Chapter 2: Problem 47
Another alternative refrigerant (see preceding problem) is \(1,1,1,2\) tetrafluoroethane (refrigerant HFC-134a). Tillner-Roth and Baehr published a compendium of thermophysical properties of this substance (R. Tillner-Roth and H.D. Baehr, \(J\). Phys. Chem. Ref. Data 23, 657 (1994) from which properties such as the Joule-Thomson coefficient \(\mu\) can be computed. (a) Compute \(\mu\) at \(0.100 \mathrm{MPa}\) and \(300 \mathrm{~K}\) from the following data (all referring to \(300 \mathrm{~K}\) ): \(\begin{array}{llll} \text { P/MPa } & 0.080 & 0.100 & 0.12 \\ \text { Specific enthalpy/(k] } \mathrm{kg}^{-1} \text { ) } & 426.48 & 426.12 & 425.76 \end{array}\) (The specific constant-pressure heat capacity is \(\left.0.7649 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}\right)\) (b) Computer \(\mu\) at \(1.00 \mathrm{MPa}\) and \(350 \mathrm{~K}\) from the following data (all referring to \(350 \mathrm{~K}\) ): \(\begin{array}{llll} \text { PIMPa } & 0.80 & 1.00 & 1.2 \\ \text { Specific enthalpy/(k] } \mathrm{kg}^{-1} \text { ) } & 461.93 & 459.12 & 456.15 \end{array}\) (The specific constant-pressure heat capacity is \(1.0392 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~kg}^{-1}\).)
Short Answer
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