Argon is expanded in an isentropic turbine from 2 MPa and 500oC to 200 kPa. Determine the outlet temperature and the work produced by this turbine per unit mass of argon. (ANSWER: 308 K and 96.8 kJ/kg)

The answer to the second part of the question on work done is wrong. The right answer is 144.72 Kj/kg

A) T2 = 308K

B) Work done per unit mass of argon = 144.72 Kj/kg

Explanation:

A) Initial temperature, T1=500∘C = 773K

Initial pressure, p1 = 2MPa = 2000KPa

Final pressure, p2=200kPa

From the ideal properties table i attached, under argo, we'll get;

R(Gas constant) = 0.2081 Kj/kg.k

Cp = 0.5203 Kj/Kg.k and Cv = 0.3122 Kj/kg.k.

Formula for adiabatic index is given as;

γ = Cp/Cv = 0.5203/0.3122 = 1.67

Now, to calculate the final temperature using temperature and pressure relationship, we have;

T2/T1 = (p2/p1)^((γ - 1)/γ)

So, T2/773 = (200/2000)^((1.67 - 1)/1.67)

T2/773 = 0.1^(0.4012)

T2/773 = 0.397

T2 = 773 x 0.397 ≈ 308K

B) Work done by turbine per unit mass of argon is given as;

W1-2 = R(T1 - T2)/(γ - 1)

W1-2 = 0.2081(773 - 306.89)/(1.67 - 1) = 96.9975/0.67 = 144.72 Kj/kg