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

Date: Wed Aug 11 19:20:41 2004
Posted By: John Christie, Faculty, Dept. of Chemistry,
Area of science: Earth Sciences
ID: 1092151599.Es
Message:
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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!)

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