Hydraulic Design of Energy Dissipators for Culverts and Channels Hydraulic Engineering Circular Number 14, Third Edition
Chapter 6: Hydraulic Jump
The hydraulic jump is a natural phenomenon that occurs when supercritical flow is forced to change to subcritical flow by an obstruction to the flow. This abrupt change in flow condition is accompanied by considerable turbulence and loss of energy. The hydraulic jump can be illustrated by use of a specific energy diagram as shown in Figure 6.1. The flow enters the jump at supercritical velocity, V_{1}, and depth, y_{1},that has a specific energy of E = y_{1} + V_{1}^{2}/(2g). The kinetic energy term, V^{2}/(2g), is predominant. As the depth of flow increases through the jump, the specific energy decreases. Flow leaves the jump area at subcritical velocity with the potential energy, y, predominant.
Figure 6.1. Hydraulic Jump
6.1 Types Of Hydraulic Jump
When the upstream Froude number, Fr, is 1.0, the flow is at critical and a jump cannot form. For Froude numbers greater than 1.0, but less than 1.7, the upstream flow is only slightly below critical depth and the change from supercritical to subcritical flow will result in only a slight disturbance of the water surface. On the high end of this range, Fr approaching 1.7, the downstream depth will be about twice the incoming depth and the exit velocity about half the upstream velocity.
The Bureau of Reclamation (USBR, 1987) has related the jump form and flow characteristics to the Froude number for Froude numbers greater than 1.7, as shown in Figure 6.2. When the upstream Froude number is between 1.7 and 2.5, a roller begins to appear, becoming more intense as the Froude number increases. This is the prejump range with very low energy loss. The water surface is quite smooth, the velocity throughout the cross section uniform, and the energy loss in the range of 20 percent.
Figure 6.2. Jump Forms Related to Froude Number (USBR, 1987)
An oscillating form of jump occurs for Froude numbers between 2.5 and 4.5. The incoming jet alternately flows near the bottom and then along the surface. This results in objectionable surface waves that can cause erosion problems downstream from the jump.
A well balanced and stable jump occurs where the incoming flow Froude number is greater than 4.5. Fluid turbulence is mostly confined to the jump, and for Froude numbers up to 9.0 the downstream water surface is comparatively smooth. Jump energy loss of 45 to 70 percent can be expected.
With Froude numbers greater than 9.0, a highly efficient jump results but the rough water surface may cause downstream erosion problems.
The hydraulic jump commonly occurs with natural flow conditions and with proper design can be an effective means of dissipating energy at hydraulic structures. Expressions for computing the before and after jump depth ratio (conjugate depths) and the length of jump are needed to design energy dissipators that induce a hydraulic jump. These expressions are related to culvert outlet Froude number, which for many culverts falls within the range 1.5 to 4.5.
6.2 Hydraulic Jump In Horizontal Channels
The hydraulic jump in any shape of horizontal channel is relatively simple to analyze (Sylvester, 1964). Figure 6.3 indicates the control volume used and the forces involved. Control section 1 is before the jump where the flow is undisturbed, and control section 2 is after the jump, far enough downstream for the flow to be again taken as parallel. Distribution of pressure in both sections is assumed hydrostatic. The change in momentum of the entering and exiting stream is balanced by the resultant of the forces acting on the control volume, i.e., pressure and boundary frictional forces. Since the length of the jump is relatively short, the external energy losses (boundary frictional forces) may be ignored without introducing serious error. Also, a channel may be considered horizontal up to a slope of 18 percent (10 degree angle with the horizontal) without introducing serious error. The momentum equation provides for solution of the sequent depth, y_{2}, and downstream velocity, V_{2}. Once these are known, the internal energy losses and jump efficiency can be determined by application of the energy equation.
Figure 6.3. Hydraulic Jump in a Horizontal Channel
The general form of the momentum equation can be used for the solution of the hydraulic jump sequentdepth relationship in any shape of channel with a horizontal floor. Defining a momentum quantity as, M = Q^{2}/(gA) + AY and recognizing that momentum is conserved through a hydraulic jump, the following can be written:
(6.1)
Q^{2}/(gA_{1}) + A_{1}Y_{1} = Q^{2}/(gA_{2}) + A_{2}Y_{2}
where,
Q = channel discharge, m^{3}/s (ft^{3}/s)
A_{1},A_{2} = crosssectional flow areas in sections 1 and 2, respectively, m^{2} (ft^{2})
Y_{1},Y_{2} = depth from water surface to centroid of crosssection area, m (ft)
The depth from the water surface to the centroid of the crosssection area can be defined as a function of the channel shape and the maximum depth: Y = Ky. In this relationship, K is a parameter representing the channel shape while y is the maximum depth in the channel. Substituting this quantity into Equation 6.1 and rearranging terms yields:
A_{1} K_{1} y_{1}  A_{2} K_{2} y_{2} = (1/A_{2}  1/A_{1})Q^{2}/g
Rearranging and using Fr_{1}^{2} = V_{1}^{2} /(gy_{1}) = Q^{2} /(A_{1}^{2} gy_{1}), gives:
A_{1} K_{1} y_{1}  A_{2} K_{2} y_{2} = Fr_{1}^{2} A_{1} y_{1} (A_{1} /A_{2} 1).
