Kreith SI - Ch-06, Notas de estudo de Cultura

Baixe Kreith SI - Ch-06 e outras Notas de estudo em PDF para Cultura, somente na Docsity! © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 513 Chapter 6 PROBLEM 6.1 To measure the mass flow rate of a fluid in a laminar flow through a circular pipe, a hot wire type velocity meter is placed in the center of the pipe. Assuming that the measuring station is far from the entrance of the pipe, the velocity distribution is parabolic, or max ( )u r U = ( )    22 1 – r D where Umax is the centerline velocity (r = 0) r is the radial distance from the pipe centerline D is the pipe diameter. (a) Derive an expression for the average fluid velocity at the cross-section. (b) Obtain an expression for the mass flow rate. (c) If the fluid is mercury at 30°C, D = 10 cm, and the measured value of Umax is 0.2 cm/s, calculate the mass flow rate from the measurement. GIVEN • Fully developed flow of mercury through a circular pipe • Parabolic velocity distribution: u(r)/Umax = 1 – (2r/D)2 • Mercury temperature (T) = 30°C • Pipe diameter (D) = 10 cm = 0.1 m • Measured center velocity (Umax) = 0.2 cm/s = 0.002 m/s FIND (a) An expression for the average fluid velocity ( u ) (b) An expression for the mass flow rate ( m ) (c) The value of the mass flow rate ( m ) SKETCH PROPERTIES AND CONSTANTS From Appendix 2, Table 25, for Mercury at 30°C: Density (ρ) = 13,555 kg/m3 SOLUTION (a) The average fluid velocity is calculated as follows u = 1 or ( ) or o u r dr where ro = 2 D © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 514 u = 2 max 1 or o o o U r dr r r   −       = 3 max 2 0 1 3 or o o U r r r r   −   u = Umax 1 1 3  −   = 2 3 Umax (b) The mass flow rate is given by m = u Ac ρ = 2 3 Umax(π ro2)ρ m = 2 3 π Umax ro2 ρ (c) Inserting the values of these quantities into this expression m = 2 3 π ( )0.002 m/s (0.05 m)2 ( )313,555 kg/m = 0.14 kg/s PROBLEM 6.2 Nitrogen at 30°C and atmospheric pressure enter a triangular duct 0.02 m on each side at a rate of 4 × 10–4 kg/s. If the duct temperature is uniform at 200°C, estimate the bulk temperature of the nitrogen 2 m and 5 m from the inlet. GIVEN • Atmospheric nitrogen flowing through a triangular duct • Bulk inlet temperature (Tb,in) = 30°C • Width of each side of the duct (w) = 0.02 m • Mass flow rate ( m ) = 4 × 1–4 kg/s • Duct temperature (Ts) = 200°C (uniform) FIND • The bulk temperature (a) 2 m from the inlet and, (b) 5 m from the inlet SKETCH w = 0 . 02 m Nitrogen = 30° C = 4 x 10 kg/s T m b,in – 4 x SOLUTION (a) Assuming the outlet temperature is 70°C, then the average bulk temperature is 50°C From Appendix 2, Table 32, for nitrogen Specific heat (cp) = 1042 J/(kg K) Thermal conductivity (k) = 0.0278 W/(m K) Absolute viscosity (μ) = 18.79 × 10–6 (Ns)/m2 Prandtl Number (Pr) = 0.71 The hydraulic diameter of the duct is DH = 4 cA P = 2 21 W4 W W 2 2 3W   −     = 4 24(1.73 10 m ) 3(0.02m) −× = 0.0115 m © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 517 Therefore, entrance effects will be neglected. From Appendix 2, Table 27, for dry air at the average bulk temperature of 49.7°C Thermal conductivity (k) = 0.0272 W/(m K) Absolute viscosity (μ) = 19.503 × 10–6 (Ns)/m2 Density (ρ) = 1.015 kg/m3 The Reynolds number is ReD = U Dρ μ ∞ = w Hm D H μ  = ( ) ( ) ( )6 2 2 0.0004 kg/s (0.016m)(0.004 m) 19.503 10 (Ns)/m (kg m)/(Ns )−× = 2051 < 2100 Therefore, the flow is laminar. The Nusselt number for this geometry is given in Table 6.1 For 2 2 b a = 0.008 0.032 = 0.25, DNu = 2H Nu = 2.93 ∴ hc = NuD H k D = 2.93 ( )0.0272 W/(m K) 0.0064 m = 12.5 W/(m2 K) The average surface temperature is Ts = 30 C 69.4 C 2 ° + ° + ( ) ( ) 2 2 500 W/m 12.5 W/(m K) = 90°C From Table 6.1 for 2b/2a = 1/4, fReD = 72.93 ∴ f = 72.93 ReD = 72.93 2051 = 0.0356 The pressure drop is given by Equation (6.13) Δp = f 2 2H c L U D g ρ = f 2 1 2 w HH c L m D g ρ       Δp = 0.0356 1m 0.0064m ( ) ( ) 2 2 3 1 0.0004 kg/s (0.016m)(0.004 m)2 (kg m)/(Ns ) 1.059 kg/m      = 102 Pa PROBLEM 6.4 Engine oil flows at a rate of 0.5 kg/s through a 2.5 cm ID tube. The oil enters 25°C while the tube wall is at 100°C. (a) If the tube is 4 m long. Determine whether the flow is fully developed. (b) Calculate the heat transfer coefficient. GIVEN • Engine oil flows through a tube • Mass flow rate ( m ) = 0.5 kg/s • Inside diameter (D) = 2.5 cm = 0.025 m • Oil temperature at entrance (Ti) = 25°C • Tube surface temperature (Ts) = 100°C • Tube length (L) = 4 m FIND (a) Is flow fully developed? (b) The heat transfer coefficient (hc) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 518 ASSUMPTIONS • Steady state SKETCH D = 2.5 cm Engine Oil = 0.5 kg/s = 25 m Ti ° C L = 4 m PROPERTIES AND CONSTANTS From Appendix 2, Table 16, for unused engine oil at the initial temperature of 25°C Density (ρ) = 885.2 kg/m3 Thermal conductivity (k) = 0.145 W/(m K) Absolute viscosity (μ) = 0.652 (Ns)/m2 Prandtl number (Pr) = 85.20 Specific heat (c) = 1091 J/(kg K) SOLUTION The Reynolds number is ReD = 4V D m D ρ μ π μ =  = ( ) ( ) ( )2 2 4 0.5kg/s (0.025m) 0.652 Ns/m kg m/(Ns )π = 39.1 Therefore, the flow is laminar. (a) The entrance length at which the velocity profile approaches its fully developed shape is given by Equation (6.7) fdx D = 0.05 ReD  xfd = 0.05 D ReD = 0.05 (0.025 m) (39.1) = 0.049 m = 4.9 cm Therefore, the velocity profile is fully developed for 98.8% of the tube length. The entrance length at which the temperature profile approaches its fully developed shape is given by Equation (6.8) fdx D = 0.05 ReD Pr  xfd = 0.05 D ReD Pr = 0.05(0.025 m) (39.1) (8520) = 416 m Therefore, the temperature profile is not fully developed. (b) Since the velocity profile is fully developed but the temperature profile is not, Figure 6.10 will be used to estimate the Nusselt number 210D Re PrD L −× = (39.1) (85.20) (0.025m) 4m × 10–2 = 0.208 Using the ‘parabolic velocity’ curve of Figure 6.12, NuD ≈ 4.8 hc = NuD k D = 4.8 ( )0.145 W/(m K) 0.025m = 27.8 W/(m2 K) COMMENTS The rate of heat transfer calculated with the heat transfer coefficient at the inlet is qmax = hc π D L (Ts – Tb) = ( )227.8 W/(m K) π (0.025 m) (4 m) (100°C – 25°C) = 656 W The outer temperature (To) is given by qmax = m c (To,max – Ti) To – Ti ≤ max q mc = ( ) ( ) ( ) 656W J/(Ws) 0.5kg/s 1091J/(kg K) = 1.2°C © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 519 This small temperature change does not warrant another iteration. If the temperature change was larger, the fluid properties would need to be re-evaluated at the average bulk temperature and a new heat transfer coefficient calculate. PROBLEM 6.5 The equation c h D Nu k = =                   2 3 0.0668 3.65 1 0.04 D RePr L D RePr L + + =     0.14 b s μ μ was recommended by H. Hausen (Zeitschr. Ver. Deut. Ing., Belherft No. 4, 1943) for forced-convection heat transfer in fully developed laminar flow through tubes. Compare the values of the Nusselt number predicted by Hausen’s equation for Re = 1000, Pr = 1, and L/D = 2, 10 and 100, respectively, with those obtained from two other appropriate equations or graphs in the text. GIVEN • Fully developed laminar flow through a tube • The Nusselt number correlation shown above • Reynolds number (Re) = 1000 • Prandtl number (Pr) = 1 • Length divided by diameter (L/D) = 2, 10, or 100 FIND • The Nusselt number (Nu) from the above correlation and two others from the text ASSUMPTIONS • μb / μs ≈ 1.0 • Constant wall temperature SKETCH Fluid Flow Pr 1ª Re = 1000 D L SOLUTION Using the Hausen correlation and L/D = 2 Nu = c h D k = 2 3 0.14 1 0.0668 (1000) (1) 23.65 1 1 0.04 (1000) (1) 2 b s μ μ         +        +        = 13.1 0.14 b s μ μ      ≈ 13.1 Similarly for the other cases For L D = 10 → Nu ≈ 7.2 For L D = 100 → Nu ≈ 4.2 © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 522 HD Nu = 0.14 0.66 0.0668(1760)(0.71) (4.44) 23.683 3.66 33.6661 0.045[(1760) (0.71) (4.44)]    +     +  = 28.3 ch = HD Nu H k D = 28.3 ( )0.0339 W/(m K) 0.1m = 9.59 W/(m2 K) (b) Applying the flow-over-flat-plate relation of Equation (6.38) HD Nu = ln 4 HD H Re Pr D L ( ) 0.50.167 1 2.654 1 H H D D LPr Re Pr         −        HD Nu = (1760) (0.71) 4 (4.44) ln 0.167 0.5 1 2.654 1 (0.71) [1760 (0.71) (4.44)]        −    = 53.3 ch = HD Nu H k D = 53.3 ( )0.0339 W/(m K) 0.1m = 18.1 W/(m2 K) COMMENTS The flat plate estimate is almost twice the previous estimate based on flow through a short duct. It should be noted that the flow-over-flat-plate relation is only applicable in the following range: [ReD Pr (D/L)] from 100 to 1500. For this problem, ReD Pr D/L = 5548. PROBLEM 6.7 Water enters a double pipe heat-exchanger at 60°C. The water flows on the inside through a copper tube 2.54 cm (1 in) ID at a velocity of 2 cm/s. Steam flows in the annulus and condenses on the outside of the copper tube at a temperature of 80°C. Calculate the outlet temperature of the water if the heat exchanger is 3m long. GIVEN • Water flow through a tube in a double pipe heat-exchanger • Water entrance temperature (Tb,in) = 60°C • Inside tube diameter (D) = 2.54 cm = 0.0254 m • Water velocity (V) = 2 cm/s = 0.02 m/s • Steam condenses at (Ts) = 0.80°C on the outside of the pipe • Length of heat exchanger (L) = 3 m FIND • Outlet temperature of the water (Tb,out) ASSUMPTIONS • Steady state • Thermal resistance of the copper pipe is negligible • Pressure in the annulus is uniform therefore, Ts is uniform • Heat transfer coefficient of the condensing steam is large (see Table 1.4) so its thermal resistance can be neglected • Outside surface of the heat exchanger is insulated © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 523 SKETCH Steam = 80°CTs Tb,in = 60°C L = 3 m Copper Pipe = 2.54 cmID Water PROPERTIES AND CONSTANTS From Appendix 2, Table 13, for water at the inlet temperature of 60°C Specific heat (c) = 4182 J/(kg K) Thermal conductivity (k) = 0.657 W/(m K) Kinematic viscosity (ν) = 0.480 × 10–6 m2/s Prandtl number (Pr) = 3.02 Density (ρ) = 982.8 kg/m3 The absolute viscosity is μb = 484 × 10–6 (Ns)/m2 at 60°C μs = 357 × 10–6 (Ns)/m2 at 80°C SOLUTION The Reynolds number is ReD = V D ν = 6 2 (0.02 m/s) (0.0254 m) 0.480 10 m /s−× = 1058 (Laminar) The thermal entrance length is given by Equation (6.8) fdx D = 0.05 ReD Pr = 0.05 (1058) (3.02) = 159.8 → xfd = 159.8 (0.0254m) = 4.06m > L Therefore, the flow is not fully developed and the Sieder and Tale correlation, Equation (6.40) will be used HD Nu = 1.86 0.140.33 b D s D Re Pr L μ μ           HD Nu = 1.86 0.33 0.140.0254 484 1058 (3.02) 3m 357               = 5.76 ch = DNu k D = 5.76 ( )0.657 W/(m K) 0.0254 m = 149.1 W/(m2 K) The outlet temperature is given by Equation (6.36) out in T T Δ Δ = ,out ,in b s b s T T T T − − = exp c p PLh mc   −   = exp 2 ( ) 4 c p D Lh V D c π πρ    −       Solving for the bulk water outlet temperature Tb,out = Ts + (Tb,in – Ts) exp c p P h L VDcρ   −   Tb,out = 80°C + (60°C – 80°C) exp ( ) ( ) ( ) ( ) ( ) 2 