Irodov 1.335: The Kink That Reveals the Kinetic Head

Problem Irodov 1.335

A viscous liquid of density $\rho = 1.0$ g/cm $^3$ drains from a wide tank $A$ through a straight horizontal tube of uniform bore, leaving the tube’s open far end as a jet. Three vertical open piezometer tubes stand on it, one equal distance $l$ apart: the first on the tube’s entrance section, where the liquid arriving out of the wide tank is still effectively at rest; the second a distance $l$ downstream of it; and the third another $l$ downstream of that, itself a further length $l$ short of the open outlet. Reading from the tank toward the outlet, the liquid stands in the three tubes at heights $h_3 = 35$ cm, $h_2 = 20$ cm and $h_1 = 10$ cm above the tube’s own axis. Find the velocity $v$ of the liquid flowing through the tube.

Liquid draining from the wide tank A along a horizontal tube of uniform bore. Its three risers, one length l apart, read h_3, h_2 and h_1; the outlet lies l beyond the third. The dashed line is the tank's free surface. — Liquid draining from the wide tank $A$ along a horizontal tube of uniform bore. Its three risers, one length $l$ apart, read $h_3$ , $h_2$ and $h_1$ ; the outlet lies $l$ beyond the third. The dashed line is the tank’s free surface.

Given

$\rho = 1.0$ g/cm $^3$
$h_1 = 10$ cm
$h_2 = 20$ cm
$h_3 = 35$ cm
the three distances $l$ along the tube are equal

Find

v

, the liquid’s flow velocity in the tube

Solution

Step 1The friction-only rate, from the two readings clear of the tank

In laminar (Poiseuille) flow through a tube of uniform bore the pressure gradient is constant along the tube, so once the flow is fully established the piezometric head must fall in equal steps over equal lengths. The two readings farthest from the tank, $h_2$ and $h_1$ , are both already in this regime and fix that rate directly, with no need to know $\eta$ , the flow rate or $l$ itself:

\Delta h_f \equiv h_2 - h_1,

the head lost to friction over one section length $l$ .

Step 2The excess at the entrance tap

Subtracting the friction-only prediction from the measured $h_3$ isolates that surplus head,

\begin{align*}\Delta h &= h_3 - \bigl(h_2+\Delta h_f\bigr) \\ &= h_3 - h_2 - (h_2-h_1)\end{align*}

\Delta h = h_3 - 2h_2 + h_1

(1)

Step 3Bernoulli across the entrance, then evaluate

Between the first riser and the second, the liquid is brought from rest up to the tube’s flow speed $v$ , and the friction line already accounts for the head lost over that length. What is left over is the head that the acceleration itself draws, so energy conservation across the entrance turns the excess (1) directly into the flow’s own kinetic energy per unit weight:

\frac{v^2}{2g} = \Delta h.

Substituting the given depths,

\Delta h = 35 - 2(20) + 10\ \text{cm} = 5\ \text{cm} = 0.05\ \text{m},

v = \sqrt{2(9.8)(0.05)}\ \text{m/s},

v = \sqrt{2g\Delta h} \approx 1.0\ \text{m/s}

(2)

Check

$\Delta h$ comes out positive, $5$ cm, and the sign is guaranteed rather than lucky: the entrance tap reads the head of liquid that has not yet been set moving, so it must stand above the line the two moving-stream taps define, never below it. Dimensionally $[g][\Delta h] = (\text{m/s}^2)(\text{m}) = \text{m}^2/\text{s}^2$ , and its square root is a speed, as required.

Remark

The liquid’s density, given in the statement, never entered the calculation: every quantity here is a piezometric height, an energy per unit weight, so $\rho$ cancels throughout and would only resurface if the question asked for a pressure, a force on the tank, or a mass flow rate instead of a velocity. (A volumetric flow rate would not bring it back either: that is the speed times the bore, and the bore is not given.)