Subfield: Thermodynamics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1

\end{array}

$$

Extra Supplementary Reading Materials:

Green Revolution ${ }^{1}$



In this problem, we will investigate a simple thermodynamic model for the conversion of solar energy into wind. Consider a planet of radius $R$, and assume that it rotates so that the same side always faces the Sun. The bright side facing the Sun has a constant uniform temperature $T_{1}$, while the dark side has a constant uniform temperature $T_{2}$. The orbit radius of the planet is $R_{0}$, the Sun has temperature $T_{s}$, and the radius of the Sun is $R_{s}$. Assume that outer space has zero temperature, and treat all objects as ideal blackbodies.
Context question:
a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma A T^{4}$.)
Context answer:
\boxed{$P=\pi \sigma T_{s}^{4} \frac{R^{2} R_{s}^{2}}{R_{0}^{2}}$}


Extra Supplementary Reading Materials:

In order to keep both $T_{1}$ and $T_{2}$ constant, heat must be continually transferred from the bright side to the dark side. By viewing the two hemispheres as the two reservoirs of a reversible heat engine, work can be performed from this temperature difference, which appears in the form of wind power. For simplicity, we assume all of this power is immediately captured and stored by windmills.
Context question:
b. The equilibrium temperature ratio $T_{2} / T_{1}$ depends on the heat transfer rate between the hemispheres. Find the minimum and maximum possible values of $T_{2} / T_{1}$. In each case, what is the wind power $P_{w}$ produced?
Context answer:
The minimum value is simply zero, $P_{w}=0$.
The maximum value is $T_{2} / T_{1}=1$. $P_{w}=0$.


Context question:
c. Find the wind power $P_{w}$ in terms of $P$ and the temperature ratio $T_{2} / T_{1}$.
Context answer:
\boxed{$P_{w}=\frac{x^{3}(1-x)}{1+x^{3}} P$}

Question: d. Estimate the maximum possible value of $P_{w}$ as a fraction of $P$, to one significant figure. Briefly explain how you obtained this estimate.

Solution:
There are many ways to get the required answer. For example, by sketching the function, one can see that there is a unique maximum at an intermediate value of $x$, and furthermore that this maximum is at $x>0.5$, because of the rapid rise of the $x^{3}$ factor. One could then compute $P_{w} / P$ with a calculator at a few trial values such as $x=0.5,0.7,0.9$, which are already enough to get the desired accuracy.



The optimum value is $x=0.69$, at which point



$$

P_{w}^{\max }=0.077 P

$$







Hence in this model, at most $7.7 \%$ of solar energy can be converted into wind energy. Any answer within $15 \%$ of this value was accepted.



Final answer: $P_{w}^{\max }=0.077 P$