The initial enthalpy and the entropy of the saturated water can be found out from the table of A-12E
i.e. [tex]$h_1= 101 \ \text{Btu/lbm}$[/tex]
[tex]$s_1 = 0.22717 \text{ Btu/lbmR}$[/tex]
Since the process mentioned above is an adiabatic compression process, the entropy will remain constant throughout the process. Therefore, we take the value of entropy and the final pressure using the table with few interpolations and also approximations to find the final enthalpy. It is given by :
[tex]$h_2= 116.09 \text{ Btu/lbm}$[/tex]
So the work input from the energy balance equation :
[tex]$\dot{W} + \dot{m}h_1 = \dot{m}h_2$[/tex]
[tex]$w=h_2 - h_1$[/tex]
= 116.09 - 101
= 15.09
Therefore, [tex]$w= 15.09 \text{ Btu/lbm}$[/tex]
