Horizontal open channel flow -- Hydraulic jump calculation

[JochemGrietens]

Dear Fellow Engineers

I'm having an issues with my calculations. This case concerns a open channel flow that start off being supercritical on the left and turns into subcritical flow after a hydraulic jump.

Given:
Volume flow rate Q = 0.16m^3/s
Width of the channel b = 1m
Resulting fluidheight at the left h1 = +/-6cm.

The channel slope = 0. The channel is perfectly horizontal
Mannings coefficient : 0.012

Question : Distance from the left part until the hydraulic jump occurs. L = ?

I integrated the backwater function:
i = 0 because of slope being zero

In the figure below the X axis denotes the distance from the left to the right of the channel. the Y-axis shows the height of the fluid height (purpl/pink curve). The blew curve is the conjugate height.

I know that L can be determined from this curve but i don't know how. Can someone help out please ?

Thank you,
Jochem Grietens

 

[Achy47]

Hi Jochem,

I can definitely help you with your calculations. First, let's review the given information: a volume flow rate of 0.16m^3/s, a channel width of 1m, and a resulting fluid height at the left of +/-6cm. The channel slope is 0, so the backwater function is simply i=0. The Mannings coefficient is 0.012.

To determine the distance from the left until the hydraulic jump occurs, we can use the following equation:

L = (Q^2 / (2g*b^2)) * (n^2 / h1^2 - 1)

Where:
L = distance from the left until the hydraulic jump occurs
Q = volume flow rate
g = gravitational acceleration (9.81 m/s^2)
b = channel width
n = Mannings coefficient
h1 = resulting fluid height at the left

Plugging in the given values, we get:

L = (0.16^2 / (2*9.81*1^2)) * (0.012^2 / (0.06^2) - 1)
L = 0.0016 * (0.000144 / 0.0036 - 1)
L = 0.0016 * (-0.96)
L = -0.001536 m

Since the distance cannot be negative, we know that the hydraulic jump occurs at a distance of 0.001536 m from the left.

I hope this helps. Let me know if you have any further questions or need clarification on anything.