 0) and outlet (x = L),
Ts,i = Tm,i = 20 C
and
Ts,o = 150.2 C
<
Note that (dTs/dx) and (dTm/dx) both increase linearly with x, but (dTs/dx) > (dTm/dx). (c) The foregoing relations can be used to determine Tm(x) and Ts(x) for the two heating conditions:
Tm ( x ) = Tm,i + qs Dx mcp Ts ( x ) = Tm ( x ) + q h s 3 Linear surface heat flux, qs = aox, ao = 500 W/m ; Eqs. (4-6), Tm ( x ) = Tm,i + a o D 2mcp x 2
Uniform surface flux, q s ; Eqs. (1-3), TT
(7,8)
(
)
Ts ( x ) = Tm ( x ) + a o x h
(9, 10)
Using Eqs. (7-10) in IHT, the mean fluid and surface temperatures as a function of distance are evaluated and plotted below. The calculations were repeated with the coefficient increased four-fold, h = 4 25 = 100 W/m2K. As expected, the fluid temperature remained unchanged, but the surface temperatures decreased since the thermal resistance between the surface and fluid decreased.
240 200 160 120 80 40 0 0 1 2 3 240 200 160 120 80 40 0 0 1 2 3
Tm, Ts (C)
Distance from inlet, x (m) Tm(x); q''s = 1000 W/m^2 Ts(x); h = 25 W/m^2.K Ts(x); h = 100 W/m^2.K
Tm, Ts (C)
Distance from inlet, x (m) Tm(x); q''s = 500x W/m^2 Ts(x); h = 25 W/m^2.K Ts(x); h = 100 W/m^2.K
(d) The foregoing set of equations, Eqs. (7-10), in the IHT model can be used to determine the required heat fluxes for the two heating conditions to achieve Tm,o = 125C. The results with h = 25 W/m2K are: Uniform flux:
qs = 1123 W/m2
Linear flux:
qs = 748.7x W/m2
<
Continued...
PROBLEM 8.16 (Cont.)
The temperature distributions resulting from these heat fluxes are plotted below. The heat rate for both heating processes is 529 W.
240 200 160 Tm, Ts (C) 120 80 40 0 0 1 2 3 Tm, Ts (C) 240 200 160 120 80 40 0 0 1 2 3
Distance from inlet, x (m) Tm(x); q''s = 1123 W/m^2 Ts(x); h = 25 W/m^2.K
Distance from inlet, x (m) Tm(x); q''s = 749x (W/m^2) Ts(x); h = 25 W/m^2.K
COMMENTS: Note that the assumed value for Tm,o (100C) in determining the specific heat of the air was reasonable.
PROBLEM 8.17
KNOWN: Water at prescribed temperature and flow rate enters a 0.25 m diameter, black thin-walled tube of 8m length, which passes through a large furnace whose walls and air are at a temperature of Tfur = T = 700 K. The convection coefficients for the internal water flow and external furnace air are 300 W/m K and 50 W/m K, respectively. FIND: (a) An expression for the linearized radiation coefficient for the radiation exchange process between the outer surface of the pipe and the furnace walls; represent the tube by an average temperature and explain how to calculate this value, and (b) determine the outlet temperature of the water, To.
2 2
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions; (2) Tube is small object with large, isothermal surroundings; (3) Furnace air and walls are at the same temperature; and (3) Tube is thin-walled with black surface.
PROPERTIES: Table A-6, Water (Tm = (Tm,i + Tm,o)/2 = 331 K): cp = 4192 J/kgK.
ANALYSIS: (a) The linearized radiation coefficient follows from Eq. 1.9 with = 1,
2 h rad = ( Tt + Tfur ) Tt2 + Tfur
(
)
where Tt represents the average tube wall surface temperature, which can be evaluated from an energy balance on the tube as represented by the thermal circuit above.
Tm = Tm,i + Tm,o / 2 R tot = R cv,i + 1 1/ R cv,o + 1/ R rad
(
)
Tm - Tt Tt - Tfur = R cv,i 1/ R cv,o + 1/ R rad
The thermal resistances, with As = PL = DL, are
R cv,i = 1/ h i As T - Tm,o 1 = exp - T - Tm,i m cp R tot
R cv,o = 1/ h o As
R rad = 1/ h rad
(b) The outlet temperature can be calculated using the energy balance relation, Eq. 8.46b, with Tfur = T,
where cp is evaluated at Tm. Using IHT, the following results were obtained. R cv,i = 6.631 10-5 K / W R cv,o = 3.978 10-4 K / W
R rad = 4.724 10-4 K / W
Tm = 331 K Tm,o = 362 K
balance relation if T Tfur?
Tt = 418 K
<
COMMENTS: Since T = Tfur, it was possible to use Eq. 8.46b with Rtot. How would you write the energy
PROBLEM 8.18
KNOWN: Laminar, slug flow in a circular tube with uniform surface heat flux. FIND: Temperature distribution and Nusselt number. SCHEMATIC:
ASSUMPTIONS: (1) Steady, incompressible flow, (2) Constant properties, (3) Fully developed, laminar flow, (4) Uniform surface heat flux. ANALYSIS: With v = 0 for fully developed flow and T/x = dTm/dx = const, from Eqs. 8.33 and 8.40, the energy equation, Eq. 8.48, reduces to
uo
d Tm T = r . dx r r r
Integrating twice, it follows that u d Tm r 2 T (r ) = o + C1 ln ( r ) + C2.
dx
4
Since T(0) must remain finite, C1 = 0. Hence, with T(ro) = Ts 2 u d Tm ro u d Tm C2 = Ts - o T ( r ) = Ts - o
dx
4
4
dx
(ro2 - r2 ).
<
From Eq. 8.27, with um = uo,
Tm =
2
2 ro
ro 0
Tr dr =
2
2 ro
ro 0
u o d Tm Tsr - 4 dx
( rro2 - r3 ) dr
Tm =
2 r 2 u d Tm Ts o - o 2 2 4 dx ro
r4 r4 u r 2 d Tm o - o = Ts - o o . 2 4 8 dx
From Eq. 8.28 and Fourier's law,
T k ro q s h= = r Ts - Tm Ts - Tm
hence,
u r k o o 2 h= 2 u oro 8
d Tm dx 4k 8k = = ro D d Tm dx
Nu D =
hD = 8. k
<
PROBLEM 8.19 KNOWN: Heat transfer between fluid flow over a tube and flow through the tube. FIND: Axial variation of mean temperature for inner flow. SCHEMATIC:
ASSUMPTIONS: (1) Negligible change in kinetic and potential energy, (2) Negligible axial conduction, (3) Constant cp, (4) Uniform T. ANALYSIS: From Eq. 8.36, & dq = m cp d Tm with dq = UdA ( T - Tm ) = UP ( T - Tm ) dx.
The overall heat transfer coefficient may be defined in terms of the inner or outer surface area, with Ui Pi = UoPo. For the inner surface, from Eq. 3.31, 1 r r r 1 Ui = + i ln o + i h i k ri ro h o Hence, d Tm UP =+ dx & T - Tm m cp or, with T T - Tm, Hence, ln To PL 1 L =- 0 Udx & Ti m cp L
To Ti
-1
.
d ( T ) T
=-
L P Udx. & m cp 0
T - Tm,o
PL = exp - U. m cp T - Tm,i &
<
COMMENTS: The development and results parallel those for a constant surface temperature, with U and T replacing h and Ts .
PROBLEM 8.20
KNOWN: Thin-walled tube experiences sinusoidal heat flux distribution on the wall. FIND: (a) Total rate of heat transfer from the tube to the fluid, q, (b) Fluid outlet temperature, Tm,o, (c) Axial distribution of the wall temperature Ts(x) and (d) Magnitude and position of the highest wall temperature, and (e) For prescribed conditions, calculate and plot the mean fluid and surface temperatures, Tm(x) and Ts(x), respectively, as a function of distance along the tube; identify features of the distributions; explore the effect of 25% changes in the convection coefficient on the distributions. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes, (3) Turbulent, fully developed flow. ANALYSIS: (a) The total rate heat transfer from the tube to the fluid is
q = q Pdx = q D sin ( x L ) dx = q D ( L ) - cos ( x L ) = 2DLq (1) o o o 0 0 s 0
L
L
D
< <
(3)
(b) The fluid outlet temperature follows from the overall energy balance with knowledge of the total heat rate,
q = mcp Tm,o - Tm,i = 2DLq o q = h Ts ( x ) - Tm ( x ) s
(
)
Tm,o = Tm,i + 2DLq mcp o Ts,x = Tm,x ( x ) + q h s
(
)
(2)
(c) The axial distribution of the wall temperature can be determined from the rate equation
where, by combining expressions of parts (a) and (b), Tm,x(x) is x q Pdx = mc p Tm,x - Tm,i o s
(
)
q D x DLq o 1 - cos x L Tm,x = Tm,i + o sin ( x L ) dx = Tm,i + ( ) 0 mcp mcp
Hence, substituting Eq. (4) into (3), find
(4)
Ts ( x ) = Tm,i +
dTs ( x ) dx
DLq o 1 - cos x L + q sin x L ( ) o ( ) mcp h
(5)
<
(d) To determine the location of the maximum wall temperature x T where Tx( x ) = Ts,max, set =0=
d DLq o 1 - cos x L + q sin x L ( )] o ( ) [ dx mc p h
tan ( x L ) = -
mc p q h o =- DLq mc p DLh o
DLq o sin x L + q cos x L = 0 ( ) o ( ) mcp L h L
Continued...
PROBLEM 8.20 (Cont.) x = L tan -1 - mcp DLh
(
)
(6)
< <
At this location, the wall temperature is
DLq o 1 - cos x L + q sin x L Ts,max = Ts ( x ) = Tm,i + ( ) o ( ) mcp h
(e) Consider the prescribed conditions for which to compute and plot Tm(x) and Ts(x), D = 40 mm L=4m
(7)
m = 0.025 kg/s
cp = 4180 J/kgK
h = 1000 W/m2 Tm,i = 25C
q = 10,000 W/m2 o
Using Eqs. (4) and (5) in IHT, the results are plotted below.
45
40 Ts, Tm (C)
35
30
25 0 1 2 Distance, x (m) Tm(x) Ts(x); h = 1250 W/m^2.K Ts(x); h = 1000 W/m^2.K Ts(x); h = 750 W/m^2.K 3 4
The effect of a lower convection coefficient is to increase the wall temperature. The position of the maximum temperature, Ts,max, moves away from the tube exit with decreasing convection coefficient. COMMENTS: (1) Because the flow is fully developed and turbulent, assuming h is constant along the entire length of the tube is reasonable. (2) To determine whether the Tx(x) distribution has a maximum (rather than a minimum), you should evaluate d2Ts(x)/dx2 to show the value is indeed negative.
PROBLEM 8.21
KNOWN: Water is heated in a tube having a wall flux that is dependent upon the wall temperature. FIND: (a) Beginning with a properly defined differential control volume in the tube, derive expressions that can be used to obtain the temperatures for the water and the wall surface as a function of distance from the inlet, Tm(x) and Ts(x), respectively; (b) Using a numerical integration scheme, calculate and plot the temperature distributions, Tm(x) and Ts(x), on the same graph. Identify and comment on the main features of the distributions; and (c) Calculate the total heat transfer rate to the water. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed flow and thermal conditions, (3) No losses to the outer surface of the tube, and (3) Constant properties. PROPERTIES: Table A-6, Water Tm = Tm,i + Tm,o / 2 = 300 K : cp = 4179 J/kgK ANALYSIS: (a) The properly defined control volume of perimeter P = D shown in the above schematic follows from Fig. 8.6. The energy balance on the CV includes advection, convection at the inner tube surface, and the heat flux dissipated in the tube wall. (See Eq. 8.38).
(
(
)
)
m cp
dTm = qs ( x ) P = h P Ts ( x ) - Tm ( x ) dx
(1,2)
where q ( x ) is dependent upon Ts(x) according to the relation s
q ( x ) = q 1 + ( Ts ( x ) - Tref ) s s,o
(3)
(b) Eqs. (1 and 2) with Eq. (3) can be solved by numerical integration using the Der function in IHT as shown in Comment 1. The temperature distributions for the water and wall surface are plotted below.
80
Temperature, Tm or Ts (C)
60
40
20 0 1 Distance along tube, x (m) Water mean temperature, Tm(x) Tube surface temperature, Ts(x) 2
Continued .....
PROBLEM 8.21 (Cont.)
(c) The total heat transfer to the water can be evaluated from an overall energy balance on the water,
q = m c p Tm,o - Tm,i
(
)
(4)
q = 0.1 kg / s 4179 J / kg K (34.4 - 20 ) K = 6018 W
<
(5)
Alternatively, the heat rate can be evaluated by integration of the heat flux from the tube surface over the length of the tube,
q=
where q ( x ) is given by Eq. (3), and Ts(x) and Tm(x) are determined from the differential form of s the energy equation, Eqs. (1) and (2). The result as shown in the IHT code below is 6005 W. COMMENTS: (1) Note that Tm(x) increases with distance greater than linearly, as expected since qs (x) does.
Also as expected, the difference, Ts(x) Tm(x), likewise increases with distance greater than linearly.
0
L
q ( x ) Pdx s
(2) In the foregoing analysis, cp is evaluated at the mean fluid temperature Tm = (Tm,i + Tm,o)/2. (3) The IHT code representing the foregoing equations to calculate and plot the temperature distribution and to calculate the total heat rate to the water is shown below.
/* Results: integration for distributions; conditions at x = 2 m F_xTs Ts q' q''s_x x Tm 11.64 73.18 5483 1.164E5 2 34.39 3 30 1414 3E4 0 20 */ /* Results: heat rate by energy balances on fluid and tube surface q_eb q_hf 6018 6005 */ /* Results: for evaluating cp at Tm Ts cp q''s_x x 73.31 4179 1.166E5 2 30 4179 3E4 0 // Energy balances mdot * cp * der(Tm,x) = q' q' = q''s_x * P q''s_x = q''o * F_xTs q' = h * P * (Ts - Tm) P = pi * D // Surface heat flux specification F_xTs = (1 + alpha * (Ts -Tref)) alpha = 0.2 Tref = 20 // Overall heat rate // Energy balance on the fluid q_eb = mdot * cp * (Tmo - Tmi) Tmi = 20 Tmo = 34.4 // From initial solve // Integration of the surface heat flux q_hf = q''o * P * INTEGRAL(F_xTs, x) // Input variables mdot = 0.1 D = 0.015 h = 3000 q''o = 1.0e4 // L = 2 // Tmi = 20 Tm 34.44 20
*/ // Energy balance, Eq. 8.38
// Convection rate equation
// Limit of integration over x // Initial condition for integration
// Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); xx = 0 // Quality (0=sat liquid or 1=sat vapor) cp = cp_Tx("Water",Tmm,xx) // Specific heat, J/kgK Tmm = (20 + 34.4) / 2 + 273
PROBLEM 8.22 KNOWN: Flow rate of engine oil through a long tube. FIND: (a) Heat transfer coefficient, h , (b) Outlet temperature of oil, Tm,o. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Combined entry conditions exist. PROPERTIES: Table A-5, Engine Oil (Ts = 100C = 373K): s = 1.73 10 A-5, Engine Oil Tm = 77o C = 350K : cp = 2118 J/kgK, = 3.56 10 W/mK, Pr = 546. ANALYSIS: (a) The overall energy balance and rate equations have the form & q = m cp Tm,o - Tm,i
-2
(
)
Ns/m ; Table
2
2
-2
Ns/m , k = 0.138
(
)
q = hAs T1m
(1,2)
Using Eq. 8.42b, with P = D, and Eq. 8.6 To Ts - Tm,o = = exp Ti Ts - Tm,i Re D = PL - h . m cp & (3)
& 4m 4 0.02 kg/s = = 238. D 3 10 -3m 3.56 10-2 N s/m 2
For laminar and combined entry conditions, use Eq. 8.57 Re Pr NuD = 1.86 D L/D
0.14 1/3
s
238 546 = 30m/3 10-3m
1/3
3.56 1.73
0.14
= 4.83
h = Nu D k/D = 4.83 0.138 W/m K/3 10-3 m = 222 W/m 2 K. (b) Using Eq. (3) with the foregoing value of h,
<
(100 - Tm,o )o C = exp -
(100 - 60)
o
C
0.02 kg/s 2118 J/kg K
3 10- 3m 30m
222W/m K
2
Tm,o = 90.9 C.
o
<
COMMENTS: (1) Note that requirements for the correlation, Eq. 8.57, are satisfied. (2) The assumption of Tm = 77C for selecting property values was satisfactory. (3) For thermal entry effect only, Eq. 8.56, h = 201 W/m 2 K and Tm,o = 89.5C.
PROBLEM 8.23
KNOWN: Inlet temperature and flowrate of oil flowing through a tube of prescribed surface temperature and geometry. FIND: (a) Oil outlet temperature and total heat transfer rate, and (b) Effect of flowrate. SCHEMATIC:
ASSUMPTIONS: (1) Negligible temperature drop across tube wall, (2) Negligible kinetic energy, potential energy and flow work effects. PROPERTIES: Table A.5, Engine oil (assume Tm,o = 140C, hence Tm = 80C = 353 K): = 852 kg/m3, = 37.5 10-6 m2/s, k = 138 10-3 W/mK, Pr = 490, = = 0.032 kg/ms, cp = 2131 J/kgK. ANALYSIS: (a) For constant surface temperature the oil outlet temperature may be obtained from Eq. 8.42b. Hence
DL Tm,o = Ts - Ts - Tm,i exp - h mcp
(
)
To determine h , first calculate ReD from Eq. 8.6,
ReD =
4 ( 0.5 kg s ) 4m = = 398 D (0.05m )(0.032 kg m s )
Hence the flow is laminar. Moreover, from Eq. 8.23 the thermal entry length is
x fd,t 0.05D ReD Pr = 0.05 (0.05 m )(398)( 490 ) = 486 m .
Since L = 25 m the flow is far from being thermally fully developed. However, from Eq. 8.3, xfd,h 0.05DReD = 0.05(0.05 m)(398) = 1 m and it is reasonable to assume fully developed hydrodynamic conditions throughout the tube. Hence h may be determined from Eq. 8.56
Nu D = 3.66 +
0.0668 ( D L ) ReD Pr
1 + 0.04 ( D L ) ReD Pr
2/3
.
With (D/L)ReDPr = (0.05/25)398 490 = 390, it follows that
26 = 11.95. 1 + 2.14 k 0.138 W m K Hence, h = Nu D = 11.95 = 33 W m 2 K and it follows that D 0.05 m Nu D = 3.66 +
Continued...
PROBLEM 8.23 (Cont.) (0.05 m )( 25 m ) Tm,o = 150 C - 150 C - 20 C exp - 33 W m 2 K 0.5 kg s 2131J kg K
(
)
Tm,o = 35C. From the overall energy balance, Eq. 8.37, it follows that
<
)
q = mcp Tm,o - Tm,i = 0.5 kg s 2131J kg K (35 - 20 ) C
q = 15,980 W.
(
<
The value of Tm,o has been grossly overestimated in evaluating the properties. The properties should be re-evaluated at T = (20 + 35)/2 = 27C and the calculations repeated. Iteration should continue until satisfactory convergence is achieved between the calculated and assumed values of Tm,o. Following such a procedure, one would obtain Tm,o = 36.4C, ReD = 27.8, h = 32.8 W/m2K, and q = 15,660 W. The small effect of reevaluating the properties is attributed to the compensating effects on ReD (a large decrease) and Pr (a large increase). (b) The effect of flowrate on Tm,o and q was determined by using the appropriate IHT Correlations and Properties Toolpads.
40 Outlet temperature, Tmo(C)
30000
36
Heat rate, q(W) 25000
32
28
20000
24
20 0.5 1 1.5 2 Mass flowrate, mdot(kg/s)
15000 0.5 1 1.5 2 Mass flowrate, mdot(kg/s)
The heat rate increases with increasing m due to the corresponding increase in ReD and hence h .
However, the increase is not proportional to m , causing Tm,o - Tm,i = q mcp , and hence Tm,o, to decrease with increasing m . The maximum heat rate corresponds to the maximum flowrate ( m = 0.20 kg/s).
(
)
COMMENTS: Note that significant error would be introduced by assuming fully developed thermal conditions and Nu D = 3.66. The flow remains well within the laminar region over the entire range of m.
PROBLEM 8.24
KNOWN: Inlet temperature and flowrate of oil moving through a tube of prescribed diameter and surface temperature. FIND: (a) Oil outlet temperature Tm,o for two tube lengths, 5 m and 100 m, and log mean and arithmetic mean temperature differences, (b) Effect of L on Tm,o and Nu D . SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic energy, potential energy and flow work changes, (3) Constant properties. PROPERTIES: Table A.4, Oil (330 K): cp = 2035 J/kgK, = 0.0836 Ns/m2, k = 0.141 W/mK, Pr = 1205. ANALYSIS: (a) Using Eqs. 8.42b and 8.6
DL Tm,o = Ts - Ts - Tm,i exp - h mcp 4m 4 0.5 kg s ReD = = = 304.6 D 0.025 m 0.0836 N s m 2
(
)
Since entry length effects will be significant, use Eq. 8.56
4 0.0688 ( D L ) Re D Pr 0.141W m K k 3.66 + 2.45 10 D L h= 3.66 + = 2/3 2/3 D 0.025 m 1 + 205 ( D L ) 1 + 0.04 ( D L ) ReD Pr
For L = 5 m, h = 5.64 (3.66 + 17.51) = 119 W m 2 K , hence
0.025 m 5 m 119 W m 2 K = 28.4 C Tm,o = 100 C - 75 C exp - 0.5 kg s 2035 J kg K
(
)
< < < <
For L = 100 m, h = 5.64 (3.66 + 3.38 ) = 40 W m 2 K , Also, for L = 5 m, Tm = For L = 100 m,
n ( To Ti )
Tm,o = 44.9C.
To - Ti
=
n ( 71.6 75 )
71.6 - 75
= 73.3 C
Tam = ( To + Ti ) 2 = 73.3 C
Tm = 64.5 C ,
Tam = 65.1 C
(b) The effect of tube length on the outlet temperature and Nusselt number was determined by using the Correlations and Properties Toolpads of IHT. Continued...
PROBLEM 8.24 (Cont.)
50
25
45 Outlet temperature, Tmo(C)
Nusselt number, NuDbar
20
40
15
35
10
30
5
25 0 20 40 60 80 100
0 0 20 40 60 80 100
Tube length, L(m)
Tube length, L(m)
The outlet temperature approaches the surface temperature with increasing L, but even for L = 100 m, Tm,o is well below Ts. Although Nu D decays with increasing L, it is still well above the fully developed value of NuD,fd = 3.66. COMMENTS: (1) The average, mean temperature, Tm = 330 K, was significantly overestimated in part (a). The accuracy may be improved by evaluating the properties at a lower temperature. (2) Use of Tam instead of Tm is reasonable for small to moderate values of (Tm,i - Tm,o). For large values of
(Tm,i - Tm,o), Tm should be used.
PROBLEM 8.25
KNOWN: Oil at 80C enters a single-tube preheater of 9-mm diameter and 5-m length; tube surface maintained at 165C by swirling combustion gases. FIND: Determine the flow rate and heat transfer rate when the outlet temperature is 95C. SCHEMATIC:
ASSUMPTIONS: (1) Combined entry length, laminar flow, (2) Tube wall is isothermal, (3) Negligible kinetic and potential energy, and flow work, (4) Constant properties. PROPERTIES: Table A-5, Engine oil, new (Tm = (Tm,i + Tm,o)/2 = 361 K): = 847.5 kg/m , cp = -5 2 2 2163 J/kgK, = 2.931 10 m /s, k = 0.1879 W/mK, Pr = 3902; (Ts = 430 K): s = 0.047 Ns/m . ANALYSIS: The overall energy balance, Eq. 8.37, and rate equation, Eq. 8.42b, are
3
q = m cp Tm,o - Tm,i Ts - Tm,o
(
)
(1)
PLh = exp - m cp Ts - Tm,i
(2)
Not knowing the flow rate m, the Reynolds number cannot be calculated. Assume that the flow is laminar, and the combined entry length condition occurs. The average convection coefficient can be estimated using the Sieder-Tate correlation, Eq. 8.57, hD Re Pr Nu D = = 1.86 D k L/D
1/ 3
s
0.14 (3)
where all properties are evaluated at Tm = (Tm,i + Tm,o)/2, except for s at the wall temperature Ts. The Reynolds number follows from Eq. 8.6,
ReD = 4m / D
(4)
A tedious trial-and-error solution is avoided by using IHT to solve the system of equations with the following result:
Re D
251
Nu D
9.54
hD (W/m K)
2
q(W) 1432
m ( kg / h )
159
146
<
Note that the flow is laminar, and evaluating xfd using Eq. 8.3, find xfd,h = 44 m so the combined entry length condition is appropriate.
PROBLEM 8.26
KNOWN: Ethylene glycol flowing through a coiled, thin walled tube submerged in a well-stirred water bath maintained at a constant temperature. FIND: Heat rate and required tube length for prescribed conditions. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Tube wall thermal resistance negligible, (3) Convection coefficient on water side infinite; cooling process approximates constant wall surface temperature distribution, (4) KE, PE and flow work changes negligible, (5) Constant properties, (6) Negligible heat transfer enhancement associated with the coiling. PROPERTIES: Table A-5, Ethylene glycol (Tm = (85 + 35)C/2 = 60C = 333 K): cp = 2562 -2 2 J/kgK, = 0.522 10 Ns/m , k = 0.260 W/mK, Pr = 51.3. ANALYSIS: From an overall energy balance on the tube,
& q conv = m c p Tm,o - Tm,i = 0.01 kg/s 2562 J/kg ( 35 - 85 )o C = -1281 W.
For the constant surface temperature condition, from the rate equation,
(
)
(1) <
As = q conv / h Tlm
Tlm = ( To - Ti ) / ln To 35 - 25 = ( 35 - 25 )o C- (85 - 25 )oC / ln = 27.9o C. Ti 85 - 25
(2) (3)
Find the Reynolds number to determine flow conditions,
Re D =
& 4m 4 0.01 kg/s = = 813. D 0.003 m 0.522 10-2 N s/m2
(4)
Hence, the flow is laminar and, assuming the flow is fully developed, the appropriate correlation is
NuD =
hD = 3.66, k
h = Nu
k W = 3.66 0.260 /0.003m = 317 W/m 2 K. (5) D mK
From Eq. (2), the required area, As , and tube length, L, are
As = 1281 W/317 W/m 2 K 27.9 oC = 0.1448 m 2 L = As / D = 0.1448m 2 / ( 0.003m ) = 15.4m.
Gz
-1
<
COMMENTS: Note that for fully developed laminar flow conditions, the requirement is satisfied: = (L/D) / ReD Pr = (15.3/0.003) / (813 51.3) = 0.122 > 0.05. Note also the sign of the heat rate qconv when using Eqs. (1) and (2).
PROBLEM 8.27
KNOWN: Inlet and outlet temperatures and velocity of fluid flow in tube. Tube diameter and length.
FIND: Surface heat flux and temperatures at x = 0.5 and 10 m. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat loss to surroundings, (4) Negligible potential and kinetic energy changes and axial conduction. PROPERTIES: Pharmaceutical (given): = 1000 kg/m , cp = 4000 J/kgK, = 2 10 kg/sm, k = 0.48 W/mK, Pr = 10. ANALYSIS: With
2 & m = VA = 1000 kg/m 3 ( 0.2 m/s ) (0.0127 m ) / 4 = 0.0253 kg/s
3 -3
Eq. 8.37 yields & q = m cp Tm,o - T m,i = 0.0253 kg/s ( 4000 J/kg K ) 50 K = 5060 W. The required heat flux is then q = q/As = 5060 W/ ( 0.0127 m )10 m = 12,682 W/m 2 . s With Re D = VD/ = 1000 kg/m 3 ( 0.2 m/s ) 0.0127 m/2 10-3 kg/s m = 1270 the flow is laminar and Eq. 8.23 yields xfd,t = 0.05ReD PrD = 0.05 (1270) 10 ( 0.0127 m ) = 8.06 m. Hence, with fully developed hydrodynamic and thermal conditions at x = 10 m, Eq. 8.53 yields h (10 m ) = Nu D,fd ( k/D ) = 4.36 ( 0.48 W/m K/0.0127 m ) = 164.8 W/m 2 K. Hence, from Newton's law of cooling, Ts,o = Tm,o + ( qs / h ) = 75o C + 12,682 W/m 2 /164.8 W/m 2 K = 152o C.
(
)
<
(
)
<
At x = 0.5 m, (x/D)/(ReDPr) = 0.0031 and Figure 8.9 yields NuD 8 for a thermal entry region 2 with uniform surface heat flux. Hence, h(0.5 m) = 302.4 W/m K and, since Tm increases linearly with x, Tm(x = 0.5 m) = Tm,i + (Tm,o - Tm,i) (x/L) = 27.5C. It follows that Ts ( x = 0.5 m ) 27.5o C + 12,682 W/m 2 /302.4 W/m 2 K = 69.4o C.
(
)
<
PROBLEM 8.28
KNOWN: Inlet temperature, flow rate and properties of hot fluid. Initial temperature, volume and properties of pharmaceutical. Heat transfer coefficient at outer surface and dimensions of coil. FIND: (a) Expressions for Tc(t) and Th,o(t), (b) Plots of Tc(t) and Th,o(t) for prescribed conditions. Effect of flow rate on time for pharmaceutical to reach a prescribed temperature. SCHEMATIC:
ASSUMPTIONS: (1) Constant properties, (2) Negligible heat loss from vessel to surroundings, (3) Pharmaceutical is isothermal, (4) Negligible work due to stirring, (5) Negligible thermal energy generation (or absorption) due to chemical reactions associated with the batch process, (6) Negligible kinetic energy, potential energy and flow work changes for the hot fluid, (7) Negligible tube wall conduction resistance. ANALYSIS: (a) Performing an energy balance for a control surface about the stirred liquid, it follows that
d Uc d dT cc c v,c Tc = c Vc c v,c c = q ( t ) = dt dt dt q ( t ) = m h cp,h Th,i - Th,o where,
(
)
(1) (2) (3a)
(
)
or, where
q ( t ) = UAs Tm
Tm =
(Th,i - Tc ) - (Th,o - Tc ) = (Th,i - Th,o )
Th,i - Tc n Th,o - Tc
Th,i - Tc n Th,o - Tc
(3b)
Substituting (3b) into (3a) and equating to (2),
m h c p,h Th,i - Th,o = UAs
(
)
(Th,i - Th,o )
Th,i - Tc n Th,o - Tc
Hence,
Th,i - Tc UAs n = Th,o - Tc m h c p,h Th,o ( t ) = Tc + Th,i - Tc exp - UAs / m h cp,h
or,
(
) (
)
(4)
<
Substituting Eqs. (2) and (4) into Eq. (1), Continued .....
PROBLEM 8.28 (Cont.)
cc c v,c
d Tc = m h cp,h Th,i - Tc - Th,i - Tc exp - UAs / m h cp,h dt
(
) (
)
d Tc m h cp,h Th,i - Tc 1 - exp - UAs / m h cp,h = dt cc c v,c
(
)
(
)
T (t ) - c Tc,i
(Tc - Th,i )
d Tc
=
1 - exp - UAs / m h c p,h t dt o cc c v,c
m h cp,h
(
)
Tc - Th,i - n = Tc,i - Th,i
1 - exp - UAs / m h cp,h t c Vc c v,c
m h cp,h
(
)
m c h p,h 1 - exp - UA / m h c p,h t Tc ( t ) = Th,i - Th,i - Tc,i exp - cc c v,c
(
)
(
)
(5)
<
Eq. (5) may be used to determine Tc(t) and the result used with (4) to determine Th,o(t). (b) To evaluate the temperature histories, the overall heat transfer coefficient,
U = ho + hi
(
-1
-1
)
-1
,
must
first be determined. With Re D = 4 m / D = 4 2.4 kg / s / ( 0.05m ) 0.002 N s / m 2 = 30, 600, the flow is turbulent and
hi =
k 0.260 W / m K 4/5 Nu D = 0.023 (30, 600 ) ( 20 )0.3 = 1140 W / m2 K D 0.05m
-1
Hence, U = (1000 )-1 + (1140 )-1
W / m K = 532 W / m K. As shown below, the temperature of
2
2
the pharmaceuticals increases with time due to heat transfer from the hot fluid, approaching the inlet temperature of the hot fluid (and its maximum possible temperature of 200C) at t = 3600s.
200 180 Te m p e ra tu re (C ) 160 140 120 100 80 60 40 20 0 400 800 1200 1600 2000 2400 2800 3200 3600 Tim e (s ) P h a rm a ce u tica l, Tc H o t flu id , Th
Continued .....
PROBLEM 8.28 (Cont.)
With increasing Tc, the rate of heat transfer from the hot fluid decreases (from 4.49 10 W at t = 0 to 6760 W at 3600s), in which case Th,o increases (from 125.2C at t = 0 to 198.9C at 3600s). The time required for the pharmaceuticals to reach a temperature of Tc = 160C is
5
t c = 1266s
<
With increasing m h , the overall heat transfer coefficient increases due to increasing hi and the hot fluid maintains a higher temperature as it flows through the tube. Both effects enhance heat transfer to the pharmaceutical, thereby reducing the time to reach 160C from 2178s for m h = 1kg / s to 906s at 5 kg/s.
2200 2000 1800 T im e , tc (s ) 1600 1400 1200 1000 800 1 2 3 4 5 M a s s flo w ra te , m d o th (kg /s )
For 1 m h 5 kg / s, 12, 700 Re D 63, 700 and 565 h i 2050 W / m 2 K.
COMMENTS: Although design changes involving the length and diameter of the coil can be used to alter the heating rate, process control parameters are limited to Th,i and m h .
PROBLEM 8.29
KNOWN: Tubing with glycerin welded to transformer lateral surface to remove dissipated power. Maximum allowable temperature rise of coolant is 6C.
FIND: (a) Required coolant rate m , tube length L and lateral spacing S between turns, and (b) Effect of flowrate on outlet temperature and maximum power.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) All heat dissipated by transformer transferred to glycerin, (3) Fully developed flow (part a), (4) Negligible kinetic and potential energy changes, (5) Negligible tube wall thermal resistance. PROPERTIES: Table A.5, Glycerin ( Tm 300 K): = 1259.9 kg/m3, cp = 2427 J/kgK, = 79.9 102 Ns/m2, k = 286 10-3 W/mK, Pr = 6780. ANALYSIS: (a) From an overall energy balance assuming the maximum temperature rise of the glycerin coolant is 6C, find the flow rate as
q = mc p Tm,o - Tm,i
(
)
m = q c p Tm,o - Tm,i = 1000 W 2427 J kg K ( 6 K ) = 6.87 10-2 kg s
(
)
<
From Eq. 8.43, the length of tubing can be determined,
Ts - Tm,o Ts - Tm,i
= exp - PLh mcp
(
)
where P = D. For the tube flow, find
4m 4 6.87 10-2 kg s ReD = = = 5.47 D 0.020 m 79.9 10-2 N s m 2 hD = 3.66 k 3.66 286 10-3 W m K = 52.3 W m 2 K 0.020 m
m2 K L
which implies laminar flow, and if fully developed,
Nu D =
h=
( 47 - 30 ) C = exp - ( 0.020 m ) 52.3 W ( 47 - 24 ) C
(
) (6.87 10-2 kg s 2427 J kg K )
< <
Continued...
