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Here is Bernoulli's equation:

$$P_{{1}}+1/2\,{\rho v_{{1}}}^{2}+\rho {\it gy}_{{1}}=P_{2}+1/2\,{\rho v _{{2}}}^{2}+\rho {\it gy}_{{2}} $$

A friend of mine was looking at some lecture notes today and there was an interesting question included, without an answer.

Let's say the equation is rearranged as follows, to solve the pressure $P_{2}$:

$$P_{2}=P_{{1}}+1/2\,{\rho v_{{1}}}^{2}+\rho {\it gy}_{{1}} -1/2\,{\rho v_{{2}}}^{2}-\rho {\it gy}_{{2}} $$

The question is: Is it possible for the pressure $P_{2}$ to be less than or equal to zero? And if so, what would that mean? Apparently there was no picture included so I cannot say much more about the scenario in the question.

So what would, say, a negative pressure indicate in this equation? I spent some time with this question but I could not think of a sure answer. Does negative pressure here indicate a negative force on the fluid element? Is the fluid flowing backwards?

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  • $\begingroup$ An ideal gas always has positive pressure, but a sting under tension has negative longitudinal stress. If you could arrange for tension in all three dimensions, you would have negative pressure. $\endgroup$ – Bert Barrois Sep 12 '18 at 11:43
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Is it possible for the pressure P2 to be less than or equal to zero?

In this equation, yes. The Bernoulli equation is essentially an energy balance for a fluid at two points.

And if so, what would that mean?

For starters, it means you aren't measuring absolute pressure. You could be measuring gauge pressure, or you could have some arbitrary pressure defined as 0. It isn't really a big deal for the Bernoulli equation; because it is the pressure difference that determines the potential of the system due to pressure.

So what would, say, a negative pressure indicate in this equation?

Nothing special. It might indicate that they should have picked a different scale for pressure to avoid this. It doesn't say much besides that the potential due to pressure is lower at point 2 than point 1.

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No, Pressure is defined positive.

The only term of pressure that has a reality in this equation here is $\Delta P = -\frac{\rho \Delta(v^2)}{2} - \rho g \Delta z$.

As an example, with most systems, Pressure is really close to atmospheric pressure $10^5 Pa$ which is really high to make the other term $P_2$ become zero or negative.

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  • $\begingroup$ Pressure is only defined positive if you're using absolute pressure. Gauge pressure can have negative values, which means it shouldn't be an issue since you point out we're really looking at the pressure difference. $\endgroup$ – JMac Sep 12 '18 at 11:41
  • $\begingroup$ @JMac Gauge pressure is a differential pressure. $\endgroup$ – PackSciences Sep 12 '18 at 17:06
  • $\begingroup$ And it's perfectly fine to use a differential pressure in this equation as long as you're consistent. $(P_2+P_A) - (P_1 + P_A) = P_2 - P_1$. It just seems a bit misleading here to say "pressure is defined as positive" when the $P_2$ and $P_1$ can be negative in this context. $\endgroup$ – JMac Sep 12 '18 at 18:31
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We know that, with certain constraints (steady state, no friction, laminar flow), this equation should be valid - it is a law.

Therefore, if all given values are real, the calculated pressure (assuming we are working with absolute pressure) should be positive. Conversely, if the calculated pressure is negative, at least some of the given values must be bogus.

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  • $\begingroup$ I disagree that if the pressure is negative the given values are bogus. If you're dealing with gauge pressure, it would be completely valid to have a negative pressure in the equation. The pressure differential is what is driving it. It doesn't matter where you define 0, you just have to keep that in mind when looking at the results. $\endgroup$ – JMac Sep 12 '18 at 11:39
  • $\begingroup$ @JMac I don't disagree with your point and I understand how it could be convenient and completely fine to use negative pressure in practice, but I've interpreted this question differently, because it appeared to me that the PO was talking about absolute pressure. $\endgroup$ – V.F. Sep 12 '18 at 12:03
  • $\begingroup$ Given the context of the question, I don't think you can assume absolute pressure is a condition here. You should at least specify that you're talking only about absolute pressure, or else this is just misleading. $\endgroup$ – JMac Sep 12 '18 at 12:07
  • $\begingroup$ @JMac Agree. I've updated the answer. Thanks for your feedback. $\endgroup$ – V.F. Sep 12 '18 at 12:13
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It is perfectly possible for the pressure to be negative, giving a liquid under tension - and this goes on regularly in trees if they are tall enough Check this out, this will surely blow your mind..

I m writing this as an answer and not as a comment, so that it maybe noticable.

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