Solution manual heat and mass transfer a practical approach 3rd edition cengel CH07 1

7-43 7-57 A steam pipe is exposed to a light winds in the atmosphere The amount of heat loss from the steam during a certain period and the money the facility will save a year as a result of insulating the steam pipe are to be determined Assumptions Steady operating conditions exist Air is an ideal gas with constant properties The plant operates every day of the year for 10 h a day The local atmospheric pressure is atm Properties The properties of air at atm and the film temperature of (Ts + T∞)/2 = (75+5)/2 = 40°C are (Table A-15) Wind V = 10 km/h k = 0.02662 W/m.°C T∞ = 5°C ν = 1.702 × 10 -5 m /s Steam pipe Pr = 0.7255 Ts = 75°C Analysis The Reynolds number is D = 10 cm VD [(10 ×1000/3600) m/s](0.1 m) ε = 0.8 Re = = = 1.632 × 10 −5 ν 1.702 × 10 m /s The Nusselt number corresponding to this Reynolds number is determined to be 0.62 Re 0.5 Pr / hD Nu = = 0.3 + 1/ k + (0.4 / Pr) / [ ] ⎡ ⎛ Re ⎞ / ⎤ ⎢1 + ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ 4/5 ⎡ 0.62(1.632 × 10 ) 0.5 (0.7255)1 / ⎢ ⎛⎜ 1.632 × 10 = 0.3 + 1+ 1/ ⎢ ⎜⎝ 282,000 + (0.4 / 0.7255) / ⎣ [ ] ⎞ ⎟ ⎟ ⎠ 5/8 ⎤ 4/5 ⎥ ⎥ ⎦ = 71.19 The heat transfer coefficient is k 0.02662 W/m.°C h = Nu = (71.19) = 18.95 W/m °C D 0.1 m The rate of heat loss by convection is As = πDL = π (0.1 m)(12 m) = 3.77 m Q& = hAs (Ts − T∞ ) = (18.95 W/m °C)(3.77 m )(75 − 5)°C = 5001 W The rate of heat loss by radiation is Q& = εA σ (T − T ) rad s s surr [ ] = (0.8)(3.77 m )(5.67 × 10 -8 W/m K ) (75 + 273 K ) − (0 + 273 K ) = 1558 W The total rate of heat loss then becomes Q& = Q& + Q& = 5001 + 1558 = 6559 W total conv rad The amount of heat loss from the steam during a 10-hour work day is Q = Q& total Δt = (6.559 kJ/s)(10 h/day × 3600 s/h ) = 2.361 × 10 kJ/day The total amount of heat loss from the steam per year is Qtotal = Q& day ( no of days) = ( 2.361 × 10 kJ/day )(365 days/yr) = 8.619 × 10 kJ/yr Noting that the steam generator has an efficiency of 80%, the amount of gas used is Qtotal 8.619 × 10 kJ/yr ⎛ therm ⎞ = ⎜⎜ ⎟⎟ = 1021 therms/yr 0.80 0.80 ⎝ 105,500 kJ ⎠ Insulation reduces this amount by 90% The amount of energy and money saved becomes Energy saved = (0.90)Q gas = (0.90)(1021 therms/yr) = 919 therms/yr Q gas = Money saved = (Energy saved)(Unit cost of energy) = (919 therms/yr)($1.05/therm) = $965 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 7-44 7-58 A steam pipe is exposed to light winds in the atmosphere The amount of heat loss from the steam during a certain period and the money the facility will save a year as a result of insulating the steam pipes are to be determined Assumptions Steady operating conditions exist Air is an ideal gas with constant properties The plant operates every day of the year for 10 h The local atmospheric pressure is atm Properties The properties of air at atm and the film temperature of Wind (Ts + T∞)/2 = (75+5)/2 = 40°C are (Table A-15) V = 10 km/h k = 0.02662 W/m.°C T∞ = 5°C ν = 1.702 × 10 -5 m /s Steam pipe Ts = 75°C D = 10 cm ε = 0.8 Pr = 0.7255 Analysis The Reynolds number is VD [(10 ×1000/3600) m/s](0.1 m) Re = = = 1.632 × 10 ν 1.702 × 10 −5 m /s The Nusselt number corresponding to this Reynolds number is determined to be 0.62 Re 0.5 Pr / hD Nu = = 0.3 + 1/ k + (0.4 / Pr) / [ ] ⎡ ⎛ Re ⎞ / ⎤ ⎢1 + ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ 4/5 ⎡ 0.62(1.632 × 10 ) 0.5 (0.7255)1 / ⎢ ⎛⎜ 1.632 × 10 = 0.3 + 1+ 1/ ⎢ ⎜⎝ 282,000 + (0.4 / 0.7255) / ⎣ [ ] ⎞ ⎟ ⎟ ⎠ 5/8 ⎤ 4/5 ⎥ ⎥ ⎦ = 71.19 The heat transfer coefficient is k 0.02662 W/m.°C h = Nu = (71.19) = 18.95 W/m °C D 0.1 m The rate of heat loss by convection is As = πDL = π (0.1 m)(12 m) = 3.77 m Q& = hAs (Ts − T∞ ) = (18.95 W/m °C)(3.77 m )(75 - 5)°C = 5001 W For an average surrounding temperature of 0°C, the rate of heat loss by radiation and the total rate of heat loss are Q& = εA σ (T − T ) rad s s surr [ ] = (0.8)(3.77 m )(5.67 × 10 -8 W/m K ) (75 + 273 K ) − (0 + 273 K ) = 1558 W Q& total = Q& conv + Q& rad = 5001 + 1588 = 6559 W If the average surrounding temperature is -20°C, the rate of heat loss by radiation and the total rate of heat loss become = εA σ (T − T ) Q& rad s s surr [ = (0.8)(3.77 m )(5.67 × 10 -8 W/m K ) (75 + 273 K ) − (−20 + 273 K ) Q& total = 1807 W = Q& + Q& conv rad ] = 5001 + 1807 = 6808 W which is 6808 - 6559 = 249 W more than the value for a surrounding temperature of 0°C This corresponds to Q& 249 W % change = difference × 100 = × 100 = 3.8% (increase) & 6559 W Q total,0°C If the average surrounding temperature is 25°C, the rate of heat loss by radiation and the total rate of heat loss become 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 7-45 Q& rad = εAs σ (Ts − Tsurr ) = (0.8)(3.77 m )(5.67 × 10 -8 W/m K ) ⎡(75 + 273 K ) − (25 + 273 K ) ⎤ ⎢⎣ ⎥⎦ = 1159 W = Q& + Q& = 5001 + 1159 = 6160 W Q& total conv rad which is 6559 - 6160 = 399 W less than the value for a surrounding temperature of 0°C This corresponds to Q& 399 W % change = difference × 100 = × 100 = 6.1% (decrease) & 6559 W Q total,0°C Therefore, the effect of the temperature variations of the surrounding surfaces on the total heat transfer is less than 6% 7-59E An electrical resistance wire is cooled by a fan The surface temperature of the wire is to be determined Assumptions Steady operating conditions exist Radiation effects are negligible Air is an ideal gas with constant properties The local atmospheric pressure is atm Properties We assume the film temperature to be 200°F The properties of air at this temperature are (Table A-15E) Air k = 0.01761 Btu/h.ft.