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Unformatted text preview: PROBLEM 17.11 KNOWN: Parallel ﬂow over the ﬂat plate shown in Example 17.3; ﬂow is tripped at the leading
edge to induce turbulence over its entire length. FIND: Total electrical power required for the ﬁrst ﬁve heaters, x = L5.
SCHEMATIC: T(JD = 25°C
um: 60 m/s l—> x Pe L5 — 250 mm ASSUMPTIONS: (1) Steadystate conditions, (2) Turbulent ﬂow over entire length of the plate, and
(3) Constant properties. PROPERTIES: Table HT—3, Air (Tf= 400 K, 1 atm): v = 26.41 x 10'6 mZ/s, k = 0.0338 W/mK, Pr
= 0.690. ANALYSIS: The electrical power required for the ﬁrst ﬁve heaters, x = L5, is given by the
convection rate equation written as Fe =qconv =El—5 (L5 W)(TS _T°0) For turbulent ﬂow over the entire plate, the correlation of Eq. 17.32 is appropriate for estimating the
average coefﬁcient, NHL =w=0037 Ref/5 Pr1/3
k where the Reynolds number is uoO L5 _ 60 m/sx0.250 m
v 26.41x10‘6 mz/s Substituting numerical values into the correlation, ﬁnd =5.667><105 ReL = — 5 4/5 1/3 NuL =0.037(5.667x10 ) (0.690) =1310 El5=£NuL =W1310=177 W/mK
L5 0250 m Hence, the required electrical power is Pe=177 W/mK(O.250 mxl m)(230—25) Pe=9085 w < Continued ..... PROBLEM 17.11 (Cont) COMMENTS: (1) IHT can be used to evaluate the thermophysical properties required of the
convection correlations for selected ﬂuid. Shown below is the code used to perform the analysis for
this problem. I* Results NuLbar ReL hbar1_5 qconv
1310 5.667E5 177.3 9085
Properties, evaluated at Tf Pr Tf k nu 0.69 400.5 0.03383 2.647E5 */ I/ Convection rate equation, first five heaters, x = L5
qconv = hbar1_5 * L5 * w * (Ts  Tinf) Il Correlation, turbulent flow NuLbar = 0.037 * ReL"O.8 * Pr"O.333 // Eq. 17.32
NuLbar = hbar1_5 * L5 / k ReL = uinf * L5 / nu I/ Input variables
Tinf = 25 uinf = 60 Ts = 230 L5 = 0.250 w = 1 ll Air property functions : From Table H3
// 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
Pr = Pr_T("Air",Tf) // Prandtl number Tf = (Ts + Tinf)/2 + 273 // Temperature units in kelvins for Properties functions (2) The Correlations Tools menu of [HT includes correlations for external ﬂow over ﬂat plates,
cylinders and spheres. The correlations are written as functions to avoid tedious entry of the
equations from the keyboard. This feature of [HT will save you considerable time, especially when
you need to perform parametric analysis, generate graphs, etc. ll Correlation, laminar or mixed; used for fully turbulent flow
NuLbar = NuL_bar_EF_FP_LM(ReL,Rexc,Pr) // Eq 7.26, 7.29, 7.30
NuLbar = hbar1_5 * L5 / k ReL = uinf * L5 / nu Rexc = 0.001 // Set Rexc ca zero to represent fully turbulent conditions /* Correlation description: Parallel external flow (EF) over a flat plate (FP), average coefﬁcient; laminar (L)
if ReL<Rexc, Eq 17.26; mixed (M) if ReL>Rexc, Eq 17.29 and 17.30; 0.6<=Pr<=60. See Table 17.3. */ PROBLEM 17.14 KNOWN: Plate dimensions and temperature. Velocity and temperature of air in parallel ﬂow over
plates. FIND: Rate of heat transfer from plate.
