Maximum Wind Power

Find the greatest fraction of absorbed starlight that an ideal day and night heat engine can convert into captured wind power, while both hemispheres radiate to space.

3 Nov 2026 About 7 min read

Heat Transfer

JEE Advanced

Physics Olympiad

Before you begin

You should know blackbody radiation, steady energy balance and Carnot efficiency. The precise maximum uses differentiation; trial values give the requested one-significant-figure estimate.

How much absorbed starlight can an ideal heat engine between a planet’s day and night sides convert into captured wind power? A colder night side raises the engine’s efficiency, but also reduces the heat it can accept. We will find the balance that gives the greatest power.

The problem

A spherical planet of radius \(R\) always presents the same face to a star of radius \(R_s\) and temperature \(T_s\). Their centres are separated by \(R_0\), much greater than either radius. The incident rays are parallel across the planet.

The day hemisphere has uniform temperature \(T_{\mathrm{h}}\) and the night hemisphere has uniform temperature \(T_{\mathrm{c}}\). Both temperatures are steady in each chosen operating state, but may change between states. Treat the star and planet as blackbodies, and the surrounding space as \(0\,\mathrm{K}\).

The hemispheres act as the hot and cold reservoirs of an ideal reversible engine. All heat transferred between them passes through the engine. Its work appears as wind power \(P_{\mathrm{w}}\), which is immediately captured and stored without reheating the planet.

For a blackbody of area \(A\),

\[ P_{\mathrm{rad}}=\sigma AT^4, \qquad \sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}. \tag{1}\]

Part a. Express the absorbed stellar power \(P\) in terms of \(R\), \(R_s\), \(R_0\), \(T_s\) and \(\sigma\).

Part b. Determine the limiting range of \(x=T_{\mathrm{c}}/T_{\mathrm{h}}\). Find the wind power at each endpoint.

Part c. Obtain \(P_{\mathrm{w}}/P\) as a function of \(x\) alone.

Part d. Estimate the largest possible fraction \(P_{\mathrm{w}}/P\) to one significant figure, with justification. We will also find a more precise optimum.

Part a The intercepted stellar power

The star emits uniformly over its spherical surface. Its luminosity is

\[ L_s=4\pi R_s^2\sigma T_s^4. \tag{2}\]

At the planet’s orbit, this power is spread over a sphere of radius \(R_0\). The irradiance is therefore

\[ I=\frac{L_s}{4\pi R_0^2} =\sigma T_s^4\frac{R_s^2}{R_0^2}. \tag{3}\]

The planet intercepts a beam with cross-sectional area \(\pi R^2\), so

\[ P=I\pi R^2=\pi\sigma T_s^4\frac{R^2R_s^2}{R_0^2}. \tag{4}\]

Parallel starlight falls on a planet with a hot day hemisphere and cooler night hemisphere. The intercepted power uses the projected disk area pi R squared, while each hemisphere radiates over two pi R squared. — Figure 1. The planet intercepts parallel starlight over projected area \(\pi R^2\). Both the day hemisphere and the night hemisphere radiate over area \(2\pi R^2\). The central line separates the two uniform-temperature regions; geometry is schematic.

The day hemisphere has surface area \(2\pi R^2\), but oblique sunlight supplies less power per unit area away from the point facing the star. The projected area determines absorption. Each hemisphere emits over its full curved area \(2\pi R^2\).

Set up the three energy balances

Define \(C=2\pi R^2\sigma\), with units \(\mathrm{W\,K^{-4}}\). Let \(\dot{Q}_{\mathrm{h}}\) be the heat rate entering the engine and \(\dot{Q}_{\mathrm{c}}\) the heat rate leaving it. Dots denote energy per unit time.

Absorbed stellar power P enters the day side. Each hemisphere radiates to space. Heat Q dot h passes from the day side into a reversible engine; rejected heat Q dot c reaches the night side. Captured work P w leaves the engine and is stored. — Figure 2. Every arrow represents a power. Day-side radiation and engine input share the absorbed power \(P\). The engine rejects heat to the night side and exports captured wind power \(P_{\mathrm{w}}\), which is stored without reheating the planet. Both hemispheres radiate to space; no heat bypasses the engine.

