Question

Estimate the Joule-Thomson coefficient of refrigerant-134a at 0.70 MPa and 50°C. Assume the second state will be selected for a pressure of 0.60 MPa. Use data from the tables.

Answer 1

Answer:

$\mu = 0.018\,\frac{^{\textdegree}C}{kPa}$

Explanation:

The Joule-Thomson coefficient is the ratio of the change of temperature to the change of pressure under isoenthalpic conditions:

$\mu = \left(\frac{\Delta T}{\Delta P}\right)_{h}$

Initial and final properties are:

$T_{1} = 50^{\textdegree}C, P_{1}=700\,kPa, h_{1}=288.54\,\frac{kJ}{kg}$. Superheated Vapor.

$T_{2} = 48.186^{\textdegree}C, P_{2}=600\,kPa, h_{2}=288.54\,\frac{kJ}{kg}$. Superheated Vapor.

The Joule-Thomson coefficient is approximately:

$\mu = \frac{50^{\textdegree}C-48.186^{\textdegree}C}{700\,kPa-600\,kPa}$

$\mu = 0.018\,\frac{^{\textdegree}C}{kPa}$