---
title: "Radiation and Heat Transport"
book: "University Earth and Space Sciences — Year 1"
subject: earth
language: en
chapter: 27
exercises: 12
source: https://one-course.com/books/earth/3/en/chapter/27-radiation-and-heat-transport
license: CC-BY-NC-SA-4.0
credit: "One Earth and Space Sciences Book, One Course (one-course.com)"
---

# Chapter 27 — Radiation and Heat Transport

From a million miles away, the sunlit face of the Earth is a disc of blue sea, brown land and white cloud. The sunlight falls most directly on the middle of the disc, near the tropics, and grazes its edges; the tropics absorb far more energy than the polar regions. Yet the tropics do not grow hotter year after year, nor the poles colder. Winds and ocean currents carry the excess towards the poles, several petawatts of it. This chapter measures the three terms of that balance: the sunlight each [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) receives, the part it keeps and the infrared it sends back to space, and the transport that makes up the difference. A large share of the transport travels in an invisible form, as water vapour, whose latent heat is released wherever it condenses into rain.

![The whole sunlit side of the Earth, photographed on 6 July 2015 by a camera on the Deep Space Climate Observatory, a million miles away; North and Central America are in the middle, with the turquoise shallow seas of the Caribbean. Two weeks after the June solstice the Arctic, at the top, is lit all round the clock. The bright clouds reflect much of the light that reaches them; the bluish tint is sunlight scattered by the molecules of the air. (NASA Goddard Space Flight Center, public domain, via Wikimedia Commons.)](https://one-course.com/images/onecourse/chapters/earth-3/b1-radiation-heat-transport/img-8b82188f61b6.jpg)

*The whole sunlit side of the Earth, photographed on 6 July 2015 by a camera on the Deep Space Climate Observatory, a million miles away; North and Central America are in the middle, with the turquoise shallow seas of the Caribbean. Two weeks after the June solstice the Arctic, at the top, is lit all round the clock. The bright clouds reflect much of the light that reaches them; the bluish tint is sunlight scattered by the molecules of the air. (NASA Goddard Space Flight Center, public domain, via Wikimedia Commons.)*

**You already know.**

[Chapter 1](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#ch-b1-spherical-astronomy) defined the [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation), the power of sunlight received per square metre of a horizontal surface at the top of the atmosphere, $I = S_0(a/r)^2\sin h_\odot$ while the Sun is up, with $S_0 = 1361\,\mathrm{W}\,\mathrm{m}^{-2}$ at the mean distance $a$; gave the Sun’s [altitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-horizontal) from its [hour angle](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) $H$, its [declination](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) $\delta$ and the [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) $\varphi$, $\sin h_\odot = \sin\varphi\sin\delta + \cos\varphi\cos\delta\cos H$; and the [hour angle](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) $H_0$ of sunset, $\cos H_0 = -\tan\varphi\tan\delta$ ([Proposition 1.16](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#prop-b1-spherical-astronomy-rising)). [Chapter 3](https://one-course.com/books/earth/3/en/chapter/3-comparative-planetology#ch-b1-comparative-planetology) defined the [Bond albedo](https://one-course.com/books/earth/3/en/chapter/3-comparative-planetology#def-b1-comparative-planetology-albedo) of a planet and its [equilibrium temperature](https://one-course.com/books/earth/3/en/chapter/3-comparative-planetology#def-b1-comparative-planetology-equilibrium-temperature). The physics volumes give the inverse-square law for light, the radiation of a black body, $\sigma T^4$ per square metre with $\sigma = 5.670 \times 10^{-8}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{K}^{-4}$ (Stefan–Boltzmann), and the latent heat of a change of state. Book 2 showed that the Earth intercepts on average a quarter of the solar constant, drew the global energy budget of the Earth and checked that it balances, and described the poleward transport of heat in words.

## 27.1 Insolation with latitude and season

The instantaneous [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) changes from minute to minute as the Sun crosses the sky. For the climate, what counts is its sum over a day.

**Definition 27.1 (Daily-mean insolation).**

The *daily-mean insolation* $\bar Q$ at a [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) on a given day is the instantaneous [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) averaged over the 24 hours of that day, nights included, the [declination](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) of the Sun and its distance being held at their values for the day. It is a power per square metre of horizontal surface at the top of the atmosphere, in $\mathrm{W}\,\mathrm{m}^{-2}$.

Holding $\delta$ and $r$ fixed for a day is harmless: the [declination](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) changes by at most $0.4{}^{\circ}$ a day, and the distance by less than $0.03\,\%$.

**Theorem 27.2 (Daily-mean insolation).**

At [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) $\varphi$, on a day when the Sun’s [declination](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) is $\delta$ and its distance $r$,

$$
\bar Q = \frac{S_0}{\pi}\left(\frac{a}{r}\right)^{2}
  \bigl(h_0\sin\varphi\sin\delta + \cos\varphi\cos\delta\sin h_0\bigr),
$$

where $h_0 \in [0, \pi]$ is the [hour angle](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) of sunset in radians: $\cos h_0 =
-\tan\varphi\tan\delta$ when $|\tan\varphi\tan\delta| \le 1$; $h_0 = \pi$ in the polar day ($\tan\varphi\tan\delta > 1$); $h_0 = 0$, and $\bar Q = 0$, in the polar night ($\tan\varphi\tan\delta < -1$).

**Proof.** The [hour angle](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) $H$ advances uniformly through $2\pi$ in a day, so the mean over the day is the mean over $H \in [-\pi, \pi]$. The Sun is up for $-h_0 < H < h_0$, where $\sin h_\odot > 0$, and the [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) is zero otherwise. Hence

$$
\bar Q = \frac{1}{2\pi}\int_{-h_0}^{h_0} S_0\left(\frac{a}{r}\right)^{2}
  \bigl(\sin\varphi\sin\delta + \cos\varphi\cos\delta\cos H\bigr)\,\mathrm dH ,
$$

which is $(S_0/2\pi)(a/r)^2\bigl(2h_0\sin\varphi\sin\delta +
2\cos\varphi\cos\delta\sin h_0\bigr)$. The bounds come from [Proposition 1.16](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#prop-b1-spherical-astronomy-rising): the [altitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-horizontal) is zero when $\cos H = -\tan\varphi\tan\delta$. If that number exceeds 1 in absolute value, the Sun never crosses the horizon: when $\tan\varphi\tan\delta > 1$, the integrand is positive for every $H$ and the integral runs over the whole day, $h_0 = \pi$; when $\tan\varphi\tan\delta < -1$, it is negative for every $H$ and the Sun never rises. ∎

Two special cases are worth knowing by heart. At the equator, $\varphi = 0$ and $h_0 = \pi/2$ whatever the season:

$$
\bar Q_\text{equator} = \frac{S_0}{\pi}\left(\frac{a}{r}\right)^{2}\cos\delta ,
$$

largest at the equinoxes, $S_0/\pi = 433.2\,\mathrm{W}\,\mathrm{m}^{-2}$ at the mean distance. At a pole in its summer, $h_0 = \pi$ and $\varphi = \pi/2$: $\bar Q_\text{pole} =
S_0(a/r)^2\sin\delta$, the Sun circling the sky at a constant [altitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-horizontal) equal to the [declination](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial).

**Proposition 27.3 (The polar summer).**

On a day of [declination](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) $\delta > 0$, the [daily-mean insolation](#def-b1-radiation-heat-transport-daily-insolation) at the North Pole exceeds that at the equator if and only if $\pi\tan\delta > 1$, that is $\delta >
17.66{}^{\circ}$. At the June solstice, $\delta = 23.44{}^{\circ}$, the ratio is $\pi\tan
\delta = 1.362$.