The steady balances for the hot hemisphere and cold hemisphere are

\[ P=CT_{\mathrm{h}}^4+\dot{Q}_{\mathrm{h}}, \qquad \dot{Q}_{\mathrm{c}}=CT_{\mathrm{c}}^4. \tag{5}\]

For the engine,

\[ P_{\mathrm{w}}=\dot{Q}_{\mathrm{h}}-\dot{Q}_{\mathrm{c}}. \tag{6}\]

A reversible engine operating between positive absolute temperatures obeys

\[ \frac{\dot{Q}_{\mathrm{c}}}{\dot{Q}_{\mathrm{h}}} =\frac{T_{\mathrm{c}}}{T_{\mathrm{h}}}=x, \qquad \eta=1-x. \tag{7}\]

Whenever the engine produces work, it transfers less heat to the cold side than it removes from the hot side.

Part b The two limiting cases

A work-producing engine requires \(T_{\mathrm{h}}>T_{\mathrm{c}}>0\). Including the ideal endpoint limits gives \(0\leq x\leq1\).

A very cold night side

The night side receives no sunlight, so its radiation determines how much heat it can accept. Combining its balance with reversibility gives

\[ \dot{Q}_{\mathrm{h}} =\frac{CT_{\mathrm{c}}^4}{x} =CT_{\mathrm{h}}T_{\mathrm{c}}^3. \tag{8}\]

As \(x\) tends to zero, the night-side temperature tends to zero. Both heat rates and \(P_{\mathrm{w}}\) then tend to zero. The day side radiates all the absorbed power, with \(T_{\mathrm{h}}\to(P/C)^{1/4}\).

The limiting Carnot efficiency approaches one, but the engine’s heat input approaches zero. There is no finite-power engine operating with a \(0\,\mathrm{K}\) reservoir in this model.

Nearly equal temperatures

As \(x\) tends to one, the engine efficiency tends to zero. With both temperatures equal to \(T\), the two radiation rates are equal:

\[ P=2CT^4,\qquad \dot{Q}_{\mathrm{h}}=\dot{Q}_{\mathrm{c}}=\frac{P}{2}, \qquad P_{\mathrm{w}}=0. \tag{9}\]

Maintaining finite heat flow while the driving temperature difference tends to zero would require unbounded thermal conductance. This endpoint is an ideal limit.

Part c The wind power for any ratio

For \(0<x<1\), put \(T_{\mathrm{c}}=xT_{\mathrm{h}}\) into the cold-side balance and the reversible-engine relation:

\[ \dot{Q}_{\mathrm{c}}=CT_{\mathrm{h}}^4x^4, \qquad \dot{Q}_{\mathrm{h}}=CT_{\mathrm{h}}^4x^3. \tag{10}\]

The hot-side balance fixes its temperature:

\[ P=CT_{\mathrm{h}}^4(1+x^3), \qquad T_{\mathrm{h}}^4=\frac{P}{C(1+x^3)}. \tag{11}\]

Subtracting the rejected heat rate from the supplied heat rate gives

\[ P_{\mathrm{w}}=CT_{\mathrm{h}}^4x^3(1-x). \tag{12}\]

Eliminate the hot-side temperature to obtain

\[ \frac{P_{\mathrm{w}}}{P}=\frac{x^3(1-x)}{1+x^3}. \tag{13}\]

Both endpoint limits give zero, as required. The denominator accounts for the day-side temperature adjustment when the operating state changes.

