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Unformatted text preview: Chap 23 Heat Exchangers 2395 Cold water is heated by hot water in a heat exchanger. The net rate of heat transfer and the heat transfer surface area of the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 The overall heat transfer coefficient is constant and uniform. 5 The thickness of the tube is negligible. Properties The specific heats of the cold and hot water are given to be 4.18 and 4.19 kJ/kg. ° C, respectively. Analysis The heat capacity rates of the hot and cold fluids are C m C C m C h h ph c c pc = = ° = ° = = ° = ° , (0.25 kg / s)(4180 J / kg. C) W/ C (3 kg / s)(4190 J / kg. C) W/ C 1045 12 570 Therefore, C C c min = = ° 1045 W/ C and C C C = = = min max , . 1045 12 570 0083 Then the maximum heat transfer rate becomes ( ) , max min , , Q C T T h in c in = = ° ° ° = (1045 W/ C)(100 C15 C) W 88 825 The actual rate of heat transfer is W 31,350 = ° ° ° = = ) C 15 C 45 )( C W/ 1045 ( ) ( , , out h in h h T T C Q Then the effectiveness of this heat exchanger becomes ε = = = Q Q max , , . 31 350 88 825 0 35 The NTU of this heat exchanger is determined using the relation in Table 235 to be 438 . 1 083 . 35 . 1 35 . ln 1 083 . 1 1 1 ln 1 1 =  × =  = C C NTU ε ε Then the surface area of the heat exchanger is determined from 2 m 0.482 = ° ° = = → = C . W/m 950 ) C W/ 1045 )( 438 . ( 2 min min U C NTU A C UA NTU 2371 Hot Water 100 ° C 3 kg/s Cold Water 15 ° C 0.25 kg/s 45 ° C Chap 23 Heat Exchangers 2396 "GIVEN" T_cw_in=15 "[C]" T_cw_out=45 "[C]" m_dot_cw=0.25 "[kg/s]" C_p_cw=4.18 "[kJ/kgC]" T_hw_in=100 "[C], parameter to be varied" m_dot_hw=3 "[kg/s]" C_p_hw=4.19 "[kJ/kgC]" "U=0.95 [kW/m^2C], parameter to be varied" "ANALYSIS" "With EES, it is easier to solve this problem using LMTD method than NTU method. Below, we use LMTD method. Both methods give the same results." DELTAT_1=T_hw_inT_cw_out DELTAT_2=T_hw_outT_cw_in DELTAT_lm=(DELTAT_1DELTAT_2)/ln(DELTAT_1/DELTAT_2) Q_dot=U*A*DELTAT_lm Q_dot=m_dot_hw*C_p_hw*(T_hw_inT_hw_out) Q_dot=m_dot_cw*C_p_cw*(T_cw_outT_cw_in) T hw, in [C] Q [kW] A [m 2 ] 60 31.35 1.25 65 31.35 1.038 70 31.35 0.8903 75 31.35 0.7807 80 31.35 0.6957 85 31.35 0.6279 90 31.35 0.5723 95 31.35 0.5259 100 31.35 0.4865 105 31.35 0.4527 110 31.35 0.4234 115 31.35 0.3976 120 31.35 0.3748 U [kW/m 2C] Q [kW] A [m 2 ] 0.75 31.35 0.6163 0.8 31.35 0.5778 0.85 31.35 0.5438 0.9 31.35 0.5136 0.95 31.35 0.4865 1 31.35 0.4622 1.05 31.35 0.4402 1.1 31.35 0.4202 1.15 31.35 0.4019 1.2 31.35 0.3852 1.25 31.35 0.3698 2372 Chap 23 Heat Exchangers 60 70 80 90 100 110 120 31 31.25 31.5 31.75 32 0.2 0.4 0.6 0.8 1 1.2 1.4 T hw,in [C] Q [kW] A [m...
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.
 Spring '08
 Chung
 Heat Transfer

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