Question

Refrigerant-134a enters a compressor as a saturated vapor at 160 kPa at a rate of 0.03 m3/s and leaves at 800 kPa. The power input to the compressor is 10 kW. If the surroundings at 20°C experience an entropy increase of 0.008 kW/K, determine (a) the rate of heat loss from the compressor, (b) the exit temperature of the refrigerant, and (c) the rate of entropy generation.

Answer 1

Answer:

a) $\dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW$

b) T_2 = 36.4 degree C

c) $\Delta S_{gen} = 0.006512 KW/K$

Explanation:

Given data:

$P_1 = 160 kPa$

volumetric flow $V_1 = 0.03 m^3/s$

$P_2 = 800 kPa$

power input $W_{in} = 10 kW$

$T_{surr} = 20 degree C$

entropy = 0.008 kW/K

from refrigerant table for P_1 = 160 kPa and x_1 = 1.0

$v_1 = 0.12355 m^3/kg$

$h_1 = 241.14 kJ/kg$

$s_1 = 0.94202 kJ/kg K$

a) mass flow rate $\dot m = \frac{V_1}{v_1}$

$\dot m = \frac{0.03}{0.12355} = 0.2428 kg/s$

heat loss$= T_{surr} \times entropy$

heat loss$\dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW$

b) from energy balance equation

$W_{in} 0 \dot Q_{out} = \dot m (h_2 -h_1}$

$10 - 2.344 = 0.2428 (h_2 - 241.14}$

$h_2 = 272.67 kJ/kg$

from refrigerant table, for P_2 = 800 kPa and h_2 = 272.67 kJ/kg

T_2 = 36.4 degree C

c) from refrigerant table P_2 = 800 kPa and h_2 = 272.67 kJ /kg

$s_2 = 0.93589 kJ/kg K$

rate of entropy

$\Delta S_R =  \dot m =(s_2 -s_1)$

$\Delta S_R = 0.2428 \times (0.93589 -0.94202) = - 0.0014884 kW/K$

rate of entropy for entire process

$\Delta S_{gen} = \Delta _S_R + \Delta_{surr}$

$\Delta S_{gen} = 0.0014884 + 0.008 = 0.006512 KW/K$