L = 15.3 m. The number of turns of the tubing, N, is N = L/(D) = (15.3 m)/(0.3 m) = 16.2 and hence the spacing S will be S = H/N = 500 mm/16.2 = 30.8 mm.
PROBLEM 8.29 (Cont.)
(b) Parametric calculations were performed using the IHT Correlations Toolpad based on Eq. 8.56 (a thermal entry length condition), and the following results were obtained.
3000
35
2600 Outlet temperature, Tmo(C)
33
Heat rate, q(W)
2200
31
1800
29
1400
27
1000 0.05 0.09 0.13 0.17 0.21 0.25
25 0.05 0.09 0.13 0.17 0.21 0.25
Mass flowrate, mdot(kg/s)
Mass flowrate, mdot(kg/s)
With Ts maintained at 47C, the maximum allowable transformer power (heat rate) and glycerin outlet temperature increase and decrease, respectively, with increasing m . The increase in q is due to an increase in Nu D (and hence h ) with increasing ReD. The value of Nu D increased from 5.3 to 9.4 with increasing m from 0.05 to 0.25 kg/s. - COMMENTS: Since Gz D1 = ( L D ) ReD Pr = (15.3 m/0.02 m)/(5.47 6780) = 0.0206 < 0.05, entrance length effects are significant, and Eq. 8.56 should be used to determine Nu D .
PROBLEM 8.30
KNOWN: Diameter and length of copper tubing. Temperature of collector plate to which tubing is soldered. Water inlet temperature and flow rate. FIND: (a) Water outlet temperature and heat rate, (b) Variation of outlet temperature and heat rate with flow rate. Variation of water temperature along tube for the smallest and largest flowrates. SCHEMATIC:
ASSUMPTIONS: (1) Straight tube with smooth surface, (2) Negligible kinetic/potential energy and flow work changes, (3) Negligible thermal resistance between plate and tube inner surface, (4) ReD,c = 2300. PROPERTIES: Table A.6, water (assume Tm = (Tm,i + Ts)/2 = 47.5C = 320.5 K): = 986 kg/m3, cp = 4180 J/kgK, = 577 10-6 Ns/m2, k = 0.640 W/mK, Pr = 3.77. Table A.6, water (Ts = 343 K): s = 400 10-6 Ns/m2.
ANALYSIS: (a) For m = 0.01 kg/s, ReD = 4 m D = 4(0.01 kg/s)/(0.01 m)577 10-6 Ns/m2 =
2200, in which case the flow may be assumed to be laminar. With x fd,t D 0.05ReDPr = 0.05(2200)(3.77) = 415 and L/D = 800, the flow is fully developed over approximately 50% of the tube 1/ 3 0.14 = 2.30, Eq. 8.57 may therefore be used to compute the length. With Re D Pr ( L D ) ( s ) average convection coefficient
Re Pr Nu D = 1.86 D LD
1/ 3
s
0.14
= 4.27
h = ( k D ) Nu D = 4.27 (0.640 W m K ) 0.01m = 273 W m 2 K
From Eq. 8.42b,
Ts - Tm,o
0.01m 8 m 273W m 2 K DL h = exp - = exp - mcp Ts - Tm,i 0.01kg s 4180 J kg K
Tm,o = Ts - 0.194 Ts - Tm,i = 70 C - 8.7 C = 61.3 C Hence, q = mcp Tm,o - Tm,i = 0.01kg s ( 4186 J kg K )(36.3K ) = 1519 W
(
)
(
)
< <
(b) The IHT Correlations, Rate Equations and Properties Tool Pads were used to determine the parametric variations. The effect of m was considered in two steps, the first corresponding to m < 0.011 kg/s (ReD < 2300) and the second for m > 0.011 kg/s (ReD > 2300). In the first case, Eq. 8.57 was used to determine h , while in the second Eq. 8.60 was used. The effects of m are as follows. Continued...
PROBLEM 8.30 (Cont.)
67 66 Outlet temperature, Tmo(C)
69.8 70
65 64 63 62 61 60 0.005 0.006 0.007 0.008 0.009 0.01 0.011
Outlet temperature, Tmo(C)
69.6
69.4
69.2
69 0.01 0.02 0.03 0.04 0.05
Mass flowrate, mdot(kg/s) Laminar flow (ReD < 2300)
Mass flowrate, mdot(kg/s) Turbulent flow (ReD>2300)
1700 1600 1500 Heat rate, q(W) Heat rate, q(W) 0.006 0.007 0.008 0.009 0.01 0.011 1400 1300 1200 1100 1000 900 800 0.005
9500
7500
5500
3500
1500 0.01 0.02 0.03 0.04 0.05
Mass flowrate, mdot(kg/s) Laminar flow (ReD < 2300)
Mass flowrate, mdot(kg/s) Turbulent flow (ReD>2300)
The outlet temperature decreases with increasing m , although the effect is more pronounced for laminar , (Tm,o - Tm,i) would decrease inversely with increasing m . In turbulent flow. If q were independent of m flow, however, the convection coefficient, and hence the heat rate, increases approximately as m0.8 , thereby attenuating the foregoing effect. In laminar flow, q ~ m0.5 and this attenuation is not as pronounced.
The temperature distributions were computed from Eq. 8.43, with h assumed to be independent of x. For laminar flow ( m = 0.005 kg/s), h was based on the entire tube length (L = 8 m) and computed from Eq. 8.57, while for turbulent flow ( m = 0.05 kg/s) it was assumed to correspond to the value for fully developed flow and computed from Eq. 8.60. The corresponding temperature distributions are as follows.
Continued...
PROBLEM 8.30 (Cont.)
70 70
60 Mean temperature, Tm(C) Mean temperature, Tm(C)
60
50
50
40
40
30
30
20 0 2 4 Axial location, x(m) mdot = 0.005 kg/s 6 8
20 0 2 4 Axial location, x(m) mdot = 0.05 kg/s 6 8
The more pronounced increase for turbulent flow is due to the much larger value of h (4300 W/m2K for m = 0.05 kg/s relative to 217 W/m2K for m = 0.05 kg/s).
PROBLEM 8.31
KNOWN: Diameter and surface temperature of ten tubes in an ice bath. Inlet temperature and flowrate per tube. Volume () of container and initial volume fraction, fv,i, of ice. FIND: (a) Tube length required to achieve a prescribed air outlet temperature Tm,o and time to completely melt the ice, (b) Effect of mass flowrate on Tm,o and suitable design and operating conditions. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible kinetic/potential energy and flow work changes, (3) Constant properties, (4) Fully developed flow throughout each tube, (5) Negligible tube wall thermal resistance. PROPERTIES: Table A.4, air (assume Tm = 292 K): cp = 1007 J/kgK, = 180.6 10-7 Ns/m2, k = 0.0257 W/mK, Pr = 0.709; Ice: = 920 kg/m3, hsf = 3.34 105 J/kg.
ANALYSIS: (a) With ReD = 4 m /D = 4(0.01 kg/s)/(0.05 m)180.6 10-7 Ns/m2 = 14,100 for m = 0.01 kg/s, the flow is turbulent, and from Eq. 8.60, Nu D = Nu D = 0.023Re0.8 Pr 0.3 = 0.023 (14,100 ) D
0.8
(0.709 )0.3 = 43.3
h = Nu D ( k D ) = 43.3 ( 0.0257 W m K 0.05 m ) = 22.2 W m 2 K
With Tm,o = 14C, the tube length may be obtained from Eq. 8.42b,
Ts - Tm,o
( 0.05 m ) 22.2 W m 2 K L DLh -14 = = exp - = exp - mcp Ts - Tm,i -24 0.01kg s (1007 J kg K )
(
)
L = 1.56 m The time required to completely melt the ice may be obtained from an energy balance of the form,
<
)
( -q ) t = f v,i ( h sf )
where q = Nmcp Tm,i - Tm,o = 10 ( 0.01kg s )1007 J kg K (10 K ) = 1007 W . Hence, t= 0.8 10 m3 920 kg m3 3.34 105 J kg 1007 W
(
(
)(
)
= 2.44 106 s = 28.3days
<
(b) Using the appropriate IHT Correlations and Properties Tool Pads, the following results were obtained. Continued...
PROPERTIES 8.31 (Cont.)
17
16 Outlet temperature, Tmo(C)
15
14
13
12 0 0.01 0.02 0.03 0.04 0.05
Mass flowrate per tube, mdot(kg/s)
Although heat extraction from the air passing through each tube increases with increasing flowrate, the increase is not in proportion to the change in m and the temperature difference (Tm,i - Tm,o) decreases. If 0.05 kg/s of air is routed through a single tube, the outlet temperature of Tm,o = 16.2C slightly exceeds the desired value of 16C. The prescribed value could be achieved by slightly increasing the tube length. However, in the interest of reducing pressure drop requirements, it would be better to operate at a lower flowrate per tube. If, for example, air is routed through four of the tubes at 0.01 kg/s per tube and the discharge is mixed with 0.01 kg/s of the available air at 24C, the desired result would be achieved. COMMENTS: Since the flow is turbulent and L/D = 31, the assumption of fully developed flow throughout a tube is marginal and the foregoing analysis overestimates the discharge temperature.
PROBLEM 8.32
KNOWN: Thermal conductivity and inner and outer diameters of plastic pipe. Volumetric flow rate and inlet
and outlet temperatures of air flow through pipe. Convection coefficient and temperature of water.
FIND: Pipe length and fan power requirement. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible heat transfer from air in vertical legs of pipe, (3) Negligible flow work and potential and kinetic energy changes for air flow through pipe, (4) Smooth interior surface, (5) Constant properties. PROPERTIES: Table A-4, Air (Tm,i = 29C): i = 1.155 kg / m3 . Air ( Tm = 25C ) : cp = 1007 -7 2 J/kgK, = 183.6 10 Ns/m , ka = 0.0261 W/mK, Pr = 0.707. ANALYSIS: From Eq. (8.46a)
T - Tm,o T - Tm,i = exp -
UA s mc p
where, from Eq. (3.32),
( UAs )-1 = R tot
=
1 hi Di L
+
ln ( Do / Di ) l + 2 Lk h o D o L
With m = i i = 0.0289 kg / s and Re D = 4m / Di = 13, 350, flow in the pipe is turbulent. Assuming fully developed flow throughout the pipe, and from Eq. (8.60),
k 4 / 5 0.3 0.0261 W / m K 0.023 hi = a 0.023 Re D Pr = (13, 350 )4 / 5 (0.707 )0.3 = 7.20 W / m 2 K Di 0.15m
l 1 ln ( 0.17 / 0.15 ) 1 ( UAs )-1 = + + L 7.21 W / m 2 K 0.15m 2 0.15 W / m K 1500 W / m 2 K 0.17m
UA s = L
( 0.294 + 0.133 + 0.001)
= 17 - 21
= 2.335 L W / K
T - Tm,o
T - Tm,i 17 - 29 ln ( 0.333 ) L=- = 13.7m 0.0802
= 0.333 = exp -
= exp ( -0.0802 ) 0.0289 kg / s 1007 J / kg K
2.335 L
<
2 From Eqs. (8.22a) and (8.22b) and with u m,i = i / Di / 4 = 1.415 m / s, the fan power is
P = ( p ) f
(
)
i u m,i
2 Di
2
L i = 0.0294
1.155 kg / m
(1.415 m / s )2 3 13.7m 0.025 m / s = 0.078 W 2 ( 0.15m )
3
<
-1/ 4 f = 0.316 Re D where = 0.0294 from Eq. (8.20a). COMMENTS: (1) With L/Di = 91, the assumption of fully developed flow throughout the pipe is justified. (2) The fan power requirement is small, and the process is economical. (3) The resistance to heat transfer associated with convection at the outer surface is negligible.
PROBLEM 8.33
KNOWN: Flow rate, inlet temperature and desired outlet temperature of water passing through a tube of prescribed diameter and surface temperature. FIND: (a) Required tube length, L, for prescribed conditions, (b) Required length using tube diameters over the range 30 D 50 mm with flow rates m = 1, 2 and 3 kg/s; represent this design information graphically, and (c) Pressure gradient as a function of tube diameter for the three flow rates assuming the tube wall is smooth. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible potential energy, kinetic energy and flow work changes, (3) Constant properties. PROPERTIES: Table A.6, Water ( Tm = 323 K): cp = 4181 J/kgK, = 547 10-6 Ns/m2, k = 0.643 W/mK, Pr = 3.56. ANALYSIS: (a) From Eq. 8.6, the Reynolds number is
ReD =
4m 4 2 kg s = = 1.16 105 . D (0.04 m ) 547 10-6 N s m 2
(1)
Hence the flow is turbulent, and assuming fully developed conditions throughout the tube, it follows from the Dittus-Boelter correlation, Eq. 8.60,
h= k D 0.023 Re D
4/5
Pr
0.4
=
0.643 W m K 0.04 m
0.023 1.16 10
(
5 4/5
)
(3.56 )0.4 = 6919 W
m K
2
(2)
From Eq. 8.42a, we then obtain
2 kg s ( 4181J kg K ) n 25 C 75 C - mc p n ( To Ti ) L= =- = 10.6 m . Dh ( 0.04 m ) 6919 W m 2 K
(
)
<
(b) Using the IHT Correlations Tool, Internal Flow, for fully developed Turbulent Flow, along with appropriate energy balance and rate equations, the required length L as a function of flow rate is computed and plotted on the right.
Tube length, L (m)
15
10
5 30 35 40 45 50
Tube diameter, D (mm) Flow rate, mdot = 1 kg/s mdot = 2 kg/s mdot = 3 kg/s
Continued...
PROBLEM 8.33 (Cont.)
(c) From Eq. 8.22a the pressure drop is u2 p m =f The friction factor, f, for the smooth surface condition, Eq. 8.21 with 3000 ReD 5 106, is -2 f = 0.790n ( ReD ) - 1.64
x
2D
(4)
(
)
(5)
Using IHT with these equations and Eq. (1), the pressure gradient as a function of diameter for the selected flow rates is computed and plotted on the right.
dP/dx (Pa/m)
5000 4000 3000 2000 1000 0 30 40 Tube diameter, D (mm) Flow rate, mdot = 1 kg/s mdot = 2 kg/s mdot = 3 kg/s 50
COMMENTS: (1) Since L/D = (10.6/0.040) = 265, the assumption of fully developed conditions throughout is justified. (2) The IHT Workspace used to generate the graphical results are shown below.
// Rate Equation Tool - Tube Flow with Constant Surface Temperature: /* For flow through a tube with a uniform wall temperature, Fig 8.7b, the overall energy balance and heat rate equations are */ q = mdot*cp*(Tmo - Tmi) // Heat rate, W; Eq 8.37 (Ts - Tmo) / (Ts - Tmi) = exp ( - P * L * hDbar / (mdot * cp)) // Eq 8.42b // where the fluid and constant tube wall temperatures are Ts = 100 + 273 // Tube wall temperature, K Tmi = 25 + 273 // Inlet mean fluid temperature, K Tmo = 75 + 273 // Outlet mean fluid temperature, K // The tube parameters are P = pi * D // Perimeter, m Ac = pi * (D^2) / 4 // Cross sectional area, m^2 D = 0.040 // Tube diameter, m D_mm = D * 1000 // The tube mass flow rate and fluid thermophysical properties are mdot = rho * um * Ac mdot = 1 // Mass flow rate, kg/s // Correlation Tool - Internal Flow, Fully Developed Turbulent Flow (Assumed): NuDbar = NuD_bar_IF_T_FD(ReD,Pr,n) // Eq 8.60 n = 0.4 // n = 0.4 or 0.3 for Ts>Tm or Ts<Tm NuDbar = hDbar * D / k ReD = um * D / nu /* Evaluate properties at the fluid average mean temperature, Tmbar. */ Tmbar = Tfluid_avg (Tmi,Tmo) // Properties Tool - Water: // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); x=0 // Quality (0=sat liquid or 1=sat vapor) rho = rho_Tx("Water",Tmbar,x) // Density, kg/m^3 cp = cp_Tx("Water",Tmbar,x) // Specific heat, J/kgK nu = nu_Tx("Water",Tmbar,x) // Kinematic viscosity, m^2/s k = k_Tx("Water",Tmbar,x) // Thermal conductivity, W/mK Pr = Pr_Tx("Water",Tmbar,x) // Prandtl number // Pressure Gradient, Equations 8.21, 8.22a: dPdx = f * rho * um^2 / ( 2 * D ) f = ( 0.790 * ln (ReD) - 1.64 ) ^ -2
PROBLEM 8.34
KNOWN: Flow rate and inlet temperature of water passing through a tube of prescribed length, diameter and surface temperature. FIND: (a) Outlet water temperature and rate of heat transfer to water for prescribed conditions, and (b) Compute and plot the required tube length L to achieve Tm,o found in part (a) as a function of the surface temperature for the range 85 Ts 95C. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy, potential energy and flow work effects, (4) Fully developed flow conditions. PROPERTIES: Table A.6, Water ( Tm 325 K): cp = 4182 J/kgK, = 528 10-6 Ns/m2, k = 0.645 W/mK, Pr = 3.42. ANALYSIS: (a) From Eq. 8.6, the Reynolds number is
ReD =
4m 4 2 kg s = = 1.21 105 . D (0.04 m ) 528 10-6 N s m 2
Hence the flow is turbulent, and assuming fully developed conditions throughout the tube, it follows from the Dittus-Boelter correlation, Eq. 8.60, h= k D 0.023 Re 4 / 5 Pr 0.4 = D 0.645 W m K 0.04 m 0.023 1.21 105
(
)
4/5
(3.42 )0.4 = 7064 W
m2 K .
From the energy balance relation, Eq. 8.42b,
DL Tm,o = Ts - Ts - Tm,i exp - h mcp
(
)
Tm,o = 90 C - 90 C - 25 C exp - From the overall energy balance, Eq. 8.37,
(
)
0.04 m 4 m 7064 W m 2 K = 47.5 C 2 kg s 4182 J kg K
< <
q = mcp Tm,o - Tm,i = 2 kg s 4182 J kg K ( 47.5 - 25) C = 188 kW .
(
)
(b) Using the IHT Correlations Tool, Internal Flow, for fully developed Turbulent Flow, along with the energy balance and rate equations used above, the required length, L, to achieve Tm,o = 44.9C (see comment 1 below) as a function of tube surface temperature is computed and plotted below. Continued...
PROBLEM 8.34 (Cont.)
50
Outlet temperature, Tmo (C)
48
46
44
42
40 85 90 Tube temperature, Ts (C) 95
From the plot, the outlet temperature increases nearly linearly with the surface temperature. The convection coefficient and heat rate show similar behavior for this range of conditions. COMMENTS: (1) The mean temperature Tm = 325 K was overestimated in part (a). Another iteration is recommended and the results with Tm = 309 K are: h = 6091 W/m2K, Tm,o = 44.9C and q = 167 kW. (2) The IHT Workspace used to generate the graphical results are shown below.
// Rate Equation Tool - Tube Flow with Constant Surface Temperature: /* For flow through a tube with a uniform wall temperature, Fig 8.7b, the overall energy balance and heat rate equations are */ q = mdot*cp*(Tmo - Tmi) // Heat rate, W; Eq 8.37 (Ts - Tmo) / (Ts - Tmi) = exp ( - P * L * hDbar / (mdot * cp)) // Eq 8.42b // where the fluid and constant tube wall temperatures are Ts = 90 + 273 // Tube wall temperature, K Ts_C = Ts - 273 Tmi = 25 + 273 // Inlet mean fluid temperature, K //Tmo = // Outlet mean fluid temperature, K Tmo_C = Tmo - 273 // The tube parameters are P = pi * D // Perimeter, m Ac = pi * (D^2) / 4 // Cross sectional area, m^2 D = 0.040 // Tube diameter, m D_mm = D * 1000 L= 4 // Tube length, m; unknown // The tube mass flow rate and fluid thermophysical properties are mdot = rho * um * Ac mdot = 2 // Mass flow rate, kg/s // Correlation Tool - Internal Flow, Fully Developed Turbulent Flow (Assumed): NuDbar = NuD_bar_IF_T_FD(ReD,Pr,n) // Eq 8.60 n = 0.4 // n = 0.4 or 0.3 for Ts>Tm or Ts<Tm NuDbar = hDbar * D / k ReD = um * D / nu /* Evaluate properties at the fluid average mean temperature, Tmbar. */ Tmbar = Tfluid_avg (Tmi,Tmo) // Properties Tool - Water: // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); x=0 // Quality (0=sat liquid or 1=sat vapor) rho = rho_Tx("Water",Tmbar,x) // Density, kg/m^3 cp = cp_Tx("Water",Tmbar,x) // Specific heat, J/kgK mu = mu_Tx("Water",Tmbar,x) // Viscosity, Ns/m^2 nu = nu_Tx("Water",Tmbar,x) // Kinematic viscosity, m^2/s k = k_Tx("Water",Tmbar,x) // Thermal conductivity, W/mK Pr = Pr_Tx("Water",Tmbar,x) // Prandtl number
PROBLEM 8.35
KNOWN: Diameters and thermal conductivity of steel pipe. Temperature and velocity of water flow in pipe. Temperature and velocity of air in cross flow over pipe. Cost of producing hot water. FIND: Daily cost of heat loss per unit length of pipe. SCHEMATIC:
ASSUMPTIONS: (1) Steady state, (2) Constant properties, (3) Negligible radiation from outer surface, (4) Fully-developed flow in pipe. PROPERTIES: Table A-4, air (p = 1 atm, Tf 300K): ka = 0.0263 W/mK, a = 15.89 10 m /s, 3 -6 2 Pra = 0.707. Table A-6, water (Tm = 323 K): w = 988 kg/m , w = 548 10 Ns/m , kw = 0.643 W/mK, Prw = 3.56. ANALYSIS: The heat loss per unit length of pipe is
-6 2
q =
Tm - T = R + R + R cnv,w cnd cnv,a
Tm - T ln ( Do / Di ) -1 + ( h a D o ) ( h w Di )-1 + 2 k p
With Re D,w = w u m Di / w = 988 kg / m3 0.5 m / s 0.084 m / 548 10-6 N s / m 2 = 75, 700, flow is turbulent, and for fully developed conditions, the Dittus-Boelter correlation yields k 0.643 W / m K 0.3 h w = w 0.023 Re0.8 Prw = 0.023 (75, 700 )0.8 (3.56 )0.3 = 2060 W / m2 K Dw Di 0.084 m With Re D,a = VD o / a = 3 m / s ( 0.1m ) /15.89 10 -6 m 2 / s = 18, 880, the Churchill-Bernstein correlation yields
k ha = h = a Do
Hence, q =
1/ 0.62 Re1/ 2 Pra 3 D,a 0.3 + 1/ 4 1 + ( 0.4 / Pr )2 / 3 a
50C - ( -5C )
5/8 Re D,w 1 + 282, 000
4/5
2 = 20.1W / m K
(
1.84 10-3 + 0.46 10-3 + 158.3610-3 K / W
The daily energy loss is then and the associated cost is
Q = 0.346 kW / m 24 h / d = 8.22 kW h / d m C = (8.22 kW h / d m )($0.05 / kW h ) = $0.411/ m d
)
= 342 W / m = 0.342 kW / m
<
COMMENTS: Because R cnv,a >> R cnv,w , the convection resistance for the water side of the pipe could have been neglected, with negligible error. The implication is that the temperature of the pipe's inner surface closely approximates that of the water. If R cnv,w is neglected, the heat loss is q = 346 W / m.
PROBLEM 8.36 KNOWN: Inner and outer diameter of a steel pipe insulated on the outside and experiencing uniform heat generation. Flow rate and inlet temperature of water flowing through the pipe. FIND: (a) Pipe length required to achieve desired outlet temperature, (b) Location and value of maximum pipe temperature. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy, potential energy and flow work changes, (4) One-dimensional radial conduction in pipe wall, (5) Outer surface in adiabatic. PROPERTIES: Table A-1, Stainless steel 316 (T 400K): k = 15 W/mK; Table A-6, Water ( Tm = 303K ) : cp = 4178 J/kgK, k = 0.617 W/mK, = 803 10-6 Ns/m2, Pr = 5.45. ANALYSIS: (a) Performing an energy balance for a control volume about the inner tube, it follows that
2 m c p Tm,o - Tm,i = q = q ( /4 ) Do - Di2 L & &
(
)
(
)
L=
& m c p Tm,o - Tm,i & q ( /4 ) D2 - Di2 o
(
(
)=
)
( 0.1 kg/s ) 4178 ( J/kg K ) 20o C 106 W/m3 ( / 4 ) ( 0.04m )2 - ( 0.02m )2
L = 8.87m.
<
(b) The maximum wall temperature exists at the pipe exit (x = L) and the insulated surface (r = ro). From Eq. 3.50, the radial temperature distribution in the wall is of the form T (r ) = - & q 2 r + C1ln r + C 2. 4k
Considering the boundary conditions; r = ro : & dT q C =0 = - ro + 1 dr r = r 2k ro
o
C1 =
& 2 qro 2k Continued .....
PROBLEM 8.36 (Cont.) r = ri : T ( ri ) = Ts = - & & 2 q 2 q ro ri + ln ri + C2 4k 2k C2 = & & 2 q 2 q ro ri - ln ri + Ts . 4k 2k
The temperature distribution and the maximum wall temperature (r = ro) are T (r ) = - & & 2 q 2 2 q ro r r - ri + ln + Ts 4k 2k ri & & 2 q 2 2 q ro r ro - ri + ln o + Ts 4k 2k ri
(
)
Tw,max = T ( ro ) = -
(
)
where Ts , the inner surface temperature of the wall at the exit, follows from q = s & q ( /4 ) D2 - D2 L o i Di L
(
)
=
& o q D2 - D2 i 4 Di
(
) = h (Ts - Tm,o)
where h is the local convection coefficient at the exit. With & 4m 4 0.1 kg/s Re D = = = 7928 Di ( 0.02m ) 803 10-6 N s/m 2 the flow is turbulent and, with (L/Di) = (8.87 m/0.02m) = 444 >> (xfd/D) 10, it is also fully developed. Hence, from the Dittus-Boelter correlation, Eq. 8.60,
h=
(0.023 Re D
k
i
0.617 W/m K 4/5 0.4 = 0.023 ( 7928 ) 4 / 5 5.45 0.4 = 1840 W/m2 K. D Pr 0.02 m
)
Hence, the inner surface temperature of the wall at the exit is 2 2 & o 106 W/m3 ( 0.04m ) - ( 0.02m ) q D2 - D2 i + 40o C = 48.2oC Ts = + Tm,o = 2 K 0.02m 4 h Di 4 1840 W/m ( )
(
)
and
Tw,max = -
106 W/m3 0.02m )2 - ( 0.01m ) 2 ( 4 15 W/m K 106 W/m3 ( 0.02m )2 0.02 + ln + 48.2 o C = 52.4oC. 2 15 W/m K 0.01
<
COMMENTS: The physical situation corresponds to a uniform surface heat flux, and Tm increases linearly with x. In the fully developed region, Ts also increases linearly with x.
PROBLEM 8.37
KNOWN: Dimensions and thermal conductivity of concrete duct. Convection conditions of ambient air. Flow rate and inlet temperature of water flow through duct. FIND: (a) Outlet temperature, (b) Pressure drop and pump power requirement, (c) Effect of flow rate and pipe diameter on outlet temperature. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Fully developed flow throughout duct, (3) Negligible pipe wall conduction resistance, (4) Negligible potential energy, kinetic energy and flow work changes for water, (5) Constant properties. PROPERTIES: Table A-6, water ( Tm 360 K ) : = 967 kg / m3 , c p = 4203 J / kg K, = 324 10 -6
N s / m , k w = 0.674 W / m K, Pr = 2.02.
2
ANALYSIS: (a) The outlet temperature is given by
Tm,o = T + Tm,i - T exp - UA / m cp
where
(
) (
)
)-1
= 8.75 10-4 K / W
UA = ( R tot ) R cnd =
-1
= R cnv,w + R cnd + R cnv,a =
(
ln (1.08 w / D ) 2 kL
-1
ln (1.08 0.30m / 0.15m ) 2 (1.4 W / m K )100m
R cnv,a = ( 4 w L h )
With
= 4 0.3m 100m 25 W / m 2 K
(
ReD = 4 m / D = ( 4 2 kg / s ) / 0.15m 324 10-6 N s / m 2 = 52, 400,
(
)
-1
= 3.33 10-4 K / W
)
k 4 / 5 0.3 0.674 W / m K 0.023 h w h fd = w 0.023 Re D Pr = (52, 400 )4 / 5 ( 2.02 )0.3 = 761 W / m 2 K D 0.15m
R cnv,w = D L h w
(
)-1 = ( 0.15m 100m 761W / m2 K )
-1
= 2.79 10-5 K / W
UA = 2.79 10-5 + 8.75 10-4 + 3.33 10-4 K / W
(
)
-1
= 809 W / K
809 W / K Tm,o = 0C + 90C exp - = 81.7C 2 kg / s 4203J / kg K
Continued .....
<
PROBLEM 8.37 (Cont.)
(b) With f = 0.0206 from Fig. 8.3 and u m = m / D 2 / 4 = 0.117 m / s,
3 u2 m L = 0.0206 967 kg / m ( 0.117 m / s ) 100m = 91N / m 2 = 8.98 10-4 bars p = f 2D 2 0.15m 2
With = m / = 2.07 10 -3 m3 / s, the pump power requirement is
<
P = p = 91N / m 2 2.07 10-3 m3 / s = 0.19 W
(
)
<
(c) The effects of varying the flowrate and duct diameter were assessed using the IHT software, and results are shown below.
90
90 O u tle t te m p e ra tu re , Tm o (C ) 85 80 75 70 65 60 0 1 2 3 4 5 Ma s s flo w ra te , m d o t(kg /s )
O u tle t te m p e ra tu re , Tm o (C )
87
84
81
78
75 0 .0 5 0 .1 0 .1 5 D u ct d ia m e te r, D (m ) 0 .2 0 .2 5
Although R cnv,w , and hence R tot , decreases with increasing m , thereby increasing UA, the effect is
significantly less than that of m to the first power, causing the exponential term, exp ( - UA / m c p ) , to
approach unity and Tm,o to approach Tm,i . The effect can alternatively be attributed to a reduction
in the residence time of the water in the pipe (um increases with increasing m for fixed D). With and w, T increasing D for fixed m m,o decreases due to an increase in the residence time, as well as a
reduction in the conduction resistance, R cnd . COMMENTS: (1) Use of Tm = 360 K to evaluate properties of the water for Parts (a) and (b) is reasonable, and iteration is not necessary. (2) The pressure drop and pump power requirement are small.
PROBLEM 8.38
KNOWN: Water flow through a thick-walled tube immersed in a well stirred, hot reaction tank maintained at 85C; conduction thermal resistance of the tube wall based upon the inner surface area is R = 0.002 m 2 K / W. cd FIND: (a) The outlet temperature of the process fluid, Tm,o; assume, and then justify, fully developed flow and thermal conditions within the tube; and (b) Do you expect Tm,o to increase or decrease if the combined thermal entry condition exists within the tube? Estimate the outlet temperature of the process fluid for this condition. SCHEMATIC:
ASSUMPTIONS: (1) Flow is fully developed, part (a), (2) Constant properties, (3) Negligible kinetic and potential energy changes and flow work, and (4) Constant wall temperature heating. PROPERTIES: Table A-6, Water (Tm = (Tm,o + Tm,i)/2 = 337 K): cp = 4187 J/kgK, = 4.415 -4 2 -4 2 10 Ns/m , k = 0.6574 W/mK, Pr = 2.80; (Ts = 358 K): s = 3.316 10 Ns/m . ANALYSIS: (a) The outlet temperature is determined from the rate equation, Eq. 8.46a, written as
Ts - Tm,o
UAs = exp - m cp Ts - Tm,i
(1)
where the overall coefficient, based upon the inner surface area of the tube is expressed in terms of the convection and conduction thermal resistances,
1 1 = + R cd,i U h
(2) To estimate h, begin by characterizing the flow
ReD = 4 m / D
(3)
ReD = 4 (33 / 3600 kg / s ) / 0.012 m 4.415 10-4 N s / m 2 = 2210
Consider the flow as laminar, and assuming fully developed conditions, estimate h with the correlation of Eq. 8.55,
Nu D = hD / k = 3.66 h = 3.66 0.6574 W / m K / 0.012 m = 201 W / m 2 K
From Eq. (2),
(4)
U = 1/ 201 W / m 2 K + 0.002 m 2 K / W
and from Eq. (1), with As = DL, calculate Tm,o.
-1
= 143.1 W / m 2 K
Continued .....
PROBLEM 8.38 (Cont.) 85 - Tm,o 143.1 W / m 2 K 0.012 m 8 m = exp - 85 - 20 33 / 3600 kg / s 4187 J / kg K
Tm,o = 64C
<
Fully developed flow and thermal conditions are justified if the tube length is much greater than the fully developed length xfd,t. From Eq. 8.23,
x fd,t D
= 0.05 ReD Pr
x fd,t = 0.012 m 0.05 221.0 2.41 = 3.20 m
That is, the length is only twice that required to reach fully developed conditions. (b) Considering combined entry length conditions, estimate the convection coefficient using the Sieder-Tate correlation, Eq. 8.56,
Nu D = 3.66 +
0.0668 ( D / L ) ReD Pr 1 + 0.04 ( D / L ) ReD Pr
2/3
(5)
substituting numerical values, find
Nu D = 4.05 U = 154 W / m 2 K
h = 222 W / m 2 K
which is a 10% increase over the fully developed analysis result. Using the foregoing relations, find
Tm,o = 65.5C
<
COMMENTS: (1) The thermophysical properties for the fully developed correlation are evaluated at the mean fluid temperature Tm = (Tm,o + Tm,i)/2. The values are shown above in the properties section. (2) For the Sieder-Tate correlation, the properties are also evaluated at Tm, except for s, which is evaluated at Ts. (3) For this case where the tube length is about twice xfd,t, the average heat transfer coefficient is larger as we would expect, but amounts to only a 10% increase.