Part d Find the maximum

Write the power fraction as \(f(x)\). Differentiation gives

\[ f'(x)=\frac{x^2(3-4x-x^4)}{(1+x^3)^2}. \tag{14}\]

An interior maximum satisfies

\[ x^4+4x-3=0. \tag{15}\]

The numerator factor \(3-4x-x^4\) decreases strictly from \(3\) to \(-2\) as \(x\) runs from zero to one. Thus the derivative changes sign exactly once, from positive to negative. Together with the zero endpoint values, this proves that the maximum is unique and global.

\[ x_*=0.69250484\ldots, \qquad \left(\frac{P_{\mathrm{w}}}{P}\right)_{\mathrm{max}} =0.07666021\ldots. \tag{16}\]

The maximum is about \(7.7\%\). To the one significant figure requested, it is \(P_{\mathrm{w,max}}\simeq0.08P\), or \(8\%\).

The captured wind-power fraction is zero at temperature ratios zero and one. It rises to a single maximum of about 7.7 percent near x equal to 0.693, then falls. — Figure 3. Calculated captured-power fraction as a function of \(x=T_{\mathrm{c}}/T_{\mathrm{h}}\). The maximum is about \(7.7\%\) at \(x\simeq0.693\). Each point represents a different steady operating state, with both temperatures adjusted by the energy balances; this is not a time-evolution graph.

Estimate without solving the quartic

Trial ratios \(x=0.50\), \(0.70\) and \(0.90\) give power fractions \(0.0556\), \(0.0766\) and \(0.0422\), respectively. A sketch and a few nearby trials place the peak close to \(0.70\), enough for the requested estimate. The derivative argument also proves that this is the only peak.

An algebraic check. At the optimum, \(x_*^4=3-4x_*\). Using this relation in the power expression gives

\[ f(x_*)=\frac{x_*^4}{3} =1-\frac{4x_*}{3} =0.07666021\ldots. \tag{17}\]

Why the maximum is small

The engine efficiency and the fraction of absorbed starlight converted into work have different denominators:

\[ \frac{P_{\mathrm{w}}}{P} =\frac{\dot{Q}_{\mathrm{h}}}{P} \frac{P_{\mathrm{w}}}{\dot{Q}_{\mathrm{h}}} =\frac{x^3}{1+x^3}(1-x). \tag{18}\]

The first factor is the fraction entering the engine; the second is its efficiency. At the optimum, only about one quarter of the incoming stellar power reaches the engine. The rest is radiated directly by the day hemisphere. The engine converts about \(30.75\%\) of its own input into work.

Energy flow at the optimum	Fraction of \(P\)	Percentage
Radiation from the day side	\(1/(1+x_*^3)\)	\(75.0695\%\)
Heat entering the engine	\(x_^3/(1+x_^3)\)	\(24.9305\%\)
Heat rejected and radiated at night	\(x_^4/(1+x_^3)\)	\(17.2645\%\)
Captured wind power	\(x_^3(1-x_)/(1+x_*^3)\)	\(7.6660\%\)

The first two rows add to \(100\%\), and the last two add to the second row. For the whole planet, the two radiation rates plus captured wind power equal the incident power. The engine input is an internal flow; do not count it again as an output.

The temperatures must adjust together

For a fixed star and orbit, \(P\) is fixed. Once \(x\) is chosen, both temperatures follow from the balances:

\[ T_{\mathrm{h}}=\left[\frac{P}{C(1+x^3)}\right]^{1/4}, \qquad T_{\mathrm{c}}=xT_{\mathrm{h}}. \tag{19}\]

We cannot maximise the work rate while treating the engine heat input as an independent constant. Making the night side colder reduces its radiation and the heat rate it can accept. This also limits how much heat the reversible engine can draw from the day side.

Points to remember

Use \(\pi R^2\) for intercepted starlight and \(2\pi R^2\) for radiation from each hemisphere.

Use absolute temperatures in the fourth-power radiation law and the engine efficiency.

Keep the two engine heat rates distinct; their difference is the captured work rate.

Apply the Carnot factor to heat entering the engine, not to all the absorbed starlight.

The \(7.7\%\) result belongs to this ideal two-temperature model. It is not a prediction for Earth’s actual wind power. Only the engine is reversible; the whole star–planet radiation process is not assumed reversible.