SCHEMATIC: Plate, T5 = 300°C
thickness, 6 = 6 mm
area,As=1mx1m «pg/g Tm = 20°C um: 10 m/s /" 1/ ASSUMPTIONS: (1) Negligible radiation, (2) Negligible effect of conveyor velocity on boundary
layer development, (3) Plates are isothermal, (4) Negligible heat transfer from sides of plate, (5) Rex,c = 5 ><105 , (6) Constant properties. PROPERTIES: Table HT3, Air (Tf= 433 K, 1 atm): v = 30.4 x 10'6 mz/s, k = 0.0361 W/mK, Pr =
0.688. ANALYSIS: The rate of heat transfer from a plate is
q =2HAS(Ti 4,0): 2hL2(Ti 400) To estimate the average convection coefficient, characterize the boundary layer ﬂow with the
Reynolds number ReL = qu/v =10m/sx1m/30.4x10_6m2 /s = 3.29x105 Hence, ﬂow is laminar over the entire surface and the appropriate convection correlation, Eq. 17.26,
has the form U3 U2
) =336 REL = 0.66411e1L/2 Pr1/3 =0.664(3.29x105 (0.688) E=(k/L)ﬁL =(0.0361W/mK/1m)336=12.1W/m2 K Hence, the heat transfer rate is
q=2><12.1W/m2K(1m)2(300—20)°C=6780W < COMMENTS: (1) It is quite likely that air ﬂow disturbances are present in the processing environment and that fully turbulent ﬂow could occur over the plate. As such, the correlation of Eq.
17.32 would be appropriate giving these results: N_uL =0.037 Reg's Pr1/3z847 h=30.6 W/m2 K q=17,120 w The effect is to double the estimate for the heat transfer rate. (2) Despite the high plate temperature and the small convection coefﬁcient, if adjoining plates are in close proximity, radiation exchange with the surroundings will be small and the assumption of
negligible radiation is justiﬁable. PROBLEM 17.18 KNOWN: Plate dimensions and freestream conditions. Maximum allowable plate temperature. FIND: Maximum allowable power dissipation for electrical components attached to bottom of plate if
TS does not exceed 350 K. SCHEMATIC: a, Plate, 1.2 x 1.2 m
———> T5 s 350 K Tm: 300 K ———>
um=15mls ——> ASSUMPTIONS: (1) Steadystate conditions, (2) Constant properties, (3) Negligible heat loss from
sides and bottom, (4) Transition Reynolds number is 5 x 105, (5) Isothermal plate. PROPERTIES: Table HT3, Air (Tf= 325 K, 1 atm): v = 18.4 x 10'6 m2/s, k = 0.028 W/mK, Pr =
0.70. ANALYSIS: The heat transfer from the plate by convection is Fe 2 C1conv = 11A5 (Ts ‘T00)
For uw = 15 m/s, the Reynolds number is ReL = uOOL _ 15m/sx1.2m V 18.41x10‘6 m2/s = 9.78x 105 > Rem. where Rex,c = 5 X 105. Hence, transition occurs on the plate and, from Eq. 17.31, ﬁnd the average
convection coefﬁcient, _ 4/5
NuL =(0.037 Rei/S —871)Pr1/3 =[0.037(9.78x105) —871:l(0.70)1/3 =1263 E=NuL§=1263W=29.7w/m2.1<
L 1.2m
The heat rate is
Clconv =297W/m2K(1.2m)2(350—300)K=2137W. < COMMENTS: The equations for the convection heat rate and the convection coefﬁcient correlation, along with the Properties functions for air, were entered into the IHT Workspace to obtain results for this
application. Continued ..... PROBLEM 17.18 (Cont) /* Results: Correlation and convection heat rate
NuLbarReL hbar qconv 1265 9.78E5 29.68 2137 Air properties. evaluated at Tf = 325 K Pr Tf k nu 0.7035 325 0.02815 1.841 E5 */ II Convection rate equation
qconv = hbar * L * w * (Ts  Tinf) // W [I Correlation, mixed flow conditions, Eq. 17.31
NuLbar = (0.037 * ReL"0.8  871) * Pr"0.333
NuLbar = hbar * L / k ReL = uinf" L / nu ll Input variables
Tinf = 300 // K uinf= 15 //m/s
Ts=350 l/K
L=1.2 //m
w= 1.2 //m ll Air property functions : From Table HT3
// Units: T(K); 1 atm pressure nu = nu_T("Air",Tf) /l Kinematic viscosity, m"2/s k = k_T("Air",Tf) // Thermal conductivity. W/mK