**Proof.** The two special cases above share the factor $(S_0/\pi)(a/r)^2$; their ratio is $\pi\sin\delta/\cos\delta = \pi\tan\delta$, which exceeds 1 when $\tan\delta >
1/\pi$, $\delta > \arctan(1/\pi) = 17.66{}^{\circ}$. With the obliquity, $\pi\tan23.44{}^{\circ} = 1.362$. ∎

With the Earth’s orbit, the June solstice falls near aphelion, $(a/r)^2 = 0.968$: the North Pole then receives $524\,\mathrm{W}\,\mathrm{m}^{-2}$ as a daily mean, the equator $385\,\mathrm{W}\,\mathrm{m}^{-2}$. Six months later the South Pole, near perihelion, receives $559\,\mathrm{W}\,\mathrm{m}^{-2}$, $6.7\,\%$ more than the North Pole had in its summer: the ratio is exactly that of the two values of $(a/r)^2$. The figure below shows $\bar Q$ over a whole year. The maxima are at the poles at the solstices; the dark wedges are the polar nights, bounded by the [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) $90{}^{\circ} -
|\delta|$; the subsolar point, where the Sun passes overhead at noon, moves between the tropics with the [declination](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial).

![Daily-mean insolation at the top of the atmosphere against the day of the year and the latitude, from with S_0 = 1361\, W\, m-2 and the Earth’s orbit. Dashed: the subsolar latitude, equal to the declination; solid: the edges of the polar nights. The summer poles receive more than any other place on the Earth; the southern summer, near perihelion, is the brighter of the two.](https://one-course.com/images/onecourse/chapters/earth-3/b1-radiation-heat-transport/fig-5a5c743ad420.svg)

*[Daily-mean insolation](#def-b1-radiation-heat-transport-daily-insolation) at the top of the atmosphere against the day of the year and the [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), from [Theorem 27.2](#thm-b1-radiation-heat-transport-daily) with $S_0 = 1361\,\mathrm{W}\,\mathrm{m}^{-2}$ and the Earth’s orbit. Dashed: the subsolar [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), equal to the [declination](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial); solid: the edges of the polar nights. The summer poles receive more than any other place on the Earth; the southern summer, near perihelion, is the brighter of the two.*

Averaging $\bar Q$ over the year gives the annual-mean [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) of each [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) (figure below): $415.6\,\mathrm{W}\,\mathrm{m}^{-2}$ at the equator, $328.3\,\mathrm{W}\,\mathrm{m}^{-2}$ at $40{}^{\circ}$, $172.4\,\mathrm{W}\,\mathrm{m}^{-2}$ at either pole. The polar summer is intense but the polar winter is dark, and over a year the equator receives $2.41$ times as much as a pole. The curve is the same in both hemispheres, although the southern summer is the brighter; [Exercise 27.10](#exo-b1-radiation-heat-transport-10) shows why.

![Daily-mean insolation against latitude at the two solstices, and its annual mean (black). The annual mean exceeds the global mean, 340.3\, W\, m-2 (dashed), between 37 S and 37 N, on 60\,\% of the Earth’s surface.](https://one-course.com/images/onecourse/chapters/earth-3/b1-radiation-heat-transport/fig-d6d8dfcbf6e7.svg)

*[Daily-mean insolation](#def-b1-radiation-heat-transport-daily-insolation) against [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) at the two solstices, and its annual mean (black). The annual mean exceeds the global mean, $340.3\,\mathrm{W}\,\mathrm{m}^{-2}$ (dashed), between $37{}^{\circ}$ S and $37{}^{\circ}$ N, on $60\,\%$ of the Earth’s surface.*

**Proposition 27.4 (The global mean).**

At every instant, the [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) averaged over the whole sphere is $\tfrac14
S_0(a/r)^2$. Averaged over the year as well, it is $S_0/(4\sqrt{1 - e^2})$, where $e$ is the eccentricity of the orbit: $340.30\,\mathrm{W}\,\mathrm{m}^{-2}$ for the Earth, within $0.02\,\%$ of $S_0/4 = 340.25\,\mathrm{W}\,\mathrm{m}^{-2}$.

**Proof.** The Earth, of radius $R$, intercepts the sunlight crossing a disc of area $\pi
R^2$, a power $S_0(a/r)^2\pi R^2$, which falls on the day hemisphere; spread over the whole surface $4\pi R^2$, it gives the first result. The time average of $(a/r)^2$ over the orbit is $1/\sqrt{1-e^2}$ ([Exercise 27.11](#exo-b1-radiation-heat-transport-11)). ∎

## 27.2 Absorbed and emitted radiation

Of the sunlight a [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) receives, it reflects a fraction straight back, from clouds, snow, ice, desert and, to a lesser degree, sea; it absorbs the rest, in the air and at the surface. It also loses energy by emitting infrared radiation to space, day and night.

**Definition 27.5 (Outgoing longwave radiation).**

The *outgoing longwave radiation* (OLR) is the power of infrared radiation leaving the top of the atmosphere towards space, per square metre. The *emission temperature* $T_e$ is the temperature of a black body that would emit the same power: $\sigma T_e^4 = \text{OLR}$.

The OLR comes partly from the surface, through the air where it is transparent to infrared, and partly from the air and the cloud tops higher up. The [emission temperature](#def-b1-radiation-heat-transport-olr) is a property of the radiation, not the temperature of a particular level: it is lower than that of the ground beneath, because the emitting layers are colder.

**Definition 27.6 (Net radiation; radiative surplus and deficit).**

The *net radiation* at the top of the atmosphere is

$$
R = \underbrace{(1 - \alpha)\,\bar Q}_{\text{absorbed solar}} - \text{OLR},
$$

where $\alpha$ is the local albedo, the fraction of the incoming sunlight reflected to space; all three quantities are annual means at one [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic). A [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) where $R > 0$ has a *radiative surplus*; one where $R < 0$ has a *radiative deficit*.

Over the whole Earth the [net radiation](#def-b1-radiation-heat-transport-net-radiation) must average almost zero, since the planet neither heats nor cools by more than a few tenths of a watt per square metre: the latest assessment puts the imbalance at $0.7\,\mathrm{W}\,\mathrm{m}^{-2}$, on an incoming $340\,\mathrm{W}\,\mathrm{m}^{-2}$ of which $100\,\mathrm{W}\,\mathrm{m}^{-2}$ are reflected (a global albedo of $0.294$) and $239\,\mathrm{W}\,\mathrm{m}^{-2}$ leave as infrared.

The zonal profiles of this chapter come from a reanalysis, that is, a weather forecasting system run over the past decades, which assimilates the observations of each day and outputs a complete, consistent state of the atmosphere. The reanalysis used here, from 1948 on, gives monthly means on a grid of 94 [latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) and 192 [longitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic); this chapter uses its averages over 1991–2020. Its radiation fluxes are model outputs, forecasts constrained by the observations. The figure below shows their zonal means against $\sin\varphi$, so that equal widths on the axis are equal areas on the sphere.

![Zonal and annual mean absorbed solar radiation (orange) and outgoing longwave radiation (red) at the top of the atmosphere, from the NCEP-NCAR Reanalysis 1, means 1991–2020 (data provided by the NOAA PSL, Boulder, Colorado, USA, from their website at https://psl.noaa.gov). Shaded: the radiative surplus of the low latitudes and the deficits of the high latitudes.](https://one-course.com/images/onecourse/chapters/earth-3/b1-radiation-heat-transport/fig-f11bad477f6d.svg)

*Zonal and annual mean absorbed solar radiation (orange) and [outgoing longwave radiation](#def-b1-radiation-heat-transport-olr) (red) at the top of the atmosphere, from the NCEP-NCAR Reanalysis 1, means 1991–2020 (data provided by the NOAA PSL, Boulder, Colorado, USA, from their website at https://psl.noaa.gov). Shaded: the [radiative surplus](#def-b1-radiation-heat-transport-net-radiation) of the low [latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) and the deficits of the high [latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic).*

The two curves have very different shapes. The absorbed sunlight falls from $288\,\mathrm{W}\,\mathrm{m}^{-2}$ near the equator to $76\,\mathrm{W}\,\mathrm{m}^{-2}$ at $81{}^{\circ}$ N, by a factor of $3.8$: the [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) drops, and the albedo rises from $0.31$ to $0.57$, with the low Sun, the snow and the sea ice. The OLR falls only from $245\,\mathrm{W}\,\mathrm{m}^{-2}$ to $192\,\mathrm{W}\,\mathrm{m}^{-2}$, by a factor of $1.27$; the [emission temperatures](#def-b1-radiation-heat-transport-olr) are $256\,\mathrm{K}$ and $241\,\mathrm{K}$. The OLR is even slightly lower at the equator, $245\,\mathrm{W}\,\mathrm{m}^{-2}$, than near $20{}^{\circ}$, $269\,\mathrm{W}\,\mathrm{m}^{-2}$: the clouds of the equatorial rain belt rise high and their cold tops emit little, while the clearer subtropical skies let more infrared escape. Wherever the absorbed sunlight exceeds the OLR, the [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) has a surplus.

