Assignment On No Kinetic And Potential Energy Change

Answer 1
1. Assumptions are- no heat transfer, no kinetic and potential energy change, reversible
process. For an isentropic process. S1 =S2
At P1 = 0.8MPa and saturated vapor condition the ν1 = 0.0256m3/Kg (specific
volume)
U1 = 246.76KJ/Kg (internal energy), s1 = 0.91865KJ/KgK
To find out mass. V/ν1=m. Therefore. 0.05/0.025621 = 1.952Kg
At P2 = 0.4MPa and S2=S1,
dryness fraction x2 = (s2-sf)/sfg = (0.91835-0.2476)/0.67929 = 0.984
based on this dryness fraction internal energy at state 2 is U2 = uf+x2*Ufg
U2 = 63.62+0.984*171.45 = 232.91KJ/Kg
Temperature at his internal energy = 8.9C
Based on energy principle, Energy in-Energy out = energy change
-W = ΔU = m(u2-u1)
W = m(u1-u2) = 1.952*(246.76-232.91) – 27.09Kj
Answer 2
2. The cycle is shown below
Process 1-2 is constant temperature process