°F V = 20 ft/s ν = 2.406 × 10 - ft /s T∞ = 85°F Pr = 0.7124 Analysis The Reynolds number is VD (20 ft/s)(0.1/12 ft) = = 692.7 Re = ν 2.406 ×10 − ft /s The proper relation for Nusselt number corresponding to this Reynolds number is hD 0.62 Re 0.5 Pr / Nu = = 0.3 + 1/ k + (0.4 / Pr) / [ ] ⎡ ⎛ Re ⎞ / ⎤ ⎢1 + ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ 4/5 5/8 0.62(692.7) 0.5 (0.7124)1 / ⎡ ⎛ 692.7 ⎞ ⎤ ⎢1 + ⎜⎜ ⎟ = 0.3 + ⎟ ⎥ 1/ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ + (0.4 / 0.7124) / The heat transfer coefficient is [ h= Resistance wire D = 0.1 in ] 4/5 = 13.34 k 0.01761 Btu/h.ft.°F Nu = (13.34) = 28.19 Btu/h.ft °F D (0.1 / 12 ft) Then the average temperature of the outer surface of the wire becomes As = πDL = π (0.1 / 12 ft )(12 ft) = 0.3142 ft Q& (1500 × 3.41214) Btu/h ⎯→ Ts = T∞ + = 85°F + = 662.9°F Q& = hAs (Ts − T∞ ) ⎯ hA (28.19 Btu/h.ft °F)(0.3142 ft ) Discussion Repeating the calculations at the new film temperature of (85+662.9)/2=374°F gives Ts=668.3°F 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 7-46 7-60 The components of an electronic system located in a horizontal duct is cooled by air flowing over the duct The total power rating of the electronic device is to be determined Assumptions Steady operating conditions exist Radiation effects are negligible Air is an ideal gas with constant properties The local atmospheric pressure is atm Properties The properties of air at atm and the film temperature of 20 cm (Ts + T∞)/2 = (65+30)/2 = 47.5°C are (Table A-15) k = 0.02717 W/m.°C 20 cm 65°C ν = 1.774 × 10 -5 m /s Air Pr = 0.7235 30°C Analysis The Reynolds number is 200 m/min 1.5 m VD [(200/60) m/s](0.2 m) Re = = = 3.758 ×10 ν 1.774 ×10 −5 m /s Using the relation for a square duct from Table 7-1, the Nusselt number is determined to be hD Nu = = 0.102 Re 0.675 Pr / = 0.102(3.758 × 10 ) 0.675 (0.7235)1 / = 112.2 k The heat transfer coefficient is k 0.02717 W/m.°C h = Nu = (112.2) = 15.24 W/m °C D m Then the rate of heat transfer from the duct becomes As = (4 × 0.2 m)(1.5 m) = 1.2 m Q& = hAs (Ts − T∞ ) = (15.24 W/m °C)(1.2 m )(65 − 30)°C = 640 W 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 7-47 7-61 The components of an electronic system located in a horizontal duct is cooled by air flowing over the duct The total power rating of the electronic device is to be determined √ Assumptions Steady operating conditions exist Radiation effects are negligible Air is an ideal gas with constant properties Properties The properties of air at atm and the film temperature 20 cm of (Ts + T∞)/2 = (65+30)/2 = 47.5°C are (Table A-15) k = 0.02717 W/m.°C 20 cm ν = 1.774 × 10 -5 m /s 65°C Air Pr = 0.7235 30°C For a location at 4000 m altitude where the 1.5 m atmospheric pressure is 61.66 kPa, only kinematic 200 m/min viscosity of air will be affected Thus, ⎛ 101.325 ⎞ −5 −5 ⎟(1.774 × 10 ) = 2.915 × 10 m /s ⎝ 61.66 ⎠ ν @ 61.66 kPa = ⎜ Analysis The Reynolds number is VD [(200/60) m/s](0.2 m) Re = = = 2.287 × 10 −5 ν 2.915 × 10 m /s Using the relation for a square duct from Table 7-1, the Nusselt number is determined to be hD = 0.102 Re 0.675 Pr / = 0.102(2.287) 0.675 (0.7235)1 / = 80.21 Nu = k The heat transfer coefficient is 0.02717 W/m.°C k h = Nu = (80.21) = 10.90 W/m °C m D Then the rate of heat transfer from the duct becomes As = (4 × 0.2 m)(1.5 m) = 1.2 m Q& = hAs (Ts − T∞ ) = (10.90 W/m °C)(1.2 m )(65 − 30)°C = 458 W 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 7-48 7-62 A cylindrical electronic component mounted on a circuit board is cooled by air flowing across it The surface temperature of the component is to be determined Assumptions Steady operating conditions exist Radiation effects are negligible Air is an ideal gas with constant properties The local atmospheric pressure is atm Properties We assume the film temperature to be 50°C The properties of air at atm and at this temperature are (Table A-15) k = 0.02735 W/m.°C Air ν = 1.798 × 10 -5 m /s V = 240 m/min Pr = 0.7228 T∞ = 35°C Q& Analysis The Reynolds number is Resistor VD (240/60 m/s)(0.003 m) = = 667.4 Re = 0.4 W − ν 1.798 ×10 m /s D = 0.3 cm The proper relation for Nusselt number L = 1.8 cm corresponding to this Reynolds number is hD 0.62 Re 0.5 Pr / Nu = = 0.3 + 1/ k + (0.4 / Pr) / [ ] ⎡ ⎛ Re ⎞ / ⎤ ⎢1 + ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ 4/5 5/8 0.62(667.4) 0.5 (0.7228)1 / ⎡ ⎛ 667.4 ⎞ ⎤ ⎢1 + ⎜⎜ ⎟ = 0.3 + ⎟ ⎥ 1/ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ + (0.4 / 0.7228) / The heat transfer coefficient is 0.02735 W/m.