**Example 27.7 (The reanalysis’s global budget).**

Averaged over the globe, the reanalysis receives $341.9\,\mathrm{W}\,\mathrm{m}^{-2}$, reflects $117.0\,\mathrm{W}\,\mathrm{m}^{-2}$ (an albedo of $0.342$), absorbs $224.8\,\mathrm{W}\,\mathrm{m}^{-2}$ and emits $237.7\,\mathrm{W}\,\mathrm{m}^{-2}$: a [net radiation](#def-b1-radiation-heat-transport-net-radiation) of $-12.9\,\mathrm{W}\,\mathrm{m}^{-2}$. The OLR is within $1.3\,\mathrm{W}\,\mathrm{m}^{-2}$ of the assessed $239\,\mathrm{W}\,\mathrm{m}^{-2}$, but the reflection is $17\,\mathrm{W}\,\mathrm{m}^{-2}$ larger than the assessed $100\,\mathrm{W}\,\mathrm{m}^{-2}$. A real imbalance of this size would cool the Earth by $6.6\,\mathrm{PW}$; it is an error of the model, not a property of the Earth, and the transport below is computed after removing the global mean.

## 27.3 Meridional heat transport

**Definition 27.8 (Meridional heat transport).**

The *meridional heat transport* $T(\varphi)$ across the parallel of [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) $\varphi$ is the energy carried northwards across that parallel per unit time, by the atmosphere and the ocean together, in the annual mean. It is counted positive northwards and measured in petawatts ($1\,\mathrm{PW}$ = $10^{15}\,\mathrm{W}$).

**Proposition 27.9 (Transport from the net radiation).**

If the energy stored in each [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) band does not change from year to year, then for an Earth of radius $a_\oplus$,

$$
T(\varphi) = 2\pi a_\oplus^2\int_{-\pi/2}^{\varphi} R(\varphi')\cos\varphi'\,
  \mathrm d\varphi' = -2\pi a_\oplus^2\int_{\varphi}^{\pi/2} R(\varphi')\cos
  \varphi'\,\mathrm d\varphi' ,
$$

and $T$ is extreme at the [latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) where $R$ changes sign.

**Proof.** Take the polar cap south of the parallel $\varphi$ as a control volume, from the bottom of the ocean to the top of the atmosphere (figure below). Energy enters or leaves it in only two ways: as radiation at the top, a net power $\int R\,\mathrm dA$, and by the transport across its rim, $-T(\varphi)$ (a northward transport leaves the southern cap). The band between $\varphi'$ and $\varphi' + \mathrm d\varphi'$ has area $\mathrm dA = 2\pi a_\oplus\cos\varphi'
\cdot a_\oplus\,\mathrm d\varphi'$. With no change of storage, the sum is zero: $\int_{-\pi/2}^{\varphi} R\,2\pi a_\oplus^2\cos\varphi'\,\mathrm d\varphi' -
T(\varphi) = 0$. Applied to the whole Earth ($\varphi = \pi/2$, no rim), the same balance gives $\int R\,\mathrm dA = 0$ over the sphere, whence the second form. Differentiating, $T'(\varphi) = 2\pi a_\oplus^2 R(\varphi)\cos\varphi$, which vanishes where $R$ changes sign. ∎

![The control volume of , drawn for the cap north of the parallel (shaded): in a steady state, what the cap loses by radiation at the top of the atmosphere (red) is what the atmosphere and the ocean bring in across its rim (orange).](https://one-course.com/images/onecourse/chapters/earth-3/b1-radiation-heat-transport/fig-34809f93b5ba.svg)

*The control volume of [Proposition 27.9](#prop-b1-radiation-heat-transport-mht), drawn for the cap north of the parallel $\varphi$ (shaded): in a steady state, what the cap loses by radiation at the top of the atmosphere (red) is what the atmosphere and the ocean bring in across its rim (orange).*

Applied to the reanalysis, with its global mean of $-12.9\,\mathrm{W}\,\mathrm{m}^{-2}$ subtracted from $R$ at every [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), the proposition gives the curve below. The transport is northward in the northern hemisphere and southward in the southern one, always towards the pole. It peaks at $4.1\,\mathrm{PW}$ near $38{}^{\circ}$ N and $4.5\,\mathrm{PW}$ near $32{}^{\circ}$ S, where $R$ minus its mean changes sign; across $40{}^{\circ}$ N it is $4.11\,\mathrm{PW}$. It is small at the equator, where the surplus of each hemisphere is exported to its own pole.

![The meridional heat transport implied by the reanalysis’s net radiation (black) and its latent-heat part (dashed green, from the evaporation and the precipitation, ); NCEP-NCAR Reanalysis 1, means 1991–2020, data provided by the NOAA PSL. Square: the peak of the atmospheric transport alone in an estimate constrained by satellite measurements of the radiation, 5.0\, PW at 43 N (Trenberth and Caron, 2001).](https://one-course.com/images/onecourse/chapters/earth-3/b1-radiation-heat-transport/fig-8cc70c28dd9c.svg)

*The [meridional heat transport](#def-b1-radiation-heat-transport-mht) implied by the reanalysis’s [net radiation](#def-b1-radiation-heat-transport-net-radiation) (black) and its latent-heat part (dashed green, from the evaporation and the precipitation, [Section 27.4](#sec-27-4)); NCEP-NCAR Reanalysis 1, means 1991–2020, data provided by the NOAA PSL. Square: the peak of the atmospheric transport alone in an estimate constrained by satellite measurements of the radiation, $5.0\,\mathrm{PW}$ at $43{}^{\circ}$ N (Trenberth and Caron, 2001).*

The square on the figure is a warning. A study of the years 1985–1989, when a satellite experiment measured the radiation at the top of the atmosphere reliably, computed the transport of the atmosphere from the winds, temperatures and moisture of two reanalyses: the atmosphere alone carries $5.0\,\mathrm{PW}$ at $43{}^{\circ}$ N, more than this reanalysis’s $4.0\,\mathrm{PW}$ for the atmosphere and the ocean together. The same study found that the model-generated fluxes of the reanalysis agree less well with independent ocean estimates than fluxes derived from the observations. Once the global mean is removed, the transport depends only on how $R$ varies with [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), and this reanalysis’s [net radiation](#def-b1-radiation-heat-transport-net-radiation) varies too little from the tropics to the poles. The shape of the curve, its peaks in the [mid-latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) and its near-symmetry, can be trusted; its height falls at least a fifth short.

Who carries the heat? The same study finds that the ocean carries more than the atmosphere only between the equator and $17{}^{\circ}$ N; at $35{}^{\circ}$, where the total transport peaks in each hemisphere, the atmosphere carries $78\,\%$ of it in the north and $92\,\%$ in the south. The ocean’s share is the subject of [Chapter 29](https://one-course.com/books/earth/3/en/chapter/29-the-circulation-of-the-ocean#ch-b1-ocean-circulation), the atmosphere’s of [Chapter 28](https://one-course.com/books/earth/3/en/chapter/28-the-general-circulation-of-the-atmosphere#ch-b1-atmospheric-circulation).

**Method 27.10 (Transport from zonal bands).**

1. Divide the cap poleward of the [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) of interest into bands; for each, take the mean [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) $\bar Q_i$ , albedo $\alpha_i$ and OLR $_i$ .
2. Compute the [net radiation](#def-b1-radiation-heat-transport-net-radiation) of each band, $R_i = (1 - \alpha_i)\bar Q_i -  \text{OLR}_i$ , and subtract the global mean of $R$ if it is not zero.
3. Compute the area of each band, $A_i = 2\pi a_\oplus^2(\sin\varphi_{i,2} -  \sin\varphi_{i,1})$ , with $a_\oplus = 6371\,\mathrm{km}$ .
4. The transport across the rim is minus the net power of the cap, $-\sum_i R_i A_i$ , positive towards the pole; divide by $10^{15}$ for petawatts.

