A hydraulic jump occurs in open channel flow whenever supercritical flow is taking place on a slope that cannot maintain the supercritical flow. The hydraulic jump provides an abrupt transition from supecritical flow to subcritical flow.
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 reveiwed briefly. Supercritical flow is characterized by high velocity, shallow depth of flow, and relatively steep channel bottom slope. Subcritical flow, on the other hand, is characterized by low 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 Fr 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
A hydraulic jump occurs wherever a supercritical flow is taking place in a channel that 
doesn't have a slope steep enough to maintain the supercritical flow. The hydraulic jump provides a transition from supercritical to subcritical 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 sluiice 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.
Equation Across a Hydraulic Jump
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.
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 velocity before the hydraulic jump, ft/sec
V2 = average 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:
hL = y1 + V12/2g - (y2 + V22/2g)
Rearranging this equation and substituting the expression for Fr1, leads to the following equation for hL in terms of the two depths of flow and the upstream Froude Number:
hL/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 velocity in the channel before the jump from V = Q/A = 20/(2x1) = 10 ft/sec. Second calculate Fr1. Then you are ready to calculate y2 from eqn 1 and hL 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: hL = 1 - 2.03/1 + (1.762/2)[1 - (1/2.03)2] = 0.143 ft