For this process determine (i) the maximum work (ii) the available energy (iii) the irreversibility.

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written 9.4 years ago by

teamques10 ★ 70k

• modified 6.2 years ago

Two kg of air at 500KPa, $80^oc$ expands adiabatically in a closed system until its volume is doubled and its temperature becomes equal to that of the surroundings which is at 100KPa, $5^oc$.

thermodynamics

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written 9.4 years ago by

teamques10 ★ 70k

M = 2kg; P1 = 500000 Pa;

T1 = 353K; V2 = 2V1; P0=P2 = 100000 Pa;

T2 = 278K

Solution:

S2-s1 = mRln(p1/p2) + m Cpln(T2/T1)

= 2 x 280 x ln(5/1) + 2 x 1000 ln(278/353)

= 422 J/K

V1 = mRT1/P1 = 2 x 280 x 353/ …

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