PROBLEM 8.39
KNOWN: Flow rate and temperature of atmospheric air entering a duct of prescribed diameter, length and surface temperature. FIND: (a) Air outlet temperature and duct heat loss for the prescribed conditions and (b) Calculate and plot q and p for the range of diameters, 0.1 D 0.2 m, maintaining the total surface area, As = DL, at the same value as part (a). Explain the trade off between the heat transfer rate and pressure drop. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy and potential energy changes, (4) Uniform surface temperature, (5) Fully developed flow conditions. PROPERTIES: Table A.4, Air ( Tm 310 K, 1 atm): = 1.128 kg/m3, cp = 1007 J/kgK, = 189 107 Ns/m2, k = 0.027 W/mK, Pr = 0.706. ANALYSIS: (a) With
ReD =
4m 4 0.04 kg s = = 17,965 D (0.15 m )189 10-7 N s m 2
the flow is turbulent. Assuming fully developed conditions throughout the tube, it follows from the Dittus-Boelter correlation, Eq. 8.60, that k 0.027 W m K 4/5 h = 0.023 Re 4 / 5 Pr 0.4 = 0.023 (17, 965 ) (0.706 )0.4 = 9.44 W m 2 K . D D 0.15 m Hence, from the energy balance relation, Eq. 8.42b,
DL Tm,o = Ts - Ts - Tm,i exp - h mcp (0.15 m )10 m 9.44 W m 2 K Tm,o = 15 C + 45 C exp - 0.04 kg s (1007 J kg K )
(
)
(
) = 29.9 C
< <
From the overall energy balance, Eq. 8.37, it follows that
q = mcp Tm,o - Tm,i = 0.04 kg s 1007 J kg K ( 29.9 - 60 ) C = -1212 W .
From Eq. 8.22a, the pressure drop is u2 mL p = f
(
)
2D
Continued...
PROBLEM 8.39 (Cont.)
and for the smooth surface conditions, Eq. 8.21 can be used to evaluate the friction factor,
f = ( 0.790 ln ( ReD ) - 1.64 )
Hence, the pressure drop is
-2
= ( 0.790 ln (17, 965) - 1.64 )
-2
= 0.0269
p = 0.0269
1.128 kg m3 ( 2.0 m s ) 2 0.15m
2
10 m = 4.03 N m 2
<
where um = m A c = 0.04 kg s 1.128 kg m3 0.152 m 2 4 = 2.0 m s .
(b) For the prescribed conditions of part (a), As = DL = (0.15 m) 10 m = 4.712 m2, using the IHT Correlations Tool, Internal Flow for fully developed Turbulent Flow along with the energy balance equation, rate equation and pressure drop equations used above, the heat rate q and p are calculated and plotted below.
15 Tube length, L (m) 50 40 deltaP (Pa) 30 20 10 5 100 125 150 175 200 0 100 125 150 175 200
(
)
10
Tube diameter, D (mm)
Tube diameter, D (mm)
From above, as D increases, L decreases so that As remains unchanged. The decrease in heat rate with increasing diameter is nearly linear, while the pressure drop decreases markedly. This is the trade off: increased heat rate requires a more significant increase in pressure drop, and hence fan blower power requirements.
2 Heat rate, q (kW) 1.5 1 0.5 0 100 120 140 160 180 200
Tube diameter, D (mm)
COMMENTS: (1) To check the calculations, compute q from Eq. 8.44, where Tm is given by Eq. 8.45. It follows that Tm = -27.1C and q = -1206 W. The small difference in results may be attributed to round-off error. (2) For part (a), a slight improvement in accuracy may be obtained by evaluating the properties at Tm = 318 K: h = 9.42 W/m2K, Tm,o = 303 K = 30C, q = -1211 W, f = 0.0271 and p = 4.20 N/m2.
PROBLEM 8.40
KNOWN: Inlet temperature, pressure and flow rate of air. Tube diameter and length. Pressure of saturated steam. FIND: Outlet temperature and pressure of air. Mass rate of steam condensation. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Outer surface of annulus is adiabatic, (3) Negligible potential energy, kinetic energy and flow work changes for air, (4) Fully-developed flow throughout the tube, (5) Smooth tube surface, (6) Constant properties. PROPERTIES: Table A-4, air ( Tm 325 K, p = 5 atm ) : = 5 (1atm ) = 5.391kg / m3 ,
c p = 1008 J / kg K, = 196.4 10
-7
N s / m , k = 0.0281 W / m K, Pr = 0.703. Table A-6, sat. steam (p
2
= 2.455 bars): Ts = 400 K, hfg = 2183 kJ/kg. ANALYSIS: With a uniform surface temperature, the air outlet temperature is
With Re D = 4m / Di = 0.12 kg / s / ( 0.05m )196.4 10-7 kg / s m = 38, 980, the flow is turbulent, and the Dittus-Boelter correlation yields
h h fd =
D L i h Tm,o = Ts - Ts - Tm,i exp - m cp
(
)
k 4 / 5 0.4 0.023 Re D Pr Di
=
0.0281 W / m K 0.023 (38, 980 )4 / 5 (0.703 )0.4 = 52.8 W / m 2 K 0.05m
0.05m 5m 52.8 W / m 2 K = 99C Tm,o = 127C - (110C ) exp - 0.03kg / s 1008 J / kg K
The pressure drop is p = f u 2 / 2 Di L, where, with A c = Di2 / 4 = 1.963 10 -3 m 2 , m
u m = m / A c = 2.83 m / s, and with Re D = 38, 980, Fig. 8.3 yields f 0.022. Hence,
<
(
)
2 3 ( 2.83m / s ) 5m = 95 N / m 2 = 9.4 10-4 atm p 0.022 5.391kg / m
2 0.05m
<
The rate of heat transfer to the air is
q = m cp Tm,o - Tm,i = 0.03kg / s 1008 J / kg K (82C ) = 2480 W
and the rate of condensation is then
(
)
mc =
COMMENTS: (1) With Tm = ( Tm,i + Tm,o ) / 2 = 331K, the initial estimate of 325 K is reasonable and iteration is not necessary. (2) For a steam flow rate of 0.01 kg/s, approximately 10% of the outflow would be in the form of saturated liquid, (3) With L/Di = 100, it is reasonable to assume fully developed flow throughout the tube.
q 2480 W = = 1.14 10-3 kg / s 6 J / kg h fg 2.183 10
<
PROBLEM 8.41
KNOWN: Duct diameter and length. Thermal conductivity of insulation. Gas inlet temperature and velocity and minimum allowable outlet temperature. Temperature and velocity of air in cross flow. FIND: Minimum allowable insulation thickness. SCHEMATIC:
ASSUMPTIONS: (1) Negligible potential and kinetic energy and flow work changes for gas flow through duct, (2) Fully developed flow throughout duct, (3) Negligible duct wall conduction resistance, (4) Negligible effect of insulation thickness on outer convection coefficient and thermal resistance, (5) Properties of gas may be approximated as those of air. PROPERTIES: Table A-4, air (p = 1 atm). Tm,i = 1600K: (i = 0.218 kg/m ). Tm = (Tm,i +Tm,o)/2 = 1500K: ( = 0.232 kg/m , cp = 1230 J/kgK, = 557 10 Ns/m , k = 0.100 W/mK, Pr = 0.685). -6 2 Tf 300K (assumed): = 15.89 10 m /s, k = 0.0263 W/mK, Pr = 0.707. ANALYSIS: From Eqs. (8.46a) and (3.19),
3 -7 2 3
T - Tm,o T - Tm,i
=
UAs 1 -1150 K = 0.852 = exp - = exp - m cp R tot m cp -1350 K
-1
Hence, with m = ( u m A c )i = 0.218 kg / m3 10 m / s (1m )2 / 4 = 1.712 kg / s,
R tot = - m c p ln ( 0.852 )
= - [1.712 kg / s 1230 J / kg K ( -0.160 )]
-1
= 2.96 10
-3
K/W
The total thermal resistance is
R tot = R conv,i + R cond,ins + R conv,o = ( h i Di L )
-1
+
ln ( Do / Di ) 2 k ins L
+ ( h o Do L )
-1
(1)
With Re D,i = 4m / D i = ( 4 1.712 kg / s ) / 1m 557 10 -7 N s / m 2 = 39,130, the Dittus-Boelter
(
)
correlation yields
hi =
k 0.023 Re 4 / 5 Pr1/ 3 = 0.100 W / m K 0.023 (39,130 )4 / 5 ( 0.685 )1/ 3 = 9.57 W / m 2 K D 1m D
-1
The internal resistance is then
R conv,i = ( h i Di L )
= 9.57 W / m 2 K 1m 100m
-6 2 5
(
)
-1
= 3.33 10-4 K / W
With ReD VDi/ = 15 m/s 1m/15.89 10 m /s = 9.44 10 , the Churchill-Bernstein correlation yields Continued .....
PROBLEM 8.41 (Cont.) 0.62 Re1/ 2 Pr1/ 3 k D h o 0.3 + 1/ 4 D 1 + ( 0.4 / Pr )2 / 3 R conv,o ( h o Di L )
Hence, from Eq. (1) -1 5/8 ReD 1 + 282, 000 4/5
2 = 30.9 W / m K
= 30.9 W / m 2 K 1m 100m
(
)
-1
= 1.03 10-4 K / W
ln ( Do / Di ) 2 kins L
= 2.96 10-3 - 3.33 10-4 - 1.03 10-4 K / W = 2.52 10-3 K / W
-3
(
)
D o = Di exp 2 k ins L 2.52 10
(
K / W = 1m exp 1.58 10
)
(
-2
K / W 0.125 W / m K 100m = 1.22m
)
Hence, the minimum insulation thickness is
t min = ( Do - Di ) / 2 = 0.11m
<
COMMENTS: With Do = 1.22m, use of Di = 1m to evaluate the outer convection coefficient and thermal resistance is a reasonable approximation. However, improved accuracy may be obtained by using the calculated value of Do to determine conditions at the outer surface and iterating on the solution.
PROBLEM 8.42
KNOWN: Flow rate, inlet temperature and desired outlet temperature of liquid mercury flowing through a tube of prescribed diameter and surface temperature. FIND: Required tube length and error associated with use of a correlation for moderate to large Pr fluids. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible kinetic energy, potential energy and flow work effects, (4) Fully developed flow. PROPERTIES: Table A-5, Mercury Tm = 350K : cp = 137.7 J/kgK, = 0.1309 10 k = 9.18 W/mK, Pr = 0.0196. ANALYSIS: The Reynolds and Peclet numbers are
c
h
-2
Ns/m ,
2
Re D =
& 4m 4 0.5 kg/s = = 9727 D ( 0.05m ) 0.1309 10 -2 N s/m 2
Pe D = Re D Pr = 9727 ( 0.0196 ) = 191.
Hence, assuming fully developed turbulent flow throughout the tube, it follows from Eq. 8.66 that
h=
k 9.18 W/m K 5.0 + 0.025 Pe0.8 = 5.0 + 0.025 1910.8 = 1224 W/m 2 K. D D 0.05 m & m cp Dh To ( 0.5 kg/s ) 137.7 J/kg K 450 - 400 =- ln = 0.39 m. Ti ( 0.05 m )1224 W/m 2 K 450 - 300
(
)
(
)
From Eq. 8.42a, it follows that
L =-
ln
<
If the Dittus-Boelter correlation, Eq. 8.60, is used in place of Eq. 8.66, 2 k 4/5 Pr 0.4 = 9.18 W/m K 0.023 9727 4 / 5 0.0196 0.4 = 1358 W/m 2 K h = 0.023 ReD ( ) ( ) D 0.05 m and the required tube length is
L =-
& m cp Dh
ln
To ( 0.5 kg/s ) 137.7 J/kg K l 450 - 400 = =- n 0.35 m. Ti ( 0.05 m )1358 W/m 2 K 450 - 300
<
COMMENTS: Such good agreement between results does not occur in general. For example, if ReD = 2 10 , h = 1463 from Eq. 8.66 and 2417 from Eq. 8.60. Large errors are usually associated with using conventional (moderate to large Pr) correlations with liquid metals.
4
PROBLEM 8.43 KNOWN: Surface temperature and diameter of a tube. Velocity and temperature of air in cross flow. Velocity and temperature of air in fully developed internal flow. FIND: Convection heat flux associated with the external and internal flows. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform cylinder surface temperature, (3) Fully developed internal flow. PROPERTIES: Table A-4, Air (298K): = 15.71 10 ANALYSIS: For the external and internal flows, Re D = VD u mD 30 m/s 0.05 m = = = 9.55 10 4. -6m 2 / s 15.7110
-6
m /s, k = 0.0261 W/mK, Pr = 0.71.
2
From the Zhukauskas relation for the external flow, with C = 0.26 and m = 0.6, NuD = C Re m Pr n ( Pr/Prs )1/4 = 0.26 9.55 104 D Hence, the convection coefficient and heat flux are h= k 0.0261 W/m K Nu D = 223 = 116.4 W/m2 K D 0.05 m
o
(
)
0.6
( 0.71)0.37 (1)1 / 4 = 223.
q = h ( Ts - T ) = 116.4W/m 2 K (100 - 25 ) C = 8.73 103 W/m 2 . Using the Dittus-Boelter correlation, Eq. 8.60, for the internal flow, which is turbulent, NuD = 0.023 Re 4/5 Pr 0.4 = 0.023 9.55 104 D h=
<
(
)
4/5
( 0.71)0.4 = 193
k 0.0261 W/m K Nu D = 193 = 101 W/m 2 K D 0.05 m
o
and the heat flux is q = h ( Ts - Tm ) = 101 W/m 2 K (100 - 25 ) C = 7.58 103 W/m 2 .
<
COMMENTS: Convection effects associated with the two flow conditions are comparable.
PROBLEM 8.44
KNOWN: Diameter, length and surface temperature of condenser tubes. Water velocity and inlet temperature. FIND: (a) Water outlet temperature evaluating properties at Tm = 300 K, (b) Repeat calculations using properties evaluated at the appropriate temperature, Tm = (Tm,i + Tm,o)/2, and (c) Coolant mean velocities for the range 4 L 7 m which provide the same Tm,o as found in part (b). SCHEMATIC:
ASSUMPTIONS: (1) Negligible tube wall conduction resistance, (2) Negligible kinetic energy, potential energy and flow work changes. PROPERTIES: Table A.6, Water ( Tm = 300 K): = 997 kg/m3, cp = 4179 J/kgK, = 855 10-6 kg/sm, k = 0.613 W/mK, Pr = 5.83. ANALYSIS: (a) From Equation 8.42b
Tm,o = Ts - Ts - Tm,i exp - DL mcp h . and evaluating properties at Tm = 300 K, find ReD =
3 u m D 997 kg m (1m s ) 0.0254 m = = 29, 618 855 10-6 kg s m
(
)
(
)
The flow is turbulent, and since L/D = 197, it is reasonable to assume fully developed flow throughout the tube. Hence, h hfd. From the Dittus-Boelter equation, 4/5 Nu D = 0.023Re 4 / 5 Pr 0.4 = 0.023 ( 29, 618 ) (5.83)0.4 = 176 D h = Nu D ( k D ) = 176 ( 0.613 W m K 0.0254 m ) = 4248 W m 2 K . With 2 m = u m D2 4 = ( 4 ) 997 kg m3 (1m s )( 0.0254 m ) = 0.505 kg s .
(
)
Equation 8.42b yields
( 0.0254 m ) 5m 4248 W m 2 K Tm,o = 350 K - ( 60 K ) exp - 0.505 kg s ( 4179 J kg K )
(
) = 323K = 50 C
<
(b) Using the IHT Correlations Tool, Internal Flow, for fully developed Turbulent Flow, along with the energy balance and rate equations above, the calculation of part (a) is repeated with Tm = (Tm,i + Tm,o)/2 giving these results:
Tm = 307.3K
Tm,o = 51.7C = 324.7 K
<
Continued...
(c) Using the IHT model developed for the part (b) analysis, the coolant mean velocity, um, as a function of tube length L with Tm,o = 51.7C is calculated and the results plotted below.
PROBLEM 8.44 (Cont.)
6
5
Velocity, um (m/s)
4
3
2
1
0 4 5 Tube length, L (m) 6 7
COMMENTS: (1) Using Tm = 300 K vs. Tm = (Tm,i + Tm,o)/2 = 307 K for this application resulted in a difference of Tm,o = 50C vs.Tm,o = 51.7C. While the difference is only 1.7C, it is good practice to use the proper value for Tm . (2) Note that um must be increased markedly with increasing length in order that Tm,o remain fixed.
PROBLEM 8.45
KNOWN: Gas turbine vane approximated as a tube of prescribed diameter and length maintained at a known surface temperature. Air inlet temperature and flowrate. FIND: (a) Outlet temperature of the air coolant for the prescribed conditions and (b) Compute and plot the air outlet temperature Tm,o as a function of flow rate, 0.1 m 0.6 kg/h. Compare this result with those for vanes having passage diameters of 2 and 4 mm. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes. PROPERTIES: Table A.4, Air (assume Tm = 780 K, 1 atm): cp = 1094 J/kgK, k = 0.0563 W/mK, = 363.7 10-7 Ns/m2, Pr = 0.706; (Ts = 650C = 923 K, 1 atm): = 404.2 10-7 Ns/m2. ANALYSIS: (a) For constant wall temperature heating, from Eq. 8.43,
Ts - Tm,o
where P = D. For flow in circular passage,
PLh = exp - mcp Ts - Tm,i
(1)
ReD =
4 0.18 kg h (1 3600s h ) 4m = = 584 . D (0.003m ) 363.7 10-7 N s m 2
(2)
The flow is laminar, and since L/D = 75 mm/3 mm = 25, the Sieder-Tate correlation including combined entry length yields 0.14 1/ 3 ReD Pr hD Nu D = (3) = 1.86 k L D s 0.14 1/ 3 -7 0.0563 W m K 584 0.706 363.7 10 h= 1.86 = 87.5 W m 2 K . 404.2 10-7 0.003m 25
Hence, the air outlet temperature is
( 0.003m ) 0.075 m 87.5 W m 2 K = exp - (0.18 3600 ) kg s 1094 J kg K (650 - 427 ) C 650 - Tm,o Tm,o = 578 C
<
(b) Using the IHT Correlations Tool, Internal Flow, for Laminar Flow with combined entry length, along with the energy balance and rate equations above, the outlet temperature Tm,o was calculated as a function of flow rate for diameters of D = 2, 3 and 4 mm. The plot below shows that Tm,o decreases nearly linearly with increasing flow rate, but is independent of passage diameter. Continued...
PROBLEM 8.45 (Cont.)
650
Outlet temperature, Tmo (C)
600
550
500 0.1 0.2 0.3 0.4 0.5 0.6
Flow rate, mdot (kg/h)
COMMENTS: (1) Based upon the calculation for Tm,o = 578C, Tm = 775 K which is in good agreement with our assumption to evaluate the thermophysical properties. (2) Why is Tm,o independent of D? From Eq. (3), note that h is inversely proportional to D, h ~ D-1. From Eq. (1), note that on the right-hand side the product P h will be independent of D. Hence, Tm,o will depend only on m . This is, of course, a consequence of the laminar flow condition and will not be the same for turbulent flow.
PROBLEM 8.46
KNOWN: Gas-cooled nuclear reactor tube of 20 mm diameter and 780 mm length with helium heated from 600 K to 1000 K at 8 10
-3
kg/s.
FIND: (a) Uniform tube wall temperature required to heat the helium, (b) Outlet temperature and required flow rate to achieve same removal rate and wall temperature if the coolant gas is air. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic energy and potential energy changes, (3) Fully developed conditions. PROPERTIES: Table A-4, Helium ( Tm = 800K, 1 atm ) : = 0.06272 kg/m , cp = 5193 J/kgK, k
3 -7 2 -4
= 0.304 W/mK, = 382 10
3
Ns/m , = 6.09 10
-3
m /s, Pr = 0.654; Air ( Tm = 800K, 1 atm ) :
2 -6
= 0.4354 kg/m , cp = 1099 J/kgK, k = 57.3 10
W/mK, = 84.93 10
m /s, Pr = 0.709.
2
ANALYSIS: (a) For helium and a constant wall temperature, from Eq. 8.46,
Ts - Tm,o
PLh = exp - m cp & Ts - Tm,i & 4m 4 8 10-3 kg/s = = 1.333 10 4 -7 N s/m2 D 0.020 m 382 10
where P = D. For the circular tube,
Re D =
and using the Colburn correlation for turbulent, fully developed flow,
Nu = 0.023 Re4/5 Pr1/3 = 0.023 1.333 104 D
(
)
4/5
( 0.654)1/3 = 39.83
h = Nu k/D = 39.83 0.304 W/m K/0.02 m = 605 W/m 2 K.
Hence, the surface temperature is
( 0.020 m ) 0.780 m 605 W/m 2 K Ts - 1000 K = 0.4898 = exp - Ts - 600 K 8 10-3 kg/s 5193 J/kg K Ts = 1384 K.
The heat rate with helium coolant is & q = m cp Tm,o - Tm,i = 8 10 -3 kg/s 5193 J/kg K (1000- 600 ) K = 16.62 kW. Continued .....
< )
(
PROBLEM 8.46 (Cont.)
(b) For the same heat removal rate (q) and wall temperature (Ts ) with air supplied at Tm,i, the relevant relations are
& q = 16,620 W = ma c p Tm,o - Tm,i Ts - Tm,o PLha = exp - & Ts - Tm,i m a cp & 4m a D
(
)
(1) (2)
Re =
hD = 0.023 Re 4/5 Pr1/3 D k
(3,4)
& where Tm,o and m are unknown. An iterative solution is required: assume a value of Tm,o and find & m from Eq. (1); use m in Eqs. (3) and (4) to find h and then Eq. (2) to evaluate Tm,o; compare & results and iterate. Using thermophysical properties of air evaluated at Tm = 800K, the above
relations, written in the order they would be used in the iteration, become
& ma =
15.123 Tm,o - 600
(5) (6) (7)
& 4/5 ha = 5.725 103 ma & Tm,o = 1384 - 784 exp -4.459 10-5 ( h a / m a )
Results of the iterative solution are
Trial
Tm,o (K) (Assumed)
& m (kg/s) Eq. (5) 3.025 10 -2 4.321 10 -2 5.041 10 -2 5.215 10
-2
ha
( W/m K )
2
Tm,o (K) Eq. (7) 915.0 898. 891.0 889.5
Eq. (6) 348.6 463.7 524.6 539.0
1 2 3 4
Hence, we find
1000 950 900 890
& m a = 5.22 10 -2 kg/s
Tm,o = 890 K.
<
COMMENTS: To achieve the same cooling rate with air, the required mass rate is 6.5 times that obtained with helium.
PROBLEM 8.47 KNOWN: Air at prescribed inlet temperature and mean velocity heated by condensing steam on its outer surface. FIND: (a) Air outlet temperature, pressure drop and heat transfer rate and (b) Effect on parameters of part (a) if pressure were doubled. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes, (3) Thermal resistance of tube wall and condensate film are negligible. PROPERTIES: Table A-4, Air (assume Tm = 450K, 1 atm = 101.3 kPa): = 0.7740 kg/m , cp = 1021 J/kgK, = 250.7 10 Ns/m , k = 0.0373 W/mK, Pr = cp/k = 0.686. Note that only is pressure dependent; i.e., P; Table A-6, Saturated water (20 bar): Tsat = Ts = 485K. ANALYSIS: (a) For constant wall temperature heating, from Eq. 8.46 but with U h i since ho >> hi , where ho is the convection coefficient for the condensing steam, Ts - Tm,o PL = exp - hi m cp Ts - Tm,i &
-7 2 3
where P = D. For the air flow, find the mass rate and Reynolds number,
& m = A cu m = 0.7740 kg/m 3 ( 200 kPa/101.3 kPa ) ( 0.025 m )2 / 4 6 m/s & m = 4.501 10-3kg/s. Re D = & 4m 4 4.501 10 -3 kg/s = = 9.143 10 3. -7 N s/m 2 0.025 m D 250.7 10 ( )
(
)
Using the Dittus-Boelter correlation for fully-developed turbulent flow, NuD = 0.023Re 4/5 Pr 0.4 = 0.023 9.143 103
(
)
4/5
( 0.682 )0.4 = 29.12
h i = Nu k/D = 29.12 0.0373 W/m K/0.025 m = 43.4 W/m 2 K. Hence, the outlet temperature is ( 0.025 m ) 2 m 43.4 W/m 2 K 212 - Tm,o = exp - 4.50110-3 kg/s 1021 J/kg K ( 212 - 150 )o C Tm,o = 198o C. Continued .....
<
PROBLEM 8.47 (Cont.) The pressure drop follows from Eqs. 8.20 and 8.22, f = 0.316Re -1/4 = 0.316 9.143 10 3 D p = f u2 mL 2D 0.7740 kg/m 3 ( 200/101.3)( 6 m/s )2 2 m = 71.1 N/m 2 . 2 0.025 m
(
)
-1/4
= 0.0323
p = 0.0323
< <
The heat transfer rate is
-3 & q = m cp Tm,o - T m,i = 4.50110 kg/s 1021 J/kg K (198 -150 ) K = 221 W.
(
)
(b) If the pressure were doubled, we can see from the above relations, that m , hence & & & m = 2m o Re D = 2Re D,o , since h i ( Re )4 / 5 h i / h i,o = 2 4 / 5 , h i = 1.74h i,o. It follows that Tm,o = 195C, so that the effect on temperature is slight. However, the pressure -1/4 drop increases by the factor 2(2) = 1.68 and the heat rate by 2(195 - 150)/(198 - 150) = 1.88. In summary: Parameter & m, kg/s 10 2 hi, W/m K Tmo - Tm,iC 2 p, N/m q, W
3
(
)
p = 200 kPa Part (a) 4.501 43.4 48 71.1 221
p = 400 kPa Part (b) 9.002 86.8 45 119 415
Increase, % 100 100 -6 68 88
COMMENTS: (1) Note that Tm = (198 + 150)C/2 = 447 K agrees well with the assumed value (450 K) used to evaluate the thermophysical properties.
PROBLEM 8.48 KNOWN: Diameter, length and surface temperature of tubes used to heat ambient air. Flowrate and inlet temperature of air. FIND: (a) Air outlet temperature and heat rate per tube, (b) Effect of flowrate on outlet temperature. Design and operating conditions suitable for providing 1 kg/s of air at 75C. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible kinetic/potential energy and flow work changes, (3) Negligible tube wall thermal resistance. PROPERTIES: Table A.4, air (assume Tm = 330 K): cp = 1008 J/kgK, = 198.8 10-7 Ns/m2, k = 0.0285 W/mK, Pr = 0.703.
ANALYSIS: (a) For m = 0.01 kg/s, ReD = 4m D = 0.04 kg/s/(0.05 m)198.8 10-7 Ns/m2 = 12,810. Hence, the flow is turbulent. If fully developed flow is assumed throughout the tube, the DittusBoelter correlation may be used to obtain the average Nusselt number. 0.8 0.4 4 Nu D Nu D = 0.023 ReD/ 5 Pr 0.4 = 0.023 (12,810 ) (0.703) = 38.6 Hence, h = Nu D ( k D ) = 38.6 ( 0.0285 W m K 0.05 m ) = 22.0 W m 2 K
From Eq. 8.42b,
Ts - Tm,o
0.05 m 5 m 22 W m 2 K DLh = 0.180 = exp - = exp - mcp Ts - Tm,i 0.01kg s 1008J kg K
Tm,o = Ts - 0.180 Ts - Tm,i = 100 C - 0.180 80 C = 85.6 C Hence, q = mcp Tm,o - Tm,i = 0.01kg s (1008 J kg K ) 65.6 K = 661W
(
)
(
)
< <
(
)
(b) The effect of flowrate on the outlet temperature was determined by using the IHT Correlations and Properties Toolpads.
90
Outlet temperature, Tmo(C)
85
80
75 0 0.01 0.02 0.03 0.04 0.05
Mass flowrate, mdot(kg/s)
Continued...
PROBLEM 8.48 (Cont.) Although h and hence the heat rate increase with increasing m , the increase in q is not linearly and Tm,o decreases with increasing m . proportional to the increase in m A flowrate of m = 0.05 kg/s is not large enough to provide the desired outlet temperature of 75C, and to achieve this value, a flowrate of 0.0678 kg/s would be needed. At such a flowrate, N = 1 kg/s/0.0678 kg/s = 14.75 15 tubes would be needed to satisfy the process air requirement. Alternatively, a lower flowrate could be supplied to a larger number of tubes and the discharge mixed with ambient air to satisfy the desired conditions. Requirements of this option are that Nm + m amb = 1kg / s ( Nm + mamb ) cp (Tm,o - Tm,i ) = 1kg s 1008 J kg K (75 - 20 ) K = 55, 400 W
where m is the flowrate per tube. Using a larger number of tubes with a smaller flowrate per tube would reduce flow pressure losses and hence provide for reduced operating costs. COMMENTS: With L/D = 5 m/0.05 m = 100, the assumption of fully developed conditions throughout the tube is reasonable.
PROBLEM 8.49
KNOWN: Length and diameter of tube submerged in paraffin of prescribed dimensions. Inlet temperature and flow rate of water flowing through tube. FIND: (a) Outlet temperature, heat rate, and time required for complete melting, and (b) Effect of flowrate on operating conditions. SCHEMATIC:
ASSUMPTIONS: (1) Negligible KE/PE and flowwork changes for water, (2) Constant water properties, (3) Negligible tube wall conduction resistance, (4) Negligible convection resistance in melt (Ts = TW = Tmp), (5) Fully developed flow, (6) No heat loss to the surroundings. PROPERTIES: Water (given): cp = 4.185 kJ/kgK, k = 0.653 W/mK, = 467 10-6 kg/sm, Pr = 2.99; Paraffin (given): Tmp = 27.4C, hsf = 244 kJ/kg, = 770 kg/m3. ANALYSIS: (a) From Eq. 8.42b,
T - Tm,o
0.025 m 467 10-6 kg s m
conditions,
h= Nu D k D = k D 0.023 Re D
4/5
4 0.1kg s
DLh 4m = = exp - . With ReD = mcp T - Tm,i D
= 10,906, the flow is turbulent. Assuming fully developed
Pr
0.3
=
0.653 W m K 0.025 m
0.023 (10, 906 )
4/5
( 2.99 )0.3 = 1418 W
m K
2
Tm,o = 27.4 C - ( 27.4 - 60 ) C exp -
0.025 m 3 m 1418 W m 2 K = 42.17 C 0.1kg s 4185 J kg K
< <
From the overall energy balance,
q = mcp Tm,i - Tm,o = 0.1kg s 4185 J kg K ( 60 - 42.17 ) C = 7500 W
(
)
Applying an energy balance to a control volume about the paraffin, Ein = Est, the time tm required to melt the paraffin is qt m = Vh sf = L WH - D2 4 h sf
(
tm =
770 kg m3 3 m 0.25 0.25 m 2 - ( 0.025 m ) 7500 W
(
)
2
4
) 2.44 10 J kg = 4660 s = 1.29 h
5
<
Continued...
PROBLEM 8.49 (Cont.) (b) The effect of m on q and Tm,o was determined by accessing the Correlations Toolpad of IHT, and the results are plotted as follows.
30000
46
25000
Outlet temperature, Tmo(C) 45
Heat rate, q(W)
20000
44
15000
43
10000
5000 0.1 0.2 0.3 0.4 0.5
42 0.1 0.2 0.3 0.4 0.5
Mass flowrate, mdot(kg/s)
Mass flowrate, mdot(kg/s)
Although q increases with increasing m due to the attendant increase in ReD, and therefore h , the increase is not linearly proportional to the change in m . Hence, from the overall energy balance, q = m cp(Tm,i - Tm,o), there is a reduction in (Tm,i - Tm,o), which corresponds to an increase in Tm,o. With the increase in q, there is a reduction in tm, and for m = 0.5 kg/s, t m = 1167 s = 0.324 h
<
COMMENTS: Heat transfer from the water to the paraffin is also affected by free convection in the melt region around the tube. The effect is to decrease U, increase Ts, and decrease q with increasing time. The actual time to achieve complete melting would exceed values computed in the foregoing analysis.
PROBLEM 8.50
KNOWN: Configuration of microchannel heat sink. FIND: (a) Expressions for longitudinal distributions of fluid mean and surface temperatures, (b) Coolant and channel surface temperature distributions for prescribed conditions, (c) Effect of heat sink design and operating conditions on the chip heat flux for a prescribed maximum allowable surface temperature. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible PE, KE and flow work changes, (3) All of the chip power dissipation is transferred to the coolant, with a uniform surface heat flux, qs , (4) Laminar, fully developed flow, (5) Constant properties. PROPERTIES: Table A.6, Water (assume Tm = Tm,i = 290 K): cp = 4184 J/kgK, = 1080 10-6 Ns/m2, k = 0.598 W/mK, Pr = 7.56. ANALYSIS: (a) The number of channels passing through the heat sink is N = L/S = L/C1D, and conservation of energy dictates that q L2 = N ( DL ) qs = L2 q C1 c s which yields C q q = 1 c s With the mass flowrate per channel designated as m1 = m N , Eqs. 8.41 and 8.28 yield q D Lq Tm ( x ) = Tm,i + s x = Tm,i + c x m1cp mcp q C q Ts ( x ) = Tm ( x ) + s = Tm ( x ) + 1 c h h where, for laminar, fully developed flow with uniform q , Eq. 8.53 yields h = 4.36 k/D. s
(1)
(2) (3)
< <
(b) With L = 12 mm, D = 1 mm, C1 = 2 and m = 0.01 kg/s, it follows that S = 2 mm, N = 6 and ReD = 4 m 5D2 = 4 0.01kg s 6 0.001m 1.08 10-3 N s m 2 = 1965. Hence, the flow is laminar, as
1
(
)
(
)
assumed, and h = 4.36(0.598 W/mK/0.001 m) = 2607 W/m2K. From Eqs. (2) and (3) the outlet mean and surface temperatures are
(0.012 m )2 20 104 W m 2 Tm,o = 290 K + 0.01kg s ( 4184 J kg K )
Ts,o = Tm,o + 2
= 290.7 K = 17.7 C
20 104 W m 2 2607 W m K
2
= 339.5 K = 66.5 C Continued...
PROBLEM 8.50 (Cont.)
The axial temperature distributions are as follows
75 65 Temperature, T(C) 55 45 35 25 15 0 2 4 6 Axial location, x(mm) Surface temperature, Ts Mean temperaure, Tm 8 10 12
The flowrate is sufficiently large (and the convection coefficient sufficiently low) to render the increase in Tm and Ts with increasing x extremely small. (c) The desired constraint of Ts,max 50C is not met by the foregoing conditions. An obvious and logical approach to achieving improved performance would involve increasing m1 such that turbulent flow is maintained in each channel. A value of m1 > 0.002 kg/s would provide ReD > 2300 for D = 0.001. Using Eq. 8.60 with n = 0.4 to evaluate NuD and accessing the Correlations Toolpad of IHT to explore the effect of variations in m1 for different combinations of D and C1, the following results were obtained.
