A weir measures open-channel discharge by forcing flow over a crest. Head $H$ is measured above the crest, upstream from the drawdown zone.
Use the formula form and coefficient stated in the problem; review sheets may use different constants for metric and English units.
For contracted rectangular weirs, use effective length. For V-notch weirs, discharge varies with $H^{5/2}$.
When water level changes with time, integrate the storage change against the weir discharge relation.
A suppressed rectangular weir is 1.20 m long with head 0.40 m. Use $Q=1.84LH^{3/2}$.
Answer: $Q=0.558\text{ m}^3/\text{s}$.
A rectangular sharp-crested weir is 1.50 m long with two end contractions. The measured head is 0.30 m. Use Francis formula $Q=1.84(L-0.1NH)H^{3/2}$, where $N$ is the number of contractions.
Answer: $Q=0.435\text{ m}^3/\text{s}$.
A 90-degree triangular weir carries water under a head of 0.25 m. Use $Q=1.417H^{5/2}$ for a right-angle V-notch weir. Find the discharge.
Answer: $Q=0.0443\text{ m}^3/\text{s}$.
A Cipolletti weir is 0.90 m long and must pass 0.60 m3/s. Use $Q=1.86LH^{3/2}$. Find the required head.
Answer: Required head is about 0.504 m.
A suppressed rectangular weir 2.0 m long has a measured head of 0.35 m. The actual discharge measured by a volumetric method is 0.420 m³/s. Using the general weir formula $Q = C_w L H^{3/2}$, determine the experimental discharge coefficient $C_w$ and compare it to the standard value of 1.84 for a sharp-crested weir.
Wait — recheck using SI standard form. The formula $Q = C_w L H^{3/2}$ with $C_w \approx 1.84$ already absorbs $\frac{2}{3}C_d\sqrt{2g}$. Check: $\frac{2}{3}(0.62)\sqrt{2(9.81)} = \frac{2}{3}(0.62)(4.429) = 1.826$, confirming $C_d \approx 0.62$ for standard weir.
Hmm — let us recompute $(0.35)^{3/2}$ carefully: $\sqrt{0.35} = 0.5916$, so $(0.35)^{3/2} = 0.35 \times 0.5916 = 0.2071$. Then $C_w = 0.420 / (2.0 \times 0.2071) = 0.420/0.4142 \approx 1.014$.
Rechecking with $C_w = 1.84$: $Q_{theory} = 1.84(2.0)(0.2071) = 0.762$ m³/s, which is much larger. This means the measured 0.420 m³/s corresponds to a head lower than 0.35 m, OR the weir is submerged. Let us instead find the head that gives Q = 0.420 using standard $C_w = 1.84$:
Answer: With standard $C_w = 1.84$, the actual head for Q = 0.420 m³/s over a 2.0 m weir is H = 0.235 m — not 0.35 m. If the stated head of 0.35 m is correct, the back-calculated coefficient is $C_w = 0.420/(2.0 \times 0.2071) \approx 1.014$, significantly lower than 1.84. A low coefficient indicates a broad-crested weir or submerged conditions rather than a sharp-crested free-flow weir.
A rectangular reservoir has a constant plan area of 1200 m². It discharges over a suppressed sharp-crested weir 3.0 m long ($C_w = 1.84$). The initial head over the weir crest is 0.60 m and the final head is 0.15 m. Determine the time in minutes for the water level to drop from the initial head to the final head. The weir crest is at the base of the pool — as the head drops, discharge decreases.
For an unsteady reservoir with weir outflow: $A_s \frac{dH}{dt} = -Q_{weir} = -C_w L H^{3/2}$. Separating variables and integrating:
Answer: The reservoir level takes about 9.36 minutes to drop from 0.60 m to 0.15 m head above the weir crest. Note the weir emptying formula uses $1/\sqrt{H}$ terms (inverse square root), unlike the orifice emptying formula which uses $\sqrt{H}$ terms — because weir flow varies as $H^{3/2}$ not $H^{1/2}$.
A 90° triangular (V-notch) weir must discharge 0.250 m³/s. Using the formula $Q = 1.417 H^{5/2}$ (metric, in m³/s), find the required head $H$ over the notch vertex. Then find the percentage increase in discharge if the head increases by 10%.
Compute: $(0.1764)^{0.4}$. Use $\ln(0.1764) = -1.733$, then $0.4 \times (-1.733) = -0.693$, so $e^{-0.693} = 0.500$ m.
Verify: $Q = 1.417(0.500)^{5/2} = 1.417(0.1768) = 0.250$ m³/s ✓
For a 10% increase in head: $H' = 1.10H$. Since $Q \propto H^{5/2}$:
Answer: Required head is H = 0.500 m. A 10% increase in head produces a 26.9% increase in discharge — the V-notch weir is very sensitive to head changes because of the $H^{5/2}$ relationship, making it ideal for measuring small discharges accurately.
A compound weir consists of a central V-notch (90°, $Q_v = 1.417 H_v^{5/2}$) flanked on both sides by rectangular suppressed sections. The rectangular sections together have a total effective length of 4.0 m and use $Q_r = 1.84 L H_r^{3/2}$. The V-notch vertex is 0.30 m below the rectangular crest. When total head above the rectangular crest is $H = 0.45$ m, find the total discharge through the compound weir.
The V-notch head is measured from its vertex (0.30 m lower), so $H_v = H + 0.30 = 0.45 + 0.30 = 0.75$ m. The rectangular sections see head $H_r = 0.45$ m.
Answer: Total discharge is 2.91 m³/s. The rectangular sections dominate at higher heads. Compound weirs combine the low-flow sensitivity of a V-notch with the high-flow capacity of a rectangular weir, making them practical for streams with highly variable flows.
A sharp-crested suppressed weir 2.5 m long has an upstream head $H_1 = 0.60$ m and a downstream tailwater head $H_2 = 0.40$ m above the weir crest (both measured from the weir crest). Using the Villemonte submergence formula $Q_s = Q_{free}\left[1 - \left(\frac{H_2}{H_1}\right)^{3/2}\right]^{0.385}$ and the standard free-flow formula $Q_{free} = 1.84 L H_1^{3/2}$, compute the submerged discharge and the percentage reduction from free-flow conditions.
Answer: Submerged discharge is 1.578 m³/s, a 26.2% reduction from the free-flow rate of 2.138 m³/s. When the tailwater ratio $H_2/H_1 \geq 0.67$, submergence effects become significant and must be accounted for in weir design and flow measurement.