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assignment10 - PROBLEM 17.11 KNOWN Parallel flow over the...

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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) Steady-state 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/m-K, 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 El-5=£NuL =W1310=177 W/m-K L5 0250 m Hence, the required electrical power is Pe=177 W/m-K(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 2.647E-5 */ 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 H-3 // Units: T(K); 1 atm pressure nu = nu_T("Air",Tf) // Kinematic viscosity, m"2/s k = k_T("Air",Tf) // Thermal conductivity, W/m-K 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 HT-3, Air (Tf= 433 K, 1 atm): v = 30.4 x 10'6 mz/s, k = 0.0361 W/m-K, 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/m-K/1m)336=12.1W/m2 -K Hence, the heat transfer rate is q=2><12.1W/m2-K(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) Steady-state 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 HT-3, Air (Tf= 325 K, 1 atm): v = 18.4 x 10'6 m2/s, k = 0.028 W/m-K, 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 =29-7W/m2-K(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 E-5 */ 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 HT-3 // 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/m-K 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/m-K, 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/m-K/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) Steady-state conditions, (2) One-dimensional 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 HT-I, Copper (350 K): k = 399 W/m-K; Table HT—3, Air (Tf z325 K, 1 atm): v = 18.41 x 10-6 m2/s, k = 0.0282 W/m-K, 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/m-K] 50K=2.15W 1/2 m = (hP/kAc )“2 = (4x235W/m2 -K/399 W/m-Kx0.002m) = 34.3 m‘1 mL = 34.3 m“1(0.012m)= 0.412 (h/mk)=(235W/m2 -K/34.3m‘1x399w/m-K)= 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) Steady-state conditions, (2) One-dimensional conduction in rod, (3) Negligible contact re51stance, (4) Negligible radiation, (5) Constant properties. PROPERTIES: Table HT-I, Copper (T z 80°C = 353 K): k = 398 W/m-K; 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/m-K, 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 Churchill-Bernstein 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/m-K/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/m-Kx0.010 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/m-K(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) Steady-state conditions, (2) Uniform surface temperature. HuflTRTflB:HMkHTiSMmWfimeUQ=2%K)p=9%kgm{u=lm7xw6Nsm2 k = 0.603 W/m-K, Pr = 7.00; (Ts = 333 K): p. = 467 X 10'6 N-s/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 N-s/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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