Yeah I think but I hope this helps
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 = .......................1
m1 =
m1 = 1.2353 kg/s
and here energy balance equation is express as
0 = Qcv - Wcv + m × [ ( h1-h2) + + g (z1-z2) ] ....................2
so here Qcv will be
Qcv = m × [ ] ......................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
Answer:
right answer is option no d
Answer:
The efficiency of the engine is 22.5%.
Explanation:
Efficiency = power output ÷ power input
power output = 55 kW
power input = specific energy×volumetric flow rate×density
specific energy = 44,000 kJ/kg
volumetric flow rate = 27.4 L/h = 27.4 L/h × 1000 cm^3/1 L × 1 h/3600 s = 7.61 cm^3/s
density = 0.73 g/cm^3 = 0.73 g/cm^3 × 1 kg/1000g = 7.3×10^-4 kg/cm^3
power input = 44,000 kJ/kg × 7.61 cm^3/s × 7.3×10^-4 kg/cm^3 = 244.4332 kJ/s = 244.4332 kW
Efficiency = 55 ÷ 244.4332 = 0.225 × 100 = 22.5%