Reversible work of compression

[Radion]

Homework Statement

Air enters a compressor at 100 [Kpa] and 10 [c], the air leave the compressor at 12[Mpa]
the isentropic efficiency η is 87%, calculate:

• work done
• reversible work
• irreversibility

Homework Equations

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T₂=T₁(P₂/P₁)^(1.4-1)/1.4
W = (C_p(T₂-T₁))/η
W_rev=Δh - T₀Δs
Δs = c_pln(T₂/T₁) - Rln(P₂/P₁)

The Attempt at a Solution

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Work done can be calculated with:
T₂=T₁(P₂/P₁)^(1.4-1)/1.4
W = (C_p(T₂-T₁))/η
T₁ , P₁ , P₂ are given.
Δs = c_pln(T₂/T₁) - Rln(P₂/P₁)
I tried to find W_rev with Δh - T₀Δs
and Δs = c_pln(T₂/T₁) - Rln(P₂/P₁)

the correct answers are :

• work : -338.2 kj/kg
• reversibility : -317.4 kj/kg
• irreversibility : 20.8 kj/kg

 

[nilesh_pat]

but I can't figure out how to solve this.Can someone please help me with the steps?thanks in advance!