Dividing this by A_{1} y_{1} provides:
(6.2)
K_{2} A_{2} y_{2} /(A_{1} y_{1})  K_{1} = Fr_{1}^{2} (1  A_{1} /A_{2})
This is a general expression for the hydraulic jump in a horizontal channel. The constants K_{1} and K_{2} and the ratio A_{1}/A_{2} have been determined for rectangular, triangular, parabolic, circular, and trapezoidal shaped channels by Sylvester (1964). The relationships for rectangular and circular shapes are summarized in the following sections.
6.2.1 Rectangular Channels
For a rectangular channel, substituting K_{1} = K_{2} = 1/2 and A_{1} /A_{2} = y_{1} /y_{2} into Equation 6.2, the expression becomes:
y_{2}^{2} /y_{1}^{2} 1 = 2Fr_{1}^{2} (1  y_{1} /y_{2})
If y_{2} /y_{1} = J, the expression for a hydraulic jump in a horizontal, rectangular channel becomes Equation 6.3, which is plotted as Figure 6.4.
(6.3)
The length of the hydraulic jump can be determined from Figure 6.5. The jump length is measured to the downstream section at which the mean water surface attains the maximum depth and becomes reasonably level. Errors may be introduced in determining length since the water surface is rather flat near the end of the jump. This is undoubtedly one of the reasons so many empirical formulas for determining jump length are found in the literature. The jump length for rectangular basins has been extensively studied.
Stilling basin design is a common application for hydraulic jumps in rectangular channels (see Chapter 8). Free jump basins can be designed for any flow conditions; but because of economic and performance characteristics they are, in general, only employed in the lower range of Froude numbers. Flows with Froude numbers below 1.7 may not require stilling basins but may require protection such as riprap and wingwalls and apron. For Froude numbers between 1.7 and 2.5, the free jump basin may be all that is required. In this range, loss of energy is less than 20 percent; the conjugate depth is about three times the incoming flow depth; and, the length of basin required is less than about 5 times the conjugate depth. Many highway culverts operate in this flow range. At higher Froude numbers, the use of baffles and sills make it possible to reduce the basin length and stabilize the jump over a wider range of flow situations.
Figure 6.4. Hydraulic Jump  Horizontal, Rectangular Channel
Figure 6.5. Length of Jump for a Rectangular Channel
Design Example: Hydraulic Jump in a Horizontal, Rectangular Channel (SI)
Determine the height and length of a hydraulic jump in a box culvert with a 2.134 m span. Also, estimate the range of flows for which a jump would be triggered as discharged to a trapezoidal channel. Given:
 S = 0.2%
 Q = 11.33 m^{3}/s
 V_{1 } = 5.79 m/s
 y_{1 } = 0.914 m
 Fr = 1.9
For the trapezoidal channel:
 B = 3.04 m
 Side slopes = 1V:2H
 n = 0.03
 S = 0.04%
Solution
Step 1. Find the conjugate depth from Figure 6.4.
J = y_{2} /y_{1} = 2.2
y = 2.2(0.914) = 2.011 m
Step 2. Find the Length of jump from Figure 6.5
L /y_{1} = 9.0
L = 0.914(9.0) = 8.226 m
Step 3. Calculate the after jump velocity
V_{2} = Q/A_{2} = 11.33/ [2.134(2.011)] = 2.64 m/s
Velocity reduction is (5.79  2.64)/5.79 = 54.4%.
Step 4. Develop a Q vs. stage curve for the downstream trapezoidal channel using either HDS No. 3 (FHWA, 1961) or Table B.1 to determine the relationship with conjugate depth (see below).
Step 5. Review sequent depth requirements. The plot shows that excess tailwater depth is available in the downstream channel for discharges up to approximately 13.6 m^{3}/s. For larger discharges, the jump would begin to move downstream. The assumption in this example is that normal depth in the downstream channel is obtained as soon as the flow leaves the culvert. In a real case, a stilling basin (see Section 8.1) will normally be required to generate enough tailwater depth to cause a jump to occur or the culvert will need to be designed as a brokenback culvert (see Chapter 7).
