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Unformatted text preview: PROBLEM 17.50 KNOWN: Flow rate, inlet temperature of atmospheric air passing through a duct of prescribed
diameter and length maintained with a constant heat ﬂux; average convection coefﬁcient. FIND: (a) The heat transfer rate, q, and the mean temperature of outlet air Tmp; (b) Surface temperature at the tube inlet TSJ and outlet T53; (0) Sketch the axial variations of TS and Tm with distance from the inlet x; also sketch their variations for the more realistic case in which the local
convection coefﬁcient varies with x. SCHEMATIC: 53E) q§=1000 WIm2 Tm,Ts
Tm,i=20°C HHH HHHl p = 1 atm l
rh = 0.005 kg/s —°
F = 25 W/m2K Ts,i Constant h —>x ASSUMPTIONS: (1) Steadystate conditions, (2) Negligible kinetic and potential energy effects; no
shaft work, (3) Fully developed ﬂow conditions, and (4) Constant properties. D=50mm L=3m PROPERTIES: Table HT —3, Air (assume Tm =340 K, 1 atm): cp = 1009 J/kgK. ANALYSIS: (a) From an overall energy balance on the tube, the total heat transfer provided by the
applied surface heat ﬂux is q=q§AS =qgnDL=1OOOW/m2(nxo.050 m)x3m=471.2W <
From an overall energy balance on the ﬂuid, Eq. 17.48,
q = mop (Tm, 4m) T =T +q/mc =20°C+47l.2W/0.005kg/sx1009J/kgK
m,o m,1 p Tm, =113.4°C < (b) Since the convection coefﬁcient is uniform within the tube, the difference between the tube surface
and mean ﬂuid temperatures will be constant q; = 3(Ts 'Tm)
(TS — Tm)=q's'/h = 1000 W/m2 /25 W/m2 K = 40°C
Hence the tube surface temperatures at the inlet and outlet are T5,, = 60°C Tsyo =153.4°c < (c) The axial variations for Tm and Ts shown in the sketch above are both linear with distance x.
However, if h is no longer constant because of entrance region effects, the surface temperature distribution is no longer linear so that TS — Tm is smaller near the inlet and increases with distance
until ﬁilly developed conditions are achieved. PROBLEM 17.51 KNOWN: Flow rate, inlet and outlet temperatures of atmospheric air passing through a duct of
prescribed diameter, length, and surface temperature.
FIND: (a) The heat transfer rate, (b) Log mean temperature difference, ATlm; (0) Average convection coefﬁcient for the air ﬂow; and ((1) Sketch axial variations of TS and Tm with distance from the inlet;
comment on key features. SCHEMATIC: D=50mm (LEE) rh = 0.04 kg/s
Tm'i = 60°C
p = 1 atm ASSUMPTIONS: (l) Steadystate conditions, (2) Negligible kinetic and potential energy effects; no
shaft work, (3) Fully developed ﬂow conditions, and (4) Constant properties. PROPERTIES: Table HT3, Air (Tm = (Tm; + Tm, )/2 = 318 K, 1 atm): cp = 1008 J/kgK ANALYSIS: (a) From the overall energy balance on the duct, Eq. 17.48, the heat transfer rate from
the air to the tube is q = mop (Tm, — Tm,i)=0.04 kg/s x1008 J/kgK(30 — 60)°C = —1210 w < (b) The log mean temperature difference, Eq. 17.58, is _ . _ O _ _1 O
ATim = ATO ATl = (30 15) C (60 5) C =2730C <
1n(ATo/ATi) ln(15/45) (c) From the rate equation, Eq. 17.57, and knowledge of the heat rate and ATlm, ﬁnd the average
convection coefﬁcient, Clconv = hAs AT1m With AS = 1rDL, and substituting numerical values, 3— qconv _ 1210w ————————=9.4W/m2K <
7rDL ATlm ”(0.150 m)10m x 27.3 K (d) The temperature distributions for TS and Tm(x) as a function of distance from the inlet are shown in
the Schematic above. The tube wall temperature is constant, however, the ﬂuid temperature changes
exponentially. The logmean temperature represents the mean difference between the wall and ﬂuid
temperatures. PROBLEM 17.52 KNOWN: Ethylene glycol ﬂowing through a coiled, thin walled tube submerged in a wellstirred
water bath maintained at a constant temperature. FIND: Heat rate and required tube length for prescribed conditions. SCHEMATIC:
J71h=0.01kg/s
T = 55°C T
Tm,,=85°c c. ’"I" "’6"
Thinwalled Wellsﬂrred T
'I‘ube, D=5mm wafer ba‘l‘h, 5 7;, =25 °C ASSUMPTIONS: (l) Steadystate conditions, (2) Tube wall thermal resistance negligible, (3)
Convection coefﬁcient on water side inﬁnite; cooling process approximates constant wall surface temperature distribution, (4) Negligible kinetic and potential energy effects; no shaft work, (5)
Constant properties, (6) Negligible heat transfer enhancement associated with the coiling. PROPERTIES: Table HT 4, Ethylene glycol (Tm = (85 + 35)°C/2 = 60°C = 333 K): cp = 2562
J/kgK, u = 0.522 X 102 N‘s/m2, k = 0.260 W/m‘K, Pr = 51.3.