## 27.4 Latent and sensible heat

The surface of the Earth absorbs most of the sunlight, but the atmosphere radiates most of the infrared to space. The surface hands energy to the air in three ways: by its own infrared radiation, by warming the air in contact with it, and by evaporating water, whose latent heat is released wherever the vapour condenses. That place may be thousands of kilometres away: the latent heat travels with the wind.

**Definition 27.11 (Latent and sensible heat fluxes; Bowen ratio).**

The *latent heat flux* $LE$ is the power per square metre that the surface loses by evaporation: the evaporation rate $E$, in $\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$, times the latent heat of vaporisation $L$. The *sensible heat flux* $H$ is the power per square metre that the surface passes to the air by conduction and turbulence, warming it. Both are counted positive upwards. Their ratio $B = H/LE$ is the *Bowen ratio*.

Near $15\,{}^{\circ}\mathrm{C}$, $L = 2.4655\,\mathrm{MJ}\,\mathrm{kg}^{-1}$. The [Bowen ratio](#def-b1-radiation-heat-transport-surface-fluxes) is small over the sea and over wet ground, where water is plentiful, and large over dry land. In the reanalysis it is $0.09$ at the equator, $0.37$ at $25{}^{\circ}$ N, the [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) of the great hot deserts, and negative at $70{}^{\circ}$ N, where the snow and ice surfaces are colder than the air above them in the annual mean, so that the sensible [heat flows](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#def-b1-heat-budget-heat-flow) downwards. Averaged over the globe, the reanalysis gives $LE =
82.5\,\mathrm{W}\,\mathrm{m}^{-2}$ and $H = 15.2\,\mathrm{W}\,\mathrm{m}^{-2}$, a [Bowen ratio](#def-b1-radiation-heat-transport-surface-fluxes) of $0.18$; the assessed values are $82\,\mathrm{W}\,\mathrm{m}^{-2}$ and $21\,\mathrm{W}\,\mathrm{m}^{-2}$, a ratio of $0.26$.

Book 2 turned the [latent heat flux](#def-b1-radiation-heat-transport-surface-fluxes) into a depth of water evaporated each year. The converse turns a rainfall, which a gauge can collect, into an energy flux.

**Proposition 27.12 (Precipitation as an energy flux).**

A precipitation rate of $1\,\mathrm{m}$ of water per year releases in the atmosphere, where its vapour condensed, a mean power of

$$
\frac{1000\,\mathrm{kg}\,\mathrm{m}^{-2} \times 2.4655 \times 10^{6}\,\mathrm{J}\,\mathrm{kg}^{-1}}{3.156 \times 10^{7}\,\mathrm{s}}
  = 78.1\,\mathrm{W}\,\mathrm{m}^{-2} .
$$

**Proof.** A layer of water $1\,\mathrm{m}$ deep has a mass of $1000\,\mathrm{kg}$ per square metre. Condensing the vapour that forms it releases $1000L$; a year of $365.25$ days lasts $3.156 \times 10^{7}\,\mathrm{s}$. ∎

**Example 27.13 (The water cycle as an energy flux).**

The global water cycle brings $373\,000$ cubic kilometres of rain and snow to the ocean every year and $113\,000$ to the land: spread over the Earth’s $5.10 \times 10^{8}\,\mathrm{km}^{2}$, $0.953\,\mathrm{m}$ per year, which releases $74\,\mathrm{W}\,\mathrm{m}^{-2}$ by [Proposition 27.12](#prop-b1-radiation-heat-transport-rain). The reanalysis rains $2.82\,\mathrm{mm}$ a day, $1.03\,\mathrm{m}$ a year, $80\,\mathrm{W}\,\mathrm{m}^{-2}$. Both are close to the assessed [latent heat flux](#def-b1-radiation-heat-transport-surface-fluxes) of $82\,\mathrm{W}\,\mathrm{m}^{-2}$, whose uncertainty range is $72\text{ to }85\,\mathrm{W}\,\mathrm{m}^{-2}$: over the years, the water that evaporates comes down again, and the latent heat taken from the surface is all given to the air.

The balance holds for the globe, not for each [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), and the difference is a transport (figure below). In the subtropics the sea evaporates more water than falls there: $LE$ exceeds $LP$, the latent heat of the precipitation. The equatorial rain belt and the [mid-latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) receive more rain than they evaporate. Between the two, the vapour is carried by the wind, and with it latent heat. By the argument of [Proposition 27.9](#prop-b1-radiation-heat-transport-mht), applied to the integral of $LE - LP$, the reanalysis carries $1.4\,\mathrm{PW}$ of latent heat northwards across $40{}^{\circ}$ N, a third of its total transport there; the latent transport is drawn dashed on the transport figure above.

![Zonal and annual means of the latent heat flux LE (green), of the latent heat released by the precipitation, LP (blue), and of the sensible heat flux H (red, dashed), from the NCEP-NCAR Reanalysis 1, means 1991–2020 (data provided by the NOAA PSL). Where LE > LP, in the subtropics, water vapour is exported; where LP > LE, under the rain belts, it is imported.](https://one-course.com/images/onecourse/chapters/earth-3/b1-radiation-heat-transport/fig-48b0621dfda2.svg)

*Zonal and annual means of the [latent heat flux](#def-b1-radiation-heat-transport-surface-fluxes) $LE$ (green), of the latent heat released by the precipitation, $LP$ (blue), and of the [sensible heat flux](#def-b1-radiation-heat-transport-surface-fluxes) $H$ (red, dashed), from the NCEP-NCAR Reanalysis 1, means 1991–2020 (data provided by the NOAA PSL). Where $LE > LP$, in the subtropics, water vapour is exported; where $LP > LE$, under the rain belts, it is imported.*

## 27.5 The surface energy budget

The global budget drawn in Book 2 can be read as one equation for the surface.

**Proposition 27.14 (Surface energy balance).**

Let $R_n$ be the [net radiation](#def-b1-radiation-heat-transport-net-radiation) at the surface, absorbed sunlight plus infrared received from the air minus infrared emitted, and $G$ the power per square metre that goes into the ground or the ocean below. Then, per square metre,

$$
R_n = LE + H + G .
$$

In the global mean, $R_n = 160 + 342 - 398 = 104\,\mathrm{W}\,\mathrm{m}^{-2}$, of which $82\,\mathrm{W}\,\mathrm{m}^{-2}$ go into evaporation, $21\,\mathrm{W}\,\mathrm{m}^{-2}$ into the [sensible heat flux](#def-b1-radiation-heat-transport-surface-fluxes), and $G \approx 1\,\mathrm{W}\,\mathrm{m}^{-2}$, within the rounding of the figures, into the ocean and the ground.

**Proof.** Take as a system a thin layer at the surface, with no heat capacity of its own. Energy reaches it only as radiation ($R_n$, net) and leaves it only as the latent and [sensible heat fluxes](#def-b1-radiation-heat-transport-surface-fluxes) upwards and the conduction or mixing downwards; the balance is the equation. The numbers are the assessed global means. ∎

![The global mean surface energy balance in W\, m-2, from the values of the IPCC Sixth Assessment Report, Working Group I, Figure 7.2. The net radiation absorbed by the surface, 104\, W\, m-2, is handed to the atmosphere mostly as latent heat; G, the small remainder, warms the ocean.](https://one-course.com/images/onecourse/chapters/earth-3/b1-radiation-heat-transport/fig-99fa72e82ed4.svg)

*The global mean surface energy balance in $\mathrm{W}\,\mathrm{m}^{-2}$, from the values of the IPCC Sixth Assessment Report, Working Group I, Figure 7.2. The [net radiation](#def-b1-radiation-heat-transport-net-radiation) absorbed by the surface, $104\,\mathrm{W}\,\mathrm{m}^{-2}$, is handed to the atmosphere mostly as latent heat; $G$, the small remainder, warms the ocean.*

The [latent heat flux](#def-b1-radiation-heat-transport-surface-fluxes), four fifths of $R_n$, is the larger of the two non-radiative terms by far: over the globe, the surface cools mainly by evaporating water. The small $G$ is the climate system’s heating, the $0.7\,\mathrm{W}\,\mathrm{m}^{-2}$ of the budget at the top of the atmosphere; nine tenths of it is stored in the ocean.

