# Answer to Question #83273 in Mechanical Engineering for muhammad abdul hadi

The work W done in an isobaric process is given by:

W=p∆V, (1)

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:

Q=∆H. (2)

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_1=v_L+x(v_V-v_L ),

v_L=0.001108 m^3/kg,v_V=0.27277 m^3/kg,

v_1=0.001108+0.9∙(0.27277-0.001108)=0.2456 m^3/kg,

h_1=h_L+x(h_V-h_L ),

h_L=697.0 kJ/kg,h_V=2762.8 kJ/kg,

h_1=697+0.9∙(2762.8-697)=2556.2 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,

Q=2845.3-2556.2=289.1 kJ/kg.

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