ANALYSIS: From an overall energy balance on the tube, qconv = m cp (Tmp ‘Tm,i ) = 0.01 kg/sx2562 J/kg(35 —85)° C = —1281 W. (1) < For the constant surface temperature condition, from the rate equation, As = (Iconv / HATﬂm (2)
AT 0 o 35 — 25 0
AT = AT AT M 0=[35—25 C— 85—25 C]/€ =27.9 C. 3
6m ( o 1) nATi ( ) ( ) 1185—25 ()
Find the Reynolds number to determine ﬂow conditions,
' x . k
ReD= 4m _ 4 001 g/s 2813. (4) 7r Dﬂ _ 7r><0.003 mX0.522x10'2Ns/m2 Hence, the ﬂow is laminar and, assuming the ﬂow is fully developed, the appropriate correlation is ﬁp=h—D=3.66, h=Nu£=3.66x0.260i/0.003m=317W/m2K. (5)
k D mK From Eq. (2), the required area, AS, and tube length, L, are
As = 1281 W/317 W/m2 ~Kx27.9°C = 0.1448 m2
L=AS/7r D=0.1448m2/7r(0.003m)=15.4m. < COMMENTS: Note that for fully developed laminar ﬂow conditions, the requirement of Eq. 17.41
is satisﬁed: (L/D) / ReD Pr = (153/0003) / (813 X 51.3) = 0.122 > 0.05. Note also the sign ofthe heat rate ‘lconv when using Eqs. (1) and (2). PROBLEM 17.65 KNOWN: Diameter, length and surface temperature of condenser tubes. Water velocity and inlet
temperature. FIND: Water outlet temperature evaluating properties at Tm = 300 K. SCHEMATIC:
L = s m “ﬁr”’9'
D = 25.4 mm Tmp
1* H
Um = 1 m/s
Tm,i = 290 K
a ' T5 = 350 K ASSUMPTIONS: (1) Negligible tube wall conduction resistance, (2) Negligible kinetic and potential
energy effects; no shaft work. PROPERTIES: Table HT5, Water (assume Tm = 300 K): p = 997 kg/m3, cp = 4179 J/kgK, u = 855 x
10'6 kg/sm, k = 0.613 W/mK, Pr = 5.83. ANALYSIS: From the energy balance relation, Eq. 17.55b,
Tm, = Ts — (TS —Tm’i )exp [—(nDL/r'ncp )h].
Evaluating properties at Tm = 300 K, ﬁnd the Reynolds number as _ pumD _ 997kg/m3(1m/s)0.0254m _ RCD
H 855x10‘6 kg/s~m 29,618 The ﬂow is turbulent, and since L/D = 197, it is reasonable to assume fully developed ﬂow throughout
the tube. From the DittusBoelter correlation, Eq. 17.64 with n = 0.4 since Ts > Tm, NuD =o.023 Re4D/ 5 Pro‘4 = 0.023(29,618)4/5 (5.83)0‘4 = 176
E = NuD (k/D)=176(0.613W/mK/0.0254m)= 4248 w/m2 K. The mass ﬂow rate is
. _ 2 / _ 3 2 _
m — pum ED 4 — (7r/4)997 kg m (lm/s)(0.0254m) — 0.505 kg/s.