**Remark 27.15 (Local and global).**

At a single place, $G$ is far from small: it is large and positive by day and in summer, negative by night and in winter, and it averages out only over a day or a year. Over the ocean, the heat stored in the [mixed layer](https://one-course.com/books/earth/3/en/chapter/26-structure-of-the-atmosphere-and-the-ocean#def-b1-fluid-envelopes-structure-ocean-layers) can also be carried away sideways by the currents: the local balance must then include the ocean’s transport, as the balance of a [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) band includes the atmosphere’s.

## 27.6 Exercises

**Exercise 27.1 ★.**

(a) Compute $S_0/\pi$ with $S_0 = 1361\,\mathrm{W}\,\mathrm{m}^{-2}$. (b) On the day of the March equinox, the [daily-mean insolation](#def-b1-radiation-heat-transport-daily-insolation) at the equator is $436.7\,\mathrm{W}\,\mathrm{m}^{-2}$. Deduce $(a/r)^2$ and $r/a$ on that day. (c) At the June solstice, $(a/r)^2 = 0.9683$ and $\delta = 23.44{}^{\circ}$. Compute the [daily-mean insolation](#def-b1-radiation-heat-transport-daily-insolation) at the equator.

**Solution of Exercise 27.1.**

(a) $S_0/\pi = 1361/\pi = 433.2\,\mathrm{W}\,\mathrm{m}^{-2}$. (b) At the equator on an equinox $\bar Q = (S_0/\pi)(a/r)^2$, so $(a/r)^2 = 436.7/433.2 = 1.0080$ and $r/a =
1.0080^{-1/2} = 0.9960$: the Earth is $0.4\,\%$ closer to the Sun than its mean distance. (c) $\bar Q = 433.2 \times 0.9683 \times \cos23.44{}^{\circ} =
384.9\,\mathrm{W}\,\mathrm{m}^{-2}$.

**Exercise 27.2 ★.**

The annual-mean [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) is $415.6\,\mathrm{W}\,\mathrm{m}^{-2}$ at the equator and $172.4\,\mathrm{W}\,\mathrm{m}^{-2}$ at the poles; the global mean is $340.3\,\mathrm{W}\,\mathrm{m}^{-2}$. (a) By what factor does the equator exceed the poles? (b) The annual mean equals the global mean at $37.0{}^{\circ}$ N and S. What fraction of the Earth’s surface lies between these two parallels? (c) Why does this make the annual-mean curve look more favourable to the high [latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) than the Earth’s surface really is?

**Solution of Exercise 27.2.**

(a) $415.6/172.4 = 2.41$. (b) The zone between the parallels $\pm\varphi$ has the area $2\pi a_\oplus^2 \times 2\sin\varphi$, a fraction $\sin\varphi$ of the sphere: $\sin37.0{}^{\circ} = 0.602$, $60\,\%$. (c) On an axis in [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), the bands near the poles look as wide as those near the equator, but they cover much less area: the zone poleward of $60{}^{\circ}$ is only $1 - \sin60{}^{\circ} =
13.4\,\%$ of the surface. Against $\sin\varphi$, the tropics fill most of the axis.

**Exercise 27.3 ★.**

In the reanalysis the OLR is $244.9\,\mathrm{W}\,\mathrm{m}^{-2}$ near the equator, $192.3\,\mathrm{W}\,\mathrm{m}^{-2}$ at $81{}^{\circ}$ N and $237.7\,\mathrm{W}\,\mathrm{m}^{-2}$ in the global mean. (a) Compute the three [emission temperatures](#def-b1-radiation-heat-transport-olr). (b) Over the same [latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), the absorbed sunlight falls from $287.7\,\mathrm{W}\,\mathrm{m}^{-2}$ to $75.8\,\mathrm{W}\,\mathrm{m}^{-2}$. Compare the two ratios. (c) What does the comparison imply for the [net radiation](#def-b1-radiation-heat-transport-net-radiation)?

**Solution of Exercise 27.3.**

(a) $T_e = (\text{OLR}/\sigma)^{1/4}$: $256.4\,\mathrm{K}$, $241.3\,\mathrm{K}$ and $254.5\,\mathrm{K}$. (b) Absorbed sunlight: $287.7/75.8 = 3.80$; OLR: $244.9/192.3 =
1.27$. (c) The [net radiation](#def-b1-radiation-heat-transport-net-radiation) follows the absorbed sunlight: positive and large in the tropics, strongly negative at high [latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic). The OLR, which varies little, does not offset the fall of the absorbed sunlight.

**Exercise 27.4 ★.**

The reanalysis’s global mean precipitation is $2.82\,\mathrm{mm}$ a day. (a) Convert it to metres per year. (b) Compute the latent heat it releases, in $\mathrm{W}\,\mathrm{m}^{-2}$, with [Proposition 27.12](#prop-b1-radiation-heat-transport-rain). (c) Compare with the reanalysis’s own global [latent heat flux](#def-b1-radiation-heat-transport-surface-fluxes), $82.5\,\mathrm{W}\,\mathrm{m}^{-2}$, and explain why the two should agree over many years.

**Solution of Exercise 27.4.**

(a) $2.82 \times 365.25 = 1030\,\mathrm{mm}$, $1.03\,\mathrm{m}$ a year. (b) $1.030 \times
78.1 = 80.4\,\mathrm{W}\,\mathrm{m}^{-2}$. (c) The reanalysis’s $LE$ is $2\,\mathrm{W}\,\mathrm{m}^{-2}$ larger. Over many years the atmosphere cannot keep accumulating water vapour: all the water evaporated comes down again, so $LE$ and $LP$ must be equal in the global mean. The difference is an imbalance of the model, like its radiative imbalance.

**Exercise 27.5 ★★.**

(a) At the June solstice ($\delta = 23.44{}^{\circ}$, $(a/r)^2 = 0.9683$), compute the [daily-mean insolation](#def-b1-radiation-heat-transport-daily-insolation) at the North Pole. (b) At the December solstice, $(a/r)^2 =
1.0334$; compute it at the South Pole and the ratio of the two. (c) The Sun stands only $23.4{}^{\circ}$ above the horizon at the pole: explain how the pole can still beat the equator.

**Solution of Exercise 27.5.**

(a) $\bar Q = S_0(a/r)^2\sin\delta = 1361 \times 0.9683 \times 0.3978 =
524.2\,\mathrm{W}\,\mathrm{m}^{-2}$. (b) $1361 \times 1.0334 \times 0.3978 =
559.5\,\mathrm{W}\,\mathrm{m}^{-2}$; the ratio is $1.0334/0.9683 = 1.067$. (c) The pole has sunlight 24 hours out of 24 at a constant [altitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-horizontal), so its mean $\sin h_\odot$ over the day is $\sin\delta = 0.398$. At the equator the Sun is high at noon but down half the time; the mean of $\sin h_\odot$ over the 24 hours is $\cos\delta/\pi =
0.292$. The ratio is $\pi\tan\delta = 1.362$ ([Proposition 27.3](#prop-b1-radiation-heat-transport-polar-summer)).

**Exercise 27.6 ★★.**

At $40{}^{\circ}$ N, compute the [hour angle](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) of sunset, the length of the day and the [daily-mean insolation](#def-b1-radiation-heat-transport-daily-insolation) (a) at the June solstice, with $(a/r)^2 = 0.9683$, $\delta =
23.44{}^{\circ}$; (b) at the December solstice, with $(a/r)^2 = 1.0334$, $\delta =
-23.44{}^{\circ}$. (c) What share of the ratio between the two comes from the distance to the Sun?

