Question

Refrigerant-134a enters a compressor at 180 kPa as a saturated vapor with a flow rate of 0.35 m3/min and leaves at 900 kPa. The power supplied to the refrigerant during the compression process is 2.2 kW. What is the temperature of R-134a at the exit of the compressor?

Answer 1

Answer:

$T = 49.641\,^{\textdegree}C$

Explanation:

The compressor is modelled by means of the First Law of Thermodynamics:

$\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate can be found by using the following expression:

$\dot m = \frac{\dot V}{\nu_{in}}$

The properties of the refrigerant at inlet are:

State - Saturated Vapor

$P = 180\,kPa$

$T_{sat} = 12.73\,^{\textdegree}C$

$\nu_{in} = 0.11049\,\frac{m^{3}}{kg}$

$h_{in}= 242.90\,\frac{kJ}{kg}$

The mass flow rate is:

$\dot m = \frac{(0.35\,\frac{m^{3}}{min} )\cdot(\frac{1\,min}{60\,s} )}{0.11049\,\frac{m^{3}}{kg} }$

$\dot m = 0.053\,\frac{kg}{s}$

The specific enthalpy at outlet is:

$h_{out} = h_{in}+\frac{\dot W_{in}}{\dot m}$

$h_{out} = 242.90\,\frac{kJ}{kg} + \frac{2.2\,kW}{0.053\,\frac{kg}{s} }$

$h_{out} = 284.409\,\frac{kJ}{kg}$

The state of the fluid at outlet has the characteristics of superheated vapor. The temperature associated the specific enthalpy is:

$T = 49.641\,^{\textdegree}C$