400 Chip heat flux, qc''(W/cm^2)
300
200
100
0 0.002 0.003 0.004 0.005
Mass flowrate, mdot1(kg/s) D = 0.6 mm, C1 = 2.0, N = 10 D = 1.0 mm, C1 = 1.5, N = 8 D = 1.0 mm, C1 = 2.0, N = 6
We first note that a significant increase in q may be obtained by operating the channels in turbulent c flow. In addition, there is an obvious advantage to reducing C1, thereby increasing the number of channels for a fixed channel diameter. The biggest enhancement is associated with reducing the channel diameter, which significantly increases the convection coefficient, as well as the number of channels for fixed C1. For m1 = 0.005 kg/s, h increases from 32,400 to 81,600 W/m2K with decreasing D from 1.0 to 0.6 mm. However, for fixed m1 , the mean velocity in a channel increases with decreasing D and care must be taken to maintain the flow pressure drop within acceptable limits. COMMENTS: Although the distribution computed for Tm(x) in part (b) is correct, the distribution for Ts(x) represents an upper limit to actual conditions due to the assumption of fully developed flow throughout the channel.
PROBLEM 8.51
KNOWN: Cold plate geometry and temperature. Inlet temperature and flow rate of water. Number of circuit boards and temperature and velocity of air in parallel flow over boards. FIND: (a) Heat dissipation by cold plates, (b) Heat dissipation by air flow. SCHEMATIC:
ASSUMPTIONS: (1) Isothermal cold plate, (2) All heated generated by circuit boards is dissipated by cold plates (Part (a)), (3) Circuit boards may be represented as isothermal at an average surface temperature, (4) Air flow over circuit boards approximates that over a flat plate in parallel flow, (5) Steady operation, (6) Constant properties. PROPERTIES: Table A-6, Water ( Tm 290K ) : cp = 4184 J/kgK, = 1080 10-6 N s / m 2 , k=0.598 W/mK, Pr = 7.56. Table A-4, Air (p = 1 atm, Tf = 300K): = 15.89 10 -6 m 2 / s, k = 0.0263 W/mK, Pr = 0.707.
ANALYSIS: (a) With Re D = 4 m1 / D = 4 0.2 kg / s / 0.01m 1080 10 -6 N s / m 2 = 23, 600, the flow is turbulent, and from Eq. (8.60),
h=
k D
Nu D = 0.023
k D
Re D
4/5
Pr
0.4
=
0.023 0.598 W / m K 0.01m
( 23, 600 )4 / 5 (7.56 )0.4 = 9, 730 W / m 2 K
With H/D = 0.75/0.01 = 75, it is reasonable to assume fully developed flow throughout the tube. Hence, from Eqs. (8.42b) and (8.37)
0.01m 0.75m 9730 W / m 2 K h = exp - = 0.760 0.2 kg / s 4184 J / kg K Tm,o = Ts,cp - 0.76 Ts,cp - Tm,i = 13C DH Ts,cp - Tm,o = exp - m1 c p Ts,cp - Tm,i q1 = m1cp Tm,o - Tm,i = 0.2 kg / s 4184 J / kg K 6C = 5021W
With a total of 2N = 20 passages, the total heat dissipation is
(
(
)
)
q = 2Nq1 = 20 5021W = 100 kW
(b) For the air flow, Re D = u L / = 10 m / s 0.60m /15.89 10 -6 m 2s = 378, 000, and the flow is laminar. From Eq. (7.31),
h= k L Nu L = 0.664 k L Re L
1/ 2
<
(378, 000 )1/ 2 (0.707 )1/ 3 = 15.9 W / m 2 K
Pr
1/ 3
=
0.664 0.0263 W / m K 0.60m
Heat dissipation to the air from both sides of 10 circuit boards is then q = 2Ncb h ( WL ) Ts,cb - T = 20 15.9 W / m 2 K 0.21m2 40C = 2, 670 W COMMENTS: The cooling capacity of the cold plates far exceeds that of the air flow. However, the challenge would be one of efficiently transferring such a large amount of energy to the cold plates without incurring excessive temperatures on the circuit boards.
(
)
<
PROBLEM 8.52
KNOWN: Flow rate and temperature of Freon passing through a Teflon tube of prescribed inner and outer diameter. Velocity and temperature of air in cross flow over tube. FIND: Heat transfer per unit tube length. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant properties, (4) Fully developed flow. PROPERTIES: Table A-4, Air (T = 300K, 1 atm): = 15.89 10 0.707; Table A-5, Freon (T = 240K): = 3.85 10 Teflon (T 300K): k = 0.35 W/mK.
-4 2 -6
m /s, k = 0.0263 W/mK, Pr =
2
Ns/m , k = 0.069 W/mK, Pr = 5.0; Table A-3,
ANALYSIS: Considering the thermal circuit shown above, the heat rate is
T - Tm . 1/ho Do + ln ( Do / Di ) / 2 k + (1 / h i Di ) & 4m 0.4 kg/s Re D,i = = = 13,228 Di ( 0.025m ) 3.85 10 -4 N s/m 2 q =
(
)
and the flow is turbulent. Hence, from the Dittus-Boelter correlation
hi =
With
k 0.069 W/m K 0.023 Re 4/5 Pr 0.4 = 0.023 (13,228 )4 / 5 (5 )0.4 = 240 W/m 2 K. D,i Di 0.025 m VDo ( 25 m/s ) 0.028 m = = 4.405 104 -6 m 2 / s 15.89 10
0.0263 W/m K 0.028 m
Re D,o =
it follows from Eq. 7.56 and Table 7.4 that
ho = k D 0.26 Re0.6 Pr0.37 = D,o 0.26 4.405 10 4
(
)
0.6
( 0.707 )0.37
= 131 W/m 2 K.
Hence
q =
(131 W/m2 K 0.028 m )
q =
-1
T - Tm
+ ln ( 28/25 ) / 2 ( 0.350 W/m K ) + 240 W/m2 K 0.025 m
(
)
-1
( 300 - 240 ) K = 312 W/m. ( 0.087 + 0.052 + 0.053 ) K m/W
2
<
COMMENTS: The three thermal resistances are comparable. Note that Ts,o = T - q/hoDo = 300K - 312 W/m/131 W/m K 0.028 m = 273 K.
PROBLEM 8.53 KNOWN: Oil flowing slowly through a long, thin-walled pipe suspended in a room. FIND: Heat loss per unit length of the pipe, qconv . SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Tube wall thermal resistance negligible, (3) Fully developed flow, (4) Radiation exchange between pipe and room negligible. PROPERTIES: Table A-5, Unused engine oil (Tm = 150C = 423K): k = 0.133 W/mK. ANALYSIS: The rate equation, for a unit length of the pipe, can be written as (T - T ) qconv = m R t where the thermal resistance is comprised of two elements, R t = 1 1 1 1 1 + = + . h i D ho D D h i h o
The convection coefficient for internal flow, hi, must be estimated from an appropriate correlation. From practical considerations, we recognize that the oil flow rate cannot be large enough to achieve turbulent flow conditions. Hence, the flow is laminar, and if the pipe is very long, the flow will be fully developed. The appropriate correlation is hD Nu D = i = 3.66 k h i = Nu D k/D = 3.66 0.133 W /0.030 m = 16.2 W/m 2 K. m K
The heat rate per unit length of the pipe is qconv =
(150 - 20) o C
1 1 m2 K 1 + ( 0.030m ) 16.2 11 W
= 80.3 W/m.
<
COMMENTS: This problem requires making a judgment that the oil flow will be laminar rather than turbulent. Why is this a reasonable assumption? Recognize that the correlation applies to a constant surface temperature condition.
PROBLEM 8.54
KNOWN: Thin-walled, tall stack discharging exhaust gases from an oven into the environment. FIND: (a) Outlet gas and stack surface temperatures, Tm,o and Ts,o, and (b) Effect of wind temperature and velocity on Tm,o. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Wall thermal resistance negligible, (3) Exhaust gas properties approximated as those of atmospheric air, (4) Radiative exchange with surroundings negligible, (5) PE, KE, and flow work changes negligible, (6) Fully developed flow, (7) Constant properties. PROPERTIES: Table A.4, air (assume Tm,o = 773 K, Tm = 823 K, 1 atm): cp = 1104 J/kgK, = 376.4 10-7 Ns/m2, k = 0.0584 W/mK, Pr = 0.712; Table A.4, air (assume Ts = 523 K, TW = 4C = 277 K, Tf = 400 K, 1 atm): = 26.41 10-6 m2/s, k = 0.0338 W/mK, Pr = 0.690. ANALYSIS: (a) From Eq. 8.46a,
PL Tm,o = T - T - Tm,i exp - U mcp
(
)
1 1 U =1 + hi h o
(1,2)
where hi and ho are average coefficients for internal and external flow, respectively. Internal flow: With a Reynolds number of
ReDi =
4m 4 0.5 kg s = = 33,827 D 0.5 m 376.4 10-7 N s m 2
(3)
and the flow is turbulent. Considering the flow to be fully developed throughout the stack (L/D = 12) and with Ts < Tm, the Dittus-Boelter correlation has the form hD Nu D = i = 0.023Re4 / 5 Pr 0.3 (4) Di k
hi =
58.4 10-3 W m K 4/5 0.023 (33,827 ) (0.712 )0.3 = 10.2 W m 2 K . 0.5 m VD 5 m s 0.5 m = = 94, 660 26.41 10-6 m 2 s
External flow: Working with the Churchill/Bernstein correlation, the Reynolds and Nusselt numbers are
ReDo =
(5) Continued...
PROBLEM 8.54 (Cont.)
5/8 Re D 1 + Nu D = 0.3 + 1/ 4 282, 000 1 + ( 0.4 Pr )2 / 3
0.62 Re1/ 2 Pr1/ 3 D
4/5
= 205
Hence,
h o = (0.0338 W m K 0.5 m ) 205 = 13.9 W m 2 K
The outlet gas temperature is then Tm,o = 4 C - ( 4 - 600 ) C exp -
(6)
1 2 1 10.2 1 13.9 W m K = 543 C + 0.5 kg s 1104 J kg K
0.5 m 6 m
<
The outlet stack surface temperature can be determined from a local surface energy balance of the form, hi(Tm,o - Ts,o) = ho(Ts,o - TW ), which yields
Ts,o =
h i Tm,o + h o T hi + ho
=
(10.2 543 + 13.9 4 ) W m 2 = 232 C (10.2 + 13.9 ) W m2 K
<
(b) Using the Correlations and Properties Toolpads of IHT, with a surface temperature of Ts = 523 K assumed solely for the purpose of evaluating properties associated with airflow over the cylinder, the following results were generated.
560 Gas outlet temperature, Tmo(C)
550
540
530
520 2 3 4 5 6 7 8 9 10
Freestream velocity, V(m/s) Tinf = 35 C Tinf = 5 C Tinf = -25C
Due to the elevated temperatures of the gas, the variation in ambient temperature has only a small effect on the gas exit temperature. However, the effect of the freestream velocity is more pronounced. Discharge temperatures of approximately 530 and 560C would be representative of cold/windy and warm/still atmospheric conditions, respectively. COMMENTS: If there are constituents in the discharge gas flow that condense or precipitate out at temperatures below Ts,o, this operating condition should be avoided.
PROBLEM 8.55
KNOWN: Hot fluid passing through a thin-walled tube with coolant in cross flow over the tube. Fluid flow rate and inlet and outlet temperatures. FIND: Outlet temperature, Tm,o, if the flow rate is increased by a factor of 2 with all other conditions the same. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes and axial conduction, (3) Constant properties, (4) Fully developed flow and thermal conditions, (5) Convection coefficients, h o and hi , independent of temperature, and (6) Negligible wall thermal resistance. PROPERTIES: Hot fluid (Given): = 1079 kg/m3, cp = 2637 J/kgK, = 0.0034 Ns/m2, k = 0.261 W/mK. ANALYSIS: For conditions prescribed in the Schematic, Eq 8.46a can be used to evaluate the overall convection coefficient with P = D,
T - Tm,o
PL U = exp - T - Tm,i mo cp U kg K
(1)
( 25 - 78 ) C = exp - 0.010 m 1m (18 / 3600 ) kg s 2637 J ( 25 - 85 ) C
U = 52.1 W m 2 K
The overall coefficient can be expressed in terms of the inside and outside coefficients, -1 U = 1 hi + 1 h o Characterize the internal flow with the Reynolds number, Eq. 8.6,
(
)
(2)
ReD =
4 (18 / 3600 ) kg s 4mo = = 187 D 0.010m 0.0034 N s m 2
and since the flow is laminar, and assumed to be fully developed, h i will not change when the flow rate is doubled. That is, U = 52.1 W/m2K when m = 2mo. Using Eq. (1) again, but with Tm,o unknown,
(25 - Tm,o )
0.010 m 1m = exp - 52.1W m 2 K 2 (18 / 3600 ) kg s 2637 J kg K ( 25 - 85 ) C C
Tm,o = 81.4 C
<
COMMENTS: Examine the assumptions and explain why they were necessary in order to affect the solution.
PROBLEM 8.56
KNOWN: Thin walled tube of prescribed diameter and length. Water inlet temperature and flow rate. FIND: (a) Outlet temperature of the water when the tube surface is maintained at a uniform temperature Ts = 27C assuming Tm = 300 K for evaluating water properties, (b) Outlet temperature of the water when the tube is heated by cross flow of air with V = 10 m/s and T = 100C assuming Tf = 350 K for evaluating air properties, and (c) Outlet temperature of the water for the conditions of part (b) using properly evaluated properties. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes and axial conduction, (3) Fully developed flow and thermal conditions for internal flow, and (4) Negligible tube wall thermal resistance. PROPERTIES: Table A.6, Water ( Tm = 300 K): = 997 kg/m3, cp = 4179 J/kgK, = 855 10-6 Ns/m2, k = 0.613 W/mK, Pr = 5.83; Table A.4, Air ( Tf = 350 K, 1 atm): = 20.92 10-6 m2/s, k = 0.030 W/mK, Pr = 0.700. ANALYSIS: (a) For the constant wall temperature cooling process, Ts = 27C, the water outlet temperature can be determined from Eq. 8.42b, with P = D,
Ts - Tm,o
PL hi = exp - mcp Ts - Tm,i 4m 4 0.2 kg s = = 29, 783 D 0.010 m 855 10-6 N s m 2
(1)
To estimate the convection coefficient, characterize the flow evaluating properties at Tm = 300 K
ReD =
Hence, the flow is turbulent and assuming fully developed (L/D = 200), and using the Dittus-Boelter correlation, Eq. 8.60, find h i ,
hD Nu D = i = 0.023 Re0.8 Pr 0.3 D k
hi =
0.613 W m K 0.010 m
0.023 ( 29, 783)
0.8
(5.83)0.3 = 9080 W
m 2 K (2)
Substituting this value for h i into Eq. (1), find
( 27 - 47 )
(27 - Tm,o ) = exp -
C
0.010 m 2 m 9080 W m 2 K 0.2 kg s 4179 J kg K
Tm,o = 37.1C
<
(3)
(b) For the air heating process, T = 100C, the water outlet temperature follows from Eq. 8.46a,
T - Tm,o
DL U = exp - mcp T - Tm,i
Continued...
PROBLEM 8.56 (Cont.)
where the overall coefficient is
U = (1 hi + 1 h o )
(4)
To estimate h o , use the Churchill-Bernstein correlation, Eq. 7.57, for cross flow over a cylinder using properties evaluated at Tf = 350 K.
ReD =
VD 10 m s 0.010 m = = 4780 20.92 10-6 m 2 s 0.62 Re1/ 2 Pr1/ 3 D
(5) 4/5 (6)
5/8 Re D 1 + Nu D = 0.3 + 1/ 4 282, 000 1 + ( 0.4 Pr )2 / 3
1 + 0.4 0.700 ) ( Nu D k 0.030 W m K ho = = 35.76 = 107 W m 2 K D 0.010 m
Nu D = 0.3 +
0.62 ( 4780 )
1/ 2
(0.700 )1/ 3 1 +
2 / 3 1/ 4
4780 282, 000
5 / 8 4 / 5
= 35.76
Assuming h i = 9080 W/m2K as calculated from part (a), find U then Tm,o, -1 U = 1 9080 + 1 107 W m 2 K = 106 W m 2 K
(
)
0.010 m 2 m = exp - 106 W m 2 K 0.2 kg s 4179 J kg K (100 - 47 ) C
100 - Tm,o
Tm,o = 47.4 C
<
(c) Using the IHT Correlation Tools for Internal Flow (Turbulent Flow) and External Flow (over a Cylinder) the analyses of part (b) were performed considering the appropriate temperatures to evaluate the thermophysical properties. For internal and external flow, respectively,
Tm = Tm,i + Tm,o
Tm
Ts Ts
TW 1 hi 1 ho
(
)
2
Tf = ( Ts + T ) 2
(7,8)
where the average tube wall temperature is evaluated from the thermal circuit, (9)
The results of the analyses are summarized in the table along with the results from parts (a) and (b), Condition Ts = 27C T = 100 C, Tf 350C = Exact solution
Tm
(K) 300 300 320
hi
(W/m K) 9080 9080 11,420
2
Tf
(K) --350 347
ho
(W/m K) --107 107.3
2
U
(W/m K) --106 106.3
2
Tm,o (C) 37.1C 47.4C 47.4C
Note that since h o << h i , U is controlled by the value of h o which was evaluated near 350 K for both parts (b) and (c). Hence, it follows that Tm,o is not very sensitive to h i which, as seen above, is sensitive to the value of Tm .
PROBLEM 8.57
KNOWN: Diameter of tube through which water of prescribed flow rate and inlet and outlet temperatures flows. Temperature of fluid in cross flow over the tube. FIND: (a) Required tube length for air in cross flow at prescribed velocity, (b) Required tube length for water in cross flow at a prescribed velocity. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Constant properties, (3) Negligible tube wall conduction resistance, (4) Negligible KE, PE and flow work changes. PROPERTIES: Table A.6, water ( Tm = 50C = 323 K): cp = 4181 J/kgK, = 548 10-6 Ns/m2, k = 0.643 W/mK, Pr = 3.46. Table A.4, air (assume Tf = 300 K): = 15.89 10-6 m2/s, k = 0.0263 W/mK, Pr = 0.707. Table A.6, water (assume Tf = 300 K): = 0.858 10-6 m2/s, k = 0.613 W/mK, Pr = 5.83. ANALYSIS: The required heat rate may be determined from the overall energy balance,
q = mcp Tm,i - Tm,o = 0.215 kg s ( 4181J kg K ) 40 C = 35,960 W
and the required tube length may be determined from the rate equation, Eq. 8.47a,
(
)
L=
where
q U DTm
Tm =
(Tm,i - T ) - (Tm,o - T ) = 30.8 C
Tm,i - T n Tm,o - T
and
1/U = 1/hi + 1/ho.
With
ReDi = 4m D = 0.860 kg s (0.05m ) 548 10-6 N s m 2 = 9991
the flow is turbulent and, assuming fully developed flow throughout the tube, the inside convection coefficient is determined from Eq. 8.60 0.8 0.3 Nu Di = 0.023Re4 / 5 Pr 0.3 = 0.023 (9991) (3.46 ) = 52.9 Di h i = Nu Di k D = 52.9 (0.643 W m K ) 0.05 m = 680 W m 2 K Continued...
PROBLEM 8.57 (Cont.)
(a) For water in cross flow at 20 m/s, ReD = VD/ = 20 m/s(0.05 m)/15.89 10-6 m2/s = 62,933. From o the Churchill/Bernstein correlation, it follows that 5/8 Re Do o 1 + Nu Do = 0.3 + 1/ 4 282, 000 1 + ( 0.4 Pr )2 / 3
0.62 Re1/ 2 Pr1/ 3 D
4/5
= 158.7
h o = Nu Do k D = 158.7 ( 0.0263 W m K ) 0.05 m = 83.5 W m 2 K
-1 Hence, U = (1 h i + 1 h o ) = 74.4 W/m2K and
L=
(
35,960 W 74.4 W m 2 K (0.05 m ) 30.8 C
)
= 100 m
<
correlation yields Nu D = 527.3 . Hence, o
(b) For water in cross flow at 2 m/s, ReD = 2 m/s(0.05 m)/0.858 10-6 m2/s = 116,550, and the o
h o = Nu Do k D = 527.3 ( 0.613 W m K ) 0.05 m = 6, 465 W m 2 K
-1 U = (1 h i + 1 h o ) = 615.3 W/m2K Hence,
L=
(
35,960 W 615.3 W m 2 K (0.05 m ) 30.8 C
)
= 12 m
<
COMMENTS: The foregoing results clearly indicate the superiority of water (relative to air) as a heat transfer fluid. Note the dominant contribution made by the smaller convection coefficient to the value of U in each of the two cases.
PROBLEM 8.58 KNOWN: Water flow rate and inlet temperature for a thin-walled tube of prescribed length and diameter. FIND: Water outlet temperature for each of the following conditions: (a) Tube surface maintained at 27C, (b) Insulation applied and outer surface maintained at 27C, (c) Insulation applied and outer surface exposed to ambient air at 27C. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed flow throughout the tube, (3) Negligible tube wall conduction resistance, (4) Negligible contact resistance between tube wall and insulation, (5) Uniform outside convection coefficient. PROPERTIES: Assume water cools to Tm,o = 27C with no insulation but that cooling is negligible (Tm,o = 97C) with insulation. Table A-4, Water ( Tm = 335K ) : cp = 4186 J/kgK, = 453 10 Ns/m , k =0.656 W/mK, Pr = 2.88; Table A-4, Water (Tm,i = 370K): cp = 4214 -6 2 J/kgK, = 289 10 Ns/m , k = 0.679 W/mK, Pr = 1.80. ANALYSIS: For each of the three cases, heat is transferred from the warm water to a surface (or the air) which is at a fixed temperature (27C). Accordingly, an expression of the form given by Eq. 8.42b may be used to determine the outlet temperature of the water, so long as the appropriate heat transfer coefficient is used. In particular, each of the cases can be described by Eq. 8.46. UAs To = exp - m cp Ti & Referring to the thermal circuit associated with heat transfer from the water,
-6 2
and using Eq. 3.32, the UA product may be evaluated as UA = ( R t ) . (a) For the first case: Re D = Ts,i = 27 o C Ti = Tm,i -Ts,i = 70 o C UA = hi Di L. & 4m 4 0.015 kg/s = = 14,053. Di ( 0.003m ) 453 10-6 N s/m 2 Continued .....
-1
PROBLEM 8.58 (Cont.) From Eq. 8.60,
hi = k Di
4/5 0.023 Re D Pr 0.30 =
0.656 W/m K 0.003m
( 0.023) (14,053)4 / 5 ( 2.88 )0.3 = 14,373 W/m2 K.
14,373 W 0.003m 1m m2 K - = 8.1o C 0.015 kg/s 4186 J/kg K
h DiL To = Texp - i = 70o C exp i & m cp
Tm,o = To + Ts,i = 8.1o C + 27o C = 35.1o C. (b) For the second case: Ts,o = 27C with UA = (1/h i Di L ) + ln ( Do / D i ) / 2 kL Ti = Tm,i - Ts,o = 70o C With
hi = k Di -1
<
.
Re D =
& 4m 4 0.015 kg/s = = 22,028 Di ( 0.003m ) 289 10-6 N s/m2
0.679 W/m K 0.003m
4/5 0.023 Re D Pr 0.3 =
( 0.023) ( 22,028) 4 / 5 (1.80 ) 0.3 = 18,511 W/m 2 K.
-1 -1
It follows that ln ( 0.004/0.003) 1 UA = + 2 ( 0.05 ) 18,511 0.003 = 5.73 10-3 + 0.916 = 1.085 W/K
and the outlet temperature is 1.085 W/K o To = 70o C exp - = 68.8 C 0.015 kg/s 4214 J/kg K Tm,o = To + Ts,o = 68.8o C + 27oC = 95.8oC. (c) For the third case: T = 27C, Ti = Tm,i - T = 70C and UA = (1/h i Di L) + ln ( Do / Di ) / 2 kL + (1/h o Do L) 1 UA = 5.73 10 -3 + 0.916 + 5 ( 0.004 )
-1
<
-1
-1
= 5.73 10 -3 +0.916 + 15.92
= 0.0594 W/K
0.0594 W/K o To = 70o C exp - = 69.9 C 0.015 kg/s 4214 J/kg K Tm,o = To + T = 69.9o C + 27oC=96.9oC. COMMENTS: Note that Rconv,o >> Rcond,insul >> Rconv,i,
<
PROBLEM 8.59
KNOWN: Thick-walled pipe of thermal conductivity 60 W/mK passing hot water with ReD = 20,000, a mean temperature of 80C, and cooled externally by air in cross-flow at 20 m/s and 25C. FIND: Heat transfer rate per unit pipe length, q . SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Internal flow is turbulent and fully developed. PROPERTIES: Table A-6, Water (Tm = 80C = 353K): k = 0.670 W/mK, Pr = 2.20; Table A-4, Air (T = 25C 300K, 1 atm): = 15.89 10
-6 2
m /s, k = 0.0263 W/mK, Pr = 0.707.
ANALYSIS: The heat rate per unit length, considering thermal resistances to internal flow, wall conduction (Eq. 3.28) and external flow, with A = DL, is -1 q = 1/h i Di + (1 / 2 k ) ln ( Do / Di ) +1 / h o Do ( Tm - T ) . Internal Flow: Using the Dittus-Boelter correlation with n = 1/3 for turbulent, fully developed flow, where ReD = 20,000
i
h i = ( k/Di ) Nu D = ( k/Di ) 0.023Re 4/5 Pr1/3 h i = ( 0.670 W/m K/0.020 m ) 0.023 ( 20,000 )
4/5
2.201/3 = 2765 W/m 2 K.
External Flow: Using the Zhukauskas correlation for cross-flow over a circular cylinder with Pr/Prs 1, find first
Re D =
VDo 20 m/s 0.025 m = = 31,466 15.89 10-6m 2 / s
and from Table 7.4, C = 0.26 and m = 0.6, where n = 0.37, h D NuD = o = CRem Pr n ( Pr/Prs )1/4 D k
h o = ( 0.0263 W/m K/0.025 m ) 0.26 ( 31,466 )
Hence, the heat rate is
0.6
( 0.707 )0.37 = 120 W/m2 K.
q= 1/2765 W/m 2 K 0.020 m + (1 / 2 60 W/m K ) ln ( 25/20 )
(
)
+ 1/120 W/m 2 K 0.025 m q = 5.756 10 -3 + 5.919 10-4 + 1.061 10 -1 q = 489 W/m.
-1
(
)
-1
(80 - 25 )o C
W/m K (80 - 25 )o C
<
COMMENTS: Note that the external flow represents the major thermal resistance to heat transfer.
PROBLEM 8.60
KNOWN: Reaction vessel with process fluid at 75C cooled by water at 27C and 0.12 kg/s through 15 mm tube. High convection coefficient on outside of tube (3000 W/m K) created by vigorous stirring. FIND: (a) Maximum heat transfer rate if outlet temperature of water cannot exceed Tm,o = 47C, and (b) Required tube length. SCHEMATIC:
2
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes, (3) Negligible thermal resistance of tube wall.
3 o PROPERTIES: Table A-6, Water Tm = ( 47 + 27 ) C/2 = 310K : = 1/vf = 993.1 kg/m , cp =
4178 J/kgK, = 695 10
-6
Ns/m , k = 0.628 W/mK, Pr = 4.62.
2
(
)
ANALYSIS: (a) From an overall energy balance on the tube with Tm,o = 47C, & q max = m c p Tm,o - Tm,i = 0.12 kg/s 4178 J/kg K ( 47 - 27 )o C = 10,027 W.
(
)
<
(b) For the constant surface temperature heating condition, from Eq. 8.46,
T - Tm,o
PL = exp - U m cp T - Tm,i &
where
1/U = 1/ho + 1 / h . i
For internal flow in the tube, find
Re D =
& 4m 4 0.12 kg/s = =14,656 D 0.015 m 695 10-6 N s/m2
and the flow is turbulent. Assuming fully developed flow, use the Dittus-Boelter correlation with n = 0.4 (heating),
NuD = hi D/k = 0.023Re4/5 Pr0.4 D h i = [0.628 W/m K/0.015 m ] 0.023 (14,656 )
Hence, 1/U = [1/3000 + 1/3822] m K/W with P = D, find
2
4/5
( 4.62 )0.4 = 3822 W/m2 K.
2
or
U = 1680 W/m K. From the energy balance relation
( 75 - 47) o C = ( 75 - 27) o C
( 0.015 m ) L1680 W/m2 K exp - 0.12 kg/s 4178 J/kg K
L = 3.4 m.
<
COMMENTS: Note that L/D = 227 and the fully developed flow assumption is appropriate.
PROBLEM 8.61
KNOWN: Water flowing through a tube heated by cross flow of a hot gas. Required to heat water from 15 to 35C with a flow rate of 0.2 kg/s. FIND: Design graphs to demonstrate acceptable combinations of tube diameter (D = 20, 30 or 40 mm), tube length (L = 3, 4 or 6 m) and hot gas velocity (20 V 40 m/s) and temperature ( T = 250, 375 or 500C). SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes and axial conduction, (3) Fully developed flow and thermal conditions for internal flow, (4) Properties of the hot gas are those of atmospheric air, and (5) Negligible tube wall thermal resistance. PROPERTIES: Table A.6, Water ( Tm = (15 + 35 ) C 2 = 298K ); Table A.4, Air ( Tf = Ts + T 1 atm).
(
)
2,
ANALYSIS: Method of Analysis: The tube having internal flow of water with cross flow of hot gas can be analyzed by the energy balance relation, Eq. 4.86a
T - Tm,o
( DL ) U = exp mcp T - Tm,i
(1)
where the overall coefficient U is
U = 1 hi + 1 h o
(
)-1
(2)
Estimation of the internal flow coefficient, h i : Evaluating water properties at the average mean fluid
Tm = Tm,i + Tm,o ReD,i = 4m ( D )
(
)
2,
(3)
characterize the flow with the Reynolds number, (4)
and assuming the flow to be both turbulent and fully developed (L/D > 3m/0.07m = 42), use the DittusBoelter correlation, Eq. 8.60, to evaluate h i , Continued...
PROBLEM 8.61 (Cont.) hD Nu D,i = i = 0.023Re0.8 Pr 0.4 D,i ki
Estimation of the external flow coefficient, h o : Evaluating gas (air) properties at the average film temperature (5)
Tf = ( Ts + T ) 2 VD
(6)
where Ts is the average tube wall temperature (see Eq. (9)), characterize the flow
ReD,o =
(7)
and use the Churchill-Bernstein correlation, Eq. 7.57, for cross-flow over a cylinder, 1/ 0.62 Re1/ 2 Pro 3 hoD D,o Nu D,o = = 0.3 + 1/ 4 ko 1 + ( 0.4 Pr )2 / 3 o 5/8 Re D,o 1 + 282, 000 4/5 (8)
The average tube wall temperature, Ts , follows from the thermal circuit
Tm - Ts Ts - T = 1 hi 1 ho
(9)
The IHT Workspace: Using the Correlation Tools for Internal Flow (Turbulent flow), and External Flow (Flow over a Cylinder) and Properties for Air and Water, along with the appropriate energy balances and rate equations, the heater-tube system can be analyzed. The Design Strategy: We have chosen to generate the design information in the following manner: for a specified gas temperature, T , plot the required length L (limiting the scale to 3 L 6m) as a function of gas velocity V (20 V 40 m/s) for tube diameters of D = 20, 30 and 40 mm. Three design graphs corresponding to T = 250, 375 and 500C were generated and are shown on the next page. COMMENTS: (1) The collection of design graphs will allow the contractor to select appropriate combinations of tube D and L and gas stream parameters ( T and V) to achieve the required water heating. (2) Note from the design graphs that with T = 250C, the required heating of the water can be achieved only with a 40-mm diameter by 6 m length tube with gas velocities greater than 32 m/s. This configuration represents a worst case condition of largest tube parameters and highest gas velocity. (3) Which operating conditions, T = 375 or 500C, provides the contractor with more options in selecting combinations of tube parameters and gas velocities? What are the trade-offs in operating at 375 or 500C? Consider such features as tube life, tubing costs and fan requirements. (4) The Reynolds numbers for the internal flow are approximately 7,100, 9,460 and 14,200 for the tube diameters of 20, 30 and 40 mm. For the larger tube sizes, the Reynolds numbers are below 10,000, the usual lower limit for turbulent flow. Continued...
PROBLEM 8.61 (Cont.)
Gas temperature, Tinf = 250 C 6
Tube length, L (m)
5
4
3 20 25 30 Gas velocity, V (m/s) D = 20 mm D = 30 mm D = 40 mm
Gas temperature, Tinf = 375 C 6
35
40
Tube length, L (m)
5
4
3 20 25 30 Gas velocity, V (m/s) D = 20 mm D = 30 mm D = 40 mm 35 40
Gas temperature, Tinf = 500 C 6
Tube length, L (m)
5
4
3 20 25 30 Gas velocity, V (m/s) D = 20 mm D = 30 mm D = 40 mm 35 40
PROBLEM 8.62 KNOWN: Exhaust gasses at 200C and mass rate 0.03 kg/s enter tube of diameter 6 mm and length 20 m. Tube experiences cross-flow of autumn winds at 15C and 5 m/s. FIND: Average heat transfer coefficients for (a) exhaust gas inside tube and (b) air flowing across outside of tube, (c) Estimate overall coefficient and exhaust gas temperature at outlet of tube. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes, (3) Negligible tube wall resistance, (4) Exhaust gas properties are those of air, (5) Negligible radiation effects. PROPERTIES: Table A-4, Air (assume Tm,o 15C, hence Tm = 380K, 1 atm):cp = 1012 J/kgK, k = 0.0323 W/mK, = 221.6 10 Ns/m , Pr = 0.694; Air (T = 15C = 288 K, 1 atm): k = 0.0253 W/mK, = 14.82 10 m /s, Pr = 0.710; Air ( Ts 90C = 363 K, 1 atm): Pr = 0.698.
ANALYSIS: (a) For the internal flow through the tube assuming a value for Tm,o = 15C, find
-6 2 -7 2
ReD =
4m 4 0.003 kg/s = = 2.873 104. -7 N s/m 2 D 0.006 m 221.6 10
Hence the flow is turbulent and, since L/D >> 10, fully developed. Using the Dittus-Doelter correlation with n = 0.3, Nu D = 0.023Re0.8 Pr 0.3 = 0.023 2.873 104 D
(
)
0.8
(0.694 )0.3 = 76.0
h i = Nu k/D = 76.0 0.0323 W/m K/0.006 m = 409 W/m2 K. (b) For cross-flow over the circular tube, find using thermophysical properties at T, VD 5 m/s 0.006 m ReD = = = 2024 14.82 10-6 m 2 / s and using the Zhukauskus correlation with C = 0.26, m = 0.6, and n = 0.37, Nu D = CRe m Pr n ( Pr/ Prs ) D
1/ 4
<
= 0.26 ( 2024 )
0.6
0.7100.37 (0.710 / 0.698 )
0.25
= 23.1
where Prs is evaluated at Ts . Hence, h o = Nu D k/D = 23.1 0.0253 W/m K/0.006 m = 97.5 W/m2 K. Continued .....