**Solution of Exercise 27.6.**

(a) $\cos h_0 = -\tan40{}^{\circ}\tan23.44{}^{\circ} = -0.3638$, $h_0 = 111.33{}^{\circ} =
1.943\,\mathrm{rad}$; the day lasts $24 \times 111.33/180 = 14.84\,\mathrm{h}$; $\bar Q =
(1361/\pi) \times 0.9683 \times (1.943 \times 0.6428 \times 0.3978 + 0.7660 \times
0.9175 \times 0.9315) = 483.0\,\mathrm{W}\,\mathrm{m}^{-2}$. (b) $\cos h_0 = +0.3638$, $h_0 =
68.67{}^{\circ}$, $9.16\,\mathrm{h}$; $\bar Q = 155.9\,\mathrm{W}\,\mathrm{m}^{-2}$. (c) The ratio is $483.0/155.9 = 3.10$. The distance contributes $0.9683/1.0334 = 0.937$: it works against the northern summer. The geometry, the [altitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-horizontal) of the Sun and the length of the day, contributes $3.10/0.937 = 3.31$, all of the seasonal contrast and more.

**Exercise 27.7 ★★.**

The reanalysis gives the following zonal means ($\mathrm{W}\,\mathrm{m}^{-2}$):

| [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) | $0{}^{\circ}$ | $25{}^{\circ}$ N | $45{}^{\circ}$ N | $70{}^{\circ}$ N |
| --- | --- | --- | --- | --- |
| $LE$ | 96.0 | 96.4 | 52.4 | 23.7 |
| $H$ | 8.9 | 36.1 | 17.7 | $-3.6$ |

(a) Compute the [Bowen ratio](#def-b1-radiation-heat-transport-surface-fluxes) at each [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic). (b) Which [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) has the largest ratio? What kind of surface explains it? (c) Interpret the sign of $H$ at $70{}^{\circ}$ N.

**Solution of Exercise 27.7.**

(a) $B = H/LE$: $0.093$ at the equator, $0.374$ at $25{}^{\circ}$ N, $0.338$ at $45{}^{\circ}$ N, $-0.152$ at $70{}^{\circ}$ N. (b) $25{}^{\circ}$ N: the zonal mean includes the great hot deserts, dry land with little water to evaporate, where the heated ground warms the air instead. (c) $H < 0$ means that [heat flows](https://one-course.com/books/earth/3/en/chapter/10-the-earths-heat-budget#def-b1-heat-budget-heat-flow) from the air to the surface: in the annual mean the snow and sea ice are colder than the air above them, which is warmed by air brought from lower [latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic).

**Exercise 27.8 ★★.**

The assessed global budget at the top of the atmosphere is $340\,\mathrm{W}\,\mathrm{m}^{-2}$ incoming, $100\,\mathrm{W}\,\mathrm{m}^{-2}$ reflected, $239\,\mathrm{W}\,\mathrm{m}^{-2}$ emitted; the reanalysis gives $341.9\,$, $117.0\,$ and $237.7\,\mathrm{W}\,\mathrm{m}^{-2}$. (a) Compute the two albedos and the two [net radiations](#def-b1-radiation-heat-transport-net-radiation). (b) Which term explains most of the reanalysis’s imbalance? (c) Show that if the global mean of $R$ were not removed, [Proposition 27.9](#prop-b1-radiation-heat-transport-mht) would leave a transport of $-6.6\,\mathrm{PW}$ at the North Pole, and explain why that is impossible.

**Solution of Exercise 27.8.**

(a) Assessed: $\alpha = 100/340 = 0.294$, $R = 340 - 100 - 239 =
+1\,\mathrm{W}\,\mathrm{m}^{-2}$ (the $0.7$ rounded). Reanalysis: $\alpha = 117.0/341.9 =
0.342$, $R = 341.9 - 117.0 - 237.7 = -12.8\,\mathrm{W}\,\mathrm{m}^{-2}$ with these rounded terms, $-12.9\,\mathrm{W}\,\mathrm{m}^{-2}$ with the unrounded ones of [Example 27.7](#ex-b1-radiation-heat-transport-global). (b) The reanalysis absorbs $224.8$ against $240$, $15.2\,\mathrm{W}\,\mathrm{m}^{-2}$ less, because it reflects $17\,\mathrm{W}\,\mathrm{m}^{-2}$ more; its OLR is only $1.3\,\mathrm{W}\,\mathrm{m}^{-2}$ lower. The excess reflection explains the imbalance. (c) Integrating $R$ from the South Pole to the North Pole gives the global integral, $-12.9 \times 4\pi a_\oplus^2 = -12.9 \times
5.10\times10^{14} = -6.6\,\mathrm{PW}$. A transport across the North Pole is impossible: there is nothing beyond it, the parallel is a point. The proposition holds only if $\int R\,\mathrm dA = 0$, which the steady state requires.

**Exercise 27.9 ★★.**

In the reanalysis, the belt between $30{}^{\circ}$ S and $30{}^{\circ}$ N has a mean [net radiation](#def-b1-radiation-heat-transport-net-radiation), after removal of the global mean, of $+33\,\mathrm{W}\,\mathrm{m}^{-2}$. (a) Compute the area of the belt and its total surplus in petawatts. (b) If the two hemispheres exported the surplus equally, how much would cross each of the parallels $30{}^{\circ}$ N and $30{}^{\circ}$ S? (c) The reanalysis gives $4.00\,\mathrm{PW}$ northwards across $30{}^{\circ}$ N and $4.50\,\mathrm{PW}$ southwards across $30{}^{\circ}$ S. What does the difference say?

**Solution of Exercise 27.9.**

(a) The area is $2\pi a_\oplus^2 \times 2\sin30{}^{\circ} = 2\pi a_\oplus^2 =
2.55 \times 10^{14}\,\mathrm{m}^{2}$, half the Earth; the surplus is $33 \times 2.55\times10^{14} =
8.4\,\mathrm{PW}$. (b) $4.2\,\mathrm{PW}$ across each parallel. (c) The two add up to $8.5\,\mathrm{PW}$, the surplus, but they are unequal: the southern hemisphere exports $0.5\,\mathrm{PW}$ more than the northern one. The surplus is not shared at the equator: in this reanalysis $0.36\,\mathrm{PW}$ of the northern tropics’ surplus crosses the equator southwards.

**Exercise 27.10 ★★★.**

The [daily-mean insolation](#def-b1-radiation-heat-transport-daily-insolation) depends on the [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), the [declination](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) and the factor $(a/r)^2$. (a) With Kepler’s second law, $r^2\,\mathrm d\lambda/\mathrm dt = h$ constant, where $\lambda$ is the Sun’s [ecliptic](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-ecliptic) [longitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), show that the energy a [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) receives while $\lambda$ advances by $\mathrm d\lambda$ does not depend on $r$. (b) Deduce that the annual-mean [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) at [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) $\varphi$ equals that at $-\varphi$, although the southern summer is brighter. (c) In the orbit model of this chapter, the Sun takes $186.4$ days from the March to the September equinox and $178.8$ days back. Explain how these numbers fit (b).

**Solution of Exercise 27.10.**

(a) Write the daily mean as $\bar Q = (a/r)^2 F(\varphi, \delta)$, with $F$ the bracket of [Theorem 27.2](#thm-b1-radiation-heat-transport-daily) times $S_0/\pi$. During $\mathrm dt$ the energy received per square metre is $\bar Q\,\mathrm dt$, and $\mathrm dt = r^2\,\mathrm d\lambda/h$, so $\bar Q\,\mathrm dt = (a^2/h)
F(\varphi, \delta(\lambda))\,\mathrm d\lambda$: the factor $r^2$ cancels. (b) Over a year the energy is $(a^2/h)\int_0^{2\pi} F(\varphi,
\delta(\lambda))\,\mathrm d\lambda$. The formula is unchanged when $\varphi$ and $\delta$ change sign together, $F(-\varphi, \delta) = F(\varphi, -\delta)$, and $\delta(\lambda + \pi) = -\delta(\lambda)$, since $\sin\delta = \sin\varepsilon
\sin\lambda$. Shifting $\lambda$ by $\pi$ in the integral shows that $-\varphi$ receives the same energy as $\varphi$. (c) The northern spring and summer last $7.6$ days longer, because the Earth is near aphelion and moves slowly then; the southern summer is brighter but shorter. By (a), the two effects cancel exactly.

