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-13, “Superheated refrigerant-134a”. The enthalpy ( h 1 ) corresponding to the pressure of 0.8 MPa and the temperature of 60 ° C is 296.82 kJ/kg . 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 = 296.82 kJ/kg At state 2: Consider the pressure of 0.7 MPa that is the next lowest pressure listed in Table A-13. Refer Table A-13, “Superheated refrigerant-134a”. Obtain the temperature ( T 2 ) corresponding to the pressure of 0.7 MPa and the enthalpy of 296.82 kJ/kg -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-13 as in below table. S.No. x y Enthalpy ( h ) ,in kJ/kg Temperature ( T ) , in °C 1 288

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Chapter 12.6, Problem 106FEP

To determine

The Joule-Thomson coefficient of the refrigerant-134a at 0.8 MPa and 60 °C from the following:

(a) 0

(b) 5 °C/MPa

(c) 11 °C/MPa

(d) 16 °C/MPa

(e) 25 °C/MPa

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Chapter 12.6, Problem 105FEP

Chapter 13.3, Problem 1P

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4. Block A and B are two different pieces of wood. Determine the minimum dimension for "a", if the shear stress of the wood is 50Mpa. The thickness of the wood is 30cm. 600N A

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CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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The Joule-Thomson coefficient of the refrigerant-134a at 0.8 MPa and 60 ° C from the following: (a) 0 (b) 5 ° C/MPa (c) 11 ° C/MPa (d) 16 ° C/MPa (e) 25 ° C/MPa

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The Joule-Thomson coefficient of the refrigerant-134a at 0.8 MPa and 60 °C from the following:

(a) 0

(b) 5 °C/MPa

(c) 11 °C/MPa

(d) 16 °C/MPa

(e) 25 °C/MPa

Want to see the full answer?

Chapter 12 Solutions

The Joule-Thomson coefficient of the refrigerant-134a at 0.8 MPa and 60 ° C from the following: (a) 0 (b) 5 ° C/MPa (c) 11 ° C/MPa (d) 16 ° C/MPa (e) 25 ° C/MPa

Solution Summary: The author explains the Joule-Thomson coefficient of the refrigerant-134a at 0.8MPa and 60

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The Joule-Thomson coefficient of the refrigerant-134a at 0.8 MPa and 60 °C from the following: (a) 0 (b) 5 °C/MPa (c) 11 °C/MPa (d) 16 °C/MPa (e) 25 °C/MPa

Want to see the full answer?

Chapter 12 Solutions

The Joule-Thomson coefficient of the refrigerant-134a at 0.8 MPa and 60 °C from the following:

(a) 0

(b) 5 °C/MPa

(c) 11 °C/MPa

(d) 16 °C/MPa

(e) 25 °C/MPa