What is the partial derivative of pressure (bar) with respect to time?

is it m^3/s?

 

If 1 bar = 100kPa where the Pascal [Pa] is the SI unit of pressure then presumably the unit for the time derivative (rate of change) of pressure is \(P_as^{-1}\)

\(1 \ P_a=1 \ kgm^{-1}s^{-2}\)

So presumably the unit of the time derivative of pressure in terms of the fundamental units is

\(kgm^{-1}s^{-3}\)

 

Actually, I think it is \(\frac{\partial P(bar)}{\partial t}=100k \cdot (-2) kgm^{-1}s^{-3}\)
given that the formula for the pressure is as you say.

 

If the question was actually about conversion between units then perhaps more succinctly

\(1 \ Bars^{-1} \ <=> \ 100 \ kPas^{-1} \ <=> \ 10^5 \ kgm^{-1}s^{-3} \)

I doubt the OP has any real ongoing interest in this matter.

 

But the partial derivative of a function in respect to the time is
\(\frac{df}{dt}\)
not
\(\frac{f}{s}\)

One of us has a really brain-dead moment. By the end of the thread we 'll see who it is. (Please let it not be me!)

Now I saw the units reference. It was me who had the brain dead moment after all. Sigh...