Explanation Write the expression for Clapeyron equation. ln ( p sat, 2 p sat, 1 ) = − h fg R [ 1 T 2 − 1 T 1 ] (I) Here, saturation pressure at 10 ° F is p sat, 1 , saturation pressure at 20 ° F is p sat, 2 , universal gas constant is R , specific enthalpy of the wet refrigerant is h fg , saturation temperature at 10 ° F is T 1 and saturation temperature at 20 ° F is T 2 . Write the expression for percentage change in saturation pressure from the value of saturation pressure in the table. E % = p sat, 2 − p sat,2T p sat, 2 (II) Here, saturation pressure at 20 ° F is p sat, 2 and saturation pressure at 20 ° F from the table is p sat, 2T . Conclusion: Refer to the Table A-10E “Properties of Saturated Refrigerant 134a (Liquid–Vapor) Temperature Table” to obtain the value of specific enthalpy in wet region for the refrigerant as 88.53 Btu / lb at temperature of 20 ° F and saturated pressure at the temperature 20 ° F as 33.132 lbf / in 2 of the 134a refrigerant. Substitute 88.53 Btu / lb for h fg , 0.01642 ( Btu lb ⋅ °R ) for R , 20 ° F for T 2 and 10 ° F for T 1 in Equation (I). ln ( p sat, 2 p sat,1 ) = − ( 88.53 Btu / lb ) 0

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 11.9, Problem 34P

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The saturation pressure at 20°F.

The accuracy of the estimated pressure.

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Chapter 11.9, Problem 33P

Chapter 11.9, Problem 35P

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Fundamentals of Engineering Thermodynamics

Ch. 11.9 - Prob. 1E Ch. 11.9 - Prob. 2E Ch. 11.9 - 3. What is an advantage of using the Redlich–Kwong...Ch. 11.9 - To determine the specific volume of superheated...Ch. 11.9 - Prob. 5E Ch. 11.9 - Prob. 6E Ch. 11.9 - Prob. 7E Ch. 11.9 - Prob. 8E Ch. 11.9 - Prob. 9E Ch. 11.9 - Prob. 10E

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The saturation pressure at 20 ° F . The accuracy of the estimated pressure.

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The saturation pressure at 20°F. The accuracy of the estimated pressure.

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The saturation pressure at 20°F.

The accuracy of the estimated pressure.