HW6_solution - Homework 6 solution 4.11 -20C is -4F, and...

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4.11 -20C is -4F, and 50C is 122F Use subscript I to refer to interstage state. There is no interstage cooler or economizer between the compressors. Stream numbers from figure 4.9 in text. (a) condenser (P sat at 122F) = 1.7 MPa =P 3 Evaporator (P sat at -4F) = 0.24 MPa =P 2 Interstage (P I /P 2 ) = (P 3 /P I ) so P I = sqrt(1.7*0.24) =0.64 MPa (b) cooling in evaporator, Q C =H 2 -H 1 = 877-665 = 212 kJ/kg m = (10,000,000 kJ/hr)(kg/212 kJ) = 47,200 kg/hr flow must be the same through both compressors. (c) Take efficiency as per stage, W’ compressorI = H’ I -H 2 =945-877 = 68 kJ/kg, T’ I = 82F W compressorI = 68/0.8 = 85 kJ/kg, H I = 877+85 =962 kJ/kg, T I = 105F W I =(85 kJ/kg)(47,200 kg/hr) = (4.012 kJ/hr)(hr/3600s) = 1,114 kW = 1500 hp W’ compressorII = 1020-962 =58 kJ/kg W compressorI =58/0.8=72.5 kJ/kg, H 3 =962+72.5=1035 kJ/kg, T 3 =192F W II = (72.5)(47,200)=(3.422e6 kJ/hr)(hr/3600s)=950 kW (d) H 4 =665 kJ/kg Q H =m(H 4 -H 3 )47,200(665-1035)=-1.75e7 kJ/hr COP=1e7/((4.0+3.42)e6)=1.35 4.12 Stream numbers from figure 4.11 in text. Solved with propane chart ini the appendix. -20C is -4F, -50C is 122F (a) condenser (P sat at 122F) = 1.7 MPa=P 5 Evaporator (P sat at -4F) = 0.24 MPa = P 2 Interstage (P 3 /P 2 ) = (P 5 /P 3 ), so P 3 =sqrt(1.7(0.24))=0.64 MPa Shaded values from problem statement. State T(F) P(MPa) H(kJ/kg) S(BTU/lb-R) m(kg/hr) 1 -4 0.24 565 32,051 2 -4 0.24 877 1.36 3’ 0.64 924 1.36 3 0.64 936 4 0.64 929 1.37 45,787 5’ 1.7 977 1.37 5 1.7 989 6 122 1.7 665 45,787 7 0.64 912 13,736 8 0.64 565 32,051 (b) Fill in states on table for H 6 , H 2 , H 8 (H 1 ), H 7 Cooling duty of evaporator, Q
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This note was uploaded on 05/10/2008 for the course CHBE 2110 taught by Professor Gallivan during the Spring '08 term at Georgia Institute of Technology.

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HW6_solution - Homework 6 solution 4.11 -20C is -4F, and...

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