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
Answer:
Yes
Explanation:
Given Data
Temprature of source=750°c=1023k
Temprature of sink =0°c=273k
Work produced=3.3KW
Heat Rejected=4.4KW
Efficiency of heat engine(η)=
and
Heat Supplied 
η=
η=42.85%
Also the maximum efficiency of a heat engine operating between two different Tempratures i.e. Source & Sink
η=1-
η=1-
η=73.31%
Therefore our Engine Efficiency is less than the maximum efficiency hence the given claim is valid.
Answer:
the branch of science and technology concerned with the design, building, and use of engines, machines, and structures. Anything that involves engines, wires, etc. basically
