Question

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)

Answer 1

Answer:308 K and 96.8 kJ/kg)

Explanation:

This is an isentropic process.

From the question,

Initial temperature,

T1=500∘C

p1=2MPa

Final pressure,

p2=200kPa

Checking this values from the Properties Table of Ideal Gases:

R=0.2081kJ/kg.K

Cp=0.5203kJ/kg.K

Cv=0.3122kJ/kg.K

R=0.2081 kJ/kg.KCp=0.5203 kJ/kg.KCv=

For the adiabatic value index.

γ=CpCv

=0.5203

0.3122

γ=1.67

γ​=​C​p​C​v​=​0.5203​0.3122​γ​=​1.67

(a) the outlet temperature

T2sT1=(p2p1)(γ−1γ)T2s773=(2002000)(1.67−11.67)T2s=308.K(approximately)

(b) the workdone per kg of argon during the process.

w1−2=R(T1−T2s)γ−1=0.2081(773−306.9)1.67−1w1−2=144.77kJ/kg

w​1​−​2​=​R​(​T​1​−​T​2​s​)​γ​−​1​=​0.2081​(​773​−​306.9​)=96kj/kg

Answer 2

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

Answer:

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