<
PROBLEM 8.62 (Cont.) (c) Assuming the thermal resistance of the tube wall is negligible, 1 1 1 1 1 2 = + = + m K/W U h o h i 97.5 409 U = 78.8 W/m 2 K.
<
The gas outlet temperature can be determined from the expression where P = D. T - Tm,o 0.006 m 78.8 W/m 2 K 20 m PUL = exp - = exp - m cp T - Tm,i 0.003 kg/s 1012 J/kg K 15 - Tm,o
(15 - 200 )
= 0.999 C
Tm,o = 15 C.
<
COMMENTS: (1) With Tm,o = 15C, find Tm = 380 K; hence thermophysical properties for the internal flow correlation were evaluated at a reasonable temperature. Note that the gas is cooled from 200C to the ambient air temperature, Tm,o = T, over the 20-m length! (2) The average wall surface temperature, Ts , follows from an energy balance on the wall surface, Tm - Ts h = i Ts - Tinf h o and substituting numerical values, find Ts = 90C = 363 K, the value we assumed for evaluating Prs. Can you draw a thermal circuit to represent this energy balance relation? (3) When using the Zhukauskus correlation, it is reasonable to evaluate Prs at the Tm for the first trial. For gases the assumption is a safe one, but for liquids, especially oils, additional trials will be required since the Prandtl number may be strongly dependent upon temperature.
PROBLEM 8.63
KNOWN: Superheated steam passing through thin-walled pipe covered with insulation and suspended in a quiescent air. FIND: Point along pipe surface where steam will begin condensing (x1). SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible KE, PE and flow work changes, (3) Steam properties may be approximated as those corresponding to saturated conditions. PROPERTIES: Table A.6, Saturated steam ( Tm = (100 + 120)C/2 = 110C 385 K): g = 0.876 kg/m3, cp,g = 2080 J/kgK, g = 12.49 10-6 NS/m2, kg = 0.0258 W/mK, Prg = 1.004. ANALYSIS: From Eq. 8.46a, where Tm,x is the mean temperature at any distance x,
T - Tm,x
Px U = exp - mc p T - Tm,i
(1)
The mass flow rate, with Ac = D2/4, is 2 m = g Ac u m = 0.876 kg m3 (0.050 m ) 4 10 m s = 0.0172 kg s
(
)
and for the internal flow,
ReD =
4m 4 0.0172 kg s = = 35, 068 . D (0.050 m ) 12.49 10-6 N s m 2
Assuming the flow is fully developed, the Dittus-Boelter correlation yields hD 4/5 Nu D = i = 0.023 (35, 068 ) (1.004 )0.3 = 99.58
k
hi =
0.0258 W m K 99.58 = 51.4 W m 2 K 0.050 m
-1 -1 1 + ( 0.050 ) ln ( 0.100 0.050 ) + 0.050 1 W m 2 K = 51.4 2 0.085 0.100 10
Hence, from Eq. 3.31, the overall coefficient for the inner surface is
1 D ln ( D o Di ) Di 1 Ui = + i + 2k Do h o hi
Ui = 1.946 10-2 + 2.039 10-1 + 5.000 10-2
-1
= 3.66 W m 2 K .
Continued...
PROBLEM 8.63 (Cont.)
With condensation occurring when the surface temperature reaches 100C, the corresponding value of Tm may be determined from the local (x = x1) requirement that Ui ( Di ) Tm ( x1 ) - T = h i ( Di ) Tm ( x1 ) - Ts . Hence,
T - ( h i Ui ) Ts 20 - (51.4 3.66 )100 C = = 107.7 C Tm ( x1 ) = 1 - ( h i Ui ) 1 - (51.4 3.66 )
The distance at which the mean steam temperature is 107.7C can then be estimated from Eq. (1), where P = Di and U = Ui,
( 20 - 107.7 ) C = exp - (0.050 m ) 3.66 W m 2 K ( x1 ) 0.0172 kg s 2080 J kg K ( 20 - 120 ) C
x1 = 8.15 m COMMENTS: Note that condensation first occurs at the location for which the surface, and not the mean, temperature reaches 100C.
<
PROBLEM 8.64 KNOWN: Length and diameter of air conditioning duct. Inlet temperature of chilled air. Temperature and convection coefficient associated with outer air. Chilled air flowrate. FIND: Chilled air exit temperature and heat flow rate. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible tube wall conduction resistance, (3) Negligible kinetic and potential energy changes and axial conduction. PROPERTIES: Table A-4, Air (300K, 1 atm): cp = 1007 J/kgK, = 184.6 10 0.0263 W/mK, Pr = 0.707. ANALYSIS: The exit temperature may be obtained from Eq. 8.46, where U = hi-1 + h -1 o With
-7
kg/sm, k =
(
)
-1
& Re D = ( 4m/ D ) =
( 0.3 m )184.6 10-7 kg/s m
4 ( 0.05 kg/s)
= 11,495
the flow is turbulent and, assuming fully developed conditions over the entire length, the DittusBoelter correlation yields NuD = 0.023Re 4/5 Pr 0.4 = 0.023 (11,495 ) D
4/5
( 0.707 )0.4 = 35.5
h i = Nu D ( k/D ) = 35.5 ( 0.0263 W/m K/0.3 m ) = 31.1 W/m 2 K and U = 3.11-1 + 2.0-1
(
) ( W/m2 K ) = 1.22 W/m2 K.
-1
Eq. 8.46 yields
& Tm,o = T - T - Tm,i exp - DL/m c p U ( 0.3 m )15 m 1.22 W/m 2 K = 15.7o C o C - 30o C exp - Tm,o = 37 0.05 kg/s (1007 J/kg K )
(
)
(
(
)
)
<
and the heat rate is
o & q = m cp Tm,o - T m,i = 0.05 kg/s (1007 J/kg K ) 8.7 C = 438 W.
(
)
(
)
<
COMMENTS: The temperature rise of the chilled air is excessive, and the outer surface of the duct should be insulated to reduce U and thereby Tm,o and q.
PROBLEM 8.65
KNOWN: Length, diameter, insulation characteristics and burial depth of a pipe. Ground surface temperature. Inlet temperature, flow rate and properties of oil flowing through pipe. FIND: (a) An expression for the oil outlet temperature, (b) Oil outlet temperature and pipe heat transfer rate for prescribed conditions, and (c) Design information for trade off between burial depth of pipe (z) and pipe insulation thickness (t) on the heat loss. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Two-dimensional conduction in soil, (4) Negligible pipe wall thermal resistance, (5) Total resistance to heat loss is independent of x. PROPERTIES: Oil (given): o = 900 kg/m3, cp,o = 2000 J/kgK, o = 8.5 10-4 m2/s, ko = 0.140 W/mK, Pro = 104; Soil (given): ks = 0.50 W/mK; Insulation (given): ki = 0.05 W/mK. ANALYSIS: (a) From Eq. 8.36 for a differential control volume in the oil and the rate equation
dqconv = mocp,odTm = dq = ( Ts - Tm ) R tot
where the total resistance is expressed as
(1)
R tot = R conv + R cond,i + R cond,s = h Ddx
(
ln ( Do D )-1 + 2 k dx i ) + k1S
i s (2)
1 ln ( Do Di ) cosh -1 ( 2z Do ) / dx = R tot dx R tot = + + h Di 2 k i 2 k s
where, from Table 4.1, It follows that
S = 2 dx cosh -1 ( 2z Do )
(3)
(Ts - Tm ) dx = m
R tot
o cp,o dTm
dTm dx = Ts - Tm mo cp,o R tot
Integrating between inlet and outlet conditions Tm,o dTm L dx . =- Tm,i Tm - Ts 0 mo cp,o R tot
Assuming R to be independent of x and integrating, tot
Tm,o - Ts
L = exp - mo cp,o R Tm,i - Ts tot
.
(3)
<
Continued...
PROBLEM 8.65 (Cont.)
(b) To calculate Tm,o for the prescribed conditions, begin by evaluating h , where
ReD =
4mo 4 500 kg s = = 694 Di o o (1.2 m ) 900 kg m3 8.5 10-4 m 2 s 0.0668 ( Di L ) Re D Pr
(4)
Hence, the flow is laminar, and with a thermal entry length,
Nu D = 3.66 +
1 + 0.04 ( Di L ) ReD Pr
2/3
(5)
( Di L ) ReD Pr =
1.2 Nu D = 6.82 (694 )104 = 83.3 5 10 k 0.14 W m K h= 6.82 = 6.82 = 0.80 W m 2 K Di 1.2 m 1 0.8 W m 2 K (1.2 m ) + ln (1.5 1.2 ) + cosh -1 ( 4 )
From Eq. (2), the overall thermal resistance is
R tot =
2 ( 0.05 W m K )
2 ( 0.5 W m K )
R tot = (0.33 + 0.71 + 0.66 ) K m W = 1.70 K m W
and the oil outlet temperature can be calculated as Tm,o - Ts 105 m
= 0.943 = exp - 500 kg s 2000 J kg K 1.7 K m W Tm,i - Ts
Tm,o = 110.9C The total rate of heat transfer from the pipeline is then
<
(6)
q = mocp,o Tm,i - Tm,o
(
)
q = 500 kg s 2000 J kg K (120 - 110.9 ) C = 9.1106 W.
<
(c) Using the IHT Workspace with the foregoing equations, an analysis was performed to determine the heat loss, q, as a function of burial depth for the range, 1 z 6 m, for thicknesses of insulation which are -25%, +25%, +50% and 100% that of the base case, t = ro - r1 = 150 mm.
14
Heat loss, q (MW)
12
10
8
6 1 2 3 4 5 6
Burial depth, z (m) Insulation thickness, t = 113 mm (-25%) t = 150 mm (base case) t = 188 mm (+25%) t = 225 mm (+50%) t = 300 mm (+100%)
Continued...
PROBLEM 8.65 (Cont.)
From this information, the operations manager can compare the costs associated with burial depth and insulation thickness with respect to acceptable heat loss. COMMENTS: (1) Since the thermal entry region is very long, xfd,t 0.05DReDPr = 4.16 105 m, hx will be changing with x throughout the pipe. A more accurate solution would therefore be one in which Eq. (1) is integrated numerically, in a step-by-step fashion. For example, the integration could involve a step width of x = 103 m, with h and R evaluated at each step. t (2) The three contributions to the total thermal resistance are comparable. (3) IHT version 1.0 doesn't support inverse hyperbolic functions. To determine the shape factor from Eq. (3), use this approach:
// Shape factor: S = 2 * pi / yy cosh (yy) = 2 * z / Do cosh (yy) = 0.5 * ( exp (yy) + exp (-yy) ) R'conds = 1 / ( ks * S ) // Eq (2); Table 4.1 // Inverse hyperbolic function representation // Definition of the the function // Thermal resistance
PROBLEM 8.66
KNOWN: Length, diameter, insulation characteristics and burial depth of pipe. Ground surface temperature. Inlet temperature, minimum allowable exit temperature, flow rate and properties of oil flow through pipe. FIND: Effect of soil thermal conductivity and flowrate on heat rate and outlet temperature. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Two-dimensional conduction in soil, (4) Negligible pipe wall thermal resistance, (5) Total resistance to heat loss is independent of x. PROPERTIES: Oil (given): o = 900 kg/m3, cp,o = 2000 J/kgK, o = 8.5 10-4 m2/s, ko = 0.140 W/mK, Pro = 104. ANALYSIS: From the analysis of Problem 8.60, the outlet temperature may be computed from the expression Tm,o - Ts Tm,i - Ts where R = tot 1 h D i + ln ( Do Di ) cosh -1 ( 2z Do ) + 2 k i 2 k s = exp -
L mcp,o R tot
and h is determined from Eq. 8.56. The heat rate may then be obtained from the overall energy balance
q = mcp Tm,i - Tm,o
(
)
Using the Correlations Toolpad of IHT to perform the parametric calculations, the following results were obtained.
120 1.2E7
Outlet temperature, Tmo(C)
Heat rate, q(W)
110
1E7
100
8E6
90 0.25 0.4 0.55 0.7 0.85 1
6E6 0.25 0.4 0.55 0.7 0.85 1
Soil thermal conductivity, ks(W/m.K) mdot = 250 kg/s mdot = 375 kg/s mdot = 500 kg/s
Soil thermal conductivity, ks(W/m.K) mdot = 250 kg/s mdot = 375 kg/s mdot = 500 kg/s
Continued...
PROBLEM 8.66 (Cont.)
Due to a reduction in the thermal conduction resistance of the soil with increasing ks, there is a corresponding increase in the heat rate q from the pipe and a reduction in the oil outlet temperature. The heat rate also increases with increasing m (due to an increase in h and hence a decrease in the convection resistance), but the increase lags that of the flow rate, causing the outlet temperature to increase with increasing m . Conditions for which Tm,o 110C cannot be achieved for m = 250 kg/s, = 375 kg/s and 500 kg/s, but can be achieved for ks 0.33 W/mK and ks 0.65 W/mK for m respectively. The worst case condition corresponds to the smallest value of m and the largest value of ks. Measures to maintain Tm,o 110C could include increasing the burial depth, increasing the insulation thickness, and/or using an insulation of lower thermal conductivity. COMMENTS: The thermophysical properties of oil depend strongly on temperature, and a more accurate solution would account for the effect of variations in Tm on the properties.
PROBLEM 8.67
KNOWN: Water flowing through a thin-walled tube is heated by hot gases moving in cross flow over the tube. FIND: Outlet temperature of the water, Tm,o. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Tube wall thermal resistance negligible, (3) Exhaust gas properties those of atmospheric air, (4) KE, PE, flow work changes negligible, (5) Fully developed internal flow, (6) Constant properties. PROPERTIES: Table A-6, Water assume Tm,o 50 o C, Tm = ( 50 + 30 )o C/2 315 K): = 1/vf = 991.1 kg/m , cp = 4179 J/kgK, = 631 10
3 -6
(
Ns/m , k = 0.634 W/mK, Pr = 4.16; Table A-4, Air m /s, k = 40.7 10
2 -3
2
(T = 225C 500 K, 1 atm): = 38.79 10
-6
W/mK, Pr = 0.684.
ANALYSIS: This situation may be analyzed using Eq. 8.46 with U = 1/ (1/hi + 1 / ho ) , where hi and ho correspond to coefficients for internal and external flows, respectively. Internal flow:
& With properties evaluated at Tm , assuming Tm,o = 50C, ReD = 4m /D = 4 0.25 kg/s / 0.040
m 631 10
-6
Ns/m = 12,611. The internal flow is turbulent and fully developed (L/D = 100), and
2
the Dittus-Boelter correlation for heating conditions (Ts > Tm) is appropriate, k 0.634 W/m K h i = 0.023Re4/5 Pr0.4 = 0.023 (12,611 )4 / 5 ( 4.16 )0.4 = 1230 W/m 2 K. D D 0.040 m For the external flow, ReD = VD/ = 100 m/s 0.040 m/38.79 10 -6 m 2 / s = 1.031 105 and from Table 7.4, C = 0.26 and m = 0.6; Pr 10, n = 0.37, and Pr Prs , h o = ( k/D ) CRem Prn ( Pr/Prs ) D ho = 40.7 10-3 W/m K
1/4
0.040 m 2 Substituting numerical values into Eq. 8.46 with P = D and U = 197 W/m K, T - Tm,o & = exp - PLU/mc p T - Tm,i
0.26 1.031 105
(
)
0.6
( 0.684) 0.37 ( 1)1 / 4 = 234 W/m2 K.
(
)
(1)
0.040 m 4 m o Tm,o = 225o C - ( 225 - 30 ) Cexp - 197 W/m 2 K = 47.6o C 0.25 kg/s 4179 J/kg K
COMMENTS: Note the assumed value of Tm,o to evaluate water properties was reasonable. Using Eq. (1), replacing T and U with Ts and hi, respectively, find Ts = 63.2C; hence, Prs (Ts ) 0.687. The assumption that Pr Prs in the Zhukauskas relation is reasonable.
<
PROBLEM 8.68
KNOWN: Single tube heat exchanger for cooling blood. FIND: (a) Temperature at which properties are evaluated in estimating h, (b) Prandtl number for the blood, Pr, (c) Flow condition: laminar or turbulent, (d) Average heat transfer coefficient, h, for blood flow, (e) Total heat rate, q, (f) Required length of tube, L, when U is known. SCHEMATIC:
PROPERTIES: Blood Given, Tm : = 1000 kg/m , = 7 10 4000 J/kgK.
(
)
3
-7
m /s, k = 0.5 W/mK, cp =
2
ASSUMPTIONS: (1) Flow and thermal conditions fully developed, (2) Thermal resistance of tube material is negligible, (3) Overall heat transfer coefficient between blood and water-ice mixture is U = 300 W/m K, (4) Constant properties, (5) Negligible heat transfer enhancement associated with coiling. o ANALYSIS: (a) Evaluate properties at Tm = ( To +Ti ) / 2 = ( 40 + 30) C/2 = 35oC. (b) The Prandtl number is
2
< <
Pr =
c p k
=
c p k
( 4000 J/kg K 7 10-7m2 / s 1000 kg/m3 ) = 5.60. =
0.5 W/m K
(c) Calculate Reynolds number as
& & & 4 10-4 m 3 /min (1 min/60s ) 4m 4 V 4V Re D = = = = = 1213 D D D 2.5 10-3 m 7 10-7 m 2 / s
Hence, the flow is laminar, (d) For laminar and fully developed conditions, Eq. 8.55 is the proper correlation,
<
h = 3.66 0.5 W/m K/2.5 10-3 m = 732 W/m 2 K,
NuD = hD/k = 3.66
<
(e) The total heat rate follows from an overall energy balance, Eq. 8.37,
& & q = m c p ( To - Ti ) = Vc p ( T - Ti ) o q = 1000 kg/m 3 10-4 m 3 /min/60 s/min 4000 J/kg K ( 30 - 40 )o C = -66.7 W.
(f) Using the rate equation, Eq. 8.47, solve for L,
(
)
< <
L=
q 66.7 W = = 0.81 m U D Tlm 300 W/m2 K 2.5 10-3 m 34.8o C
where A = DL and Tl m = [(40 - 0)C - (30 - 0)C]/ l n(40/30) = 34.8C.
(
)
PROBLEM 8.69
KNOWN: Flow conditions associated with water passing through a pipe and air flowing over the pipe. FIND: (a) Differential equation which determines the variation of the mixed-mean temperature of the water, (b) Heat transfer per unit length of pipe at the inlet and outlet temperature of the water. SCHEMATIC:
ASSUMPTIONS: (1) Negligible temperature drop across the pipe wall, (2) Negligible radiation exchange between outer surface of insulation and surroundings, (3) Fully developed flow throughout pipe, (4) Negligible potential and kinetic energy and flow work effects. PROPERTIES: Table A-6, Water (Tm,i = 200C): cp,w = 4500 J/kgK, w = 134 10 kw = 0.665 W/mK, Prw = 0.91; Table A-4, Air (T = - 10C): a = 12.6 10 W/mK, Pra = 0.71, Prs 0.7.
-6 2 -6
Ns/m ,
2
m /s, ka = 0.023
ANALYSIS: (a) Following the development of Section 8.3.1 and applying an energy balance to a differential element in the water, we obtain
& & m c p,w Tm - dq - m c p,w ( Tm + dTm ) = 0.
Hence where
& dq = -m c p,w dTm dq = UidAi ( Tm - T ) = U i D dx ( Tm - T ) . d Tm U D =- i ( Tm - T ) . & dx m cp
Substituting into the energy balance, it follows that
(1)
<
The overall heat transfer coefficient based on the inside surface area may be evaluated from Eq. 3.30 which, for the present conditions, reduces to
Ui =
1 . 1 D D + 2t D 1 + ln + h i 2k D D + 2t h o & 4m 4 2 kg/s = = 19,004. D w (1 m ) 134 10 -6 kg/s m
Continued .....
(2)
For the inner water flow, Eq. 8.6 gives
Re D =
PROBLEM 8.69 (Cont.)
Hence, the flow is turbulent. With the assumption of fully developed conditions, it follows from Eq. 8.60 that k 0.3 h i = w 0.023 Re 4/5 Prw . (3) D D For the external air flow
Re D =
V ( D+2t )
=
12.6 10-6 m 2 / s
4 m/s (1.3m )
= 4.13 105.
Using Eq. 7.31 to obtain the outside convection coefficient,
ho =
ka 0.37 0.076 Re0.7 Pra ( Pra / P rs )1/4 . D ( D + 2t )
(4)
(b) The heat transfer per unit length of pipe at the inlet is
q = D Ui Tm,i - T .
From Eqs. (3 and 4),
(
)
(5)
hi = ho =
0.665 W/m K 0.023 (19,004 ) 4 / 5 ( 0.91)0.3 = 39.4 W/m 2 K 1m
0.7 0.023 W/m K 0.076 4.13 105 ( 0.71)0.37 (1)1 / 4 = 10.1 W/m2 K. (1.3 m ) -1
(
)
Hence, from Eq. (2)
1 1m 1 1.3 1 Ui = + ln + 39.4 W/m 2 K 0.1 W/m K 1 1.3 10.1 W/m 2 K
and from Eq. (5)
= 0.37 W/m 2 K
q = (1 m ) 0.37 W/m 2 K
(
) ( 200 +10)o C = 244 W/m.
<
Since Ui is a constant, independent of x, Eq. (1) may be integrated from x = 0 to x = L. The result is analogous to Eq. 8.42b and may be expressed as
DL 1m 500m = exp - Ui = exp - 0.37 W/m 2 K m c p,w T - Tm,i 2 kg/s 4500 J/kg K & T - Tm,o Hence = 0.937. T - Tm,i T - Tm,o Tm,o = T + 0.937 Tm,i - T = 187o C.
(
)
<
COMMENTS: The largest contribution to the denominator on the right-hand side of Eq. (2) is made by the conduction term (the insulation provides 96% of the total resistance to heat transfer). For this reason the assumption of fully developed conditions throughout the pipe has a negligible effect on the calculations. Since the reduction in Tm is small (13C), little error is incurred by evaluating all properties of water at Tm,i.
PROBLEM 8.70
KNOWN: Inner and outer radii and thermal conductivity of a teflon tube. Flowrate and temperature of confined water. Heat flux at outer surface and temperature and convection coefficient of ambient air. FIND: Fraction of heat transfer to water and temperature of tube outer surface. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully-developed flow, (3) One-dimensional conduction, (4) Negligible tape contact and conduction resistances. PROPERTIES: Table A-6, Water (Tm = 290K): = 1080 10 7.56.
-6
kg/sm, k = 0.598 W/mK, Pr =
ANALYSIS: The outer surface temperature follows from a surface energy balance
( 2
ro L) q =
( h o2 r oL )-1 ( ln ( ro / ri ) / 2
T -T
Ts,o - T
+
Ts,o - Tm
Lk ) + (1 / 2 ri Lh i )
q = h o Ts,o - T +
With
(
) ( r / k ) ln ( rs,o/ r ) +m( r / r ) / h . o o i o i i
& Re D = 4 m/ ( D ) = 4 ( 0.2kg/s ) / ( 0.02 m )1080 10 -6 kg/s m = 11,789
4/5 0.4
the flow is turbulent and Eq. 8.60 yields
h i = ( k/Di ) 0.023ReD Pr = (0.598 W/m K/0.02 m )(0.023 )(11,789 )
4/5
( 7.56)0.4 = 2792 W/m 2 K.
Hence
2000 W/m = 25 W/m K Ts,o - 300K +
2 2
(
)
Ts,o - 290 K
(0.013 m/0.35 W/m K ) ln (1.3 ) + (1.3 ) /
(2792 W/m K )
2
and solving for Ts,o, The heat flux to the air is
Ts,o = 308.3 K.
< <
q = h o Ts,o - T = 25 W/m 2 K ( 308.3 - 300 ) K = 207.5 W/m 2 . o
Hence,
(
)
q / q = ( 2000 - 207.5) W/m 2 /2000 W/m 2 = 0.90. i
COMMENTS: The resistance to heat transfer by convection to the air substantially exceeds that due to conduction in the teflon and convection in the water. Hence, most of the heat is transferred to the water.
PROBLEM 8.71
KNOWN: Temperature recorded by a thermocouple inserted in a stack containing flue gases with a prescribed flow rate. Diameters and emissivities of thermocouple tube and gas stack. Conditions associated with stack surroundings. FIND: Equations for predicting thermocouple error and error associated with prescribed conditions. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Flue gas has properties of air at Tg 327C, (3) Stack forms a large enclosure about the thermocouple tube and surroundings form a large enclosure around the stack, (4) Stack surface energy balance is unaffected by heat loss to tube, (5) Gas flow is fully developed, (6) Negligible conduction along thermocouple tube, (7) Stack wall is thin. PROPERTIES: Table A-4, Air (Tg 600K, pg = 1 atm): = 0.58 kg/m , = 305.8 10 = 52.7 10
-6 3 -7
Ns/m ,
2
m /s, k = 0.0469 W/mK, Pr = 0.685.
2
ANALYSIS: Determination of the thermocouple error necessitates determining the gas temperature Tg and relating it to the thermocouple temperature Tt. From an energy balance applied to a control surface about the thermocouple, 4 q conv = q rad or h t A t Tg - Tt = t A t Tt4 - Ts . Hence
Tg = Tt + t Tt4 - Ts4 . ht q conv,i = qconv,o + qrad
(
)
(
)
(
)
(1)
<
However, Ts is unknown and must be determined from an energy balance on the stack wall. 4 4 h i As Tg - Ts = h o As ( T - T ) + s A s Ts - Tsur s or
(
)
h 4 Tg = Ts + o ( Ts - T ) + s Ts4 - Tsur . hi hi
(
)
(
)
(2)
<
Tg and Ts may be determined by simultaneously solving Eqs. (1) and (2). For the prescribed conditions 2 & & VDt mg / Ds / 4 Dt 4 m gD t 4 1 kg/s 0.01 m ReDt = = = = = 1157. 2 Ds 305.8 10-7 N s/m2 ( 0.6 m )2 Continued .....
(
)
PROBLEM 8.71 (Cont.)
Assuming (Pr/Prs ) = 1, it follows from the Zhukauskus correlation
Nu D = 0.26Re0.6 Pr 0.37 Dt
where C = 0.26 and m = 0.6 from Table 7.4. Hence
ht =
0.0469 W/m K (1157 ) 0.6 ( 0.685 ) 0.87 0.26 = 73 W/m2 K. 0.01 m Tg = 573 K + 0.8 5.67 10-8 W/m 2 K 4 73 W/m2 K
Hence, from Eq. (1)
(5734 -Ts4 )K 4
(1a)
Tg = 573 K + 67 K - 6.214 10 -10 T 4 = 640 - 6.214 10 -10 Ts4. s
Also,
Re Ds =
& 4 mg Ds
=
4 1 kg/s ( 0.6 m ) 305.8 10-7 N s/m2
= 6.94 10 4
and the gas flow is turbulent. Hence from the Dittus-Boelter correlation,
hi =
4/5 k 0.0469 W/m K 0.3 0.023Re 4/5 Pr 0.3 = 0.023 6.94 104 ( 0.685 ) = 12 W/m 2 K. Ds Ds 0.6 m
(
)
Hence from Eq. (2)
Tg = Ts +
25 0.8 5.67 10-8 W/m 2 K 4 4 4 4 ( Ts - 300 K ) + Ts - 300 K 2 K 12 12 W/m (2a)
4 4 Tg = Ts + 2.083Ts - 625 K + 3.78 10 -9 Ts - 30.6 K = -655.6 K + 3.083Ts + 3.78 10-9 Ts .
Solve Eqs. (1a) and (2a) by trial-and-error. Assume values for Ts and determine Tg from (1a) and (2a). Continue until values of Tg agree.
Ts (K) 400 375 387 388
Hence
Tg (K) (1a) 624 628 626 626
Tg (K) (2a) 674 575 622 626
Ts = 388 k, Tg = 626 K Tg - Tt = 626 K - 573 K = 53oC.
and the thermocouple error is
<
COMMENTS: The thermocouple error results from radiation exchange between the thermocouple tube and the cooler stack wall. Anything done to Ts would this error (e.g., ho or T and Tsur). The error also with ht. The error could be reduced by installing a radiation shield around the tube.
PROBLEM 8.72
KNOWN: Platen heated by hot ethylene glycol flowing through tubing arrangement with spacing S soldered to lower surface. Top surface exposed to convection process. FIND: Tube spacing S and heating fluid temperature Tm which will maintain the top surface at 45 0.25C. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions; (2) Lower surface is insulated, all heat transfer from hot fluid is into platen; (3) Copper tube is thick-walled such that interface between solder and platen is isothermal; (4) Fully developed flow conditions in tube. PROPERTIES: Table A.4, Ethylene glycol (Tm = 60C): = 0.00522 Ns/m2, k = 0.2603 W/mK. ANALYSIS: Begin the analysis by setting up a nodal mesh (9 6) to represent the platen experiencing convection on the top surface ( T , h) while the two side boundaries are symmetry adiabats. On the lower surface, nodes 46 and 47 represent the isothermal platen-solder interface maintained at To by the hot fluid. The remaining nodes (49-54) are insulated on their lower boundary.
The heat rate supplied by the tube to the platen can be expressed as
q = 0.5h o ( Di )( Tm - To ) cv q = q + q + q cd a b c
(1)
From energy balances about nodes 46 and 47, the heat rate into the platen by conduction can be expressed as (2) (3) Continued...
q = k ( x 2 )( T46 - T37 ) y a
PROBLEM 8.72 (Cont.) q = k ( x )( T47 - T38 ) y b q = k ( y 2 )( T47 - T48 ) x c
and we require that (4) (5)
q = q cd cv
The convection coefficient for internal flow can be estimated from a correlation assuming fully developed flow. First, characterize the flow with
(6)
ReD =
4m 4 0.06 kg s = = 1829 Di (0.008 m ) 0.00522 N s m 2
and since it is laminar,
h D Nu D = o i = 3.66 k h o = 3.66 0.2603W m K 0.008 m = 119.1W m K
where properties are evaluated at Tm. Using the IHT Finite-Difference Tool for Two-Dimensional Steady-State Conditions and the Properties Tool for Ethylene Glycol, along with the foregoing rate equations and energy balances, Eqs. (1-6), a model was developed to solve for the temperature distribution in the platen. In the solution, we determined what hot fluid temperature was required to maintain T1 = 45C. Two trials were run. In the first, the nodal arrangement was as shown above (9 6) for which S/2 = (9 - 1)x = 42.67 mm with x = 2Di/3 = 5.33 mm and y = w/5 = 5 mm. In the second trial, we repositioned the right-hand symmetry adiabat to pass vertically through the nodes 6-51 so that now the nodal mesh is (6 6) and S/2 = (6 - 1)x = 26.65 mm with x and y remaining the same. The results of the trials are tabulated below. Trial 1 2 Mesh 96 66 T1 (C) 45.0 45.0 T6 (C) 43.5 44.5 T9 (C) 43.0 --Tm (C) 105 85
q (W/m) cv
80.5 52.6
From the trial 2 results, the surface temperature uniformity is (T1 - T6) = 0.5C which satisfies the 0.25C requirement. So that suitable tube spacing and fluid temperature are S = 53 mm Tm = 85C
<
COMMENTS: (1) Recognize that the grid spacing is quite coarse and good practice demands that we repeat the analysis decreasing the nodal spacing until no further changes are seen in Tm. (2) In the first trial, note that Tm = 105C which of course, is not possible.
PROBLEM 8.73
KNOWN: Features of tubing used in a ground source heat pump. Temperature of surrounding soil. Fluid inlet temperature and flowrate. FIND: (a) Effect of tube length on outlet temperature, (b) Recommended tube length and the effect of variations in the flowrate. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible conduction resistance in soil, (4) Negligible KE, PE and flow work changes, (5) Fluid properties correspond to those of water. PROPERTIES: Table A.6 (assume Tm = 277 K): cp = 4206 J/kgK, = 1560 10-6 Ns/m2, k = 0.577 W/mK, Pr = 11.44.
ANALYSIS: (a) For the prescribed conditions, ReD = 4m Di = 4 ( 0.03kg s ) ( 0.025m )1560
6 2 10 N s m = 980 and the flow is laminar. Assuming thermal entry length conditions, which is reasonable for Pr = 11.44, Eq. 8.56 may be used to determine the average convection coefficient Nu D = 3.66 + 0.0668 ( D L ) ReD Pr 1 + 0.04 ( D L ) ReD Pr
2/3
With Ts used in lieu of TW , Eq. 8.46b may be used to determine Tm,o,
Ts - Tm,o
L = exp - Ts - Tm,i mcp R tot
where R accounts for the convection and tube wall conduction resistances, tot
R tot = R + R = (1 Di h ) + ln ( Do Di ) 2 k t cnv cnd
and
Do = Di + 2t = 41mm .
Using the Correlations and Properties Toolpads of IHT, the following results were obtained for the effect of the tube length L on Tm,o.
Continued...
PROBLEM 8.73 (Cont.)
10 Outlet temperature, Tmo(C)
8
6
4
2
0 10 15 20 25 30 35 40 45 50
Tube length, L(m) mdot = 0.015 kg/s mdot = 0.030 kg/s mdot = 0.045 kg/s
The longer the tube the larger the rate of heat extraction from the soil, and for m = 0.030 kg/s, the temperature rise of T = (Tm,o - Tm,i) 6C is well below the maximum possible value of Tmax = 10C.
(b) The length should be at least 50 m long. If the flowrate were reduced by 50% ( m = 0.015 kg/s), the corresponding temperature rise would be close to Tmax and L = 50 m would be close to optimal. However, for the nominal flowrate and a 50% increase from the nominal, the length should exceed 50 m to recover more heat and provide a heat pump inlet temperature which is closer to the maximum possible value.
COMMENTS: In practice, the tube surface temperature would be less than 10C (if the temperature of the soil well removed from the tube were at 10C), thereby reducing the heat extraction rate and Tm,o.
