Air is compressed adiabatically from p1 1 bar, T1 300 K to p2 15 bar, v2 0.1227 m3 /kg. The air is then cooled at constant volume to T3 300 K. Assuming ideal gas behavior, and ignoring kinetic and potential energy effects, calculate the work for the first process and the heat transfer for the second process, each in kJ per kg of air. Solve the problem each of two ways:

Work done for the adiabatic process = -247873.6 J/kg = - 247.9 KJ/kg

Heat transfer for the constant volume process = - 244.91 KJ/kg

Explanation:

For the first State,

P₁ = 1 bar = 10⁵ Pa

T₁ = 300 K

V₁ = ?

Second state

P₂ = 15 bar = 15 × 10⁵ Pa

T₂ = ?

V₂ = 0.1227 m³/kg

Third state

P₃ = ?

T₃ = 300 K

V₃ = ?

We require the workdone for step 1-2 (which is adiabatic)

And heat transferred for steps 2-3 (which is isochoric/constant volume)

Work done for an adiabatic process is given by

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats = 1.4 for air since air is mostly diatomic

K = PVʸ

Using state 2 to calculate for k

K = P₂V₂ʸ = (15 × 10⁵)(0.1227)¹•⁴ = 79519.5

We also need V₁

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

P₁V₁ʸ = K

(10⁵) (V₁¹•⁴) = 79519.5

V₁ = 0.849 m³/kg

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

W = 79519.5 [(0.1227)⁻⁰•⁴ - (0.849)⁻⁰•⁴]/(1 - 1.4)

W = (79519.5 × 1.247)/(-0.4) = - 247873.6 J/kg = - 247.9 KJ/kg

To calculate the heat transferred for the constant volume process

Heat transferred = Cᵥ (ΔT)

where Cᵥ = specific heat capacity at constant volume for air = 0.718 KJ/kgK

ΔT = T₃ - T₂

We need to calculate for T₂

Assuming air is an ideal gas,

PV = mRT

T = PV/mR

At state 2,

V/m = 0.1227 m³/kg

P₂ = 15 bar = 15 × 10⁵ Pa

R = gas constant for air = 287.1 J/kgK

T₂ = 15 × 10⁵ × 0.1227/287.1 = 641.1 K

Q = 0.718 (300 - 641.1) = - 244.91 KJ/kg