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If you have a fluid with with pressure field $p$ what is the physical significance of the the volume integral of the pressure $\int_V p dV$?

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  • $\begingroup$ Can you give some more context about where this quantity shows up? Kyle's answer below covers one interpretation but it is for changes in volume between two states. But it could also just be a way to define a mean value... $\endgroup$
    – tpg2114
    Nov 28, 2015 at 3:41
  • $\begingroup$ $pdV$ shows the elementary work done on the fluid. The term $dV$ represents elementary change in the volume of the fluid under pressure $P$ $\endgroup$ Nov 28, 2015 at 3:42

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The equation for work W that you have written comes from the relationship we learned in freshman physics, $W=\int{Fdx}$. If A the area over which the force acts, then the pressure is $p=F/A$, and the differential change in volume is dV=Adx. So,

$$W=\int{Fdx}=\int{\left(\frac{F}{A}\right)Adx}=\int{pdV}$$

In terms of interpretation of the integral as an average, we can define the average pressure as $\bar{p}=W/\Delta V$. So, the average pressure during the volume change is given by:$$\bar{p}=\frac{\int{pdV}}{\Delta V}$$

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The significance is that it is the work done by the fluid.

Moreover, $$ W=\int_{V_i}^{V_f} P\,dV $$ where $W$ denotes the work done by the system during the whole of the reversible process.

This relation appears regularly in thermodynamics, usually as the first law of thermodynamics $$ dU=\delta Q+\delta W=\delta Q-pdV $$ (the minus sign represents the work done on the system, rather than by the system)

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  • $\begingroup$ This was what I thought first also -- but that's only for changes in volume right? A volume integral of any quantity in a volume is really just... the average value in the volume times the volume... Perhaps we need more context? $\endgroup$
    – tpg2114
    Nov 28, 2015 at 3:37
  • $\begingroup$ Hmm, I suppose it could be the mean value of the pressure in the domain as well. Guess it would depend on OP's context; hopefully they'll respond to your comment on their post. $\endgroup$
    – Kyle Kanos
    Nov 28, 2015 at 11:31

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