A Carnot heat engine absorbs 235 KW of heat from a heat source and rejects 164 KW to the atmosphere. Determine the thermal effic
1 answer:
Answer:
43.2%
Explanation:
Given that,
Heat absorbed by a carnot heat engine,
Heat rejected to the atmosphere,
We need ti find the thermal efficiency of the heat engine. It is equal to the ratio of output work to the energy supplied. Its mathematical form is given by :
or
The egative value of efficiency shows work is done by the engine.
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Answer:
The final temperature of water is 381.39 °C.
Explanation:
Given that
Mass of water = 5 kg
Heat transfer at constant pressure Q = 2960 KJ
Initial temperature = 240 °C
We know that heat transfer at constant pressure given as follows
We know that for water
Lets take final temperature of water is T
So
T=381.39 °C
So the final temperature of water is 381.39 °C.
Answer:
A) 1040 steel is not a possible candidate for this application
B) 35.94%
Explanation:
Initial length = 100 mm = 0.1 m
Initial diameter ( d ) = 7.5 mm = 0.0075 m
Tensile load ( p ) = 18,000 N
Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m
<u>A) would the 1040 steel be a possible candidate for this application</u>
<em>Yield strength of 1040 steel < stress ( in order to be a possible candidate )</em>
stress = p / A0 = ( 18000 ) / ( ) * 0.0075^2
= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa
Yield strength of 1040 steel = 450 MPa
stress = 407.424 MPa
∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )
Therefore 1040 steel is not a possible candidate for this application
<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>
Area1 = ( ) ( 0.006 )^2 = 2.83 * 10^-5 m^2
therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%