Explanation Write the formula for volume expansivity ( β ) . β = 1 v ( ∂ v ∂ T ) P (I) Write the formula for isothermal compressibility ( α ) . α = − 1 v ( ∂ v ∂ P ) T (II) Here, the specific volume is v , the pressure is P , the temperature is T and the symbol ∂ indicates the partial derivative of variables. Refer Table A-13, “Superheated refrigerant-134a”. Consider the two temperatures 20 ° C and 40 ° C (at constant pressure 200 kPa ), these temperatures can give the average temperature of 30 ° C . Also consider the two pressures 180 kPa and 240 kPa (at constant temperature 30 ° C ), these pressures can give the average pressure of 210 kPa (close to the value of 200 kPa ). Rewrite the Equations (I) and (II) by approximating the partial differentials about the specified state. β ≅ 1 v ( Δ v Δ T ) P = 200 kPa = 1 v ( v 40 ° C − v 20 ° C 40 ° C − 20 °C ) P = 200 kPa (III) α ≅ − 1 v ( Δ v Δ P ) T = 30 ° C = − 1 v ( v 240 kPa − v 180 kPa 240 kPa − 180 kPa ) T = 30 ° C (IV) Refer Table A-13, “Superheated refrigerant-134a”. The specific volume ( v ) corresponding to the pressure of 200 kPa ( 0.2 MPa ) and the temperature of 30 ° C is 0.11874 m 3 /kg . And also obtain the specific volumes corresponding pressure of 200 kPa ( 0.2 MPa ) and the temperatures 20 ° C and 40 ° C . v 40 ° C = 0.12322 m 3 /kg v 20 ° C = 0.11418 m 3 /kg Refer Table A-13, “Superheated refrigerant-134a”. Obtain the specific volumes corresponding to the temperature of 30 ° C at the pressures 180 kPa ( 0.18 MPa ) and 240 kPa ( 0

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Chapter 12.6, Problem 40P

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The volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and 30 °C.

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Chapter 12.6, Problem 39P

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The volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and 30 ° C .

Solution Summary: The author explains the volume expansivity and isothermal compressibility of refrigerant-134a at 200kPa, and 30°

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The volume expansivity β and the isothermal compressibility α of refrigerant-134a at 200 kPa and 30 °C.

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