Suppose I have a have a pump of inlet area $A_1$, outlet area of $A_2$ and pressure head $h_p$. Ambient pressure is atmospheric pressure. The inlet and outlet are at the same elevation $z_1 = z_2 = 0$. Inlet and outlet velocities are denoted by $v_1$ and $v_2$ respectively. The flow is assumed to be incompressible.

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The Bernoulli equation is given by $$\frac{P_1}{\gamma}+\frac{{v_1}^2}{2g}+z_1 + h_p = \frac{P_2}{\gamma}+\frac{{v_2}^2}{2g}+z_2$$ And mass conservation demands $A_1 v_1 = A_2 v_2$. Since ambient pressure is the same at outlet and inlet $P_1 = P_2 = P_{atm}$. Plugging in and rearranging, we obtain $$v_2 = \sqrt{\frac{2gh_p}{1-(A_2/A_1)^2}}$$ Now if the areas $A_1$ and $A_2$ are the same, the solution breaks down. For very close values of $A_1 \approx A_2$, the value of $v_2$ becomes extremely large. In fact, for exactly equal areas we find $v_1 = v_2$, just by mass continuity, but with no prediction of the value of $v_1$ or $v_2$ from the Bernoulli equation. So predicting the velocity very sensitively depends on the exact area. This makes zero sense. I have probably made a false assumption, but where? I doubt that engineers select different values of $A_1$ and $A_2$ over concerns of predictivity of the Bernoulli equation.

  • $\begingroup$ The Bernoulli equation omits viscous frictional drag in the pipe. $\endgroup$ Jan 30, 2023 at 12:16

1 Answer 1


The pressure head $h_p$ must be zero otherwise the fluid indeed accelerates without limit (assuming no frictional losses).

Mathematically speaking, one is not allowed to solve explicitly for $v_2$ when $A_1=A_2$ since there is division by zero. One must leave the equation as: $$ 2gh_p=v_2^2(1-A_2^2/A_1^2) $$ And see that $h_p=0$ for flow to have constant velocity.


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