Motivation #

In this article we will derive the orifice discharge equation from Bernoulli’s equation and see what changes when we alter the opening shape or the water conditions on either side.

Most hydraulic engineers know Q_o=C_dA_o\sqrt{2gH} . The less obvious question is which head belongs in it. For an opening discharging into air, we measure H from the upstream water surface to the opening. If the downstream water rises over the opening, we need the difference between the two water levels instead. Use the wrong head and the arithmetic can be flawless while the discharge is wrong.

We will start with the steady-flow energy equation, take the ideal Bernoulli case, and then put the real outlet conditions back in.

General equations #

Energy and continuity #

For steady, incompressible flow from section 1 to section 2, we write the energy equation per unit weight as

\displaystyle \frac{p_1}{\gamma}+z_1+\alpha_1\frac{V_1^2}{2g}=\frac{p_2}{\gamma}+z_2+\alpha_2\frac{V_2^2}{2g}+h_L.

Every term is a head, measured in metres of water. Here p is pressure, \gamma=\rho g is the specific weight of water, z is elevation above a datum, V is mean velocity, and g is gravitational acceleration. The factor \alpha allows for a non-uniform velocity distribution across a section. Friction and local disturbances between the sections appear as the head loss h_L .

We also need continuity to turn velocity into discharge:

\displaystyle Q=AV,

Here A is the flow area. The energy equation gives the velocity available from the head, and continuity gives the corresponding flow rate.

Baseline: a circular outlet discharging into air #

First, take a wide, open pond with a circular hole in its outlet wall. We put point 1 at the pond surface and point 2 in the free jet at the opening centre.

To find the ideal outlet velocity, we assume:

• Equal pressure: both points are exposed to air, so p_1=p_2=p_{atm} and their pressure heads cancel. On a gauge-pressure scale, both are zero.
• Negligible surface velocity: the pond surface area is much larger than the opening area, so continuity gives V_1\approx0 .
• No losses in the ideal calculation: we set h_L=0 and take \alpha_1\approx\alpha_2\approx1 . A real sharp-edged opening still has jet contraction and turbulence; we will account for those below.

The energy equation then reduces to

\displaystyle H=z_1-z_2=\frac{V_{2,ideal}^2}{2g},\qquad V_{2,ideal}=\sqrt{2gH}.

Here H is the vertical distance from the pond surface to the opening centre. The resulting V_{2,ideal}=\sqrt{2gH} is Torricelli’s velocity.

Multiply velocity by opening area to get ideal discharge. For a real opening, we include a discharge coefficient:

\displaystyle Q_o=C_dA_o\sqrt{2gH}.

The dimensionless C_d is actual discharge divided by ideal discharge for the stated opening and head. It captures jet contraction and other departures from the ideal velocity, and comes from hydraulic data or the prescribed design method. We set h_L=0 to derive the ideal velocity; C_d then represents the real opening. If a particular loss is already included in C_d , adding it again would count the same effect twice.

The US Bureau of Reclamation’s Water Measurement Manual, Chapter 2 §8 also derives ideal orifice flow and discusses jet contraction.

Geometrical cases #

We will change the opening shape while keeping the same open pond, negligible approach velocity and free jet. Both sides of the opening are at atmospheric pressure. The height through which the jet falls after leaving the wall does not enter the head; for free discharge, we measure head to the opening. If it is nearly uniform over that opening, shape enters through area:

\displaystyle Q_o=C_dA_o\sqrt{2gH}.

Circular opening: diameter sets the area #

For the circular opening of diameter D , area and discharge under nearly uniform head are

\displaystyle A_o=\frac{\pi D^2}{4},\qquad Q_o=C_d\frac{\pi D^2}{4}\sqrt{2gH}.

This familiar calculation treats head as nearly constant from the top of the circle to the bottom.

Rectangular opening: keep its area and head variation together #

For a rectangle of width b and height a , the opening area is A_o=ba . We can use that area with one head if the head barely changes over its height. For a tall opening, the bottom has more head than the top. Let y be height above the bottom edge and H_b the head from the pond surface to that edge. A strip at y has head h(y)=H_b-y . The opening shown is completely below the upstream water surface, so H_b>a .