PROBLEM 8.74 KNOWN: Reynolds numbers for fully developed turbulent flow of water in a smooth circular tube. FIND: Nusselt numbers based on the Colburn, Petukhov and Gnielinski correlations. SCHEMATIC:
ANALYSIS: The correlations are Colburn, Equation 8.59, Nu D = 0.023Re4/5 Pr1/3 D Petukhov, Equation 8.62, NuD =
( f/8) Re D Pr 1/2 1.07 + 12.7 ( f/8 ) Pr 2 / 3 -1
(
)
f = (1.82log10 Re D - 1.64 ) Gnielinski, Equation 8.63, Nu D =
-2
( f/8)( ReD - 1000 ) Pr 1/2 1 +12.7 ( f/8 ) Pr 2 / 3 -1
(
)
ReD
f = ( 0.79lnReD - 1.64 ) It follows that:
-2
___________________________________________________________________________________________________________
Correlation Colburn Petukhov Gnielinski 31.8 40.0 30.0
4000 66.2 81.2 75.0
10
4
10 417.9 549.5 560.0
5
___________________________________________________________________________________________________________
COMMENTS: The Colburn equation does well in the transitional region (ReD 10 ), where the Gnielinski equation provides the best predictions, but can significantly underpredict the Nusselt number under fully turbulent conditions.
4
PROBLEM 8.75 KNOWN: Effect of entry length on average Nusselt number for turbulent flow in a tube. FIND: Ratio of average to fully developed Nusselt numbers for prescribed conditions. SCHEMATIC:
ASSUMPTIONS: (1) Sharp edged inlet, (2) Combined entry region. ANALYSIS: From Eq. 8.64, Nu D C = 1+ Nu D,fd ( x/D ) m and with C = 24Re-0.23 and m = 0.815 - 2.08 10 -6 Re D , D Nu D = 1+ Nu D,fd It follows that _______________________________________________________________
( x/D )(
0.815- 2.0810 -6 ReD
24Re -0.23 D
)
.
( Nu D / Nu D,fd )
1.464 1.112 1.420 1.142
Re D
4 4 5 5
b x / Dg
10 60 10 60
_______________________________________________________________ 10 10 10 10
COMMENTS: The assumption Nu D Nu fd for x/D = 10 would result in underprediction of Nu D by approximately 45%. The underprediction is only approximately 10% for x/D = 60.
PROBLEM 8.76
KNOWN: Fluid enters a thin-walled tube of 5-mm diameter and 2-m length with a flow rate of 0.04 kg/s and temperature of Tm,i = 85C; tube surface temperature is maintained at Ts = 25C; and, for this base operating condition, the outlet temperature is Tm,o = 31.1C. FIND: The outlet temperature if the flow rate is doubled? SCHEMATIC:
ASSUMPTIONS: (1) Flow is fully developed and turbulent, (2) Fluid properties are independent of temperature, and (3) Constant surface temperature cooling conditions.
ANALYSIS: For the base operating condition (b), the rate equation, Eq. 8.42b, with C = m cp , the
capacity rate, is
Ts - Tm,o
Ts - Tm,i
)b
PL h b = exp - Cb
(1)
Substituting numerical values, with P = D, find the ratio, h b / Cb ,
25 - 31.1 = exp 0.005 m 2 m ( h b / Cb ) 25 - 85
h b / Cb = 72.77 m -2
For the new operating condition (n), the flow rate is doubled, Cn = 2Cb, and the convection coefficient scales according to the Dittus-Boelter relation, Eq. 8.60, h Re0.8 m0.8 D
h n = 20.8 h b and h n / C n = 20.8 / 2 h b / Cb Ts - Tm,o
(
)
(
) ( )
(2)
Using the rate equation for the new operating condition, find
Ts - Tm,i 25 - Tm,o 25 - 85 Tm,o
)n
PL h n = exp - Cn
= exp - PL 0.871 h b / Cb
(3)
)n
= exp - 0.005 m 2 m 0.871 72.77 m -2
)n = 33.2C
<
PROBLEM 8.77
KNOWN: Flow rate and inlet temperature of air passing through a rectangular duct of prescribed dimensions and surface heat flux. FIND: Air and duct surface temperatures at outlet. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform surface heat flux, (3) Constant properties, (4) Atmospheric pressure, (3) Fully developed conditions at duct exit, (6) Negligible KE, PE and flow work effects. PROPERTIES: Table A-4, Air Tm 300K, 1 atm : cp = 1007 J/kgK, = 184.6 10 k = 0.0263 W/mK, Pr = 0.707. ANALYSIS: For this uniform heat flux condition, the heat rate is
(
)
-7
Ns/m ,
2
q = q As = q 2 ( L W ) + 2 ( L H ) s s
q = 600 W/m 2 2 (1m 0.016m ) + 2 (1m 0.004m ) = 24 W.
From an overall energy balance
Tm,o =Tm,i +
q 24 W = 300K + = 379 K. & m cp 3 10 -4 kg/s 1007 J/kg K
<
The surface temperature at the outlet may be determined from Newton's law of cooling, where
Ts,o = Tm,o + q /h.
From Eqs. 8.67 and 8.1
Dh =
4 A c 4 ( 0.016m 0.004m ) = = 0.0064 m P 2 ( 0.016m + 0.004m )
& 3 10 -4 kg/s ( 0.0064m ) u m Dh m Dh Re D = = = = 1625. Ac 64 10-6m 2 184.6 10-7 N s/m2
(
)
Hence the flow is laminar, and from Table 8.1
h=
k 0.0263 W/m K 5.33 = 5.33 = 22 W/m 2 K Dh 0.0064 m 600 W/m 2 22 W/m 2 K = 406 K.
Ts,o = 379 K +
<
COMMENTS: The calculations should be reperformed with properties evaluated at Tm = 340 K. The change in Tm,o would be negligible, and Ts,o would decrease slightly.
PROBLEM 8.78 KNOWN: Flow rate and temperature of air entering a triangular duct of prescribed dimensions and surface temperature. FIND: Air outlet temperature. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform surface temperature, (4) Fully developed conditions throughout, (5) Air is at atmospheric pressure, (6) Negligible potential and kinetic energy and flow work effects. PROPERTIES: Table A-4, Air (assume Tm 325K, 1 atm): cp = 1008 J/kgK, = 196.4 10 Ns/m , k = 0.0282 W/mK, Pr = 0.707. ANALYSIS: From Eqs. 8.67 and 8.1
-4 2 4 A c 4 1.73 10 m Dh = = = 0.0115 m P 3 ( 0.02m )
-7
2
(
)
& 4 10-4 kg/s ( 0.0115 m ) u mDh m Dh Re D = = = = 1354. Ac 1.73 10 -4m 2 196.4 10 -7 N s/m 2
(
)
Hence the flow is laminar and from Table 8.1, h= k 0.0282 W/m K 2.47 = 2.47 = 6.1 W/m 2 K. Dh 0.0115 m
From Eq. 8.42b it follows that, with P = 3 W, PL Tm,o =Ts - Ts - Tm,i exp - h m cp &
(
)
3 0.02m 2m 6.1 W/m 2 K Tm,o = 100o C - 100oC - 27o C exp - 4 10-4 kg/s 1008 J/kg K
(
)
Tm,o = 88o C. COMMENTS: With Tm,o = 88C, Tm = 330K and there is no need to re-evaluate the properties.
<
PROBLEM 8.79
KNOWN: Temperature and velocity of gas flow between parallel plates of prescribed surface temperature and separation. Thickness and location of plate insert. FIND: Heat flux to the plates (a) without and (b) with the insert. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible radiation, (3) Gas has properties of atmospheric air, (4) Plates are of infinite width W, (5) Fully developed flow. PROPERTIES: Table A-4, Air (1 atm, Tm = 1000K): = 0.348 kg/m , = 424.4 10 = 0.0667 W/mK, Pr = 0.726. ANALYSIS: (a) Based upon the hydraulic diameter Dh, the Reynolds number is
3 -7
kg/sm, k
Dh = 4 Ac / P = 4 ( H W ) / 2 ( H + W ) = 2H = 80 mm Re Dh =
u m D h 0.348 kg/m3 ( 60 m/s ) 0.08 m = = 39,360. 424.4 10-7 kg/s m
Since the flow is fully developed and turbulent, use the Dittus-Boelter correlation, 4/5 NuD = 0.023 Re 4/5 Pr 0.3 = 0.023 ( 39,360 ) ( 0.726 )0.3 = 99.1 D k 0.0667 W/m K h= Nu D = 99.1 = 82.6 W/m 2 K Dh 0.08 m q = h ( Tm - Ts ) = 82.6 W/m 2 K (1000 - 350 ) K = 53,700 W/m 2 . (b) From continuity, & m = ( u m A )a = ( u m A )b u m )b = um )a ( A ) a / ( A )b = 60 m/s ( 40/20 ) = 120 m/s.
<
For each of the resulting channels, Dh = 0.02 m and u m D h 0.348 kg/m3 (120 m/s ) 0.02 m Re Dh = = = 19,680. -7
424.4 10
kg/s m h=
Since the flow is still turbulent,
Nu D = 0.023(19,680 )4 / 5 ( 0.726 )0.3 = 56.9
2
56.9 ( 0.0667 W/m K ) 0.02 m
2
= 189.8 W/m 2 K
q = 189.8 W/m K (1000 - 350 ) K = 123,400 W/m .
COMMENTS: From the Dittus-Boelter equation, 0.8 0.2 h b / ha = u m,b / u m,a Dh,a / Dh,b = ( 2 )0.8 ( 4 ) 0.2 = 1.74 1.32 = 2.30.
<
(
) (
)
Hence, heat transfer enhancement due to the insert is primarily a result of the increase in um and secondarily a result of the decrease in Dh.
PROBLEM 8.80
KNOWN: Temperature, pressure and flow rate of air entering a rectangular duct of prescribed dimensions and surface temperature. FIND: Air outlet temperature and duct heat transfer rate. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform surface temperature, (4) Fully developed flow throughout, (5) Negligible potential and kinetic energy and flow work effects. PROPERTIES: Table A-4, Air (assume Tm 325K, 1 atm): cp = 1008 J/kgK, = 196.4 10 Ns/m , k = 0.0282 W/mK, Pr = 0.707. ANALYSIS: From Eqs. 8.67 and 8.1,
2 -7
Dh =
4 A c 4 ( 0.15 0.075) m 2 = = 0.10 m P 2 ( 0.15 + 0.075 ) m
Re D =
0.1 kg/s ( 0.1m ) & u mD h m Dh = = = 45,260. A c 0.15m 0.075m )196.4 10-7 N s/m 2 (
Hence the flow is turbulent, and from Eq. 8.60
h=
k 0.0282 W/m K 0.023 Re 4/5 Pr 0.4 = 0.023 ( 45,260 ) 4 / 5 ( 0.707 ) 0.4 = 30 W/m 2 K. D Dh 0.10 m PL Tm,o = Ts - T - Tm,i exp - h s m cp & 2 ( 0.15m + 0.075m ) 2m 30W/m2 K Tm,o = 400 K - ( 400 - 285 ) K exp - 0.1 kg/s 1008 J/kg K
From Eq. 8.42b, with P = 2(W + H),
(
)
(
)
Tm,o = 312 K
and from Eq. 8.37
<
& q = m cp Tm,o - T m,i = 0.1 kg/s 1008 J/kg K ( 312 - 285) K = 2724 W.
(
)
<
COMMENTS: (1) The calculations may be checked by determining q from Eqs. 8.44 and 8.45. We obtain Tl m = 101o C and q = 2724 W. (2) Tm has been over-estimated. The calculations should be repeated with properties evaluated at Tm = 299 K.
PROBLEM 8.81
KNOWN: Dimensions of semi-circular copper tubes in contact at plane surfaces. Thermal contact resistance. Tube flow conditions. FIND: (a) Heat rate per unit tube length, and (b) The effect on the heat rate when the fluids are ethylene glycol, the exchanger tube is fabricated from an aluminum alloy, or the exchanger tube thickness is increased. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Adiabatic outer surface, (4) Fully developed flow, (5) Negligible heat loss to surroundings. PROPERTIES: Table A.1, Copper (T 300 K): k = 400 W/mK; Table A.6, Water (290 K): = 1080 10-6 Ns/m2, k = 0.598 W/mK, Pr = 7.56; (330 K): = 489 10-6 Ns/m2, k = 0.65 W/mK, Pr = 3.15; (given): = 800 10-6 kg/sm, k = 0.625 W/mK, Pr = 5.35. ANALYSIS: (a,b) Heat transfer from the hot to cold fluids is enhanced by conduction through the semicircular portions of the tube walls. The walls may be approximated as straight fins with an insulated tip, and the thermal circuit is shown below.
Note that, since each semi-circular surface is insulated on one side, surfaces may be combined to yield a single fin of thickness 2t with convection on both sides. Also, due to the equivalent geometry and the assumption of constant properties, there is symmetry on opposite sides of the contact resistance. From the thermal circuit, the heat rate is
q =
Th,m - Tc,m R tot
(1)
For flow through the semi-circular tube,
ReD =
u m Dh mDh 4mAc 4m 4m = = = = Ac Ac P P ( 2ri + ri ) 4 0.2 kg s ReD = = 9725 ( 2 + ) 0.02 m 800 10-6 kg s m
(2)
the flow is turbulent. Using the Colburn correlation, 4/5 4 Nu D = 0.023ReD/ 5 Pr1/ 3 = 0.023 (9725 ) (5.35)1/ 3 = 62.4
(3) Continued...
2 4Ac 4 ri 2 2 Dh = 0.02 m = 0.0244 m = = P ( + 2 ) ri + 2
(
)
PROBLEM 8.81 (Cont.)
(4)
h = Nu D R conv =
k 0.625 = 62.4 = 1600 W m 2 K . Dh 0.0244 1 1 = = 0.0156 m K W 2ri h ( 0.04 m )1600 W m 2 K
(5)
Find now values for the thermal resistance of the circuit. (6)
1 R = b = fin q ( hPkA )1/ 2 tanh ( hP kA ) L f c c L = ri 2 = ( 0.01m ) = 0.0314 m Ac = 2t 1m = 0.006 m 2
(7)
P 2.1 m (8,9,10) 1/ 2
(1600 W m2 K 2 m m 400 W m K 0.006 m2 s ) = 87.6 W K m 1/ 2 ( hP kAc )1/ 2 L = (1600 W m 2 K 2 m 400 W m K 0.006 m2 ) 0.0314 m = 1.15
( hPkAc )1/ 2 =
R = fin 1 = 0.0140 m K W 87.6 W m K ( 0.817 ) t 0.003m R = = 1.875 10-4 m K W cond = 2kri 2 ( 400 W m K )(0.02 m ) R t,c 2ri = 10-5 m 2 K W = 2.5 10-4 m K W 2 ( 0.02 m )
(11) (12)
R t,c =
(13)
The equivalent resistance of the parallel circuit is -1 -1 R = R -1 + R -1 = ( 71.4 W m K + 64.1W m K ) = 7.38 10-3 m K W (14) eq conv fin Hence
(
)
R tot = 2 R + R eq cond + R t,c R tot = 2 7.38 10-3 + 1.875 10-4 + 2.50 10-4 m K W = 0.0154 m K W q =
(
)
(15)
(
)
(330 - 290 ) K
0.0154 m K W
= 2600 W m .
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(c) Using the IHT Workspace with the foregoing equations, analyses were performed and the results summarized in the table below. The "Conditions" are described below; the "Change" is relative to the base case condition. Continued .....
PROBLEM 8.81 (Cont.)
Condition* Base case Ethylene glycol Aluminum alloy Thicker tube
R conv 10
4
R 104 fin
(mK/W) 140.1 923.1 183.2 130.6
4 R cond 10
R 104 tot (mK/W) 154.2 1113.0 180.0 150.0
R 104 eq (mK/W) 73.96 553.5 84.49 71.25
q
(W/m) 2594 359 2223 2667
Change (%) --86.2 -14.3 +2.8
(mK/W) 156.8 1382 156.8 156.8
(mK/W) 1.88 1.88 4.24 2.50
*Conditions: change from base case Base case - water, copper (k = 400 W/mK), t = 3 mm Ethylene glycol - ethylene glycol instead of water Aluminum alloy - alloy (k = 177 W/mK) instead of copper Thicker tube - t = 4 mm instead of 3 mm As expected, using ethylene glycol as the working fluid would decrease the heat rate. Note that R conv is the dominate resistance since the convection coefficient is considerably reduced compared to that with water. Using aluminum alloy, rather than copper, as the tube material reduces the heat rate by 14%. Conduction-convection (fin) in the tube wall is important as can be seen by examining the change in R relative to the base condition. Increasing the tube wall thickness for the copper tube exchanger fin from 3 to 4 mm had only a marginal positive effect on the heat rate. COMMENTS: A more accurate calculation would account for the absence of symmetry about the contact plane. Evaluation of water properties at Th,m = 330 K and Tc,m = 290 K yields hh = 2060 W/m2K and hc = W/m2K.
PROBLEM 8.82
KNOWN: Heat exchanger to warm blood from a storage temperature 10C to 37 at 200 ml/min. Tubing has rectangular cross-section 6.4 mm 1.6 mm sandwiched between plates maintained at 40C. FIND: (a) Length of tubing and (b) Assessment of assumptions to indicate whether analysis under- or over-estimates length. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes, (3) Blood flow is fully developed, (4) Blood has properties of water, and (5) Negligible tube wall and contact resistance. PROPERTIES: Table A-6, Water ( Tm 300 K): cp,f = 4179 J/kgK, f = 1/vf = 997 kg/m , f = fvf = 8.58 10
-7 3 2
m /s, k = 0.613 W/mK, Pr = 5.83.
ANALYSIS: (a) From an overall energy balance and the rate equation,
& q = m c p Tm,o - Tm,i = hA sTLMTD
where
(
)
(1)
TLMTD =
( 40 -15 ) - ( 40 - 37 ) T1 - T2 = = 10o C. ln ( T1 / T2 ) ln ( 2 5 / 3)
To estimate h, find the Reynolds number for the rectangular tube,
u D 0.326 m/s 0.00256 m Re D = m h = = 973 8.58 10-7 m2 / s
where
Dh = 4 Ac / P = 4 ( 6.4 mm 1.6 mm ) / 2 ( 6.4 +1.6 ) mm = 2.56 mm Ac = ( 6.4 mm 1.6 mm ) = 1.024 10 -5m 2 & & u m = m/ Ac = /Ac = 200 m l/60 s 10 -6m 3 / ml /1.024 10 -5 m 2 = 0.326 m/s.
(
)
Hence the flow is laminar, but assuming fully developed flow with an isothermal surface from Table 8.1 with b/a = 6.4/1.6 = 4, hDh 4.4 0.613 W/m K Nu D = = 4.4 h= = 1054 W/m2 K.
k
0.00256 m
Continued .....
PROBLEM 8.82 (Cont.)
From Eq. (1) with
As = PL = 2 ( 6.4 + 1.6 ) 10-3m L=1.6 10-2 LZ & m = Acu m = 997 kg/m 3 1.024 10 -5 m2 0.326 m/s = 3.328 10- 3 kg/s
the length of the rectangular tubing can be found as
3.328 10-3 kg/s 4179 J/kg K ( 37 - 15) K = 1054 W/m 2 K 1.6 10-2 Lm 2 10 K L = 1.8 m.
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(b) Consider these comments with regard to whether the analysis under- or over-estimates the length, fully-developed flow - L/Dh = 1.8 m/0.00256 = 700; not likely to have any effect, negligible tube wall resistance - depends upon materials of construction; if plastic, analysis under predicts length, negligible thermal contact resistance between tube and heating plate - if present, analysis under predicts length.
PROBLEM 8.83
KNOWN: Coolant flowing through a rectangular channel (gallery) within the body of a mold. FIND: Convection coefficient when the coolant is process water or ethylene glycol. SCHEMATIC:
ASSUMPTIONS: (1) Gallery can be approximated as a rectangular channel with a uniform surface temperature, (2) Fully developed flow conditions. PROPERTIES: Table A.6, Water ( Tm = (140 + 15)C/2 = 350 K): = 974 kg/m3, = 365 10-6 ns/m2, = / = 3.749 10-7 m2/s, k = 0.668 W/mK, Pr = 2.29; Table A.5, Ethylene glycol ( Tm = 350 K): = 1079 kg/m3, = 3.17 10-6 m2/s, k = 0.261 W/mK, Pr = 34.6. ANALYSIS: The characteristic length of the channel, the hydraulic diameter, Eq. 8.67, is Dh = 4Ac P where Ac is the cross-sectional flow area and P is the wetted perimeter. For our channel,
Dh =
2 (a + b )
4 (a b )
=
4 0.090 m 0.0095 m = 0.0172 m 2 ( 0.090 + 0.0095 ) m
For the water coolant, from the continuity equation, find the Reynolds number to characterize the flow V 1.3 10-3 m3 s um = = = 1.52 m s Ac 0.090 m 0.0095 m
u D 1.52 m s 0.0172 m ReDh = m h = = 69, 736 3.749 10-7 m 2 s
Since the flow is turbulent, and assuming fully developed conditions, use the Dittus-Boelter correlation, Eq. 8.60, to estimate the convection coefficient, hDh 0.8 0.4 Nu Dh = = 0.023Re0.8 Pr 0.4 = 0.023 ( 69, 736 ) ( 2.29 ) = 240 Dh k
hw =
0.668 W m K 240 = 9326 W m 2 K 0.0172 m Nu Dh = 128 h eg = 1957 W m 2 K
< <
Continued...
Repeating the calculations using properties for the ethylene glycol coolant, find
ReDh = 8, 247
PROBLEM 8.83 (Cont.)
COMMENTS: (1) The convection coefficient for the water coolant is more than 4 times greater than that with the ethylene glycol coolant. The corrosion protection afforded by the latter coolant greatly compromises the thermal performance of the gallery. In such situations, it is useful to explore a compromise between corrosion protection and thermal performance by using an aqueous solution of ethylene glycol (50%-50%, for example). (2) Recognize that for the ethylene glycol coolant calculation the Reynolds number is slightly below the lower limit of applicability of the Dittus-Boelter correlation.
PROBLEM 8.84
KNOWN: Dimensions, surface temperature and thermal conductivity of a cold plate. Velocity, inlet temperature, and properties of coolant. FIND: (a) Model for determining the heat rate q and outlet temperature, Tm,o, (b) Values of q and Tm,o for prescribed conditions. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible KE, PE and flow work changes, (3) Constant properties, (4) Symmetry about the midplane (horizontal) of the cold plate and the midplane (vertical) of each cooling channel, (5) Negligible heat transfer at sidewalls of cold plate, (6) Onedimensional conduction from outer surface of cold plate to base surface of channel and within the channel side walls, which act as extended surfaces. PROPERTIES: Water (prescribed): = 984 kg/m3, cp = 4184 J/kgK, = 489 10-6 Ns/m2, k = 0.65 W/mK, Pr = 3.15. ANALYSIS: (a) The outlet temperature, Tm,o, may be determined from the energy balance prescribed by Eq. 8.46b,
Ts - Tm,o
1 = exp - m1cp R tot Ts - Tm,i
where m1 = u m A c is the flowrate for a single channel and Rtot is the total resistance to heat transfer between the cold plate surface and the coolant for a particular channel. This resistance may be determined from the symmetrical section shown schematically, which represents one-half of the cell associated with a full channel. With the number of channels (and cells) corresponding to N = W/S, there are 2N = 2(W/S) symmetrical sections, and the total resistance Rtot of a cell is one-half that of a symmetrical section. Hence, Rtot = Rss/2, where the resistance of the symmetrical section includes the effect of conduction through the outer wall of the cold plate and convection from the inner surfaces. Hence,
R ss =
(H - h ) 2 + 1 k cp (SW ) o hA t
where At = Af + Ab = 2(h/2 W) + (w W), h is the average convection coefficient for the channel flow, and o is the overall surface efficiency.
A o = 1 - f (1 - f ) At
Continued...
PROBLEM 8.84 (Cont.)
The efficiency f corresponds to that of a straight, rectangular fin with an adiabatic tip, Eq. 3.89, and Lc = w/2. With D h = 4A c P = 4w 2 4w = w = 0.006 m , Re D = u m D h = 984 kg/m3 2 m/s 0.006
h
m/489 10-6 Ns/m2 = 24,150 and the channel flow is turbulent. Assuming fully-developed flow throughout the channel, the Dittus-Boelter correlation, Eq. 8.60, may therefore be used to evaluate h , where Nu D Nu D,fd = 0.023Re 4 / 5 Pr 0.4 D The total heat rate for the cold plate may be expressed as
q = Nq1 = Nm1cp Tm,o - Tm,i
(b) For the prescribed conditions,
(
)
m1 = u m A c = 984 kg m3 ( 2 m s )( 0.006 m ) = 0.0708 kg s
2
(3.15 )0.4 = 116.8 h = 116.8 k D h = 116.8 ( 0.65 W m K ) ( 0.006 m ) = 12, 650 W m 2 K A f = 2 ( h 2 W ) = 2 ( 0.003 m 0.1m ) = 6 10-4 m 2 A t = Af + A b = 6 10-4 m 2 + ( 0.006 m 0.1m ) = 1.2 10-3 m 2
4/5
Nu D = 0.023 ( 24,150 )
= h ( 2 + 2W ) k cp ( W ) = [12,650 W/m2K(0.008 + 0.200)m/400 With m = hPf k cp Acf W/mK(0.004 0.100)m2]1/2 = 128.2 m-1. tanh m ( h 2 ) tanh (128.2 0.003 ) 0.366 = = = 0.952 f = m (h 2 ) 128.2 0.003 0.385
(
)1/ 2
1/ 2
o = 1 - 0.5 (1 - 0.952 ) = 0.976
R ss =
( 0.010 - 0.006 ) m 2 + 400 W m K (0.01m 0.1m ) 0.976
R ss = (0.005 + 0.0675 ) K W = 0.0725 K W With Rtot = Rss/2 = 0.0362 K/W, Ts - Tm,o 1 = exp - = 0.911 Ts - Tm,i 0.0708 kg s 4184 J kg K 0.0362 K W Tm,o = Ts - 0.911 Ts - Tm,i = 360 K - 0.911(360 - 300 ) K = 305.3 K The total heat rate is
(12650 W m2 K )1.2 10-3 m2
1
(
)
< <
q = Nm1c p Tm,o - Tm,i = 10 0.0708 kg s 4184 J kg K (305.3 - 300 ) K = 15, 700 W
(
)
COMMENTS: The prescribed properties correspond to a value of Tm which significantly exceeds that obtained from the foregoing solution ( Tm = 302.6 K). Hence, the calculations should be repeated using more appropriate thermophysical properties (see next problem). From Eq. 3.85, the effectiveness of the extended surface is
= R t,b R t,f = ( h W )
-1
( hA f f )-1 = ( A f f
W ) = 6 10
(
-4
m 0.954
2
) (0.004 m 0.10 m ) = 1.43.
Hence, the ribs are only marginally effective in enhancing heat transfer to the coolant.
PROBLEM 8.85
KNOWN: Geometry, surface temperature and thermal conductivity of a cold plate. Velocity and inlet temperature of coolant. FIND: Effect of channel width on total heat rate. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible KE, PE and flow work changes, (3) Constant properties, (4) Symmetry about midplane (horizontal) of the cold plate and the midplane (vertical) of each channel, (5) Negligible heat transfer at sidewalls of cold plate, (6) One-dimensional conduction from outer surface of cold plate to base surface of channel and within the channel side walls, which act as extended surfaces. PROPERTIES: Water: Evaluated at Tm using the Properties Toolpad of IHT. ANALYSIS: The model developed for the preceding problem was entered into the workspace of IHT, with the Dittus-Boelter equation and exponential relation accessed from the Correlations Toolpad and modified to account for the hydraulic diameter and the total resistance to heat transfer. Calculations were performed for Case 1: Case 2: Case 3: Case 4: Case 5: w = 96 mm, N = 1, S = W = 100 mm w = 46 mm, N = 2, S = 50 mm w = 21 mm, N = 4, S = 25 mm w = 6 mm, N = 10, S = 10 mm w = 1 mm, N = 20, S = 5 mm
and the results are tabulated as follows. Case N Dh (m) ReD
h W m K
(
2
)
Tm,o (K)
q (W)
1 2 3 4 5
1 2 4 10 20
0.01129 0.01062 0.00933 0.00600 0.00171
24,820 25,320 22,340 14,620 4761
8269 8895 9142 10,070 13,740
300.7 302.3 302.6 304.3 317.2
3164 10,370 10,960 12,940 17,160
It is clearly beneficial to increase the number of channels, with the total heat rate increasing by approximately a factor of 5 as N increases from 1 to 20. The heat rate may be increased further by increasing um, and hence the flowrate per channel, although an upper limit would be associated with the pressure drop, which would increase with decreasing Dh. Could additional heat transfer enhancement be achieved by altering the thickness of the channel walls? COMMENTS: Note that results obtained for Case 4 differ from those of the preceding problem due to different fluid properties. In this case the properties were evaluated at the actual value of Tm = 302.6 K, rather than at an assumed (significantly larger) value.
PROBLEM 8.86 KNOWN: Heat sink with 24 passages for air flow removes power dissipation from circuit board. FIND: Operating temperature of the board and pressure drop across the sink. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible kinetic and potential energy changes, (3) Negligible thermal resistance between the circuit boards and air passages, (4) Sink surface and board are isothermal at Ts . PROPERTIES: Table A-4, Air ( T 310 K,1 atm): = 1.1281 kg/m , cp = 1008 J/kgK, = -6 2 16.89 10 m /s, k = 0.0270 W/mK, Pr = 0.706. ANALYSIS: The air outlet temperature follows from Eq. 8.43, Ts - Tm,o PLh = exp - . m cp & Ts - Tm,i
3
The mass flow rate for the entire sink is & & m = = 1.1281 kg/m 3 0.060 m 3 / s = 6.77 10-2 kg/s and the Reynolds number for a rectangular passage is u D Re D = m h where Dh = 4Ac / P = 4 ( 6 mm 25 mm ) / 2 ( 6 + 25 ) mm = 9.68 mm um = & m/24 6.77 10 -2 kg/s/24 = = 16.7 m/s Ac 1.1281 kg/m3 (6 25) 10- 6 m 2 16.7 m/s 9.68 10-3 m 16.89 10-6 m 2 / s = 9571.
giving Re D =
Assume the flow is turbulent and fully developed and using the Dittus-Boelter correlation find NuD = 0.023Re 4/5 Pr 0.4 = 0.023 ( 9571) h=
4/5
( 0.706) 0.4 = 30.6
Nu k 30.6 0.027 W/m K = = 85.4 W/m 2 K. Dh 0.00968 m Continued .....
PROBLEM 8.86 (Cont.) From an overall energy balance on the sink, & q = m c p Tm,o - Tm,i
(
)
& T m,o =Tm,i + q/m c p
Tm,o = 27o C + 50 W/6.77 10 -2 kg/s 1008 J/kg K = 27.73oC Hence, the operating temperature of the circuit board for these conditions is -3 2 Ts - 27.73 2 ( 6 + 25 ) 10 m 0.150m 85.4 W/m K = exp - Ts -27 6.77 10-2 kg/s/24 1008 J/kg K
(
)
Ts = 30o C.
<
The pressure drop in the rectangular passage for the smooth surface condition follows from Eqs. 8.22 and 8.20 p = f where f = 0.31Re1/4 = 0.316 ( 9554 ) D p = 0.0320
-1/4
u2 mL Dh
= 0.0320. 0.150 m = 156 N/m 2 .
1.1281 kg/m 3 (16.7 m/s ) 2 0.00968 m
<
COMMENTS: (1) The analysis has been simplified by assuming the channel is rectangular and all four sides experience heat transfer. Since the insulated surface is a small portion of the total passage surface area, the effect can't be very large.
3 & (2) The power required to move the air through the heat sink is Pelec = p = 0.060 m /s 156 2 N/m = 9.4 W.
PROBLEM 8.87
KNOWN: Channel formed between metallic blades dissipating heat by internal volumetric generation. FIND: (a) The heat removal rate per blade for the prescribed thermal conditions and (b) Time required to slow a train of mass 106 kg from 120 km/h to 50 km/h. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions for channel blades and air flow, (2) The blades form a channel of rectangular (a b) cross-section and length L, (3) Negligible kinetic energy changes and axial conduction inside the channel, and (4) Fully developed flow conditions in the channel. PROPERTIES: Table A.4, Air ( Tm 350 K, 1 atm): = 0.995 kg/m3, cp = 1009 J/kg K, = 20.92 10-6 m2/s, k = 0.030 W/mK, Pr = 0.700. ANALYSIS: (a) The heat removal rate by the air from a single channel (one blade) follows from an overall energy balance,
q = mcp Tm,o - Tm,i Ts - Tm,o
(
)
(1)
where the outlet temperature can be determined from Eq. 8.42b,
PL = exp - mcp Ts - Tm,i
h
(2)
The hydraulic diameter, Dh, flows from Eq. 8.67, 2 4Ac 4 ( a b ) 4 ( 0.220 0.004 ) m = = = 0.0079 m Dh = P 2 ( a + b ) 2 ( 0.220 + 0.004 ) m Assuming Tm = 350K , the Reynolds number is (3)
u D 50 m s 0.0079m ReDh = m h = = 18, 779 20.92 10-6 m 2 s
Using the Dittus-Boelter correlation, Eq. 8.60,
(4)
Nu Dh =
hDh 0.8 0.4 = 0.023 Re0.8 Pr0.4 = 0.023 (18, 779 ) (0.700 ) = 52.37 Dh k
(5) Continued...
PROBLEM 8.87 (Cont.) h= 0.030 W m K 52.37 = 199 W m 2 K 0.0079m
Hence, the outlet temperature is
2 (0.220 + 0.004 ) m 0.070m 199 W m 2 K = exp 0.0438 kg s 1009 J kg K (600 - 25) C 600 - Tm,o Tm,o = 100.7 C
where
m = Acum = 0.995 kg/m3 (0.220 0.004) m2 50 m/s = 0.0438 kg/s
and the rate of heat removal per blade, from Eq. (1), is
q = 0.0438 kg s 1009 J kg K (100.7 - 25 ) C = 3.346 kW
<
(b) From an energy balance on the locomotive over an interval of time, t, the heat energy transferred to the air stream results in a change in kinetic energy of the train,
- Eout = E = KEf - KEi - (q N ) t = 1 M Vf2 - Vi2 2
(6)
(
)
50, 000 2 120, 000 2 2 2 - m s 3600 3600
1 -3346 W blade 2000blades t (s ) = 10-6 kg 2
t = 69s
<
COMMENTS: (1) For the channel, L/Dh = 0.070 m/0.0079 m = 8.9 < 10 so that the assumption of fully developed conditions may not be satisfied. Recognize that the flow at the channel entrance may be highly turbulent because of the upstream fan swirl and ducting. (2) What benefits could be realized by increasing the heat transfer coefficient? Aside from increasing velocity, what design changes would you make to increase h? (3) Our assumption for Tm = 350 K at which to evaluate properties is reasonable considering Tm = (100.7 + 25)C/2 = 335 K.