Normal Channel Depth  Conjugate Depth Relationship
Design Example: Hydraulic Jump in a Horizontal, Rectangular Channel (CU)
Determine the height and length of a hydraulic jump in a box culvert with a 7 ft span. Also, estimate the range of flows for which a jump would be triggered as discharged to a trapezoidal channel. Given:
 S = 0.2%
 Q = 400 ft^{3}/s
 V_{1 } = 19 ft/s
 y_{1 } = 3.0 ft
 Fr = 1.9
For the trapezoidal channel:
 B = 10 ft
 Side slopes = 1V:2H
 n = 0.03
 S = 0.04%
Solution
Step 1. Find the conjugate depth in a rectangular basin from Figure 6.4.
J = y_{2} /y_{1} = 2.2
y_{2} = 2.2(3.0) = 6.6 ft
Step 2. Find the Length of jump from Figure 6.5
L /y_{1} = 9.0
L = 3.0(9.0) = 27 ft
Step 3. Calculate the after jump velocity
V_{2} = Q/A_{2} = 400/ [7(6.6)] = 8.7 ft/s
Velocity reduction is (19  8.7)/19 = 54.2%.
Step 4. Develop a Q vs. stage curve for the downstream trapezoidal channel using either HDS 3 (FHWA, 1961) or Table B.1 to determine the relationship with conjugate depth (see below).
Step 5. Review sequent depth requirements. The plot shows that excess tailwater depth is available in the downstream channel for discharges up to approximately 480 ft^{3}/s. For larger discharges, the jump would begin to move downstream. The assumption in this example is that normal depth in the downstream channel is obtained as soon as the flow leaves the culvert. In a real case, a stilling basin (see Section 8.1) will normally be required to generate enough tailwater depth to cause a jump to occur or the culvert will need to be designed as a brokenback culvert (see Chapter 7).
Normal Channel Depth  Conjugate Depth Relationship
6.2.2 Circular Channels
Circular channels are divided into two cases: where y_{2} is greater than the diameter, D, and where y_{2} is less than D. For y_{2} less than D:
(6.4)
(K_{2} y_{2} C_{2} /(y_{1}C_{1}))  K_{1 }= Fr_{1}^{2 }(1  C_{1} /C_{2})
For y_{2} greater than or equal to D:
(6.5)
(y_{2}C_{2}/(y_{1}C_{1}))  0.5 (C_{2} D/(C_{1} y_{1}))  K_{1} = Fr_{1}^{2} (1  C_{1}/C_{2})
C and K are functions of y/D and may be evaluated from the Table 6.1.
Table 6.1. Coefficients for Horizontal, Circular Channels
Y/D  0.1  0.2  0.3  0.4  0.5  0.6  0.7  0.8  0.9  1.0 
K  0.410  0.413  0.416  0.419  0.424  0.432  0.445  0.462  0.473  0.500 
C  0.041  0.112  0.198  0.293  0.393  0.494  0.587  0.674  0.745  0.748 
C'  0.600  0.800  0.917  0.980  1.000  0.980  0.917  0.800  0.600  
In Equations 6.4 and 6.5, Fr_{1} is computed using the maximum depth in the channel. Figure 6.6 may be used as an alternative to these equations.
Alternatively, the designer may calculate a Froude number based on hydraulic depth, Fr_{m}.= V/(gy_{m})^{1/2}. Where y_{m} = (C/C')D or y_{m} = A/T. For the first expression, C' is taken from Table 6.1. For the second expression, A is the crosssectional area of flow and T is the water surface width. Figure 6.7 is the design chart for horizontal, circular channels using the hydraulic depth in computing the Froude number.
The length of the hydraulic jump is generally measured to the downstream section at which the mean water surface attains the maximum depth and becomes reasonably level. The jump length in circular channels is determined using Figure 6.8. This curve is for the case where y_{2} is less than D. For the case where y_{2} is greater than D, the length should be taken as seven times the difference in depths, i.e., L_{J} = 7(y_{2} y_{1}).
