fundamentals of heat and mass transfer solutions manual phần 5 ppsx

Hence, () 1 2 22 3 t,o 2 0.0858m R 4236 W / m K 0.0913m 1 0.776 9.55 10 K / W 0.0913m − − =⋅×− =×        and () () belecbt,o T T P R R 17 C 1200W 0.0187K / W 39.5 C ∞ =+ + =°+ = ° (1) The boundary layer thickness at the trailing edge of the fin is () 2 1/2 2w 5w / Re δ = () 0.84 mm S t . =<<− Hence, the assumption of parallel flow over a flat plate is reasonable. (2) If a finned heat sink is not employed and heat transfer is simply by convection from the 22 ww × base surface, the corresponding convection resistance would be 0.0195 K/W, which is only twice the resistance associated with the fin array. The small enhancement by the array is attributable to the large value of h and the correspondingly small value of f . η Were a fluid such as air or a dielectric liquid used as the coolant, the much smaller thermal conductivity would yield a smaller h, a larger f η and hence a larger effectiveness for the array. Plate dimensions and freestream conditions. Maximum allowable plate temperature. (a) Maximum allowable power dissipation for electrical components attached to bottom of plate, (b) Effect of air velocity and fins on maximum allowable power dissipation. (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat loss form sides and bottom, (4) Transition Reynolds number is 5 × 10 5 , (5) Isothermal plate. Table A.1, Aluminum (T ≈ 350 K): k ≈ 240 W/m ⋅ K; Table A.4, Air (T f = 325 K, 1 atm): ν = 18.4 × 10 -6 m 2 /s, k = 0.028 W/m ⋅ K, Pr = 0.70. (a) The heat transfer from the plate by convection is () elec s s PqhATT ∞ == − . For u ∞ = 15 m/s, 5 Lx,c 62 u L 15m s 1.2m Re 9.78 10 Re 18.41 10 m s ν ∞ − × == =×> × . Hence, transition occurs on the plate and ( ) ( ) () 4/5 1/3 4/5 1/3 5 L L Nu 0.037Re 871 Pr 0.037 9.78 10 871 0.70 1263  =−=×− =   2 L k 0.028W m K h Nu 1263 29.7W m K L 1.2m ⋅ == = ⋅ The heat rate is ()( ) 2 2 q 29.7 W m K 1.2m 350 300 K 2137 W =⋅ −= . (b) The effect of the freestream velocity was considered by combining the Correlations Toolpad for the average coefficient associated with flow over a flat plate with the Explore and Graph options of IHT. 5 10 15 20 25 Freestream velocity, uinf(m/s) 0 500 1000 1500 2000 2500 3000 3500 4000 Power dissipation, q(W) Continued The effect of increasing u ∞ is significant, particularly following transition at u ∞ ≈ 7.7 m/s. A maximum heat rate of q = 3876 W is obtained for u ∞ = 25 m/s, which corresponds to h ≈ 54 W/m 2 ⋅ K and Re L = 1.63 × 10 6 . The Extended Surfaces Model for an Array of Straight Rectangular Fins was used with the Correlations Toolpad to determine the effect of adding fins, and a copy of the program is appended. With L f = 25 mm, w = 1.2 m, t = 0.005 m, S = 0.015 m, N = 80 and u ∞ = 25 m/s, the solution yields q = 16,480 W which is more than a four-fold increase relative to the unfinned case. (1) With a fin efficiency of η f = 0.978, there is significant latitude for yet further enhancement in heat transfer, as, for example, by increasing the fin length, L f . (2) The IHT code below includes the model for the Extended Surface, Array of Straight Fins and the Correlation for the convection coefficient of a flat plate with mixed flow conditions. /* Fin analysis results, uinf = 25 m/s Ab Acb Af Ap At Aw etaf etaoc m qt R''tc 0.96 0.006 0.066 0.0001375 6.24 1.44 0.978 0.9814 9.471 1.648E4 0 */ /* Correlation results and air thermophysical properties at Tf NuLbarPr ReL Tf hLbar k nu uinf 2294 0.7035 1.63E6 325 53.82 0.02815 1.841E-5 25 */ // IHT Model, Extended Surfaces, Array of Straight Rectangular Fins /* Model: Fin array with straight fins of rectangular