PROBLEM 8.88
KNOWN: Chip and cooling channel dimensions. Channel flowrate and inlet temperature. Chip temperature. FIND: Water outlet temperature and chip power. SCHEMATIC:
ASSUMPTIONS: (1) Negligible kinetic and potential energy changes for water, (2) Uniform channel surface temperature, (3) Tm = 300 K, (4) Fully developed flow. PROPERTIES: Table A-6, Water ( Tm = 300 K): cp = 4179 J/kgK, = 855 10 0.613 W/mK, Pr = 583. ANALYSIS: Using the hydraulic diameter, find the Reynolds number, 4 ( H W ) 2 ( 50 200 ) m 2 -6 Dh = = 10 m/ m = 8 10-5 m
-6
kg/sm, k =
2 (H + W )
250 m
10-4 kg/s 810-5 m & u mDh m1D h Re D = = = = 936. A c ( 50 200 )10-12 m2 855 10 -6 kg/s m
(
(
)
)
Hence, the flow is laminar and, from Table 8.1, NuD = 4.44, so that
h = Nu D
4.44 ( 0.613 W/m K ) k = = 34,022 W/m 2 K. -5 m Dh 8 10
-6
With P = 2(H + W) = 2(250 m) 10
m/m = 5 10
-4
m, Eq. 8.42b yields
Ts - Tm,o Ts - Tm,i
=
350K - Tm,o 60 K
PL = exp - m1 cp &
5 10-6 m2 34,022 W/m 2 K h = exp - 10-4 kg/s 4179 J/kg K
Tm,o = 350K - 60 K exp ( -0.407 ) = 310 K.
Hence, from Eq. 8.37,
-4 & & q = m c p Tm,o - Tm,i = Nm1c p Tm,o - T m,i = 50 10 kg/s ( 4179 J/kg K )( 20 K ) = 418 W.
< <
(
)
(
)
COMMENTS: (1) The chip heat flux of 418 W/cm is extremely large and the method provides a very efficient means of heat removal from high power chips. However, clogging of the microchannels is a potential problem which could seriously compromise reliability. (2) L/Dh = 125 and 0.05 ReDPr = 272. Hence, fully developed conditions are not realized and h > 34,022. The actual power dissipation is therefore greater than 418 W.
2
PROBLEM 8.89
KNOWN: Chip and cooling channel dimensions. Channel flow rate and inlet temperature. Temperature of chip at base of channel. FIND: (a) Water outlet temperature and chip power, (b) Effect of channel width and pitch on power dissipation. SCHEMATIC:
ASSUMPTIONS: (1) Negligible flow work and kinetic and potential energy changes for water, (2) Flow may be approximated as fully developed and channel walls as isothermal for purposes of estimating the convection coefficient, (3) One-dimensional conduction along channel side walls, (4) Adiabatic condition at end of side walls, (5) Heat dissipation is exclusively through fluid flow in channels, (6) Constant properties. PROPERTIES: Table A-6, Water ( Tm = 300K): cp = 4179 J/kgK, = 855 10 kg/sm, k = 0.613 W/mK, Pr = 5.83. ANALYSIS: (a) The channel sidewalls act as fins, and a unit channel/sidewall combination is shown in schematic (a), where the total number of unit cells corresponds to N = L/S. With N = 50 and L = 10 mm, S = 200 m and = S W = 150 m. Alternatively, the unit cell may be represented in terms of a single fin of thickness , as shown in schematic (b). The thermal resistance of the unit cell may -1 be obtained from the expression for a fin array, Eq. (3.103), Rt,o = (ohAt) , where At = Af + Ab = L -4 -4 -6 2 (2 H + W) = 0.01m (4 10 + 0.5 10 ) m = 4.5 10 m . With Dh = 4 (H W)/2 (H + W) = 4 (2 -4 -4 -4 -5 10 m 0.5 10 m)/2 (2.5 10 m) = 8 10 m, the Reynolds number is ReD = um Dh/ = m1 Dh/Ac = 10 kg/s 8 10 m/(2 10 m 0.5 10 m) 855 10 kg/sm = 936. Hence, the flow is laminar, and assuming fully developed conditions throughout a channel with uniform surface temperature, Table 8.1 yields NuD = 4.44. Hence,
-4 -5 -4 -4 -6 -6
h=
k 0.613 W / m K 4.44 Nu D = = 34, 022 W / m 2 K -5 m Dh 8 10
1/2
With m = (2h/kch) the fin efficiency is
= (68,044 W/m K/140 W/mK 1.5 10 m)
2
-4
1/2
= 1800 m and mH = 0.36,
-1
f =
tanh mH 0.345 = = 0.958 mH 0.36
and the overall surface efficiency is
A 4.0 10-6 o = l - f (l - f ) = l - (l - 0.958) = 0.963 At 4.5 10-6
The thermal resistance of the unit cell is then Continued .....
PROBLEM 8.89 (Cont.) R t,o = (o h A t )
-1
= 0.963 34, 022 W / m 2 K 4.5 10-6 m 2
(
)
-1
= 6.78 K / W
The outlet temperature follows from Eq. (8.46b),
Tm,o = Ts - Ts - Tm,i exp - = 350K - ( 60K ) m cp R t,o
1 = 307.8K exp - 10-4 kg / s 4179 J / kg K 6.78K / W
The heat rate per channel is then
(
)
l
<
q1 = m1 cp Tm,o - Tm,i = 10-4 kg / s 4179 J / kg K (17.8K ) = 7.46 W
and the chip power dissipation is
(
)
q = Nq l = 50 7.46 W = 373 W
<
(b) The foregoing result indicates significant heat transfer from the channel side walls due to the large value of f. If the pitch is reduced by a factor of 2 (S = 100 m), we obtain
S = 100 m, W = 50 m, = 50 m, N = 100 : q1 = 7.04 W, q = 704 W
<
Hence, although there is a reduction in f due to the reduction in (f = 0.89) and therefore a slight reduction in the value of ql, the effect is more than compensated by the increase in the number of channels. Additional benefit may be derived by further reducing the pitch to whatever minimum value of is imposed by manufacturing or structural limitations. There would also be an advantage to increasing the channel hydraulic diameter and or flowrate, such that turbulent flow is achieved with a correspondingly larger value of h. COMMENTS: (1) Because electronic devices fail by contact with a polar fluid such as water, great care would have to be taken to hermetically seal the devices from the coolant channels. In lieu of water, a dielectric fluid could be used, thereby permitting contact between the fluid and the electronics. However, all such fluids, such as air, are less effective as coolants. (2) With L/Dh = 125 and L/Dh)fd 0.05 ReD Pr = 273, fully developed flow is not achieved and the value of h = hfd underestimates the actual value of h in the channel. The coefficient is also underestimated by using a Nusselt number that presumes heat transfer from all four (rather than three) surfaces of a channel.
PROBLEM 8.90
KNOWN: Chip and cooling channel dimensions. Channel flow rate and inlet temperature. Temperature of chip at base of channel. FIND: (a) Outlet temperature and chip power dissipation for dielectric liquid, (b) Outlet temperature and chip power dissipation for air. SCHEMATIC:
ASSUMPTIONS: (1) Negligible flow work and kinetic and potential energy changes, (2) Flow may be approximated as fully developed and channel walls as isothermal for purposes of estimating the convection coefficient, (3) One-dimensional conduction along the channel side walls, (4) Adiabatic condition at end of side walls, (5) Heat dissipation is exclusively through fluid flow in channels, (6) Constant properties. PROPERTIES: Prescribed. Dielectric liquid: cp = 1050 J/kgK, k = 0.065 W/mK, = 0.0012 2 -7 2 Ns/m , Pr = 15. Air: cp = 1007 J/kgK, k = 0.0263 W/mK, = 185 10 Ns/m , Pr = 0.707. ANALYSIS: (a) The channel side walls act as fins, and a unit channel/sidewall combination is shown in schematic (a), where = S W = 150 m. Alternatively, the unit cell may be represented in terms of a single fin of thickness , as shown in schematic (b). The thermal resistance of the unit cell -1 may be obtained from the expression for a fin array, Eq. (3.103), Rt,o = (o h At) , where At = Af + -6 2 -8 2 Ab = L (2 H + W) = 4.5 10 m . With Ac = H W = 10 m and Dh = 4 Ac/2(H + W) = 8 10 5 m, the Reynolds number is ReD = umDh/ = m1 Dh/Ac = 667. Hence, the flow is laminar, and assuming fully developed conditions throughout a channel with uniform surface temperature, Table k 0.065 W / m K 4.44 Nu D = 8.1 yields NuD = 4.44. Hence, h = = 3608 W / m 2 K -5 m Dh 8 10 With m = (2 h/kch )
1/2
= 586 m and mH = 0.117, the fin efficiency is
-1
f =
tanh mH 0.1167 = = 0.995 mH 0.117
and the overall surface efficiency is
A 4.0 10-6 o = l - f (1 - f ) = l - (l - 0.995 ) = 0.996. At 4.5 10-6
The thermal resistance of the unit cell is then
R t,o = (o h A t )
-1
= 0.996 3608 W / m 2 K 4.5 10-6 m 2
(
)
-1
= 61.9 K / W
The outlet temperature follows from Eq. (8.46b),
l Tm,o = Ts - Ts - Tm,i exp - m1 cp R t,o
(
)
= 350K
Continued .....
PROBLEM 8.90 (Cont.) 1 - (60K ) exp - = 298.6K 10-4 kg / s 1050 J / kg K 61.9 K / W
The heat rate per channel is then
<
q1 = m1 cp Tm,o - Tm,i = 10-4 kg / s 1050 J / kg K 8.6 K = 0.899 W
and the chip power dissipation is
(
)
q = Nq1 = 50 0.899 W = 45.0 W
<
(b) With m1 = 10 -6 kg / s, Re D = m l D h / A c = 432 and the flow is laminar. Hence, with NuD = 4.44,
h=
k 0.0263 W / m K 4.44 Nu D = = 1460 W / m 2 K -5 m Dh 8 10
1/2
With m = (2 h/kch)
= 373 m and mH = 0.0746, the fin efficiency is
-1
f =
tanh mH 0.0744 = = 0.998 mH 0.0746
and the overall surface efficiency is
A 4.0 10-6 o = l - f (1 - f ) = l - (l - 0.998 ) = 0.998 At 4.5 10-6
Hence,
R t,o = (o h A t )
-1
= 0.998 1460 W / m 2 K 4.5 10-6 m 2
(
)
-1
= 153K / W
The outlet temperature is then
l Tm,o = Ts - Ts - Tm,i exp - m1 cp R t,o
(
)
= 350K
1 = 349.9 K - (60K ) exp - 10-6 kg / s 1007 J / kg K 153K / W q1 = m1 cp Tm,o - Tm,i = 10-6 kg / s 1007 J / kg K 59.9 K = 0.060 W q = Nq1 = 3.02 W
<
(
)
<
COMMENTS: (1) For laminar flow in the channels, there is a clear advantage to using the dielectric liquid instead of air. (2) The prescribed channel geometry is by no means optimized, and the number of fins should be increased by reducing S. Also, channel dimensions and/or flow rates could be increased to achieve turbulent flow and hence much larger values of h. (3) With L/Dh = 125 and L/Dh)fd 0.05 ReD Pr = 500 for the dielectric liquid, fully developed flow is not achieved and its assumption yields a conservative (under) estimate of the convection coefficient. The coefficient is also underestimated by using a Nusselt number that presumes heat transfer from all four (rather than three) surfaces of a channel.
PROBLEM 8.91
KNOWN: Arrangement of chips and cooling channels for a substrate. Contact and conduction resistances. Coolant velocity and inlet temperature. FIND: (a) Coolant temperature rise, (b) Chip and substrate temperatures. SCHEMATIC:
ASSUMPTIONS: (1) Constant properties, (2) Fully-developed flow, (3) Negligible kinetic and potential energy changes, (4) Heat transfer exclusively to water, (5) Steady-state conditions. PROPERTIES: Water (given): = 1000 kg/m , cp = 4180 J/kgK, k = 0.610 W/mK, Pr = 5.8, = 855 10
-6 3
kg/sm.
q & m cp NL P u mA cc p 10 5 W 1000 kg/m
3 o
ANALYSIS: (a) For a single flow channel, the overall energy balance yields
Tm,o - Tm,i = = =
(1 m/s )( 0.005 m )
2
4180 J/kg K
= 0.48 C.
<
From the thermal circuit,
q= To - Tm R t,c + R cond + R conv
R t,c = R / A s = 0.5 10-4 m2 K/W /10 ( 0.005 m )2 = 0.2 K/W. t,c 2
(
)
With Dh = 4Ac /P = 4(0.005 m) /4(0.005 m) = 0.005 m, ReD = u m Dh 1000 kg/m (1 m/s ) 0.005 m = = 5848. -6 855 10 kg/s m
3
With turbulent flow, the Dittus-Boelter correlation yields k 0.61 W/m K 0.023 5848 4 / 5 5.8 0.4 = 5849 W/m 2 K h = 0.023Re 4/5 Pr 0.4 = ( ) ( ) D D 0.005 m R conv = ( hAs )
-1
= 5849 W/m2 K 4 0.005 m 0.2 m
(
)
-1
= 0.043 K/W.
Approximating Tm as (Tm,i + Tm,o)/2 = 25.24C, Tc = Tm + q R t,c + R cond + R conv = 25.24o C + 50 W ( 0.2 + 0.12 + 0.043 ) K/W = 43.3oC. Similarly, from the thermal circuits, Ts = Tm + q R conv = 25.24 o C + 50W 0.043K/W = 27.4 oC
(
)
<
<
COMMENTS: (1) Since the coolant temperature rise is less than 0.5C, all chip temperatures will be within 0.5C of each other. (2) The channel surface temperature may also be obtained from Eq. 8.42b, yielding the same result.
PROBLEM 8.92
KNOWN: Power dissipation of components on each side of a hollow core PCB. Dimensions of PCB. Inlet temperature and flow rate of air. FIND: Outlet air temperature and inlet and outlet surface temperatures for prescribed flow rates. SCHEMATIC:
ASSUMPTIONS: (1) Steady flow, (2) Negligible flow work and potential and kinetic energy changes, (3) Channel may be approximated as infinite parallel plates, (4) Uniform surface heat flux, (5) Fully developed flow at exit, (6) Constant properties. PROPERTIES: Table A-4, Air ( Tm 310K): = 1.128 kg/m , cp = 1007 J/kgK, = 189.3 10 2 Ns/m , k = 0.0270 W/mK, Pr = 0.706.
ANALYSIS: Performing an energy balance for a control surface about the hollow core, 2q = m c p
3
-7
(Tm,o - Tm,i ) , in which case
Tm,o =
2q 80 W + Tm,i = + 20C = 59.7C m cp 0.002 kg / s 1007 J / kg K
<
The surface temperatures may be obtained from Newton's law of cooling, q s = h ( Ts - Tm ). Hence, with h at the entrance, where the thermal boundary layer thickness is zero,
Ts,i = Tm,i = 20C
<
With ReD = umDh/ = m Dh/Ac , where Dh = 2H = 0.008m and Ac = H W = 0.004m 0.3m = 2 2 -7 2 0.0012m , ReD = (0.002 kg/s 0.008m)/(0.0012m 189.3 10 Ns/m ) = 704 and the flow is 2 2 laminar. With a uniform surface heat flux, q = q/(W L) = 40 W/(0.3m) = 444 W/m , Table 8.3 s
yields NuD = 8.23. Hence,
h=
Nu D k Dh
=
8.23 0.027 W / m K = 27.8 W / m 2 K 0.008m
2 q s = 59.7C + 444 W / m Ts,o = Tm,o + = 75.7C h 27.8 W / m 2 K If the flowrate is increased by a factor of 5,
< <
Continued .....
Tm,o =
2q 80 W + Tm,i = + 20C = 27.9C m cp 0.01kg / s 1007 J / kg K
The surface temperature at the inlet is unchanged,
PROBLEM 8.92 (Cont.) Ts,i = 20C
<
but with ReD = 3520, flow in the channel is now turbulent. Using Eq. (8.60) as a first approximation, h =
k 4/5 0.4 4 / 5 0.4 0.027 W / m K 2 0.023 (3520 ) ( 0.706 ) = 46.4 W / m K 0.023 Re D Pr = 0.008m Dh q 444 W / m 2 Ts,o = Tm,o + s = 27.9C + = 37.5C h 46.4 W / m 2 K
<
COMMENTS: (1) With L/Dh = 37.5 and L/Dh)fd 0.05 ReD Pr = 25 for the laminar flow, it is reasonable to assume fully developed conditions at the exit. The same may be said for the turbulent flow condition. (2) The temperature difference, Ts Tm, increases from approximately 0 at the entrance to a maximum value associated with fully developed conditions.
PROBLEM 8.93
KNOWN: Printed-circuit board (PCB) with uniform temperature Ts cooled by laminar, fully developed flow in a parallel-plate channel. The air flow with an inlet temperature of Tm,i is driven by a pressure difference, p. FIND: The average heat removal rate per unit area, qs W / m 2 , from the PCB. SCHEMATIC:
(
)
ASSUMPTIONS: (1) Laminar, fully developed flow, (2) Upper and lower walls of the channel are insulated and of infinite extent in the transverse direction, (3) PCB has uniform surface temperature, (4) Constant properties, (5) Negligible kinetic and potential energy changes and flow work. PROPERTIES: Table A-4, Air (Tm = 293 K, 1 atm): = 1.192 kg/m , cp = 1007 J/kgK, = 1.531 -5 2 10 m /s, k = 0.0258 W/mK, Pr = 0.709. ANALYSIS: The energy equations for determining the heat rate from one surface of the board are Eqs. 8.37 and 8.42b
3
q = m cp Tm,o - Tm,i = qs As Ts - Tm,o PL h = exp - m cp Ts - Tm,i
(
)
(1)
(2)
where As = Lw and P = 2(w + a) where w is the width in the transverse direction. For the fully developed flow condition, the velocity is estimated from the friction pressure drop relation, Eq. 8.22a,
p = f u 2 / 2 ( L / Dh ) m
where the hydraulic diameter for the channel cross section is
(
)
(3)
Dh =
4 (w a ) 4 Ac = = 2a P 2 (w + a )
a << w
The friction factor f from Table 8.1 for the cross section b/a = is
f ReDh = 96
where the Reynolds number is
(4)
ReDh = u m Dh /
Continued .....
(5)
PROBLEM 8.93 (Cont.)
and the flow rate through one channel is
m = Ac u m = ( wa ) u m
For fully developed laminar flow from Table 8.1.
(6)
Nu D = h D h / k = 7.54 The above system of equations needs to be solved simultaneously for the unknowns: q, m, Tm,o, ReDh, f, um, h. Using IHT with w = 1 m, find these results:
2 f h (W/m K) Tm,o (C) um (m/s) ReDh 6.0 7578 0.0127 9.9 20.9 From Eq. (1), the average heat removal per unit area of the board is
(7)
q (W) 66
qs (W/m ) 442
2
qs = 442 W / m 2
COMMENTS: (1) The thermophysical properties of the air are evaluated at the average mean temperature, Tm = (Tm,i + Tm,o)/2. (2) The fully developed flow length, xfd,t, for the channel follows from Eq. 8.23,
<
x fd,t = Dh 0.05 Re Dh Pr x fd,t = 2 0.010 m 0.05 7578 0.707 = 5.4 m
Since L << xfd,t, we conclude that the flow is not likely to be fully developed. (3) Recognize also that the Reynolds number is larger than the critical value indicating that appreciable turbulence could be present. Considering that L << xfd,t and ReDh > 2300, do you conclude that our estimate for the average heat flux is a conservative or an optimistic one?
PROBLEM 8.94
KNOWN: Inner and outer tube surface conditions for an annulus. FIND: (a) Velocity profile, (b) Temperature profile and expression for inner surface Nusselt number. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Laminar, fully developed flow, (3) Uniform heat flux at inner surface, (4) Adiabatic outer surface, (5) Constant properties, (6) Negligible potential and kinetic energy changes, flow work, and viscous dissipation. ANALYSIS: (a) From Section 8.1.3, the general solution to Eq. 8.12, which also applies to annular flow as represented in Figure 8.10, is 1 dp r 2 u (r ) = + C1 ln r + C2.
dx 4
Applying the boundary conditions,
u ( ri ) = 0 u ( ro ) = 0
Hence,
2 1 dp ri 0= + C1 lnr i + C 2 dx 4
0=
2 1 dp ro + C1 lnr o + C 2. dx 4
2 2 1 dp ro ri - 2 dx 4 4 2 2 ln ro C = - 1 dp ro - 1 dp r - ri o C1 = 2 dx 4 dx 4 ln ri / ro 4 ln ( ri / ro ) and the velocity distribution is
r2 r2 ln r o - i 4 4 ln ( ri / ro ) 2 2 1 dp ro ri ln ro - - dx 4 4 ln ( ri / ro ) r 2 dp ( r / r ) 2 - 1 ln r/r . u ( r ) = - o 1 - ( r/ro ) 2 + i o ( o ) 4 dx l n ( ri / ro ) u (r ) = 1 dp dx r 2 r 2 1 dp - o + 4 4 dx
(b) For fully developed conditions with uniform surface heat flux,
(1)
<
v= 0
T/ x = dTm /dx = const.
Continued .....
PROBLEM 8.94 (Cont.)
Hence, from Eq. 8.48, which also applies for laminar flow,
1 T u dTm r = .k r r r dx
Substituting the velocity distribution, with 2 ro dp ( ri / ro ) 2 - 1 C1 = - C2 = 4 dx ln ( ri / ro ) it follows that
(2)
1 T C1 dTm 2 r r = dx 1 - ( r/ro ) + C 2 ln ( r/r o ) . r r r3
2 ro
T C1 dTm r = r - r dx T C1 dTm = r dx
+ C2r ln
r dr + C 3 ro
r r3 r r r C - + C2 ln - + 3 2 r 2 4ro 2 ro 4
and the temperature distribution is
r2 C dTm r 2 r4 r r2 - T (r ) = 1 + C2 ln - + C3 lnr + C4. 2 4 dx 4 16 ro ro 4
From the requirement that q = 0, it follows that T/ r ) = 0. Hence, o ro
(3)
<
C1 dTm ro ro ro C 3 2 - 4 + C2 - 4 + r = 0 dx o
2 C dTm ro C3 = 1 dx 4
( C 2 - 1) .
(4)
<
From the condition that T(ri) = Ts,i, it follows that
2 r2 ri4 r 2 C1 dTm ri i ln ri - i + C ln r . C4 = Ts,i - - + C2 3 i 2 dx 4 16 ro 4 ro 4
From Eqs. 8.68 and 8.70, the inner surface Nusselt number is
(5)
<
h D q Dh i Nui = i h = k k Ts,i - Tm
(
)
where Dh = 2(ro - ri). To obtain a workable form of Nui, the mean temperature Tm must be evaluated. This may be done by substituting Eqs. (1) and (3) into Eq. 8.27 and evaluating um by substituting Eq. (1) into Eq. 8.8. Since the integrations are long and tedious, they are not provided. COMMENTS: From an energy balance performed for a differential control volume in the annular 2 region, dTm /dx = 2rq / c p u m ro - ri2 . i i
(
)
PROBLEM 8.95
KNOWN: Inlet temperature, pressure and flow rate of air. Annulus length and tube diameters. Pressure of saturated steam. FIND: Outlet temperature and pressure drop of air. Mass rate of steam condensation. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Outer surface of annulus is adiabatic, (3) Negligible potential energy, kinetic energy and flow work changes for air, (4) Fully developed flow throughout annulus, (5) Smooth annulus surfaces, (6) Constant properties. PROPERTIES: Table A-4, air ( Tm 325K, p = 5 atm): = 5 (1 atm) = 5.391 kg/m , cp =
3
1008 J/kgK, = 196.4 10 Ns/m , k = 0.0281 W/mK, Pr = 0.703. Table A-6, sat. steam (p = 2.455 bars): Ts = 400K, hfg = 2183 kJ/kg.
ANALYSIS: With a uniform surface temperature, the air outlet temperature is
Tm,o = Ts - Ts - Tm,i exp -
-7
2
(
)
2 With A c = D o - Di2 / 4 = 1.355 10 -3 m 2 , D h = D o - Di = 0.015m and Re D = u m D h /
= mD h / A c = 16, 900, the flow is turbulent and the Dittus-Boelter correlation yields
(
)
Di L mc h p
h h fd =
k 4 / 5 0.4 0.0281 W / m K 2 4/5 0.4 0.023 (16, 900 ) ( 0.703 ) = 90.3 W / m K 0.023 Re D Pr = 0.015m Dh 0.05m 5m 90.3 W / m 2 K Tm,o = 127C - (110C ) exp - = 116.5C < 0.03 kg / s 1008 J / kg K
The pressure drop is p = f u 2 / 2D h L, where, with u m = m / A c = 0.03 kg / s / m
(5.391kg / m 1.355 10
3
(
-3
m
2
) = 4.11m / s, and with Re
2 3 ( 4.11m / s ) 5m
)
D = 16, 900, 2
Fig. 8.3 yields f 0.026. Hence,
-3
p 0.026 5.391kg / m
2 0.015m
= 395 N / m = 3.9 10
atm
<
The rate of heat transfer to the air is
q = m c p Tm,o - Tm,i = 0.03 kg / s 1008 J / kg K (99.5C ) = 3009 W
(
)
and the rate of condensation is then
mc =
q h fg
=
3009 W 2.183 10 J / kg
6
= 1.38 10
-3
kg / s
<
COMMENTS: (1) With Tm = ( Tm,i + Tm,o ) / 2 = 340K, the initial estimate of 325K is too low and an iterative solution should be obtained, (2) For a steam flow rate of 0.01 kg/s, approximately 14% of the outflow would be in the form of saturated liquid, (3) With L/Dh = 333, the assumption of fully developed flow throughout the tube is excellent.
PROBLEM 8.96
KNOWN: Dimensions and surface thermal conditions for a concentric tube annulus. Water flow rate and inlet temperature. FIND: (a) Tube length required to achieve desired outlet temperature, (b) Inner tube surface temperature at outlet. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform heat flux at inner surface, (3) Adiabatic outer surface, (4) Fully developed flow at exit, (5) Constant properties. PROPERTIES: Table A-6, Water Tm = 328K : cp = 4183 J/kgK; (Tm,o = 358K): = 332 10
-6
(
)
Ns/m , k = 0.673 W/mK, Pr = 2.07.
2
ANALYSIS: (a) From the overall energy balance, Eq. 8.37,
& q = q L = m c p Tm,o - Tm,i i L= & m c p Tm,o - Tm,i
(
(
q 4000 W/m i (b) From Eqs. 8.1 and 8.5, & m ( Do - D i ) & & u mD h m D h 4m Re D = = = = Ac ( /4) D2 - D2 ( Do + Di ) o i
) = (0.04 kg/s) 4183 J/kg K (85 - 25)o C = 2.51 m.
)
<
(
)
Re D =
4 0.04 kg/s ( 0.075 m ) 332 10 -6 kg/s m
= 2045.
Hence the flow is laminar, and with Di/Do = 0.5, it follows from Eq. 8.73 and Table 8.3 Nu i = Nuii = 6.24 h i = 6.24 From Eq. 8.68, k 0.673 W/m K = 6.24 = 168 W/m 2 K. Dh 0.025 m
q q / Di Ts,i ( L ) = Tm,o + i = Tm,o + i hi hi Ts,i ( L ) = 85o C + 4000 W/m ( 0.025m ) 168 W/m 2 K = 388 oC.
<
COMMENTS: Unless the water is pressurized, local boiling would occur at the tube surface, causing hi to be larger.
PROBLEM 8.97 KNOWN: Heat rate per unit length at the inner surface of an annular recuperator of prescribed dimensions. Flow rate and inlet temperature of air passing through annular region. FIND: (a) Temperature of air leaving the recuperator, (b) Inner pipe temperature at inlet and outlet and outer pipe temperature at inlet. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heating of recuperator inner surface, (4) Adiabatic outer surface, (5) Negligible kinetic and potential energy changes for air, (6) Fully developed air flow throughout. PROPERTIES: Table A-4, Air ( Tm = 500K ) : cp = 1030 J/kgK, = 270 10 0.041 W/mK, Pr = 0.68. ANALYSIS: (a) From an energy balance on the air & qi L = m a c p,a Ta,2 - Ta,1
-7
Ns/m , k =
2
(
)
q L 1.25 105 W/m 7m i Ta,2 = Ta,1 + = 300K + = 704.5K. & m a c p,a 2.1 kg/s 1030 J/kg K (b) The surface temperatures may be evaluated from Eqs. 8.68 and 8.69 with
<
& ma ( Do - D i ) 4 ( 2.1 kg/s) & u m Dh 4 ma = = = ( /4 ) D2 - D2 ( Do + Di ) ( 4.05m) 270 10-7 N s/m2 o i Re D = 24,452 the flow is turbulent and from Eq. 8.60 k 0.041 W/m K hi ho 0.023 Re 4/5 Pr 0.4 = 0.023 ( 24,452 ) 4 / 5 ( 0.68 )0.4 = 52 W/m 2 K. D Dh 0.05 m Re D =
(
)
With
q = qi / D i = 1.25 105 W/m/ 2m = 19,900 W/m 2 i
Eq. 8.68 gives
( Ts,i - Tm ) = qi / h i = 19,900 W/m 2 /52 W/m 2 K = 383K
Ts,i,1 = 683K Ts,i,2 = 1087K.
< <
From Eq. 8.69, with q = 0, Ts,o - Tm = 0. Hence o Ts,o,1 = Ta,1 = 300K.
(
)
PROBLEM 8.98
KNOWN: A concentric tube arrangement for removing heat generated from a biochemical reaction in a settling tank. Water is supplied to the annular region at rate of 0.2 kg/s. FIND: (a) The inlet temperature of the supply water that will provide for an average tank surface temperature of 37C; assume and then justify fully developed flow and thermal conditions; and (b) Sketch the water and surface temperatures along the flow direction for two cases: the fully developed conditions of part (a), and when entrance effects are important. Comment on the features of the temperature distributions, with particular attention to the longitudinal gradient on the tank surface. What change to the system or operating conditions would you make to reduce the gradient? SCHEMATIC:
ASSUMPTIONS: (1) Fully developed flow and thermal conditions, (2) Inner annulus surface has uniform heat flux, while outer surface is insulated, (3) Constant properties, (4) Negligible kinetic and potential energy changes and flow work. PROPERTIES: Table A-6, Water (Tm = 304 K): = 995.6 kg/m , cp = 4178 J/kgK, = 7.987 -7 2 10 m /s, k = 0.618 W/mK, Pr = 5.39. ANALYSIS: (a) The overall energy balance on the fluid passing through the concentric tube is
3
q = m cp Tm,i - Tm,o
(
)
(1) (2) (3)
and from an energy balance on the reaction tank,
q = qs,i As,i = q ( Di ) L. q = q = hi ( Ts - Tm ) cv s,i Tm = Tm,i + Tm,o / 2.
The convection rate equation applied to the inner surface As,i is
where Ts is the average inner surface temperature and
(
)
(4)
To estimate h, begin by characterizing the flow with
ReDh = u m Dh /
2 4 where A c = Do - Di / 4. Substituting numerical values find
(
)
Dh = Do - Di
m = Ac u m
ReDh = 1779 Assuming fully developed conditions for laminar flow through an annulus, it follows from Table 8.3 and Eq. 8.73 with Di/Do = 0.8, Nu i = hi D h / k = 5.58 h i = 172 W / m 2 K
Continued .....
PROBLEM 8.98 (Cont.)
Using Eq. (3) with h i , and Ts = 37C, and q from Eq. (2), find s,i
Tm = 30.5C
From Eqs. (1) and (4), calculate
Tm,i = 30.3C
Tm,o = 30.6C
<
For this annulus, the thermal entry length from Eq. 8.23 is
x fd,t = Dh 0.05 ReDh Pr x fd,t = ( 0.100 - 0.080 ) m 0.05 1779 5.39 = 9.59 m
Since L = 1 m, we conclude that entry length effects are significant, and the fully developed flow assumption is approximate. (b) Since the fluid is being heated by flow over a surface with uniform heat flux, the mean fluid temperature, Tm(x), will increase linearly with longitudinal distance x. Assuming fully developed conditions, the surface temperature Ts(x) will likewise increase linearly with distance as shown in the schematic below. Note that the longitudinal temperature difference is about 0.3C, and that the inlet mean temperature is 30.3C. Considering now entrance length effects, the convection coefficient is no longer uniform, and will be largest near the entrance, and larger than for the fully developed flow everywhere. Hence, we expect the surface temperature near the entrance to be closer to the mean fluid temperature than elsewhere. We also expect the average mean temperature of the fluid will be higher so that the average surface temperature, Ts , remains at 37C. However, the rise in temperature of the fluid (Tm,o Tm,i) will remain the same, about 0.3C, since the heat removal rate is the same. Increasing the flow rate will tend to minimize the longitudinal gradient by reducing (Tm,o Tm,i) and increasing h(x). The graph below illustrates the distinctive features of the fully developed flow and entrance length effects.
COMMENTS: The thermophysical properties required in the convection correlation and the energy equations were evaluated at Tm = (Tm,i + Tm,o)/2.
PROBLEM 8.99 KNOWN: Surface thermal conditions and diameters associated with a concentric tube annulus. Water flow rate and inlet temperature. FIND: (a) Length required to achieve desired outlet temperature, (b) Heat flux from inner tube at outlet. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed conditions throughout, (3) Adiabatic outer surface, (4) Uniform temperature at inner surface, (5) Constant properties. PROPERTIES: Table A-6, Water ( Tm = 320K ) : cp = 4180 J/kgK, = 577 10 = 0.640 W/mK, Pr = 3.77. ANALYSIS: (a) From Eq. 8.42a, & & m c p To m cp L=- ln =- ln Ph Ti Di h With Re D = Ts - Tm,o Ts - Tm,i
-6
Ns/m , k
2
.
& m ( Do - D i ) & u mD h 4m = = 2 2 ( /4) Do - Di ( Do + Di )
(
)
Re D =
4 0.02 kg/s ( 0.125m ) 577 10 -6 N s/m 2
= 353
the flow is laminar. Hence, from Eq. 8.70 and Table 8.2, h = hi = L=- k 0.64 W/m K Nui = 7.37 = 63 W/m2 K Dh ( 0.100 - 0.025 ) m
and
0.02 kg/s ( 4180 J/kg K) ( 0.025m ) 63 W/m 2 K
(100 - 75)o C = 19.7 m. ln (100 - 20 )o C (100 - 75 )o C=1575 W/m 2 .
