1 kg of water initially at 100 kPa is contained in a pistonc

1 kg of water initially at 100 kPa is contained in a piston-cylinder device where the piston is sitting on a set of stops. The volume at this state is 0.8 m^3. Heat is now added to the device until the piston starts rising at a pressure of 200 kPa. Heat is continued until volume is doubled. Calculate (a) the final temperature, (b) the change in internal energy of the system, (c) work, and (d) heat added. Also plot the T-v and P-v diagrams with respect to the saturation lines.

Solution

State 1: P1= 100 kPa, V1=0.8 m³ and T1 = 20^oC (Assumed)

State 2: P2= 200 kPa, V2=V1 then T2:

Assuming ideal gas conditions

P1/P2 = T1/T2 therfore, T2= 2xT1 = 2x20 = 40^oC.

Heat added during this porcess Q_1-2 = m c(T2-T1) = 1x4.187(40-20) =83.74 kJ, where c-specific heat of water =4.187 kj/kg K

State 3: P3 =P2= 200 kPa , V3=2x0.8 =1.6 m³ then final temperature T3:

V2/V3 = T2/T3 gives, T3 = 2xT2 = 2x40 = 80^oC.

Heat added Q_2-3= mc(T3-T2) = 1x4.187x(80-40) = 167.48 kJ

d) Total heat added 1-3 = 83.74+167.48 =251.22 kJ.

c) Workdone: Work is done only during process 2-3 = P2(V3-V2) = 200(1.6-0.8) =160 kJ

b) Change in Internal Energy = Q-W = 251.22-160 = 91.22 kJ