CIVE 210: Hydraulics - Open Channel Flow & Hydraulic Jump
Open Channel Flow - Hydraulic Jump
Overview
Phenomenon of hydraulic jump
Sequent depths
Basic calculations for hydraulic jumps
Examples of hydraulic jump calculations
Calculation methods for complicated surface profiles involving hydraulic jumps
Summary of key concepts
What is Hydraulic Jump?
Definition: A hydraulic jump is a rapid varied flow (RVF) phenomenon characterized by a sudden rise in water surface over a short distance.
It occurs when flow transitions from supercritical to subcritical.
Importance:
Critical in the design of hydraulic structures.
Acts as an energy dissipator.
A natural physical phenomenon.
Example:
A hydraulic jump is shown with an upstream depth of 30 cm and a velocity of V = 16 m/s.
A water level diagram illustrates the transition in gradually varied flow (GVF) over a distance, with the hydraulic jump causing a sudden change.
Hydraulic Jump Examples
Circular hydraulic jumps.
Hydraulic jumps following flow over a weir.
Hydraulic Jump Practice Characteristics
High turbulence, which leads to energy dissipation.
Erosive effect on the bed.
Stilling basins (aprons) are used to manage the erosive effects.
Hydraulic Jump in Practice
Context: Hydraulic jumps often occur downstream of spillways.
Key considerations:
How does water depth change across the jump?
Where does the jump occur?
How can jumps be used for energy dissipation?
Aprons are used to protect the dam structure from erosion caused by the hydraulic jump.
Without proper design, severe erosion can undermine the dam.
Types of Hydraulic Jumps: Upstream Froude Number
Based on upstream Froude number Fr = V/\sqrt{gD}, where:
V is the flow velocity.
g is the acceleration due to gravity.
D is the flow depth.
Types of jumps based on Fr:
(a) Fr = 1.0 to 1.7: Undular jumps.
(b) Fr = 1.7 to 2.5: Weak jumps.
(c) Fr = 2.5 to 4.5: Oscillating jumps.
(d) Fr = 4.5 to 9.0: Steady jumps.
(e) Fr > 9.0: Strong jumps.
Energy dissipation increases from type (a) to (e); strong jumps can dissipate up to 85% of the energy.
Types of Hydraulic Jumps: Tailwater Depth
Tailwater depth (y_t): The depth of the downstream flow after a hydraulic structure.
Upstream depth (y1) & Sequent depth (y2): Depths upstream and downstream of the jump, respectively.
Jump happens right after the vena contractor (free jump).
Types of Hydraulic Jumps: Tailwater Depth cont.
Tailwater depth (yt), Upstream depth (y1), Sequent depth (y_2).
Jump travels downstream.
M3 or H3 backwater curves can arise before the jump.
Jump is pushed upstream.
Submerged jump occurs.
Less energy dissipation occurs in submerged jumps.
Energy in Hydraulic Jump
Diagram showing the energy grade line (EGL) and head loss (h_L) across the hydraulic jump.
Note that there is a lot of viscous dissipation (= head loss) within the hydraulic jump. Therefore, E1 > E2.
The energy equation cannot be directly applied to the jump, and the head loss hL cannot be determined unless V1, V2, y1, y_2, and Q (flow rate) are known.
Momentum in Hydraulic Jump
Conservation of mass requires: Q1 = Q2.
Momentum equation: \Sigma F = \rho Q (V2 - V1) \Rightarrow F1 - F2 = \rho Q (V2 - V1)
Where:
\rho is the density of water.
Q is the flow rate.
V1 and V2 are the velocities before and after the jump.
F1 = \rho g \bar{y}1 A1, F2 = \rho g \bar{y}2 A2 \Rightarrow \rho g (\bar{y}1 A1 - \bar{y}2 A2) = \rho Q (V2 - V1)
\bar{y}1 & \bar{y}2 are the depths to the centroids of Sections 1 & 2.
Control surface diagram illustrating hydrostatic pressure at sections 1 and 2.
Jump in a Rectangular Channel
Continuity equation: y1 b V1 = y2 b V2 = Q (where b is the width of the channel).
