Explanation The figure below shows the process on Refrigerant 134a from state 1 to state 2 on a ( T-S ) chart. Figure-(1) Write the expression for specific entropy of a substance in Liquid-vapour mixture. s 2 = s f + x ( s g − s f ) (I) Here, the specific entropy of liquid content is s f , the specific entropy of vapour content is s g , the fraction of vapour content in mixture is x . Write the expression for entropy balance in a closed system. m ( s 2 − s 1 ) = ∫ 1 2 ( δ Q T ) + σ (II) 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 ( s 2 − s 1 ) = σ m (III) Write the expression for change in specific entropy between two states. s 2 − s 1 = { s f , 2 + x 2 ( s g , 2 − s f , 2 ) } − { s f , 1 + x 1 ( s g , 1 − s f , 1 ) } (IV) Here, liquid specific entropy at point 2 is s f , 2 , vapor specific entropy at point 2 is s g , 2 , liquid specific entropy at point 1 is s f , 1 , vapor specific entropy at point 1 is s g , 1 , dryness fraction at point 2 is x 2 and dryness fraction at point 1 is x 1 . Conclusion: Refer to Table A-11E “Properties of Saturated Refrigerant 134a (Liquid–Vapor): Pressure Table” to obtain value of liquid specific entropy at state 1 using interpolation. 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 denote by x and y are temperature and liquid specific entropy. Show the liquid specific entropy at 58.35 ° F and 65.93 ° F as in Table (1). Temperature, ° F Liquid specific entropy, Btu / lb ⋅ ° R 58.35 ° F 0 .0638 60 ° F ? 65.93 ° F 0.0686 Substitute 58.35 ° F for x 1 , 60 ° F for x 2 , 65.93 ° F for x 3 , 0 .0638 Btu / lb ⋅ ° R for y 1 , and 0.0686 Btu / lb ⋅ ° R for y 3 in Equation (V). y 2 = [ ( 60 ° F − 58.35 ° F ) ( 0.0686 Btu / lb ⋅ ° R − 0

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In physics, entropy means the amount of disorder or the degree of randomness associated with a thermodynamic system. The system is not self-organized when the entropy of the system is high. For instance, when a system is at an elevated temperature, its m…

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Chapter 6.13, Problem 47P

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The change in specific entropy of refrigerant in Btu/lb⋅°R.

Whether the process can be achieved adiabatically.

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Chapter 6.13, Problem 46P

Chapter 6.13, Problem 48P

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The change in specific entropy of refrigerant in Btu / lb ⋅ ° R . Whether the process can be achieved adiabatically.

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The change in specific entropy of refrigerant in Btu/lb⋅°R.

Whether the process can be achieved adiabatically.