Re: How can I convert Pascal/sec to meters/sec?

Hello Anna.

As you are well aware, this is a very technical and specialized question. But it is a type of question 
that crops up a lot in advanced science -- particularly in developing mathematical models for 
various situations.

Ultimately, you have to think your way through the detail of the meaning of the various quantities 
and equations involved in your model. Here is a short-cut strategy for finding AN answer as a 
starting point for this sort of problem. Unfortunately there is no guarantee that it will provide the 
correct answer -- it might work, or it might be quite wrong, and without putting in a couple of 
weeks on the detail of your problem I cannot tell.

You have two quantities that you want to equate. One is in pascal/sec, the other is in (vertical) 
metre/sec. To do the conversion, there must be some quantity that is central to the situation, that 
can be expressed in pascal/(vertical) metre. The obvious quantity in this case is the vertical rate of 
pressure fall-off: dp(ambient)/dx(vertical). It may or may not be the correct one.

We might be able to do a bit better. In the case of your particular problem, there is some good 
news. Your complicated expression is in metre/sec; it is only the zonal mean vertical velocity that 
is in pascal/sec. But metre/sec is an obvious velocity unit for vertical movement of air parcels, and 
it should be a fairly familiar and straightforward conversion that does not involve any of the 
complications of the other part of your expression.

Why should a vertical velocity be expressed in Pa/s? An obviously related quantity is the vertical 
flux of gas, and a natural set of units for that quantity is kg/m^2/s. Perhaps there is a reason (e.g. 
calculating aerodynamic forces) for dealing in weights of gas rather than masses of gas. In that 
case,  1 kg/m^2/s becomes 9.8 newton/m^2 = 9.8 Pa/s. Are vertical fluxes of gas typically 
measured as weight fluxes?

If the Pa/s units refer to a vertical flux, then you must divide by g=9.8 to get kg/m^2/s. The 
density of air in kg/m^3 is given by the ideal gas equation pV=nRT, where m = nM/1000, and 
M= 28.96 g/mol is the average molar mass of air.

Density = m/V = Mp/1000RT

The mean velocity in m/s can then be given by Flux/Density. So if this analysis is correct,
mean velocity -- v m/s -- is related to weight flux -- f Pa/s -- by

v = 1000 RT f / Mgp

I suspect that this is quite a different answer from the one I suggested a few lines earlier.

I wish you success! (You can probably detect that I am not one of the world's leading theoretical 
meteorologists!)