Answer to Question #215321 in Molecular Physics | Thermodynamics for Wavie

Question #215321

A single stage reciprocating compressor takes 1 m3 of air per minute at 1.013 bar and 15°C and delivers it at 7 bar. Assuming that the law of compression ispv1.35=c , and that clearance volume is negligible, calculate the indicated power.



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Expert's answer
2021-07-09T08:31:37-0400

p1=1.013  barT1=288  KV1=1  m3/minp2=7  barpv1.35=cma=p1V1RT1=1.013×102×1/600.287×288=0.02042  kg/sp_1=1.013 \;bar \\ T_1=288 \;K \\ V_1= 1 \;m^3/min \\ p_2= 7 \;bar \\ p_v^{1.35}=c \\ m_a= \frac{p_1V_1}{RT_1} \\ = \frac{1.013 \times 10^2 \times 1/60}{0.287 \times 288} = 0.02042 \;kg/s

Delivery temperature

T2T1=(p2p1)n1nT2=288(71.013)0.351.35=475.2  K\frac{T_2}{T_1}=(\frac{p_2}{p_1})^{\frac{n-1}{n}} \\ T_2=288(\frac{7}{1.013})^{\frac{0.35}{1.35}} = 475.2\;K

The indicated power =nn1mR(T2T1)= \frac{n}{n-1}mR(T_2-T_1)

=1.350.35×0.02042×0.287×(475.2288)=4.23  kW= \frac{1.35}{0.35} \times 0.02042 \times 0.287 \times (475.2-288) =4.23 \;kW


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