Wednesday, June 23, 2010

Venus temperatures and the adiabatic pump.

Science of Doom discussed recent theorizing on the causes of high temperatures on Venus, as did Chris Colose. On both blogs, Leonard Weinstein proposed a thought experiment in which a shell was placed over Venus - he believed that the high surface temperatures would remain. SoD took up this idea here, and there was further discussion. This post sets out my ideas on the problem, with some background.

It follows my earlier post on Venus and lapse rates, and a post from a while ago on the adiabatic heat pump.


Heat transfer mechanisms

This may seem elementary, but in the atmosphere, some aspects of heat transfer are different:
Conduction
Molecular conduction carries very minor flux, in accordance with Fourier's Law.
Radiation
Transmission of thermal range IR in an atmosphere like Venus is not like transmission in a vacuum. At most wavelengths the absorption length L is fairly small relative to the depth of the atmosphere. However, heat balance requires that absorption is balanced by emission. This gas-to-gas transmission is very dependent on the temperature gradient.

The Rosseland model of IR transmission works for high opacity gas. If you simplify by neglecting scattering, then
F/F0 = (16/3)*(L/*T0) ∇ T
where F0 is BB (black body) emission at ambient T0, F is flux.
This is a Fourier Law with a conductivity that can be quite high. For example, if L is 1 km, T0=400K, the flux is about 0.13 times BB flux.

You can visualise how this transmission works. Imagine yourself with IR vision in such an atmosphere with a lapse rate. The gas below is hotter than the gas above. If you hold out your hand, it's warmer below than above. There is a nett heat flux upward.

How much? Well, it's proportional to the lapse rate, which determines the temperature difference that you see. But it depends on the absorption length too. If that is higher, you can see further, and see hotter (and colder) gas, because of the linear lapse rate. This effect is proportional to L, hence the Rosseland expression.

As L increases further, things change. Looking down, the gas gets denser, so you can't see as far. The hotness increases less than linearly. Looking up, you see further. The coolness increases more than linearly. They cancel somewhat, so changes to the flux are second order in L. But eventually there is a deviation from Fourier's Law. And if you see far enough, the ground will have an effect too.

Eventually, when L is large relative to the depth of the troposphere, a substantial fraction of IR is transmitted without any absorption. This is an atmospheric window, and the relation between transmission and lapse rate diminishes.

But up to that point, Fourier's Law is a good way of thinking about IR transmission.

Convection

This also has a wrinkle. On a room scale, turbulent convection is also often thought of as obeying a Fourier Law. Gravity-forced changes of pressure are negligible. The adiabatic lapse rate is about 0.01 C/m.

However, on the atmospheric scale, compression with gravity is important in modifying the temperature. Compression warms. Fourier's Law applies not to the gradient of temperature, but of potential temperature:
θ=T(P/P0)^(-ν)
where ν=R/cp and P0 is a reference pressure.

The important thing here is that unlike with conductive and radiative transport, convective transport is zero at the dry adiabat lapse rate, rather than zero gradient, and is proportional to the difference between the actual lapse rate and the adiabat. Below the adiabat, the gas is convectively stable, and energy is used making it go up and down. Above the adiabat, it is unstable, and the temperature gradient adds energy to the motion, and accelerates the transport.


Atmospheric fluxes


Now that we've outlined the modes, and agreed to ignore conduction, we have a downflux of sunlight on Venus, after allowing for albedo, of about 160 W/m2. Not much of that actually reaches the surface - it is absorbed at various depths. But once absorbed, the heat has to get out again, and the modes available for it to reach the tropopause to be radiated away are IR transfer and convection.

Suppose the convective component were small. The lapse rate would then be determined by a Fourier Law, as given by the Rosseland model. This might be more or less than the adiabat.

If more, then the gas would be convectively unstable. Motions would be induced which would increase convective transport. Since the overall flux is determined by the sunlight, that means the proportion carried by IR would reduce, lowering the lapse rate. This argument, made by DeWitt in the first SoD thread, gives a mechanism whereby the adiabat lapse rate cannot be much exceeded.

But there is an analogous argument from below. If the IR flux caused a temperature gradient less than the adiabat, then the gas would be convectively stable, and it could remain at that. However, there are likely to be other effects inducing motion. Polar regions (and the long nights on Venus) emit more heat than they receive - sunny regions emit less. This heat has to be transported by the atmosphere, and the temperature difference drives a heat engine. The resulting circulation conveys the heat.

But the motion then affects vertical convection. I described earlier how motion pumps heat downward. This then augments the heat that must come up conveyed by IR. This in turn increases the lapse rate towards to adiabat. In doing so, it extracts KE from the air to drive the pump.