Momentum equation:
A = by & \bar{y} = \frac{1}{2} y
\frac{1}{2} \rho g b (y1^2 - y2^2) = \rho Q (V2 - V1)
Eliminating V_2 from the momentum equation using the continuity equation gives:
\frac{y1^2}{2} - \frac{y2^2}{2} + \frac{Q^2}{gb^2y1} - \frac{Q^2}{gb^2y2} = 0
\frac{gy1^2}{2} - \frac{gy2^2}{2} + \frac{V1^2 y1}{y2} - V1^2 y_1 = 0
\frac{y2}{y1} = \frac{1}{2} (\sqrt{1 + 8Fr_1^2} - 1)
Fr1 = \frac{V1}{\sqrt{gy_1}}
Solving this quadratic equation produces:
\frac{y2}{y1} = \frac{1}{2} (\sqrt{1 + 8Fr_1^2} - 1)
Similarly, \frac{y1}{y2} = \frac{1}{2} (\sqrt{1 + 8Fr_2^2} - 1)
Fr2 = \frac{V2}{\sqrt{gy_2}}
Sequent or conjugate depths:
y2 is called the sequent/conjugate depth of y1, and vice versa.
Characteristics of Hydraulic Jump
Once velocities and depths are obtained at both the upstream and downstream locations, the jump length (L_j) can be estimated:
Lj = \alpha hj, where \alpha = 4.5 \sim 7
hj=y2-y_1
Proof of Energy Loss
Energy loss or dissipation after the jump:
\Delta E = \left(y1 + \frac{V1^2}{2g}\right) - \left(y2 + \frac{V2^2}{2g}\right)
\Delta E = (y1 - y2) + \frac{V1^2}{2g} - \frac{V2^2}{2g} = (y1 - y2) + \frac{Q^2}{2g} (\frac{1}{y1^2} - \frac{1}{y2^2})
\Delta E = (y1 - y2) + \frac{V1^2 y1^2}{2gy1^2y2^2} (y2^2 - y1^2) = (y1 - y2) + \frac{V1^2}{2g} (\frac{y2 - y1}{y1 y2})(y2 + y_1)
\Delta E = (y2 - y1) - \frac{(y2 + y1)}{2y1y2} (y2 - y1) = \frac{(y2 - y1)}{4y1y2} (y2 + y1)
\Delta E = (y2 - y1) \frac{(y2 + y1)^2}{4y1 y2}
\Delta E = \frac{(y2 - y1)^3}{4y1 y2}
Characteristics…
Diagram illustrating different types of hydraulic jumps based on the upstream Froude number (F_1).
The jump length, surface profile, turbulence, and performance characteristics vary with F_1.
Undular jumps occur at very low F1, while strong jumps occur at high F1.
Characteristics…
Graphs illustrating the relationships between depth, momentum, and energy relative to critical depth in open-channel flow.
Shows the location of critical depth, and how a sluice gate affects the flow profile.
Example 1
Problem: A hydraulic jump forms in a rectangular open channel with a width of 2 m, where water is flowing at 4 m³/s. The depth just downstream of the jump is 1.25 m. Calculate the theoretical depth upstream of the jump and estimate the energy loss as a percentage of the initial energy.
Example 2
Water flows in a horizontal rectangular channel at a depth of 0.15 m and a velocity of 2.4 m/s. Determine the two possible depths at a location slightly downstream, assuming viscous effects are negligible.
Solution to Example 2
Given:
V_1 = 2.4 m/s
y_1 = 0.15 m
Calculate the Froude number:
Fr1 = \frac{V1}{\sqrt{gy_1}} = \frac{2.4}{\sqrt{9.81 * 0.15}} = 1.978
Since Fr1 > 1, a hydraulic jump could occur with y2 > y_1.
Calculate the sequent depth y_2:
y2 = \frac{y1}{2} (\sqrt{1 + 8Fr_1^2} - 1)
y_2 = \frac{0.15}{2} (\sqrt{1 + 8 * 1.978^2} - 1) = 0.35 m
Hence, either y2 = 0.15 m (no jump) or y2 = 0.35 m (with jump).
Location of Hydraulic Jump
Discussing how to determine where a hydraulic jump will occur in a channel with mild slope.
Key steps:
Determine critical and normal depth.
Find out control points.
Sketch surface variation profiles: GVF and HJ.
Where does the HJ start?
What is upstream depth (y1) and the sequent depth (y2)?
…Location
A detailed method is outlined for determining the location of a hydraulic jump within a gradually varied flow profile.
Plot the M3 profile from point A to B.
Plot the M2 profile from point D to C.
Determine the sequent depth (depth after the HJ) of the AB profile to plot A’B.
Find the intersection of A’B and DC, which is F’ in this case.
The completed surface profile, ignoring jump length, is AHF’D.
The jump occurs between the depths y1 and y2.
