Solution manual heat and mass transfer a practical approach 3rd edition cengel CH10 2

10-56 Review Problems 10-79 Water is boiled at Tsat = 100°C by a spherical platinum heating element immersed in water The surface temperature is Ts = 350°C The boiling heat transfer coefficient is to be determined Assumptions Steady operating conditions exist Heat losses from the heater and the boiler are negligible Properties The properties of water at the saturation temperature of 100°C are (Table A-9) h fg = 2257 × 10 J/kg ρ l = 957.9 kg/m 350°C The properties of water vapor at (350+100)/2 = 225°C are (Table A-16) Water 100°C ρ v = 0.444 kg/m μ v = 1.749 × 10 −5 kg/m ⋅ s c pv = 1951 J/kg ⋅ °C k v = 0.03581 W/m ⋅ °C Analysis The film boiling occurs since the temperature difference between the surface and the fluid The heat flux in this case can be determined from q& film [ ⎡ gk v3 ρ v ( ρ l − ρ v ) h fg + 0.4c pv (Ts − Tsat ) = 0.67 ⎢ μ v D(Ts − Tsat ) ⎢⎣ ]⎤⎥ / (T [ ⎥⎦ s − Tsat ) ⎡ (9.81)(0.03581) (0.444)(957.9 − 0.444) 2257 × 10 + 0.4(1951)(350 − 100) = 0.67 ⎢ (1.749 × 10 −5 )(0.15)(350 − 100) ⎢⎣ ]⎤⎥ ⎥⎦ 1/ (350 − 100) = 27,386 W/m The boiling heat transfer coefficient is ⎯→ h = q& film = h(Ts − Tsat ) ⎯ q& film 27,386 W/m = = 109.5 W/m ⋅ C Ts − Tsat (350 − 100)°C PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 10-57 10-80 Water is boiled at Tsat = 120°C in a mechanically polished stainless steel pressure cooker whose inner surface temperature is maintained at Ts = 130°C The time it will take for the tank to empty is to be determined Assumptions Steady operating conditions exist Heat losses from the heater and the boiler are negligible Properties The properties of water at the saturation temperature of 120°C are (Tables 10-1 and A-9) ρ l = 943.4 kg/m ρ v = 1.121 kg/m σ = 0.0550 N/m h fg = 2203 × 10 J/kg μ l = 0.232 × 10 −3 kg/m ⋅ s 120°C c pl = 4244 J/kg ⋅ °C Water Prl = 1.44 130°C Also, C sf = 0.0130 and n = 1.0 for the boiling of water on a mechanically polished stainless steel surface (Table 10-3) Note that we expressed the properties in units specified under Eq 10-2 in connection with their definitions in order to avoid unit manipulations Heating Analysis The excess temperature in this case is ΔT = Ts − Tsat = 130 − 120 = 10°C which is relatively low (less than 30°C) Therefore, nucleate boiling will occur The heat flux in this case can be determined from Rohsenow relation to be q& nucleate ⎡ g(ρ l − ρ v ) ⎤ = μ l h fg ⎢ ⎥ σ ⎣ ⎦ 1/ ⎛ ⎞ ⎜ c p ,l (Ts − Tsat ) ⎟ n ⎜ C h Pr ⎟ ⎝ sf fg l ⎠ ⎡ 9.8(943.4 - 1.121) ⎤ = (0.232 × 10 )(2203 × 10 ) ⎢ ⎥ 0.0550 ⎣ ⎦ −3 1/2 ⎛ ⎞ 4244(130 − 120) ⎜ ⎟ ⎜ 0.0130(2203 × 10 )1.44 ⎟ ⎝ ⎠ = 228,400 W/m The rate of heat transfer is Q& = Aq& nucleate = π (0.20 m) (228,400 W/m ) = 7174 W The rate of evaporation is Q& 7174 W m& evap = = = 0.003256 kg/s h fg 2203 × 10 kJ/kg Noting that the tank is half-filled, the mass of the water and the time it will take for the tank to empty are m= [ ] 1 ρ lV = (943.4 kg/m ) π (0.20 m) / × (0.30 m) = 4.446 kg 2 t = m& evap = 4.446 kg m = = 1365 s = 22.8 m& evap 0.003256 kg/s PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 10-58 10-81 Saturated ammonia vapor at a saturation temperature of Tsat = 25°C