A horizontal strip has area b\,dy and ideal speed \sqrt{2g(H_b-y)} . We add the discharges from y=0 at the bottom to y=a at the top:

$latex \displaystyle Q_o=C_db\sqrt{2g}\int_0^a\sqrt{H_b-y}\,dy =\frac{2}{3}C_db\sqrt{2g}\left[H_b^{3/2}-(H_b-a)^{3/2}\right]. $

This gives free discharge through the whole rectangle while allowing head to vary with height. If the opening is short compared with its head, we can instead use the centre head H_c=H_b-a/2 :

\displaystyle Q_o\approx C_dba\sqrt{2gH_c}.

Other outlines: use their area or strip width #

For nearly uniform head, a slot uses width times height. An ellipse with full axes a and b has area A_o=\pi ab/4 . Other outlines follow the same area relation when we use an appropriate C_d .

If a vertical circle is tall enough for head to vary appreciably, area alone is insufficient. Let r be its radius, and measure height \eta upward from its centre. With centre head H_c , a strip at \eta has head H_c-\eta . Its half-width, vertical offset and radius make a right triangle: (w/2)^2+\eta^2=r^2 . Thus the full chord width is w(\eta)=2\sqrt{r^2-\eta^2} . We assume the whole circle is below the water surface, H_c>r .

At height \eta , a thin horizontal strip has width w(\eta) and thickness d\eta . Its area is

\displaystyle dA=w(\eta)\,d\eta=2\sqrt{r^2-\eta^2}\,d\eta.

Torricelli’s relation gives the ideal speed at that strip from its local head:

\displaystyle v_{ideal}(\eta)=\sqrt{2g(H_c-\eta)}.

Taking C_d as uniform over the opening, the strip discharge is

$latex \displaystyle dQ=C_dv_{ideal}(\eta)\,dA =C_d\,2\sqrt{r^2-\eta^2}\sqrt{2g(H_c-\eta)}\,d\eta. $

Now add the strips from the bottom of the circle, \eta=-r , to the top, \eta=r :

$latex \displaystyle Q_o=\int_{-r}^{r}dQ =C_d\sqrt{2g}\int_{-r}^{r}2\sqrt{r^2-\eta^2}\sqrt{H_c-\eta}\,d\eta. $

Both chord width and head change as we move up the circle. When H_c is much larger than r , the head is nearly uniform and the integral approaches Q_o=C_d\pi r^2\sqrt{2gH_c} . For another outline, we use its width at each height in the same strip calculation.

Physical boundary conditions #

We return to the circular opening of diameter D and area A_o=\pi D^2/4 . When driving head is uniform, shape enters through area and C_d ; it does not change ideal velocity. We will use a vertical rectangle only for the partly submerged case, where tailwater cuts across the opening. First, let us change the water level downstream. Then we will consider trapped air, a surcharged pipe and a fast approach to the outlet.

Fully submerged circular opening: upstream and downstream levels both act #

Suppose the upstream pond and receiving drain both rise above the circular opening. Their free surfaces are open to air, but the water pressure at the opening on the downstream side is now hydrostatic. We take both pools as slowly moving, so the upstream surface velocity is negligible.

Put point 1 at the upstream free surface and point 2 at the opening centre on the downstream side. If z_o is the opening-centre elevation, then z_1=z_{WS,u} , z_2=z_o and p_1=p_{atm} . At point 2, the downstream water exerts p_2=p_{atm}+\gamma(z_{WS,d}-z_o) . For the ideal calculation, we set h_L=0 and \alpha_1\approx\alpha_2\approx1 . Bernoulli between the two points becomes

$latex \displaystyle \frac{p_1}{\gamma}+z_1+\frac{V_1^2}{2g} =\frac{p_2}{\gamma}+z_2+\frac{V_2^2}{2g}. $

At the upstream surface, V_1\approx0 , and the pressure-plus-elevation head is

\displaystyle \frac{p_1}{\gamma}+z_1=\frac{p_{atm}}{\gamma}+z_{WS,u}.

