Open Channel Flow Basics - Hydraulic Jump Calculations

Written by: Harlan Bengtson • Edited by: Lamar Stonecypher
Updated Sep 12, 2010

A hydraulic jump occurs in open channel flow where supercritical flow (Froude number > 1) is taking place on a slope that cannot maintain supercritical flow. The hydraulic jump is an abrupt transition to subcritical from supercritical flow. The head loss across the hydraulic jump can be calculated.

Review of Supercritical flow and Subcritical Flow

In order to discuss the hydraulic jump, it's necessary to refer to supercritical flow and subcritical flow, so these two types of open channel flow will be reviewed briefly. Supercritical flow is characterized by high flow velocity, shallow depth of flow, and relatively steep channel bottom slope. Subcritical flow, on the other hand, is characterized by low flow velocity, high depth of flow, and relatively small channel bottom slope. Supercritical flow occurs for Froude Number (Fr) greater than 1, and subcritical flow occurs for Froude number less than 1, where Fr = V/(gy)^1/2, and V is the average velocity in the channel, y is the depth of flow, and g is the acceleration due to gravity.

Where Does a Hydraulic Jump Occur

click to enlarge

A hydraulic jump occurs wherever a supercritical flow is taking place in a channel that

click to enlarge

doesn't have a slope steep enough to maintain the supercritical flow. The hydraulic jump provides a transition to subcritical from supercritical flow.Two physical situations that would give rise to a hydraulic jump are shown in the diagrams. The one on the left is a transition from a steep channel slope with supercritical flow to a smaller slope that will only sustain a subcritical flow. The diagram on the right shows a supercritical flow coming under a sluice gate on a slope that will only sustain a subcritical flow. In both cases a hydraulic jump is present to provide an abrupt transition from supercritical to subcritical flow. Open channel flow can transition smoothly from subcritical to supercritical flow, but there is no way for a smooth transition from supercritical to subcritical flow to take place in open channel flow.

Equation Across a Hydraulic Jump

click to enlarge

Through the use of the three fluid mechanics conservation equations (the energy equation, the continuity equation, and the momentum equation), the following equation can be derived relating the conditions before a hydraulic jump to the condition after the jump. The diagram at the left shows the parameters in the equation.

click to enlarge

y₂/y₁ = (1/2)[-1 + (1 + 8Fr₁²)^1/2] (eqn 1)

where Fr₁ = V₁/(y₁g)^1/2, and

y₁ = depth of flow before the hydraulic jump, ft

y₂ = depth of flow after the hydraulic jump, ft

V₁ = average flow velocity before the hydraulic jump, ft/sec

V₂ = average flow velocity after the hydraulic jump, ft/sec

Head Loss Across a Hydraulic Jump

Due to the turbulence in the hydraulic jump, there will be frictional head loss across the jump. The amount of head loss may sometimes be of interest and is given by the following equation:

h_L = y₁ + V₁²/2g - (y₂ + V₂²/2g)

Rearranging this equation and substituting the expression for Fr₁, leads to the following equation for h_L in terms of the two depths of flow and the upstream Froude Number:

h_L/y₁ = 1 - y₂/y₁ + (Fr₁²/2)[1 - (y₁/y₂)²] (eqn 2)

Example Calculations

Consider a hydraulic jump in a 2 ft wide rectangular channel. The flow rate through the channel is 20 cfs and the depth of flow before the jump is 1 ft. What will be the depth of flow after the hydraulic jump and what will be the head loss across the jump?

Solution: First calculate the flow velocity in the channel before the jump from V = Q/A = 20/(2x1) = 10 ft/sec. Second calculate the upstream Froude number, Fr₁. Then you are ready to calculate y₂ from eqn 1 and head loss, h_L, from eqn 2.

Fr₁ = 10/(1x32.2)^1/2 =1.76

From eqn 1: y₂ = (1)(1/2)[-1 +(1 + 8*1.76²)^1/2] = 2.03 ft

From eqn 2: h_L = 1 - 2.03/1 + (1.76²/2)[1 - (1/2.03)²] = 0.143 ft

For an Excel spreadsheet template for making hydraulic jump calculations, see the article, "Use of Excel Formulas for Hydraulic Jump Calculations."

References and Image Credits

References for more information:

1. Munson, B. R., Young, D. F., & Okiishi, T. H., Fundamentals of Fluid Mechanics, 4^th Ed., New York: John Wiley and Sons, Inc, 2002.

