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2 answers:
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Answer:
engine B is more efficient.
Explanation:
We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.
We also kn ow that the efficiency of Carnot cycle given as follows
Here temperature should be in Kelvin.
For engine A
For engine B
So from above we can say that engine B is more efficient.
Answer:
The heat is transferred is at the rate of 752.33 kW
Solution:
As per the question:
Temperature at inlet, = 273 + 20 = 293 K
Temperature at the outlet, = 273 + 200 = 473 K
Pressure at inlet,
Pressure at outlet,
Speed at the outlet,
Diameter of the tube,
Input power,
Now,
To calculate the heat transfer, , we make use of the steady flow eqn:
where
= specific enthalpy at inlet
= specific enthalpy at outlet
= air speed at inlet
= specific power input
H and H' = Elevation of inlet and outlet
Now, if
and H = H'
Then the above eqn reduces to:
(1)
Also,
Area of cross-section, A =
Specific Volume at outlet,
From the eqn:
Now,
Also,
Now, using these values in eqn (1):
Now, rate of heat transfer, q:
q = mQ =