Let's say a pipe runs along the seafloor with velocity $v_1$, then exits at the surface with velocity $v_2$. I want to use Bernoulli's equation to find $v_2$, but I am confused about the pressures in the problem. I know Bernoulli's equation is

$P_1 + \frac{1}{2} \rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g y_2$

If I solve for $v_2$, I get $P_1 - P_2$ on the left hand side, as well as $\rho g (y_1 - y_2)$, but aren't these the same thing since the fluid pressure is equal to $\rho g h$? Would I just have 2 factors of $P_1 - P_2$ that add together then?


1 Answer 1


$P_2-P_1=\rho gh$, or more generally $P_2-P_1=\int_{z_1}^{z_2} \rho(z)g(z)zdz$ is only true in hydrostatics.

As the statics in the name implies, this equation is true when the fluid is not moving: $v_1=v_2=0$

Plug that in the Bernoulli equation and you do get $P_1-P_2=\rho g(y_2-y_1)$ right away.

So this equation is a simplification of Bernoulli for the case of a static fluid. If the fluid is moving , there is no reason to assume that the equation holds true. Bernoulli is instead used to account for the effect of the speeds.

In short, $P_2-P_1 \neq \rho gh$ in general


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