2. Chow, V. T., Open Channel Hydraulics, New York: McGraw-Hill, 1959.

3. Bengtson, Harlan H. Open Channel Flow II - Hydraulic Jumps and Supercritical and Nonuniform Flow - An online, continuing education course for PDH credit.

Image credits:

1. Hydraulic Jump drawings - Drawn by H. Bengtson

2. Hydraulic Jump picture - Wikipedia Commons, StAnthonyFalls apron

Harlan Bengtson Dec 15, 2011 8:25 PM

RE: Open Channel Flow Basics - Hydraulic Jump Calculations

The efficiency of a hydraulic jump is simply the E2/E1, where E2 is the energy after the jump [E2 = y2 + V2^2/(2g)] and E1 is the energy after the jump [E1 = y1 + V1^2/(2g)]. There are other possible expressions for the efficiency because there are relationships between the conditions before the jump and conditions after the jump. Derivation of any of the other equations for efficiency should start with Eff = E2/E1. For another article and spreadsheets for hydraulic jump calculations see: "Hydraulic Jump Calculations with Excel Spreadsheets" at <br>http://www.engineeringexcelspr...

Abirthecolour Dec 15, 2011 4:42 PM

i want the detail derivation of efficiency equation of hydraulic jump

Harlan Bengtson Apr 3, 2011 4:50 PM

EGL Calculation with Hydraulic Jump

For uniform open channel flow the energy grade line is parallel to the channel bottom at a height above the liquid surface of V^2/2g. This would be the case for the steep channel before the hydraulic jump and for the mild channel after the hydraulic jump. The difference in the energy grade line before the jump and after the jump will be equal to the head loss across the jump.

GCH Apr 3, 2011 1:55 PM

EGL calculation with Hydraulic Jump

How can I calculate the EGL of a rectangular Channel (box pipe) from the tailwater depth at connection, going up in a mild slope (S=0.003) let's say at about 40 meters then breaks at a steep slope (S=0.10) going upstream? There is an occurrence of Hydraulic Jump at mild slope. Do I need to locate the HJ in order for me to calculate the EGL for the whole system using the Energy Continuity Equation?

kerry Mar 28, 2011 11:11 AM

hyrogeology

im doing final year in hydrology(hons)and am struggling to come up with a research topic based in hyrogeology. am asking for any idea

Harlan Bengtson Mar 7, 2011 11:24 PM

Hydrology - Equation to use

Knowing y1 and y2, you can calculate Fr1 from the equation for y2/y1 in this article. Then, from the definition for Fr1 (in this article), you can calculate V1. Now that you know V1 and Q, you can calculate the width, b, from the equation: Q = (V1)(A1) = (V1)(by1).

You can now calculate the head loss across the jump from the equation for hL/y1 in this article.

kerry Mar 7, 2011 5:39 PM

hydrology

i have a problem with which formulars to use ;y1 and y2 are given and Q.The task is to determine the width of the flow and energy loss due to the jump

Harlan Bengtson Mar 6, 2011 9:04 AM

the equations

Yes, the equations given in the article are for a rectangular channel only.

stephan stys Mar 5, 2011 11:56 PM

I think the equations for Y2/Y1 and hl/Y1 is only for horizontal rectangular channels. Is that right.

Harlan Bengtson Feb 13, 2011 8:23 PM

Sequent Depth

Sequent depth is a term used for the depth of subcritical flow after a hydraulic jump. It is often represented by the symbol y2, as in this article.

anjali Feb 13, 2011 2:36 AM

sequent depth

what is sequent depth? please define it?

Harlan Bengtson Feb 5, 2011 3:38 AM

Why check on whether flow is supercritical

One reason for checking on whether a given flow is supercritical or subcritical is to determine whether a hydraulic jump will occur. A hydraulic jump will occur when supercritical flow is taking place on a subcritical slope.

Aman Feb 5, 2011 12:36 AM

Hydraulics

Why we need to check weather the flow is supercritical or subcritical

Harlan Bengtson Jan 25, 2011 9:32 PM

Location of Hydraulic Jump

I don't know of any very straight forward method of determining the location of a hydraulic jump. Here is a link to an example calculation: http://www.efm.leeds.ac.uk/CIVE/CIVE2400/Handouts/backwater.pdf

Arnout Jan 25, 2011 6:17 AM

location jump

Dear Harlan,

is there a method to calculate the location of the jump? I mean the distance behind the sluice gate in your drawing. I think it will be determined by tehe thickness of the boundary layer of the fluid building up behind the sluice gate.
Is the jump also possible in laminair conditions or only in turbulent flow?

thank you for helping me!

