20 kg of steam initially at 2000 kPa and 400°C expansion to a final pressure of...

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Answer 1

Answer #1

Answer:

We know that for adiabatic processes, PV^γ = constant, where, γ = C_P/C_v = 5/3 for an ideal gas.

Also for ideal gases we can write the equation as PV = nRT

∴ (V₁/V₂) = P₂/P₁ and (V₁/V₂) = (T₁/T₂) (P₂/P₁)

or (T₁/T₂) = (P₂/P₁)^1-γ/γ

Given: T₁ = 400°C = (400+273) K = 673 K, P₁ = 2000 kPa and P₂= 500 kPa, γ = 5/3 (say)

Putting all above values we can calculate T₂ as-

T₂ = T₁(P₂/P₁)^γ-1/γ

∴ T₂ = 673 × (500/2000)^{1.666-1/1.666} = 673 × (500/2000)^0.4 = 673 × 0.574 = 386.3 K = 113.3°C

Work done = m Cv ∆T = 20 kg × 1.4108 kJ/kg.K (386.3-673) K = -8089.52 kJ

[Considering for steam Cv = 1.4108 kJ/kg.K and m = 20 kg (given)]

Final Temperature	386.3 K or 113.3°C
Work done by steam	-8089.52 kJ

Answer 2

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