Pr = Pr_T("Air",Tf) // Prandtl number Tf = (Ts + Tinf) / 2 // Film temperature, K PROBLEM 17.29 KNOWN: Heat rate from and surface temperature of top surface of an oven under quiescent room air
conditions (Case A). FIND: (a) Thermal resistance due to oven wall and internal convection Rm associated with the quiescent
room air condition; represent this condition (Case A) by a thermal circuit and label all elements; (b) Determine the heat transfer from top surface under forced convection conditions (u00 = 20 m/s); represent this condition (Case B) by a thermal circuit; assume Rt,i remains unchanged; and (c) Estimate the surface
temperature achieved for the Case B condition. SCHEMATIC:
GA = 40 W Ts,A = 47°C ‘
0.5x0.5 m @
Quiescent *9
a" Oven air ——>
0 Ti = 150°C um = 20 m/s
T0047" Tw=17°C
Case A Case B K’—“ L = 0.5 m ——>l ASSUMPTIONS: (1) Surface has uniform temperature under both conditions, (2) Negligible radiation
effects, (3) Air is blown parallel to edge, (4) Thermal resistance due to oven wall and internal convection are the same for both conditions Rt,i’ and (5) Transition Reynolds number is Rex,c = 5 x 105. PROPERTIES: Table HT—3, Air (Assume Tf= (TS + Too)/2 = 300 K, 1 atm): v = 15.89 x 10'6 mZ/s, k =
0.0263 W/mK, Pr = 0.707. ANALYSIS: (a) The thermal circuit representing the quiescent room air condition, A, is shown below. Note the labeling for the thermal resistances, heat rate, and temperatures. The thermal resistance due to
the wall conduction and interior convection is _Ti—TS’A _(150—47)°C Rt i = 2.575 K/ W <
’ q A 40 W
Note that this quantity remains constant for the Case B conditions.
Tm = 17°C Tm
Rt,o,A Rt,o,B
T5, A = 47°C Ts,B
Ru Rm
Ti = 150°C Ti
Thermal circuits: Case A T CIA = 40 W Case B T qB (b) The thermal circuit representing the forced convection conditions, Case B, is shown above with all
elements properly labeled. The heat rate can be expressed as T —T
QB=—_c °°
Rt,i+Rt,o,B Continued ..... PROBLEM 17.29 (Cont) where Rt,o,B represents the thermal resistance due to forced convection on the top surface, 1
R = _
t,o,B hL As where AS = L x L = 0.25 m2 and EL can be estimated from an appropriate correlation. Find ReL = 1100 L =m=6293x105 v 15.89x10_6 mZ/s Assuming Rex,c = 5 x 105, ﬂow conditions are mixed and from Eq. 17.31 NuL =hL—L=(0.037 Ref/5 —871)Pr“3
k
5 4/5 1/3
NuL= 0.037(6.293x10 ) —871 (0.707) =660.0 EL =660.0x 0.0263 W/mK/0.5 m=34.7 W/m2 ~K
Hence, the forced convection thermal resistance is Rt,o,B=——l——=O.115 K/W 34.7 W/m2 Kx0.25 m2 and substituting numerical values into the rate equation (150—17)°C
qB=( ————~—=49.4 w <
2.575+0.115)K/W (c) From the thermal circuit, Case B, write an energy balance to determine TS’B,
_ TS ~T00
Rt,o,B QB Ts =Too +QB Rt,o,B
Ts =17°C+49.4 Wx0.115 K/W=(17+5.7)°C=22.7°C COMMENTS: (1) Based upon the result TS = 227°C, the ﬁlm temperature is Tf= (22.7 + 17)°C/2 =
293 K, which is sufﬁciently close to the assumed value of 300 K. (2) The IHT code to solve this problem includes the analysis of the thermal circuit for Case A to obtain
Rm, estimate the average convection coefﬁcient for Case B, and from analysis of the thermal circuit for Case A, to calculate the surface temperature T53, and the heat rate, qB. Note for the correlation
calculation, we ﬁrst use a guessed value for Tf, then load the results as initial guesses, and solve for the correct Tf. Continued... PROBLEM 17.34 KNOWN: Dimensions of chip and pin ﬁn. Chip temperature. Air coolant velocity and temperature.
