Kvant Physics Problem 11
The system consists of three open barrels containing water.
Verified: yes
Verdicts: PASS + PASS
Solve time: 2m56s
Source on kvant.digital
Problem
Three open barrels are filled with water and placed at different heights (see Fig. 3). Tubes are led upward from each barrel and joined together. The tubes are also filled with water. In which direction will the water flow through the tubes if valves $K_1$, $K_2$, and $K_3$ are opened simultaneously?
Fig. 3
Setup and Assumptions
The system consists of three open barrels containing water. The free surfaces of the water in the barrels are located at different heights. Each barrel is connected to a common junction by a tube that first rises upward and then joins the other tubes. The tubes are completely filled with water. Valves $K_1$, $K_2$, and $K_3$ initially isolate the barrels from one another and are opened simultaneously.
The unknown is the direction of water flow immediately after the valves are opened.
Let the heights of the free water surfaces above an arbitrary horizontal reference level be $h_1$, $h_2$, and $h_3$, corresponding to the left, middle, and right barrels. From the figure,
$$h_1 > h_2 > h_3.$$
The water is treated as an incompressible liquid of density $\rho$ (kg/m$^3$). Atmospheric pressure is $p_0$ (Pa). Viscous losses, surface tension effects, and transient wave phenomena are neglected. The analysis concerns the pressure distribution immediately after the valves are opened.
Physical Principles
The solution rests on the law of hydrostatic pressure.
For a liquid at rest,
$$p+\rho g z=\text{const},$$
where $p$ is the pressure, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $z$ is the height coordinate.
Since every barrel is open to the atmosphere, the pressure at each free surface equals
$$p_0.$$
Consequently, at any point connected to a given barrel by a continuous column of water,
$$p=p_0+\rho g(h-z),$$
where $h$ is the height of the free surface in that barrel.
Fluid flows from regions of higher pressure toward regions of lower pressure.
Derivation
Consider the common junction where the three tubes meet. Let its height above the reference level be $H$.
Before the valves are opened, each tube contains a stationary water column connected only to its own barrel. The pressure that would exist at the junction end of each tube is determined by the hydrostatic relation.
For the tube connected to barrel 1,
$$p_1=p_0+\rho g(h_1-H).$$
For the tube connected to barrel 2,
$$p_2=p_0+\rho g(h_2-H).$$
For the tube connected to barrel 3,
$$p_3=p_0+\rho g(h_3-H).$$
Subtracting these pressures gives
$$p_1-p_2=\rho g(h_1-h_2),$$
and
$$p_2-p_3=\rho g(h_2-h_3).$$
Since
$$h_1>h_2>h_3,$$
it follows that
$$p_1>p_2>p_3.$$
The height $H$ of the junction cancels from the pressure differences. The fact that the tubes rise upward does not alter the comparison of pressures. Only the heights of the free surfaces matter.
Immediately after the valves are opened, the highest pressure is supplied by barrel 1 and the lowest pressure by barrel 3. Water must move away from barrel 1 through the junction. At the same time, water enters both barrels 2 and 3 because their pressures are lower than that supplied by barrel 1.
The pressure difference between barrels 2 and 3 is also positive:
$$p_2-p_3=\rho g(h_2-h_3)>0.$$
Hence water tends to move from barrel 2 toward barrel 3 as well.
The resulting motion is that water from the highest barrel feeds the two lower barrels, with the ultimate tendency for water levels to equalize.
Result
The pressure at the junction associated with a barrel whose free surface is at height $h_i$ is
$$p_i=p_0+\rho g(h_i-H).$$
Since
$$h_1>h_2>h_3,$$
the pressures satisfy
$$p_1>p_2>p_3.$$
Thus the water flows away from the barrel with the highest water level and toward the barrels with lower water levels.
The qualitative flow directions are
$$1 \rightarrow 2, \qquad 1 \rightarrow 3, \qquad 2 \rightarrow 3.$$
There are no numerical values to substitute.
$$\boxed{\text{Water flows from the highest barrel toward the other two barrels, ultimately from higher levels to lower levels.}}$$
Sanity Checks
The hydrostatic expression $p=p_0+\rho g(h-z)$ has dimensions of pressure because $\rho g h$ has units
$$\left(\frac{\text{kg}}{\text{m}^3}\right) \left(\frac{\text{m}}{\text{s}^2}\right) (\text{m}) = \frac{\text{kg}}{\text{m},\text{s}^2} = \text{Pa}.$$
If all three free surfaces were at the same height, then
$$h_1=h_2=h_3,$$
which would give
$$p_1=p_2=p_3.$$
No flow would occur, which agrees with physical intuition.
If the junction were raised even higher, the quantity $H$ would increase in every pressure expression by the same amount. The pressure differences would remain unchanged, so the flow direction would not change. This confirms that the upward bends of the tubes do not determine the motion.
The most common error is to assume that water cannot flow upward through a tube. The correct comparison is between the hydrostatic pressures generated by the free surfaces. The term $H$ cancels when pressure differences are formed, and the barrel with the highest free surface always produces the greatest pressure at the junction.