Explanation Write the Clapeyron equation. ( d P d T ) sat = h f g T v f g (I) Here, the vaporization enthalpy is h f g , the temperature is T , the slope on the saturation curve of P − T diagram is ( d P d T ) sat , the specific volume at saturated liquid + saturated vapor condition is v f g . Rewrite the Equation (I) with finite difference approximations of pressure and temperature. ( d P d T ) sat ≈ ( P 2 − P 1 T 2 − T 1 ) sat (II) Here, the pressure is P and the temperature is T ; the subscripts 1 and 2 indicates the two states. Rewrite the Equation (II) to obtain the pressure P 2 . d P d T = P 2 − P 1 T 2 − T 1 P 2 = P 1 + d P d T ( T 2 − T 1 ) (III) The properties of refrigerant-134a at the temperature ( T 1 ) of − 40 ° C are as follows. P 1 = 51.25 kPa h f g = 225.86 kJ/kg v f g = 0.35993 m 3 /kg Conclusion: Substitute 225.86 kJ/kg for h f g , − 40 ° C for T and 0.35993 m 3 /kg for v f g in Equation (I). ( d P d T ) sat = 225.86 kJ/kg ( − 40 ° C ) ( 0.35993 m 3 /kg ) = 225.86 kJ/kg × kPa ⋅ m 3 1 kJ ( − 40 + 273 K ) ( 0.35993 m 3 /kg ) = 2.6932 kPa/K When, the temperature T 2 = − 50 ° C

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Chapter 12.6, Problem 32P

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The saturation pressure of the refrigerant-134a at the temperature of −50 °C and −30 °C.

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Chapter 12.6, Problem 31P

Chapter 12.6, Problem 33P

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The saturation pressure of the refrigerant-134a at the temperature of − 50 ° C and − 30 ° C .

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