The tank B (3 bar, 50°C) has more stored energy.
The availability of the system is
a=m((u1−u0)−T0(s1−s0))=m(cv(T1−T0)−T0Δs)
ΔsA=cplnT0T1−Rlnp0p1=1.005ln15+27350+273−0.287ln11=0.115kg⋅KkJ
aA=m(cv(T1−T0)−T0ΔsA)=1(0.718(323−288)−288⋅0.115)=−7.99kJ
ΔsB=cplnT0T2−Rlnp0p2=1.005ln15+27350+273−0.287ln13=−0.2kg⋅KkJ
aB=m(cv(T2−T0)−T0ΔsB)=1(0.718(323−288)−288⋅(−0.2))=82.73kJ
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