3 4 149.1 W/(m K) (3m) 982.8 kg/m 0.02 m/s (0.0254 m) 4182 J/(kg K) Ws/J   −   = 71.5°C © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 524 Performing a second iteration using the water properties at the average temperature of 66°C c = 4186 J/(kg K) ν = 0.434 × 10–6 m2/s ReD = 1171 ρ = 988.1 kg/m3 Pr = 2.71 DNu = 5.68 k = 0.662 W/(m K) μb = 440.9 × 10–6 (Ns)/m2 ch = 147.9 W/(m 2 K) Tb,out = 71.4°C COMMENTS The negligible change of Tb,out in the second iteration could be expected because the changes in the water properties are small. PROBLEM 6.8 An electronic device is cooled by passing air at 27°C through six small tubular passages in parallel drilled through the bottom of the device as shown below. The mass flow rate per tube is 7 × 10–5 kg/s. Air Air in 27 °C 7 × 10 – kg/s5 5.0 mm Surface Temperature-353 K Single Tubular Passage Air Out 44° C 10 cm Heat is generated in the device resulting in approximately uniform heat flux to the air in the cooling passage. To determine the heat flux, the air outlet temperature is measured and found to be 77°C. Calculate the rate of heat generation, the average heat transfer coefficient, and the surface temperature of the cooling channel at the center and at the outlet. GIVEN • Air flow through small tubular passages as shown above • Air temperature  Entrance (Tb,in) = 27°C  Exit (Tb,out) = 77°C • Mass flow rate per passage ( m )= 7 × 10–5 kg/s • Number of passages (N) = 6 FIND (a) The rate of heat generation ( GQ ) (b) The average heat transfer coefficient ( ch ) (c) Cooling channel surface temperature at the center (Ts,c) (d) Cooling channel surface temperature at the outlet (Ts,out) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 527 SOLUTION The Reynolds number for oil flow inside the pipe is ReD = 4 4 0.25 0.051 0.017b b VD m D ρ μ π μ π ×= = × ×  = 367.1  Laminar flow The thermal entrance length is given by Equation (6.8) for laminar flow, and it can be calculated as xfd = 0.05 Re Pr 0.05 0.051 367 276DD = × × × = 258 m 9L = m Hence, the temperature profile is NOT fully developed, or the flow is thermally developing. Because there is a large variation in the oil viscosity at the pipe wall temperature and the bulk temperature, the effect of property (viscosity) variation has to be considered. From Section 6.3.3 either the Hausen correlation of Equation (6.41) or the Sieder and Tate correlation of Equation (6.42) could be used because (μb/μs) = 3.1 (< 9.75; the limit for Equation (6.42) to calculate the Nusselt number. Thus, using the more simpler Sieder and Tate correlation NuD = 0.14 Re Pr 1.86 bD s D L μ μ           0.14367 276 0.051 0.0171 1.86 1251 9 0.00552 × ×   = =        ch = 0.137 Nu 1251 3361 0.051D k D = = W/(m2 K) The outlet temperature can now be calculated by Equation (6.36) as out in T T Δ Δ = exp c p h PL mc   −    ( ), , exp cb out s b in s p h PL T T T T mc   = + − −   ∴ Tb, out = ( ) 3361 0.051 9 150 100 150 exp 0.25 2219 π× × × + − −  × = 149.9 ≈150°C COMMENTS The oil flow attains the tube wall (or the condensing steam) temperature at the outlet of the 9-m-long pipe. Also, because of the 50°C temperature difference between the inlet and the outlet, the above calculation should be repeated after evaluating the properties at the average temperature between the inlet and outlet. PROBLEM 6.10 Determine the rate of heat transfer per foot length to a light oil flowing through a 1-in.-ID, 0.6 m copper tube at a velocity of 0.03 m/s. The oil enters the tube at 15°C and the tube is heated by steam condensing on its outer surface at atmospheric pressure with a heat transfer coefficient of 11.3 kW/(m2 K). The properties of the oil at various temperatures are listed in the accompanying tabulation T(°C) 15 30 40 65 100 ρ (kg/m3) 938.8 938.8 922.3 906 889.4 c (kJ/(kg K)) 1.8 1.84 1.92 2.0 2.13 k (W/(m K)) 0.133 0.133 0.131 0.129 0.128 μ (kg/ms) 0.089 0.0414 0.023 0.00786 0.0033 Pr 1204 573 338 122 55 © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 528 GIVEN • Oil flowing through a copper tube with atmospheric pressure steam condensing on the outer surface • Oil properties listed above • Inside diameter (D) = 2.5 cm • Tube length (L) = 0.6 m • Oil velocity (V) = 0.03 m/s • Inlet oil temperature (Tb,in) = 16°C • Heat transfer coefficient on outside of pipe ( ,c oh )= 11.3 × 10 3 W/(m2 K) FIND • The rate of heat transfer (q) to the oil ASSUMPTIONS • Steady state • The thermal resistance of the copper tube is negligible • Constant wall temperature • The tube wall is thin SKETCH Steam Oil Tb,in = 15°C V = 0.03 m/s Copper Tube ID = 2.5 cm L = 0.6 m PROPERTIES AND CONSTANTS At atmospheric pressure, steam condenses at a temperature (Ts) of 100°C. SOLUTION The Reynolds number for the oil flowing through the pipe is ReD = V D ρ μ Using the oil properties at the inlet temperature of 15°C ReD = –2(0.03m/s) (2.5 10 m)(938.8kg/m) (0.089kg/ms) × = 7.91 (Laminar) The thermal entrance length is given by Equation (6.8) fdx D = 0.05 ReD Pr = 0.05 (7.91) (1204) = 476  xfd = 476 (0.025 m) = 11.9 m >> L Therefore, the temperature profile is not fully developed and the Hausen correlation of Equation (6.39) will be used (assuming the wall temperature ≈ Ts for μs) Nu = 0.14 0.66 0.0668 3.66 1 0.045 H H D b s D D Re Pr L D Re Pr L μ μ          +        +        © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 529 Nu = –2 0.14 0.66 2.5 10 0.0668(7.91) (1204) 0.6 0.089 3.66 0.00330.025 1 0.045 (7.91) (1204) 0.6    ×       +       +      = 26.51 3.66 (1.586) 2.335  +   = 23.8 ch = D k Nu D = 23.8 0.133W/(m K) 0.025m = 126.6 W/(m2 K) The thermal circuit for heat flow from the steam to the oil is shown below R =c,o RK = CO To R =c,o 1 h Aci i Ts 1 h Aco o If the tube wall is thin, Ao ≈ Ai = πDL = π(0.025 m)(0.6 m) = 0.0471 m2 and the thermal resistance is A Rco = 2 1 11.3kW/(m K) = 8.85 × 10–5 (m2 K)/W A Rco = 2 1 126.6 W/(m K) = 7.9 × 10–3 (m2 K)/W A Rtotal = A Rco + A Rci = (0.0885 + 7.9) × 10–3 (m2K)/W = 7.99 × 10–3 (m2 K)/W The outlet temperature can be calculated by replacing hcA by 1/ARtotal in Equation (6.36) out in T T Δ Δ = ,out ,in b s b s T T T T − − = exp total( ) p P L A R m c   −   = exp total( ) ( ) p DL A R VA c π ρ   −   Tb,out = Ts + (Tb,in – Ts) exp total 4 ( ) p L A R D VA cρ   −   Tb,out = 100 + (15 – 100) exp –3 2 3 4(0.6m) – (7.99 10 (m K)/W)(0.025m)(938.8 kg/m )(0.03 m/s)(1800J/kg K)    ×   Tb,out = 100 – 85 exp 2.4 – 10.126      = 32.9°C The mean temperature of oil = 15 32.9 2 +  24°C. Hence the properties used above may change significantly at this temperature. This is a significant change in the oil temperature and warrants another iteration using the properties of the oil at the average bulk temperature of 24°C. Interpolating the oil properties from the given data ρ = 938.8 kg/m3 Re = 11.93 c = 1840 J/(kg K) DNu = 21.6 k = 0.133 W/(m K) A Rtotal = 8.15 × 10–3 ((m2 K)/W) μb = 0.059 kg/ms Tb,out = 32.3°C Pr = 816 © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 532 • Steam temperature (Ts) = 149°C • Air velocity (V) = 6 m/s • Air bulk temperature (Tb) = 38°C FIND • The heat transfer coefficient ( ch ) ASSUMPTIONS • Steady state • Steam temperature is constant and uniform • Heat transfer to the outer surface is negligible • Air temperature given is the average air temperature • Thermal resistance of inner tube wall and condensing steam is negligible (Inner tube wall surface temperature = Ts) SKETCH Do = 38 cmDi = 25 cm Air = 38°C = 6 m/s Tb V Steam = 149°CTs PROPERTIES AND CONSTANTS From Appendix 2, Table 27, for dry air at 38°C Density (ρ) = 1.099 kg/m3 Thermal conductivity (k) = 0.0264 W/(m K) Absolute viscosity (μb) = 19.0 × 10–6 (Ns)/m2 Prandtl number (Pr) = 0.71 At the surface temperature of 149°C μs = 23.7 × 10–6 (Ns)/m2 SOLUTION As shown in Equation (6.3), the hydraulic diameter of the annulus is given by DH = Do – Di = 0.38 m – 0.25 m = 0.13 m The Reynolds number based on this diameter is ReD = HV D ρ μ = ( ) ( ) ( ) 3 6 2 2 (6 m/s) (0.13m) 1.099 kg/m 19.035 10 (Ns)/m (kg m)/(s N)−× = 4.50 × 104 (Turbulent) Applying the Seider-Tale correlation of Equation (6.64) DNu = 0.027 ReD 0.8 Pr0.3 0.14 b s μ μ      = 0.027 (4.50 × 104)0.8 (0.71)0.3 0.1419.0 23.7      = 125 ch = DNu k D = 125 ( )0.0264 W/(m K) 0.13m = 25.4 W/(m2 K) PROBLEM 6.13 If the total resistance between the steam and the air (including the pipe wall and scale on the steam side) in Problem 6.12 is 0.05 m2 K/W, calculate the temperature difference between the outer surface of the inner pipe and the air. Show the thermal circuit. From Problem 6.12: In a long annulus (25 cm ID, 38 cm OD), atmospheric air is heated by steam condensing at 149°C on the inner surface. The velocity of the air is 6 m/s and its bulk temperature is 38°C. © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 533 GIVEN • Atmospheric flow through an annulus with steam condensing in inner tube • Diameters Inside  (Di) = 25 cm = 0.25 m  Outside (Do) = 38 cm = 0.38 m • Steam temperature (Ts) = 149°C • Air velocity (V) = 6 m/s • Total resistance between the steam and air (At Rtot) = 0.05 (m2 K)/W • Air bulk temperature (Tb) = 38°C • From Problem 6.12 heat transfer coefficient on the outer surface of the inner pipe ( ch ) = 25.4 W/(m2 K) FIND • The temperature difference between the outer surface of the inner pipe and the air (ΔT) ASSUMPTIONS • Steady state • Steam temperature is constant and uniform • Heat transfer to the outer surface is negligible • Air temperature given is the average air temperature • Thermal resistance of inner tube wall and condensing steam is negligible (Inner tube wall surface temperature = Ts) SKETCH Do = 38 cmDi = 25 cm Air = 38°C = 6 m/s T V b Steam = 149°CTs SOLUTION The thermal circuit for the heat transfer between the steam and the air is shown below Ts Rc,s Tb Rk,s Rk,p Rc,a where Rc,s = Convective thermal resistance on the steam side Rk,s = Conductive thermal resistance of scaling on the steam side Rk,p = Conductive thermal resistance of the pipe wall Rc,a = Convective thermal resistance on the air side = 1/At ch RTot = Rc,s + Rk,s + Rk,p + Rc,a Rca = 1 t cA h → At Rca = 1 ch = ( )2 1 25.4 W/(m K) = 0.0394 (m2 K)/W The total rate of heat transfer must equal the rate of convective heat transfer from the pipe wall to the air total s bT T R − = ca T R Δ ΔT = total caR R (Ts – Tb) = total t ca t A R A R (Ts – Tb) = 0.0394 0.05      (149°C – 38°C) = 87.4°C © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 534 COMMENTS Note that 79% of the thermal resistance is the convective resistance on the air side. PROBLEM 6.14 Atmospheric air at a velocity of 61 m/s and a temperature of 16°C enters a 0.61-m-long square metal duct of 20 × 20 cm cross section. If the duct wall is at 149°C, determine the average heat transfer coefficient. Comment briefly on the L/Dh effect. GIVEN • Atmospheric air flow through a square metal duct • Air velocity (V) = 61 m/s • Inlet air