Answer:
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
