(a) Explanation Write the expression for specific entropy of a substance in Liquid-vapour mixture at state 1. s 1 = s f , 1 + x 1 ( s g , 1 − s f , 1 ) (I) Here, the specific entropy at point 1 is s 1 , liquid specific entropy at point 1 is s f , 1 , vapor specific entropy at point 1 is s g , 1 and dryness fraction at point 1 is x 1 . Write the expression for specific entropy of a substance in Liquid-vapour mixture at state 2. s 2 = s f , 2 + x 2 ( s g , 2 − s f , 2 ) (II) Here, the specific entropy at point 2 is s 2 , liquid specific entropy at point 2 is s f , 2 , vapor specific entropy at point 2 is s g , 2 and dryness fraction at point 2 is x 2 . Write the expression for entropy balance in a closed system. m ( s 2 − s 1 ) = ∫ 1 2 ( δ Q T ) + σ (III) Here, the mass of refrigerant is m, the specific entropy at state 2 is s 2 , the specific entropy at state 1 is s 1 , the heat transfer through the boundary of system is δ Q , the absolute temperature of system is T , the entropy generation within the system is σ . For an adiabatic process as there is no heat transfer across the boundary of system, hence δ Q in the Equation (II) will be zero. Substitute 0 for δ Q in Equation (II) m ( s 2 − s 1 ) = ∫ 1 2 ( 0 T ) + σ m ( s 2 − s 1 ) = σ (IV) Write the expression for specific internal energy of a substance in Liquid-vapour mixture. u = u f + x ( u g − u f ) (V) Here, the specific internal energy of liquid content is u f , the specific internal energy of vapour content is u g , the fraction of vapour content in mixture is x . Write the expression for energy balance in a system. Δ U + Δ K .E . + Δ P .E . = Q − W (VI) Here, the change in internal energy is Δ U , the change in kinetic energy is Δ K .E . , the change in potential energy is Δ P .E . , heat supplied to the system is Q and the work done by system is W . By assuming change in kinetic and potential energy of the system negligible and for an adiabatic process no heat transfer across the boundary of system. Substitute, zero for Δ K .E . , Δ P .E . and Q in Equation (V). Δ U + 0 + 0 = 0 − W − m ( u 2 − u 1 ) = W W m = − ( u 2 − u 1 ) (VII) Conclusion: Refer to Table A-11E “Properties of Saturated Refrigerant 134a (Liquid–Vapor): Pressure Table” to obtain value of liquid specific entropy as − 0.009 Btu / lb ⋅ ° R and vapor specific entropy as 0.2311 Btu / lb ⋅ ° R at pressure of 5 lbf / in 2 and temperature of − 53.48 ° F . Substitute − 0.009 Btu / lb ⋅ ° R for s f , 2 , 0.2311 Btu / lb ⋅ ° R for s g , 2 and 0 (b) To determine Whether the process can accomplished adiabatically. The work done if process is adiabatic, else the heat flow direction.

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ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

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Chapter 6.13, Problem 48P

(a)

To determine

Whether the process can accomplished adiabatically.

The work done if process is adiabatic, else the heat flow direction.

(b)

To determine

Whether the process can accomplished adiabatically.

The work done if process is adiabatic, else the heat flow direction.

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Chapter 6.13, Problem 47P

Chapter 6.13, Problem 49P

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