Respuesta :
Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
[tex]P_{1}[/tex] = 5 Mpa
[tex]T_{1}[/tex] = 1623°C
= 1896 K
[tex]V_{1}[/tex] = 0.05 [tex]m^{3}[/tex]
Also given [tex]\frac{V_{2}}{V_{1}} = 20[/tex]
Therefore, [tex]V_{2}[/tex] = 1 [tex]m^{3}[/tex]
R = 0.27 kJ / kg-K
[tex]C_{V}[/tex] = 0.8 kJ / kg-K
Also given : [tex]P_{1}V_{1}^{1.25}=C[/tex]
Therefore, [tex]P_{1}V_{1}^{1.25}[/tex] = [tex]P_{2}V_{2}^{1.25}[/tex]
[tex]5\times 0.05^{1.25}=P_{2}\times 1^{1.25}[/tex]
[tex]P_{2}[/tex] = 0.1182 MPa
a). Work transfer, δW = [tex]\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}[/tex]
[tex]\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}[/tex]
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
= [tex]\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}[/tex]
=[tex]\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W[/tex]
=[tex]\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200[/tex]
= 197.7 kJ