temperature (Tb,in) = 16°C • Duct dimensions: 20 cm × 10 cm × 0.61 m = 0.2 m × 0.2 m × 0.61 m • Duct wall surface temperature (Ts) = 149°C FIND • The average heat transfer coefficient ( ch ) ASSUMPTIONS • Steady state • Constant and uniform wall surface temperature SKETCH L = 0.61 m 20 cm 20 cm Air V = 61 m/s Tb,in = 16°C PROPERTIES AND CONSTANTS From Appendix 2, Table 27, for dry air at the inlet temperature of 16°C Density (ρ) = 1.182 kg/m3 Thermal conductivity (k) = 0.0248 W/(m K) Absolute viscosity (μb) = 18.08 × 10–6 (Ns)/m2 Prandtl number (Pr) = 0.71 Specific heat (c) = 1012 J/(kg K) At the wall temperature of 149°C μs = 23.8 × 10–6 (Ns)/m2 SOLUTION The hydraulic diameter of the duct is given by Equation (6.2) DH = 4 cA P = 4(0.2m)(0.2m) 4(0.2m) = 0.2 m  = H L D = 0.61m 0.2m = 3.05 The Reynolds number based on the hydraulic diameter is ReD = HV D ρ μ = ( ) ( ) ( ) 3 6 2 2 (61 m/s) (0.2 m) 1.182 kg/m 18.08 10 (Ns)/m (kg m)/(Ns )−× = 7.97 × 105 (Turbulent) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 537 3. Using the Petukhov-Popov correlation of Equation (6.66) NuD = 1 2 2 3 1 2 8 ( 1) 8 D f Re Pr f K K Pr        + −   where f = (1.82 log(ReD) – 1.64) –2 = (1.82 log(7.69 × 104) – 1.64)–2 = 0.0190 K1 = 1 + 3.4 f = 1 + 3.4(0.019) = 1.065 K2 = 11.7 + ( )13 1.8 Pr = 11.7 + ( )13 1.8 9.5 = 12.55 NuD = 4 1 2 2 3 0.019 (7.69 10 ) (9.5) 8 0.019 1.065 12.55 (9.5) 1 8   ×      + −        = 543 hc = NuD k D = 543 ( )0.577 W/(mK) 0.025m = 12,530 W/(m2 K) (b) The friction factor correlation of Equation (6.54) is good only for 1 × 105 < ReD. Therefore, the friction factor will be estimated from the bottom curve of Figure 6.18: For Re = 7.69 × 104, f ≈ 0.0188 (Note that this is in good agreement with the friction factor, f in the Petukhov-Popov correlation). The pressure drop per unit length can be calculated from Equation (6.13) Δ P D = 2 2 c f V D g ρ = ( ) ( ) ( ) 23 2 2 0.0188 999.7 kg/m 4 m/s 0.025m 2 (Nm )/Pa (kg m)/(Ns ) = 6014 Pa COMMENTS The heat transfer coefficients vary around the average of 11,345 W/(m2 K) by a maximum of 10%. This is within the accuracy of empirical correlations. PROBLEM 6.16 Water at 80°C is flowing through a thin copper tube (15.2 cm ID) at a velocity of 7.6 m/s. The duct is located in a room at 15°C and the heat transfer coefficient at the outer surface of the duct is 14.1 W/(m2 K). (a) Determine the heat transfer coefficient at the inner surface. (b) Estimate the length of duct in which the water temperature drops 1°C. GIVEN • Water flowing through a thin copper tube in a room • Water temperature (Tb) = 80°C • Inside diameter of tube (D) = 15.2 cm = 0.152 cm • Water velocity (V) = 7.6 m/s • Room air temperature (T∞) = 15°C • Outer surface heat transfer coefficient ( coh ) = 14.1 W/(m 2 K) FIND (a) The heat transfer coefficient at the inner surface ( cih ) (b) Length of duct (L) for temperature drop of 1°C © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 538 ASSUMPTIONS • Steady state • Thermal resistance of the copper tube is negligible • Fully developed flow SKETCH D = 15.2 cm T• = 15°C Water = 80°C = 7.6 m/s T V b PROPERTIES AND CONSTANTS From Appendix 2, Table 13, for water at 80°C Density (ρ) 971.6 kg/m3 Thermal conductivity (k) = 0.673 W/(m K) Absolute viscosity (μ) = 356.7 × 10–6 (Ns)/m2 Prandtl number (Pr) = 2.13 Specific heat (c) = 4194 J/(kg K) SOLUTION The Reynolds number is ReD = V D ρ μ = ( ) ( ) ( ) 3 6 2 2 (7.6 m/s) (0.152 m) 971.6kg/m 356.7 10 (N s)/m (kg m)/(s N)−× = 3.15 × 106 (Turbulent) (a) Applying the Dittus-Boelter correlation of Equation (6.63) DNu = 0.023 ReD 0.8 Prn where n = 0.3 for cooling DNu = 0.023 (3.15 × 10 6)0.8 (2.13)0.3 = 4555 cih = DNu k D = 4555 ( )0.673 W/(m K) 0.152 m = 20,170 W/(m2 K) (b) Since the pipe wall is thin, Ao = Ai and the overall heat transfer coefficient is 1 U = 1 cih + 1 coh = 1 1 20,170 14.1  +   (m2 K)/W = 0.071 (m2 K)/W  U = 14.1 W/(m2 K) = coh The length can be calculated using Equation (6.63) out in T T Δ Δ = ,out ,in b co b co T T T T − − = exp co p P L h mc −     = exp 2 4 co p D L h D Vcπ π ρ  −      Solving for the length L = – ,out ,in ln 4 p b co b coc D V c T T T Th ρ −   −  L = – ( ) ( ) ( ) ( ) ( ) 3 2 (0.152 m) 971.6kg/m 7.6m/s 4194 J/(kg K) (Ws)/J 4 14.11W/(m K) ln 79°C 15°C 80°C 15°C −    − = 1294 m For these conditions, it would take over a kilometer for a 1°C temperature drop. This is largely the result of the small natural convection heat transfer coefficient over the outer surface. © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 539 PROBLEM 6.17 Mercury at an inlet bulk temperature of 90°C flows through a 1.2-cm-ID tube at a flow rate of 4535 kg/h. This tube is part of a nuclear reactor in which heat can be generated uniformly at any desired rate by adjusting the neutron flux level. Determine the length of tube required to raise the bulk temperature of the mercury to 230°C without generating any mercury vapor, and determine the corresponding heat flux. The boiling point of mercury is 355°C. GIVEN • Mercury flow in a tube • Inlet bulk temperature (Tb,in) = 90°C • Inside tube diameter (D) = 1.2 cm = 0.012 m • Flow rate ( m ) = 4535 kg/h = 1.26 kg/s • Outlet bulk temperature (Tb,out) = 230°C • Boiling point of mercury = 355°C FIND (a) The length of tube (L) required to obtain Tb,out without generating mercury vapor (b) The corresponding heat flux (q/A) ASSUMPTIONS • Steady state • Fully developed flow SKETCH Tb.out = 230° C Uniform Heat Flux, q/A D = 1.2 cmMercury = 90°C = 4335 kg/h T m b,in PROPERTIES AND CONSTANTS From Appendix 2, Table 25, for mercury at the average bulk temperature of 160°C Density (ρ) = 13,240 kg/m3 Thermal conductivity (k) 11.66 W/(m K) Absolute viscosity (μ) = 11.16 × 10–4 (Ns)/m2 Prandtl number (Pr) = 0.0130 Specific heat (c) = 140.6 J/(kg K) SOLUTION The Reynolds number is ReD = 4V D m D ρ π π μ =  = ( ) ( ) ( )4 2 2 4 1.26kg/s (0.012 m) 11.16 10 (Ns)/m (kg m)/(Ns )π −× = 1.2 × 102 (Turbulent) ReD Pr = 1.2 × 105 (0.013) = 1557 > 100 Therefore, Equation (6.76) can be applied to calculate the Nusselt Number DNu = 4.82 + 0.0185(ReD Pr) 0.827 = 4.82 + 0.0185 (1557)0.827 = 12.9 ch = DNu k D = 12.9 ( )11.66 W/(mK) 0.012 m = 1.25 × 104 W/(m2 K) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 542 and Nu = fdNu 1 bL a D   +      where a = 24/ReD 0.23 = 24/(3.98 × 104)0.23 = 2.10 b = 2.08 × 10–6 ReD – 0.815 = 2.08 × 10–6 (3.98 × 104) – 0.815 = –0.732 Nu = 0.023 (3.98 × 104)0.8 (0.71)0.4 [1 + 2.10(16)–0.732] = 122.5 ch = Nu k D = 122.5 ( )0.0472 W/(m K) 0.5m = 11.6 W/(m2 K) U = 2 1 1 1 (m K)/W 16 11.6  +  = 6.7 W/(m2 K) ∴ Tb,out = 280 K + (800 K – 280 K) exp ( ) ( ) ( ) ( ) 26.7 W/(m K) (0.5m)(8m) 0.5kg/s 1056J/(kg K) (Ws)/J π  −    = 723 K Another iteration using the same procedure yields Average bulk temperature = 762 K Specific Heat (cp) = 1074 J/(kg K) Thermal conductivity (k) = 0.0534 W/(m K) Absolute viscosity (μ) = 35.460 × 10–6 (Ns)/m2 Prandtl number (Pr) = 0.72 Reynolds number (ReD) = 3.59 × 104 Heat transfer coefficient ( c ih ) = 12.1 W/(m 2 K) Outlet temperature (Tb,out) = 722 K The outlet gas temperature = 722 K PROBLEM 6.19 Water at an average temperature of 27°C is flowing through a smooth 5.08-cm-ID pipe at a velocity of 0.91 m/s. If the temperature at the inner surface of the pipe is 49°C, determine (a) the heat transfer coefficient, (b) the rate of heat flow per meter of pipe, (c) the bulk temperature rise per meter, and (d) the pressure drop per meter. GIVEN • Water flowing through a smooth pipe • Average water temperature (Tw) = 27°C • Pipe inside diameter (D) = 5.08 cm = 0.0508 m • Water velocity (V) = 0.91 m/s • Inner surface temperature of pipe (Ts) = 49°C FIND (a) The heat transfer coefficient ( ch ) (b) The rate of heat flow per meter of pipe (q/L) (c) The bulk temperature rise per meter of pipe (ΔTw/L) (d) The pressure drop per meter of pipe (Δp/L) ASSUMPTIONS • Steady state • Fully developed flow © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 543 SKETCH D = 5.08 cm Ts = 49°C Water V = 0.91 m/s Tw = 27°C PROPERTIES AND CONSTANTS From Appendix 2, Table 13, for water at 27°C Density (ρ) = 996.5 kg/m3 Specific Heat (cp) = 4178 J/(kg K) Thermal conductivity (k) = 0.608 W/(m K) Absolute viscosity (μ) = 845.3 × 10–6 (Ns)/m2 Kinematic viscosity (ν) = 0.852 × 10–6 m2/s Prandtl number (Pr) = 5.8 At the surface temperature of 49°C Absolute viscosity (μs) = 565.1 × 10–6 (Ns)/m2 SOLUTION The Reynolds number for this flow is ReD = V D ν = 6 2 (0.91m/s) (0.0508m) 0.852 10 m /s−× = 5.42 × 104 > 2000 Therefore, the flow is turbulent. The variation in property values is accounted for by using Equation (6.64) to calculate the Nusselt number DNu = 0.027ReD 0.8 Pr0.3 0.14 b s μ μ      = 0.027(5.42 × 104)0.8 (5.8)0.3 0.14845.3 565.1      = 296 ch = DNu k D = 296 ( )0.608W/(m K) 0.0508m = 3543 W/(m2 K) (b) The rate of convective heat transfer is given by q = ch At (Ts – Tw) = hc π D L (Ts – Tw) q L = ( )23543 W/(m K) π (0.0508 m) (49°C – 27°C) = 12,438 W/m (c) This rate of heat transfer will lead to a temperature rise in the water given by q = pmc ΔTw = 2 4 V D πρ    cp ΔTw ∴ w T L Δ = 2 4 pV D cρ π q L      w T L Δ = ( ) ( ) ( ) ( ) ( )23 4 996.5kg/m 0.91m/s 0.0508m 4178J/(kg K) (Ws)/Jπ ( )12,438 W/m = 1.6 K/m © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 544 (d) From Table 6.4, the friction factor for fully developed turbulent flow through smooth tubes is given by Equation (6.59) f = 0.184 ReD –0.2 = 0.184 (5.42 × 104)–0.2 = 0.0208 The pressure drop is given by Equation (6.13) p L Δ = 2 2 f V D ρ = ( ) ( ) ( ) ( ) 23 2 2 0.0208 996.5kg/m 0.91m/s 2 (0.0508m) (kg m)/(s N) N/(Pa m ) = 169 Pa/m PROBLEM 6.20 An aniline-alcohol solution is flowing at a velocity of 3 m/s through a long, 2.5 cm-ID thin-wall tube. On the outer surface of the tube, steam is condensing at atmospheric pressure, and the tube-wall temperature is 100°C. The tube is clean, and there is no thermal resistance due to a scale deposit on the inner surface. Using the physical properties tabulated below, estimate the heat transfer coefficient between the fluid and the pipe by means of Equations (6.63) and (6.64), and compare the results. Assume that the bulk temperature of the aniline solution is 20°C and neglect entrance effects. Physical properties of the aniline solution Temperature Viscosity Thermal Specific Specific Heat (°C) (kg/ms) Conductivity Gravity (kJ/(kg K)) (W/(m K)) 20 0.0051 0.173 1.03 2.09 60 0.0014 0.169 0.98 2.22 100 0.0006 0.164 2.34 GIVEN • An aniline-alcohol solution flowing through a thin-walled tube • Tube is clean with no scaling on inner surface • Velocity (V) = 3 m/s • Inside diameter of tube (D) = 2.5 × 10–2 m • Tube wall surface temperature (Ts) = 100°C • Solution has the properties listed above • Solution bulk temperature (Tb = 20°C) FIND • The heat transfer coefficient ( ch ) using: (a) Equation (6.63) (b) Equation (6.64) ASSUMPTIONS • Steady state • Entrance