°C k h = Nu = (13.17) = 120.0 W/m °C 0.003 m D Then the surface temperature of the component becomes [ ] 4/5 = 13.17 As = πDL = π (0.003 m)(0.018 m) = 0.0001696 m Q& 0.4 W ⎯→ Ts = T∞ + = 35 °C + = 54.6°C Q& = hAs (Ts − T∞ ) ⎯ hA (120.0 W/m °C)(0.0001696 m ) The film temperature is (54.6+35)/2=44.8°C, which is sufficiently close to the assumed value of 50°C Therefore, there is no need to repeat calculations 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 7-49 7-63 A cylindrical hot water tank is exposed to windy air The temperature of the tank after a 45-min cooling period is to be estimated Assumptions Steady operating conditions exist Radiation effects are negligible Air is an ideal gas with constant properties The surface of the tank is at the same temperature as the water temperature The heat transfer coefficient on the top and bottom surfaces is the same as that on the side surfaces Properties The properties of water at 80°C are (Table A-9) ρ = 971.8 kg/m3 Water tank c p = 4197 J/kg.°C D =50 cm L = 95 cm The properties of air at atm and at the anticipated film temperature of 50°C are (Table A-15) k = 0.02735 W/m.°C ν = 1.798 × 10 -5 m /s Pr = 0.7228 Analysis The Reynolds number is ⎞ ⎛ 40 × 1000 m/s ⎟(0.50 m) ⎜ VD ⎝ 3600 ⎠ Re = = 3.090 × 10 = −5 ν 1.798 × 10 m /s The proper relation for Nusselt number corresponding to this Reynolds number is 5/8 0.62 Re 0.5 Pr / ⎡ ⎛ Re ⎞ ⎤ ⎢1 + ⎜⎜ Nu = 0.3 + ⎟ ⎟ ⎥ 1/ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ + (0.4 / Pr )2 / [ Air V =40 km/h T∞ = 18°C 4/5 ] 4/5 5/8 0.62(3.090 × 10 ) 0.5 (0.7228)1 / ⎡⎢ ⎛⎜ 3.090 × 10 ⎞⎟ ⎤⎥ 1+ = 0.3 + = 484.8 1/ ⎢ ⎜⎝ 282,000 ⎟⎠ ⎥ + (0.4 / 0.7228)2 / ⎦ ⎣ The heat transfer coefficient is k 0.02735 W/m.°C h = Nu = (484.8) = 26.52 W/m °C D 0.50 m The surface area of the tank is D2 = π (0.5)(0.95) + 2π (0.5) / = 1.885 m As = πDL + 2π The rate of heat transfer is determined from ⎞ ⎛ 80 + T2 (Eq 1) Q& = hAs (Ts − T∞ ) = (26.52 W/m °C)(1.885 m )⎜⎜ − 18 ⎟⎟°C ⎠ ⎝ where T2 is the final temperature of water so that (80+T2)/2 gives the average temperature of water during the cooling process The mass of water in the tank is D2 m = ρV = ρπ L = (971.8 kg/m )π (0.50 m) (0.95 m)/4 = 181.3 kg The amount of heat transfer from the water is determined from Q = mc p (T2 − T1 ) = (181.3 kg)(4197 J/kg.°C)(80 − T2 )°C [ ] Then average rate of heat transfer is Q (181.3 kg)(4197 J/kg.°C)(80 − T2 )°C (Eq 2) Q& = = 45 × 60 s Δt Setting Eq to be equal to Eq we obtain the final temperature of water (181.3 kg)(4197 J/kg °C)(80 − T2 )°C ⎛ 80 + T2 ⎞ Q& = ( 26.52 W/m °C)(1.885 m )⎜⎜ − 18 ⎟⎟°C = 45 × 60 s ⎝ ⎠ T2 = 69.9°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 7-50 7-64 EES Prob 7-63 is reconsidered The temperature of the tank as a function of the cooling time is to be plotted Analysis The problem is solved using EES, and the solution is given below "GIVEN" D=0.50 [m] L=0.95 [m] T_w1=80 [C] T_infinity=18 [C] Vel=40 [km/h] time=45 [min] "PROPERTIES" Fluid$='air' k=Conductivity(Fluid$, T=T_film) Pr=Prandtl(Fluid$, T=T_film) rho=Density(Fluid$, T=T_film, P=101.3) mu=Viscosity(Fluid$, T=T_film) nu=mu/rho T_film=1/2*(T_w_ave+T_infinity) rho_w=Density(water, T=T_w_ave, P=101.3) C_p_w=CP(Water, T=T_w_ave, P=101.3)*Convert(kJ/kg-C, J/kg-C) T_w_ave=1/2*(T_w1+T_w2) "ANALYSIS" Re=(Vel*Convert(km/h, m/s)*D)/nu Nusselt=0.3+(0.62*Re^0.5*Pr^(1/3))/(1+(0.4/Pr)^(2/3))^0.25*(1+(Re/282000)^(5/8))^(4/5) h=k/D*Nusselt A=pi*D*L+2*pi*D^2/4 Q_dot=h*A*(T_w_ave-T_infinity) m_w=rho_w*V_w V_w=pi*D^2/4*L Q=m_w*C_p_w*(T_w1-T_w2) Q_dot=Q/(time*Convert(min, s)) Tw2 [C] 73.06 69.86 66.83 63.96 61.23 58.63 56.16 53.8 51.54 49.39 47.33 45.36 43.47 41.65 39.91 38.24 36.63 35.09 33.6 75 70 65 60 T w [C] time [min] 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 55 50 45 40 35 30 50 100 150 200 250 300 tim e [m in] 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 7-51 7-65 Air flows over a spherical tank containing iced water The rate of heat transfer to the tank and the rate at which ice melts are to be determined Assumptions Steady operating conditions exist Radiation effects are negligible Air is an ideal gas with constant properties The local atmospheric pressure is atm Properties The properties of air at atm pressure and the free stream temperature of 25°C are (Table A-15) k = 0.02551 W/m.°C ν = 1.562 × 10 -5 m /s μ ∞ = 1.849 × 10 −5 μ s , @ 0°C = 1.729 × 10 −5 Air V = m/s T∞ =25°C kg/m.s kg/m.s D =1.8 m Pr = 0.7296 Iced water Analysis The Reynolds number is 0°C (7 m/s)(1.8 m) VD 067 10 Re = = = × ν 1.562 × 10 −5 m /s The proper relation for Nusselt number corresponding to this Reynolds number is [ ] ⎛μ hD Nu = = + 0.4 Re 0.5 + 0.06 Re / Pr 0.4 ⎜⎜ ∞ k ⎝ μs [ = + 0.4(8.067 × 10 ) 0.5 ⎞ ⎟ ⎟ ⎠ 1/ + 0.06(8.067 × 10 ) 2/3 ](0.7296) 0.4 ⎛ ⎜ 1.849 × 10 −5 ⎜ 1.729 × 10 −5 ⎝ ⎞ ⎟ ⎟ ⎠ 1/ = 790.1 The heat transfer coefficient is k 0.02551 W/m.°C (790.1) = 11.20 W/m °C h = Nu = D m Then the rate of heat transfer is determined to be As = πD = π (1.8 m) = 10.18 m Q& = hA (T − T ) = (11.20 W/m °C)(10.18 m )(25 − 0)°C = 2850 W s s ∞ The rate at which ice melts is Q& 2.85 kW ⎯→ m& = = = 0.00854 kg/s = 0.512 kg/min Q& = m& h fg ⎯ h fg 333.7 kJ/kg 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 7-52 7-66 A cylindrical bottle containing cold water is exposed to windy air The average wind velocity is to be estimated Assumptions Steady operating conditions exist Radiation effects are negligible Air is an ideal gas with constant properties Heat transfer at the top and bottom surfaces is negligible Properties The properties of water at the average temperature of (T1 + T2)/2=(3+11)/2=7°C are (Table A-9) ρ = 999.8 kg/m c p = 4200 J/kg.