At the opening, the downstream hydrostatic pressure and the elevation of point 2 combine as

$latex \displaystyle \frac{p_2}{\gamma}+z_2 =\frac{p_{atm}}{\gamma}+(z_{WS,d}-z_o)+z_o =\frac{p_{atm}}{\gamma}+z_{WS,d}. $

Substitute both expressions into Bernoulli. The atmospheric-pressure heads cancel:

$latex \displaystyle \frac{p_{atm}}{\gamma}+z_{WS,u} =\frac{p_{atm}}{\gamma}+z_{WS,d}+\frac{V_2^2}{2g} \quad\Longrightarrow\quad \Delta H=z_{WS,u}-z_{WS,d}=\frac{V_2^2}{2g}. $

The ideal outlet velocity is therefore

\displaystyle V_{2,ideal}=\sqrt{2g\Delta H}.

For the circular opening, A_o=\pi D^2/4 . Multiply ideal velocity by this area and use C_d for the real opening:

$latex \displaystyle Q_o=C_dA_o\sqrt{2g\Delta H} =C_d\frac{\pi D^2}{4}\sqrt{2g\Delta H}. $

The difference between the two water levels sets velocity; the circular diameter sets area. Equal levels give no net flow. If the downstream level rises above the upstream level, flow tends to reverse unless a flap gate stops it. The USBR submerged-orifice guidance likewise uses upstream and downstream heads.

Partly submerged rectangular opening: split at tailwater #

Now let tailwater cover only the lower part of a tall rectangular opening. Measure both depths from the opening bottom: H_u upstream and H_d downstream. Let the opening be a high and b wide. The diagram shows 0<H_d<a<H_u .

Below tailwater, the submerged area is bH_d and the driving head is \Delta H=H_u-H_d . That part carries

\displaystyle Q_{sub}=C_dbH_d\sqrt{2g(H_u-H_d)}.

The upper part discharges freely into air. At height y above the bottom, its local head is H_u-y , so

\displaystyle Q_{free}=C_db\sqrt{2g}\int_{H_d}^{a}\sqrt{H_u-y}\,dy.

Add the lower and upper discharges, then evaluate the integral:

$latex \displaystyle Q_o=C_db\sqrt{2g}\left[ H_d\sqrt{H_u-H_d} +\frac{2}{3}\left((H_u-H_d)^{3/2}-(H_u-a)^{3/2}\right) \right]. $

The first term belongs to the submerged lower portion; the second belongs to the free upper portion. They need different treatments because the pressure conditions differ across the opening. Here we have taken C_d as uniform over both portions.

An open pond stays at atmospheric pressure, even near its crest #

An open pond has atmospheric pressure p_{atm} at its water surface, or zero gauge pressure. Even if the water comes close to the wall crest, the small remaining freeboard does not pressurise it. Raising the surface increases the head above a low outlet. Water overtopping the wall follows another flow path; the orifice calculation still concerns the low opening.

A sealed chamber or surcharged pipe can add pressure head #

A roof can trap air over the water. A surcharged pipe can also carry pressure above or below atmospheric. We must keep the pressure reference clear in either case. Absolute pressure p_{abs} is measured from a perfect vacuum; gauge pressure p_g is measured relative to local atmospheric pressure, so p_g=p_{abs}-p_{atm} . A pressure gauge normally reads p_g . If a sensor gives p_{abs} , subtract local atmospheric pressure before using gauge head.

Let z_u and z_d be the elevations of the two water surfaces, or of the sections chosen in a pressurised pipe. Their gauge pressures are p_{g,u} and p_{g,d} . Subtract the downstream piezometric head from the upstream one to get the static driving head:

$latex \displaystyle H_{static}=\left(z_u+\frac{p_{g,u}}{\gamma}\right)-\left(z_d+\frac{p_{g,d}}{\gamma}\right) =(z_u-z_d)+\frac{p_{g,u}-p_{g,d}}{\gamma}. $

For example, near sea level take p_{atm}\approx101.3\;\text{kPa} . An upstream gauge reading of p_{g,u}=20\;\text{kPa} means an absolute pressure of about 121.3\;\text{kPa} . If the downstream side is open to air, that pressure difference adds 20\;\text{kN/m}^2\div9.81\;\text{kN/m}^3\approx2.0\;\text{m} of water head. If both chambers have the same air pressure, the gauge-pressure terms cancel instead.

