Need Help Solving This Open-Channel Problem

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Timmy!

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I've already began studying for the October '09 exam (Civil, Water Resources).

I spent about three hours today trying to solve the attached problem, to no avail. Can someone please give me some insight as to how to determine the total discharge "Q" at outlet elevation 120.0 feet? Once I know that, the rest of the problem is easy.

I tried to use Bernoulli's principle but I cannot get it to work out. Working backwards from the correct answer of r=1.80 it appears that the discharge into the semicircular flume should be about 60 CFS, whereas I came up with about 69.7 CFS.

 
To help you get started, the questions relates the flow into the channel which the depth of flow in the channel. Once you have gone through the calcs to convert the head loss to velocity at the discharge you would use the manning equation to get the depth of flow.

 
To help you get started, the questions relates the flow into the channel which the depth of flow in the channel. Once you have gone through the calcs to convert the head loss to velocity at the discharge you would use the manning equation to get the depth of flow.
assume potential head+pressure head at D is H

for C->D, you have 241.4-H=K1*Q1^1.852

for A->D, similarly, 213.9-H=K2*Q2^1.852

and D->B, H-120=K3*(Q1+Q2)^1.852

where, K=4.73*L/(C^1.852*D^4.865)

three unknowns and three eqs, sovle by trial and error: Q1=42.68cfs,Q2=19.02cfs,H=146.27ft

in B,

Q1+Q2=A*C*SQRT(R*S)

where A=2*3.14*R^2

and R=r/2

finally, r=2*R=1.8ft

 
How do you solve this system of equations?

please

assume potential head+pressure head at D is H
for C->D, you have 241.4-H=K1*Q1^1.852

for A->D, similarly, 213.9-H=K2*Q2^1.852

and D->B, H-120=K3*(Q1+Q2)^1.852

where, K=4.73*L/(C^1.852*D^4.865)

three unknowns and three eqs, sovle by trial and error: Q1=42.68cfs,Q2=19.02cfs,H=146.27ft

in B,

Q1+Q2=A*C*SQRT(R*S)

where A=2*3.14*R^2

and R=r/2

finally, r=2*R=1.8ft
 
How do you solve this system of equations?please
HP 35S is what I used. enter the eq correctly then solve for the desired variable. It's good for complex eq. + you can edit eqns fairly quick. It will save you precious min in the exam. With practice and a good review class I solved all the WR probs for the Breath in 6 min or less. Time mgmnt is a big key for the exam that I learned from others.

As long as you comprehend the question the rest is UNIT conversion. The Wiki from this board has a good summary of WR eqns for the breath.

The 35S works the same way for AM transpo questions. :sharkattack:

 
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assume potential head+pressure head at D is H
for C->D, you have 241.4-H=K1*Q1^1.852

for A->D, similarly, 213.9-H=K2*Q2^1.852

and D->B, H-120=K3*(Q1+Q2)^1.852

where, K=4.73*L/(C^1.852*D^4.865)

three unknowns and three eqs, sovle by trial and error: Q1=42.68cfs,Q2=19.02cfs,H=146.27ft

in B,

Q1+Q2=A*C*SQRT(R*S)

where A=2*3.14*R^2

and R=r/2

finally, r=2*R=1.8ft
I used excel, assume a H(a reasonable value, say 180, I will always try a value below 213.9),then get Q1,Q2, based on the calculated Q1 and Q2,you get a new H from the third equation, compare this calcualted H with your assumed H, repeat above.

it is time consuming I know.

 
Thanks so very much for the insight!

I avoided the simultaneous equations by using substitution, as shown.

One question: not being a civil engineer I tried to use the following equations to find the discharge from the tanks:

hf = f(L/D)(v^2/g) where f=(8g)/C^2

Why didn't the above approach work?

 
I used excel, assume a H(a reasonable value, say 180, I will always try a value below 213.9),then get Q1,Q2, based on the calculated Q1 and Q2,you get a new H from the third equation, compare this calcualted H with your assumed H, repeat above.
it is time consuming I know.
I found it easiest to choose an H and then compute the error (delta Q = Qad + Qcd - Qdb)... and adjusting H to minimize the delta Q - it converges pretty quickly. I used a table to track the numbers.

 
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