Question
Asked Nov 19, 2019
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The figure shows the schematic of a vapor power plant in which 100 kg/s of water circulates through the four components operating at steady state. The water flows through the boiler and condenser at constant pressure and through the turbine and pump adiabatically. Kinetic and potential energy effects can be ignored. Process data follow:
Process 4–1: constant-pressure at 1 MPa from saturated liquid to saturated vapor
Process 2–3: constant-pressure at 20 kPa from x2= 88% to x3= 18%
Determine the cycle thermal efficiency. Determine σcycle , in kW/K. Determine if the cycle is internally reversible, irreversible, or impossible.

Lin
4
Boiler
Turbine
Pump
Condenser
3
2
Qout
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Lin 4 Boiler Turbine Pump Condenser 3 2 Qout

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Expert Answer

Step 1

The p-v diagram for the given power plant is shown below,

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P x 88 % хз %3D 18 % т 3 100 kg /s P-1MPа TH 4 Р- 20 kPа

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Step 2

From steam table,

At 1 MPa the property of water vapor at saturated state is,

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= T., = 180°C = 453 K Sat h h 762.7 kJ/kg h 2777.2 kJ/kg s, S2.13187 kJ/kg K sS 6.5863 kJ/kg K At 20 kPa, TtT 60°C 333 K h, 251.4 kJ/kg h 2608.95 kJ/kg h2357.55 kJ/kg s 0.8320 kJ/kg K 7.9085kJ/kg K sat

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Step 3

The enthalpy at 2 & 3 c...

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h, 251.40.88x 2357.55 h,2326.044 kJ/kg Simliraly h = h +x,hg h 251.4 0.18x 2357.55 h 675.76 kJ/kg

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