FIND: (a) Average pin convection coefﬁcient, (b) Pin heat transfer rate, and (c) Total heat rate.
SCHEMATIC: ——» T Pin ﬁn, D=2mm ASSUMPTIONS: (1) Steadystate conditions, (2) Onedimensional conduction in pin, (3) Constant
properties, (4) Convection coefﬁcients on pin surface (tip and side) and chip surface correspond to single
cylinder in cross ﬂow, (5) Negligible radiation. PROPERTIES: Table HTI, Copper (350 K): k = 399 W/mK; Table HT—3, Air (Tf z325 K, 1 atm): v
= 18.41 x 106 m2/s, k = 0.0282 W/mK, Pr = 0.704. ANALYSIS: (a) With uGO = 10 m/s and D = 0.002 m, u00 D _ 10m/sx0.002m _ =1087
V 18.41x10‘6 m2/s ReD = Using the Churchill and Bernstein correlations, Eq. 17.35, 4/5
=16.7 1/2 1/3 5/8
— 0.62R P R
NuD=0.3+—————e—Q—r— 1+[ 613 1/ 4 282,000
[1+(0.4/Pr)2/3] E = (ITuDk/D) = (16.7 x 0.0282 W/m  K/0.002m) = 235 w/m2 K < (b) For the ﬁn with tip convection, using Eq. 16.74, _ 2 1/2 2 3 1/2
M=(h7er7rD /4) 0b=(7r/2):235W/m K(0.002m) 399W/mK] 50K=2.15W 1/2 m = (hP/kAc )“2 = (4x235W/m2 K/399 W/mKx0.002m) = 34.3 m‘1 mL = 34.3 m“1(0.012m)= 0.412 (h/mk)=(235W/m2 K/34.3m‘1x399w/mK)= 0.0172.
The ﬁn heat rate is
sinh mL + (h/mk)cosh mL _—=0.868W. <
cosh mL + (h/mk)sinh mL Qf=M (c) The total heat rate is that from the base and through the ﬁn, q =% +qf = h(W2 —7rD2/4)6b +qf =(0.151+0.868)W =1.019w. < PROBLEM 17.35 KNOWN: Diameter and length of a copper rod, with ﬁxed end temperatures, inserted in an airstream
of prescribed velocity and temperature. FIND: (a) Midplane temperature of rod, (b) Rate of heat transfer from the rod. SCHEMATIC:
Copper rod, D = 10 mm
Tb = 90°C Tb = 90°C Qf qf ? um: 25 m/s ‘_> X T f L = 50 mm ASSUMPTIONS: (1) Steadystate conditions, (2) Onedimensional conduction in rod, (3)
Negligible contact re51stance, (4) Negligible radiation, (5) Constant properties. PROPERTIES: Table HTI, Copper (T z 80°C = 353 K): k = 398 W/mK; Table HT—3. Air (assume TS = 80°C; hence 1f: (TS + Tag/2 = 326 K, 1 atm): v = 1.846 x 10'5 mz/s, k = 0.02819
W/mK, Pr = 0.703. ANALYSIS: (a) The copper rod is a pin ﬁn, Fig. l6.l7b, having a base temperature Tb = 90°C with an adiabatic tip, L = 50 mm, Case B. To determine the midplane temperature T(L), use Eq. 16.69,
with x = L,
6 _T(L)—Too_cosh m(L—x)_ 1 9b Tb——T(,O cosh(mL) _cosh(L) — 1/2 — 1/2
Where ﬁn parameter, with P = ND and Ac = nD2/4, is m = (hP/ kAc) =(4h/ kD) The average convection coefﬁcient is estimated using the ChurchillBernstein correlation, Eq. 17.35, —l/ 4 4/5
— 2/3 5/8
— hD . R
NuD =—=0.3+ 0.62 Reg2 Pr1/3 1+ E 1+ —eD——
k Pr 282, 000
Assuming a ﬁlm temperature (see Properties), the Reynolds number is
ReD zuooD : 25 m/sx05.0102m 213,550
V 1.846x10_ m /s
and substituting numerical values on the correlation, ﬁnd NED and h.
