Answer to Question #83273 in Mechanical Engineering for muhammad abdul hadi
The work W done in an isobaric process is given by:
where p is pressure, and ΔV is the change in the volume during the process.
The heat Q supplied is equal to the change in the enthalpy ΔH:
From the steam table (see https://www.nist.gov/sites/default/files/documents/srd/NISTIR5078-Tab3.pdf), we can find the specific volume and enthalpy of the steam.
For the initial state (saturated steam at 7 bar):
v_L=0.001108 m^3/kg,v_V=0.27277 m^3/kg,
h_L=697.0 kJ/kg,h_V=2762.8 kJ/kg,
For the final state (superheated steam at 7 bar and 200oC):
v_2=0.300 m^3/kg,h_2=2845.3 kJ/kg,
Substitute into (1) and (2):
W=7∙〖10〗^5 (0.3-0.2456)=38077 kJ/kg,