Explanation The Joule-Thomson coefficient is expressed as follows. μ JT = ( ∂ T ∂ P ) h (I) Here, the Joule-Thomson coefficient is μ JT , the temperature is T , the pressure is P , the enthalpy is h and the change of temperature with respect to pressure at constant enthalpy is ( ∂ T ∂ P ) h . Rewrite the Equation (I) by using finite difference approximations of temperature and pressure. μ JT ≅ ( Δ T Δ P ) h = T 2 − T 1 P 2 − P 1 (II) Here, the subscripts 1 and 2 indicates the two states 1 and 2. Consider the given state of refrigerant as state 1. At state 1: Refer Table A-13E, “Superheated refrigerant-134a”. The enthalpy ( h 1 ) corresponding to the pressure of 30 psia and the temperature of 20 ° F is 106.27 Btu/lbm . Here, the Joule-Thomson coefficient is calculated at constant enthalpy. Hence, the enthalpy at state 1 and state 2 is equal. h 1 = h 2 = 106.27 Btu/lbm At state 2: Consider the pressure of 20 psia that is the next lowest pressure listed in Table A-13E. Refer Table A-13E, “Superheated refrigerant-134a”. Obtain the temperature ( T 2 ) corresponding to the pressure of 20 psia and the enthalpy of 106.27 Btu/lbm -using interpolation method. 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 (III) Show the temperature and enthalpy values from the Table A-13E as in below table. S.No. x Y Enthalpy ( h ) ,in Btu/lbm Temperature ( T ) , in °F 1 103

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Chapter 12.6, Problem 61P

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The Joule-Thomson coefficient of the refrigerant-134a at 30 psia and 20 °F.

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Chapter 12.6, Problem 60P

Chapter 12.6, Problem 62P

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