Learn: Rectangular sharp-crested weir
A rectangular sharp-crested weir estimates discharge from upstream head above its crest. The width of flowing water and jet speed both contribute to the head exponent.
What the formula is saying
Raise H to the power 3/2, multiply by crest width b and √(2g), then multiply by (2/3)Cd. The fractional power comes from adding the discharge of horizontal strips through the opening.
Read the symbols in plain language
- Cd
- Discharge coefficient
Discharge coefficient. Combine the head, geometry, gravity and discharge-coefficient contributions.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- b
- Weir width
Weir width. Use the opening width that belongs to the supplied discharge coefficient.
mMetres measure length; 1 m = 1000 mm.
- g
- Gravity
Gravity. Combine the head, geometry, gravity and discharge-coefficient contributions.
m/s²Use m/s² as the base unit shown here. H and b are in m, g is m/s² and Cd is dimensionless. b is the effective rectangular opening width, not the channel depth. The result is m³/s.
- H
- Head over crest
Head over crest. Apply the correct fractional exponent before multiplying by the remaining factors.
mMetres measure length; 1 m = 1000 mm.
- Q
- Result to find
Rectangular sharp-crested weir. Combine the head, geometry, gravity and discharge-coefficient contributions.
m³/s
Sort out the units first
H and b are in m, g is m/s² and Cd is dimensionless. b is the effective rectangular opening width, not the channel depth. The result is m³/s.
Assumptions before calculating
Assume a sharp-crested free-flow weir with a ventilated nappe, negligible approach-velocity correction in this simplified form, and a discharge coefficient appropriate to the installation and measurement range.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find Q and explain the result in the stated output unit.
- Cd · Discharge coefficient
- 0.62
- b · Weir width
- 1 m
- g · Gravity
- 9.81 m/s²
- H · Head over crest
- 0.3 m
Identify the effective rectangular width
Use the opening width that belongs to the supplied discharge coefficient.
(1) = 1 mEvaluate the head power
Apply the correct fractional exponent before multiplying by the remaining factors.
(0.3)^1.5 ≈ 0.1643167673 m^(3/2)Calculate the weir discharge
Combine the head, geometry, gravity and discharge-coefficient contributions.
(2 ÷ 3) × (0.62) × (1) × √(2 × (9.81)) × (0.1643167673) ≈ 0.3008373913 m³/s
Does this worked answer make sense?
Doubling H multiplies predicted flow by 2^(3/2), about 2.83 at unchanged geometry and Cd. This strong sensitivity makes accurate head measurement important.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- Cd · Discharge coefficient
- 0.6
- b · Weir width
- 1.5 m
- g · Gravity
- 9.81 m/s²
- H · Head over crest
- 0.4 m
Identify the effective rectangular width
Use the opening width that belongs to the supplied discharge coefficient.
(1.5) = 1.5 mEvaluate the head power
Apply the correct fractional exponent before multiplying by the remaining factors.
(0.4)^1.5 ≈ 0.2529822128 m^(3/2)Calculate the weir discharge
Combine the head, geometry, gravity and discharge-coefficient contributions.
(2 ÷ 3) × (0.6) × (1.5) × √(2 × (9.81)) × (0.2529822128) ≈ 0.6723427697 m³/s
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- Cd · Discharge coefficient
- 0.63
- b · Weir width
- 0.8 m
- g · Gravity
- 9.81 m/s²
- H · Head over crest
- 0.25 m
Find: Learn: Rectangular sharp-crested weir
A hint, not the answer
Raise H to the power 3/2, multiply by crest width b and √(2g), then multiply by (2/3)Cd. The fractional power comes from adding the discharge of horizontal strips through the opening.
H and b are in m, g is m/s² and Cd is dimensionless. b is the effective rectangular opening width, not the channel depth. The result is m³/s.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Identify the effective rectangular width
Use the opening width that belongs to the supplied discharge coefficient.
(0.8) = 0.8 mEvaluate the head power
Apply the correct fractional exponent before multiplying by the remaining factors.
(0.25)^1.5 = 0.125 m^(3/2)Calculate the weir discharge
Combine the head, geometry, gravity and discharge-coefficient contributions.
(2 ÷ 3) × (0.63) × (0.8) × √(2 × (9.81)) × (0.125) ≈ 0.1860367706 m³/s
Avoid the common trap
Do not measure H at the falling nappe where drawdown occurs; use the prescribed upstream measurement location. Do not use the V-notch exponent 5/2 or omit the crest-width multiplier.
When this method applies — and when it does not
This is a coefficient-based teaching equation, not a substitute for a calibrated installation standard. End contractions, crest thickness and approach conditions affect the applicable coefficient and effective width. Submergence, debris, unventilated flow and very small heads can invalidate the assumptions.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Rectangular sharp-crested weir. Water-measurement principles; for discharge devices, read the orifice/weir chapters and the installation and head-measurement conditions, not only the coefficient formula.
- U.S. Bureau of Reclamation — Water Measurement Manual, 3rd edition (1997; revised reprint 2001)
- Dawei Han, University of Bristol — Concise Hydraulics (2008, Ventus Publishing)
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.