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Diurnal Temperature Profiles

The Dynamics of the Temperature of a Physical Body

The temperature of a body irradiated with light increases as a result of the absorption of the energy in that light. A heated body radiates energy in an amount that is proportional to the fourth power of its absolute temperature. This means that the dynamics of the temperature of the body is given by an equation of the form

C(dT/dt) = aF(t) − cT4

where T is the absolute temperature, C is the heat capacity of the body and F(t) is the intensity of the radiation impinging upon the body, a cyclical function. The parameters a and c represent such factors as the reflection of the incoming radiation (a) and capture of the outgoing thermal radiation (c).

A mean level of temperature is established that balances the incoming and outgoing energy flows over a cycle. Let Fmean be the average energy inflow over a cycle. Then a mean temperature Tmean is such that

0 = aFmean − cT4mean

Tmean will be close to but not necessarily equal to the actual mean temperature; it is the fourth root of the mean of the fourth power of the temperature.

The last equation can be subtracted from the next to the last equation to give

C(dΔT/dt) = af(t) − c[T4 − T4mean]

where f(t)=F(t)−Fmean and ΔT = [T(t)−Tmean].

The quantity [T4 − T4mean] can be approximated by the quantity 4T3meanΔT. The dynamics of temperature is then given by

dΔT/dt = af(t) − bΔT

where b=c(4T3mean).

A pertinent form for F(t) is where F(t)=0 for t=0 to 6 (hours) and for t=18 to t=24 and F(t)=cos((t-12)/K) for t=6 to t=18, where K=12/π. This is where the Sun rises at 6 AM (0600) and sets at 6 PM (1800). Then for f(t)=(F(t)−Fmean), the shapes of F(t) and f(t) are shown below.

A solution of the equation for ΔT can be obtained numerically for any specific values of a and b, such as the ones shown below for b=0.1 and b=0.2. The value of a is 1.0 for both curves. For computational convenience the temperatures are deviations from the midnight temperatures. The midnight temperatures are different for the two values of b.

In order to obtain a strictly cyclic solution a quantity had to be deducted from F(t) that was a bit larger than the mean value of F(t). The units of temperature are just arbitrary.

The peak energy input is at noon (t=12). For the value of b=0.2 the peak temperature occurs at t=15.5 (3:30 PM). For b=0.2 the lowest temperature occurs at 6:30 AM. For b=0.1 the peak temperature is at t=16.0 (4:00 PM). The minimum temperature is at t=7.0 (7 AM).

After nightfall at 6 PM the temperature falls as a negative exponential function of time, T(6PM)e-bt. It continues to fall even after sunrise at 6 AM because initially the energy input from the Sun is less than the radiated outflow of energy. The low point for the temperature comes at the time when the input from the Sun exactly balances the radiated outflow. Likewise the temperature continues to rise in the afternoon until the diminished inflow falls to the level of the radiated outflow.

The larger the value of b the sharper the decline in temperature at night and hence the lower the minimum temperature.

An Analytical Solution to
the Ordinary Differential Equation
for the Temperature

Consider the equation for the diurnal dynamics of temperature

dΔT/dt = af(t) − bΔT

First the equation should be put in the form

dΔT/dt + bΔT = af(t)
and then multiplied by ebt to give
ebt(dΔT/dt) + bebtΔT = d/dt(ebtΔT) = aebtf(t)

Integrating this last equation from 0 to t gives:

ebtΔT(t) − ΔT(0) = a∫0tebsf(s)ds

Solving for ΔT(t) gives:

ΔT(t) = e-btΔT(0) + ae-bt0tebsf(s)ds
which is equivalent to
ΔT(t) = e-btΔT(0) + a∫0te-b(t-s)f(s)ds

The variable of integration may be changed from s to z where z=t-s and hence s=t-z and ds=-dz. The result is

ΔT(t) = e-btΔT(0) + a∫t0e-bzf(t-z)(-dz)
which reduces to
ΔT(t) = e-btΔT(0) + a∫0te-bzf(t-z)dz

If f(t) is cyclical with period Φ then ΔT(t) will also be cyclical with period Φ. This means that

ΔT(Φ) = e-bΦΔT(0) + a∫0Φe-bzf(Φ-z)dz = ΔT(0)
and hence
(1−e-bΦ)ΔT(0) = a ∫0Φe-bzf(Φ-z)dz
or, equivalently
ΔT(0) = a [∫0Φe-bzf(Φ-z)dz/(1−e-bΦ)]

Let the above ratio within the brackets be denoted as S. Then ΔT(0)=aS and the general solution for ΔT(t) is

ΔT(t) = a e-btS + a∫0te-bzf(t-z)dz

The coefficient a may be factored out to give

ΔT(t) = a[e-btS + ∫0te-bzf(t-z)dz]

The coefficient a is merely a scale variable that determines the amplitude of the cycles. The details of the cycle are determined by b and f(t).

(To be continued.)

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