condenses on the outer surfaces of a tube bank in which cooling water flows The rate of condensation of ammonia, the overall heat transfer coefficient, and the tube length are to be determined Assumptions Steady operating conditions exist The tubes are isothermal The thermal resistance of the tube walls is negligible Properties The properties of ammonia at the saturation temperature of 25°C are hfg = 1166×103 J/kg and ρv = 7.809 kg/m3 (Table A-11) We assume that the tube temperature is 20°C Then, the properties of liquid ammonia at the film temperature of T f = (Tsat + Ts ) / = (25 + 20)/2 = 22.5°C are (Table A-11) ρ l = 606.5 kg/m μ l = 1.479 × 10 − kg/m ⋅ s c pl = 4765 J/kg ⋅ °C k l = 0.4869 W/m ⋅ °C The water properties at the average temperature of (14+17)/2 = 15.5°C are (Table A-9) ρ = 999.0 kg/m c p = 4185 J/kg ⋅ °C μ = 1.124 × 10 −3 kg/m ⋅ s k = 0.590 W/m ⋅ °C Pr = 7.98 Analysis (a) The modified latent heat of vaporization is h *fg = h fg + 0.68c pl (Tsat − Ts ) = 1166 ×10 J/kg + 0.68 × 4765 J/kg ⋅ °C(25 − 20)°C = 1182 × 10 J/kg The heat transfer coefficient for condensation on a single horizontal tube is h = hhorizontal ⎡ gρ l ( ρ l − ρ v )h *fg k l3 ⎤ ⎥ = 0.729 ⎢ ⎢⎣ μ l (Tsat − Ts ) D ⎥⎦ 1/ ⎡ (9.8 m/s )(606.5 kg/m )(606.5 − 7.809 kg/m )(1182 × 10 J/kg )(0.4869 W/m ⋅ °C) ⎤ = 0.729 ⎢ ⎥ (1.479 × 10 − kg/m ⋅ s)(25 − 20)°C(0.025 m) ⎣⎢ ⎦⎥ 1/ = 9280 W/m ⋅ °C Then the average heat transfer coefficient for a 4-pipe high vertical tier becomes ho = hhoriz, N tubes = N 1/ hhoriz, tube = 1/ (9280 W/m ⋅ °C) = 6562 W/m ⋅ °C The rate of heat transfer in the condenser is m& = 16 ρ AcV = 16 (999 kg/m )π ( 25 )( 025 m ) ( m/s ) = 15 69 kg/s Q& = m& c p (T out − T in ) = (15 69 kg/s )( 4185 J/kg ⋅ °C )(17 − 14 ) = 970 × 10 W Then the rate of condensation becomes m& cond = Q& h *fg = 970 × 10 W 1182 × 10 J/kg = 0.167 kg/s PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 10-59 (b) For the calculation of the heat transfer coefficient on the inner surfaces of the tubes, we first determine the Reynolds number Re = VD ρ μ = ( m/s)(0.025 m)(999.0 kg/m ) 1.124 × 10 - kg/m ⋅ s = 44 , 440 which is greater than 10,000 Therefore, the flow is turbulent Assuming fully developed flow, the Nusselt number and the heat transfer coefficient are determined to be Nu = 023 Re 0.8 Pr 0.4 = 023 ( 44 , 440 ) 0.8 ( 98 ) 0.4 = 275 hi = ( 590 W/m ⋅ °C) k Nu = ( 275 ) = 6511 W/m D 0.025 m ⋅ °C Let us check if the assumed value for the rube temperature was reasonable hi ΔTi = ho ΔTo ( 6511 )(T tube − 15 5) = ( 6562 )( 25 − T tube ) T tube = 20 °C which is sufficiently close to the assumed value of 20°C Disregarding thermal resistance of the tube walls, the overall heat transfer coefficient is determined from ⎛ 1 ⎞ ⎟ + U = ⎜⎜ ⎟ h h o ⎠ ⎝ i −1 ⎞ ⎛ =⎜ + ⎟ ⎝ 6511 6562 ⎠ −1 = 3268 W/m ⋅ °C (c) The tube length may be determined from Q& = UA Δ T ⎡ ⎤ 970 × 10 W = (3268 W/m ⋅ C)(16) π (0.025 m) L ⎢ 25 − (14 + 17 ) ⎥ ⎣ ⎦ L = 5.05 m PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 10-60 10-82 Steam at a saturation temperature of Tsat = 40°C condenses on the outside of a thin horizontal tube Heat is transferred to the cooling water that enters the tube at 25°C and exits at 35°C The rate of condensation