Respuesta :
Answer:
a. W = - 108.89 kJ/kg
b. q = - 75.846 kJ/kg
Given:
n = 1.2
Pressure, P = 120 kPa
Temperature, T = [tex]10^{\circ}[/tex] = 283 K
Pressure, P' = 800 kPa
Solution:
(a) Work Produced:
Now, using the relation:
W = [tex]\frac{RT}{1 - n}[(\frac{P'}{P})^(1 - \frac{1}{n}) - 1][/tex]
where
R = Rydberg's constant
Now,
W = [tex]\frac{283\times0.208}{1 - 1.2}[(\frac{800}{120})^(1 - \frac{1}{1.2}) - 1][/tex]
W = [tex]\frac{283\times0.208}{- 0.2}[(6.67)^(0.16) - 1] = - 108.89 kJ/kg[/tex]
Now,
[tex]\frac{T'}{T}= (\frac{P'}{P})^{1 - \frac{1}{n}}[/tex]
[tex]T'= T(\frac{800}{120})^{1 - \frac{1}{1.2}}[/tex]
[tex]T'= 1.37\times 283 = 388.48 K[/tex]
(b) Heat tranferred, q = [tex]C_{v}(T' - T) + W[/tex]
q = [tex]0.3122(388.84 - 283) - 108.89 = - 75.846 kJ/kg[/tex]
