Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
With a high temperature of 1000k and a cold temperature of 308K, The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
[tex]\frac{w}{Q1} = \frac{35}{100}[/tex]
W = [tex]1.2*\frac{35}{100}*1000kj/s[/tex]
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
What is the workdone by this engine?
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
[tex]\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ[/tex]
Cournot efficiency = W/Q1
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: https://brainly.com/question/6364271
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.