°C Air The properties of air at atm and the film temperature Bottle V of (Ts + T∞)/2 = (7+27)/2 = 17°C are (Table A-15) D =10 cm T = 27°C ∞ k = 0.02491 W/m.°C L = 30 cm ν = 1.488 × 10 -5 m /s Pr = 0.7317 Analysis The mass of water in the bottle is D2 m = ρV = ρπ L = (999.8 kg/m )π (0.10 m) (0.30 m)/4 = 2.356 kg Then the amount of heat transfer to the water is Q = mc p (T2 − T1 ) = (2.356 kg)(4200 J/kg.°C)(11- 3)°C = 79,162 J The average rate of heat transfer is Q 79,162 J Q& = = = 29.32 W Δt 45 × 60 s The heat transfer coefficient is As = πDL = π (0.10 m)(0.30 m) = 0.09425 m Q& = hA (T − T ) ⎯ ⎯→ 29.32 W = h(0.09425 m )(27 − 7)°C ⎯ ⎯→ h = 15.55 W/m °C conv ∞ s s The Nusselt number is hD (15.55 W/m °C)(0.10 m) Nu = = = 62.42 k 0.02491 W/m.°C Reynolds number can be obtained from the Nusselt number relation for a flow over the cylinder Nu = 0.3 + 0.62 Re 0.5 Pr / [1 + (0.4 / Pr ) ] / 1/ ⎡ ⎛ Re ⎞ / ⎤ ⎢1 + ⎜⎜ ⎟⎟ ⎥ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ 4/5 5/8 0.62 Re 0.5 (0.7317)1 / ⎡ ⎛ Re ⎞ ⎤ ⎢1 + ⎜⎜ 62.42 = 0.3 + ⎟ ⎟ ⎥ 1/ ⎢⎣ ⎝ 282,000 ⎠ ⎥⎦ + (0.4 / 0.7317 )2 / [ ] 4/5 ⎯ ⎯→ Re = 12,856 Then using the Reynolds number relation we determine the wind velocity V (0.10 m) VD ⎯ ⎯→ 12,856 = ⎯ ⎯→ V = 1.91 m/s Re = ν 1.488 ×10 −5 m /s Flow across Tube Banks 7-67C In tube banks, the flow characteristics are dominated by the maximum velocity V max that occurs within the tube bank rather than the approach velocity V Therefore, the Reynolds number is defined on the basis of maximum velocity 7-68C The level of turbulence, and thus the heat transfer coefficient, increases with row number because of the combined effects of upstream rows in turbulence caused and the wakes formed But there is no significant change in turbulence level after the first few rows, and thus the heat transfer coefficient remains constant There is no change in transverse direction 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 7-53 7-69 Combustion air is preheated by hot water in a tube bank The rate of heat transfer to air and the pressure drop of air are to be determined Assumptions Steady operating conditions exist The surface temperature of the tubes is equal to the temperature of hot water Properties The exit temperature of air, and thus the mean temperature, is not known We evaluate the air properties at the assumed mean temperature of 20°C (will be checked later) and atm (Table A-15): k = 0.02514 W/m⋅K ρ = 1.204 kg/m3 cp =1.007 kJ/kg⋅K Pr = 0.7309 -5 μ = 1.825×10 kg/m⋅s Prs = Pr@ Ts = 90°C = 0.7132 Also, the density of air at the inlet temperature of 15°C (for use in the mass flow rate calculation at the inlet) is ρi = 1.225 kg/m3 Analysis It is given that D = 0.021 m, SL = ST = 0.05 m, Ts=90°C and V = 3.8 m/s Then the maximum velocity and the SL Reynolds number based on the maximum velocity V=3.8 m/s become Ti=15°C ST 0.05 V max = V= (3.8 m/s) = 6.552 m/s ST − D 0.05 − 0.021 ST ρV max D (1.204 kg/m )(6.552 m/s)(0.021 m) Re D = = = 9077 μ 1.825 × 10 −5 kg/m ⋅ s The average Nusselt number is determined using the proper relation from Table 7-2 to be Nu D = 0.27 Re 0D.63 Pr 0.36 (Pr/ Prs ) 0.25 D = 0.27(9077) 0.63 (0.7309) 0.36 (0.7309 / 0.7132) 0.25 = 75.60 This Nusselt number is applicable to tube banks with NL > 16 In our case the number of rows is NL = 8, and the corresponding correction factor from Table 7-3 is F = 0.967 Then the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, N L = FNu D = (0.967)(75.60) = 73.1 73.1(0.02514 W/m ⋅ °C) = 87.5 W/m ⋅ °C D 0.021 m The total number of tubes is N = NL ×NT = 8×8 = 64 For a unit tube length (L = m), the heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are As = NπDL = 64π (0.021 m)(1 m) = 4.222 m h= Nu D , N L k = m& = m& i = ρ iV ( N T S T L) = (1.225 kg/m )(3.8 m/s)(8)(0.05 m)(1 m) = 1.862 kg/s Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become ⎛ Ah⎞ ⎛ (4.222 m )(87.5 W/m ⋅ °C) ⎞ ⎟ = 28.41°C Te = Ts − (Ts − Ti ) exp⎜ − s ⎟ = 90 − (90 − 15) exp⎜ − ⎜ (1.862 kg/s)(1007 J/kg ⋅ °C) ⎟ ⎜ m& c p ⎟ ⎝ ⎠ ⎝ ⎠ (Ts − Ti ) − (Ts − Te ) (90 − 15) − (90 − 28.41) ΔTln = = = 68.08°C ln[(Ts − Ti ) /(Ts − Te )] ln[(90 − 15) /(90 − 28.41)] Q& = hA ΔT = (87.5 W/m ⋅ °C)(4.222 m )(68.08°C) = 25,150 W s ln For this square in-line tube bank, the friction coefficient corresponding to ReD = 9077 and SL/D = 5/2.1 = 2.38 is, from Fig 7-27a, f = 0.22 Also, χ = for the square arrangements Then the pressure drop across the tube bank becomes ⎞ ρV max (1.204 kg/m )(6.552 m/s) ⎛⎜ 1N ⎟ = 45.5 Pa ΔP = N L fχ = 8(0.22)(1) ⎜ kg ⋅ m/s ⎟ 2 ⎝ ⎠ Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (15 + 28.4)/2 = 21.7°C, which is fairly close to the assumed value of 20°C Therefore, there is no need to repeat calculations 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 7-54 7-70 Combustion air is preheated by hot water in a tube bank The rate of heat transfer to air and the pressure drop of air are to be determined Assumptions Steady operating conditions exist The surface temperature of the tubes is equal to the temperature of hot water Properties The exit temperature of air, and thus the mean temperature, is not known We evaluate the air properties at the assumed mean temperature of 20°C (will be checked later) and atm (Table A-15): k = 0.02514 W/m⋅K ρ = 1.204 kg/m3 cp =1.007 kJ/kg⋅K Pr = 0.7309 -5 μ = 1.825×10 kg/m⋅s Prs = Pr@ Ts = 90°C = 0.7132 Also, the density of air at the inlet temperature of 15°C (for use in the mass flow rate calculation at the inlet) is ρi = 1.225 kg/m3 Analysis It is given that D = 0.021 m, SL = ST = 0.06 SL Ts=90°C m, and V = 3.8 m/s Then the maximum velocity and the Reynolds number based on the maximum V=3.8 m/s velocity become Ti=15°C ST 0.06 V max = V= (3.8 m/s) = 5.846 m/s ST − D 0.06 − 0.021 S since S D > ( S T + D ) / Re D = T ρV max D (1.204 kg/m )(5.846 m/s)(0.021 m) = = 8099 μ 1.825 × 10 −5 kg/m ⋅ s The average Nusselt number is determined using the proper relation from Table 7-2 to be Nu D = 0.35( S T / S L ) 0.2 Re 0D.6 Pr 0.36 (Pr/ Prs ) 0.25 D = 0.35(0.06 / 0.06) 0.2 (8099) 0.6 (0.7309) 0.36 (0.7309 / 0.7132) 0.25 = 69.63 This Nusselt number is applicable to tube banks with NL > 16 In our case the number of rows is NL = 8, and the corresponding correction factor from Table 7-3 is F = 0.967 Then the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, N L = FNu D = (0.967)(69.63) = 67.33 67.33(0.02514 W/m ⋅ °C) = = 80.60 W/m ⋅ °C D 0.021 m The total number of tubes is N = NL ×NT = 8×8 = 64 For a unit tube length (L = m), the heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are As = NπDL = 64π (0.021 m)(1 m) = 4.222 m h= Nu D , N L k m& = m& i = ρ iV ( N T S T L) = (1.225 kg/m )(3.8 m/s)(8)(0.06 m)(1 m) = 2.234 kg/s Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become ⎛ Ah⎞ ⎛ (4.222 m )(80.60 W/m ⋅ °C) ⎞ ⎟ = 25.53°C Te = Ts − (Ts − Ti ) exp⎜ − s ⎟ = 90 − (90 − 15) exp⎜ − ⎜ (2.234 kg/s)(1007 J/kg ⋅ °C) ⎟ ⎜ m& c p ⎟ ⎝ ⎠ ⎝ ⎠ (Ts − Ti ) − (Ts − Te ) (90 − 15) − (90 − 25.53) ΔTln = = = 69.60°C ln[(Ts − Ti ) /(Ts − Te )] ln[(90 − 15) /(90 − 25.53)] Q& = hA ΔT = (80.60 W/m ⋅ °C)(4.222 m )(69.60°C) = 23,690 W s ln For this staggered tube bank, the friction coefficient corresponding to ReD = 8099 and ST/D = 6/2.1 = 2.86 is, from Fig 7-27b, f = 0.30 Also, χ = for the square arrangements Then the pressure drop across the tube bank becomes ⎞ ρV max (1.204 kg/m )(5.846 m/s) ⎛⎜ 1N ⎟ = 49.4 Pa ΔP = N L fχ = 8(0.30)(1) ⎜ ⎟ 2 ⎝ kg ⋅ m/s ⎠ Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (15 +25.5)/2 = 20.3°C, which is fairly close to the assumed value of 20°C Therefore, there is no need to repeat calculations 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 7-55 7-71 Combustion air is heated by condensing steam in a tube bank The rate of heat transfer to air, the pressure drop of air, and the rate of condensation of steam are to be determined Assumptions Steady operating conditions exist The surface temperature of the tubes is equal to the temperature of steam Properties The exit temperature of air, and thus the mean temperature, is not known We evaluate the air properties at the assumed mean temperature of 35°C (will be checked later) and atm (Table A-15): k = 0.02625 W/m⋅K ρ = 1.145 kg/m3 cp =1.007 kJ/kg⋅K Pr = 0.7268 -5 μ = 1.895×10 kg/m⋅s Prs = Pr@ Ts = 100°C = 0.7111 Also, the density of air at the inlet temperature of 20°C (for use in the mass flow rate calculation at the inlet) is ρi = 1.204 kg/m3 The enthalpy of vaporization of water at 100°C is hfg = 2257 kJ/kg-K (Table A9) Analysis (a) It is given that D = 0.016 m, SL = ST = 0.04 Ts=100°C SL m, and V = 5.2 m/s Then the maximum velocity and the Reynolds number based on the maximum velocity become V=5.2 m/s Ti=20°C ST 0.04 V max = V= (5.2 m/s) = 8.667 m/s ST − D 0.04 − 0.016 ST since S D > ( S T + D ) / Re D = ρV max D (1.145 kg/m )(8.667 m/s)(0.016 m) = = 8379 μ 1.895 × 10 −5 kg/m ⋅ s The average Nusselt number is determined using the proper relation from Table 7-2 to be D Nu D = 0.35( S T / S L ) 0.2 Re 0D.6 Pr 0.36 (Pr/ Prs ) 0.25 = 0.35(0.04 / 0.04) 0.2 (8379) 0.6 (0.7268) 0.36 (0.7268 / 0.7111) 0.25 = 70.87 Since NL =20, which is greater than 16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, N L = Nu D = 70.87 h= Nu D , N L k D = 70.87(0.02625 W/m ⋅ °C) = 116.3 W/m ⋅ °C 0.016 m The total number of tubes is N = NL ×NT = 20×10 = 200 For a unit tube length (L = m), the heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are As = NπDL = 200π (0.016 m)(1 m) = 10.05 m m& = m& i = ρ iV ( N T S T L) = (1.204 kg/m )(5.2 m/s)(10)(0.04 m)(1 m) = 2.504 kg/s Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become ⎛ Ah⎞ ⎛ (10.05 m )(116.3 W/m ⋅ °C) ⎞ ⎟ = 49.68°C Te = Ts − (Ts − Ti ) exp⎜ − s ⎟ = 100 − (100 − 20) exp⎜ − ⎜ (2.504 kg/s)(1007 J/kg ⋅ °C) ⎟ ⎜ m& c p ⎟ ⎝ ⎠ ⎝ ⎠ (Ts − Ti ) − (Ts − Te ) (100 − 20) − (100 − 49.68) ΔTln = = = 64.01°C ln[(Ts − Ti ) /(Ts − Te )] ln[(100 − 20) /(100 − 49.68)] Q& = hAs ΔTln = (116.3 W/m ⋅ °C)(10.05 m )(64.02°C) = 74,830 W 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 7-56 (b) For this staggered tube bank, the friction coefficient corresponding to ReD = 8379 and ST/D = 4/1.6 = 2.5 is, from Fig 7-27b, f = 0.33 Also, χ = for the square arrangements Then the pressure drop across the tube bank becomes ΔP = N L fχ ρVmax = 20(0.33)(1) (1.145 kg/m )(8.667 m/s) 2 ⎛ 1N ⎜ ⎜ kg ⋅ m/s ⎝ ⎞ ⎟ = 284 Pa ⎟ ⎠ (c) The rate of condensation of steam is ⎯→ m& cond = Q& = m& cond h fg @ 100°C ⎯ Q& h fg @ 100°C = 74.83 kW = 0.0332 kg/s = 1.99 