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A two stage single acting reciprocating compressor takes in air at the rate of 0.2 $m^3$/s. The intake temperature and pressure of air at 0.1 MPa and $16^o$C.

The air is compressed to a final pressure of 0.7 MPa. The intermediate pressure is ideal and cooling is perfect. The compression index in both the stages is 1.25 and the compressor runs at 600 r.p.m Neglect clearance. Determine:

  1. Intermediate pressure
  2. Total volume of each cylinder
  3. Power required to drive the compressor
  4. Rate of heat rejection in the intercooler
1 Answer
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Given:

Two stage single acting reciprocating comp $V_1=0.2 m^3/s, T_1=16^o, P_1=0.1M,P_a=1 ×10^5 Pa$

$P_3=0.7mPa=7×10^5 Pa,n=1.25 , N=600 rpm $

To Find:$ P_2= ?,V_{LP}=?,V_{HP}=?,Power,Q_(\frac{i}{c})$

Solution:

Assume optimum intermediate pressure

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$P_2=\sqrt(P_1.P_3 )=\sqrt(0.1 ×0.7) $

=$0.264 MPa=2.64×10^5 Pa$

$V_{LP}=V_1=0.2 m^3/s$

$V_{LP}=V_1=\frac{0.2 ×60}{600}=0.02 m^3$

PV=mRT

$\frac{PV}{RT}=m →constant $

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

$\frac{1 ×10^5 ×0.02}{287 ×289}=\frac{2.64 ×10^5 ×V_5}{287 ×289}$

$V_5=7.57 ×10^{-3} m^3=0.07575m^3/s$

$V_{HP}=V_5=7.57 ×10^{-3} m^3$

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

=$\frac{1.25}{0.25} [{10^5×0.2[(\frac{2.64}{1})^\frac{0.25}{1.25}- 1]+ 2.64 ×10^5×0.07575[(\frac{7}{2.64})^\frac{0.25}{1.25}- 1]}]$

=42.857 KW

Heat rejected by air in inter cooler

$Q_(\frac{i}{c})=mC_p (T_2-T_1) $

$m=\frac{P_1 V_1}{RT_1}=\frac{10^5×0.2}{287 ×289}=0.2411 kg/s$

$\frac{T_2}{T_1} =(\frac{P_2}{P_1})^\frac{n-1}{n}$

$T_2=T_1 \frac{P_2}{P_1}^\frac{n-1}{n}=289\frac{2.64}{1}^\frac{0.25}{1.25}$

$T_2=350.93 k$

$ Q_{\frac{i}{c}}=0.2411 ×1005 ×(350.93-289)$

$Q_{\frac{i}{c}}=15.006 KW $

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