HW4_solution

# HW4_solution - Homework 4 solution 3.18(a inlet Hin=3446...

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Homework 4 solution 3.18 (a) inlet H in =3446, S in =7.0922, outlet is two phase, q’=(7.0922-1.303)/(7.359-1.303)=0.956 H’=417.5+0.956(2675-417.5)=2575 kJ/kg W S = H’=2575-3446=-871 kJ/kg (b) Ws’= H= η H’=0.8*(-871)=-696kJ/kg 3.23 We can assume five states and their properties in the following table. Shaded numbers known at start. state T( o C) P(MPa) H S q 1 550 8 3521.8 6.8799 2’ 5 3361.3 6.8799 2 5 3393 6.9228 3’ 0.8 2892 6.9228 3 0.8 2992 State 2’ is isentropic process from state 1 to P 2 =5 MPa, state 3’ is also isentropic process from state 2 to P 3 =0.8 MPa. (2) state 2 between 450 and 500 o C, H 2 ’=3317.2+(6.8799-6.821)/(6.9781-6.821)*(3434.7-3317.2)=3361.3 kJ/kg H’=-160.547, W S = H=0.8(-160.5)=-128 kJ/kg, H 2 =3393kJ/kg S 2 =6.821+(3393-3317.2)/(3434.7-3317.2)*(6.9781-6.821)=6.9228 kJ/kgK W S =300*(-128)=-38400kJ/hr b) now step across next turbine, outlet is superheated since S>S satV at 0.8MPa, will be between 200 and 250 o C H 2 ’=2939.7+(6.9228-6.8179)/(7.0401-6.8179)*(2950.4-2839.7)=2892 kJ/kg

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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 Tech.

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HW4_solution - Homework 4 solution 3.18(a inlet Hin=3446...

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