Irodov 1.317: A Dense Liquid Reads a Light Gas – Amit & Deepali Sinha Foundation

Problem Irodov 1.317

Gas of density $\rho$ flows along the axis of a pipeline of cross-sectional area $S$ , with viscosity negligible. A Pitot tube mounted on the axis has one opening facing directly into the oncoming gas (reading the stagnation pressure) and a second opening reading the local static pressure of the gas; the two are connected to a U-tube manometer filled with a liquid of density $\rho_0$ , and the two liquid columns differ in height by $\Delta h$ . Find the volume of gas $Q$ flowing across the pipe’s section per unit time.

Side view of the gas pipe. A straight stub reads the static pressure; a bent probe facing upstream reads the higher stagnation pressure. Both connect to a U-tube of denser liquid whose columns differ by Δ h, shorter on the stagnation side. — Side view of the gas pipe. A straight stub reads the static pressure; a bent probe facing upstream reads the higher stagnation pressure. Both connect to a U-tube of denser liquid whose columns differ by $\Delta h$ , shorter on the stagnation side.

Given

$S$
$\Delta h$
$\rho_0$ (manometer liquid)
$\rho$ (gas)

Find

Find

Q

, the volume of gas flowing per unit time

Solution

Step 1Bernoulli from the free stream to the stagnation point

Follow a streamline in the gas from a point far enough from the probe that the flow is undisturbed, at the pipe’s bulk speed $v$ and static pressure $p$ , to the mouth of the probe facing upstream, where the gas is brought to rest ( $v=0$ ) and the pressure rises to the stagnation value $p_0'$ . Since the pipe is effectively horizontal over this short distance, Bernoulli’s equation gives

p + \frac{1}{2}\rho v^2 = p_0'. \tag{1}

Step 2The manometer balances that excess with its own liquid column

The static tap and the stagnation mouth are the two ends of one U-tube filled with the denser liquid $\rho_0$ ; the pressure difference $p_0'-p$ that the gas applies at the top of the two arms is balanced by the hydrostatic weight of the liquid column difference $\Delta h$ between them:

p_0' - p = \rho_0\, g\, \Delta h. \tag{2}

Step 3Solve for the gas speed

Equating the two expressions (1) and (2) for the pressure excess and solving for $v$ ,

\begin{align*}\frac{1}{2}\rho v^2 &= \rho_0\, g\, \Delta h \\ v^2 &= \frac{2 g\, \Delta h\, \rho_0}{\rho}.\end{align*}

v = \sqrt{\dfrac{2g\,\Delta h\,\rho_0}{\rho}}

(3)

Step 4The volume flow rate

The gas crosses the pipe’s constant section $S$ at this speed, so the volume flowing per unit time is the product of the two, using (3):

Q = Sv = S\sqrt{\dfrac{2g\,\Delta h\,\rho_0}{\rho}}

(4)

Check

Dimensionally $2g\,\Delta h\,\rho_0/\rho$ is $(\text{m}/\text{s}^2)(\text{m})$ times a pure ratio, a squared speed, as it must be. And setting $\rho_0=\rho$ collapses the result to $v=\sqrt{2g\Delta h}$ , the ordinary Pitot formula for a liquid read by a standpipe of its own liquid, so the factor $\rho_0/\rho$ is exactly what rescales that familiar formula when the flowing fluid and the manometer liquid differ. That limit checks the algebra rather than the drawn apparatus: a U-tube whose liquid matches the gas has no interface left to read a height difference at, and its exact balance $(\rho_0-\rho)g\,\Delta h$ correctly returns nothing.

Remark

Because $\rho_0\gg\rho$ for a gas read by a liquid manometer, a modest, easily read $\Delta h$ corresponds to a gas dynamic pressure that would otherwise demand an impractically tall column of the gas itself; this density amplification is why liquid manometers are standard instrumentation on wind tunnels and ventilation ducts. The stagnation-pressure principle underneath it is shared with a light aircraft’s airspeed indicator, which reads the same difference with an aneroid capsule and no liquid column at all, and with the bent tube of problem 1.320, where water reads its own dynamic pressure directly.