Answer:
a) [tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]
b) T_2 = 36.4 degree C
c) [tex]\Delta S_{gen} = 0.006512 KW/K[/tex]
Explanation:
Given data:
[tex]P_1 = 160 kPa[/tex]
volumetric flow [tex]V_1 = 0.03 m^3/s[/tex]
[tex]P_2 = 800 kPa[/tex]
power input [tex]W_{in} = 10 kW[/tex]
[tex]T_{surr} = 20 degree C[/tex]
entropy = 0.008 kW/K
from refrigerant table for P_1 = 160 kPa and x_1 = 1.0
[tex] v_1 = 0.12355 m^3/kg[/tex]
[tex]h_1 = 241.14 kJ/kg[/tex]
[tex]s_1 = 0.94202 kJ/kg K[/tex]
a) mass flow rate [tex] \dot m = \frac{V_1}{v_1}[/tex]
[tex]\dot m = \frac{0.03}{0.12355} = 0.2428 kg/s[/tex]
heat loss[tex] = T_{surr} \times entropy[/tex]
heat loss[tex] \dot Q_{out} = (20 + 273} \times 0.008 = 2.344 kW[/tex]
b) from energy balance equation
[tex]W_{in} 0 \dot Q_{out} = \dot m (h_2 -h_1}[/tex]
[tex]10 - 2.344 = 0.2428 (h_2 - 241.14}[/tex]
[tex]h_2 = 272.67 kJ/kg[/tex]
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
[tex]s_2 = 0.93589 kJ/kg K[/tex]
rate of entropy
[tex]\Delta S_R = \dot m =(s_2 -s_1)[/tex]
[tex]\Delta S_R = 0.2428 \times (0.93589 -0.94202) = - 0.0014884 kW/K[/tex]
rate of entropy for entire process
[tex]\Delta S_{gen} = \Delta _S_R + \Delta_{surr}[/tex]
[tex]\Delta S_{gen} = 0.0014884 + 0.008 = 0.006512 KW/K[/tex]