At the free surface of an open pond, p_g=0 . In a sealed chamber, we need to measure the trapped-air pressure or calculate it from the air state. For a fixed mass of air changing volume, use absolute pressure in p_{abs}V^n=p_{abs,0}V_0^n . The choice n=1 approximates slow, isothermal compression; n\approx1.4 approximates rapid, adiabatic compression. Subtract p_{atm} afterward to obtain p_g . Water level alone cannot tell us the trapped-air pressure without the air volume and compression process.

For a surcharged pipe, read pressure from a gauge or the modelled hydraulic grade line (HGL): p_g/\gamma=HGL-z . Then H_{static}=HGL_u-HGL_d . If this head is positive and nearly uniform across the opening, with negligible approach speed and no additional losses beyond those represented by C_d , use Q_o=C_dA_o\sqrt{2gH_{static}} . Where velocity or other losses matter, keep their terms in the energy equation and count each loss once.

Fast approach flow and a nearby screen change the energy balance #

For a broad pond, we could neglect the surface speed. A narrow forebay or approach channel may carry appreciable velocity, while a screen or valve near the opening consumes head. With equal atmospheric pressures and \alpha\approx1 , we retain those terms:

\displaystyle H+\frac{V_1^2}{2g}=\frac{V_2^2}{2g}+h_L,\qquad H=z_1-z_2.

Write one local loss as h_L=K V_2^2/(2g) , with K referenced to outlet speed. The balance gives

\displaystyle V_2=\sqrt{\frac{2gH+V_1^2}{1+K}}.

The approach velocity contributes energy and the restriction consumes some of it. If the calibrated discharge coefficient already includes that restriction, do not add its loss again.

Summary of cases #

The table collects the relations we have just derived. For free discharge under nearly uniform head, H runs from the upstream free surface to the opening centre. For a fully submerged opening, \Delta H is the difference between upstream and downstream water levels. Read each formula together with the conditions in its row.

Why is the stage–discharge relationship not always unique? #

We call the relationship between water level (stage) and discharge a rating curve. For a free outlet with fixed geometry and C_d , Q_o=C_dA_o\sqrt{2gH} gives one discharge for each upstream head H . Plot the results from a MiTS hydraulic simulation, however, and the same upstream level may appear with two discharges. What else has changed?

Check the downstream level. Once tailwater covers the opening, the driving head is \Delta H=z_{WS,u}-z_{WS,d} . Two observations can have the same upstream level and different tailwater levels, so they have different driving heads. Higher tailwater reduces discharge; if it rises above the upstream level, the flow tends to reverse.

To put numbers on it, measure the upstream and downstream heads above the opening centre: H_u=z_{WS,u}-z_o and H_d=z_{WS,d}-z_o . When both sides cover the whole opening and H_u>H_d , we have

\displaystyle Q_o=C_dA_o\sqrt{2g(H_u-H_d)}.

Take a circular opening with D=0.50\;\text{m} and C_d=0.60 . Keep the upstream head at H_u=2.50\;\text{m} . A downstream head of H_d=0.50\;\text{m} gives Q_o\approx0.74\;\text{m}^3/\text{s} . Raise H_d to 1.50\;\text{m} , still covering the opening, and discharge falls to about 0.52\;\text{m}^3/\text{s} . The two curves hold tailwater at those respective levels. At H_u=2.50\;\text{m} , the vertical line meets the curves at the two calculated discharges. If the downstream head becomes greater than the upstream head, the direction reverses unless there is a flap gate.

Time enters the plot only because tailwater can differ between observations. Fix both water levels, the opening and C_d , and the submerged-orifice equation gives one discharge. Plotting against upstream level alone conceals that second water level.

Conclusion #

For a freely discharging outlet under our baseline assumptions, Q_o\propto\sqrt{H} . Before using it, we need to ask where the water levels are, whether either side is pressurised, how fast the water approaches, and what losses C_d already covers. Tailwater or surcharge can change the head; a fast approach or nearby restriction changes the rest of the energy balance.

The HEC-22 and MSMA handbook formulas are useful when their conditions match the layout. In a connected drainage or storage system, tailwater, backwater, stored water and surcharged pipes may change together, so we may have to solve velocity, depth and discharge through time. A neat step-by-step outlet calculation cannot represent every such layout. If we apply it without checking the conditions, we can misestimate the flow and end up with an undersized structure.