Figure 6.6. Hydraulic Jump  Horizontal, Circular Channel (actual depth)
Figure 6.7. Hydraulic Jump  Horizontal, Circular Channel (hydraulic depth)
Figure 6.8. Jump Length Circular Channel with y_{2} < D
Design Example: Hydraulic Jump in a Horizontal, Circular Channel (SI)
Determine the height and length of a hydraulic jump in an RCP culvert with a 2.134 m diameter. Given:
 S = 2%
 Q = 5.664 m^{3}/s
 V_{1} = 5.182 m/s
 y_{1 } = 0.732 m
 Fr_{1} = 1.9
Solution
Step 1. Find the conjugate depth in a circular channel.
y_{1}/D = 0.732/2.134 = 0.34 (use 0.4)
J = y_{2}/ y_{1} = 2.3 from Figure 6.6
y_{2} = 2.3(0.732) = 1.684 m and y_{2}/D = 0.78 (use 0.8)
(Using Equation 6.4 with C_{1} = 0.293, K_{1} = 0.419, C_{2} = 0.674, K_{2} = 0.462 yields the same result.)
Step 2. Find the Length of jump from Figure 6.8
L_{j} /y_{1} = 19
L_{j} = 0.732 (19) = 13.9 m
Step 3. Calculate the after jump velocity
For y_{2}/D = 0.78, A/D^{2} = 0.6573 from Table 3.3 and A = 2.99 m^{2}
V_{2} = Q/A_{2} = 5.664/2.99 = 1.89 m/s
Velocity reduction is (5.182  1.89)/ 5.182 = 63.5%.
Design Example: Hydraulic Jump in a Horizontal, Circular Channel (CU)
Determine the height and length of a hydraulic jump in an RCP culvert with a 7 ft diameter. Given:
 S = 2%
 Q = 200 ft^{3}/s
 V_{1} = 17 ft/s
 y_{1 } = 2.4 ft
 Fr_{1} = 1.9
Solution
Step 1. Find the conjugate depth in a circular channel.
y_{1}/D = 2.4/7 = 0.34 (use 0.4)
J = y_{2}/ y_{1} = 2.3 from Figure 6.6
y_{2} = 2.3(2.4) = 5.5 ft and y_{2}/D = 0.78 (use 0.8)
(Using Equation 6.4 with C_{1} = 0.293, K_{1} = 0.419, C_{2} = 0.674, K_{2} = 0.462 yields the same result.)
Step 2. Find the Length of jump from Figure 6.8
L_{j} /y_{1} = 19
L_{j} = 2.4 (19) = 46 ft
Step 3. Calculate the after jump velocity
For y_{2}/D = 0.78, A/D^{2} = 0.6573 from Table 3.3 and A = 32.2 ft^{2}
V_{2} = Q/A_{2} = 200/32.2 = 6.2 ft/s
Velocity reduction is (17  6.2)/ 17 = 63.5%.
6.2.3 Jump Efficiency
A general expression for the energy loss (H_{L}/H_{1}) in any shape channel is:
(6.6)
H_{L}/H_{1} = 2  2(y_{2}) + Fr_{m}^{2} [1  A_{1}^{2} /A_{2}^{2}] / (2 + Fr^{2})
where,
Fr_{m} = upstream Froude number at section 1, Fr_{m}^{2} = V^{2}/(gy_{m})
y_{m} = hydraulic depth, m (ft)
This equation is plotted for the various channel shapes as Figure 6.9. Even though Figure 6.9 indicates that the nonrectangular sections are more efficient for the higher Froude numbers, it should be remembered that these sections also involve longer jumps, stability problems, and a rough downstream water surface.
Figure 6.9. Relative Energy Loss for Various Channel Shapes
6.3 Hydraulic Jump In Sloping Channels
Figure 6.10 (Bradley, 1961) indicates a method of delineating hydraulic jumps in horizontal and sloping channels. Horizontal channels (case A) were discussed in the previous section. Sloping channels are discussed in this section. If the channel bottom is selected as a datum, the momentum equation becomes:
(6.7)
where,
γ = unit weight of water, N/m^{3} (lb/ft^{3})
φ = angle of channel with the horizontal
B = channel bottom width (rectangular channel), m (ft)
w = weight of water in jump control volume, N (lb)
The momentum equation used for the horizontal channels cannot be applied directly to hydraulic jumps in sloping channels since the weight of water within the jump must be considered. The difficulty encountered is in defining the water surface profile to determine the volume of water within the jumps for various channel slopes. This volume may be neglected for slopes less than 10 percent and the jump analyzed as a horizontal channel.
The Bureau of Reclamation (Bradley, 1961) conducted extensive model tests on case B and C type jumps to define the length and depth relationships. This reference should be consulted if a hydraulic jump in a sloping rectangular channel is being considered. Model tests should be considered if other channel shapes are being considered.
Figure 6.10. Hydraulic Jump Types Sloping Channels (Bradley, 1961)
