Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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Chapter 8, Problem 8.1P
Interpretation Introduction

Interpretation:

The thermal efficiency of the cycle and the turbine efficiency for the simple power plant shown should be determined.

Concept introduction:

The steam power plant has the turbine operating adiabatically with the given inlet steam pressure and temperature. We used the values shown in the figure to determine the thermal efficiency and turbine efficiency. Turbine efficiency is determined to be the ratio of enthalpy change in the process to the enthalpy change in the reversible process while thermal efficiency is obtained by taking the ratio of energy released by the system to the heat applied to the system.

Expert Solution & Answer
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Answer to Problem 8.1P

The thermal efficiency of the cycle is determined to be 31.1 % and the turbine efficiency is 79 %.

Explanation of Solution

Given information:

The basic cycle for a steam power plant is shown by Fig. 8.1.

Introduction to Chemical Engineering Thermodynamics, Chapter 8, Problem 8.1P

Suppose the turbine operates adiabatically with inlet steam at 6800 kPa and 550°C, and the exhaust steam enters the condenser at 50°C with a quality of 0.96. Saturated liquid water leaves the condenser, and is pumped to the boiler. Neglecting pump work and kinetic- and potential-energy changes, we need to determine the thermal efficiency of the cycle and the turbine efficiency.

Turbine efficiency is determined to be the ratio of enthalpy change in the process to the enthalpy change in the reversible process while thermal efficiency is obtained by taking the ratio of energy released by the system to the heat applied to the system.

Calculation:

Turbine efficiency is given by the formula

  ηturb=ΔHprocessΔHreversible=H3H2H3'H2

Where H3'=(1x')Hliq+x'Hvap

Here x'is obtained by the formula

S2=(1x')Sliq+x'Svap=>6.9636=(1x')0.7035+(x')(8.0776)=>x'=0.845

We know that

x=0.96represents the vapour fraction of steam and the enthalpy of steam is H2=3531.5kJ/kg

H3=(1x)Hliq+xHvap=(10.96)209.3+(0.96)(2592.2)=2496.9kJ/kg

We know that

H3'=(1x')Hliq+x'Hvap = (10.845)209.3+(0.845)(2592.2)=2222.9kJ/kg

Enthalpy of steam after neglecting work done is H2=Hliq=209.3kJ/kg .

Therefore, the turbine efficiency is

ηturb=ΔHprocessΔHreversible=H3H2H3'H2 = 2496.93531.52222.93531.5=79%

And the thermal efficiency of the cycle is

ηtherm=H2H3H2H1=3531.52496.93531.5209.3=31.1%

Conclusion

The thermal efficiency of the cycle is determined to be 31.1 % and the turbine efficiency is 79 %.

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