kg/min 2257 kJ/kg ⋅ °C Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (20 + 49.7)/2 = 34.9°C, which is very close to the assumed value of 35°C Therefore, there is no need to repeat calculations 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 7-57 7-72 Combustion air is heated by condensing steam in a tube bank The rate of heat transfer to air, the pressure drop of air, and the rate of condensation of steam are to be determined Assumptions Steady operating conditions exist The surface temperature of the tubes is equal to the temperature of steam Properties The exit temperature of air, and thus the mean temperature, is not known We evaluate the air properties at the assumed mean temperature of 35°C (will be checked later) and atm (Table A-15): k = 0.02625 W/m⋅K ρ = 1.145 kg/m3 cp =1.007 kJ/kg⋅K Pr = 0.7268 -5 μ = 1.895×10 kg/m⋅s Prs = Pr@ Ts = 100°C = 0.7111 Also, the density of air at the inlet temperature of 20°C (for use in the mass flow rate calculation at the inlet) is ρi = 1.204 kg/m3 The enthalpy of vaporization of water at 100°C is hfg = 2257 kJ/kg-K (Table A9) Analysis (a) It is given that D = 0.016 m, SL = ST = 0.06 Ts=100°C m, and V = 5.2 m/s Then the maximum velocity and the SL Reynolds number based on the maximum velocity V=5.2 m/s become Ti=20°C ST 0.06 V max = V= (5.2 m/s) = 7.091 m/s ST − D 0.06 − 0.016 ST Re D = ρV max D (1.145 kg/m )(7.091 m/s)(0.016 m) = = 6855 μ 1.895 × 10 −5 kg/m ⋅ s The average Nusselt number is determined using the proper relation from Table 7-2 to be Nu D = 0.27 Re 0D.63 Pr 0.36 (Pr/ Prs ) 0.25 D = 0.27(6855) 0.63 (0.7268) 0.36 (0.7268 / 0.7111) 0.25 = 63.17 Since NL =20, which is greater than 16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, N L = Nu D = 63.17 h= Nu D , N L k D = 63.17(0.02625 W/m ⋅ °C) = 103.6 W/m ⋅ °C 0.016 m The total number of tubes is N = NL ×NT = 20×10 = 200 For a unit tube length (L = m), the heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are As = NπDL = 200π (0.016 m)(1 m) = 10.05 m m& = m& i = ρ iV ( N T S T L) = (1.204 kg/m )(5.2 m/s)(10)(0.06 m)(1 m) = 3.756 kg/s Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become ⎛ Ah⎞ ⎛ (10.05 m )(103.6 W/m ⋅ °C) ⎞ ⎟ = 39.25°C Te = Ts − (Ts − Ti ) exp⎜ − s ⎟ = 100 − (100 − 20) exp⎜ − ⎜ (3.756 kg/s)(1007 J/kg ⋅ °C) ⎟ ⎜ m& c p ⎟ ⎝ ⎠ ⎝ ⎠ (Ts − Ti ) − (Ts − Te ) (100 − 20) − (100 − 39.25) ΔTln = = = 69.93°C ln[(Ts − Ti ) /(Ts − Te )] ln[(100 − 20) /(100 − 39.25)] Q& = hAs ΔTln = (103.6 W/m ⋅ °C)(10.05 m )(69.93°C) = 72,810 W = 72.81 kW 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 7-58 (b) For this in-line arrangement tube bank, the friction coefficient corresponding to ReD = 6855 and SL/D = 6/1.6 = 3.75 is, from Fig 7-27a, f = 0.12 Note that an accurate reading of friction factor does not seem to be possible in this case Also, χ = for the square arrangements Then the pressure drop across the tube bank becomes ΔP = N L fχ ρV max = 20(0.12)(1) (1.145 kg/m )(7.091 m/s) 2 ⎛ 1N ⎜ ⎜ kg ⋅ m/s ⎝ ⎞ ⎟ = 69.1 Pa ⎟ ⎠ (c) The rate of condensation of steam is ⎯→ m& cond = Q& = m& cond h fg @ 100°C ⎯ Q& h fg @ 100°C = 72.81 kW = 0.0323 kg/s = 1.94 kg/min 2257 kJ/kg ⋅ °C Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (20 + 43.4)/2 = 31.7°C, which is fairly close to the assumed value of 35°C Therefore, there is no need to repeat calculations 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 7-59 7-73 Water is preheated by exhaust gases in a tube bank The rate of heat transfer, the pressure drop of exhaust gases, and the temperature rise of water are to be determined Assumptions Steady operating conditions exist For exhaust gases, air properties are used Properties The exit temperature of air, and thus the mean temperature, is not known We evaluate the air properties at the assumed mean temperature of 250°C (will be checked later) and atm (Table A-15): k = 0.04104 W/m⋅K ρ = 0.6746 kg/m3 cp =1.033 kJ/kg⋅K Pr = 0.6946 μ = 2.76×10-5 kg/m⋅s Prs = Pr@ Ts = 80°C = 0.7154 The density of air at the inlet temperature of 300°C (for use in the mass flow rate calculation at the inlet) is ρi = 0.6158 kg/m3 The specific heat of water at 80°C is 4.197 kJ/kg.°C (Table A-9) Analysis (a) It is given that D = 0.021 m, SL = ST = 0.08 Ts=80°C m, and V = 4.5 m/s Then the maximum velocity and the Reynolds number based on the maximum velocity SL V=4.5 m/s become Ti=300°C ST 0.08 V max = V= (4.5 m/s) = 6.102 m/s ST − D 0.08 − 0.021 S T Re D ρV max D (0.6746 kg/m )(6.102 m/s)(0.021 m) = = = 3132 μ 2.76 × 10 −5 kg/m ⋅ s The average Nusselt number is determined using the proper relation from Table 7-2 to be Nu D = 0.27 Re 0D.63 Pr 0.36 (Pr/ Prs ) 0.25 D = 0.27(3132) 0.63 (0.6946) 0.36 (0.6946 / 0.7154) 0.25 = 37.46 Since NL =16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, N L = Nu D = 37.46 h= Nu D , N L k D = 37.46(0.04104 W/m ⋅ °C) = 73.2 W/m ⋅ °C 0.021 m The total number of tubes is N = NL ×NT = 16×8 = 128 For a unit tube length (L = m), the heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are As = NπDL = 128π (0.021 m)(1 m) = 8.445 m m& = m& i = ρ iV ( N T S T L) = (0.6158 kg/m )(4.5 m/s)(8)(0.08 m)(1 m) = 1.774 kg/s Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer become ⎛ Ah⎞ ⎛ (8.445 m )(73.2 W/m ⋅ °C) ⎞ ⎟ = 237.0°C