A Carnot engine operates between two heat reservoirs at temperatures TH and TC. An inventor proposes to increase the efficiency by running one engine between TH and an intermediate temperature T and a second engine between T and TC using as input the heat expelled by the first engine. Compute the efficiency of this composite system and compare it to that of the original engine ?

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

false.

explanation:

this statement is false. after falling for 5 seconds, its speed will not be 100 m/s.

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.