Solution to Assignment 4

# Solution to Assignment 4 - SOLUTIONTOHOMEWORK#4 5.1(, 5.4...

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SOLUTION TO HOMEWORK #4 5.1 (although the temperatures are slightly different, the procedure to solve the problem is the same).

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5.4

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5.8

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Problem 5: Flowsheet: Work: To begin with, we know how to deal with the rankine cycle, except that, we will have different flowrates in different places (Pump and boiler will have a 90 Kg/sec flowrate and the turbine and condenser 50 Kg/sec. Thus, look at section 5.2 where all the detailed calculations are done and just be careful with the flowrate you use. You can use 3 MPa in the boiler (evaporator) as in illustration 5.2-1. If we do that, the work in the turbine of the cycle will be (-947.6 Kj/kg * 50 Kg/sec) in the rankine cycle turbine (see page 157 for the origin of the number -947.6 kJ/Kg). As for the other turbine, we use 3 ˆˆ ˆ () SR WH H =− . Now 3 ˆ H = 3682.3 Kj/Kg (from illustration 5.2-1) and ˆ SR H (the enthalpy of the steam entering steam reforming), is obtained by looking for the enthalpy of a steam at P=1 MPA and the same entropy as 3 ˆ S =7.5085 KJ/(Kg K). From steam tables (page 922) for 1 MPa we get (after

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Solution to Assignment 4 - SOLUTIONTOHOMEWORK#4 5.1(, 5.4...

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