Question

Refrigerant R-134a enters a compressor at 0.14MPa and - 10 degree

Answer 1

Answer:

Explanation:

R-134a enters the compressor of a refrigerator as super heated vapor at 0.14 MPa and -10 degrees C at a rate of 0.12 kg/s, and it leaves at 0.7 MPa and 50 degrees C. The refrigerant is cooled in the condenser to 24 degrees C and 0.65 MPa, and it is throttled to 0.15 MPa.

Answer 2

Isentropic Efficiency is 0.939

Explanation:

The Complete question is as under-

Refrigerant-134a enters compressor as superheated vapor at 0.14 MPa and -10°C at a  rate of 0.05 kg/s and leaves at 0.8 MPa and 50°C. The refrigerant is cooled in condenser to  26°C and 0.72 MPa and is throttled to 0.15 MPa. Disregarding any heat transfer and pressure drops in the connecting lines between the components, determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor and, (b)  the isentropic efficiency of the compressor. (Figure Attached)

Step by Step Explanation-

$P_1$ = 0.14 MPa

$T_1$ = -10C

So, $h_1 = 246.36 \frac {kj}{kg}$

$s_1 = 0.9724\ \frac {kj}{kg}$

$P_2 = 0.8 MPa$

$T_2 = 50$°C

So, $h_2 = 286.69\ \frac {kj}{kg}$

$P_3 = 0.72$ MPa

T_3 = 26C

So, $h_3$ similar $h_{f@26C} = 87.83\ \frac {kj}{kg}$

a.) Rate of heat removal from the refrigerated space and power input to compressor

are:

$\dot{Q}_L = \dot{m}(h_1 - h_4) = 0.05\ \frac {kg}{s}[(246.36 - 87.83)\ \frac {kg}{s}]=$ 7.93 kW

$\dot{W}_{in} = \dot{m}(h_2 - h_1) = 0.05\ \frac {kg}{s}[(286.69 - 246.36)\ \frac {kg}{s}]$

= 2.02 kW

b.) The isentropic efficiency of the compressor

$P_{2s}$ = 0.8 MPa

$s_{2s} = s_1 = 0.9724\ \frac {kj}{kg}.K$

$h_{2s} = 284.21\ \frac {kj}{kg}$

$\eta = \frac {h_{2s}- h_1}{h_2 - h_1}$

= $\frac {284.21 - 246.36}{286.69 - 246.36}$= 0.939 @ 93.95%