Irodov 1.328: Only the Turned Flux Has Leverage – Amit & Deepali Sinha Foundation

Problem Irodov 1.328

A large tank of water drains through a tube of uniform inside radius $r = 0.50$ cm attached to its side. The tube runs horizontally away from the tank for a length $l = 22$ cm, then turns through a right angle and runs vertically downward, and the water leaves through its open lower end. The volume flow rate is $Q = 0.50$ liters per second. Let $O$ be the point where the tube meets the tank, at the upstream end of the horizontal run. Find the moment about $O$ of the reaction forces that the flowing water exerts on the walls of the tube.

Water leaves the tank along a horizontal run of length l, turns through a right angle, and falls from the open lower end with speed v. Here l runs from the joint O to the vertical run's axis. — Water leaves the tank along a horizontal run of length $l$ , turns through a right angle, and falls from the open lower end with speed $v$ . Here $l$ runs from the joint $O$ to the vertical run’s axis.

Given

$r = 0.50$ cm
$l = 22$ cm
$Q = 0.50$ liters/s
$\rho = 1.00\times10^3$ kg/m $^3$

Find

Find

N

, the moment about

O

of the reaction forces the flowing water exerts on the tube’s walls

Solution

Step 1Speed of the water in the tube

The bore is the same everywhere along the tube, so the same volume crosses every section each second and one continuity statement fixes the speed.

Q = Sv = \pi r^2 v.

v = \dfrac{Q}{\pi r^2}

(1)

The control volume: the water in the tube between the inlet at O and the open outlet. The incoming flux ρ Q v_1 lies on a line through O; the outgoing flux ρ Q v_2 lies a distance l away. — The control volume: the water in the tube between the inlet at $O$ and the open outlet. The incoming flux $\rho Q\vect v_1$ lies on a line through $O$ ; the outgoing flux $\rho Q\vect v_2$ lies a distance $l$ away.

Step 2Moment-of-momentum balance for the water in the tube

Take as control volume all the water between the inlet section at $O$ and the open outlet, and write the moment-of-momentum theorem about $O$ for it: the total moment of the external forces equals the rate at which angular momentum accumulates inside plus the rate at which it is carried out through the boundary.

\begin{aligned} \sum\vect M_O &= \frac{\partial}{\partial t}\int_V (\vect r\times\rho\vect u)\,dV + \oint_S (\vect r\times\vect u)\,\rho\,\vect u\cdot d\vect S, \\ \sum\vect M_O &= \rho Q\,(\vect r_2\times\vect v_2) - \rho Q\,(\vect r_1\times\vect v_1). \end{aligned} \tag{2}

The first term drops because the flow is steady, and the surface integral collects nothing along the walls, where the water does not cross the boundary, leaving only the two end sections, each carrying mass at the rate $\rho Q$ with the uniform velocity of its own section.

Step 3Only the outgoing flux carries an arm about

O

Put the origin at $O$ , with $\vect i$ along the horizontal run pointing away from the tank and $\vect j$ vertically upward, so that $\vect k = \vect i\times\vect j$ points out of the page. The inlet section sits at $O$ itself; the outlet sits a distance $l$ along $\vect i$ and a distance $h$ below, $h$ being the length of the vertical run.

\begin{align*}\vect r_1 &= \vect 0, & \vect v_1 &= v\,\vect i, \\ \vect r_2 &= l\,\vect i - h\,\vect j, & \vect v_2 &= -v\,\vect j.\end{align*}

Forming the two cross products term by term gives the angular momentum carried in and out.

\begin{align*}\vect r_1\times\vect v_1 &= \vect 0, \\ \vect r_2\times\vect v_2 &= (l\,\vect i - h\,\vect j)\times(-v\,\vect j) \\ &= -lv\,(\vect i\times\vect j) + hv\,(\vect j\times\vect j) \\ &= -lv\,\vect k.\end{align*}

Substituting both into (2),

\sum\vect M_O = -\rho Q l v\,\vect k.

Step 4The wall reaction is the only moment left

The external forces on the water in the control volume come in three kinds: the pressure on the two end sections, the water’s own weight, and the push of the walls. Measuring pressures from atmospheric, the force on the inlet section is uniform over that section, so its resultant passes through $O$ and has no arm, and the open outlet carries no such force at all; the weight contributes a static moment that contains no $Q$ . What survives is the moment the walls exert on the water. Set aside, with the weight, the static part of that wall moment which balances it, and write $\vect M_w$ for the flow-borne part that remains; the balance and Newton’s third law then read

\begin{align*}\vect M_w &= -\rho Q l v\,\vect k, \\ \vect N &= -\vect M_w.\end{align*}

\vect N = \rho Q l v\,\vect k

(3)

Out of the page is counterclockwise in the figure’s plane, so the reaction tends to swing the bent end of the tube upward about $O$ , which is the tube’s share of turning the stream downward.

Step 5Evaluate

Substituting the speed of (1) into the magnitude of (3), with every quantity converted to SI units and the density of water taken from the tables,

N = (1.00\times10^3)(0.22)\, \frac{(0.50\times10^{-3})^2}{\pi\,(0.50\times10^{-2})^2}\ \text{N}\cdot\text{m}.

N = \dfrac{\rho l Q^2}{\pi r^2} \approx 0.70\ \text{N}\cdot\text{m}

(4)

Check

The units come out right: $(\text{kg/m}^3)(\text{m})(\text{m}^3/\text{s})^2/\text{m}^2 = \text{kg}\cdot\text{m}^2/\text{s}^2$ , a newton meter. The number checks by a second route as well. The stream is turned from horizontal to vertical at the bend, so the vertical part of the force the water delivers there is $\rho Qv$ , which is $3.2$ N here, and its arm about $O$ is the whole horizontal run of $22$ cm; the product of those two is $0.70$ N $\cdot$ m, the answer above.

Remark

The answer cannot be separated from the point it is taken about, and among the points of the tube itself the bend is the one where it collapses. Seen from the corner, the incoming momentum flux runs along the horizontal leg and the outgoing flux runs along the vertical leg, so each position vector is parallel to its own velocity, both cross products vanish, and the elbow feels a force but exerts no moment about itself. The whole of $N$ is therefore the arm $l$ that the tank joint lends to the vertical, turned part of that force, while the horizontal part acts along the tube’s own axis, straight through $O$ , and contributes nothing at all. Away from the tube the vanishing is not unique: the dynamic moment is zero at every point of the resultant’s own line of action, the $45$ degree line through the corner along which $\rho Qv(\vect i+\vect j)$ acts, and the bend is simply where that line crosses the tube. Both statements lean on this particular geometry, the outlet directly below the bend and the inlet flux along the horizontal axis; tilt either leg and each of them has to be redrawn.