Introduction
A V-shaped trough narrows to a point at its bottom. As water flows in, the water’s surface gets wider as well as higher. This changing width means the cross-sectional area is not simply a fixed width multiplied by the depth.
Similar triangles describe how the width changes with depth. Once that width is expressed in terms of depth, the cross-sectional area and the volume follow. Differentiating the volume relation then connects the inflow rate to the rate at which the water depth changes.
Geometry of the Trough
Consider a straight trough with a V-shaped, triangular cross-section. Let its length be \(L\). Choose a reference depth \(H\), and let \(B\) be the width of the trough at that depth. The trough has the same cross-section along its entire length. At time \(t\), let \(h(t)\) denote the water depth, where \(0\le h(t)\le H\).
The water occupies a smaller triangle similar to the full reference triangle. Its height is \(h\), while the reference triangle’s height is \(H\). If \(w\) is the width of the water’s surface, the ratios of corresponding sides are equal.
Solving for the water-surface width gives \(w=(B/H)h\). The cross-section of the water is triangular, so its area is one-half the product of its base and height.
Because this cross-section extends uniformly along a trough of length \(L\), its volume is the cross-sectional area multiplied by \(L\).
The dimensions \(L\), \(B\), and \(H\) are fixed. The volume \(V\) changes because the depth \(h\) changes. In this model, \(L\), \(B\), and \(H\) are lengths, so \(LB/(2H)\) has units of length and \(V\) has units of volume.
Relating Inflow to Depth Change
Let \(Q(t)\) denote the volume of water flowing into the trough per unit time. When there is no outflow or other source of volume change, \(Q(t)=dV/dt\). Its units are volume per unit time, such as cubic metres per second.
At a positive water depth, differentiate the volume formula with respect to time. Since \(V\) depends on time through \(h(t)\), the chain rule applies.
The equation shows why the same inflow does not produce the same depth rate at every level. The factor \(h\) multiplies the depth rate. A deeper cross-section has more width and area available to receive the incoming volume, so the depth rises more slowly for the same positive inflow.
Why the Rate Equation Works
The relation between volume and depth is derived from the geometry, not guessed from the shape of the container. The derivation below also explains exactly where the positive-depth condition enters.
At depth \(h\), the water width is \(w=(B/H)h\), since the water triangle and reference triangle are similar.
The water cross-section has area \(\frac12wh=\frac{B}{2H}h^2\). Multiplying by the constant trough length gives \(V=\frac{LB}{2H}h^2\).
With fixed \(L\), \(B\), and \(H\), the chain rule gives \(\frac{dV}{dt}=\frac{LB}{H}h\frac{dh}{dt}\). Equating \(dV/dt\) to the inflow \(Q(t)\) gives the rate equation.
To isolate \(dh/dt\), divide by \((LB/H)h\). This division requires \(h>0\), along with positive fixed dimensions \(L\), \(B\), and \(H\).
Worked Example: Finding the Rate of Rise
Worked Example: Finding the Rate of Rise
A trough is \(6\) m long. Its width is \(2\) m at a reference depth of \(1.5\) m. Water is entering at \(0.012\) m\(^3\)/s when the water depth is \(0.5\) m. Find the depth rate at that instant.
First identify the fixed dimensions and the current values:
The water depth is positive, so the rate equation can be solved for \(dh/dt\):
Substitute the values, showing the denominator first:
Then calculate the rate:
Check: Substitute this rate into the original rate equation. The geometric factor is \((LB/H)h=(6)(2)(0.5)/(1.5)=4\) m\(^2\), so
The depth is increasing at \(0.003\) m/s at that instant.
Worked Example: Finding the Depth After a Given Time
Worked Example: Finding the Depth After a Given Time
A trough has \(L=10\) m, \(B=3\) m, and \(H=2\) m. It starts empty and receives a constant inflow of \(Q=0.015\) m\(^3\)/s. Find the water depth after \(180\) s, and then find the depth rate at that time.
Since the inflow is constant and there is no outflow, the volume after \(t\) seconds is \(V=Qt\). Equate this to the geometric volume formula:
Solving for the nonnegative depth gives \(h=\sqrt{2HQt/(LB)}\). At \(t=180\) s, the numerator inside the square root is
The denominator is \(LB=(10)(3)=30\). Therefore,
Now use the positive-depth rate equation to find the instantaneous rate:
The approximation is rounded to three significant figures. To check the depth, substitute \(h=0.6\) m in the volume formula. The coefficient is \(LB/(2H)=(10)(3)/(2(2))=7.5\) m, so \(V=(7.5)(0.6)^2=(7.5)(0.36)=2.7\) m\(^3\). The inflow over \(180\) s is \(Qt=(0.015)(180)=2.7\) m\(^3\), as required.