**Exercise 27.11 ★★★.**

(a) Using Kepler’s second law in the form $r^2\,\mathrm d\lambda/\mathrm dt = h$ and the area of the ellipse, $\pi a b = hP/2$ over a period $P$, show that the time average of $(a/r)^2$ over an orbit is $a/b = 1/\sqrt{1 - e^2}$. (b) Complete the proof of [Proposition 27.4](#prop-b1-radiation-heat-transport-global). (c) With $e = 0.0167$, compute the global annual mean and its difference from $S_0/4$.

**Solution of Exercise 27.11.**

(a) $\langle (a/r)^2\rangle = \frac1P\int_0^P \frac{a^2}{r^2}\,\mathrm dt =
\frac1P\int_0^{2\pi}\frac{a^2}{r^2}\frac{r^2}{h}\,\mathrm d\lambda = \frac{2\pi
a^2}{hP} = \frac{2\pi a^2}{2\pi a b} = \frac ab$, and $b = a\sqrt{1-e^2}$. (b) The instantaneous global mean $\frac14 S_0(a/r)^2$ averages to $\frac14 S_0
\langle(a/r)^2\rangle = S_0/(4\sqrt{1-e^2})$. (c) $340.25/\sqrt{1 - 0.0167^2} =
340.30\,\mathrm{W}\,\mathrm{m}^{-2}$, $0.05\,\mathrm{W}\,\mathrm{m}^{-2}$ or $0.014\,\%$ above $S_0/4$.

**Exercise 27.12 ★★★.**

The reanalysis gives a total transport of $4.10\,\mathrm{PW}$ across $35{}^{\circ}$ N and $4.02\,\mathrm{PW}$ across $43{}^{\circ}$ N. (a) Using the partition of Trenberth and Caron at $35{}^{\circ}$ N (atmosphere $78\,\%$), split the reanalysis’s total into an atmospheric and an oceanic part. (b) The same study finds $5.0\,\mathrm{PW}$ for the atmosphere alone at $43{}^{\circ}$ N. What does this say about the reanalysis? (c) Removing the global mean of $R$ does not cure this error. Why not?

**Solution of Exercise 27.12.**

(a) Atmosphere $0.78 \times 4.10 = 3.20\,\mathrm{PW}$, ocean $0.22 \times 4.10 =
0.90\,\mathrm{PW}$. (b) The atmosphere alone is found to carry more, $5.0\,\mathrm{PW}$, than the reanalysis’s atmosphere and ocean together, $4.02\,\mathrm{PW}$: the reanalysis’s total transport is short by at least a fifth ($1 - 4.02/5.0 = 0.20$). Its [net radiation](#def-b1-radiation-heat-transport-net-radiation) varies too little between the tropics and the poles. (c) Removing a constant from $R$ makes its global integral zero but leaves its variation with [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) unchanged; the error lies in that variation, in the albedo of the model’s clouds and surfaces at each [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic), not only in the global mean.

## 27.7 Problem: How much heat crosses the 40th parallel?

**Problem 27.1.**

Weekend problem — the energy budget of the cap north of $40{}^{\circ}$ N

The cap of the Earth north of $40{}^{\circ}$ N, which holds most of Europe, Russia and Canada, radiates more energy to space than it absorbs from the Sun. This problem measures the deficit, and so the heat that the atmosphere and the ocean bring across the 40th parallel. Use $S_0 = 1361\,\mathrm{W}\,\mathrm{m}^{-2}$, $a_\oplus = 6371\,\mathrm{km}$, $\sigma = 5.670 \times 10^{-8}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{K}^{-4}$, $L =
2.4655 \times 10^{6}\,\mathrm{J}\,\mathrm{kg}^{-1}$. The band data are the reanalysis’s means for 1991–2020.

**Part I — Sunlight north of $40{}^{\circ}$ N.** On the June solstice $\delta = 23.44{}^{\circ}$ and $(a/r)^2 = 0.9683$; on the December solstice $\delta =
-23.44{}^{\circ}$ and $(a/r)^2 = 1.0334$; on the March equinox $\delta = 0$ and $(a/r)^2 = 1.0081$.

1. Compute the [daily-mean insolation](#def-b1-radiation-heat-transport-daily-insolation) at $40{}^{\circ}$ N on the three days.
2. Compute it at $80{}^{\circ}$ N on the same three days.
3. In June, $80{}^{\circ}$ N receives more than $40{}^{\circ}$ N. Explain why, with the [hour angles](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-equatorial) of sunset.
4. By integrating the theorem over the year, one finds annual means of $306.9\,$ , $260.3\,$ , $213.7\,$ , $185.6\,$ and $173.8\,\mathrm{W}\,\mathrm{m}^{-2}$ at $45{}^{\circ}$ , $55{}^{\circ}$ , $65{}^{\circ}$ , $75{}^{\circ}$ and $85{}^{\circ}$ N. Compare with the band [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) of the table in Part II and give two reasons for the differences.
5. The reanalysis’s global mean [insolation](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-insolation) is $341.9\,\mathrm{W}\,\mathrm{m}^{-2}$ . What value of $S_0$ does it imply?
6. Compute the area of the cap north of $40{}^{\circ}$ N and the fraction of the Earth’s surface it covers.

**Part II — Absorbed and emitted.** The reanalysis gives, for five bands of the cap:

| band | $40{}^{\circ}$–$50{}^{\circ}$ | $50{}^{\circ}$–$60{}^{\circ}$ | $60{}^{\circ}$–$70{}^{\circ}$ | $70{}^{\circ}$–$80{}^{\circ}$ | $80{}^{\circ}$–$90{}^{\circ}$ |
| --- | --- | --- | --- | --- | --- |
| $\bar Q$ ($\mathrm{W}\,\mathrm{m}^{-2}$) | 308.4 | 262.4 | 217.3 | 188.7 | 176.1 |
| albedo $\alpha$ | 0.358 | 0.393 | 0.426 | 0.510 | 0.576 |
| OLR ($\mathrm{W}\,\mathrm{m}^{-2}$) | 226.1 | 210.3 | 201.3 | 197.4 | 191.6 |

7. Compute the absorbed solar radiation of each band.
8. Compute the [net radiation](#def-b1-radiation-heat-transport-net-radiation) of each band. Which band has the largest deficit?
9. Compute the [emission temperature](#def-b1-radiation-heat-transport-olr) of each band.
10. From $40{}^{\circ}$ – $50{}^{\circ}$ to $80{}^{\circ}$ – $90{}^{\circ}$ , by what factor does the absorbed sunlight fall, and by what factor the OLR?
11. Give two reasons why the albedo increases towards the pole.
12. Why does the OLR vary so much less than the absorbed sunlight?

**Part III — The net radiation of the cap.**

13. Compute the area of each band and check that they add up to the area of Question 6.
14. Compute the net radiative power of each band and of the whole cap, in petawatts.
15. Compute the mean albedo of the cap, weighting each band by its incoming sunlight, and compare it with the reanalysis’s global albedo of $0.342$ .
16. The reanalysis’s global mean [net radiation](#def-b1-radiation-heat-transport-net-radiation) is $-12.90\,\mathrm{W}\,\mathrm{m}^{-2}$ . What would this mean if it were real? Compare with the assessed imbalance of $+0.7\,\mathrm{W}\,\mathrm{m}^{-2}$ .
17. Subtract this global mean from the [net radiation](#def-b1-radiation-heat-transport-net-radiation) of each band and compute again the net power of the cap.
18. Explain, with [Proposition 27.9](#prop-b1-radiation-heat-transport-mht) , why the corrected deficit of the cap must be brought in across the 40th parallel.

**Part IV — The transport and its latent share.** Over the cap the reanalysis gives a mean [latent heat flux](#def-b1-radiation-heat-transport-surface-fluxes) of $42.4\,\mathrm{W}\,\mathrm{m}^{-2}$ and a mean precipitation of $1.95\,\mathrm{mm}$ a day; over the globe, the mean of $LE - LP$ is $2.07\,\mathrm{W}\,\mathrm{m}^{-2}$.

