Explanation Write the given equation of state of gas. P v = Z R T (I) Here, the pressure is P , the specific volume is v , the compressibility factor is Z , the universal gas constant is R , the temperature is T and the constants are a and b . The compressibility factor is the function of pressure and temperature alone. Z = Z ( P , T ) Write the general relation for Joule-Thomson coefficient. μ JT = ( ∂ T ∂ P ) h = − 1 c P [ v − T ( ∂ v ∂ T ) P ] = 1 c P [ T ( ∂ v ∂ T ) P − v ] (II) Here, the Joule-Thomson coefficient is μ JT , the temperature is T , the pressure is P , the enthalpy is h , the specific heat at constant pressure is c P and the change of specific volume with respect to temperature at constant pressure is ( ∂ v ∂ T ) P . Conclusion: Rearrange the Equation (I) to obtain v as follows. v = Z R T P (III) Partially differentiate the Equation (III) with respect to the temperature ( T ) by keeping the pressure ( P ) as constant. ( ∂ v ∂ T ) P = ∂ ∂ T ( Z R T P ) = R P ∂ ∂ T ( Z T ) = R P [ T ( ∂ Z ∂ T ) P + Z ] The inversion curve is passes the through the points of zero slopes or zero JT coefficients on the T − P plane. Equate the equation (II) is equal to zero. μ JT = 0 Substitute 0 for μ JT and Z R T P for v in Equation (II)

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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ISBN: 9781260048636

Author: CENGEL

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Chapter 12.6, Problem 90RP

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To show that the position of the Joule-Thomson coefficient inversion curve on the T−P plane is given by the equation (∂z∂T)P=0.

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Chapter 12.6, Problem 89RP

Chapter 12.6, Problem 91RP

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To show that the position of the Joule-Thomson coefficient inversion curve on the T − P plane is given by the equation ( ∂ z ∂ T ) P = 0 .

Solution Summary: The author explains that the position of the Joule-Thomson coefficient inversion curve on the T-P plane is given by the equation.

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To show that the position of the Joule-Thomson coefficient inversion curve on the T−P plane is given by the equation (∂z∂T)P=0.

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Chapter 12 Solutions