profile, thickness t, width w and length L. Array has N fins with spacing S. */ /* Find: Array heat rate and performance parameters */ /* Assumptions:(1) Steady-state conditions, (2) One-dimensional conduction along the fin, (3) Constant properties, (4) Negligible radiation exchange with surroundings, (5) Uniform convection coefficient over fins and base, (6) Insulated tip, Lc = L + t / 2 */ // The total heat rate for the array qt = (Tb - Tinf) / (Rtoc) // Eq 3.104 /* where the fin array thermal resistance, including thermal contact resistance, R''tc, at the fin base is */ Rtoc = 1 / (etaoc * h * At) // The overall surface efficiency is etaoc = 1 - (N * Af / At) * (1 - etaf / C1) // Eq 3.105 C1 = 1 + etaf * h * Af * (R''tc / Acb) // where N is the total number of fins, and the surface area of a single fin is Af = 2 * w * Lc // where the equivalent length, accounting for the adiabatic tip, is Lc = Lf + (t / 2) /* The surface area associated with the fins and the exposed portion of the base (referred to also as the prime surface, Ab) is */ At = N * Af + Ab Ab = Aw - N * Acb // The total area of the base surface follows from the schematic Aw = w * N * S // where S is the fin spacing. The base area for a single fin is Acb = t * w // The fin efficiency for a single fin is: etaf = (tanh(m * Lc)) / (m * Lc) // Eq 3.89 // where m = sqrt(2 * h / (kf * t)) /* The input (independent) values for this system are: Fin characteristics */ Tb = 350 // base temperature, K t = 0.005 // thickness, m w = 1.2 // spacing width, m Lf = 0.025 // length, m S = 0.015 // fin spacing, m N = 80 // number of fins kf = 240 // thermal conductivity, W/m·K // Convection conditions Tinf = 300 // fluid temperature, K h = hLbar // convection coefficient,W/m^2·K /* Thermal contact resistance per unit area at fin base. Set equal to zero if not present. */ R''tc = 0 // thermal resistance per unit area, K·m^2/W // Correlation, External flow, Flate Plate, Laminar or Mixed Flow NuLbar = NuL_bar_EF_FP_LM(ReL,Rexc,Pr) // Eq 7.31, 7.39, 7.40 NuLbar = hLbar * L / k ReL = uinf * L / nu Rexc = 5.0E5 // Evaluate properties at the film temperature, Tf. Tf = (Tinf + Tb) / 2 /* Correlation description: Parallel external flow (EF) over a flat plate (FP), average coefficient; laminar (L) if ReL<Rexc, Eq 7.31; mixed (M) if ReL>Rexc, Eq 7.39 and 7.40; 0.6<=Pr<=60. See Table 7.9. */ // Air property functions : From Table A.4 // 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 // Input variables, correlation uinf = 25 // freestream velocity, m/s L = 1.2 // plate width, m PROBLEM 7.32 KNOWN: Operating power of electrical components attached to one side of copper plate. Contact resistance. Velocity and temperature of water flow on opposite side. FIND: (a) Plate temperature, (b) Component temperature. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat loss from sides and bottom, (4) Turbulent flow throughout. PROPERTIES: Water (given): ν = 0.96 × 10 -6 m 2 /s, k = 0.620 W/m⋅K, Pr = 5.2. ANALYSIS: (a) From the convection rate equation, s TTq/hA ∞ =+ where q = Nq c = 2500 W and A = L 2 = 0.04 m 2 . The convection coefficient is given by the turbulent flow correlation ( ) ( ) L 4/51/3 L hNuk/L0.037RePrk/L== where ( ) ( ) 625 L ReuL/2 m/s0.2m/0.9610m/s4.1710ν − ∞ ==××=× and hence ( ) ( ) ( ) 4/5 1/3 52 h0.0374.17105.20.62 W/mK/0.2 m6228 W/mK. =×⋅=⋅ The plate temperature is then ( ) ( ) 2 2 s T17C2500 W/6228 W/mK0.20 m27C. =+⋅= oo < (b) For an individual component, a rate equation involving the component’s contact resistance can be used to find its temperature, ( ) ( ) ( ) ccst,ccst,cc qTT/RTT/R/A ′′ =−=− ( ) -4242 csct,cc TTqR/A27C25 W210 mK/W/10 m − ′′=+=+×⋅ o c T77C.