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PersonPerson13260
03.03.2022 •
Physics
Why is newton's version of kepler's third law so useful to astronomers?.
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Ответ:
A) Efficiency = 1 - (TC/TH)
B) Comparing this with the efficiency of the original carnot engine, the efficiency is the same
Explanation:
The formula for efficiency of an original carnot engine is;
e = 1 - T(C) /T(H) ——— eq 1
In like manner, for a composite engine, the efficiency is;
e(12) = (W1 + W2)/Q(H1)
Where W1 is work done by 1st engine; W2 is work done by second engine and Q(H1) is the heat input to the first engine.
Now the total work done is;
W = Q(H) + Q(C)
Where Q(H) is the heat input and Q(C) is the heat released.
Thus,
e(12) = [Q(H1) + Q(C1) + Q(H2) + Q(C2)] / Q(H1)
Now, from the earlier e(12) equation compared to this, QH2 = -QC1
Thus;
e(12) = [Q(H1) + Q(C1) - Q(C1) + Q(C2)] / Q(H1)
So e(12) = [Q(H1) + Q(C2)] / Q(H1)
So e(12) = 1 + [Q(C2)/Q(H1)] ———eq 2
Also,
Q(C2) /Q(H2) = (-Tc/T')
Where T' is intermediate temperature.
So, simplifying that,
Q(C2) = -Q(H2) (Tc/T')
This is also equal to Q(C1) (TC/T')
But Q(C1) is also equal to;
-Q(H1) (T'/TH)
Thus; Q(C2) is now written as;
Q(C2) = -Q(H1) (T'/TH)(TC/T')
So T' will cancel out to remain;
Q(C2) = -Q(H1)(TC/TH)
Replacing this with Q(C2) in eq 2 to obtain;
e(12) = 1 + [[-Q(H1)(TC/TH)] /Q(H1)]
e(12) = 1 - TC/TH