Leonard Weinstein's problem



At SoD, Leonard Weinstein proposed a thought experiment where a shell was placed around Venus at about the altitude where IR is currently emitted to space. This is high in the troposphere, where the temperature is about 230K. The shell is opaque to all radiation, but a good conductor. He contended that the temperature profile in the atmosphere would remain much the same. My initial thought was that it would become isothermal, as did others.

I now think a reasonably quantitative analysis is possible. About 160W/m2 sunlight penetrates the atmosphere, and is balanced by an IR and a convective flux. I don't know how much is radiative, but the total is known.

The shell blocks that 160W/m2. If the atmosphere remains the same, then the same IR transport upwards must continue, because it is determined by the unchanged temperature gradient. The convection would also continue, since it is determined by the gradient and the motion. But there's no longer heat supplied, so something has to change. Net upflux must go to zero.

The lapse rate must drop well below the adiabat, to ensure that convective flow is downward, by the heat pumping mechanism. If it does, there will be a balance when the downflux matches the IR upflux, giving the required zero. But how low would the lapse rate go?

Heat pump arithmetic


In this scenario, the heat pump must pump downward about 160 W/m2 than it did before the shell. It has to pump it to depths comparable to where sunlight formerly penetrated. Figures get rough here, but let's say the average depth is where the temperature has risen from 230K to 460K.

Thermodynamics tells us the energy needed to do that. It's 160*(δ T)/T, or about 80 W/m2. That comes from the KE in the gas, which must in turn have come from a heat engine somewhere.

But the equator to pole differential, which is a promising source of energy for a heat engine, is much too small. The actual flux of heat redistributed horizontally can only be a small fraction of the 160 W/m2 arriving. And the temperature differential, even with reduced circulation, is never going to match the factor of 2 difference between surface and depth ( without a shell, it's very small).

In fact, there just isn't that sort of energy available anywhere. It's half of total average incoming solar.

So what would happen?


If the convective heat pump can't replace the blocked sunlight, the lapse rate must just reduce until the IR Fourier's Law flux can be matched by the very limited energy available from a realistic heat engine. This needs a rather elaborate but doable calculation which would consider differential insolation over a sphere, and a plausible temperature differential around the shell. The resulting regional discrepancy between sunlight in and IR out would determine the horizontal fluxes, and the temperatures would determine the energy available as KE to the atmosphere. Then some unknown fraction of that could pump heat down, making possible an IR upflux and a positive lapse rate.

So a SWAG? Dunno, but surface temperatures warmer than the shell, but Earth-like rather than Venus-like.


14 comments:

  1. Nick,

    Some sunlight has to reach the surface or, as you point out, you don't get much of (I would argue none at all) a greenhouse effect below the point of complete absorption. The simple slab model says that if the short wave absorptivity is equal to the long wave absorptivity, the surface flux up is equal to the the solar flux down at the top. The value of the absorptivity doesn't matter.

    As far as conduction goes, in a LW opaque atmosphere a packet of air moved up or down when the lapse rate is not equal to the adiabatic rate would transfer heat to the surrounding air by radiation. That would not be true in a transparent atmosphere. You would have to rely on conduction or turbulent mixing. But turbulent mixing would be a relatively short range process and would be just as effective for diffusion along the temperature gradient.

    The rate of heat transfer will be proportional to the vertical velocity of air movement and the difference between the actual lapse rate and the adiabatic rate. For the Earth, the maximum vertical velocity observed is near the equator at ~0.2 Pa/sec at 500 hPa. (See this figure) For a tropical atmosphere, the pressure gradient at 500 hPa is about 8.3 Pa/m so the maximum velocity would be less than 0.003 m/sec. That's the maximum, the average is far less. Obviously local vertical velocity could be higher, but not everywhere all the time. For Venus, the denser atmosphere should mean a much lower average vertical velocity because the driving force of direct solar radiation at the surface is so much smaller, zero for an opaque shell.

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  2. DeWitt,
    "As far as conduction goes, in a LW opaque atmosphere a packet of air moved up or down when the lapse rate is not equal to the adiabatic rate would transfer heat to the surrounding air by radiation. That would not be true in a transparent atmosphere. You would have to rely on conduction or turbulent mixing."
    I identify this up-down motion with turbulent mixing. It happens on the full scale of turbulent motion. As a heat pump, it has diminishing effect with small eddies, because the adiabatic aspect fails.

    "Adiabatic" is a time-relative term. It means that conductive (including radiative) processes are slow relative to the pure convective transport. GHG with radiative transport make the conductive processes faster, so small eddies become inefficient heat pumps. The Grashof (or Rayleigh) numbers characterise the dimensional requirements. They include thermal conductivity, and with GHG, that would be the Rosseland value.

