The work required is W = 20.2 BTU per lbm
The value of entropy & enthalpy at initial conditions are
[tex]$\begin{aligned}&h_{1}=103.1 \\&\mathrm{~S}=0.225\end{aligned}$[/tex]
Final enthalpy
[tex]$h_{2}=123.3$[/tex]
Therefore work done
[tex]$\begin{aligned}&\mathrm{W}=h_{1}-h_{2} \\&\mathrm{~W}=103.1-123.3 \\&\mathrm{~W}=-20.2 \text { BTU per Ibm }\end{aligned}$[/tex]
Therefore the work required is [tex]$\mathrm{W}=20.2 \mathrm{BTU}$[/tex]per Ibm
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