Answer:
[tex]\mu = 0.018\,\frac{^{\textdegree}C}{kPa}[/tex]
Explanation:
The Joule-Thomson coefficient is the ratio of the change of temperature to the change of pressure under isoenthalpic conditions:
[tex]\mu = \left(\frac{\Delta T}{\Delta P}\right)_{h}[/tex]
Initial and final properties are:
[tex]T_{1} = 50^{\textdegree}C, P_{1}=700\,kPa, h_{1}=288.54\,\frac{kJ}{kg}[/tex]. Superheated Vapor.
[tex]T_{2} = 48.186^{\textdegree}C, P_{2}=600\,kPa, h_{2}=288.54\,\frac{kJ}{kg}[/tex]. Superheated Vapor.
The Joule-Thomson coefficient is approximately:
[tex]\mu = \frac{50^{\textdegree}C-48.186^{\textdegree}C}{700\,kPa-600\,kPa}[/tex]
[tex]\mu = 0.018\,\frac{^{\textdegree}C}{kPa}[/tex]
