Answer:
a). Work transfer = 527.2 kJ
b). Heat Transfer = 197.7 kJ
Explanation:
Given:
= 5 Mpa
= 1623°C
= 1896 K
= 0.05 
Also given 
Therefore,
= 1 
R = 0.27 kJ / kg-K
= 0.8 kJ / kg-K
Also given : 
Therefore,
= 

= 0.1182 MPa
a). Work transfer, δW = 
![\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B5%5Ctimes%200.05-0.1182%5Ctimes%201%7D%7B1.25-1%7D%20%20%5Cright%20%5D%5Ctimes%2010%5E%7B6%7D)
= 527200 J
= 527.200 kJ
b). From 1st law of thermodynamics,
Heat transfer, δQ = ΔU+δW
= 
=![\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cfrac%7B%5Cgamma%20-n%7D%7B%5Cgamma%20-1%7D%20%5Cright%20%5D%5Ctimes%20%5Cdelta%20W)
=![\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200](https://tex.z-dn.net/?f=%5Cleft%20%5B%20%5Cfrac%7B1.4%20-1.25%7D%7B1.4%20-1%7D%20%5Cright%20%5D%5Ctimes%20527.200)
= 197.7 kJ
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Answer:
Heat required (q) = 471.19kj/kg
Explanation:
Find attached below solution to problem