In a single acting, two stage reciprocating air compressor 4.5kg of air per minute are compressed from 1.013 bar and 15 deg C through a pressure ratio of 9 to 1. Both stages have the same pressure ratio and the law of compression and expansion in both stages is following P1.3=constant. An intercooler is fitted in between stages and the cooling process is complete. The clearance volume of both stages is 5% of their respective swept volumes and the compressor speed is at 300rpm.
a)Sketch the schematic diagram of the compressor and label clearly the TDC, BDC, Vs1, Vs2, Vc, d and L.
b) Sketch the P-V diagram and label clearly the TDC, BDC, Vs1, Vs2, Vc, and Vd.
c) Calculate the indicated power (kW)
d) Determine the swept volume for each stage required (liter)
e) Calculate the volumetric efficiency for the 1st stage (%)
f) Determine the isothermal power (kW)
g) Calculate the isothermal efficiency (%)
*R = 0.287 kJ/kg.K
Given data,
Air mass flow rate(m)= 4.5kg/min= 0.015kg/cycle
No. of stages(n)=2
Compressor speed, N= 300RPM
Initial condition: P1= 1.013bar, T1= 15°C= 288K
Final pressure(P2) = 9P1
Clearance volume(Vc)= 5% of Vs,
k= 1.3
R = 0.287 KJ/kgK
Compression ratio of each stage for perfect intercooling (r) = 3
P1V1 = mRT1
101m3 × Vs1 = 0.015 ×0.297 × 288
Vs1 = 0.01224 m^3
Vs1 = 12.24 L
Now, temperature of air after compression,
T2 = Ta × r^(k-1/k)
T2 = 288 × 3^(1.3-1/1.3)
T2 = 371.10 K
Vs2 = mR×(T2/P2)
Vs2 = 0.015×0.287×(371.01/911.7)
Vs2 = 1.752 L
Indiacted Power,
IP = (300×2×1.3/0.3×60) × (0.015×0.287×288) × {9^(0.3/2.6) -1}
IP = 15.50kW
Volumentric Efficiency For Stage 1,
ηV1 = (1 + C – C (r)^1/k)
ηV1 = (1 + 0.05 – 0.05 ×(3)^1/1.3) × (100)
ηV1 = 0.9336 × 100
ηV1 = 93.36 %
Now, if compression is isothermal then work done for each stage,
Wiso = mRT1 × ln (r)
Wiso = (4.5/60) × 0.287 × 288 × ln (3)
Wiso = 6.81 kN
Hence, Isothermal Power Consumption for both stage,
Piso = 2× Wiso
Piso = 2 × 6.81
Piso = 13.62 kW
Isothermal Efficiency,
ηiso = (Piso/Pactual) × 100
ηiso = (13.62/15.50) × 100
ηiso = 87.88 %