Learn: Orifice discharge
An orifice is a small opening through which a pressure or water-level difference drives flow. The ideal jet speed follows from converting head into kinetic energy; a discharge coefficient corrects the ideal estimate.
What the formula is saying
Compute ideal speed √(2gH), multiply by opening area A, and then by Cd. The coefficient includes contraction and velocity effects for the selected opening arrangement.
Read the symbols in plain language
- Cd
- Discharge coefficient
Discharge coefficient. The empirical coefficient adjusts the ideal flow for the stated orifice configuration.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- A
- Orifice area
Orifice area. Area times velocity converts a speed into a volume rate.
m²Square metres measure area; square the length conversion factor.
- g
- Gravity
Gravity. Taking the square root of twice gravity times head produces the ideal velocity.
m/s²Use m/s² as the base unit shown here. Use area in m², driving head in m and gravity in m/s². The square root gives m/s and multiplying by area gives m³/s. Cd is dimensionless, not a percent unless explicitly converted.
- H
- Head
Head. Taking the square root of twice gravity times head produces the ideal velocity.
mMetres measure length; 1 m = 1000 mm.
- Q
- Result to find
Orifice discharge. The empirical coefficient adjusts the ideal flow for the stated orifice configuration.
m³/s
Sort out the units first
Use area in m², driving head in m and gravity in m/s². The square root gives m/s and multiplying by area gives m³/s. Cd is dimensionless, not a percent unless explicitly converted.
Assumptions before calculating
Assume quasi-steady incompressible flow through an opening small enough for one representative head, with a Cd appropriate to the geometry and head range. For a submerged opening use the appropriate head difference.
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
- A · Orifice area
- 0.01 m²
- g · Gravity
- 9.81 m/s²
- H · Head
- 2 m
Convert driving head into ideal jet speed
Taking the square root of twice gravity times head produces the ideal velocity.
√(2 × (9.81) × (2)) ≈ 6.264183905 m/sMultiply speed by opening area
Area times velocity converts a speed into a volume rate.
(0.01) × (6.264183905) ≈ 0.06264183905 m³/sApply the discharge coefficient
The empirical coefficient adjusts the ideal flow for the stated orifice configuration.
(0.62) × (0.06264183905) ≈ 0.03883794021 m³/s
Does this worked answer make sense?
Four times the head gives twice the discharge at fixed area and coefficient. Twice the opening area gives twice the discharge in the same head conditions.
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
- A · Orifice area
- 0.015 m²
- g · Gravity
- 9.81 m/s²
- H · Head
- 1.5 m
Convert driving head into ideal jet speed
Taking the square root of twice gravity times head produces the ideal velocity.
√(2 × (9.81) × (1.5)) ≈ 5.424942396 m/sMultiply speed by opening area
Area times velocity converts a speed into a volume rate.
(0.015) × (5.424942396) ≈ 0.08137413594 m³/sApply the discharge coefficient
The empirical coefficient adjusts the ideal flow for the stated orifice configuration.
(0.6) × (0.08137413594) ≈ 0.04882448156 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
- 0.64
- A · Orifice area
- 0.008 m²
- g · Gravity
- 9.81 m/s²
- H · Head
- 2.5 m
Find: Learn: Orifice discharge
A hint, not the answer
Compute ideal speed √(2gH), multiply by opening area A, and then by Cd. The coefficient includes contraction and velocity effects for the selected opening arrangement.
Use area in m², driving head in m and gravity in m/s². The square root gives m/s and multiplying by area gives m³/s. Cd is dimensionless, not a percent unless explicitly converted.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Convert driving head into ideal jet speed
Taking the square root of twice gravity times head produces the ideal velocity.
√(2 × (9.81) × (2.5)) ≈ 7.003570518 m/sMultiply speed by opening area
Area times velocity converts a speed into a volume rate.
(0.008) × (7.003570518) ≈ 0.05602856414 m³/sApply the discharge coefficient
The empirical coefficient adjusts the ideal flow for the stated orifice configuration.
(0.64) × (0.05602856414) ≈ 0.03585828105 m³/s
Avoid the common trap
Do not use opening diameter as area, confuse head above the opening centroid with head above its top, or omit Cd when comparing with measured discharge.
When this method applies — and when it does not
The equation does not determine Cd or reservoir-emptying time. Large openings with appreciably varying pressure over their height, free-surface drawdown, air effects and viscosity may require a different or integrated model.
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: Orifice discharge. 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.