A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng
ine is from a reservoir at 550K, and the heat transfer from the engine is released to the surrounding air at 300K. a. Draw a schematic illustrating the heat engine and show the engineering model. (5 points) b. Determine the heat transfer to the engine and the heat transfer from the engine to the air (Actual engine). (10 points) c. If the same heat engine is replaced by a Carnot heat engine that produces the same work output, find the thermal efficiency, the heat-transfer to the heat engine, and the heat transfer from the engine to the low temperature reservoir. (10 points)
1 answer:
Answer:
a) The schematic illustrating is attached
b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ
c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ
Explanation:
b) The heat transfer to the engine and the heat transfer from the engine to the air is:
Where
W = 750 kJ
n = 35% = 0.25
Replacing:
c) The efficiency of Carnot engine is:
The heat transfer to the heat engine is:
The heat transfer from the heat engine is:
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Let's solve the problem in the three different steps
1) Uniformly accelerated motion, with acceleration
and for a total time of 
. The body is initially at rest, so the distance covered is given by
Calling
and 
the final and initial velocity, and since the 
because the body starts from rest, we can use
to find the final velocity after this first leg:
And the average velocity in this first leg is
2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg:
. This is also the average velocity of the second leg.
The total time of this second leg is
. The distance covered is given by
3) Uniformly decelerated motion, with constant deceleration of
and for a total time of 
. Here, the initial velocity of the body is the final velocity of the previous leg, i.e. 
. Therefore, the distance covered in this leg is given by
The final velocity in this leg is given by
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
4) Finally, the total distance covered in the motion is
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
Let's solve the problem in the three different steps
1) Uniformly accelerated motion, with acceleration
Calling
to find the final velocity after this first leg:
And the average velocity in this first leg is
2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg:
The total time of this second leg is
3) Uniformly decelerated motion, with constant deceleration of
The final velocity in this leg is given by
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
4) Finally, the total distance covered in the motion is
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
Answer:
The time interval is
Explanation:
From the question we are told that
The length of the string is
The mass of the string is
The tension on the string is
The velocity of the pulse is mathematically represented as
Where is the linear density which is mathematically evaluated as
substituting values
Thus
The time taken is evaluated as
substituting values