effects are negligible • Thermal resistance of the tube is negligible • Tube wall temperature is constant and uniform • Fully developed flow SKETCH © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 547 k = ( ) ( ) ( ) ( ) ( ) 1 2 0.6 0.40.8 2 2 0.025m 16,500W/(m K) 0.023(95,313) 3768J/(kg K) 0.0016(Ns)/m (kg m)/(s N) (Ws)/J           = 0.852 W/(m K) The Prandtl number is Pr = c k μ = ( ) ( ) ( ) ( ) ( ) 2 23768J/(kg K) 0.0016(Ns)/m (kg m)/(s N) 0.852 W/(m K) J/(Ws) = 7.08 The Reynolds number for the new velocity is twice the original Reynolds number: ReD = 190,626. From Equation (6.63): For fully developed flow DNu = 0.023 (190,626) 0.8 (7.08)0.4 = 843 ch = DNu k D = 843 ( )0.852 W/(m K) 0.025m = 28,733 W/(m2 K) The temperature after one meter is given by Equation (6.36) out in T T Δ Δ = ,out ,in b co b co T T T T − − = exp c P L h mc   −   = exp 4 ch L V D cρ   −   ,out ,in b s b s T T T T − − = exp ( ) ( ) ( ) ( ) ( ) 2 3 4 38,733W/(m K) (1m) 1000kg/m 12.2 m/s (0.025m) 3768J/(kg K) (Ws)/J   −   = 0.9048 (per m length) ΔTb = Tb,out – Tb,in = ( ),out ,in ,in b s b s s b s T T T T T T T −   − +  −   – Tb,in ΔTb = [0.9043 (–1°C – 18.3°C) + 18.3°C] + 1°C = 1.84°C per meter length PROBLEM 6.22 Derive an equation of the form hc = f(T, D, V) for turbulent flow of water through a long tube in the temperature range between 20° and 100°C. GIVEN • Turbulent water flow through a long tube • Water temperature range (Tb) = 20°C to 100°C FIND • An expression of the form ch = f(T, D. V) ASSUMPTIONS • Steady state • Variation of properties with temperature can be approximated with a power law • Fully developed flow • Water is being heated © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 548 SKETCH D Water V T, b PROPERTIES AND CONSTANTS From Appendix 2, Table 13, for water Temperature (°C) 20 100 Temperature (K) 293 373 Density, ρ (kg/m3) 998.2 958.4 Thermal conductivity, k ( )W/(m K) 0.597 0.682 Absolute viscosity, ( )2(Ns) m/ 993 × 10–6 277.5 × 10–6 Prandtl number, Pr 7.0 1.75 SOLUTION Applying the Dittus-Boelter expression of Equation (6.63) for the Nusselt number DNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating DNu = 0.023 0.8 DVρ μ      Prn ch = DNu k D = 0.023 0.8 0.4 0.8 kPrρ μ D–0.2 V 0.8 To put this in the required form, the fluid properties must be expressed as a function of temperature. Assuming the power law variation Property = ATR where A and n are constant evaluated from the property values. For density ρ(293) = 998.2 kg/m3 = A(293)n ρ(373) = 958.2 kg/m3 = A(373)n Solving these simultaneously A = 2613 n = – 0.1694 Therefore, ρ(T) = 2613 T –0.1694 Applying a similar analysis for the remaining properties yields the following relationships k (T) = 0.02605 T 0.5514 μ (T) = 1.058 × 1010 T –5.281 Pr (T) = 1.026 × 1015 T –5.7426 Substituting these into the expression for the heat transfer coefficient ch = 0.023 0.1694 0.8 15 5.7426 0.4 0.5514 10 5.281 0.8 (2612 ) (1.026 10 ) (0.02605 ) (1.058 10 ) T T T T − − − × × D–0.2 V0.8 ch = 0.0031 T 2.34 D–0.2 V 0.8 © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 549 COMMENTS Note that in equations of the type derived, the coefficient has definite dimensions. Hence, the use of such equations is limited to the conditions specified and are not recommended. PROBLEM 6.23 The intake manifold of an automobile engine can be approximated as a 4 cm ID tube, 30 cm in length. Air at a bulk temperature of 20°C enters the manifold at a flow rate of 0.01 kg/s. The manifold is a heavy aluminum casting and is at a uniform temperature of 40°C. Determine the temperature of the air at the end of the manifold. GIVEN • Air flow through a tube • Tube inside diameter (D) = 4 cm = 0.04 m • Tube length (L) = 30 cm = 0.30 m • Inlet bulk temperature (Tb,in) = 20°C • Air flow rate ( m ) = 0.01 kg/s • Tube surface temperature (Ts) = 40°C FIND • Outlet bulk temperature (Tb,out) ASSUMPTIONS • Steady state • Constant and uniform tube surface temperature SKETCH Air = 20°C = 0.01 kg/s T m b,in Aluminum Manifold = 40°CTs D = 4 cm L = 30 cm Tb,out = ? PROPERTIES AND CONSTANTS From Appendix 2, Table 27, for dry air at the inlet bulk temperature of 20°C Thermal conductivity (k) = 0.0251 W/(m K) Absolute viscosity (μ) = 18,240 × 10–6 (Ns)/m2 Prandtl number (Pr) = 0.71 Specific heat (c) = 1012 J/(kg K) SOLUTION The Reynolds number is ReD = V D ρ μ = 4m Dπ μ  = ( ) ( )6 2 2 4(0.01kg/s) (0.04 m) 18.240 10 (Ns)/m (kg m)/(Ns )π −× = 17,451 (Turbulent) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 552 From (6.69) fd Nu Nu = , , c L c fd h h = 1+ 6 D L      = 1 + 6(0.015) 0.3      = 1.300 The average of the two values is ,, /c c fdh L h = 1.27 ∴ ,c Lh = 1.27 ( )210,265W/(m K) = 13,037 W/(m2 K) (b) The bulk temperature can be calculated from Equations (6.36) out in T T Δ Δ = ,out ,in s b s b T T T T − − = exp c p P L h mc   −   = exp 4 ch L V D cρ   −   Tb,out = Ts – (Ts – Tb,in) exp = 4 ch L V D cρ   −   Tb,out = 204 °C – (204°C – 93°C) exp ( ) ( ) ( ) ( ) ( ) ( ) 2 3 4 13,037W/(m K (0.3m) 963.0kg/m 1.5m/s 0.015m 4205J/(kg K) (Ws)/J   −   = 111°C The bulk temperature rise is ΔTb = Tb,out – Tb,in = 111°C – 93ºC = 18°C PROBLEM 6.25 Suppose an engineer suggests that air is to be used instead of water in the tube of Problem 6.24 and the velocity of the air is to be increased until the heat transfer coefficient with the air equals that obtained with water at 1.5 m/s. Determine the velocity required and comment on the feasibility of the engineer’s suggestion. Note that the speed of sound in air at 100°C is 387 m/s. From Problem 6.24: Water at a bulk inlet temperature of 93°C is flowing with a velocity of 1.5 m/s through a 0.015-m-diameter tube, 0.3 m long. If the tube wall temperature is 204°C, determine the average heat transfer coefficient and estimate the bulk temperature rise of the water. GIVEN • Air flow through a tube • Bulk inlet air temperature (Tb,in) = 93°C • Tube diameter (D) = 0.015 m • Tube length (L) = 0.3 m • Tube surface temperature (Ts) = 204°C • From Problem 6.23: ,c Lh = 13,037 W/(m 2 K) FIND • The velocity (V) required to obtain ,c Lh = 13,037 W/(m 2 K) ASSUMPTIONS • Steady state • Constant and uniform tube temperature © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 553 SKETCH Air = 93°CTb,in D = 0.015 m Ts = 204°C L = 0.3 m PROPERTIES AND CONSTANTS From Appendix 2, Table 27, for dry air at the inlet bulk temperature of 93°C Thermal conductivity (k) = 0.0302 W/(m K) Kinematic viscosity (ν) = 22.9 × 10–6 m2/s Prandtl number (Pr) = 0.71 SOLUTION The flow must be turbulent, therefore, the heat transfer coefficient of the fully developed case must be 13,037 W/(m2 K) as shown in Problem 6.24. Therefore, the Nusselt number is fdNu = ,c fdh D k = ( ) ( ) ( ) 213,037 W/(m K) 0.015m 0.0302 W/(m K) = 6475 Applying the Dittus-Boelter correlation of Equation (6.63) fdNu = 0.023 ReD 0.8 Prn = 5099 where n = 0.4 for heating Solving for the Reynolds number ReD = V D ν = 1.25 0.40.023 fdNu Pr      = 1.25 0.4 6475 0.023(0.71)      = 7.70 × 106 Solving for the velocity V = ReD D ν = 7.70 × 106 ( )6 222.9 10 m /s 0.015m −× = 11,749 m/s This velocity is obviously unrealistic because it corresponds to a Mach number of 30. Under such conditions when the speed of sound is reached, a shock wave will form and choke the flow. PROBLEM 6.26 Atmospheric air at 10°C enters a 2 m long smooth rectangular duct with a 7.5 cm × 15 cm cross-section. The mass flow rate of the air is 0.1 kg/s. If the sides are at 150°C, estimate (a) the heat transfer coefficient, (b) the air outlet temperature, (c) the rate of heat transfer, and (d) the pressure drop. GIVEN • Atmospheric air flow through a rectangular duct • Inlet bulk temperature (Tb,in) = 10°C • Duct length (L) = 2 m • Cross-section = 7.5 cm × 15 cm = 0.075 m × 0.15 m • Mass flow rate (m) = 0.1 kg/s • Duct surface temperature (Ts) = 150°C © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 554 FIND (a) The heat transfer coefficient ( )ch (b) The air outlet temperature (Tb,out) (c) The rate of heat transfer (q) (d) The pressure drop (Δp) ASSUMPTIONS • Steady state • The duct is smooth SKETCH Air = 10°CTb,in Ts = 150°C L = 2 m 0.15 m Tb,out 0.075 m SOLUTION The hydraulic diameter of the duct is DH = 4 cA P = 4(0.15 m) (0.075 m) 2(0.15 m) (0.075 m) = 0.10m For the first iteration, let Tb,out = 50°C. For dry air at the average bulk temperature of 30°C Density (ρ) = 1.128 kg/m3 Thermal conductivity (k) = 0.0258 W/(m K) Absolute viscosity (μ) = 18.68 × 10–6 (Ns)/m2 Prandtl number (Pr) = 0.71 Specific heat (cp) = 1013 J/(kg K) ReD = H VD ν = H mD A  μ = ( ) ( ) ( ) ( ) ( ) ( )6 2 2 0.1kg/s 0.10m 0.15m 0.075m 18.68 10 (Ns)/m (kg m)/(Ns )−× = 47,585 > 10,000 (Turbulent) H L D = 2 m 0.1 m = 20 (a) Therefore, entrance effects may be significant — the correction Equations (6.68) and (6.69) will be applied to the Dittus Boelter correlation, Equation (6.63). From Equation (6.63) fdNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating fdNu = 0.023(47,585) 0.8 (0.71)0.4 = 111 ,c fdh = fdNu k D = 111 ( )0.0258W/(m K) 0.1m = 28.56 W/(m2 K) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 557 where a = 24/ReD 0.23 = 24/(24,510)0.23 = 2.347 b = 2.08 × 10–6 ReD – 0.815 = 2.08 × 10–6 (24,510) – 0.815 = – 0.7640 , , c L c fd h h = 1 + 2.347 0.76400.1 0.0125 −      = 1.480 ∴ ,c Lh = 1.480 ( )2129.2 W/(m K) = 191.1 W/(m2 K) The outlet temperature is given be Equation (6.36) out in T T Δ Δ = out in s b, s b, T T T T − − = exp c p PLh mc   −   = exp 4 ch L VDcρ   −   Tb,out = Ts – (Ts – Tb,in) exp 4 ch L VDcρ   −   Tb,out = 100°C – (100°C – 16°C) exp ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 4 191.1W/(m K) 0.1m 1.182 kg/m 30m/s 0.0125m 1012J/(kg K) (Ws)/J   −   = 29.2°C Performing another iteration Tb,avg = 22.6°C c = 1012 J/(kg K) ρ = 1.155 kg/m3 Re = 23,584 k = 0.0253 W/(m K) fdNu = 63.1 ν = 15.9 × 10–6 m2/s ,c Lh = 189.4 Pr = 0.71 Tb,out = 29.3°C (Tb,avg = 22.7°C) (b) The friction factor, from Equation (6.59) is f = 0.2 0.184 DRe = 0.2 0.184 (23,584) = 0.0246 Δp = f H L D 2 2 c V g ρ = 0.0246 0.1 0.0125      ( ) ( ) ( ) ( ) 23 2 2 1.155kg/m 30m/s 2 N/(m Pa) (kg m)/(s N) = 102.1 Pa For L = 1.02 m, L/D = 1.02m/0.0125 m = 81.6 > 60. Therefore, the analysis is the same as above except that the L/D correction of Equation (6.68) does not need to be applied. From the first iteration, the heat transfer coefficient (hc,fd) = 129.2 W/(m 2 K). ∴ Tb,out = 100°C – (100°C – 16°C) exp ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 4 129.2W/(m K) 1.02 m 1.182 kg/m 30m/s 0.0125m 1012J/(kg K) (Ws)/J   −   Tb,out = 74.1°C Performing another iteration Tb,avg = 45.0°C c = 1015 J/(kg K) ρ = 1.075 kg/m3 ReD = 20,718 k = 0.0270 W/(m K) DNu = 56.9 ν = 18.1 × 10–6 m2/s ch = 123.0 W/(m 2 K) Pr = 0.71 Tb,out = 75.4°C (Tb,avg = 45.7°C) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 558 From Equation (6.59) f = 0.2 0.184 (20,178) = 0.0252 From Equation (6.13) Δp = 0.0252 0.2 0.0125      ( ) ( ) ( ) ( ) 23 2 2 1.075kg/m 30m/s 2 N/(m Pa) (kg m)/(s N) = 995.0 Pa COMMENTS Note that by increasing the length of the pipe by a factor of 10 leads to a temperature rise increase of about 350% and a pressure drop increase of about 875% PROBLEM 6.28 The equation Nu = 0.116 ( 2 3Re – 125) 1 3Pr 2 0.14 3 1 +                b s D L μ μ has been proposed by Hausen for the transition range (2300 < Re < 8000) as well as for higher Reynolds numbers. Compare the values of Nu predicated by Hausen’s equation for Re = 3000 and Re = 20,000 at D/L = 0.1 and 0.01 with those obtained from appropriate equations or charts in the text. Assume the fluid is water at 15°C flowing through a pipe at 100°C. GIVEN • Water flowing through a pipe • The Hausen correlation given above • Water temperature = 15°C • Pipe temperature = 100°C FIND • The Nusselt number using the Hausen correlation and appropriate equations and charts in the text for Re = 3000 and 20,000 and D/L = 0.1 and 0.01 ASSUMPTIONS • Steady state • Constant and uniform pipe temperature SKETCH Ts = 100°C Water = 15°CT L D PROPERTIES AND CONSTANTS From Appendix 2, Table 13, for water at 15°C Absolute viscosity (μb) = 1136 × 10–6 (Ns)/m2 Prandtl number (Pr) = 8.1 At 100°C μs = 277.5 × 10–6 (Ns)/m2 © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 559 SOLUTION For Re = 3000, D/L = 0.1 the flow is in the transition region. In addition, L/D = 10. Therefore, the flow is not fully developed. The “short duct approximation” curve of Figure 6.12 in the text will be used to estimate the Nusselt number ReD Pr D L × 10–2 = 3000(8.1) (0.1) × 10–2 = 24.3 From Figure 6.12, NuD= 23. For Re = 3000, D/L = 0.01, the flow is fully developed and the Nusselt number will be estimated by the laminar correlation of Sieder-Tate, Equation (6.40) NuD = 1.86 0.33 D D Re Pr L      0.14 b s μ μ      NuD = 1.86 [3000(8.1) (0.01)] 0.33 0.141136 277.5      = 13.88 For Re = 20,000, D/L = 0.1, the flow is turbulent, not fully developed. The fully developed Nusselt number can be estimated from Equation (6.63) NuD = 0.023 ReD 0.8 Prn where n = 0.4 for heating NuD = 0.023 (20,000) 0.8 (8.1)0.4 = 146.5 Correcting this for the entrance effect using Equation (6.68) fd Nu Nu = ,c L c h h = 1 + a bL D      where a = 24/ReD 0.23 = 24/(20,000)0.23 = 2.459 b = 2.08 × 10–6 Re – 0.815 = 2.08 × 10–6 (20,000) – 0.815 = – 0.7734 fd Nu Nu = 1 + 2.459 (10)–0.7734 = 1.41 ∴ Nu = 1.41 (146.5) = 206.6 For Re = 20,000, D/L = 0.01, the entrance effect can be neglected NuD = 146.5 The Hausen correlation yields Nu = 0.116 2 3(3000) – 125) 1 3(8.1) 2 31 (0.1)   +     0.141136 277.5      = 28.63 Applying the Hausen correlation to the remaining cases and comparing them to the results from the text yields the following Case 1 2 3 4 Re 3000 3000 20,000 20,000 D/L 0.1 0.01 0.1 0.01 Nu from text 23 13.88 206.6 146.5 Nu from Hausen 28.63 24.65 211.0 181.7 Percent Difference 20% 44% 12% 14% © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 562 Tb,out = 30° – (30°C – 20°) exp ( ) ( ) ( ) ( ) ( ) ( ) 2212.3W/(m K) 0.0191m 0.57 m 0.003kg/s 4182J/(kg K) (Ws)/J π −   Tb,out = 24.4°C The error in outlet temperature is Error = 26 – 24.4 = 1.6°C PROBLEM 6.30 A solar thermal central receiver generates heat by focusing sunlight with a field of mirrors on a bank of tubes through which a coolant flows. Solar energy absorbed by the tubes is transferred to the coolant which can then deliver useful heat to a load. Consider a receiver fabricated from multiple horizontal tubes in parallel. Each tube is 1 cm ID and 1 m long. The coolant is molten salt which enters the tubes at 370°C. Under start-up conditions, the salt flow is 10 gm/s in each tube and the net solar flux absorbed by the tubes is 104 W/m2. The tube wall material will tolerate temperatures up to 600°C. Will the tubes survive start-up? What is the salt outlet temperature? GIVEN • Molten salt flowing through a horizontal tube that is absorbing solar energy • Tube inside diameter (D) = 1 cm = 0.01 m • Tube length (L) = 1 m • Entering salt temperature (Tb,in) = 370°C • Start-up mass flow rate (m) = 10 gm/s = 0.01 kg/s • Net solar energy absorbed by the tube (qs) = 104 W/m2 • Maximum tube wall temperature (Ts) = 600°C FIND (a) Salt outlet temperature (Tb,out) (b) Will the tubes survive start-up? SKETCH PROPERTIES AND CONSTANTS From Appendix 2, Table 23, for molten salt at 370° Specific heat (c) = 1629 J/(kg K) SOLUTION (a) By the conservation of energy qs A = m c(Tb,out – Tb,in) ∴ Tb,out = Tb,in + ( )sq DL mc π = 370°C + ( ) ( ) ( ) ( ) ( ) ( ) 4 210 W/m 0.01m 1.0m 0.01kg/ s 1629J/(kg K) (Ws)/J π = 389°C © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 563 Evaluating the molten salt properties at the average bulk temperature of 380°C Density (ρ) = 1849 kg/m3 Absolute viscosity (μ) = 1970 × 10–6 (Ns)/m2 Thermal expansion coefficient (β) = 3.55 × 10–4 1/K Thermal conductivity (k) = 0.516 W/(m K) Kinematic viscosity (ν) = 1.065 × 10–6 m2/s Prandtl number (Pr) = 6.18 (b) The Reynolds number is ReD = VDρ μ = 4m D  π μ = ( ) ( ) ( )6 2 2 4 0.01kg/s (0.01m) 1970 10 (Ns)/m (kg m)/(Ns )π −× = 646 (Laminar) With laminar flow and the high temperature differences possible, natural convection may be important. Since we do not know the tube wall temperature needed to evaluate the Grashof number, an iterative procedure must be used. For the first iteration, let the average tube wall temperature be 10°C above the average bulk salt temperature (Ts = 390°C). From Appendix 2, Table 23, at the tube temperature of 390°C μs = 1882 × 10–6 (Ns)/m2 The Graetz number is Gz = 4 π ReD Pr D L = 4 π (646.3)(6.18) 0.01m 1m      = 31.37 The Grashof number based on the diameter, from Table 4.3 is GrD = 3 2 ( )s bg T T Dβ ν − = ( ) ( ) ( ) 2 4 3 26 2 9.8m /s 3.55 10 1/K (390 C 380 C)(0.01m) 1.065 10 (Ns)/m − − × ° − ° × = 3.07 × 104 GrD Pr D L = 3.07 × 104 (6.18) (0.01) = 1.9 × 103 For this value and ReD = 6.5 × 102, Figure 6.12a indicates the flow is in the mixed convection regime, therefore, Equation (6.46) will be used to estimate the Nusselt number. Note that this will be a rough estimate since Equation (6.46) is technically only for isothermal tubes. DNu = 1.75 0.14 b s μ μ      1 1 3 0.36 0.883 + 0.12 ( )DGz Gz Gr Pr        DNu = 1.75 0.141970 1882      1 0.881 3 4 0.36331.37.1 0.12 (31.37) (3.07 10 ) (6.18)     + ×       = 8.76 ch = DNu k D = 8.76 ( )0.516W/(m K) 0.01m = 452 W/(m2 K) The rate of heat transfer to the molten salt is qc = ch At (Ts – Tb) = sq ′′ At ∴ Ts – Tb = s c q h ′′ = 4 2 2 10 W/m 452 W/(m K) = 22.1°C © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 564 Further iterations are necessary. However, the fluid properties will not change appreciably. Therefore, ReD, Pr, and Gz will not change. Iteration # 2 3 Ts (°C) 402 401 μs × 106 1791 1798 Ts – Tb (°C) 22.1 20.7 GrD × 10–4 6.78 6.35 DNu 9.36 9.31 ch ( )2W/(m K) 483 481 Ts – Tb (°C) 20.7 20.8 The maximum tube wall temperature is therefore Tb,out + (Ts – Tb) = 389°C + 21°C = 410°C which is well below the tube melting point. The tube will have no problems surviving the start-up in good shape. PROBLEM 6.31 Determine the heat transfer coefficient for liquid bismuth flowing through an annulus (5 cm ID, 6.1 cm OD) at a velocity of 4.5 m/s. The wall temperature of the inner surface is 427°C and the bismuth is at 316°C. It may be assumed that heat losses from the outer surface are negligible. GIVEN • Liquid bismuth flowing through an annulus • Annulus diameters  Di = 5 cm = 0.05 m  Do = 6.1 cm = 0.061 m • Bismuth velocity (V) = 4.5 m/s • Temperature  Inner wall surface (Tsi) = 427°C  Bismuth (Tb) = 316°C FIND • The heat transfer coefficient ASSUMPTIONS • Steady state • Heat losses from outer surfaces are negligible SKETCH Di = 5 cm Tsi = 427°C Do = 6.1 cm Bismuth = 45 m/s = 316°C V Tb PROPERTIES AND CONSTANTS From Appendix 2, Table 24, for bismuth at the bulk temperature of 316°C Thermal conductivity (k) = 16.44 W/(m K) Kinematic viscosity (ν) = 1.57 × 10–7 m2/s Prandtl number (Pr) = 0.014 © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 567 ASSUMPTIONS • Steady state • Uniform and constant surface temperature • Losses from the heater are negligible SKETCH Ts = 500°C D = 0.05 m L = ? Tb,out = 477°C Bismuth = 377°C = 4 kg/s or 8 kg/s T m b,in PROPERTIES AND CONSTANTS From Appendix 2, Table 24, for Bismuth at the average bulk temperature of 427°C Specific heat (c) = 150 J/(kg K) Thermal conductivity (k) = 15.58 W/(m K) Absolute viscosity (μ) = 13.39 × 10–4 (Ns)/m2 Prandtl number (Pr) = 0.013 SOLUTION At m = 4 kg/s the Reynolds number is Re = VDρ μ = 4m D  π μ = ( ) ( ) ( ) ( )4 2 2 4 4 kg/s 0.05m 13.39 10 (Ns)/m (kg m)/(Ns )π −× = 76,070 Re Pr = 76,070 (0.013) = 989 Therefore, Equation (6.78) can be used. The resulting L/D should be greater than 30 DNu = 5.0 + 0.025 (ReD Pr) 0.8 = 5.0 + 0.025 (989)0.8 = 11.23 ch = DNu k D = 11.23 ( )15.58W/(m K) 0.05m = 3498 W/(m2 K) Equation (6.36) can be used to find the length required out in T T Δ Δ = out in s b, s b, T T T T − − = exp c p PLh mc   −   = exp c h DL mc π −   L = – c mc h D  π ln out in s b, s b, T T T T −   −  = – ( ) ( ) ( ) ( ) ( )2 4 kg/s 150J/(kg K) (Ws)/J 3498W/(m K) 0.05mπ ln 500 C – 477 C 500 C – 377 C ° °    ° ° = 1.83 m L/D = (1.83 m)/(0.05 m) = 37 Repeating the analysis for m = 8 kg/s yields the following Re = 152,140 RePr = 1978 DNu = 15.84 ch = 4935 W/(m 2 K) L = 2.60 m L D = 52 © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 568 PROBLEM 6.34 Liquid sodium is to be heated from 500 K to 600 K by passing it at a flow rate of 5.0 kg/s through a 5 cm ID tube whose surface is maintained at 620 K. What length of tube is required? GIVEN • Liquid sodium flow in a tube • Bulk temperatures  Inlet (Tb,in) = 500 K  Outlet (Tb,out) = 600 K • Inside tube diameter (D) = 5 cm = 0.05 m • Tube surface temperature (Ts) = 620 K • Mass flow rate (m) = 5.0 kg/s FIND • The length of tube (L) required ASSUMPTIONS • Surface temperature is constant and uniform SKETCH L = ? D = 5 cm Ts = 620 K T =b,out 600 K Sodium = 500 K = 5 kg/s T m b,in PROPERTIES AND CONSTANTS From Appendix 2, Table 26, for liquid sodium at the average bulk temperature of 550 K Specific heat (c) = 1322 J/(kg K) Thermal conductivity (k) = 76.9 W/(m K) Absolute viscosity (μ) = 3.67 × 10–4 (Ns)/m2 Prandtl number (Pr) = 0.0063 SOLUTION The Reynolds number is ReD = U Dρ μ ∞ = 4m D  π μ = ( ) ( ) ( ) ( )4 2 2 4 5kg/s 0.05m 3.67 10 (Ns)/m (kg m)/(Ns )π −× = 3.47 × 105 ReD Pr = 3.47 × 105 (0.0063) = 2186 This is within the range of Equation (6.78) DNu = 5.0 + 0.025(ReD Pr) 0.8 = 5.0 + 0.025(2186)0.8 = 16.7 ch = DNu k D = 16.7 ( )76.9 W/(m K) 0.05m = 2.57 × 104 W/(m2 K) Solving Equation (6.36) for the length L = p c mc h D  π ln out in s b, s b, T T T T −   −  = ( ) ( ) ( ) ( ) ( )4 2 5kg/ s 1322J/(kg K) 2.57 10 W/(m K) 0.05m J/(Ws)π× ln 620K – 600K 620K – 500K      = 2.93 m Note that