Answer:
18.75in
Explanation:
Modulus of elasticity = Stress/Strain
Since stress = Force/Area
Given
Force = 1000lb
Area = 0.75sqin
Stress = 1000/0.75
Stress = 1333.33lbsqin
Strain
Strain = Stress/Modulus of elasticity
Strain = 1333.33/5,000,000
Strain = 0.0002667
Also
Strain = extension/original length
extension = 0.005in
Original length = extension/strain
Original length = 0.005/0.0002667
Original length = 18.75in
Hence the original length of the rectangular bar is 18.75in
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: (a) +/- 7.5° (b) +/- 3.75°
Explanation:
See attachment
Answer:
The work of the cycle.
Explanation:
The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.
The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.
