refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a mass flow rate of 0.108kg/s. As the refrigerant passes through the compressor, heat transfer to the surroundings occurs at a rate of 0.32 kJ/s. Determine at steady state the power input to the compressor, in kW.

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ...)

through prior knowledge of two other properties such as pressure and temperature.)

h1[quality=1, P=0.12Mpa) = 237KJ/Kg

h2 (P=1.2Mpa, t=70C) = 300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m (h1) + W=Q+m (h2)

solving for W

W=Q+m (h2-h1)

W=0.32+0.108 (300.6-237) = 7.19Kw

the power input to the compressor is 7.19Kw