Kvant Physics Problem 11

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m02s
Source on kvant.digital

Problem

Three open barrels are filled with water and are installed at different heights (see Fig. 3). From each barrel, pipes are led upward and connected together. The pipes are also filled with water. Where will the water flow through the pipes if taps $K_1$, $K_2$ and $K_3$ are opened simultaneously?

Fig. 3

Setup and Assumptions

Three large open barrels contain water and have free surfaces located at fixed heights $H_1$, $H_2$, $H_3$ above a common reference level, with $H_1 > H_2 > H_3$, measured in meters. Each barrel is open to the atmosphere so the pressure at every free surface equals $p_0 = 1.01 \times 10^5\ \text{Pa}$. From each barrel, a pipe is attached below the free surface and then routed upward, after which all pipes are connected together at a common junction. The pipes are completely filled with water of density $\rho = 1000\ \text{kg/m}^3$ and are initially at rest.

Taps $K_1$, $K_2$, and $K_3$ are initially closed and are opened simultaneously at time $t = 0$. The pipes are assumed rigid, of negligible diameter variation, and losses due to viscosity and turbulence are neglected except insofar as they determine the direction of flow. The fluid is incompressible and the flow is quasi-steady after the initial transient. The gravitational acceleration is $g = 9.81\ \text{m/s}^2$. The problem reduces to determining the direction of flow in each pipe immediately after opening and during the subsequent redistribution.

Physical Principles

The hydrostatic pressure in a connected static fluid column is given by $p = p_0 + \rho g (H - z)$, where $H$ is the free surface height of the connected reservoir and $z$ is the elevation of the point considered.

Bernoulli’s equation for steady, incompressible, inviscid flow along a streamline is

$p + \frac{1}{2}\rho v^2 + \rho g z = \text{const}.$

At the initial moment of opening, velocities are negligible, so the pressure field is governed by hydrostatic head differences. Fluid flows from regions of higher total hydraulic head $p/\rho g + z$ toward regions of lower hydraulic head.

At a junction where multiple pipes meet, continuity of pressure is established rapidly, and the direction of flow in each branch is determined by comparison of the hydraulic heads imposed by each reservoir.

Derivation

Each barrel imposes at its pipe connection a hydraulic head equal to its free surface height, since at the free surface $p = p_0$ and therefore the total head is

$\frac{p_0}{\rho g} + H_i.$

At the common junction, denote the instantaneous pressure as $p_J$. The corresponding hydraulic head at the junction is

$\frac{p_J}{\rho g} + z_J,$

where $z_J$ is the elevation of the junction point.

A flow from barrel $i$ into the junction occurs if the total head supplied by barrel $i$ exceeds the head at the junction:

$H_i > \frac{p_J}{\rho g} + z_J - \frac{p_0}{\rho g}.$

Since all pipes are initially filled and connected, the junction pressure adjusts such that mass conservation is satisfied, and the system evolves toward a configuration where no net flow occurs. In that final configuration, all connected points share the same hydraulic head, so the equilibrium condition is

$H_1^{(\text{final})} = H_2^{(\text{final})} = H_3^{(\text{final})}.$

During the transient, the ordering of heads is preserved, so the direction of flow is determined solely by comparison of $H_i$. For $H_1 > H_2 > H_3$, the pressure supplied by barrel 1 exceeds the junction pressure, so water flows from barrel 1 into the pipe network. The junction distributes this inflow toward barrels with lower heads. Since barrel 3 has the lowest head, its pipe corresponds to the lowest hydraulic potential, so flow enters barrel 3. Barrel 2 has intermediate head, so it receives flow from barrel 1 and simultaneously discharges to barrel 3 depending on instantaneous junction pressure, but the net direction is from barrel 1 toward barrel 3.

Thus the system behaves as a hydraulic network in which barrel 1 acts as a source, barrel 3 acts as a sink, and barrel 2 serves as an intermediate node whose net flow direction depends on the instantaneous junction pressure but is not the ultimate source.

Result

The flow direction is determined by the ordering of free surface heights. Water flows from the barrel with the highest free surface level to the barrel with the lowest free surface level through the connecting pipes. The intermediate-height barrel receives water from the highest barrel and delivers water to the lowest barrel until pressure equilibration is reached.

For $H_1 > H_2 > H_3$, the net flow pattern is

$1 \rightarrow 2 \rightarrow 3.$

In hydraulic head terms, the driving potential difference is

$\Delta H_{13} = H_1 - H_3,$

so the dominant transfer is from barrel 1 to barrel 3.

$$\boxed{\text{Water flows from the highest barrel to the lowest barrel: } 1 \rightarrow 3,\ \text{with barrel 2 acting as an intermediate conduit.}}$$

Sanity Checks

Dimensional consistency follows from the fact that all driving quantities reduce to differences in height, each measured in meters, so the hydraulic head difference $H_i - H_j$ has dimension length and corresponds to pressure difference $\rho g (H_i - H_j)$ in pascals.

If all barrels are placed at equal height so that $H_1 = H_2 = H_3$, the pressure at all connections becomes identical, eliminating any driving head difference and producing zero flow, consistent with the equilibrium condition.

If one barrel is raised infinitely high, $H_1 \to \infty$, the pressure contribution $\rho g H_1$ dominates all finite terms, forcing flow from that barrel into the network regardless of junction height, which matches the physical interpretation of hydrostatic head dominance.

The most sensitive error arises from confusing local pipe elevation with free surface height; only the free surface elevation determines the effective hydraulic head because pressure at the free surface is fixed at atmospheric pressure.