Arnout

rajahassan Jan 17, 2011 1:37 PM

Hydraulic jump after a dam

I need to calculate the length of a stilling basin.By calculating y1 and y2 I have calculated the length of the hydraulic jump.To calculate the distance from the toe of the spillway to the point of jump initiation I need either the depth at the toe of the spillway or the velocity of water coming from the dam.I have calculated discharge from the dam.The width of the stilling basin is known.

Harlan Bengtson Jan 17, 2011 11:26 AM

Hydraulic Jump after a Dam

If you know the depth and velocity downstream (after the hydraulic jump), then you can calculate y1 from the following equation, which is equation 1 from the article with subscripts 1 and 2 exchanged:

y1/y2 = (1/2)[-1 + (1 + 8(Fr2)^2)^(1/2)]

rajahassan Jan 16, 2011 11:25 PM

I have a problem with an assignment.I am trying to find out the flow velocity at the toe of a spillway.The discharge is known but the depth of flow at the toe of the dam is unknown.Is it any other way to find the velocity.

Harlan Bengtson Dec 26, 2010 9:41 AM

Length of the Jump

I believe there would be turbulence at the transition, but there shouldn't be a full hydraulic jump with only 3 ft of subcritical slope before the dropoff.

SS Dec 25, 2010 11:54 PM

Will a jump occur if the length of the subcritcal channel is less than the jump length?

I calculated the jump length to be 12' feet, however the d/s horizontal channel is only 3' long after which the water drops off a cliff. It is basically a ski jump without the bucket and the lip.

I want to be certain that a hydraulic jump does not occur at the transition.

Appreciate the responses on this matter.
SS

Harlan Bengtson Aug 27, 2010 5:53 PM

Venturi Flume Upstream and Downstream Depths

MK,
I believe in most cases the downstream depth will be determined by the normal depth in the downstream channel that the flume discharges into. There may be a non-uniform flow transition after the flume, but it should adjust to the normal depth for the downstream channel.

Mk Aug 27, 2010 4:41 PM

venturi flume upstream and downstream depths

Harlan,

How is the downstream depth of a flume determined if critical depth is known at the throat. Usually the upstream depth at the entrance is about 97% of the Energy at the throat (Ecrit. x 1.5) but not quite sure about the downstram depth just after the throat. any ideas.

MK

Harlan Bengtson Aug 27, 2010 12:32 PM

head loss in terms of velocity

The equation given in the article: hL = y1 + V1^2/2g - (y2 + V2^2/2g) gives the head loss in terms of the upstream and downstream velocities and the upstream and downstream depths of flow.

hag Aug 27, 2010 11:26 AM

asking

head loss enterms of velocity

Harlan Bengtson Aug 17, 2010 8:55 AM

length of drawdown curve, using direct step method

MK,
The uniform depth of 1.95 m doesn't enter into the direct step calculations, except in determining the endpoint of the stepwise calculations. For the nonuniform surface profile that you've described, the flow will be at the uniform depth at some point upstream from the weir if the channel is long enough. Thus the surface profile depth will range from 2.4 m at the weir to 1.95 m at the point upstream, where the flow is at the uniform flow depth. If you are using 0.2 m increments of depth, your last one should only be 2.0 - 1.95 m (0.05 m) because the depth will only go down to 1.95 m.

Harlan Bengtson

MK Aug 17, 2010 6:35 AM

Dear Harlan,
I am trying to determine the approx. surface profile of a curve using the direct step method,
It is a rectangular channel 4m wide, i know manning n = 0.019 and the slope is 1 in 400. The uniform depth is 1.95m when the flow rate is 20m^3.sec. A weir is placed downstream end which causes the depth upstream of the weir to rise to 2.4m. I am taking depth increments of 0.2 m to determine the upstream surface profile of the weir.