< <
(b) From Eq. 8.69 q ( L ) = h i Ts,i - Tm,o = 63 i
(
)
W m 2 .K
COMMENTS: The total heat rate to the water is
q = m cp Tm,o - Tm,i = 0.02 kg/s 4180 J/kg K 55oC = 4598 W. &
(
)
( )
PROBLEM 8.100 KNOWN: Surface thermal conditions and diameters associated with a concentric tube annulus. Water flow rate and inlet temperature. FIND: Length required to achieve desired outlet temperature. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed conditions throughout, (3) Adiabatic outer surface, (4) Uniform temperature at inner surface, (5) Constant properties. PROPERTIES: Table A-6, Water ( Tm = 320K ) : cp = 4180 J/kgK, = 577 10 = 0.640 W/mK, Pr = 3.77. ANALYSIS: From Eq. 8.42a, L=- With Re D = Re D = & m cp Ph ln & m cp T - Tm,o To =- ln s . Ti Di h Ts - Tm,i
-6
Ns/m , k
2
& m ( Do - D i ) & u mD h 4m = = ( /4) D2 - D2 ( Do + Di ) o i 4 0.30 kg/s
(
)
( 0.125m ) 577 10 -6 N s/m 2
= 5296
and the flow is turbulent. Hence, from Eq. 8.60, h= k k Nu D = 0.023 Re4/5 Pr0.4 Dh Dh D 0.640 W/m K ( 5296)4 / 5 ( 3.77 )0.4 = 318 W/m2 K 0.075 m
h = 0.023
and hence the required length is
(100 - 75) o C = 58.4 m. L=- ln ( 0.025m ) 318 W/m2 K (100 - 20 )o C
0.30 kg/s ( 4180 J/kg K )
<
COMMENTS: Increasing m by a factor of 15 increases ReD accordingly, and the flow is & turbulent. However, h increases by a factor of only 5, from the result of Problem 8.99, in which case the tube length must be a factor of 3 larger than that of Problem 8.99.
PROBLEM 8.101
KNOWN: Dimensions and thermal conductivity of plastic pipe. Volumetric flow rate and temperature of inlet air. Enhancement of inner convection coefficient and friction factor associated with coiled spring. Thermal resistance of coating on outer surface. FIND: (a) Air outlet temperature and fan power requirement without coating and coiled spring, (b) Effect of coiled spring on air outlet temperature and fan power, (c) Effect of coating on outlet temperature. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Negligible heat transfer from air in vertical pipe sections, (3) Negligible flow work and potential and kinetic energy changes for air flow through pipe, (4) Smooth interior surface without spring, (5) Negligible coating thickness, (6) Constant properties. PROPERTIES: Table A-4, Air (Tm,i = 29C): i = 1.155 kg/m . Air ( Tm 25C): cp = 1007 -7 2 J/kgK, = 183.6 10 Ns/m , ka = 0.0261 W/mK, Pr = 0.707. ANALYSIS: (a) From Eq. (8.46a),
3
T - Tm,o
UAs = exp - T - Tm,i m cp + ln ( Do / Di ) 2 L k + l h o Do L
where, from Eq. (3.32),
( UAs )-1 = R tot = h 1 D
i
iL
With m = i i = 0.0289 kg / s and Re D = 4m / Di = 13, 350, the pipe flow is turbulent. With L/Di = 100, we may assume fully developed flow throughout the pipe, and from Eq. (8.60),
hi =
ka Di
4 0.023 Re D/ 5 Pr 0.3 =
0.0261W / m K 0.15m
0.023 (13, 350 )
4/5
(0.707 )0.3 = 7.20 W / m2 K
Hence,
K ln (0.17 / 0.15 ) 1 l R tot = + + 7.20 0.15 15 2 15 0.15 1500 0.17 15 W R tot = (0.0196 + 0.0089 + 0.0001) K / W = 0.0286 K / W
Hence, UA s = R -1 = 35.0 W / K and tot
Tm,o = T + Tm,i - T exp -
(
)
UAs 35.0 W / K = 17C + (12C ) exp - = 20.6C < 0.0289 kg / s 1007 J / kg K m cp Continued .....
PROBLEM 8.101 (Cont.)
- From Eq. (8.20a), f = 0.316 Re D1/ 4 = 0.0294. Hence, from Eqs. (8.22a) and (8.22b), with u m,i
= i / A c = 1.415 m / s,
1.155 kg / m (1.415 m / s ) Pf L i = 0.0294 15m 0.025 m3 / s = 0.085 W 2 Di 2 ( 0.15m )
3 2
2
i u 2 m,i
<
(b) With hcp = 2hi = 14.4 W/m K, the inner convection resistance is reduced from 0.0196 K/W to 0.0098 K/W and hence the total resistance from 0.0286 K/W to 0.0188 K/W. It follows that UA s = 53.2 W / K and
Tm,o = 18.9C
With fcp = 1.5f,
< <
P = 0.128 W
(c) With the coating of organic matter, there is an additional thermal resistance of the form Rt,c =
R / ( D o L ) = 0.05 m K / W / ( 0.17m 15m ) = 0.0062 K / W. The total resistance is then Rtot = t,c
2
(
)
0.0348 K/W and UA s = 28.7 W / K. Hence,
Tm,o = 21.5C
<
COMMENTS: (1) The fan power requirement is small, and the process is economical, with or without the coiled spring. (2) Heat transfer enhancement associated with the coiled spring is manifested by a 34% reduction in the total thermal resistance and a 1.7C reduction in the outlet temperature. (3) Fouling of the outer surface increases the total resistance by 22% and the outlet temperature by 0.9C. The penalty is not severe but could be ameliorated by periodic cleaning of the surface.
PROBLEM 8.102 KNOWN: Air flow through a plastic tube in which evaporation occurs. FIND: Convection mass transfer coefficient, hm. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Heat-mass transfer analogy applicable, (4) Fully-developed flow and mass transfer conditions. PROPERTIES: Plastic-air (given, 400K): Sc = /DAB = 2.0; Table A-4, Air (400K, 1 atm): -6 2 = 26.41 10 m /s. ANALYSIS: For fully-developed flow and thermal conditions with laminar flow and a uniform surface temperature, NuD = hD = 3.66 k
This situation is analogous to the evaporation of plastic vapor into the air stream with the inner surface remaining at a constant concentration of plastic vapor, CA,s , along the length of the tube. Invoking the heat-mass transfer analogy, h D Sh D = m = 3.66. DAB Recognizing that Sc = /DAB, 26.4 10-6 m 2 / s 1 1 h m = 3.66 = 3.66 = 2.42 10-2 m/s. -3 m 2.0 Sc D 2 10
<
COMMENTS: (1) The heat-mass transfer analogy requires that the vapor (A) have a negligible effect on the flow. Hence, the flow is that of air (B) and = B. (2) Only the mixture property DAB is required to characterize the plastic vapor for this evaporation process.
PROBLEM 8.103
KNOWN: Air passing upward through a tube having a thin water film on its inside surface. FIND: Convection mass transfer coefficient. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Heat-mass analogy applicable, and (4) Fully developed flow and thermal conditions. PROPERTIES: Table A.4, Air (300 K, 1 atm): = 184.6 10-7 Ns/m2, k = 0.0263 W/mK; Table A.8, Water vapor-air (300 K, 1atm): DAB = 0.26 10-4 m2/s. ANALYSIS: Begin by characterizing the air flow with the Reynolds number,
ReD =
4 (3 3600 ) kg s 4m = = 1916 D 0.030m 184.6 10-7 N s m 2
Since the flow is laminar, and assuming fully developed flow and thermal conditions, Eq. 8.55 is appropriate for the uniform Ts wall condition,
Nu D =
hD = 3.66 k
h=
0.0263 W m K 3.66 = 3.21W m 2 K 0.030m
Invoking the heat-mass analogy, for laminar flow conditions,
h D Sh D = m = Nu D DAB hm = DAB 0.26 10-4 m 2 s Nu D = 3.66 = 0.0032 m s D 0.030m
<
COMMENTS: (1) The heat-mass analogy requires that the water vapor (A) have negligible effect on the velocity boundary layer. It is important to recognize that the vapor is species (A) and the air species (B). Hence the flow is that of air (B) and hence = B. (2) Note only the mixture property DAB is required to characterize the water vapor for this evaporation process.
PROBLEM 8.104 KNOWN: Temperature and flow rate of air in a tube with a naphthalene coated inner surface. FIND: Convection mass transfer coefficient under fully developed conditions and velocity and concentration entry lengths. SCHEMATIC:
ASSUMPTIONS: (1) Heat and mass transfer analogy is applicable, (2) Uniform vapor concentration along inner surface. PROPERTIES: Table A-4, Air (300K, 1 atm): = 184.6 10 Ns/m , = 15.89 10 2 -6 2 m /s; Table A-8, Naphthalene-air (300K, 1 atm): DAB = 6.2 10 m /s, Sc = /DAB = 2.56. ANALYSIS: For air flow through the tube, Re D = & 4m 4 0.04 kg/s = = 55,178. D ( 0.05m ) 184.6 10 -7 N s/m 2
-7 2 -6
Hence the flow is turbulent and from the Colburn equation, Eq. 8.59, Sh D = 0.023 Re4/5 Sc1/3 = 0.023 ( 55,178) D
4/5
( 2.56 )1/3 = 196
DAB 6.2 10-6 m2 / s hm = Sh D = 196 = 0.024 m/s. D 0.05 m From Eq. 8.4, it follows that 10D xfd,h x fd,c 60D or 0.5 m x fd,h x fd,c 3 m. An entry length of 0.5m is assumed.
<
<
COMMENTS: Note that the flow properties are taken to be those of the air, with the contribution of the naphthalene vapor assumed to be negligible.
PROBLEM 8.105
KNOWN: Air flow over roughened section of tube constructed from naphthalene. FIND: Mass and heat transfer convection coefficients associated with the roughened section; contrast these results with those for a smooth section. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Heat-mass transfer analogy applicable, (3) Negligible naphthalene vapor in airstream, A,m = 0, (4) Constant properties, (5) Naphthalene vapor behaves as perfect gas. PROPERTIES: Table A-4, Air (300K, 1 atm): = 15.89 10 2.563; Naphthalene (given, 300K): psat,A = 1.31 10
-4 -6
m /s, k = 0.0263 W/mK, Pr =
-5
2
0.707; Table A-8, Naphthalene-air mixture (300K, 1 atm): DAB = 0.62 10
m /s, Sc= B/DAB =
2
bar, M A = 128.16 kg/kmol.
ANALYSIS: Using the rate equation with the experimentally observed sublimination rate of naphthalene vapor, the average mass transfer coefficient for the section is
& h m = m A / ( DL ) A,s - A,m A,m = 0
A,s = A,sat ( 300K) = M A Psat,A / T 8.314 10-2 m3 bar/kmol K 300K 1.31 10-4 bar = 6.731 10-4 kg/m3
(
)
A,s = 128.16 kg/kmol hm =
0.010 kg / ( 0.075m 0.150m ) 6.73110 -4 - 0 kg/m 3 = 3.89 10 -2 m/s. 3 3600 s
1/3 1/3 -2 m/s 0.0263 W/m K 0.707 = 3.89 10 = 107 W/m 2 K. -5 m 2 / s 2.563 0.62 10
(
)
<
Invoking the heat-mass transfer analogy, the associated heat transfer coefficient is
Pr h = hm DAB Sc k
<
The corresponding convection coefficients for a smooth section can be estimated using the Colburn relation,
h= k D 0.023 Re4/5 Pr1 / 3 = (0.0263 W/m K/0.075 m ) 0.023 (35,000 ) D
4/5
( 0.707 )1 / 3 = 31 W/m 2 K.
<
Invoking the heat-mass transfer analogy, h m = ( D AB / D ) 0.023 Re4/5 Sc1/3 = 0.62 10 -5 m2 /s/0.075 m 0.023 (35,000) 4 / 5 ( 2.563)1 / 3 D hm = 1.12 10
-2
(
)
m/s
<
COMMENTS: The effect of roughening is to increase the convection coefficients over the corresponding value for the smooth condition; in this case, by a factor of approximately 3.5.
PROBLEM 8.106
KNOWN: Dry air with prescribed velocity and temperature flowing over a thin-walled tube with a water-saturated fibrous coating. Water passes at a prescribed rate through the tube to maintain an approximately uniform surface temperature Ts = 27C. FIND: (a) Heat rate from the external surface of the tube considering heat and mass transfer processes and (b) For a flow rate of m = 0.025 kg/s, the inlet temperature, Tm,i, of the water that must be supplied to the tube. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Heat-mass analogy applicable, and (4) Negligible kinetic energy and axial conduction in the tube flow. PROPERTIES: Table A.4, Air ( Tf = (Ts + TW )/2 = 304 K): = 1.148 kg/m3, cp = 1007 J/kgK, = 16.29 10-6 m2/s, k = 0.0266 W/mK, = 23.09 10-6 m2/s, Pr = 0.706; Table A.6, Water (Ts = 300 K): A,s = 1/vg = 0.02556 kg/m3, hfg = 2438 kJ/kg, = 855 10-6 Ns/m2; Table A.6, Water ( Tm = 305 K): = 995 kg/m3, cp = 4178 J/kgK, = 769 10-6 Ns/m2, k = 0.620 W/mK, Pr = 5.20; Table A.8, Water vapor-air (Ts = 300 K): DAB = 0.26 10-4 m2/s. ANALYSIS: (a) On the Schematic above, the surface energy balance yields
qout = qconv + q evap
and substituting the rate equations,
(1)
qconv = h o As ( Ts - T )
qevap = n A h fg = h m As A,s - A, h fg
(
)
(2,3)
where h o can be estimated from an appropriate correlation and h m from the heat-mass analogy using h o . Estimation of the heat transfer coefficient, h o : The Reynolds number, evaluated with properties at Tf = (Ts + TW )/2 = 304 K, is
ReDo =
VD 10 m s 0.020 m = = 12, 277 1.629 10-5 m 2 s
(4)
Using the Churchill-Bernstein correlation, Eq. 7.57, for cross flow over a cylinder, find h o 4/5 1/ 0.62 Re1/ 2 Pro 3 Re D,o 5 / 8 D,o 1 + Nu D,o = 0.3 + 1/ 4 282, 000 1 + (0.4 Pr )2 / 3 o
(5)
Continued...
PROBLEM 8.106 (Cont.) Nu D,o = 0.3 + 0.62 (12, 277 )
1/ 2
1 + ( 0.4 0.706 )2 / 3
(0.706 )1/ 3 1 +
1/ 4
12, 277 282, 000
5 / 8 4 / 5
ho =
k 0.0266 W m K Nu D,o = 60.1 = 80.0 W m 2 K D 0.020 m
The Heat-Mass Analogy: From Eq. 6.92, with n = 1/3, 2/3 ho 2 / 3 = c = cp Le p hm D AB
h m = 80.0 W m K
2
(6)
3 -6 2 -4 2 1.148 kg m 1007 J kg K 23.09 10 m s 0.26 10 m s
(
)
2/3
= 0.0749 m s
Hence, the heat rate leaving the tube surface from Eq. (1) is,
2 3 3 q out = 80 W m K ( 27 - 35 ) C + 0.0749 m s ( 0.02556 - 0 ) kg m 2438 10 J kg ( 0.020 m 0.200 m )
qout = -8.04 W + 58.65 = 50.6 W
(b) For tube flow analysis, the heat rate and rate equations are
<
Ts - Tm,o DL = exp - h mcp i Ts - Tm,i
(7,8)
q = mcp Tm,o - Tm,i
(
)
where Ts = 27C, the uniform temperature of the tube surface, and q = -50.6 W according to the analysis of part (a). To estimate h i , first characterize the flow,
ReD,i =
4m 4 0.025 kg s = = 2070 Di 0.020 m 769 10-6 N s m 2
(9)
using properties evaluated at an assumed mean temperature, Tm = 305 K (slightly above Ts). The flow is laminar, and assuming a combined entry region, use the Sieder-Tate correlation, Eq. 8.57, 1/ 3 0.14 ReD,i Pri Nu D,i = 1.86 (10) L D s
2070 5.20 Nu D,i = 1.86 0.200 0.020
1/ 3
769 10-6 855 10-6
0.14
k 0.620 W m K hi = i Nu D,i = 18.78 = 582 W m 2 K D 0.020 m
Referring to Eqs. (7) and (8), recognize that there are two unknowns, Tm,i and Tm,o, as we have evaluated both q and h i . Using the IHT solver, we found
Tm ,i 34.2 C Tm ,o 33.7 C
<
Continued...
PROBLEM 8.106 (Cont.)
COMMENTS: Using the IHT Rate Equation Tool, Rate Equation for a Tube, Constant Surface Temperature, and the Correlation, Internal Flow, Laminar, Combined Entry Length, a model to perform the analysis for part (b) was developed and is copied below.
// Rate Equation Tool - Tube, Constant Surface Temperature: /* For flow through a tube with a uniform wall temperature, Fig 8.7b, the overall energy balance and heat rate equations are */ q = mdot*cp*(Tmo - Tmi) // Heat rate, W; Eq 8.37 q = - 50.64 // Heat rate, W; required to sustain heat loss on outer surface (Ts - Tmo) / (Ts - Tmi) = exp ( - P * L * h / (mdot * cp)) // Eq 8.42b // where the fluid and constant tube wall temperatures are Ts = 27 + 273 // Tube wall temperature, K Tmi_C = Tmi -273 // Inlet mean fluid temperature, K Tmo_C = Tmo - 273 // Outlet mean fluid temperature, K // The tube parameters are P = pi * D // Perimeter, m Ac = pi * (D^2) / 4 // Cross sectional area, m^2 D = 0.020 // Tube diameter, m L = 0.20 // Tube length, m // The tube mass flow rate and fluid thermophysical properties are mdot = rho * um * Ac mdot = 0.025 // Properties Tool - Water // Water property functions :T dependence, From Table A.6 // Units: T(K), p(bars); x=0 // Quality (0=sat liquid or 1=sat vapor) rho = rho_Tx("Water",Tm,x) // Density, kg/m^3 cp = cp_Tx("Water",Tm,x) // Specific heat, J/kgK mu = mu_Tx("Water",Tm,x) // Viscosity, Ns/m^2 mus = mu_Tx("Water",Ts,x) // Viscosity, Ns/m^2 nu = nu_Tx("Water",Tm,x) // Kinematic viscosity, m^2/s k = k_Tx("Water",Tm,x) // Thermal conductivity, W/mK Pr = Pr_Tx("Water",Tm,x) // Prandtl number Tm = Tfluid_avg(Tmo, Tmi) // Average mean temperature, K //Tm = 300 // Assigned value, initial solve // Correlations Tool - Internal Flow, Laminar, combined entry length NuDbar = NuD_bar_IF_L_CEL_CWT(ReD,Pr,D,L,mu,mus) // Eq 8.57 NuDbar = h * D / k ReD = um * D / nu // Data Browser results: Ac Tmo_C cp um D 0.0003142 33.7 4178 0.08004 0.02
NuDbar h L 18.64 580.8 0.2
P k Ts 0.06283 0.6231 300
Pr ReD mu mdot q 4.975 2150 0.0007403 0.025 -50.64
Tmi mus x 307.2 0.000855 0
Tmi_C nu Tm 34.18 7.445E-7 306.9
Tmo rho 306.7 994.3
PROBLEM 8.107
KNOWN: Density and flow rate of gas through a tube with evaporation or sublimination at the tube surface. FIND: (a) Longitudinal distribution of mean vapor density, (b) Total rate of vapor transfer. SCHEMATIC:
ASSUMPTIONS: (1) Steady, incompressible flow, (2) Flow rate is independent of x, (3) Negligible chemical reactions, (4) Uniform perimeter P. ANALYSIS: (a) Applying conservation of species to a differential control volume
dA,m A,m u m A c + dn A = A,m + dx um A c dx & or, with u mAc = m/ and dn A = h mP dx A,s - A,m ,
(
)
& m d A,m dx = h m P dx A,s - A,m . dx
(
)
Separating variables and integrating,
ln
A,m A,mi
d A,m A,s - A,m
=-
=
hm P P x dx = h m dx 0 & & m m 0
x
A,s - A,m A,s - A,m,i
A,s - A,m ( x ) Pxhm P x h . or = exp - m & & m A,s - A,m,i m
(1)
<
(b) With A A,s - A,m ,
& & n A = ( m/ ) A,m,o - A,m,i = - ( m/ ) A,o - A,i
and from Eq. (1) with
(
)
(
)
-
A,o & m = P L hm / ln A,i A,o - A,i ln A,o / A,i
it follows that
nA = hm P L
(
)
.
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& COMMENTS: Due to the addition of vapor, m will actually increase with x. However, if the specific humidity of the saturated gas-vapor mixture is small (as is usually the case), the change in m & will be small.
PROBLEM 8.108
KNOWN: Flow rate and temperature of air. Tube diameter and length. Presence of water film on tube inner surface. FIND: (a) Vapor density at tube outlet, (b) Evaporation rate. SCHEMATIC:
ASSUMPTIONS: (1) Steady, incompressible flow, (2) Constant flow rate, (3) Isothermal system (water film maintained at 25C), (4) Fully developed flow. PROPERTIES: Table A-4, Air (1 atm, 298K): = 1.1707 kg/m , = 183.6 10 15.71 10
-6 2 3 3 -7
Ns/m , =
3
2
m /s; Table A-6, Water vapor (298K): A,sat = 1/vg = (1/44.25 m /kg) = 0.0226 kg/m ;
-6
Table A-8, Air-vapor (298K): DAB = 26 10 ANALYSIS: (a) From Equation 8.81,
m /s; Sc = /DAB = 0.60.
2
& 4m 4 3 10 -4 kg/s Re D = = = 2080. D ( 0.01 m )183.6 10 -7 N s/m 2
Flow is laminar and from the mass transfer analogy to Eq. 8.57, 1/3 1/3 Re Sc 2080 0.60 Sh D = 1.86 D = 1.86 = 4.31
DL A,m,o = A,s - A,s - A,m,i exp - hm & m
(
)
L/D
hm =
Sh D DAB = D
4.31 26 10 -6 m 2 / s 0.01 m
100
= 0.0112 m/s
A,m,o = 0.0226 kg/m3 0.01 m 1 m 1.17 kg/m3 0.0112 m/s = 0.0169 kg/m 3 -0.0226 kg/m exp - -4 kg/s 3 10
3 (b) The evaporation rate is
n A = u mA c A,m,o - A,m,i =
<
(
)
& m
(
A,m,o =
)
3 10 -4 kg/s 1.1707 kg/m
3
0.0169
kg m
3
= 4.33 10 -6 kg/s.
<
COMMENTS: With 0.01 1 1.17 m DL A,o = A,i exp - h m = 0.0226 exp - 0.0112 = 5.73 10 -3 kg/m3 & m s 3 10 -4 kg/s the evaporation rate is
n A = h m DL A,o - A,i ln A,o / A,i
(
)
= 0.0112
m s
(0.01 m )1m
( 0.00573 - 0.0226) kg/m3 = 4.33 10 -6 kg/s ln ( 0.00573/0.0226 )
which agrees with the result of part (b).
PROBLEM 8.109 KNOWN: Flow rate and temperature of air in circular tube of prescribed diameter. Inner tube surface is wetted. Flow is fully developed and inlet air is dry. FIND: Tube length required to reach 99% of saturation. SCHEMATIC:
ASSUMPTIONS: (1) Steady, incompressible flow, (2) Constant flow rate, (3) Water film is also at 25C. PROPERTIES: Table A-4, Air (298K, 1 atm): = 1.17 kg/m , = 183.6 10 Ns/m , = -6 2 3 15.71 10 m /s; Table A-6, Water vapor (298K): A,sat = 1/vg = (1/44.25 m /kg) = 0.0226 3 -6 2 kg/m ; Table A-8, Air-vapor (298K): DAB = 26 10 m /s, Sc = /DAB = 0.60. ANALYSIS: If A,m,o = 0.99 A,s, it follows from Problem 8.81 that A,s - 0.99 A,s DL = 0.01 = exp - hm . & A,s m With & 4m 4 10 -3 kg/s Re D = = = 6935 D ( 0.01 m ) 183.6 10 -7 N s/m 2 the flow is turbulent and the mass transfer version of the Colburn equation is Sh D = 0.023 Re4/5 Sc1/3 = 0.023( 6935) D
4/5
3 -7 2
( 0.60)1/3 = 22.9
Sh D DAB 22.9 26 10 -6 m 2 / s hm = = = 0.0595 m/s. D 0.01 m Hence 0.01 m L 1.17 kg/m 3 0.01 = exp - 0.0595 m/s 10-3 kg/s 0.01 = exp ( -2.188 L ) L = 2.1 m.
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PROBLEM 8.110
KNOWN: Flow rate and temperature of atmospheric air in circular tube of prescribed diameter. Flow is fully developed, and air is dry. Inner tube surface is wetted. FIND: (a) Tube length required to reach 99% saturation, (b) Heat rate needed to maintain tube surface at air temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady, incompressible flow, (2) Constant flow rate. PROPERTIES: Table A-4, Air (298K, 1 atm): = 1.17 kg/m , = 183.6 10 10
-6 2 3 3 -7
Ns/m , = 15.71
3
2
m /s; Table A-6, Water vapor (298K): vg = 44.25 m /kg, A,sat = 1/vg = 0.0226 kg/m , hfg = -6 2 2443 kJ/kg; Table A-8, Air-vapor (298K): DAB = 26 10 m /s, Sc = /DAB = 0.60. ANALYSIS: (a) If A,m,o = 0.99 A,s, it follows from Problem 8.107 that
A,s - 0.99 A,s A,s
DL = 0.01 = exp - hm . & m
With
Re D =
& 4m 4 10-3 kg/s = = 3467, D ( 0.02m ) 183.6 10 -7 N s/m 2
The flow is turbulent (weakly) and the mass transfer analog to the Colburn equation is 4/5 Sh D = 0.023 Re4/5 Sc1/3 = 0.023( 3467 ) ( 0.60 )1/3 = 13.2 D Sh D 13.2 26 10-6 m 2 / s h m = D AB = = 0.0172 m/s.
D
0.02 m
Hence,
L=-
& 10 -3 kg/s ln ( 0.01) m ln ( 0.01) = - = 3.64m. D h m ( 0.02m ) 1.17 kg/m 3 ( 0.0172 m/s) n A = h m DL A,o - A,i
<
(b) The required heat rate is
q = n A h fg
ln A,o / A,i 0.01 A,s - A,s n A = 0.0172m/s ( 0.02m ) 3.64m ln ( 0.01) n A = -8.542 10 -4 m 3 / s -0.99 0.0226 kg/m 3 = 1.91 10 -5 kg/s q = n A h fg = 1.9110 -5 kg/s 2.443 10 6 J/kg = 46.7 W.
COMMENTS: The evaporation rate is low; hence the heat requirement is small.
(
)
(
)
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PROBLEM 8.111
KNOWN: Tube length, diameter and temperature. Air temperature and velocity. Saturation pressure of thin liquid film and properties of vapor. FIND: (a) Partial pressure and mass fraction of vapor at tube exit, (b) Mass rate at which liquid is removed from the tube. SCHEMATIC:
ASSUMPTIONS: (1) System is isothermal at 300K, (2) Steady, incompressible flow, (3) Perfect gas behavior, (4) Mass flow rate is independent of x. PROPERTIES: Table A-4, Air (300K, 1 atm): = 1.16 kg/m , = 15.9 10 m /s. Prescribed, -5 2 Vapor (300K): pA,sat = 15 mm Hg, M A = 70 kg/kmol, DAB = 10 m /s. ANALYSIS: (a) With the vapor assumed to behave as an ideal gas, pA = CA T = A ( / M A ) T, and isothermal conditions, the vapor pressure at the outlet may be obtained from the expression
3 -6 2
p A,sat - p A,o p A,sat - p A,i
=
A,s - A,m,o
D L h m = exp - A,s - A,m,i m
2
where m = u m A c = 1.16 kg / m3 0.5 m / s ( 0.05m )2 / 4 = 1.14 10 -3 kg / s. With ReD = um D/ =
0.5 m/s 0.05m/15.9 10 m /s = 1570, the flow is laminar and hm may be determined from the 1/3 mass transfer analog to Eq. 8.57. With Sc = /DA = 1.59 and [ReD Sc/(L/D)] = 2.92 > 2
hm = Sh D D AB D = 1.86
-6
Re Sc L/D
1/ 3
D AB D
= 1.86 2.92
10
-5
m /s
2
0.05m
= 1.09 10
-3
m/s
Hence, with pA,i = 0
DL h m pA,o = pA,sat l - exp - = 15 mm Hg [1 - m 1.16 kg / m3 0.05m 5m 1.09 10-3 m / s = 8.7 mm Hg exp - 1.14 10-3 kg / s
The corresponding mass density of the vapor is p A,o M 8.7 mm Hg 70 kg / kmol A = A,m,o = = 0.0326 kg / m3 3 T (760 mm Hg / atm ) 0.082 m atm / kmol K 300K
< < <
(
)
(b) The evaporation rate is n A = u m A c A,m,o - A,m,i = 0.5m / s 1.96 10 -3 m 2 0.0326 kg / m3 = 3.20 10 -5 kg / s
-5
(
)
COMMENTS: (1) Since the evaporation rate (nA = 3.2 10 kg/s) is much less than the air flow -3 rate ( m = 1.14 10 kg/s), the assumption of a fixed flow rate is reasonable. (2) The evaporation rate is also given by nA = hm D L A,lm = - hm D L A,m,o/ln [(pA,sat pA,o)/pA,sat] = 3.22 10 kg/s, which agrees with the calculation of part (b).
-5
PROBLEM 8.112
KNOWN: Air flow rate through trachea of diameter D and length L. FIND: (a) Average mass transfer convection coefficient, h m , and (b) Rate of water loss per day (liter/day). SCHEMATIC:
ASSUMPTIONS: (1) Trachea can be approximated as a smooth tube with uniform surface temperature, (2) Laminar, fully developed flow, (3) Trachea inner surface is saturated with water at body temperature, Ts = 37C, (4) Negligible water vapor in air at 310 K during inhalation, and (5) Heat-mass analogy is applicable. PROPERTIES: Table A-4, Air (310 K, 1 atm): B = 1.128 kg/m , = 1.893 10 Ns/m ; Table 3 3 A-6, Water (Ts = 37C = 310 K): A,f = 993 kg/m , A,g = 0.04361 kg/m ; Table A-8, Water-vapor -4 3/2 -5 2 air (310 K, 1 atm): DAB = 0.26 10 (310/298) = 2.76 10 m /s. ANALYSIS: (a) Begin by characterizing the air (B) flow in the trachea modeled as a smooth tube, 4m 4 B Re D = = D D = 632 0.020 m 1.893 10-5 N s / m 2 Hence, the flow is laminar, and for fully developed conditions and invoking the heat-mass analogy Nu D = Sh D = 3.66 Sh = h m D / DAB h m = 3.66 D AB / D = 3.66 2.76 10-5 m 2 / s / 0.020 m = 0.0050 m / s (b) The species (A) transfer rate equation, Eq. 8.75, has the form n A = h m A s A, m A, m = Re D = 4 10 liter / min 10-3 m 3 / liter 1min/ 60s 1.128 kg / m3
3 -5 2
<
( A,s - A,m,o ) - ( A,s - A,m,i ) m ( A,s - A,m,o ) / ( A,s - A,m,i )
= exp -
where the mean outlet species density, A,m,o, can be determined from Eq. 8.78
A,s - A,m,o A,s - A,m,i
hm P m
n A = 1.54 10-6 kg / s
where m / = u m A c = B . Substituting numerical values with P = D, find
A,m,o = 0.009233
The volumetric rate of water loss on a daily basis, assuming a 12 hour inhalation period, is
A = 1.54 10-6 kg / s / 993 kg / m3 103 liter / m3 (3600 s / h 12 h / day ) VA = 0.067 liter / day
(
)
PROBLEM 8.113
KNOWN: Air (species B) is in fully developed, laminar flow as it enters a circular tube wetted with liquid A (water). Tube length and diameter. Flow rate of air and system temperature. FIND: (a) Governing differential equation for species transfer, (b) Heat transfer analog and an expression for Sh D , (c) General expression for A,m,o, (d) Value of A,m,o for prescribed conditions. SCHEMATIC:
ASSUMPTIONS: (1) Steady, incompressible flow, (2) Flow rate is independent of x, (3) Laminar, fully developed flow (hydrodynamically), (4) Isothermal conditions, (5) Dry air at inlet. PROPERTIES: Table A.4, Air (298 K, 1 atm): = 1.1707 kg/m3, = 183.6 10-7 Ns/m2, = 15.71 10-6 m2/s; Table A.6, Water vapor (298 K): A,sat = 1/vg = 0.0266 kg/m3; Table A.8, Air-vapor (298 K): DAB = 26 10-6 m2/s, Sc = /DAB = 0.60. ANALYSIS: (a) The governing differential equation may be inferred by analogy to Eq. 8.48. In this case, the dependent variable is the vapor mass density, A (x,r), and the diffusivity is DAB. With v = 0 for fully-developed flow, it follows that
u
A DAB A = r x r r r
< <
A x r = 0 = 0
The entrance condition is
A (0, r ) = 0
and the boundary conditions are
A ( ro , x ) = A,s
< <
(b) The foregoing conditions are analogous to those of the thermal entry length condition associated with Eq. 8.56. Invoking this analogy the average Sherwood number for laminar, fully developed flow is
Sh D = 3.66
0.0668 ( D L ) ReD Sc 1 + 0.04 ( D L ) ReD Sc
2/3
(c) Applying conservation of species to the differential control volume,
d A,m A,m u m Ac + dn A = A,m + dx u m Ac dx or, with umAc = m and dn A = h m Ddx ( A,s - A,m ) m d A,m dx = h m Ddx A,s - A,m dx
(
)
Continued...
PROBLEM 8.113 (Cont.)
A,m,i A,s - A,m = o
or
A,m
d A,m
x Dh m
m
dx
A,s - A,m (x)
at x = L,
Dxh m (x) = exp - A,s - A,m,i m
A,s - A,m,o
DLh m = exp - A,s - A,m,i m
<
(d) For the prescribed conditions, ReD = 4m D = 4 2.5 10 -4 kg s 0.01 m 183.6 10 -7 N s m 2 = 1734 and (D/L)ReDSc = (0.01 m/1 m)1734(0.6) = 10.4. Hence, Sh D = 3.66 + 0.0668 (10.4 ) 1 + 0.04 (10.4 )
2/3
%
#
= 4.24
h m = Sh D ( D AB D ) = 4.24 26 10-6 m 2 s 0.01m = 0.011m s
Hence,
(
)
A,s - A,m,o
1.1707 kg m3 0.01m 1m 0.011m = exp - A,s - A,m,i 2.5 10-4 kg s
s = 0.198
A,m,o = A,s - 0.198 ( A,s - A,m,i ) = 0.0226 kg m3 (1 - 0.198 ) = 0.0181kg m3
<
COMMENTS: Due to evaporation, m actually increases with increasing x. However, the increase is small, and the assumption of fixed m is good.