Answer:
The final temperature = 1007.26 K
The work done = 12016
Explanation:
For isentropic compression, we have;
[tex]\dfrac{p_{1}}{p_{2}} =\left (\frac{T_{_{1}}}{T_{2}} \right )^{\dfrac{k}{k-1}}[/tex]
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×[tex]c_p[/tex]×(T₂ - T₁)
The work done per kilogram of steam = [tex]c_p[/tex]×(T₂ - T₁)
Taking the average [tex]c_p[/tex] value, we have;
[tex]c_p[/tex] 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