My question is what part if any does the uniform depth play in this question and is the distance between the surface profile between 1.95m and 2.4m

thanks

MK

Harlan Bengtson Jul 14, 2010 3:28 PM

Calculating Critical Depth

Here is a way to calculate critical depth if you know both the upstream and downstream depths for a hydraulic jump in a rectangular channel: 1) from eqn 1 in this article: y2/y1 = (1/2)[-1 + (1 + 8(Fr1)^2)^1/2], calculate Fr1 for the given values of y1 and y2. 2) Using Fr1 = V1/[(y1*g)^1/2], calculate V1, the upstream velocity. 3) For a rectangular channel: q = Q/b = V(by)/b = Vy. So q = V1y1. 4) From another article in this series, Open Channel Flow Basics 2 - Supercritical Flow, yc = (q^2/g)^1/3. You now know q, so you can calculate yc from this equation.

SOUTOM PAL Jul 14, 2010 1:08 PM

calculating the critical depth

sir,

its given in one problem that the supercritical upstream depth and downstram depth is 0.5 and 2.5 m how can we calculate the cr. depth???

Harlan Bengtson Jul 8, 2010 11:09 AM

Finding Depth y2

MK,
The method you've described doesn't quite give the correct value for y2, because E2 is not equal to E1 due to the head loss across the jump. In this case the calculated value of y2 is pretty close, because the head loss across the jump is small. If the upstream Froude number were very high (very high velocity), then the head loss would be more significant and E2 would be less than E1 by a greater amount, and the procedure you described based on E1 = E2 would be wrong by a greater amount.

Harlan Bengtson

MK Jul 8, 2010 7:14 AM

Finding depth y2

Harlan
Refering the example calculation is it possible to use the following eqation and method to determine the depth y2:
E1-E2 therefore y1+Q^2/2g(A1^2)=y2+Q^2/2g(A2^2)

1+400/257.6 = y2+400/257.6 * y2
2.552= y2 + 400/257.6*y2
by iterating y2 = 2.24ft. this gives a slightly higher number than the froud eqaution.
Thanks

MK

Harlan Bengtson Jul 1, 2010 11:44 AM

Hydraulic Jump Calculation

MK,
I used U.S. units in the example, so g = 32.2 ft/sec/sec, V is in ft/sec, and y is in ft. Using S.I. units, g = 9.81 m/sec/sec, V is in m/sec, and y is in m. If you convert the velocity of 10 ft/sec to m/sec and convert the depth of 1 ft to meters, and then calculate the Froude number with g = 9.81 m/sec/sec, you will get the same value for the Froude number as in the example. The Froude Number is a dimensionless number, so any consistent set of units can be used to calculate its value.

Harlan Bengtson

MK Jul 1, 2010 9:36 AM

hydraulic jump calculation

Harlan,

I noticed in the example that when calculating the Fr. number its states a value of 32.2. How have you got this number or should it be 10/(1x9.81)^1/2
Fr1 = 10/(1x32.2)1/2 =1.76
where Fr1 = V1/(y1g)1/2

Thanks

Mk

Harlan Bengtson Apr 21, 2010 5:51 PM

Effect of Width

Increasing the width of a channel for a given flow rate through the channel would cause a decrease in the flow depth. It might have a slight effect on the velocity, but the velocity would not change very much. A decrease in upstream depth of flow, y1, would increase the Froude No., because y1 is in the denominator. An increase in upstream Froude No., would cause an increase in the ratio, y2/y1.

Harlan Bengtson

HAILEAB TEKESTE Apr 20, 2010 3:41 PM

EFFECT OF width

what is the effect of width to the hydraulic jump, what i mean is that what if we increase the width of the channel , how can affect the turbulence , jump,energy , velocity?

Harlan Bengtson Mar 17, 2010 12:45 PM

A hydraulic jump will occur whenever there is supercritical flow in a channel with a mild slope (that is a slope that will have subcritical flow as its normal flow). This is normally the case at the bottom of a spillway, because of the high velocity, supercritical flow coming down the spillway, which has to slow down to a subcritical flow on the low slope channel at the end of the spillway. The transition from superciritcal to subcritical flow is always by a hydraulic jump.

sanhita das Mar 17, 2010 12:30 PM

how is hydraulic jump made possible in the bottom of spillways... i mean is it by use of any structure

a.vengatesan Jan 27, 2010 10:31 AM

channel

i have more information in this topic then derivation?

Open Channel Flow Basics - Hydraulic Jump Calculations

Review of Supercritical flow and Subcritical Flow

Where Does a Hydraulic Jump Occur

Equation Across a Hydraulic Jump

Head Loss Across a Hydraulic Jump

Example Calculations

References and Image Credits

Comments