Irodov 1.316: Two Manometers Time a Venturi – Amit & Deepali Sinha Foundation

Problem Irodov 1.316

Water flows steadily through a horizontal pipe whose cross-sectional area changes from $S_1$ , at a narrower section, to $S_2$ , at a wider section. An open vertical manometric tube is mounted on the pipe at each of the two sections, and the water in each tube stands at a height set by the local pressure there. The two water columns differ in height by $\Delta h$ . Find the volume of water $Q$ flowing across any section of the pipe per unit time.

Side view of the horizontal pipe, widening in flow order from the narrow section S_1 to the wide section S_2, each with its own open manometer tube. The column at S_1 stands lower than the column at S_2 by Δ h. — Side view of the horizontal pipe, widening in flow order from the narrow section $S_1$ to the wide section $S_2$ , each with its own open manometer tube. The column at $S_1$ stands lower than the column at $S_2$ by $\Delta h$ .

Given

$S_1$ , $S_2$
$\Delta h$

Find

Find

Q

, the volume flow rate through the pipe

Solution

Step 1Bernoulli between the two sections

The pipe is horizontal, so a streamline running along its axis from $S_1$ to $S_2$ carries no elevation change, and the flow is steady and (water being essentially incompressible) ideal enough for Bernoulli’s equation to connect the two sections directly:

p_1+\frac{1}{2}\rho v_1^2 = p_2+\frac{1}{2}\rho v_2^2. \tag{1}

Step 2Continuity fixes both speeds in terms of

Q

The same volume flow rate $Q$ crosses every section of the pipe, so

v_1 = \frac{Q}{S_1}, \qquad v_2 = \frac{Q}{S_2}. \tag{2}

Step 3The manometers read the pressure difference directly

Each open tube is in hydrostatic equilibrium above the point where it joins the pipe, so the height of its column above that point is the local (gauge) pressure divided by $\rho g$ . With the column at $S_2$ standing $\Delta h$ above the column at $S_1$ ,

p_2 - p_1 = \rho g \,\Delta h. \tag{3}

Step 4Solve for

Q

Rearranging the Bernoulli relation (1) and inserting the pressure difference (3),

\begin{align*}p_2-p_1 &= \frac{1}{2}\rho\left(v_1^2-v_2^2\right) \\ \rho g\,\Delta h &= \frac{1}{2}\rho\left(v_1^2-v_2^2\right).\end{align*}

Substituting the continuity expressions (2) for $v_1$ and $v_2$ and solving for $Q$ ,

\begin{align*}2g\,\Delta h &= \frac{Q^2}{S_1^2}-\frac{Q^2}{S_2^2} = Q^2\,\frac{S_2^2-S_1^2}{S_1^2S_2^2} \\ Q^2 &= \frac{2g\,\Delta h\, S_1^2S_2^2}{S_2^2-S_1^2}.\end{align*}

Taking the positive square root,

Q = S_1S_2\sqrt{\dfrac{2g\,\Delta h}{S_2^2-S_1^2}}

(4)

Check

The formula requires $S_2>S_1$ for the radicand to stay positive, matching the setup: the narrower section is always the one with the faster, lower-pressure flow, so $\Delta h$ (defined as the wide column minus the narrow one) is always positive when $Q>0$ . Dimensionally, $[S_1S_2]=\mathrm{m}^4$ , while $[g\,\Delta h]=\mathrm{m}^2\,\mathrm{s}^{-2}$ against $[S_2^2-S_1^2]=\mathrm{m}^4$ makes the radicand $\mathrm{m}^{-2}\,\mathrm{s}^{-2}$ and its square root $\mathrm{m}^{-1}\,\mathrm{s}^{-1}$ , so the product is $\mathrm{m}^3/\mathrm{s}$ , a volume flow rate.

Remark

As $S_1\to S_2$ the denominator $S_2^2-S_1^2\to0$ , so any nonzero $Q$ would force $\Delta h\to0$ : two equal sections carry equal speeds and develop no pressure difference at all, however fast the flow, which is exactly the limit in which a venturi meter loses its ability to read a flow rate and a genuine taper becomes essential to the device.