Te = Ts − (Ts − Ti ) exp⎜ − s ⎟ = 80 − (80 − 300) exp⎜ − ⎜ (1.774 kg/s)(1033 J/kg ⋅ °C) ⎟ ⎜ m& c p ⎟ ⎝ ⎠ ⎝ ⎠ (Ts − Ti ) − (Ts − Te ) (80 − 300) − (80 − 237) ΔTln = = = 186.7°C ln[(Ts − Ti ) /(Ts − Te )] ln[(80 − 300) /(80 − 237)] Q& = hAs ΔTln = (73.2 W/m ⋅ °C)(8.445 m )(186.7°C) = 115,430 W 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 7-60 (b) For this in-line arrangement tube bank, the friction coefficient corresponding to ReD = 3132 and SL/D = 8/2.1 = 3.81 is, from Fig 7-27a, f = 0.18 Also, χ = for the square arrangements Then the pressure drop across the tube bank becomes ΔP = N L fχ ρV max = 16(0.18)(1) (0.6746 kg/m )(6.102 m/s) 2 ⎛ 1N ⎜ ⎜ kg ⋅ m/s ⎝ ⎞ ⎟ = 36.2 Pa ⎟ ⎠ (c) The temperature rise of water is ⎯→ ΔTwater = Q& = m& water c p , water ΔTwater ⎯ Q& 115.43 kW = = 4.6°C m& water c p , water (6 kg/s)(4.197 kJ/kg ⋅ °C) Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (300 + 237)/2 = 269°C, which is sufficiently close to the assumed value of 250°C Therefore, there is no need to repeat calculations 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 7-61 7-74 Water is heated by a bundle of resistance heater rods The number of tube rows is to be determined Assumptions Steady operating conditions exist The surface temperature of the rods is constant Properties The properties of water at the mean temperature of (15°C +65°C)/2=40°C are (Table A-9): k = 0.631 W/m⋅K ρ = 992.1 kg/m3 cp =4.179 kJ/kg⋅K Pr = 4.32 μ = 0.653×10-3 kg/m⋅s Prs = Pr@ Ts = 90°C = 1.96 Also, the density of water at the inlet temperature of 15°C (for use in the mass flow rate calculation at the inlet) is ρi =999.1 kg/m3 Analysis It is given that D = 0.01 m, SL = 0.04 m and ST Ts=90°C = 0.03 m, and V = 0.8 m/s Then the maximum velocity SL and the Reynolds number based on the maximum V=0.8 m/s velocity become Ti=15°C ST 0.03 V max = V= (0.8 m/s) = 1.20 m/s ST − D 0.03 − 0.01 S T ρV max D (992.1 kg/m )(1.20 m/s)(0.01 m) Re D = = = 18,232 μ 0.653 × 10 −3 kg/m ⋅ s The average Nusselt number is determined using the proper relation from Table 7-2 to be Nu D = 0.27 Re 0D.63 Pr 0.36 (Pr/ Prs ) 0.25 D = 0.27(18,232) 0.63 (4.32) 0.36 (4.32 / 1.96) 0.25 = 269.3 Assuming that NL > 16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, N L = Nu D = 269.3 269.3(0.631 W/m ⋅ °C) = = 16,994 W/m ⋅ °C D 0.01 m Consider one-row of tubes in the transpose direction (normal to flow), and thus take NT =1 Then the heat transfer surface area becomes As = N tube πDL = (1× N L )π (0.01 m)(4 m) = 0.1257 N L h= Nu D , N L k Then the log mean temperature difference, and the expression for the rate of heat transfer become (Ts − Ti ) − (Ts − Te ) (90 − 15) − (90 − 65) ΔTln = = = 45.51°C ln[(Ts − Ti ) /(Ts − Te )] ln[(90 − 15) /(90 − 65)] Q& = hAs ΔTln = (16,994 W/m ⋅ °C)(0.1257N L )(45.51°C) = 97,220 N L The mass flow rate of water through a cross-section corresponding to NT =1 and the rate of heat transfer are m& = ρAcV = (999.1 kg/m )(4 × 0.03 m )(0.8 m/s) = 95.91 kg/s Q& = m& c p (Te − Ti ) = (95.91 kg/s)(4179 J/kg.C) (65 − 15)°C = 2.004 × 10 W Substituting this result into the heat transfer expression above we find the number of tube rows Q& = hA ΔT → 2.004 ×10 W = 97,220 N → N = 206 s ln L L 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 7-62 7-75 Air is cooled by an evaporating refrigerator The refrigeration capacity and the pressure drop across the tube bank are to be determined Assumptions Steady operating conditions exist The surface temperature of the tubes is equal to the temperature of refrigerant Properties The exit temperature of air, and thus the mean temperature, is not known We evaluate the air properties at the assumed mean temperature of -5°C (will be checked later) and atm (Table A-15): k = 0.02326 W/m⋅K ρ = 1.317 kg/m3 cp =1.006 kJ/kg⋅K Pr = 0.7375 -5 μ = 1.705×10 kg/m⋅s Prs = Pr@ Ts = -20°C = 0.7408 Also, the density of air at the inlet temperature of 0°C (for use in the mass flow rate calculation at the inlet) is ρi = 1.292 kg/m3 Analysis It is given that D = 0.008 m, SL = ST = 0.015 m, Ts=-20°C and V = m/s Then the maximum velocity and the SL Reynolds number based on the maximum velocity become V=4 m/s ST 0.015 V max = V= (4 m/s) = 8.571 m/s Ti=0°C ST − D 0.015 − 0.008 Re D = ρV max D (1.317 kg/m )(8.571 m/s)(0.008 m) = = 5296 μ 1.705 × 10 −5 kg/m ⋅ s ST The average Nusselt number is determined using the proper relation from Table 7-2 to be Nu D = 0.27 Re 0D.63 Pr 0.36 (Pr/ Prs ) 0.25 = 0.27(5296) 0.63 (0.7375) 0.36 (0.7375 / 0.7408) 0.25 = 53.62 Since NL > 16 the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, N L = FNu D = 53.62 D 53.62(0.02326 W/m ⋅ °C) = 155.9 W/m ⋅ °C D 0.008 m The total number of tubes is N = NL ×NT = 30×15 = 450 The heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are As = NπDL = 450π (0.008 m)(0.4 m) = 4.524 m h= Nu D , N L k = m& = m& i = ρiV ( NT ST L) = (1.292 kg/m3 )(4 m/s)(15)(0.015 m)(0.4 m) = 0.4651 kg/s Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer (refrigeration capacity) become ⎛ Ah⎞ ⎛ (4.524 m )(155.9 W/m ⋅ °C) ⎞ ⎟ = −15.57°C Te = Ts − (Ts − Ti ) exp⎜ − s ⎟ = −20 − (−20 − 0) exp⎜ − ⎜ (0.4651 kg/s)(1006 J/kg ⋅ °C) ⎟ ⎜ m& c p ⎟ ⎝ ⎠ ⎝ ⎠ (Ts − Ti ) − (Ts − Te ) (−20 − 0) − [−20 − (−15.57)] ΔTln = = = 10.33°C ln[(Ts − Ti ) /(Ts − Te )] ln[(−20 − 0) /( −20 + 15.57)] Q& = hA ΔT = (155.9 W/m ⋅ °C)(4.524 m )(10.33°C) = 7286 W s ln For this square in-line tube bank, the friction coefficient corresponding to ReD = 5296 and SL/D = 1.5/0.8 = 1.875 is, from Fig 7-27a, f = 0.27 Also, χ = for the square arrangements Then the pressure drop across the tube bank becomes ⎞ ρVmax (1.317 kg/m )(8.571 m/s) ⎛⎜ N ⎟ = 392 Pa ΔP = N L fχ = 30(0.27)(1) ⎜ kg ⋅ m/s ⎟ 2 ⎝ ⎠ Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (0 -15.6)/2 = -7.8°C, which is fairly close to the assumed value of -5°C Therefore, there is no need to repeat calculations 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 7-63 7-76 Air is cooled by an evaporating refrigerator The refrigeration capacity and the pressure drop across the tube bank are to be determined Assumptions Steady operating conditions exist The surface temperature of the tubes is equal to the temperature of refrigerant Properties The exit temperature of air, and thus the mean temperature, is not known We evaluate the air properties at the assumed mean temperature of -5°C (will be checked later) and atm (Table A-15): k = 0.02326 W/m⋅K ρ = 1.316 kg/m3 cp =1.006 kJ/kg⋅K Pr = 0.7375 -5 μ = 1.705×10 kg/m⋅s Prs = Pr@ Ts = -20°C = 0.7408 Ts=-20°C SL Also, the density of air at the inlet temperature of 0°C (for use in the mass flow rate calculation at the inlet) is ρi = 1.292 kg/m3 V=4 m/s Analysis It is given that D = 0.008 m, SL = ST = 0.015 m, Ti=0°C and V = m/s Then the maximum velocity and the Reynolds number based on the maximum velocity become ST ST 0.015 V max = V= (4 m/s) = 8.571 m/s ST − D 0.015 − 0.008 Re D = ρV max D (1.317 kg/m )(8.571 m/s)(0.008 m) = = 5296 μ 1.705 × 10 −5 kg/m ⋅ s D The average Nusselt number is determined using the proper relation from Table 7-2 to be Nu D = 0.35( S T / S L ) 0.2 Re 0D.6 Pr 0.36 (Pr/ Prs ) 0.25 = 0.35(0.015 / 0.015) 0.2 (5296) 0.6 (0.7375) 0.36 (0.7375 / 0.7408) 0.25 = 53.75 Since NL > 16 the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, N L = FNu D = 53.75 53.75(0.02326 W/m ⋅ °C) = 156.3 W/m ⋅ °C D 0.008 m The total number of tubes is N = NL ×NT = 30×15 = 450 The heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are As = NπDL = 450π (0.008 m)(0.4 m) = 4.524 m h= Nu D , N L k = m& = m& i = ρiV ( NT ST L) = (1.292 kg/m3 )(4 m/s)(15)(0.015 m)(0.4 m) = 0.4651 kg/s Then the fluid exit temperature, the log mean temperature difference, and the rate of heat transfer (refrigeration capacity) become ⎛ Ah⎞ ⎛ (4.524 m )(156.3 W/m ⋅ °C) ⎞ ⎟ = −15.59°C Te = Ts − (Ts − Ti ) exp⎜ − s ⎟ = −20 − (−20 − 0) exp⎜ − ⎜ (0.4651 kg/s)(1006 J/kg ⋅ °C) ⎟ ⎜ m& c p ⎟ ⎝ ⎠ ⎝ ⎠ (Ts − Ti ) − (Ts − Te ) (−20 − 0) − [−20 − (−15.59)] ΔTln = = = 10.32°C ln[(Ts − Ti ) /(Ts − Te )] ln[(−20 − 0) /( −20 + 15.59)] Q& = hA ΔT = (156.3 W/m ⋅ °C)(4.524 m )(10.31°C) = 7290 W s ln For this staggered arrangement tube bank, the friction coefficient corresponding to ReD = 5296 and SL/D = 1.5/0.8 = 1.875 is, from Fig 7-27b, f = 0.44 Also, χ = for the square arrangements Then the pressure drop across the tube bank becomes ⎞ ρV max (1.317 kg/m )(8.571 m/s) ⎛⎜ 1N ⎟ = 639 Pa ΔP = N L fχ = 30(0.44)(1) ⎜ kg ⋅ m/s ⎟ 2 ⎝ ⎠ Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (0 -15.6)/2 = -7.8°C, which is fairly close to the assumed value of -5°C Therefore, there is no need to repeat calculations 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 7-64 7-77 Air is heated by hot tubes in a tube bank The average heat transfer coefficient is to be determined Assumptions Steady operating conditions exist The surface temperature of the tubes is constant Properties The exit temperature of air, and thus the mean temperature, is not known We evaluate the air properties at the assumed mean temperature of 70°C and atm (Table A-15): k = 0.02881 W/m-K ρ = 1.028 kg/m3 cp =1.007 kJ/kg-K Pr = 0.7177 μ = 2.052×10-5 kg/m-s Prs = Pr@ Ts = 140°C = 0.7041 Ts=140°C Analysis It is given that D = 0.02 m, SL = ST = 0.06 m, and SL V = m/s Then the maximum velocity and the Reynolds V=6 m/s number based on the maximum velocity become Ti=20°C ST 0.06 V max = V= (6 m/s) = m/s ST − D 0.06 − 0.02 ST Re D = ρV max D (1.028 kg/m )(9 m/s)(0.02 m) = = 9018 μ 2.052 × 10 −5 kg/m ⋅ s The average Nusselt number is determined using the proper relation from Table 7-2 to be Nu D = 0.27 Re 0D.63 Pr 0.36 (Pr/ Prs ) 0.25 = 0.27(9018) 0.63 (0.7177) 0.36 (0.7177 / 0.7041) D 0.25 = 74.70 Since NL > 16, the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, N L = Nu D = 74.70 h= Nu D , N L k D = 74.70(0.02881 W/m ⋅ °C) = 107.6 W/m ⋅ °C 0.02 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  ... °C 0.0 21 m The total number of tubes is N = NL ×NT = 16 ×8 = 12 8 For a unit tube length (L = m), the heat transfer surface area and the mass flow rate of air (evaluated at the inlet) are As = NπDL... using this Manual, you are using it without permission 7- 51 7-65 Air flows over a spherical tank containing iced water The rate of heat transfer to the tank and the rate at which ice melts are to... °C) = 11 6.3 W/m ⋅ °C 0. 016 m The total number of tubes is N = NL ×NT = 20 10 = 200 For a unit tube length (L = m), the heat transfer surface area and the mass flow rate of air (evaluated at the