[…]The Power Behind Hurricanes and Tornadoes « the Air Vent[…]…

]]>[…]The Power Behind Hurricanes and Tornadoes « the Air Vent[…]…

]]>It’s a pity the notes fade just when it was getting interesting.

It seems to be a work in progress. When I first discovered it a year or two ago, there was a lot less, basically just the first three chapters. And parts of those were pretty rough. I should go back and look at it more often to see when there are changes. There’s probably some automated way to do that.

]]>That’s the switch between heat pump and heat engine that I have been talking through, as the lapse rate passes through 9.8 K/km.

Yes, but only if the overall lapse rate is close to critical anyway, which will be true for a radiative atmosphere like the Earth, but has yet to be established for a transparent atmosphere. If the lapse rate at altitude is far less than critical, the boundary layer will expand during the day and contract at night with little effect on the lapse rate at higher altitude where there is much, much less turbulence. Note that it does this to a considerable extent even on the Earth. The only deep convection is moist convection, and that’s mainly in the Tropics. That’s also one of the main points of contention that Gerald Browning has against computer climate models. The grid size is too coarse to resolve that sort of thing so the temperature and moisture profile of the entire grid box is rearranged whenever unstable conditions are detected.

]]>A key part from Chap 6 is:

That’s the switch between heat pump and heat engine that I have been talking through, as the lapse rate passes through 9.8 K/km. The Richardson number does seem to be important. It’s a pity the notes fade just when it was getting interesting. ]]>

I learned about thermal winds and that a meridional temperature gradient will cause a pressure gradient and induce circulation even in a perfectly transparent atmosphere. That will increase the lapse rate at the equator and decrease the lapse rate at the poles. Whether the decrease in lapse rate at the equator will go all the way to the adiabatic lapse rate, though, is still to be determined. I still think it won’t because the heat flow will be limited by how fast you can transfer heat downward at high latitudes, and you can’t do that by convection. Or at least that’s what I think right now.

Chapter 6 in Caballero’s Physical Meteorology Lecture Notes is about turbulence and heat transfer in the boundary layer. The math is a little beyond me at the moment.

]]>For me, so far an important gain has been to appreciate the possibility that convection alone could indeed make a lapse rate even if the greenhouse substances were absent. In #84 and earlier I started from a complete denial of such a possibility.

]]>#140 OK now I do see how the extra heat sink comes in, and that my use of 150 W/m2 was mis-conceived, and my answer #134 was wrong. The 150W/m2 figure is not the right one to use, because that is a steady flow through the system, exiting, not heat that has to be pumped back.

My formula in #131 is the same as Anastassia’s #136, except that I’m doing it per unit height. Th-Tc = (dT/dx) dx = L dx.

To try to bring things together, I think this is where we’ve got to:

1. I’ve been paying insufficient attention to the role of radiative TOA cooling in creating a steady flux through the air, which would create a lapse rate. I need to deal with that better

2. deWitt needs to fit the role of the dry adiabat, which definitely does need an associated heat pump, into his way of thinking about maintaining the lapse rate

It’s late here, and I now can’t think how to quantify that.

This is link to Lorenz’ works including books http://eapsweb.mit.edu/research/Lorenz/publications.htm, I once found it very helpful.

You estimate your work as W = eps Fs taking Fs as the total flux of energy at the surface. For 10 km you have 60 W/m2. I count this as an overestimate, because here one needs to bother about latent heat only, which is H = 0.5 Fs. The formula is not W = eps * H, but W = eps/(1-eps)H. This would give 37.5 W/m2. The actual gradient is 6.5 K/km, not 10 K/km. This reduces this estimate to 24 W/m2, which is essentially close to my estimate (I took a smaller height than 10 km) but in any case much higher than the circulation power.

The extra sink in the atmosphere takes away all the heat flux that would otherwise warm the upper atmosphere and destroy the lapse rate. No problem of “pumping heat” downwards in this case. So, while the actual value of the lapse rate is set by convection (6.5 K/km is an approximate mean of the dry and moist adiabatic rate), this lapse rate is maintained by transfer of thermal radiation in the atmosphere.

Anastassia

]]>A mistake there. The lapse rate L is of course not 10 but 0.01K/m in SI units, making the power per m altitude 0.006 W/m3 (or Pa/s) to maintain a lapse rate L. That is 6 W/m2/km altitude, which sounds too high relative to that Lorenz power of circulation. Maybe my estimate of 150 W/m2 is too high.

But I don’t see how an extra heat sink helps (#138).

]]>In my estimate in #129 I only accounted for the “reversal” of sensible heat. Taking the global mean flux of latent heat to be about half of absorbed solar radiation at the surface (150 W/m2), the minimum mechanical work W needed to reverse this heat flow becomes W = eps/(1-eps) H = 0.2/0.8 * 0.5 Fs = 19 W/m2. Lorenz (1967) estimated the mechanical power of general circulation to be around 1% of total solar power (225 W/m2), i.e. around 2 W/m2. This shows that, whatever is the source of mechanical power for the observed circulation, this power is by far insufficient to sustain the observed temperature lapse rate in the absence of an additional sink for heat in the atmosphere, which is radiation by greenhouse substances.

]]>The reason why I turn to the Rosselande model is that it describes the relation of the IR energy flux to the air temp gradient – it’s linear like conduction in appropriate (high OD) bands. That’s hard to figure out any other way (most people don’t try). So it’s the only way I can think of for expressing the relation between IR and the lapse rate. It doesn’t matter if it gets rough with low OD – the nature of the dependence doesn’t change that much.

Re: Anastassia Makarieva (Feb 12 02:20),

*This means your work is always smaller than leakage.*

Well, they’re hard to compare. On my calc, if the leakage is 150 W/m2.say, that’s kL; and L=10, T=250, say, so the power needed per m2 per m altitude is 6 Pa (J/m^3). I’m not sure whether that’s high or low.