It has to do with mechanical engineering
Solution :
Given :
External diameter of the hemispherical shell, D = 500 mm
Thickness, t = 20 mm
Internal diameter, d = D - 2t
= 500 - 2(20)
= 460 mm
So, internal radius, r = 230 mm
= 0.23 m
Density of molten metal, ρ = 
= 
The height of pouring cavity above parting surface is h = 300 mm
= 0.3 m
So, the metallostatic thrust on the upper mold at the end of casting is :

Area, A 




= 7043.42 N
The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%

W = 
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>
B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000

<h3>Cournot efficiency = W/Q1</h3>
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
Answer:
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Explanation:
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