what is it ill know your question??!!
Idk I just need point an you probably already solved this by now
Answer:
a) Decrease
b) Decrease
c) Decrease
d) Decrease
Explanation:
Ti= 200°F ,
Te = 170°F
Area of heat exchanger =
= 20π
<u>A) when The flow rate of the cooling fluid is increased</u>
Temperature of process stream will decrease and this is because the tube side heat transfer coefficient will increase and this will increase the rate of heat transfer thereby decreasing the temperature of the process stream.
B) <u>when There are 200 tubes that are 1.0-in. OD and 20.0ft long</u>
The temperature of the process stream will decrease and this is because the heat transfer coefficient will increase likewise the heat transfer rate
C) <u>When The number of shell passes is doubled</u>
This will cause an increase in the overall length of the shell, an increase in velocity of constant volumetric flowrate, hence the Temperature of the process steam will decrease as well
D)<u> When The tube material is changed to copper.</u>
Due to the high thermal conductivity of copper when compared to steel , switching to copper will cause a decrease in the temperature of the process steam
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:
Explanation:
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