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In a trial on a two-stage, single acting, reciprocating air compressor, following data were recorded.

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

1 Answer
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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:

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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,$ L_1=L_2=1.2 D_1$=0.3312m=331.2mm

PV=mRT

$\frac{PV}{RT}$=m →constant

$\frac{P_1 V_1}{RT_1}=\frac{P_2 V_5}{RT_5}$

$\frac{10^5 ×6}{287 ×300}=\frac{6 ×10^5 ×V_5}{287 ×300}$

$V_5=1 m^3/min$

$V_5=(π/4)*D_2^2$ .L.N

$1=(π/4) ×D_2^2 ×0.3312 ×300$

$D_2=0.1132 m=113.2 mm$

$I.Power=\frac{n}{n-1}[ [{P_1 V_1 [\frac{P_2}{P_1 }]^\frac{n-1}{n}- 1]+ P_2 V_5 [\frac{P_3}{P_2 }]^\frac{n-1}{n}- 1]}]$

$I.Power=\frac{1.3}{0.3}[[{10^5 ×\frac{6}{60} ]* [\frac{6}{1}]^\frac{0.3}{1.3}- 1]+ 6 ×10^5 ×\frac{1}{60} [\frac{30}{6}]^\frac{0.3}{1.3}-1]}]$

I.Power=41.680 KW

Power=$\frac{IP}{η_{mech}}$ =$\frac{41.680}{0.85}$=49.035 KW

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