Consider a steam power cycle in which saturated water vapor at 25 bar (State 1)

Question

Consider a steam power cycle in which saturated water vapor at 25 bar (State 1) enters an

adiabatic steam turbine. At the exit of the turbine, steam leaves at 0.1 bar with 80%

quality (State 2). Saturated liquid water at 0.1 bar (State 3) exits the condenser. Water is

pumped to 25 bar and 50?C (State 4) in an adiabatic pump.

(a) Calculate thermal efficiency (%) of the cycle.

(b) Is the Clausius inequality satisfied? Provide quantitative justification.

SOME ANSWERS ATTACHED BUT PLEASE PROVIDE PROCEDURE.

Answers: (a) 26.1%; (b) Not attached

Explanation / Answer

(a)

Assumptions:

(i) All the process are ideal and isentropic.

Here at state 1,25 Bar the specific enthalpy h1=2801.0 KJ/Kg (From steam table pressure base saturated water and steam)

At state 2, condenser pressure=0.1 bar

the specific enthalpy of liquid=191.8KJ/Kg and specific enthalpy of gas=2584.7 KJ/Kg and quality of steam(dryness fraction)=0.8

so h2=191.8+0.8*(2584.7-191.8) KJ/Kg=2106.12 KJ/Kg

h3=specific enthalpy of liquid at 0.1 bar pressure=191.8 KJ/Kg

At state 4

h4=h3+vdP=191.8+0.001010*(25-0.1)=191.825 KJ/Kg

where v is specific volume of water at 0.1 bar pressure.

Now efficiency=total work output/heat input

i.e. efficiency=(h1-h2)/(h1-h4)=(2801-2106.12)/(2801.0-191.82)=0.26=26.1%

(b)

clausius inequality

dS ? dq/T

here it is satisfied.since here T is constant in condenser.And taking the value of dS(delta S) and dQ(delta Q) and saturation temp T of condenser.This is satisfied.

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