1) Free air delivery per minute = 6$m^3$

2) Free air conditions = 1 bar,$27^o$C

3) Delivery pressure = 30 bar

4) Compressor speed = 300 rpm

5) Intermediate pressure = 6 bar

6) Temperature at the inlet of HP cylinder = 270C

7) Law of compression = P$V^{1.3}$

8) Mechnical efficiency = 85%

9) Stroke to bore ration for LP cylinder = 1.2

Calculate:

a)Cylinder diameters

b)power input, neglecting clearance volume

Given: FAD = 6 $m^3$/min, $P_1$= 1 bar, $T_1=27^o$=300k $P_3$=30 bar , N = 300 rpm, $P_2$=6 bar ,$T_5=27^o$=300k P$V^{1.3}$=C ,$η_{mech}$=0.85,$[\frac{L}{D}]_{LP}$=1.2 To Find: $D_1 ,D_2$ ,Power. Solution:

FAD =$V_1-V_4=V_1=6m^3$/min

$V_1=(π/4)*D_1^2$ L.N

But L=1.2$D_1$

$V_1=π/4 ×1.2 ×D_1^3$ .N

6=π/4 ×1.2 ×$D_1^3$ ×300

$D_1$=0.276 m=276mm

Assuming common stroke,$ …