To determine The rate of entropy production for steam generator. The rate of entropy production for turbine. The name of component that contributes more inefficient operation of the overall system. Explanation Write the expression for mass flow rate at state 1. m ˙ 1 = ( A V ) 1 ( R T 1 P 1 ) (I) Here, the mass flow rate at state 1 is m ˙ 1 , area of steam generator at state 1 is A 1 , volumetric flow at state 1 is V 1 , gas constant is R , temperature at state 1 is T 1 and pressure at state 1 is P 1 . Write the expression for rate of entropy production for steam generator. σ ˙ he = m ˙ 1 ( s 2 − s 1 ) + m ˙ 3 ( s 4 − s 3 ) (II) Here, the rate of entropy production for steam generator is σ ˙ he , entropy at state 2 is s 2 , entropy at state 1 is s 1 , entropy at state 3 is s 3 , entropy at state 4 is s 4 , mass flow rate at state 1 is m ˙ 1 and mass flow rate at state 3 is m ˙ 3 . Write the expression for rate of entropy production for turbine. σ ˙ turbine = m ˙ 3 ( s 5 − s 4 ) (III) Here, the rate of entropy production for turbine is σ ˙ turbine and entropy at state 5 is s 5 . Conclusion: The atmospheric pressure is 14.7 lbf / in 2 and gas constant for air is 53.33 ft ⋅ lbf / lb ⋅ ° R . Convert the temperature at state 1 from Fahrenheit to Rankine. T 1 = 400 ° F = ( 400 ° F + 460 ) ° R = 860 ° R Substitute ( 2 × 10 5 ft 3 / min ) for ( A V ) 1 , 53.33 ft ⋅ lbf / lb ⋅ ° R for R , 14.7 lbf / in 2 for P 1 and 860 ° R for T 1 in Equation (I). m ˙ 1 = ( 2 × 10 5 ft 3 / min ) ( 53.33 ft ⋅ lbf / lb ⋅ ° R ) ( 860 ° R ) 14.7 lbf / in 2 = ( 2 × 10 5 ft 2 / min ) ( 144 in 2 1 ft 2 ) ( 14.7 lbf / in 2 ) ( 45863.8 lbf / lb ) = 9230.809 lb / min Refer to Table A-22E “Ideal gas properties of air” to obtain enthalpy ( h 1 ) at state 1 as 206.46 Btu / lb corresponding to temperature 860 ° R and enthalpy ( h 2 ) at state 2 as 172.39 Btu / lb . Refer to Table A-2E “Properties of saturated water” to obtain enthalpy ( h 3 ) as 70 Btu / lb corresponding to state 3. Write the expression for enthalpy at state 4. h 4 = h 3 + ( m ˙ 1 m ˙ 3 ) ( h 1 − h 2 ) (IV) Here, enthalpy at state 4 is h 4 . Substitute 70 Btu / lb for h 3 , 9230.809 lb / min for m ˙ 1 , 206.46 Btu / lb for h 1 , 172.39 Btu / lb for h 2 and 275 lb / min for m ˙ 3 in Equation (IV). h 4 = ( 70 Btu / lb ) + ( 9230.809 lb / min 275 lb / min ) ( 206.46 Btu / lb − 172.39 Btu / lb ) = ( 70 Btu / lb ) + ( 1143.6133 Btu / lb ) = 1213.613 Btu / lb Refer to Table A-4E “Properties of superheated water vapor” to obtain pressure P 4 as 40 lbf / in 2 corresponding to enthalpy 1213.613 Btu / lb

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 6.13, Problem 109P

To determine

The rate of entropy production for steam generator.

The rate of entropy production for turbine.

The name of component that contributes more inefficient operation of the overall system.

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Chapter 6.13, Problem 108P

Chapter 6.13, Problem 110P

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Fundamentals of Engineering Thermodynamics

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