L/D = 2.93 m/0.05 m = 58.6 > 30. Therefore, use of Equation (6.78) is valid. © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 569 PROBLEM 6.35 A 2.54-cm-OD, 1.9-cm-ID steel pipe carries dry air at a velocity of 7.6 m/s and a temperature of –7°C. Ambient air is at 21°C and has a dew point of 10°C. How much insulation with a conductivity of 0.18 W/(m K) is needed to prevent condensation on the exterior of the insulation if h = 2.4 W/(m2 K) on the outside? GIVEN • Dry air flowing through an insulated steel pipe • Pipe diameters  Inside (Di) = 1.9 cm = 0.019 m  Outside (Do) = 2.54 cm = 0.0254 m • Air velocity (V) = 7.6 m/s • Air temperature (Ta) = –7°C • Ambient temperature (T∞) = 21°C • Ambient dew point (Tdp) = 10°C • Thermal conductivity of insulation (kI) = 0.18 W/(m K) • Heat transfer coefficient on exterior ( )ch ∞ = 2.4 W/(m2 K) FIND • Thickness of insulation (t) required to prevent codensation ASSUMPTIONS • Steady state • Flow is fully developed • Pipe surface temperature can be considered uniform and constant • Radiation heat transfer to the insulation is negligible or included in ch ∞ • Pipe is 1% carbon steel SKETCH Steel Pipe : D D i o = 1.9 cm = 2.54 cm Air = – 7°C = 7.6 m/s T V a Insulation T T> = 10°Cdp Tp = 21°C PROPERTIES AND CONSTANTS From Appendix 2, Table 27, for dry air at –7°C by extrapolation Thermal conductivity (k) = 0.0232 W/(m K) Kinematic viscosity (ν) = 13.3 × 10–6 m2/s Prandtl number (Pr) = 0.71 From Appendix 2, Table 10, the thermal conductivity of 1% carbon steel (ks) = 52 W/(m K) SOLUTION Interior heat transfer coefficient ( )cah The Reynolds number for the air flow is ReD = iVD ν = ( ) ( ) ( )6 2 7.6m/s 0.019 m 13.3 10 m /s−× = 10,860 (Turbulent) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 572 From Appendix 2, Table 13, for water at average bulk temperature of 10°C Density, ρ = 999.7 kg/m3 Thermal conductivity, k = 0.577 W/(m K) Absolute viscosity, μb = 1.296 × 10-3 (Ns)/m2 Prandtl number, Pr = 9.5 Specific heat, cp = 4195 J/(kg K) SOLUTION The Reynolds number for water flow inside the pipe is ReD = , 999.7 3.0 0.0254 0.001296 p i b VDρ μ × ×= = 58,779  turbulent flow Using the simpler Dittus-Boelter correlation for turbulent pipe flow, Equation (6.60), the average Nusselt number and hence the heat transfer coefficient for water flow can be calculated as ( ) ( )0.8 0.40.8 0.4Nu 0.023Re Pr 0.023 58,779 9.5 370D D= = = , , 0.577 Nu 370 8405 0.0254c w D p i k h D  = = = W/(m2 K) The thermal circuit for heat flow from the steam to the water can be sketched as follows Ts Rcs Tb Rks Rkc Rcw TwiTwo Here, considering the pipe length to by L, each of the four resistances can be calculated (see Chapter 1, Section 1.6.3, and Chapter 2, Section 2.3.2, for respective definitions) as follows , , , 1 1 0.00183 5700 0.0305c s c s p o R h D L L Lπ π  = = =   × × × (m K)/W ,, , 0.000118 0.00123 0.0305 k s k s p o AR R D L L Lπ π  = = =   × × (m K)/W , , , 0.0305ln ln 0.0000730.0254 2 2 398 p o p i k c c D D R k L L Lπ π          = = =   × × (m K)/W , , , 1 1 0.00149 8405 0.0254c w c w p i R h D L L Lπ π  = = =   × × × (m K)/W (a) The overall heat transfer coefficient based on the outer area of the copper pipe is ( ), , , , , 1 1 o o total p o c s k s k c c w U A R D L R R R Rπ = = + + + ∴ ( ) ( ) 1 0.0305 0.00183 0.00123 0.000073 0.00149o U π = + + + = 2257 W/(m2 K) © 2011 Cengage Learning. 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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 573 (b) The temperature of the inner surface of the pipe can be calculated by equating the rate of heat transfer between steam and water to the rate of convection to the water ( ) ( ) ( ) ( ), , ,o p o s b c w p i wi bq U D L T T h D L T Tπ π= − = − ∴ ( ) ( ), , , 2257 0.0305 40 10 10 8405 0.0254 o p o s b wi b c w p i U D T T T T h D − × −= + = + × = 19.7°C (c) The length L can now be determined from the rate of heat transfer needed to condense 0.45 kg/s of steam as follows ( ) ( ),fg o p o s bq mh U D L T Tπ= = − ∴ ( ) ( ), 0.45 2406 1000 2257 0.0305 40 10 fg o p o s b mh L U D T Tπ π × ×= = − × × −  = 167 m (d) Recognizing that with steam condensation on the outside of the copper pipe its surface temperature would be nearly constant and uniform, and hence the inlet and outlet temperatures for water flow can be calculated from Equation (6.36) as follows , ., , , exp c w p iwi w out w i w in w p h D LT T T T m c π − = − −   and , , , ,22 w in w out b w in b w out T T T T T T + =  = − Thus, if the water mass flow rate, wm is known then both Tw,in and Tw,out can be calculated. PROBLEM 6.37 Assume that the inner cylinder in Problem 6.31 is a heat source consisting of an aluminum-clad rod of uranium, 5-cm-OD and 2 m long. Estimate the heat flux that will raise the temperature of the bismuth 40°C and the maximum center and surface temperatures necessary to transfer heat at this rate. From Problem 6.31: Determine the heat transfer coefficient for liquid bismuth flowing through an annulus (5-cm-ID, 6.1-cm-OD) at a velocity of 4.5 m/s. The bismuth is at 316°C. It may be assumed that heat losses from the outer surface are negligible. GIVEN • Liquid bismuth flowing through an annulus • Annulus inside diameter (Di) = 5 cm = 0.05 m • Annulus outside diameter (Do) = 6.1 cm = 0.061 m • Bismuth velocity (V) = 4.5 m/s • Bismuth temperature (Tb) = 316°C • Inner cylinder is an aluminum clad uranium heat source • Cylinder length (L) = 2 m • From Problem 6.31: ch = 26,800 W/(m 2 K) FIND (a) The heat flux (QG/At) necessary to raise the bismuth temperature 40°C, and (b) The maximum center (Tu,o) and surface (Tu,ro) temperatures of the uranium © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 574 ASSUMPTIONS • Steady state • The Bismuth temperature given in Problem 6.31 is the bulk Bismuth temperature • Thermal resistance of the aluminum is negligible • Thickness of the aluminum is negligible SKETCH Uranium L = 2 m Bismuth Tb,in = 316°C PROPERTIES AND CONSTANTS From Appendix 2, Table 12, for uranium Thermal conductivity (ku) = 36.4 W/(m K) at 427°C From Appendix 2, Table 24, for Bismuth at 316°C Specific heat (cb) = 144.5 J/(kg K) Density (ρ) = 10,011 kg/m3 SOLUTION (a) The rate of heat transfer required to raise the Bismuth by 40°C is q = m cbΔTb = ρ VAc cb ΔTb = ρ V 4 π (Do 2 – Di 2) cb ΔTb q = 10,011 kg/m3 ( )4.5m/s 4 π [(0.061 m)2 – (0.05 m)2] ( )144.5 J/(kg K) (40°C) ( )(Ws)/J = 2.50 × 105 W Therefore, the average heat flux is G t Q A  = t q A = i q D Lπ = ( ) ( ) 52.50 10 W 0.05 m 2 mπ × = 7.95 × 105 W/m2 The temperature difference between the uranium and bismuth (ΔTub) required to transfer this heat can be calculated from t q A = ch ΔTub  ΔTub = t c q A h = ( ) ( ) 5 2 2 7.95 10 W/m 26,500W/(m K) × = 29.7 K The maximum uranium surface temperature will occur at the outlet where the bismuth temperature is Tb,max = 316°C + 0.5(ΔTb) = 336°C Tu,ro,max = Tb,max + ΔTub = 336°C + 29.7 K ≈ 366°C The rate of internal heat generation per unit volume is Gq = Volume GQ = 2 4 i q D L π = ( ) ( ) 5 2 2.50 10 W 0.05m 2m 4 π × = 6.37 × 107 W/m3 © 2011 Cengage Learning. 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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 577 Applying Equation (5.20) DNu = 0.53 1 4( )DGr Pr = 0.53 1 7 41.855 10 (0.71) ×  = 31.93 cah = a D I k Nu D = 31.93 ( )0.0258W/(m K) 0.2 m = 4.12 W/(m2 K) The thermal circuit for this problem is shown below Tw Rcw T• Rkp RKI Rc• Rr TI where Rcw = 1 cw wh A = 1 cw wh D Lπ = ( )2 1 15,713W/(m K) (0.1m)Lπ = 1 0.000203 L      (m K)/W Rkp = Thermal résistance of the pipe wall ≈ 0 RkI = ln 2 I i I D D L kπ      = ( ) 0.2 ln 0.1 2 0.059 W/(m K)Lπ      = 1 1.870 L      (m K)/W From Equation (2.39) Rr = Radiative resistance Rcw = Natural convective resistance The insulation temperature (TI) can be determined by equating the heat transfer between Tw and TI to that from TI to T∞ w I cw kI T T R R − − = qca + qra = AI [ cah (TI – T∞) + ε σ (TI 4 – T∞ 4)] ( ) 473 K 1 (0.000203+1.87) (m K)/W IT L −      = π (0.2 m) L ( ) ( )2 8 2 4 4 44.12 W/(m K) ( 293K) 0.5 5.67 10 W/(m K ) [ (293K) ]I IT T− − + × −  Checking the units then eliminating them for clarity 1.79 × 10–8 TI4 + 3.124 TI – 1142.7 = 0 By trial and error: TI = 312 K = 39°C Further iterations are not required. The rate of heat loss can be calculated from q = w I cw kI T T R R − − = ( ) 473K 312 K 1 1.87 (m K)/W L −      = 86.1 W/m COMMENTS Note that the convective resistance in the turbulent water is negligible compared to that of the insulation. © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 578 PROBLEM 6.39 In a pipe-within-a-pipe heat exchanger, water is flowing in the annulus and an aniline- alcohol solution having the properties listed in Problem 6.20 is flowing in the central pipe. The inner pipe is 1.3 cm ID, 1.6 cm OD, and the ID of the outer pipe is 1.9 cm. For a water bulk temperature of 27°C and an aniline bulk temperature of 60°C, determine the overall heat transfer coefficient based on the outer diameter of the central pipe and the frictional pressure drop per unit length of the water and the aniline for the following volumetric flow rates, (a) water rate 0.06 litres/sec, aniline rate 0.06 litres/sec, (b) water rate 0.6 litres/sec, aniline rate 0.06 litres/sec, (c) water rate 0.06 litres/sec, aniline rate 0.6 litres/sec, and (d) water rate 0.6 litres/sec, aniline rate 0.6 litres/sec (L/D = 400). Physical properties of the aniline solution Temperature Viscosity Thermal Specific Specific Heat (°C) (kg/ms) Conductivity Gravity kJ/(kg K) (W/(m K)) 20 0.0051 0.173 1.03 2.09 60 0.0014 0.169 0.98 2.22 100 0.0006 0.164 2.34 GIVEN • Pipe-within-a-pipe heat exchanger with water in the annulus and aniline-alcohol solution in the inner pipe • Solution properties listed above • Pipe diameters  Inner pipe Dii = 1.3 cm Di = 1.6 cm  Inside of outer pipe Do = 1.9 cm • Bulk temperatures  Water (Tw) = 27°C  Aniline (Ta) = 60°C • L/D = 400 FIND • The overall heat transfer coefficient (U) based on Di and the pressure drop (Δ p) for the following volumetric flow rates ( V ) Case (a) (b) (c) (d) Water flow rate (L/s) 0.06 0.6 0.06 0.6 Aniline flow rate (L/s) 0.06 0.06 0.6 0.6 ASSUMPTIONS • Steady state • Thermal resistance of the pipe is negligible • Nusselt number can be estimated from correlations for constant and uniform surface temperature • The effect of viscosity variation is negligible • The tubes are smooth • Fully developed flow (L/D = 400) PROPERTIES AND CONSTANTS From Appendix 2, Table 13, for water at 27°C Density (ρ) = 999 kg/m3 Thermal conductivity (kw) = 0.61 W/(m K) Kinematic viscosity (νw) = 8.41 × 10–7 m2/s Prandtl number (Prw) = 5.87 © 2011 Cengage Learning. 