- #1
leonmate
- 84
- 1
Ok, this question goes on a little further. I want to find the velocity of air moving from the equator to the poles due to the change in pressure caused by differing temperatures.
I've found the flux density from the sun, so I have a value of flux (hope I am using the right term here - whatever W/m^2 is) for the equator and the N pole and I factored in different albedos too
But now I'm stuck.
I have a flux on the equator, so energy/s for 1m^2 essentially. I want to find how much the surface heats up and use that to find how much that heats the column of air above it.
If we consider at sunrise the angle between the surface and the sunlight is 0°, at midday 90°, and sunset 180°. So we are going from minimum>maximum>minimum. I've drawn myself a graph, the maximum flux is 1270 W/m^2
As the sun moves across the sky, the 1m^2 of ground we are considering has a differing amount of flux on it as F = (Fmax) sinθ so we wind up with a graph that looks like the sin curve. What I want to know is how can I work out the total energy on this 1m^2 of ground over 12 hours (43200 seconds if that's easier)? I guess it's an integral but so far nothing has worked for me
I've found the flux density from the sun, so I have a value of flux (hope I am using the right term here - whatever W/m^2 is) for the equator and the N pole and I factored in different albedos too
But now I'm stuck.
I have a flux on the equator, so energy/s for 1m^2 essentially. I want to find how much the surface heats up and use that to find how much that heats the column of air above it.
If we consider at sunrise the angle between the surface and the sunlight is 0°, at midday 90°, and sunset 180°. So we are going from minimum>maximum>minimum. I've drawn myself a graph, the maximum flux is 1270 W/m^2
As the sun moves across the sky, the 1m^2 of ground we are considering has a differing amount of flux on it as F = (Fmax) sinθ so we wind up with a graph that looks like the sin curve. What I want to know is how can I work out the total energy on this 1m^2 of ground over 12 hours (43200 seconds if that's easier)? I guess it's an integral but so far nothing has worked for me