# Homework Help: Hydraulics, venturi flume questions

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1. Oct 23, 2015

### Junkwisch

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
The equation of continuity, Q=VA
The velocity head is equal to v2/2g or (for downstream section) 0.5Critical depth
critical depth,yc = (q2/g)(1/3)
q is the flow per unit width, q=Q/width
3. The attempt at a solution
First I focused on calculating the critical depth, yc=(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

2. Oct 28, 2015