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

The final temperature = 1007.26 K

The work done = 12016

Step-by-step explanation:

For isentropic compression, we have;

`(p_(1))/(p_(2)) =\left (\frac{T_{_(1)}}{T_(2)} \right )^{(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