Substituting numerical values into the energy balance relation, ﬁnd 7r(0.0254m)5m(4248W/m2 K) = 323 K = 50°C <
0.505kg/s(4179J/kgK) Tm,o = 350K—(60K)exp — COMMENTS: Finding Tm,0 = 50°C, the average mean ﬂuid temperature is Tm = (290 + 323)K/2 =
307 K vs. the assumed value of 300 K. The analysis using the proper Tm yields Tm,0 = 51.7°C. While the difference is only 1.7°C, it is good practice to use the proper value for Tm at which to evaluate
properties. PROBLEM 17.82 KNOWN: Vertical plate maintained at a uniform temperature while suspended in quiescent air. FIND: (a) Heat transfer rate from the plate by free convection, (b) Boundary layer thickness at the trailing edge of the plate, and (0) Compare results for heat transfer rate and boundary layer thickness
with those for forced convection with an air stream having a velocity of 5 m/s. SCHEMATIC:
L = 0.25 m 0’; qhonv qconv TS  . —° b — \
co — 25°C Plate Too = 2500 x / 1 L = 0.25 m Ts = 70°C u = 5 m/s q°°rw
x T on
——>Y
Free convection (a,b) Forced convection (c) ASSUMPTIONS: (1) Steadystate conditions, (2) Plate is at uniform temperature, (3) Room air is
quiescent for free convection condition, and (4) For forced convection, Rex,C = 5 x 105. PROPERTIES: Table HT—3, Air (Tf= (TS + Tag/2 = 320.5 K, 1 atm): v = 17.95 x 10'6 mZ/s, k =
0.02783 W/mK, or = 25.53 X 10'6 mz/s, 13 = 1/Tf= 0.00312 Kl, Pr = 0.704. ANALYSIS: (a) The heat transfer by free convection from both sides of the vertical plate is given by
the rate equation Clizonvz2 E L(TS_T°0) where the average convection coefﬁcient can be estimated from the ChurchillChu correlation, Eq.
17.74. M 0.387 mi“ 1% =—= 0.825+
L
k 8/27
[1+(O.492/Pr)9/16:l Find ﬁrst the Rayleigh number, Eq. 17.19 or 17.66, g/3(Ts—TOO)L3
va RaL = _9.8 m/52x0.00312 K‘1(70—25)K(0.250 m)3 7
RaL— =4.69><10
17.95x10_6 m2/sx25.53x10_6 mZ/s Substituting numerical values, ﬁnd 5,
Continued PROBLEM 17.82 (Cont) 7 1/6 2
_ 0.387(4.68x10 )
NuL= 0.825+ 9/16 3/27 248.75
[1+(0.492/0.704) ]
E=§ITIEL =Wx48.75=5.4 W/m2 K L 0.250 m and then the convection heat transfer rate per unit width of the plate, qgonv=2x5.4 W/mzKx0.250 m(70—25)°C=122 W/m < (b) Since RaL < 109 (see Eq. 17.66), the boundary layer ﬂow is laminar and the thickness may be
estimated from Eq. 17.76 6 —1/4
—=6G /4
L (m ) where GrL = RaL/Pr. Substituting numerical values, ﬁnd Ra ““4 469x107 4/4
6=6L —L =6x0.250 m '——— =23.5 m <
4 Pr 4x0.704 (c) For the forced convection case shown on the schematic, the boundary layer ﬂow condition is
characterized by the Reynolds number, ReL=u°°L= 5 m/5x0.25 m =6.96><104 v 17.95x10_6 mz/s Since ReL < Rex,c = 5 x 105, the ﬂow is laminar over the entire plate and Eq. 17.26 is the appropriate
correlation for estimating the average coefﬁcient, —HL 1/2
NuL =T=0.664 RelL/2 Pr1/3 =0.664(6.96x104) 1/3
) (0.704 =155.9 H=5ﬁ 0.02783 W/mK = x155.9=17.3 W/mZK
L L 0.250m Using the rate equation, ﬁnd qgonv=2x173 W/m2 Kx0.25 m(70—25)°C=390 W/m < Eqs. 17.21 and 17.24 are appropriate for estimating the hydrodynamic and thermal boundary layers,
respectively, —1/2
) =4.7 m < 5=5L Reil/z =5x0.250 m(6.96x104 5t =5Pr‘1/3 =4.7 mm(0.704)‘“3 :53 m <
Continued PROBLEM 17.82 (Cont) COMMENTS: (1) Note that the convection heat rate by forced convection with a modest air stream
velocity (1100 = 5 m/s) is more than three times that by free convection. The hydrodynamic boundary layer thickness by forced convection is nearly 5 times smaller than by free convection. For free
