Hydraulics, venturi flume questions

[Junkwisch]

Homework Statement

A venturi flume is formed in a horizontal rectangular channel 1.2m wide by locally constricting the channel to a width of 0.9m and raising the floor level through the constriction by a height of 0.2m.

If the flow is 0.45 m³/s and a hydraulic jump forms downstream, calculate the difference of water level upstream and the throat section of the flume.

(The downstream part has the same width as the upstream part, the flume is located in between)

Homework Equations

The equation of continuity, Q=VA
Conservation of Energy, Total Head1=Total Head2
Critical Head,H_c=1.5 Critical depth
The velocity head is equal to v²/2g or (for downstream section) 0.5Critical depth
critical depth,y_c = (q²/g)^(1/3)
q is the flow per unit width, q=Q/width
Head=(v²/2g)+y

The Attempt at a Solution

First I focused on calculating the critical depth, y_c=(q²/g)^(1/3)=((0.45/1.2)²/9.81)^(1/3)=0.243m

Thus the critical head is equal to 0.3645

Since the question is interest in finding the different in depth/height between the upstream part and the flume, I used the conservation of energy to link the head between two points (either upstream or flume to downstream head (critical)). The equation of continuity is used to write the velocity, V in term of depth, Y.

Q=VA, =====> V=Q/A ======> V=0.45(1.2*y)=0.375/y (for upstream) and V=0.45(0.9*y)=0.5/y (for flume)

Due to conservation of energy, the head at the flume will be equal to the head at the downstream section

(v²/2g)+y+0.2(due to rising floor level at the flume) = 0.3645
(v²/2g)+y=0.0645
Since Q=0.5/y, 0.0645=0.01274/y²+y
Thus the equation becomes a cubic equation, I do not know any solution that will solves this equation (since it is not given in the equation). Similarly, if I used the upstream head along with the downstream head, the equation will also result in cubic expression of , 0.3645=0.00717/y²+y

How does one solve this kind of problem? I don't think I missed any expressions

 

[KataKoniK]

.Hello there,

Thank you for your question. Your approach to using the equations of continuity and conservation of energy is correct, but there are a few things you can do to simplify the problem and solve it more easily.

First, you can use the equation of continuity to relate the velocities at the upstream and flume sections. Since the flow rate is the same at both sections, the velocities will be inversely proportional to the cross-sectional areas. This means that V_upstream = (1.2/0.9)V_flume.

Next, you can use the equation for the critical head to relate the critical depths at the upstream and flume sections. Since the channel width is the same at both sections, the critical depth will also be inversely proportional to the velocities. This means that yc_upstream = (0.9/1.2)yc_flume.

Now, you can use the equation of continuity to relate the velocities at the downstream and flume sections. Since the flow rate is the same at both sections, the velocities will be inversely proportional to the cross-sectional areas. This means that V_downstream = (1.2/0.9)V_flume.

Finally, you can use the conservation of energy equation to relate the heads at the upstream and downstream sections, using the velocities and critical depths that you have just related.

Putting all of these relationships together, you should be able to solve for the difference in water level between the upstream and flume sections.

I hope this helps. Good luck with your calculations!