19. Compute the transport across $40{}^{\circ}$ N, in petawatts.
20. Compute the latent heat $LP$ released by the cap’s precipitation, in $\mathrm{W}\,\mathrm{m}^{-2}$ , and compare it with $LE$ .
21. Removing the global mean of $LE - LP$ , compute the latent heat brought into the cap as water vapour, in petawatts.
22. What fraction of the transport across $40{}^{\circ}$ N is latent heat?
23. Taking the share of the atmosphere found by Trenberth and Caron at $35{}^{\circ}$ N, $78\,\%$ , as valid at $40{}^{\circ}$ N, split the transport into latent heat, the rest of the atmospheric transport, and the ocean.
24. The same study finds $5.0\,\mathrm{PW}$ for the atmosphere alone at $43{}^{\circ}$ N. Is your answer an upper or a lower bound of the truth?
25. How much heat crosses the 40th parallel north, in petawatts?

**Solution of Problem 27.1.**

1. June: $\cos h_0 = -0.3638$ , $h_0 = 111.33{}^{\circ}$ , $\bar Q =  483.0\,\mathrm{W}\,\mathrm{m}^{-2}$ . December: $h_0 = 68.67{}^{\circ}$ , $\bar Q =  155.9\,\mathrm{W}\,\mathrm{m}^{-2}$ . March: $h_0 = 90{}^{\circ}$ , $\bar Q = (1361/\pi)  \times 1.0081 \times \cos40{}^{\circ} = 334.6\,\mathrm{W}\,\mathrm{m}^{-2}$ .
2. June: $\tan80{}^{\circ}\tan23.44{}^{\circ} = 2.46 > 1$ , polar day, $h_0 = \pi$ : $\bar Q = 1361 \times 0.9683 \times \sin80{}^{\circ}\sin23.44{}^{\circ} =  516.3\,\mathrm{W}\,\mathrm{m}^{-2}$ . December: polar night, $\bar Q = 0$ . March: $(1361/\pi) \times 1.0081 \times \cos80{}^{\circ} = 75.8\,\mathrm{W}\,\mathrm{m}^{-2}$ .
3. At $80{}^{\circ}$ N the Sun does not set ( $h_0 = 180{}^{\circ}$ , 24 hours) while at $40{}^{\circ}$ N it shines for $14.8\,\mathrm{h}$ . Its [altitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-horizontal) is lower, between $13.4{}^{\circ}$ and $33.4{}^{\circ}$ , but the term $h_0\sin\varphi\sin\delta$ , which grows with the day length and the [latitude](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) , wins.
4. The band values exceed the central values by $0.5\,\%$ , $0.8\,\%$ , $1.7\,\%$ , $1.7\,\%$ and $1.3\,\%$ . First, the reanalysis uses a slightly larger solar irradiance (Question 5), $0.5\,\%$ more. Second, a band mean is not the value at the centre: the equatorward half of each band has more area and more sunlight than the poleward half.
5. By [Proposition 27.4](#prop-b1-radiation-heat-transport-global) , $S_0 = 4\sqrt{1-e^2}  \times 341.9 = 1367\,\mathrm{W}\,\mathrm{m}^{-2}$ , $0.47\,\%$ more than $1361\,\mathrm{W}\,\mathrm{m}^{-2}$ .
6. $A = 2\pi a_\oplus^2(1 - \sin40{}^{\circ}) = 2\pi \times (6.371\times10^6)^2  \times 0.3572 = 9.110 \times 10^{13}\,\mathrm{m}^{2}$ , a fraction $(1 - \sin40{}^{\circ})/2 =  0.179$ of the Earth.
7. $(1 - \alpha)\bar Q$ : $198.0$ , $159.3$ , $124.7$ , $92.5$ , $74.7$ $\mathrm{W}\,\mathrm{m}^{-2}$ .
8. $R$ : $-28.1$ , $-51.0$ , $-76.6$ , $-104.9$ , $-116.9$ $\mathrm{W}\,\mathrm{m}^{-2}$ . All five bands are in deficit; the polar band $80{}^{\circ}$ – $90{}^{\circ}$ has the largest.
9. $T_e = (\text{OLR}/\sigma)^{1/4}$ : $251.3$ , $246.8$ , $244.1$ , $242.9$ , $241.1$ $\mathrm{K}$ .
10. Absorbed: $198.0/74.7 = 2.65$ ; OLR: $226.1/191.6 = 1.18$ .
11. The Sun is lower, and light arriving at a grazing angle is reflected more by clouds and water; snow and sea ice, bright surfaces, cover more of the ground and the sea towards the pole.
12. The OLR goes as $T_e^4$ , and $T_e$ varies by only $10\,\mathrm{K}$ , $4\,\%$ , across the cap: the high [latitudes](https://one-course.com/books/earth/3/en/chapter/1-spherical-astronomy#def-b1-spherical-astronomy-geographic) are kept much warmer than their sunlight alone would allow, by the heat brought from the south.
13. In $\mathrm{m}^{2}$ : $3.143 \times 10^{13}$ , $2.550 \times 10^{13}$ , $1.879 \times 10^{13}$ , $1.151 \times 10^{13}$ , $0.387 \times 10^{13}$ ; the sum is $9.110 \times 10^{13}\,\mathrm{m}^{2}$ .
14. $R_iA_i$ : $-0.883$ , $-1.300$ , $-1.439$ , $-1.207$ , $-0.453$ $\mathrm{PW}$ ; the cap: $-5.28\,\mathrm{PW}$ .
15. $\alpha_\text{cap} = \sum\alpha_i\bar Q_iA_i/\sum\bar Q_iA_i = 0.400$ , against $0.342$ for the globe.
16. $-12.90 \times 5.10\times10^{14} = -6.58\,\mathrm{PW}$ : the Earth would be losing heat at that rate and cooling. The assessed imbalance is $+0.7 \times 5.10\times10^{14} = +0.36\,\mathrm{PW}$ , eighteen times smaller and of the opposite sign: the reanalysis’s imbalance is a model error.
17. Corrected $R_i + 12.90$ : $-15.2$ , $-38.1$ , $-63.7$ , $-92.0$ , $-104.0$ $\mathrm{W}\,\mathrm{m}^{-2}$ ; powers $-0.478$ , $-0.972$ , $-1.197$ , $-1.059$ , $-0.403$ $\mathrm{PW}$ ; the cap: $-4.108\,\mathrm{PW}$ , a mean deficit of $45.1\,\mathrm{W}\,\mathrm{m}^{-2}$ .
18. The cap’s energy does not change from year to year, so what it loses at the top of the atmosphere must enter it, and the only way in is across the 40th parallel: $T(40{}^{\circ}) = -2\pi a_\oplus^2\int_{40{}^{\circ}}^{90{}^{\circ}}  R\cos\varphi\,\mathrm d\varphi$ .
19. $T(40{}^{\circ}\,\text{N}) = +4.11\,\mathrm{PW}$ , northwards.
20. $LP = 2.4655\times10^6 \times 1.95/86400 = 55.6\,\mathrm{W}\,\mathrm{m}^{-2}$ , more than $LE = 42.4\,\mathrm{W}\,\mathrm{m}^{-2}$ : more water falls on the cap than evaporates from it; the difference arrives as vapour.
21. $(LE - LP) - 2.07 = 42.4 - 55.6 - 2.07 = -15.3\,\mathrm{W}\,\mathrm{m}^{-2}$ ; times the area of the cap: $-1.40\,\mathrm{PW}$ . The atmosphere brings $1.40\,\mathrm{PW}$ of latent heat across $40{}^{\circ}$ N.
22. $1.40/4.11 = 0.34$ : a third.
23. Ocean $0.22 \times 4.11 = 0.90\,\mathrm{PW}$ ; atmosphere $0.78 \times 4.11 =  3.20\,\mathrm{PW}$ , of which $1.40\,\mathrm{PW}$ latent and $1.80\,\mathrm{PW}$ carried as sensible heat and potential energy of the air.
24. A lower bound. The reanalysis’s total at $43{}^{\circ}$ N, $4.02\,\mathrm{PW}$ , is less than the atmosphere alone in the study, $5.0\,\mathrm{PW}$ ; the true total is larger.
25. About $4.11\,\mathrm{PW}$ , $4.11 \times 10^{15}\,\mathrm{W}$ , according to the reanalysis’s radiation, a third of it as latent heat; the true value is larger still.