To determine The net work developed. The rate of heat transfer in the reheater. Explanation The figure below shows the T-s diagram of the system. Figure-(1) Write the expression for dryness fraction at 2 s . x 2 = s 2 s − s f 2 s g 2 − s f 2 (I) Here, dryness fraction at 2 s is x 2 , specific entropy of saturated liquid at 2 is s f 2 and the specific entropy of saturated gas at state 2 is s g 2 . Write the expression for specific enthalpy at state 2 s . h 2 s = h f 2 + x 2 ( h g 2 − h f 2 ) (II) Here, specific enthalpy at 2 s is h 2 s , specific enthalpy of saturated liquid at 2 is h f 2 and the specific enthalpy of saturated vapor at 2 is h g 2 . Write the expression for enthalpy at state 2. h 2 = h 1 − η t 1 ( h 1 − h 2 s ) (III) Here, specific enthalpy at state 2 is h 2 , and the efficiency of first turbine is η t 1 . Write the expression for work developed by first turbine. W ˙ 1 = m ˙ 1 ( h 1 − h 2 ) (IV) Here, work developed by turbine 1 is W ˙ 1 and the mass flow rate through turbine 1 is m ˙ 1 . Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (V) Here, the variables x and y represents temperature and specific enthalpies respectively. Write the expression for dryness fraction at 4 s . x 4 = s 4 s − s f 4 s g 4 − s f 4 (VI) Here, dryness fraction at 4 s is x 4 , specific entropy of saturated liquid at 4 is s f 4 and the specific entropy of saturated gas at state 4 is s g 4 . Write the expression for specific enthalpy at state 4s. h 4 s = h f 4 + x 2 ( h g 4 − h f 4 ) (VII) Here, specific enthalpy at 4s is h 4 s , specific enthalpy of saturated liquid at 4 is h f 4 and the specific enthalpy of saturated vapor at4 is h g 4 . Write the expression for enthalpy at state 4. h 4 = h 3 − η t 2 ( h 3 − h 4 s ) (VIII) Here, enthalpy at state 4 is h 4 and the efficiency of turbine 2 is η t 2 . Write the expression for work developed by turbine 2. W ˙ 2 = m ˙ 1 ( h 3 − h 4 ) (IX) Here, work developed by turbine 2 is W ˙ 2 . Write the expression for net work developed by the turbines. W ˙ n e t = W ˙ 1 + W ˙ 2 (X) Here, net work developed by the turbine is W ˙ n e t . Write the expression for heat transfer in the re-heater. Q ˙ i n = m ˙ 1 ( h 3 − h 2 ) (XI) Here, heat transfer in reheater is Q ˙ i n . Conclusion: Refer to Table A-4E, “Properties of superheated water vapor” to obtain specific enthalpy at state 1 as 1388.5 Btu / lb and specific entropy as 1.5665 Btu / lb ⋅ °R at temperature 800 °F and pressure 1000 lbf / in 2 . Since process 1-2 s is an isentropic process, so s 1 = s 2 s . Refer to table A3-E, “Properties of saturated water (Liquid-Vapor)” to obtain specific enthalpy of liquid as 262.2 Btu / lb , specific enthalpy of liquid vapor as 1178 Btu / lb , specific entropy of liquid as 0.4273 Btu / lb °R and specific entropy of vapor as 1.6443 Btu / lb °R at pressure 60 psi and entropy 1.5665 Btu / lb °R . Substitute 1.5665 Btu / lb ⋅ °R for s 2 s , 0.4273 Btu / lb °R for s f 2 and 1.6443 Btu / lb °R for s g 2 in Equation (I). x 2 = 1.5665 Btu / lb ⋅ °R − 0.4273 Btu / lb ⋅ °R 1.6443 Btu / lb ⋅ °R − 0.4273 Btu / lb ⋅ °R = 1.1392 Btu / lb ⋅ °R 1.217 Btu / lb ⋅ °R = 0.936 Substitute 262.2 Btu / lb for h f 2 , 1178 Btu / lb for h g 2 and 0.936 for x 2 in Equation (II). h 2 s = 262.2 Btu / lb + 0.936 ( 1178 Btu / lb − 262.2 Btu / lb ) = 262.2 Btu / lb + 857.1888 Btu / lb = 1119.5 Btu / lb Substitute 1119.5 Btu / lb for h 2 s , 1388.5 Btu / lb for h 1 and 0.88 for η t 1 in Equation (III). h 2 = 1388.5 Btu / lb − 0.88 × ( 1388.5 Btu / lb − 1119.5 Btu / lb ) = 1388.5 Btu / lb − 236.72 Btu / lb = 1151.78 Btu / lb Substitute 120000 lb / h for m ˙ 1 , 1388.5 Btu / lb for h 1 and 1151.74 Btu / lb for h 2 in Equation (IV). W ˙ 1 = ( 120000 lb / h ) ( 1388.5 Btu / lb − 1151.74 Btu / lb ) = ( 120000 lb / h ) ( 236.76 Btu / lb ) = 28.41 × 10 6 Btu / h Covert the temperature at state 3 from degree centigrade to degree Fahrenheit. T 3 = 350 °C = ( 350 °C × 9 5 + 32 ) °F = ( 630+32 ) °F = 662 °F Refer to table A-4E, “Properties of superheated Water vapor” to obtain specific enthalpy as 1381.4 Btu / lb and specific entropy as 1.8609 Btu / lb °R corresponding to 700 °F . Refer to table A-4E, “Properties of superheated Water vapor” to obtain specific enthalpy as 1332.1 Btu / lb and specific entropy as 1.8165 Btu / lb °R corresponding to 600 °F . Show the values of specific enthalpies corresponding to temperature 600 °F and 700 °F as in below table

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 165P

To determine

The net work developed.

The rate of heat transfer in the reheater.

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Chapter 6.13, Problem 164P

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Fundamentals of Engineering Thermodynamics

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