= o < COMMENTS: With 5 L Re4.1710, =× the boundary layer would be laminar over the entire plate without the boundary layer trip, causing T s and T c to be appreciably larger. Air at 27 ° C with velocity of 10 m/s flows turbulently over a series of electronic devices, each having dimensions of 4 mm × 4 mm and dissipating 40 mW. (a) Surface temperature T s of the fourth device located 15 mm from the leading edge, (b) Compute and plot the surface temperatures of the first four devices for the range 5 ≤ u ∞ ≤ 15 m/s, and (c) Minimum free stream velocity u ∞ if the surface temperature of the hottest device is not to exceed 80 ° C. (1) Turbulent flow, (2) Heat from devices leaving through top surface by convection only, (3) Device surface is isothermal, and (4) The average coefficient for the devices is equal to the local value at the mid position, i.e. 4x hh = (L). Table A.4, Air (assume T s = 330 K, () s TTT 2 ∞ =+ = 315 K, 1 atm): k = 0.0274 W/m ⋅ K, ν = 17.40 × 10 -6 m 2 /s, α = 24.7 × 10 -6 m 2 /s, Pr = 0.705. (a) From Newton’s law of cooling, s conv 4 s TT q hA ∞ =+ (1) where 4 h is the average heat transfer coefficient over the 4th device. Since flow is turbulent, it is reasonable and convenient to assume that () 4x hhL15mm == .(2) To estimate h x , use the turbulent correlation evaluating thermophysical properties at f T = 315 K (assume T s = 330 K), 4/5 1/3 xx Nu 0.0296Re Pr = where x 62 u L 10m s 0.015m Re 8621 17.4 10 m s ν ∞ − × == = × giving ()( ) 4/5 1/3 x x hL Nu 0.0296 8621 0.705 37.1 k == = 2 x 4x Nu k 37.1 0.0274W m K h h 67.8W m K L 0.015m ×⋅ == = = ⋅ Hence, with A s = 4 mm × 4 mm, the surface temperature is ( ) 3 s 2 23 40 10 W T 300K 337K 64 C 67.8W m K 4 10 m − − × =+ == ⋅×× . Continued (b) The surface temperature for each of the four devices (i = 1, 2, 3 4) follows from Eq. (1), s,i conv i s TTq hA ∞ =+ (3) For devices 2, 3 and 4, i h is evaluated as the local coefficient at the mid-positions, Eq. (2), x 2 = 6.5 mm, x 3 = 10.75 mm and x 4 = 15 mm. For device 1, 1 h is the average value 0 to x 1 , where evaluated x 1 = L 1 = 4.25 mm. Using Eq. (3) in the IHT Workspace along with the Correlations Tool, External Flow, Local Coefficient for Laminar or Turbulent Flow, the surface temperatures T s,i are determined as a function of the free stream velocity. 5 7 9 11 13 15 Free stream velocity, uinf (m/s) 40 50 60 70 80 90 100 Surface temperature, Ts (C) Device 1 Device 2 Device 3 Device 4 (c) Using the Explore option on the Plot Window associated with the IHT code of part (b), the minimum free stream velocity of u ∞ = 6.6 m/s will maintain device 4, the hottest of the devices, at a temperature T s,4 = 80 ° C. (1) Note that the thermophysical properties were evaluated at a reasonable assumed film temperature in part (a). (2) From the T s,i vs. u ∞ plots, note that, as expected, the surface temperatures of the devices increase with distance from the leading edge. PROBLEM 7.34 KNOWN: Convection correlation for irregular surface due to electronic elements mounted on a circuit board experiencing forced air cooling with prescribed temperature and velocity FIND: Surface temperature when heat dissipation rate is 30 mW for chip of prescribed area located a specific distance from the leading edge. SCHEMATIC: ASSUMPTIONS: (1) Situation approximates parallel flow over a flat plate with prescribed correlation, (2) Heat rate is from top surface of chip. PROPERTIES: Table A-4, Air (assume T s ≈ 45°C, then T = (45 + 25)°C/2 ≈ 310 K, 1 atm): k = 0.027 W/m⋅K, ν = 16.90 × 10 -6 m 2 /s, Pr = 0.706. ANALYSIS: For the chip upper surface, the heat rate is ( ) chipchipssschipchips qhATT or TTq/hA ∞∞ =−=+ Assuming the average convection coefficient over the chip length to be equal to the local value at the center of the chip (x = x o ), ( ) chipxo hhx, ≈ where 0.850.33 xx Nu0.04RePr= ( ) ( ) 0.85 0.33 -62 x Nu0.0410 m/s0.120 m/16.9010 m/s0.706473.4 =××= 2 x x o Nuk473.40.027 W/mK h107 W/mK x0.120 m ×⋅ ===⋅ Hence, ( ) ( ) 2 323 s T25C3010 W/107 W/mK410m2517.5C42.5C. −− =+×⋅××=+= o oo < COMMENTS: (1) Note that the assumed value of T used to evaluate the thermophysical properties was reasonable. (2) We could have evaluated chip h by two other approaches. In one case the average coefficient is approximated as the arithmetic mean of local values at the leading and trailing edges of the chip. ( ) ( ) 2 chipx22x11 hhxhx/2107 W/mK. ≈+=⋅  The exact approach is of the form h h x h x chip x2 2 x1 1 ⋅ = ⋅ − ⋅l Recognizing that h x ~ x -0.15 , it follows that x 0 xxx 1 hhd x1.176h x =∫⋅= and h W / m K. chip 2 = ⋅108 Why do results for the two approximate methods and the exact method compare so favorably? Air at atmospheric pressure and a temperature of 25 ° C in parallel flow at a velocity of 5 m/s over a 1-m long flat plate with a uniform heat flux of 1250 W/m 2 . (a) Plate surface temperature, T s (L), and local convection coefficient, h x (L), at the trailing edge, x = L, (b) Average temperature of the plate surface, s T , (c) Plot the variation of the plate surface temperature, T s (x), and the convection coefficient, h x (x), with distance on the same graph; explain key features of these distributions. (1) Steady-state conditions, (2) Flow is fully turbulent, and (3) Constant properties. Table A-4, Air (assume T f = 325 K, 1 atm): ν = 18.76 × 10 -6 m 2 /s; k = 0.0284 W/m ⋅ K; Pr = 0.703 (a) At the trailing edge, x = L, the convection rate equation is () () scvx s qq hLTLT ∞ ′′ ′′  == −  (1) where the local convection coefficient, assuming turbulent flow, follows from Eq. 7.51. 4/5 1/3 x xx hx Nu 0.0308 Re Pr k == (2) With x = L = 1m, find 62 5 x Re u L/ 5 m /s 1 m/18.76 10 m /s 2.67 10 ν − ∞ ==× × =× () ( ) ( ) () 4/5 1/3 52 x h L 0.0284 W / m K /1m 0.0308 2.67 10 0.703 17.1 W / m K =⋅×× =⋅ Substituting numerical values into Eq. (1), () 22 s T L 25 C 1250 W / m /17.1 W / m K 98.3 C =°+ ⋅= ° (b) The average surface temperature s T follows from the expression () L L 0 0 s ss x q 1x TT TTdx dx LLkNu ∞∞ ′′ −= − = ∫ ∫ (3) where Nu x is given by Eq. (2). Using the Integral function in IHT as described in Comment (3) find s T 86.1 C. =° (c) The variation of the plate surface temperature T s (x) and convection coefficient, h x (x), shown in the graph are calculated using Eqs. (1) and (2). Continued … [...]... W m ⋅ K 0.01m     5 / 8 4 / 5 2996     282, 000     = 28.1 28.1 = 75. 5 W m 2 ⋅ K q′ = hπ D ( Ts − T∞ ) = 75. 5 W m 2 ⋅ K π ( 0.01m ) (50 − 20 ) C = 71.1W m  Fluid: Saturated Water (5 m s ) 0.01m VD = = 68, 55 2 Re D = ν 7 25 × 10−6 N ⋅ s m 2 994 kg m3 0.62 ( 68, 55 2 ) 1/ 2 Nu D = 0.3 +  ( 4. 85 )1/ 3   1+ 1/ 4 1 + ( 0.4 4. 85 )2 / 3        5 / 8 4 / 5 68, 55 2     282, 000 ... m/s and ≤ 50 s when u ∞ = 5 m/s) 60 Temperatures, Ts (C) ln(theta/thetai) 0 -1 -2 50 40 30 0 50 100 150 200 250 300 0 50 100 Elapsed time, t (s) 150 200 250 300 Elapsed time, t (s) uinf = 3 m/s unif = 5 m/s uinf = 3 m/s unif = 5 m/s Selecting two elapsed times at which to evaluate h L , the following results were obtained u ∞ (m/s) t (s) Ts (t), (°C) h L (W/m2⋅K) Nu L 3 9 100 50 44.77 45. 80 30.81 56 .7... T(K) 600 55 0 50 0 450 400 350 300 0 50 100 150 200 250 300 Time, t(s) Without radiation With radiation, eps = 0.8 Continued The agreement between predictions with and without radiation for t < 50 s implies negligible radiation However, as the heater temperature increases with time, radiation becomes significant, yielding a reduced heater temperature Steady-state temperatures correspond to 56 2.4 K and 602.8... 