Thermodynamics and Carnot engines - university exam question

1. Aug 10, 2016

danieljones123

1. The problem statement, all variables and given/known data
A block of radioactive material is ti be used as a power source for a deep space probe which may be treated ideally as a single Carnot engine. Radioactive decay generates heat in the block at a rate Pin = dQin/dt and heat is extracted from the block to operate the probe at a rate dQout/dt. For simplicity, assume that the engine is designed to operate in such a way that the temperature of the block remains constantat Tb. Waste heat is radiated from the outer surface of the probe at a temperature Ts.

(a) Use the laws of thermodynamics to deduce the mathematical relationship between dQin(t) dQout(t). [2]

(b) Write down suitable expressions to calculate (i) the efficiency n of the engine representing the probe and thus deduce (ii) the rate at which work can be done, dW/dt. [4]

(c) If the power generated by the power source decreases exponentially with time (t is greater than 0) according to Pin = P0exp(-t/r), then integrate the resut of part b(ii) over time to show that the total amount of work ∆W that the source will provide over its whole life is given by
∆W = P0r(1-Ts/Tb). [4]

(d) Taking into account the assumption that the probe is a carnot engine, show that the resulting entropy change of the universe will be given by
∆S = P0r/Tb. [4]

Any help anyone could offer would be greatly appreciated! I would be truly greatful!

2. Relevant equations

3. The attempt at a solution
I've managed to draw out the schematic diagram of the carnot cycle/system for the first part but I cant get mathematical relationship in part (a). In part (b) I've managed to get the equation for the efficiency of the engine but am not sure about the rate at which work can be done.
And then part (c) and (d) both lead on from the answer from part (b)(ii) so without this answer I cannot complete the question and these parts are where most of the marks can be awarded.

If anyone could offer me any help or any insight then it would be massively appreciated, however small!
Cheers

2. Aug 10, 2016

Fred Wright

From thermodynamics for isothermal cycle where U is the internal energy, T is temperature and S is entropy,
$\Delta U = T\Delta S$
The change in the internal energy is equal to the heat added,
$Q_b = T_b \Delta S_b$,
$\Delta S_b = \frac {Q_b } {T_b}$.
The work done is the difference between the heat added by the block and the heat taken away by outer space,
$W = Q_{in} - Q_{out}$,
$dW = dQ_{in} - dQ_{out}$.
For a Carnot engine, the change in entropy of the block must equal the change in entropy of outer space.
$\frac {Q_b} {T_b} = \frac {Q_s} {T_s}$
$W = Q_b - Q_b\frac {T_s} {T_b} = Q_b(1 - \frac{T_s} {T_b})$
which gives the maximum efficiency.
You can integrate $\Delta Q_b = P_0\int exp (\frac {-t} {r})dt$ to get the heat added for all time and since the change in entropy of the block for all time must equal the change in entropy of outer space for all time the quoted answers follow.

3. Aug 11, 2016

danieljones123

Cheers Fred, explained it all very well and I understand a lot better now.

Really appreciate the help!

Thanks!