—l/4 4/5
2/3 5/8
— 0.4 13,550
NuD =0.3+ 0.62(13,550)“2 (0.703)“3 1+£ ) 1+
0.703 282, 000
ED = 63.3
Continued PROBLEM 17.35 (Cont) h=NuDk/D=63.3x0.02819 W/mK/0.010 m=178 W/m2 K Substituting numerical values in the ﬁn parameter expression, ﬁnd 1/2
m=(4x178 W/m2 ~K/398 W/mKx0.010 m) =13.38 m‘1
and then the tip temperature,
T L —25°C
L— 1 1 =0.8116 (90‘25)°C _ cosh(13.38 m’1 x0.010 m) — cosh(0.6688) T(L)=77.8°C <
(b) The heat transfer from the rod follows from Eq. 16.70 (see Table 16.4),
q=2qf =2 Mtanh(mL)=2(thAc)1/2 9b tanh(mL) 1/2
q=2:178 W/m2 K(7rO.010 m)398 W/mK(7r0.0102 m2/4)] (90—25)°c tanh(0.6688) q=3l.8 w < COMMENTS: (1) With T(L) = 77.8°C, and Tb = 90°C, the assumed value for the average surface temperature, T5 = 80°C, was a reasonable choice. (2) Note how the symmetry about the midplane represents an adiabatic condition resulting in
identifying the rod as a Case B tip condition in Table 16.4. PROBLEM 17.41 KNOWN: Temperature and velocity of water ﬂowing over a sphere of prescribed temperature and
diameter. FIND: Rate of heat transfer. SCHEMATIC:
Sphere, D = 20 mm ——l>
Tm = 20°C c .‘ \ q
uOD = 5 m/s ASSUMPTIONS: (1) Steadystate conditions, (2) Uniform surface temperature. HuﬂTRTﬂB:HMkHTiSMmWﬁmeUQ=2%K)p=9%kgm{u=lm7xw6Nsm2
k = 0.603 W/mK, Pr = 7.00; (Ts = 333 K): p. = 467 X 10'6 Ns/mz. 9 ANALYSIS: The rate of heat transfer from the sphere follows from the rate equation, where the
surface area is AS = TEDZ, q=HAs(Ts—Too>=ﬁ(”D2)(Ts‘Twl To estimate the average convection coefﬁcient, use the Whitaker correlation, Eq. 17.36, 1/4
NuD =2+[0.4 Reg/2 +0.06Re123/3]Pr0'4[f—]
S where all properties are evaluated at the freestream temperature, except for ,us, which is
evaluated at the surface temperature. Substituting numerical values, _ puooD _ 998 kg/m3 (5 m/s) (0.02 m)
rt 1007x10'6 Ns/m2 _ 1/2 2/3 “4
NuD=2+[0.4(9.91x104) +0.06(9.91x104) ](7.0)°4[M] =673. =9.9lx104 RCD 467 Hence, the convection coefﬁcient and heat rate are ﬁzimD =W673=2Q300 w/m2.K
D 0.02m
q=h(rtD2) (TS—Tw)=20,300 EV 7r(0.02m)2 (60—20)°C=1020w. <
m K COMMENTS: Note that ReD is beyond the range of the correlation, and for a more reliable estimate
we should seek a more appropriate correlation from the literature. ...
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 Spring '07
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