Location of Hydraulic Jump
The basic method used to calculate the location of the jump.
Backwater curves AB (M3-profile) and DC (M2-profile) can be computed from control sections A and D.
Curve BA' is plotted by calculating the depths sequent to the depths of curve AB from the jump equation:
Fr1 = \frac{V1}{\sqrt{gy_1}}
The intersection of BA' and DC is at F', where the jump is expected to form.
This method does not account for the length of the jump. If the jump length is estimated (EF), the jump will occur between G and F.
Since the jump length is often small, it is usually neglected in the analysis.
Example 3
Examine the flow conditions in a L = 10 m wide open rectangular channel with n = 0.017 when the flow rate is 400 m^3/s. The channel slope is S0 = 0.015, and a weir D = 5.0 m high with Cw = 3.4 is located at the downstream end of the channel.
Assume that the hydraulic jump happens immediately, connecting with the flow at normal depth.
Given:
The flow discharge over the weir is:
Q = Cw L \left(h + \frac{V0^2}{2g}\right)^{3/2}, where h and V_0 = \frac{Q}{L(D + h)} are the water depth and velocity above the weir.
In gradually varied flow, the energy dissipation with distance is uniform and expressed as:
\frac{EA - EB}{x{AB}} = S0 - Sf, with Sf = \frac{n V{AB}}{R{AB}}^{2/3}, where V{AB} is the average velocity from A to B and R{AB} is the average hydraulic radius.
Questions:
a) What is the water depth above the weir?
b) What is the distance from the weir to the hydraulic jump?
Example 3 - solution
Question a) Water depth above the weir
Solve the water depth at the control point with Q = f(h)
With Q = Cw L \left(h + \frac{V0^2}{2g}\right)^{3/2} and V_0 = \frac{Q}{L(D + h)}, we have:
Q = C_w L \left(h + \frac{\left(\frac{Q}{L(D + h)}\right)^2}{2 g}\right)^{3/2}
Substitute Q = 400 m^3/s, C_w = 3.4, L = 10 m, and D = 5 m:
400 = 3.4 * 10 \left(h + \frac{\left(\frac{400}{10(5 + h)}\right)^2}{2 * 9.8}\right)^{3/2}
Solving for h by trial and error method, we get:
h = 4.2 m
Therefore, the water depth above the weir is 4.2 m.
Example 3 - solution…Continue
Question b) Distance from the weir to the HJ
A sketch of surface variation with normal and critical depth
Find normal depth by applying:
V = \frac{1}{n} R^{2/3} S_0^{1/2}
\frac{400}{10yn} = \frac{1}{0.017} \left( \frac{10yn}{2y_n + 10} \right)^{2/3} 0.015^{1/2}
y_n = 3.45m by trial and error method.
Find the critical depth.
y_c = (\frac{Q}{L})^2 / g )^{1/3} = (\frac{400}{10})^2 / 9.8 )^{1/3} = 5.5 m
Solve depth at section B: Sequent depth of HJ & starting depth of S1
Depth after the HJ is y_B
yC = yn = 3.45 m
V_C = \frac{400}{10 * 3.45} = 13.8 m/s
FrC = \frac{VC}{\sqrt{g y_C}} = 1.99
yB = 0.5 * yC * (\sqrt{1 + 8 * Fr_C^2} - 1) = 8.15 m
Example 3 - solution
Continue b) Distance from the weir to the HJ
Gradual variation from B (8.15m) to A (4.2+5=9.2m)
Apply \frac{EA - EB}{x{AB}} = S0 - Sf with Sf = n \frac{V{AB}}{R{AB}}^{2/3} to work out x_{AB}
V_A = \frac{Q}{L(h+D)} = \frac{400}{10 * 9.2} = 4.35 m/s
VB = \frac{Q}{LyB} = \frac{400}{10 * 8.15} = 4.9 m/s
V_{AB} = 0.5 * (4.35 + 4.9) = 4.63 m/s
R{AB} = 0.5 * (\frac{LyB}{2yB + L} + \frac{LyA}{2y_A + L}) = 3.17 m
Sf = n \frac{V{AB}}{R_{AB}}^{2/3} = 0.017 * \frac{4.63}{3.17}^{2/3} = 0.0013
EB = yB + \frac{V_B^2}{2g} = 8.15 + \frac{4.9^2}{2 * 9.8} = 9.375 m
EA = yA + \frac{V_A^2}{2g} = 9.2 + \frac{4.35^2}{2 * 9.8} = 10.17 m
$$x_{AB} = \frac{