Question

Steam undergoes an isentropic compression in an insulatedpiston–cylinder assembly from an initial state where T1= 160°C,p1= 1 bar to a final state where the pressure p2= 30bar. Determinethe final temperature, in °C, and the work, in kJ per kg of steam

Answer 1

Answer:

The final temperature = 1007.26 K

The work done = 12016

Explanation:

For isentropic compression, we have;

$\dfrac{p_{1}}{p_{2}} =\left (\frac{T_{_{1}}}{T_{2}} \right )^{\dfrac{k}{k-1}}$

Where

p₁ = 1 bar

p₂ = 30 bar

T₁ = 160°C =

T₂ = The final temperature

K for steam = 1.33

T₂ = 433.15/(1/30)^(0.33/1.33) = 1007.26 K

The work done = m×$c_p$×(T₂ - T₁)

The work done per kilogram of steam = $c_p$×(T₂ - T₁)

Taking the average $c_p$ value, we have;

$c_p$ at (1007.26 + 433.15)/2 = 2.080 + (2.113-2.080)×20.205/50 = 2.093 kJ/(kg·K)

Which gives = 2.093*(1007.26 -433.15) = 1201.6 kJ