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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 579 SOLUTION Water side heat transfer coefficient: The Reynolds number on the water side is ReD = ii w V D ν = ii c w V D A ν  = 4 ii w V Dπ ν  For V = 1 gal/min ReD = –3 3 –2 –7 2 4 (0.06 10 m /s) (1.3 10 m)(8.41 10 m /s)π × × × × = 6990 (Turbulent) Applying Equation (6.63) DNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating DNu = 0.023 (6990) 0.8 (5.87)0.4 = 55.56 ,1cwh = DNu ii k D = 55.56 –2 0.61W/(m K) (1.3 10 m)× = 2607 W/(m2 K) For V = 0.6 m/s, Re  0.70 × 105 (Turbulent) DNu = 0.023 (0.70 × 10 5)0.8 (5.87)0.4 = 351 ,10cwh = 351 –2 0.61W/(m K) (1.3 10 m)× = 16.47 kW/(m2 K) Aniline side heat transfer coefficient: The hydraulic diameter of the annulus, from Equation (6.3) is DH = Do – Di = 1.9 cm – 1.6 cm = 0.3 cm = 3 × 10–3 m From the given properties; for aniline at 60°C Density, ρ = ρH2O(s.g.) = 998 kg/m3 (0.98) = 978 kg/m3 Kinematic viscosity, νa = μ ρ = 3 0.0014kg/ms 978kg/m = 1.431 × 10–5 m2/s Prandtl number, Pr = c k μ = (2.22kJ/kg K)(0.0014kg/ms) (0.169 W/(m K)) = 18.4 for (V ) = 0.06 L/s HD Re = H H a c a V D V D Aν ν =  = 2 2 4 ( ) H o i a V D D Dπ ν−  HD Re = –3 3 –3 2 2 –4 2 –5 2 4(0.06 10 m /s)(3 10 m) (1.9 – 1.6 ) 10 m 1.431 10 mπ × × × × × = 1526 (Laminar) From Table (6.2): For Di/Do = 1.6/1.9 = 0.84: DNu ≈ 5.15 ,1cah = HD H k Nu D = 5.15 –3 (0.169 W/(m K)) 3 10 m× = 290 W/(m2 K) For V = 0.6 L/s Re = 15260 (Turbulent) © 2011 Cengage Learning. 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May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 582 GIVEN • Air flow inside and outside a plastic tube Case 1 • Tube 1 inside diameter (D1i) = 7.6 cm = 0.076 m • Tube 1 wall thickness (S1) = 1.27 cm = 0.0127 m • Plastic properties  Thermal conductivity (kp) = 1.7 W/(m K)  Density (ρ) = 2400 kg/m3  Specific heat (c) = 1675 J/(kg K) • Tube initial temperature (Tti) = 77°C • Air temperature (Ta) = 20°C • After 10 min: (Tt – Ta) = 10% of initial (Tt – Ta) Case 2 • Tube 2 inside diameter (D2i) = 15 cm = 0.15 m • Tube 2 wall thickness (S2) = 2.5 cm = 0.025 m • Same initial temperature and air temperature as Case 1 • Same interior air flow rate as Case 1 • Air velocities are such that heat transfer coefficients inside and outside are equal FIND • Time for Tt to reach 27°C in Case 2 ASSUMPTIONS • Temperature rise along the tube is negligible • Tube may be treated as a lumped capacitance (This will be checked) SKETCH Di Air = 21°C V T o a t Tt = 77°C Initially Air = 21°C V T i a PROPERTIES AND CONSTANTS From Appendix 2, Table 27, for dry air at 20°C Density (ρ) = 1.164 kg/m3 Thermal conductivity (k) = 0.0251 W/(m K) Kinematic viscosity (ν) = 15.7 × 10–6 m2/s Prandtl number (Pr) = 0.71 Absolute viscosity (μ) = 18.24 × 10–6 (Ns)/m2 SOLUTION Case 1 Assuming the tube can be treated as a lumped capacitance: Equation (2.84) can be applied ln tf a ti a T T T T −    − = – c sh A c Vρ t = – 2 2 ( ) ( ) 4 c i o o i h D D L c D D L π πρ +  −   t where Do = Di + 2s = 0.076 m + 2(0.0127 m) = 0.1014 m © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 583 Solving for the heat transfer coefficient ch = – 2 2( ) 4 ( ) o i i o c D D D D t ρ − + ln tf a ti a T T T T −    − ch = – ( ) ( ) ( ) ( ) 3 2 21675J/(kg K) 2400kg/m [(0.1014 m) (0.076m) ] 4(0.076m 0.1014 m)(50min) 60s/min J/(Ws) − + ln (0.10) = 19.6 W/(m2 K) Checking the lumped capacity assumption, the Biot number should be based on half of the tube wall thickness since convection occurs equally on the inside and outside of the tube Bi = 2 c s h s k = ( ) ( ) 219.6W/(m K) (0.0127 m) 2 1.7 W/(m K) = 0.07 < 0.1 Therefore, the lumped capacity assumption is valid. Assuming the air flow is turbulent and applying Equation (6.63) to determine the Reynolds number for the interior air flow DNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating ∴ ReD = 1.25 0.40.023Pr K c ih D      = ( ) ( ) 1.252 0.4 19.6W/(m K) (0.076m) 0.023(0.71) 0.0251W/(m K)       = 21,825 (Turbulent) Therefore, the air velocity is V = D i Re D ν = ( )6 221,825 15.7 10 m /s 0.076m −× = 4.51 m/s The mass flow rate is m = V ρ Ac = V ρ 4 π Di 2 = 4.51 m/s ( )31.1641 kg/m 4 π (0.076 m)2 = 0.024 kg/s Case 2 Applying the mass flow rate to Case 2 ReD = VD ρ μ = 4m Dπ μ  = ( ) ( ) ( )6 2 2 4 0.024 kg/s (0.15m) 18.24 10 (Ns)/m kg m/(Ns )π −× = 11,169(Turbulent) Applying Equation (6.63) DNu = 0.023 (11,169) 0.8 (0.71)0.4 = 34.72 ch = DNu k D = 34.72 ( )0.0251 W/ (m K) 0.15m = 5.81 W/(m2 K) The Biot number is Bi = 2 c s h s k = ( ) ( ) 25.81W/(m K) (0.025m) 2 1.7 W/(m K) = 0.04 < 0.1 Therefore, the internal thermal resistance can be neglected and Equation (2.84) can be applied. Solving for the time t = – 2 2( ) 4( ) o i i o c c D D D D h ρ − + ln tf a ti a T T T T −    − = – ( ) 4 o i c c D D h ρ − ln tf a ti a T T T T −    − = – (2 ) 4 c c s h ρ ln tf a ti a T T T T −    − © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 584 t = – ( ) ( ) ( ) ( ) 3 2 1675J/(kg K) 2400kg/m 2(0.025m) 4 5.81W/(m K) J/(Ws) ln 27°C 20°C 77°C 20°C −    − t = 18,137 s = 302 min ≈ 5 hours PROBLEM 6.41 Exhaust gases having properties similar to dry air enter an exhaust stack at 800 K. The stack is made of steel and is 8 m tall and 0.5 m ID. The gas flow rate is 0.5 kg/s and the ambient temperature is 280 K. The outside of the stack has an emissivity of 0.9. If heat loss from the outside is by radiation and natural convection, calculate the gas outlet temperature. GIVEN • Exhaust gas flow through a steel stack • Exhaust gas has the properties of dry air • Entering exhaust temperature (Tb,in) = 800 K • Stack height (L) = 8 m • Stack diameter (D) = 0.5 m • Gas flow rate ( m )= 0.5 kg/s • Ambient temperature (T∞) = 280 K • Stack emissivity (ε) = 0.9 FIND • The outlet gas temperature (Tb,out) ASSUMPTIONS • Steady state • The surrounding behave as a black body enclosure at the ambient temperature • Thermal resistance of the duct is negligible • Duct thickness is negligible SKETCH Tb,out = ? L = 8 m D = 0 .5 m T• = 280 K Tb,in = 800 K Exhaust Gas m = 0.5 kg/s © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 587 GIVEN • Exhaust gases flowing through an insulated long vertical cylindrical duct • Exhaust gases have the physical properties of dry air • Duct length (L) = 3.05 m • Duct inside diameter (D) = 15.2 cm = 0.152 m • Entering exhaust temperature (Tb,in) • Rock wool insulation thickness (s) = 10.2 cm = 0.102 m • Thermal conductivity of insulation (k) = 0.7 + 0.016 T (T is in °C and k in W/(m K)) • Exhaust velocity (V) = 0.61 m/s • Gas inlet temperature (Tb,in) = 316°C FIND (a) Rate of heat transfer to ambient air at (T∞) = 15.6°C (b) Outlet exhaust gas temperature (Tb,out) ASSUMPTIONS • Steady state • Thermal resistance of the duct wall is negligible • Heat transfer by radiation is negligible • Natural convection on the inside of the duct can be approximated by natural convection from a vertical plate • The interior heat transfer coefficient can be accurately estimated using uniform surface temperature correlations • The ambient air is still SKETCH T• = 15.6°C Di = 15.2 cm Exhaust = 316°C = 0.61 m/sT Vb,in L = 3.05 m Insulation, Thickness ( ) = 10.2 cms PROPERTIES AND CONSTANTS From Appendix 2, Table 27, for dry air at the inlet temperature of 316°C Thermal expansion coefficient (β) = 0.00175 1/K Thermal conductivity (k) = 0.0438 W/(m K) Kinematic viscosity (ν) = 51.7 × 10–6 m2/s Prandtl number (Pr) = 0.71 Density (ρ) = 0.582 kg/m3 Specific heat (c) = 1049 J/(kg K) Absolute viscosity (μb) = 28.869 × 10–6 Ns/m2 SOLUTION The expression for thermal conductivity is given as k = 0.7 + 0.016 T (T in °C, k in W/(m K)) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 588 Interior convection: The Reynolds number for the exhaust flow is ReD = V D ν = ( )6 2 (0.61m/s) (0.152 m) 51.7 10 m /s−× = 1793 (Laminar) For the first iteration, let the duct wall temperature (Td) = 300°C and the insulation surface temperature (TI) = 20°C. From Appendix, Table 27, the absolute viscosity at Td = 300°C is μs = 29.332 × 10–6 (Ns)/m2. Applying Equation (6.39) DNu = 0.14 0.66 0.0668 3.66 1 0.045 D b s D D Re Pr L D Re Pr L μ μ          +        +        DNu = 0.14 0.66 0.152 0.0668(1793) (0.71) 28.8693.053.66 29.3320.152 1 0.045 (1793) (0.71) 3.05         +       +       = 6.17 ,forcedch = DNu i k D = 6.17 ( )0.0438W/(m K) 0.152 m = 1.77 W/(m2 K) The interior Grashof number based on the duct length is GrL = 3 ,in 2 ( )b d a g T T Lβ ν − = ( ) ( ) ( ) 2 3 26 2 9.8m/s 0.00171(1/K) (316 C 300 C)(3.05m) 51.7 10 m /s− ° − ° × = 2.85 × 109 2 L D Gr Re = 9 2 2.85 10 (1793) × = 885 Therefore, natural convection on the inside of the duct cannot be ignored. The natural convection Nusselt number will be estimated with Equation (5.13) LNu = 0.13 1 3( )LGr Pr = 0.13 1 9 32.85 10 (0.71) ×  = 164.4 ,naturalch = LNu k L = 164.4 ( )0.0438W/(m K) 3.05m = 2.36 W/(m2 K) Combining the free and forced convection using Equation (5.49) cih = ( ) 1 3 3 3 ci cnh h+ = 1 3 3 3(1.77) (2.36) +  = 2.65 W/(m 2 K) Exterior convection: The Grashof number on the exterior of the insulation is GrL = 3 1 2 ( ) a g T T Lβ ν ∞− For the film temperature of 17.8°C β = 0.00344 1/K ν = 15.5 × 10–6 m2/s © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 589 Pr = 0.71 k = 0.0249 W/(m K) GrL = ( ) ( ) ( ) 2 3 26 2 9.8m/s 0.00344(1/K) (20 C 15.6 C)(3.05m) 15.5 10 m /s− ° − ° × = 1.75 × 1010 The Nusselt number is given by Equation (5.13) LNu = 0.13 1 3( )LGr Pr = 0.13 1 10 31.75 10 (0.71) ×  = 301.2 coh = LNu k L = 301.2 ( )0.0249 W/(m K) 3.05m = 2.46 W/(m2 K) Conduction through the insulation: The rate of heat transfer through the insulation is q = – k A dT dr 2 o i r r q dr L rπ  = i d T T kdT = – (0.7 0.016 )i d T T T dt+ where TI = exterior insulation temperature Td = duct wall temperature = interior insulation temperature 2 q Lπ ln o i r r      = – 0.7 (TI – Td) – 0.016 2 (TI 2 – Td 2) q = 2 ln o i L r r π      [0.7 (Td – TI) + 0.008 (Td 2 – TI 2)] = d I k T T R − ∴ 1 kR = 2 22 0.7 0.008 ln d I o d I i T TL r T T r π    −+  −       = 2 ln o i L r r π      [0.7 + 0.008 (TD + TI)] where ri = Di/2 = (0.0152 m)/2 = 0.076 m ro = ri + s = 0.076 m + 0.102 m = 0.178 m 1 kR = 2 (3.05m) 0.178 ln 0.076 π      [0.7 + 0.008 (300°C + 20°)] = 73.4 W/K Rk = 0.0136 K/W The thermal circuit for the problem is shown below Tb Rci T• Rk Rco Td TI Rci = 1 ci ih A = 1 ci ih D Lπ = ( )2 1 2.65W/(m K) (0.152 m)(3.05m)π = 0.259 K/W Rco = 1 co oh D Lπ = ( )2 1 2.46W/(m K) [0.152m 2(0.102 m)](3.05m)π + = 0.119 K/W © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 592 The shape factor (S) is given in Table 2.2 S = 1 2 2 cosh i dx Z D π −      ∴ Rtotal = 1 2ln cos 1 1 2 2 i po i c po i s D Z D D dx h D k kπ π π −             + +      Let U′ = total 1 dxR = U dx then dq = U′ (Tb – Ts) dx = U dx b b s dT T T− = p U mc ′  dx Integrating ,out ,in 1b b T bT b s dT T T− = 0 L p U m c ′−   dx ln ,out ,in b s b s T T T T −   −  = p U L m c ′−  Solving for the overall heat transfer coefficient U′ = p m c L  ln ,out ,in b s b s T T T T −   −  = – ( ) ( ) ( )500kg/s 2000J/(kg K) (Ws)/J 100,000m ln = 115°C 40°C 