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08-05-2011, 11:37 PM   #21
lwcamp

Join Date: Nov 2004
Location: The plutonium rich regions of Washington State
Re: Reinventing Barsoom: 2 — Parahumans and legacy genetic engineering

Quote:
 Originally Posted by whswhs That's a proposition about the rate at which heat is added to the atmosphere per unit area of planetary surface. But it doesn't seem to imply a specific residence time of heat in the atmosphere, and therefore it doesn't seem to imply a specific energy density of the atmosphere. Is there any reason to suppose that the heat will take as long to get from the ground to the edge of space on Mars as it does on Earth, or even that it will take the same amount of time per kilometer of altitude? I don't need to see a quantitative analysis; I'm just asking whether there is some basic fact of atmospheric dynamics that requires either of that and that I'm ignorant of.
How long it takes to get out is not really important here. What is important is that if you have, say, 1 kW/m^2 of sunlight incident on the ground (and assuming perfect absorption), then every square meter perpendicular to the ground, at any altitude you choose, will have 1 kW of energy being transported through it by the air on average going back out. Since the transport mechanism is turbulent convection, you can expect similar power/area values for any orientation of your test area.

Now if I go through and do the math (which I should have done to begin with - and now it is late and I need to get up early and drive to Seattle for a wedding so I don't have time to think it through properly)
E = energy
rho = air density (assumed constant)
v = air speed
V = volume
A = area
F = aerodynamic force
P = power (P/A is assumed constant, set by insolation)
h = altitude
g = gravitational acceleration
c = specific heat
K = the universal gas constant divided by the molar mass
T = temperature
p = pressure
P = (E/V) A v --- power through a cross sectional area is given by the energy density of the fluid times the velocity of the fluid
E/V = 0.5 rho v^2 + rho g h + c p = rho (0.5 v^2 + g h + K T) --- the usual formulas for kinetic energy plus gravitational potential energy plus energy to pressure relationship of a gas
F = 0.5 rho v^2 A --- the usual formula for aerodynamic forces, assuming that the drag coefficient is 1
Limiting cases: case 1 - all energy is transported as kinetic energy
P/A = 0.5 rho v^3
v = (2 P / rho)^(1/3)
F = (rho/2)^(1/3) (P/A)^(2/3) A
case 2 - all energy is transported as thermal energy
P/A = rho K T v
v = (P/A)/(rho K T)
F = 0.5 [((P/A)/(K T))^2]/rho
so we see that depending on the balance of kinetic to thermal transport, the aerodynamic forces can have weak drop with decreasing density or a fairly strong rise with decreasing density. Eh - pretty messy. If I come up with startling insight over the weekend I'll let you know. In any case, speed always rises for a given insolation as density decreases.

Luke

08-06-2011, 12:29 AM   #22
whswhs

Join Date: Jun 2005
Re: Reinventing Barsoom: 2 — Parahumans and legacy genetic engineering

Quote:
 Originally Posted by lwcamp Limiting cases: case 1 - all energy is transported as kinetic energy v = (2 P / rho)^(1/3) case 2 - all energy is transported as thermal energy v = (P/A)/(rho K T)
Okay, I'll go for that. Modulo changing P to P/A in case 1.

Either way, you seem to get a simple relation. P/A is 39% of earth's; if rho is less than 39% of earth's, you get faster atmospheric motion, all else being equal. (T will be lower, but probably not massively lower, since this is a Mars where a human being won't freeze solid overnight.) The need to have breathable pressure, especially on a planet with lower gravity, is going to force rho to be fairly high, though, right?

Bill Stoddard

 Tags bio-tech, mars, sword & planet

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