convection, the hydrodynamic and thermal boundary layer thicknesses are of the same extent. For the case of forced convection with air under laminar ﬂow conditions, they are also of the same extent, but
only because Pr is nearly unity. (2) It is important to recognize that radiation exchange between the plate and surroundings could be
signiﬁcant. If 8 = 1 and Tsur = Tm, the linearized radiation coefﬁcient from Eq. 15.9 is hrad =80(Ts +Tsur)(T52 +T521'JI')=7'2 W/m2 'K As such, the heat rate for the free convection situation is more than doubled. For the forced
convection situation, the heat rate is increased by only 50%. (3) The foregoing equations for parts (a) and (b) —free convection case  were used in the IHT
Workspace to calculate the average convection coefﬁcient, heat rate per unit width of the plate, and the boundary layer thickness. Note the use of the Properties function for air to evaluate the properties
at the ﬁlm temperature. II Free convection case, Parts (a) and (b)
/* Results, average convection coefficient correlation, heat rate, boundary layer thickness GrL NuLbar RaL delta_mm hbar q'conv
6.661E7 48.75 4.69E7 23.48 5.424 122 l* Properties, air, Tf = 320.5 K Pr Tf alpha beta k nu 0.7041 320.5 2.553E—5 0.00312 0.02782 1.795E5 */ ll Convection rate equation
q'conv = 2 * hbar * L * (Ts  Tinf) //W/m Il Vertical plate, ChurchillChu correlation NuLbar = ( 0.825 + 0.387 * RaL" (1/6) / ( 1 + (0.492 / Pr )" (9/16) )“(8/27) )"2 // Eq. 17.74
NuLbar = hbar * L / k // Rayleigh number RaL = g * beta * (Ts Tinf) * L"3 / ( nu * alpha) 9 = 9.8 // Gravitational constant, m"2 / 3 II Boundary layer thickness, Pr = 0.7, RaL <= 10"!) delta / L = 6 * (GrL / 4 )" (1/4) // Boundary layer thickness, m
delta_mm = delta * 1000 // mm
RaL = GrL * Pr // Eq. 17.19, 17.66 II Input variables Tinf = 25 + 273 // K
Ts = 70 + 273 // K
L = 0.25 // m II Air property functions : From Table HT3
// Units: T(K); 1 atm pressure nu = nu__T("Air",Tf) // Kinematic viscosity, m"2/s k = k_T("Air",Tf) // Thermal conductivity, W/mK alpha = alpha_T("Air",Tf) // Thermal diffusivity, m"2/s Pr = Pr_T("Air",Tf) // Prandtl number beta = 1/T f // Volumetric coefﬁcient of expansion, K"(1); ideal gas
Tf = (Ts + Tinf) / 2 // Film temperature, K Continued ..... PROBLEM 17.84 KNOWN: Dimensions of vertical rectangular ﬁns; temperature of ﬁns and quiescent air. FIND: Rate of heat transfer from each ﬁn by free convection. Comment on the effect of boundary
layer formation on the spacing between the ﬁns. SCHEMATIC: Quiescent
ai Too = 27°C PROPERTIES: Table HT3, Air (Tf= (TS + Tw)/2 = 325 K, 1 atm): v = 18.41 x 10'6 mZ/s, k =
0.02815 W/mK, a = 26.20 x 10'6 mz/s, 13 = 1/Tf= 0.003077 K'l, Pr = 0.7035. ASSUMPTIONS: (1) Steadystate conditions, (2) Fin surfaces are isothermal, (3) Negligible heat
transfer from ﬁn edges, and (4) Ambient air is quiescent. ANALYSIS: The convection rate equation considering both sides of the ﬁn surface is expressed as
Clconv =2h(LXH)(TS —T°0)
to estimate h, use the ChurchillChu correlation, Eq. 17.74, 2 N— 0825 0.387 RalL/6
“L” ' + 8/27 [1+(0.492/Pr)9/16:l where the Raleigh number, Eq. 17.19 or 17.66, is written with the characteristic length as H, =gﬂ(Ts_Too)L3
va RaH 9.8 m/s2x0003077 K'1(77—27)K(0.150 m)3 7
Ray; =——————————=1.055x10 18.41x10‘6 mz/sx26.20x10_6 mz/s and the average convection coefﬁcient is 7 1/6
0.387(1.055x10 ) N—u = 0825+
L )9/16]8/27 =31.66
[1+ (O.492/0.7035 — _0.02815 W/mK NuH x31.66=5.94 W/mzK
0.150m 5:5
H Continued ..... PROBLEM 17.84 (Cont) Substituting numerical values into the rate equation, ﬁnd the heat transfer rate qconv =2x5.94 W/m2 K(0.020x0.150)m2 (7727)K Clconv =178 W < If the ﬁns are too close together, the boundary layers on adjoining surfaces will coalesce and the heat