of steam, the average overall heat transfer coefficient, and the tube length are to be determined Assumptions Steady operating conditions exist The tube can be taken to be isothermal at the bulk mean fluid temperature in the evaluation of the condensation heat transfer coefficient Liquid flow through the tube is fully developed The thickness and the thermal resistance of the tube is negligible Properties The properties of water at the saturation temperature of 40°C are hfg = 2407×103 J/kg and ρv = 0.05 kg/m3 The properties of liquid water at the film temperature of T f = (Tsat + Ts ) / = (50+20)/2 = 35°C and at the Steam 40°C Cooling water 35°C 25°C bulk fluid temperature of Tb = (Tin + Tout ) / = (25 + 35)/2 = 30°C are (Table A-9), At 30°C : At 35°C : ρ l = 996.0 kg/m ρ l = 994.0 kg/m μ l = 0.720 × 10 −3 kg/m ⋅ s c pl = 4178 J/kg ⋅ °C Condensate ν l = μ l / ρ l = 0.801× 10 −6 m /s c pl = 4178 J/kg ⋅ °C k l = 0.615 W/m ⋅ °C k l = 0.623 W/m ⋅ °C Pr = 5.42 Analysis The mass flow rate of water and the rate of heat transfer to the water are m& water = ρVAc = (996 kg/m )(2 m/s)[π (0.03 m) / 4] = 1.408 kg/s Q& = m& c (T − T ) = (1.408 kg/s)(4178 J/kg ⋅ °C)(35 − 25)°C = 58,820 W p out in The modified latent heat of vaporization is h *fg = h fg + 0.68c pl (Tsat − Ts ) = 2407 ×10 J/kg + 0.68 × 4178 J/kg ⋅ °C(40 − 30)°C = 2435 ×10 J/kg The heat transfer coefficient for condensation on a single horizontal tube is ho = hhorizontal ⎡ gρ l ( ρ l − ρ v )h *fg k l3 ⎤ ⎥ = 0.729 ⎢ ⎢⎣ μ l (Tsat − Ts ) D ⎥⎦ 1/ ⎡ (9.8 m/s )(994 kg/m )(994 − 0.05 kg/m )(2435 × 10 J/kg )(0.623 W/m ⋅ °C) ⎤ = 0.729 ⎢ ⎥ (0.720 × 10 −3 kg/m ⋅ s)(40 − 30)°C(0.03 m) ⎣⎢ ⎦⎥ 1/ = 9292 W/m ⋅ °C The average heat transfer coefficient for flow inside the tube is determined as follows: Vavg D (2 m/s)(0.03 m) = 74,906 ν 0.801 × 10-6 Nu = 0.023 Re0.8 Pr 0.4 = 0.023(74,906)0.8 (5.42)0.4 = 359 Re = hi = = kNu (0.615 W/m ⋅ °C) × 359 = = 7357 W/m ⋅ °C 0.03 m D Noting that the thermal resistance of the tube is negligible, the overall heat transfer coefficient becomes U= 1 = = 4106 W/m °C / hi + / ho / 7357 + / 9292 PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 10-61 The logarithmic mean temperature difference is: ΔTln = ΔTi − ΔTe 15 − = = 9.10°C ln(ΔTi / ΔTo ) ln(15 / 5) The tube length is determined from Q& = hAs ΔTln → L = Q& 58,820 W = = 16.7 m h(πD)ΔTln (4106 W/m ⋅ °C)π (0.03 m)(9.10°C) Note that the flow is turbulent, and thus the entry length in this case is 10D = 0.3 m is much shorter than the total tube length This verifies our assumption of fully developed flow PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual, you are using it without permission 10-62 10-83 Saturated ammonia at a saturation temperature of Tsat = 25°C condenses on the outer surface of vertical tube which is maintained at 15°C by circulating cooling water The rate of heat transfer to the coolant and the rate of condensation of ammonia are to be determined Assumptions Steady operating conditions exist The tube is isothermal The tube can be treated as a vertical plate The condensate flow is turbulent over the entire tube (this assumption will be verified) The density of vapor is much smaller than the density of liquid, ρ v