Answer:
heat transfer rate is -15.71 kW
Explanation:
given data
Initial pressure = 4 bar
Final pressure = 12 bar
volumetric flow rate = 4 m³ / min
work input to the compressor = 60 kJ per kg
solution
we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is
h1 = 262.96 kJ/kg
v1 = 0.05397 m³/kg
h2 = 310.24 kJ/kg
and here mass balance equation will be
m1 = m2
and mass flow equation is express as
m1 = [tex]\frac{A1\times V1}{v1}[/tex] .......................1
m1 = [tex]\frac{4\times \frac{1}{60}}{0.05397}[/tex]
m1 = 1.2353 kg/s
and here energy balance equation is express as
0 = Qcv - Wcv + m × [ ( h1-h2) + [tex]\frac{v1^2-v2^2}{2}[/tex] + g (z1-z2) ] ....................2
so here Qcv will be
Qcv = m × [ [tex]\frac{Wcv}{m} + (h2-h1)[/tex] ] ......................3
put here value and we get
Qcv = 1.2353 × [ {-60}+ (310.24-262.96) ]
Qcv = -15.7130 kW
so here heat transfer rate is -15.71 kW