transfer rate will decrease. If the ﬁns are too far apart, the ﬁnned surface area becomes small, and the
heat transfer rate will again decrease. To estimate the spacing, calculate the boundary layer thickness at the leading edge. Since RaH < 109, the boundary layer is laminar, and the thickness at the trailing
edge can be estimated from Eq. 17.76, 3:6(Gm/4)—1/4=6(RaH/4 Pr)_1/4
1055 107 W4 5=6x0.150m ——' x =20.5 mm
4x0.7035 In order that the boundary layers not coalesce, we require S = 25 z 40 mm. However, by making S
some fraction of 25, the added number of ﬁns per unit area of the base may offset any adverse effect
of boundary layer interference. COMMENTS: (1) Radiation exchange with the surroundings may not be important in this situation
because the ﬁn surfaces mostly “see” each other. (2) The optimum spacing between ﬁn surfaces is an important consideration in the design of ﬁn
arrays. This topic is treated in advanced heat transfer courses. PROBLEM 17.88 KNOWN: Horizontal, circular grill of 0.2 m diameter with emissivity 0.9 is maintained at a uniform
surface temperature of 130°C when ambient air and surroundings are at 24°C. FIND: Electrical power required to maintain grill at prescribed surface temperature.
SCHEMATIC: Quiescent
ai Tm. = 24°C Grill surface, D = 250 mm = o .=
qconv Ts 130 ca. 0.9 \ P e ASSUMPTIONS: (1) Room air is quiescent, (2) Surroundings are large compared to grill surface, and
(3) Sides and bottom of grill are well insulated. PROPERTIES: Table HT—3, Air (n: (To, + Ts)/2 = (24 + 130)°C/2 = 350K, 1 atm): v = 20.92 x
106 mZ/s, k = 0.030 W/mK, a = 29.9 x 10*5 mZ/s, 0 = 1/Tf. ANALYSIS: The required electrical power, Fe, is equal to the heat transfer from the grill, which is due
to free convection with the ambient air and to radiation exchange with the surroundings. Pezqus[H(TS—Too)+ga(Ts4— 51”” (1)
Calculate RaL from Eq. 17.19, RaL :gﬂ(TS—TOO)L%/va where for a horizontal disc from Eq. 17.77, Lc = As/P = (nD2/4)/1tD = D/4. Substituting numerical
values, ﬁnd R 9.8m/s2(1/350 K)(130—24)K(0.25 m/4)3
aL I
20.92x10“6m2/sx29.9x10‘6m2/s Since the grill is an upper surface heated, Eq. 17.79 (Case C, Table 17.6) is the appropriate
correlation, 21.158x106. — — 1/4 61/4
NuL =hLLc/k:O.S4RaL =0.54 1.158x10 =17.72 EL :N—uLk/Lc =17.72x0.03OW/mK/(0.25 m/4) 28.50W/m2 K. (2) Substituting from Eq. (2) for E into Eq. (1), the required electrical power is P6 = £(0.25m)2 [8.50 (130— 24)K +0.9x 5.67 x10_8 2W ((130+ 273)4 —(24 + 273)4 )K4:l
4 mZK m K4 Pe :44.2W+46.0W=90.2W. < COMMENTS: Note that for this situation, free convection and radiation modes are of equal importance. If the grill surface were highly polished such that a z 0.1, the required power would be
reduced by nearly 50%. , ..._1._.i.._imr____..__..... ,.._....,T W__W_..T._.__._,W__._..m Wm...“ PROBLEM 18.3 KNOWN: Plate maintained at 200°C by imbedded electrical heating elements; exposed surface has an emissive power of 1200 W/mz, irradiation of 2500 W/mz, and reﬂectivity of 30%; exposed surface experiences air ﬂow having a freestream temperature of 20°C with a convection coefﬁcient of 15
2
W/m ~K. FIND: (a) The radiosity, J; (b) The net radiation heat ﬂux leaving the surface, qfad, in terms of the radiosity and irradiation; (c) The combined convection and radiation heat ﬂux leaving the surface; (d)
Represent a surface energy balance schematically, and label all the radiation processes; and (e) The . . ,, 2 . . . .
