# Speed of fluid

1. Jan 11, 2016

### Dilman Sidhu

1. The problem statement, all variables and given/known data: River Pascal with a volume flow rate of 4.8×105L/s joins with River Archimedes, which carries 1.04×106L/s , to form the Bernoulli River. The Bernoulli River is 190 m wide and 20 m deep.

2. Relevant equations:
What is the speed of the water in the Bernoulli River?
Express your answer using two significant figures.
Express answer in m/s.

3. The attempt at a solution
I have not yet attempted to write any answers b/c I'm unsure how to solve the equation. I believe I have to use Bernoulli's equation since that is a recent topic in class.

2. Jan 11, 2016

### BvU

Hello Dilman,

Good of you to mention there is a b and a c part in this exercise. Maybe the a part is just a warming up ?
And you can safely assume the B river carries off the sum of the flows so there is no unpleasant accumulation at the confluence....

3. Jan 11, 2016

### Dilman Sidhu

Thanks! I understood the speed of River B would have to include the values given for Rivers 1 and 2. I got an answer of 0.4 m/s which was correct but I was wondering if you could better explain the logic behind why the summation of the given values for Rivers 1 and 2 was divided by 3800 (190 m/s x 20 m/s = 3800).

4. Jan 11, 2016

### BvU

I don't know what you are referring to, but I can easily guess that the division was not by 190 m/s x 20 m/s = 3800 m2/s2 !
Always, always check your dimensions !

Write out the detailed steps of the calculation -- with the dimensions and see that it all fits nicely. Post if you want comments.

5. Jan 11, 2016

### SteamKing

Staff Emeritus
This is an application of the equation of continuity, not so much an application of the Bernoulli equation.

Here, the equation of continuity is "amount of water flowing into the river junction = the amount of water flowing out of the river junction"

In terms of algebra, Qin = Qout
and
Q = A ⋅ v, where
Q - flow in m3 / s
A - cross sectional area of the channel(s), m2
v - velocity of the flow, m/s

The area of the channel of River Bernoulli is A = 190 m × 20 m = 3800 m2

Units are included in calculations to help you make sense of the numbers. They are not intended to be a nuisance.

6. Jan 11, 2016

### Simon Bridge

It is best practise to do the algebra on the symbols, substituting the numbers at the end.
The volume rate of flow Q is given by $Q=vA$ where A is the crossection area of the pipe (in this case the river) and v (perpendicular) is the speed of the flow.
You've already worked out that the flow rate for the Bournoulli river must be the sum of the flows for the other two so $Q = Q_A+Q_P$ ... if v is to be the speed of the Bournoulli river, then A must be its crossection - roughly width times depth: $A=WD$ (you could choose a different approximation for the shape of the river if you like.)

Now all the variables in the definition are in terms of stuff you know, you can substitute and use algebra to solve the resulting equation for v, using symbols only.

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