602.8 K, with and without radiation, respectively The time required for the heater to reach 55 2.4 K (with radiation) is t ≈ 155 s (c) If the heater temperature is to be maintained at a fixed value in the face of velocity excursions, provision must be made for adjusting the heater power Using the Explore and Graph options of IHT with the model of part (a), the following results were obtained 950 Power dissipation,... obtained 950 Power dissipation, P'elec 900 850 800 750 700 650 5 6 7 8 9 10 Freestream velocity, V(m/s) For T = 2 75 C = 54 8 K, the controller would compensate for velocity reductions from 10 to 5 m/s by reducing the power from approximately 9 35 to 690 W/m Although convection heat transfer substantially exceeds radiation heat transfer, radiation is not negligible and should be included in the analysis If...h_x(x) and Ts(x) 100 80 60 40 20 0 0 0.2 0.4 0.6 0.8 1 Distance from leading edge, x (m) h_x, W/m^2-K Ts_x, C (1) To avoid performing the integration of part (b), it is reasonable to use the approximate, simpler Eqs 7 .53 a and integrating Eq 7 .51 , 4 /5 Nu L = 0.03 85 Re 4 / 5 Pr1/ 3 = 0.03 85 2.67 × 1 05 (0.703)1/ 3 = 751 L ( ) h L = Nu L k / L = 751 × 0.0284 W / m ⋅ K /1 m = 213... ( T = Tf = 350 K): k = 15. 8 ANALYSIS: (a) Maximum heat rate from fin occurs when fin is infinitely long, 1/2 q f = M = hPkAc θb from Eq 3.80 Estimate convection heat transfer coefficient for cross-flow over cylinder, VD Re D = = 5 m/s × 0.0 05 m/20.92 ×10 -6m 2 /s = 11 95 ( ) (1) ν Using the Hilpert correlation, Eq 7 .55 , with Table 7.2,find k h = CRemPr = ( 0.030W/m ⋅ K/0.005m ) 0.683( 11 95) 0.466 ( 0.700)1/3... 0.71)1 / 3 0.0 251 Using a numerical technique to evaluate the integral, 3 x −1/2 dx q = 27 .50 ∫ Pr1/3 kW ∫ W m ⋅K = 27 .50 × 1.417 = 39 W 2m ∫ L ξ x −1/2 dx 1/3 1 − (ξ /x )3 / 4      < 2 COMMENTS: Values of h with and without the unheated starting length are 3.9 and 5. 5 W/m ⋅K Prior development of the velocity boundary layer decreases h PROBLEM 7.38 KNOWN: Surface dimensions for an array of 10 silicon... number, NuLbar 350 250 150 50 20000 40000 60000 80000 Reynold's number, ReLbar exp lam turb (1) A more extensive analysis of the experimental observations would involve determining Nu L for the full range of elapsed time (rather than at two selected times) and using a fitting routine to determine values for C and m Cylinder diameter and surface temperature Temperature and velocity of fluids in cross... 2.174 ×1 05 Hence, flow is laminar over all chips without the promoter (a) For laminar flow, the minimum hx exists on the last chip Approximating the average coefficient for Chip 10 as the local coefficient at x = 95 mm, h10 = hx = 0.095m k h10 = 0. 453 Re1/2 Pr1/3 x x u x 40 m/s × 0.0 95 m Re x = ∞ = = 2.0 65 × 10 5 -6 2 ν 18.4 × 10 m / s 1/2 0.0282 W/m ⋅ K h10 = 0. 453 2.0 65 × 1 05 ( 0.703)1 / 3 = 54 .3W/m2 . at x = 95 mm, 10x0.095m hh. = = 1/21/3 10x k h0. 453 RePr x = 5 x -62 ux40 m/s0.0 95 m Re2.0 651 0 18.410 m/s ν ∞ × ===× × ( ) ( ) 1/2 1/3 52 10 0.0282 W/mK h0. 453 2.0 651 00.70 354 .3W/mK 0.0 95 ⋅ =×=⋅ (. m/s and ≤ 50 s when u ∞ = 5 m/s). 0 50 100 150 200 250 300 Elapsed time, t (s) -2 -1 0 ln(theta/thetai) uinf = 3 m/s unif = 5 m/s 0 50 100 150 200 250 300 Elapsed time, t (s) 30 40 50 60 Temperatures,. with the Explore and Graph options of IHT. 5 10 15 20 25 Freestream velocity, uinf(m/s) 0 50 0 1000 150 0 2000 250 0 3000 350 0 4000 Power dissipation, q(W) Continued The effect of increasing u ∞