electrical power requ1rement, Pelee (W/m ) to malntain the plate under these condltrons. SCHEMATIC: G = 2500 WIm2 Ta, = 20°C
h =15 W/mZK E = 1200 WIm2 P = 03 qgonv \ T5 = 200°C / f
T Pas ASSUMPTIONS: (1) Steadystate conditions, (2) Plate is opaque, diffuse; and (3) Plate backside is
perfectly insulated. ANALYSIS: (a) The radiosity is the sum of the emissive power, E, and the reﬂected irradiation, Gref,
where p is the reﬂectivity, J=E + Gref 2E + pG=1200 W/m2 +0.3x2500 W/m2 =1950 W/m2 < (b) The net radiation heat ﬂuxing leaving the surface, Qiad , in terms of the radiosity and irradiation is qlr’ad = q’r’ad,out = [Bout _ Efn :lrad : J — G q;ad=1950W/mZ—2500W/mz=—550W/m2 < (c) The combined convection and radiation heat ﬂux leaving the surface is
qcv,rad = qlcv + qlfad
qgv =h(TS —T°°)=15 W/m2 K x (200—20)°C=2700 W/m2 qcv+qrad=2700 W/m2+(—550W/m2)=2150W/m2 < (d, e) The surface energy balance is shown schematically in terms of J and G, q;ad, and qcv+rad q’r'ad \ f (13v . I The surface energy balance, Efn — Eout = 0, expressed as
I I I I I 2
Pelee=J—G+qcv=qrad+qcv=qcv+rad=2150W/m < COMMENTS: Note how the directions for the heat ﬂuxes and radiation quantities in the schematic
are consistent with the energy balance relations. PROBLEM 18.4 KNOWN: Evacuated, aluminum sphere (D = 2 m) serving as a radiation test chamber. FIND: Irradiation on a small test object when the inner surface is lined with carbon black and at 600
K. What effect will surface coating have? SCHEMATIC: Aluminum sphere, D=2m Small +651“ surface, A1 Carbon b/ack coaﬂng,
7;=600K ASSUMPTIONS: (l) Sphere walls are isothermal, (2) Test surface area is small compared to the
enclosure surface. ANALYSIS: It follows from the discussion of Section 13.3 that this isothermal sphere is an
enclosure behaving as a blackbody. For such a condition, see Fig. 18.12(c) and Eq. 18.6, the irradiation on a small surface within the enclosure is equal to the blackbody emissive power at the
temperature of the enclosure. That is G1 =Eb (TS)=OTS4 G1=5.67x10_8W/m2K4(600K)4 =7348W/m2. < The irradiation is independent of the nature of the enclosure surface coating properties. COMMENTS: (1) The irradiation depends only upon the enclosure surface temperature and is
independent of the enclosure surface properties. (2) Note that the test surface area must be small compared to the enclosure surface area. This allows
for interreﬂections to occur such that the radiation ﬁeld, within the enclosure will be uniform
(diffuse) or isotropic. (3) The irradiation level would be the same if the enclosure were not evacuated since, in gen...
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 Spring '07
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