120°C 40°C +    + = 0.317 W/(m K) 1 U ′ = 1 c poh Dπ + ln 2 i po i D D kπ      + 1 2cosh 2 i s Z D kπ −      ( ) 1 0.317 W/(m K) = ( )2 1 0.427 W/(m K) (1.2 m)π + ( ) ln 1.2m 2 0.05W/(m K) iD π      + ( ) 1 2(3m)cosh 2 0.35m W/(m K) iD π −      checking the units then eliminating them for clarity 5.571 = 7.01 n 1.2 m iD     + cosh–1 6 iD      by trial and error: Di = 2.06 m t = ( – ) 2 iD D = [ ](2.06 m) – (1.2 m) 2 = 0.43 m = 43 cm (b) The rate of heating required at each pumping station is q = pmc ΔT = (500 kg/s) ( )2000J/(kgK) (5°C) ( )(Ws)/J = 5 × 106 W = 5 MW © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 593 (c) The pumping power P, equals the product of the volumetric flow rate and the pressure drop, or P = pm Δ Incorporating Equation (6.13) for the pressure drop and Equation (6.18) for the friction factor P = 264 2d c m L U Re D g ρ ρ       = 32 2 2 4 c D m L m Dg Re Dρπ        = 3 2 2 5 512 c D L m g Re Dπ ρ  P = ( ) ( ) ( ) ( ) 3 2 22 3 5 512 100,000m 500kg/s (Ws)/(Nm) (kg m)/(s N) (693) 900kg/m (1.2 m)π = 1.46 × 106 W = 1.46 MW PROBLEM 6.44 Show that for fully developed laminar flow between two flat plates spaced 2a apart, the Nusselt number based on the ‘bulk mean’ temperature and the passage spacing is 4.12 if the temperature of both walls varies linearly with the distance x, i.e., ∂T/∂x = C. The ‘bulk mean’ temperature is defined as Tb =   – – ( ) ( ) ( ) a a a a u y T y dy u y dy GIVEN • Fully developed laminar flow between two flat plates • Spacing = 2a • ∂T/∂x = C • Bulk mean temperature as defined above FIND • Show that the Nusselt number based on the bulk mean temperature = 4.12 ASSUMPTION • Steady state • Constant and uniform property values • Fluid temperature varies linearly with x (This corresponds to a constant heat flux boundary) SKETCH Tw x( ) u = unax u y( ) T y( ) Fluid u u y= ( ) T T y= ( ) y = a y a= – x y CL © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 594 SOLUTION The solution will progress as follows 1. Derive the temperature distribution in the fluid. 2. Use the temperature distribution to obtain an expression for the bulk mean temperature. 3. Use the bulk mean temperature to derive the Nusselt number. Beginning with the laminar flow energy equation of Equation (4.7b) u T x ∂ ∂ + v T y ∂ ∂ = α 2 2 T y ∂ ∂ v = component of the velocity in the Y direction = 0 ∴ u T x ∂ ∂ = α 2 2 T y ∂ ∂ Note that ∂T/∂x = constant by assumption. The velocity profile u(y) must be substituted into this equation before the equation can be solved for the temperature distribution. The velocity profile can be derived by considering a differential element of fluid of width w as shown below Umax P d+ p y = 0 y dy+ y dx P A force balance on this element yields 2 w y [p – (p + dp)] = 2 τ w dx = μ u y ∂ ∂ – μ 2u u dy y y  ∂ ∂+ ∂ ∂  w dx Since the flow is fully developed dp dx = μ 2 2 d u dy Integrating with respect to y u = 21 2 dp y dxμ + C This is subject to the following boundary conditions u = umax at y = 0 therefore, C = umax u = 0 at y = + a therefore, umax = – 2 2 a dp dxμ Therefore, the velocity distribution is u = umax 2 1 y a   −      © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 597 ASSUMPTIONS • Steady state • Constant and uniform property values • Fluid temperature varies linearly with x (This corresponds to a constant heat flux boundary) SKETCH T( )yu( )y y a= y x CL Insulation y a= – Fluid SOLUTION The velocity profile derived in the solution to Problem 6.44 remains unchanged u = umax 2 1 y a   −      As does the energy equation 2 2 T y ∂ ∂ = z 2 1 y a   −      where z = max u T xα ∂ ∂ (a constant) The new boundary conditions are T y ∂ ∂ = 0 at y = a (due to the insulation) T = Tw (x) at y = –a Integrating the energy equation once T y ∂ ∂ = z 3 2 1 3 y y a   −   + C1 Applying the first boundary condition 0 = za – 3 23 za a + C1  C1 = – 2 3 za Integrating the energy equation again T = – 2 3 zay + 1 2 z y2 – 212 z a y4 + C2 © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 598 Applying the second boundary condition Tw(x) = – 2 3 z a2 + 1 2 z a2 – 212 z a a4 + C2  C2 = Tw(x) + 1 4 za2 Therefore, the temperature distribution is T(x,y) = Tw(x) + 1 4 za2 – 2 3 zay + 1 2 zy2 – 212 z a y4 The numerator of the bulk mean temperature expression is ( ) ( , ) a a u y T x y dy − = 2 2 2 4 max 2 1 2 1 1 ( ) 4 3 2 12 a wa y z u T x za za y z y y a a−       − + − + −             dy = umax 2 3 3 2 3 31 1 1 2 1 1 12 ( ) ( ) 4 3 30 3 4 5 42w w a T x za za za a T x za za za     + + − − + − +         ( ) ( , ) a a u y T x y dy − = umax 2 34 1 13( ) 3 4 105w a T x za za   + +     The denominator of the bulk mean temperature is 2 max 1 a a y u dy a−   −       = umax = 2 2 3 a a −   4 3 umax a ∴ Tb = 2 1 ( ) 4w T x za +   + 3 4      13 105 za2 = Tw(x) + 57 210 za2 Tw(x) – Tb = – 57 210 za2 At z = – a: T y ∂ ∂ = z(– a) – 23 z a (– a)3 – 2 3 za = – 4 3 za ∴ ch = y a b w T k y T T = − ∂− ∂ − = 2 4 3 57 210 k a z z a  − −  = 560 57 2 k a    By definition Nu = c h L k = 2ch a k = 560 57 = 9.82 PROBLEM 6.46 For fully turbulent flow in a long tube of diameter D, develop a relation between the ratio (L/ΔT)/D in terms of flow and heat transfer parameters, where L/ΔT is the tube length required to raise the bulk temperature of the fluid by ΔT. Use Equation 6.63 for fluids with Prandtl number of the order of unity or larger and Equation 6.75 for liquid metals. © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 599 GIVEN • Fully developed turbulent flow in a long tube • Diameter = D • L/ΔT = Tube length required to raise the bulk temperature by ΔT FIND A relationship for (L/ΔT)/D in terms of flow and heat transfer parameters using (a) Equation 6.63 for fluids with Pr ≈ 1 (b) Equation 6.75 for liquid metals ASSUMPTIONS • Steady state • Constant fluid properties • Uniform wall temperatures SKETCH L TD L + TDT Fluid D SOLUTION Let k = the thermal conductivity of the fluid μ = the absolute viscosity of the fluid c = the specific heat of the fluid V = the velocity of the fluid ρ = the density of the fluid Tb = Average bulk fluid temperature Tw = wall temperature (a) Using Equation (6.63) for the Nusselt number DNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating ch = D k Nu D = 0.023 k D ReD 0.8 Pr0.4 The rate of heat transfer to the fluid must equal the energy needed to raise the temperature of the fluid by ΔT q = ch π D L (Tb – Tw) = m c ΔT L TΔ = ( )c b w mc h D T Tπ −  = 2 0.8 0.4 4 0.023 ( )D b w V D C k Re Pr D T T D πρ π − But ReD = V Dρ μ and Pr = c k μ © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 602 COMMENTS The tube diameter must not become so large that the water flow becomes laminar. PROBLEM 6.48 The following thermal-resistance data were obtained on a 5000 m2 condenser constructed with 2.5 cm-OD brass tubes, 7.2 m long, 1.2 mm wall thickness, at various water velocities inside the tubes [Trans. ASME, vol. 58, p. 672, 1936]. 1/Uo × 103 Water Velocity 1/Uo × 103 Water Velocity (Km2/W) (m/s) (km2/W) (m/s) 0.364 2.11 0.544 0.90 0.373 1.94 0.485 1.26 0.391 1.73 0.442 2.06 0.420 1.50 0.593 0.87 0.531 0.89 0.391 1.91 0.368 2.14 Assuming that the heat transfer coefficient on the steam side is 11.3 kW/(m2 K) and the mean bulk water temperature is 50°C, determine the scale resistance. GIVEN • Water flowing inside a brass tube condenser • Total transfer are (At) = 5000 m2 • Tube outside diameter (D) = 2.5 cm • Tube length (L) = 7.2 m • Tube wall thickness (t) = 1.2 mm • Heat transfer coefficient on the steam side ( csh ) = 11.3 kW/(m 2 K) • Mean bulk water temperature = 50°C • Thermal resistance data shown above FIND • The scale resistance (AtRks) ASSUMPTIONS • Data were taken at steady state • The tube temperature can be considered uniform and constant • Condenser surface area is based on the tube outside diameter SKETCH L = 7.2 m One Tube D = 2.5 cm Water PROPERTIES AND CONSTANTS From Appendix 2, Table 13, for water at 50°C Thermal conductivity (k) = 0.65 W/(m K) Kinematiuc viscosity (ν) = 5.46 × 10–7 m2/s Prandtl number (Pr) = 3.55 From Appendix 2, Table 10, the thermal conductivity of brass (kb) = 111 W/(m K) © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 603 SOLUTION The inside tube diameter is Di = Do – 2t = 2.5 cm – 0.24 cm = 2.26 cm The maximum and minimum velocities in the given data are 0.87 m/s and 2.14 m/s. These correspond to the following Reynolds numbers Remin = V D ν = –2 –7 2 (0.87 m/s) (2.26 10 m) 5.46 10 m /s × × = 36,000 Remax = V D ν = –2 –7 2 (2.14m/s) (2.26 10 m) 5.46 10 m /s × × = 88,580 Therefore, the flow is turbulent in all cases. Applying Equation (6.63) to the minimum Re case DNu = 0.023 ReD 0.8 Prn where n = 0.4 for heating DNu = 0.023 (36,000) 0.8 (3.55)0.4 = 168.6 cwh = D k Nu D = 168.6 ( ) –2 0.65W/(m K) 2.26 10 m× = 4850 W/(m2 K) The thermal circuit for this problem is shown below Ts Rcs Tw RkS RkB Rcw At Rcs = 1 csh = 3 2 1 11.3 10 W/(m K)× = 8.85 × 10–5 (K m2)/W At Rks = scaling resistance For one tube At RkB = π D L ln 2 o i s D D Lkπ      At RkB = –2 2.5(2.5 10 m)(7.2m)ln 2.26 2 (7.2m)(111W/(m K)) π π  ×   = 1.13 × 10–5 (K m2)/W For the minimum Re case Ar Rcw = 1 cwh = 2 1 48.50 W/(m K) = 2.06 × 10–4 (K m2)/W These resistances are in series, therefore Ar Rtotal = 1 oU = At (Rcs + Rks + RkB + Rcw) ∴ At Rks = 1 oU – At (Rcs + RkB + Rcw) For the minimum Re case (from the given table) At Rks = 0.593 × 10–3 (K m2)/W – (0.0885 + 0.0113 + 0.206) × 10–3 (K m2)/W At Rks = 2.33 × 10–4 (K m2)/W © 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 604 Repeating this method for the rest of the data given Water Velocity (m/s) cwh W/(m 2 K) AtRks × 104 (Km2)/W 2.11 9850 1.43 1.94 9210 1.45 1.73 8405 1.50 1.50 7500 1.64 0.89 4940 1.99 2.14 9965 1.48 0.90 4950 2.13 1.26 6522 2.02 2.06 9665 2.11 0.87 4850 2.33 1.91 9100 1.58 Average: 1.79 × 10–4 (K m2)/W COMMENTS The standard deviation in the scale resistance is 24%. PROBLEM 6.49 A nuclear reactor has rectangular flow channels with a large aspect ratio (w/h)>>1 L w Sodium , Ts o( ) x q(x) . h Heat generation is equal from the upper and lower surface and uniform at any value of x. However, the rate varies along the flow path of the sodium coolant according to q′′ (x) = qo′′ sin(π x/L) Assuming that entrance effects are negligible so that the convection heat transfer coefficient is uniform (a) Obtain an expression for the variation of the mean temperature of the sodium, Tm (x). (b) Derive a relation for the surface temperature of the upper and lower portion of the channel, Ts (x). (c) Determine the distance xmax at which Ts(x) is maximum. GIVEN • Sodium flow through a rectangular flow channel with a large aspect ratio • Heat generation from each surface (upper and lower): q′′(x) = q′′o sin(π x/L) FIND (a) An expression for the variation of the mean sodium temperature, Tm(x) (b) A relationship for the upper and lower surface temperature Ts(x) (c) The distance xmax at which Ts(xmax) is maximum