To check the rate, \((LB/H)h=(30/2)(0.6)=9\) m\(^2\), and \(9(1/600)=0.015\) m\(^3\)/s, equal to the stated inflow.
Worked Example: Finding the Inflow from a Measured Rise
Worked Example: Finding the Inflow from a Measured Rise
A trough has \(L=8\) m, \(B=2.4\) m, and \(H=1.2\) m. At a moment when the water depth is \(0.3\) m, its depth is increasing at \(0.004\) m/s. Find the inflow rate.
Rearrange the rate equation to calculate \(Q\), then substitute the given values:
Compute the factors in order:
Thus,
Check: The water width at this depth is \(w=(B/H)h=(2.4/1.2)(0.3)=0.6\) m. Its cross-sectional area is \(\frac12(0.6)(0.3)=0.09\) m\(^2\), and its volume is \((8)(0.09)=0.72\) m\(^3\). The volume formula gives the same value: \(LB/(2H)=(8)(2.4)/(2(1.2))=8\) m, and \((8)(0.3)^2=0.72\) m\(^3\). Differentiating that formula gives \(dV/dt=(LB/H)h(dh/dt)=16(0.3)(0.004)=0.0192\) m\(^3\)/s, agreeing with the calculated inflow.
The Initially Empty Trough
The solved formula \(dh/dt=HQ/(LBh)\) contains division by \(h\), so it is defined only when \(h>0\). At \(h=0\), the solved formula is undefined, and for positive inflow the depth need not have a finite derivative at the empty state. Use the volume relation \(V=(LB/(2H))h^2\) there; for constant \(Q>0\) starting from empty, \(h(t)=\sqrt{2HQ t/(LB)}\). The rate equation \(Q=(LB/H)h\,dh/dt\) applies at positive depths.
This distinction is important. The volume relation remains meaningful at zero depth, while the ordinary finite-rate equation for \(dh/dt\) obtained by division does not. For the constant-inflow case, \(h(t)\) is proportional to \(\sqrt{t}\), whose right derivative at \(t=0\) is not finite. Do not substitute \(h=0\) into the formula for \(dh/dt\) or claim that it gives a finite initial depth rate.
A Common Geometric Mistake
A tempting but incorrect approach is to treat the water cross-section as a rectangle of fixed width \(B\). For example, if \(B=2\) m and \(h=0.5\) m, that approach would give an area \(Bh=(2)(0.5)=1\) m\(^2\). But \(B\) is the width at reference depth \(H\), not necessarily the width at the current water level.
If \(H=1.5\) m, similar triangles give the actual water width as \(w=(B/H)h=(2/1.5)(0.5)=2/3\) m. The cross-section is triangular, so its area is
The incorrect rectangular calculation overestimates the area by treating the water as though it extended across the trough’s full reference width from the bottom up. The correction is to find the water-surface width at depth \(h\) using similar triangles, then use the triangular-area formula.
Interpreting the Model
The factor \(h\) in \(Q=(LB/H)h(dh/dt)\) has a direct geometric meaning: as the water rises, its surface width increases, and each additional rise in depth adds more cross-sectional area than it did at lower depths. Consequently, for a fixed positive inflow, the depth rate decreases as \(h\) increases.
The model assumes a straight trough with a uniform V-shaped cross-section, fixed dimensions, and no loss or outflow. It describes the volume below the reference depth \(H\). If the actual trough changes shape along its length or has a changing cross-section, this particular volume formula does not apply.
Check Your Understanding
Use the geometry and rate relationships from this tutorial to answer the following questions.
- A trough has reference width \(B\) at depth \(H\). What is the water-surface width at depth \(h\), and which geometric fact gives that relation?
- For a uniform trough of length \(L\), derive the volume formula from the triangular cross-sectional area.
- Why must \(h>0\) to use the solved formula for \(dh/dt\)?
- A trough has \(L=4\) m, \(B=1.5\) m, and \(H=1\) m. Write the volume formula as a function of \(h\).
- For constant positive inflow starting from an empty trough, why should the depth-rate formula obtained by division not be applied at \(t=0\)?