    There is a characteristic time scale, which is the Brunt-Vaisala frequency. Caballero, in his PDF Phys Met lecture notes, talks about this a lot. The period is independent of the scale of the motion, but depends on lapse rate, and is typically minutes (infinity at the adiabat).

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  3. I forgot all about forward scattering until Leonard Weinstein asked about how the surface of Venus could average ~17 W/m2 of incoming solar radiation but we couldn't see the surface. However, it isn't pitch dark on a cloudy day yet you can't see the sky from the ground or the ground from the sky.

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  4. After some research, I have to concede that you will have some sort of positive lapse rate under most circumstances,like a completely transparent atmosphere. But one of those circumstances isn't a LW and SW opaque atmosphere. For heat to be transported down to the surface by other than SW radiation, one would need a very positive potential temperature gradient.

    For the current Venusian atmosphere, any reduction in the LW optical density without changing the SW optical density and reflectivity by the replacement of CO2 by a transparent gas would result in a lower surface temperature and a lower total atmospheric heat content. That would bring the radiative layer lower so the lapse rate would be maintained.

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  5. DeWitt
    Actually you can (just) see the surface in a near IR band

    "For heat to be transported down to the surface by other than SW radiation, one would need a very positive potential temperature gradient."
    That sounds counter-intuitive. Do you mean positive in the sense of positive lapse rate (cooling as you go up?).

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  6. Positive in the sense of potential temperature increasing with altitude.

    The other problem is that the scale height will decrease with temperature. If the effective emission altitude is held constant, that forces the surface temperature to be high, but the emission altitude is a function of the temperature, not the other way around. The scale height on Venus is not constant with altitude. It's 15.9 km at the surface, but more like 5 km in the cloud layer. Fixing Teff does not fix the surface temperature because even at a constant lapse rate, the atmosphere will be denser at lower temperature. We can see this effect on Earth because the effective emission altitude in the tropics is much higher and colder relative to the surface than at high latitudes. As a result, greenhouse forcing is higher at the equator than at the poles by about a factor of three.

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  7. If SpectralCalc using the HITEMP CO2 database can be believed, the Venusian atmosphere has significant transparency in the thermal IR at 700 K on the scale of hundreds of meters. That leaves enough room for significant net radiative transfer from the surface to the atmosphere, but little or none directly to space.

    I subscribed to SpectralCalc for a month so if you have any suggestions for calculations, let me know.

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  8. DeWitt,
    Sounds interesting. If the harmonic mean across the spectrum is hundreds of metres, then IR would carry the total required flux. Since the lapse rate is like Earth, 1km absorption length carries about 1/5th BB radiation, which at 700K is a lot, diminishing with altitude.

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  9. Another way of thinking about this starts with the observation that Venus does not have a stratosphere (or a mesosphere come to think of it).

    What the majic shield is doing is building one.

    OK, what happens here on dear old mommy Earth.

    Part of the sun's radiation is absorbed in the stratosphere which warms it, and if you work things out, since that energy does not reach the surface, but most of the light gets through, it also cools the surface. What happens when a big volcano goes off and pushes SOx into the lower stratosphere absorbing more of the light? The surface cools, actually the entire troposphere cools.

    The majic shield is a really effective stratosphere. Details are left to the reader. A good finger exercise would be to work out the cooling effect of the stratospheric ozone (and O2) absorption on the surface and tropospheric temperature. This is included in the GCMs as the stratospheric ozone forcing.

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  10. That should be most of the visible light

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  11. Nick,

    I've probably overestimated the transparency of the Venusian atmosphere. Water vapor, even at only 30 ppmv, fills most of the hole at about 1500 cm-1 and SO2 takes a big chunk out of the hole at 3000 cm-1. That leaves the hole between ~4000-4500 cm-1. Before my subscription expires at SpectralCalc, I'm going to look at just that hole from 0 to 1 km in 50 or 100 m slices with a BB source at 1 km and calculate the net upward flux at the surface. I think if I hold the pressure constant, I'll get about the same number of data points per slice. The effect of pressure is fairly small over that altitude range.

    Thermal diffusivity is inversely proportional to density, so the Venusian atmosphere at the surface would be less conductive than the Earth's atmosphere at the surface.

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  12. DeWitt,
    OK. But the ability to transfer heat by IR is increased by T as T^3. I calculated at SoD that an absorption length of 10 m at 700 K would carry 8.5 W/m2, which is comparable to insolation at the surface.

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  13. Nick,

    For 92 bar, 730 K, 0.96 CO2, 0.00003 H2O and path 10 m, the mean transmittance between 500 and 5500 cm-1 is 0.348. By absorption length, if you mean I(x)/Io= 1/e when x = 10 m, then that's certainly in the ballpark.

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  14. To put it another way, why is it cooler under an umbrella in the full sun.

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