= Solution
Use the ideal one-dimensional pipe model: constant density, steady inviscid flow, approximately uniform endpoint speeds $u_1,u_2$, and equal body-force potential at the two ends. <Conservation of mass> gives the constant <volumetric flow rate> $q=u_1S_1=u_2S_2$, and the <mass flow rate> is $m=\rho q$. The steady <Bernoulli equation> gives
$$
\Delta p=\frac\rho2(u_2^2-u_1^2)=\frac\rho2q^2\left(\frac1{S_2^2}-\frac1{S_1^2}\right).
$$
For flow from a wider to a narrower section, $S_1>S_2$ and $\Delta p>0$. Solving for the positive mass throughput gives the <Venturi effect> formula
$$
\boxed{m=S_1S_2\sqrt{\frac{2\rho\Delta p}{S_1^2-S_2^2}}.}
$$
The formula requires these hydraulic assumptions; steady incompressibility alone does not impose uniform endpoint profiles or equal potential. If $V_1\neq V_2$, the same derivation replaces $\Delta p$ by $\Delta p+\rho(V_1-V_2)$. Viscous losses would also modify the pressure balance. The equal-area case is singular in this pressure-based determination: in the ideal equal-potential model its pressure difference must vanish and does not determine the throughput.
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