# Change in speed when water goes from straight line into a curve

1. Jun 10, 2010

### AlRo

My first post, so hope it's in the right place!

If water flowing at 2 m/s in an open channel (say 2m deep by 3m wide) flows into a circle (call the entry angle 30degrees), what happens to the speed of the water? The dimensions are just examples, it's the theory I'm interested in.

I know the velocity changes (as it's a vector, and the direction in the circle is always changing), but what happens to the water speed?

Put it another way - if you put your hand in the water in the open channel, and then you put your hand in the water swirling around the circle, would it be obviously a stronger (ie faster) current?

I'm not too worried about the effect of friction, and we can also ignore the fact that the water has to go somewhere once it gets all the way around the circle to the start (ie where the straight channel joins the circle). Unless you think either of those scenarios are either relevant or interesting to consider!

I guess I'd also be interested to know how long the water takes to go around the circle.

Please include any forces (centripetal, centrifugal, AV, gravitational, upthrust) that are relevant to furthering my understanding. I've looked some of these up, but they have yet to make sufficient sense for my basic question to be answered by them!

I know it's a complicated question, but if you could please answer it in as simple a manner as possible that would be great. Please include your formulas so I can see how you derive things, and where you have inserted values.

2. Jun 10, 2010

Is the circular channel the same size as the straight channel? If so, then the speed would be the same. Consider conservation of mass. If the channels are the same size then any water coming in at some rate would have to go out at the same rate. If you change the size of the circular channel then the water speed would change in a way that preserves conservation of mass. If the circular channel is smaller, then the water would have to move faster. If it is bigger, then it would have to move slower.

3. Jun 10, 2010

### AlRo

Thank you for that - that gives me an answer that I wasn't aware of, and is very useful.

When talking about size, are we talking about the width of the circular channel as compared to the width of the straight channel? Or width and depth? I suppose another way is to say are we talking about the surface area or the volume of water?

I'm trying to understand the variables and hence how the calculations would work.

4. Jun 11, 2010

You want the volume of water right? There is no reason for the surface area to be conserved, only the volume.

Here is an example, think of a section of pipe 1 meter squared in area. Let 1 cubic meter of water go into the pipe per second. This means there must be 1 cubic meter of water leaving the pipe per second. Now how fast is it at the beginning where it enters? One cubic meter enters per second, that's the volume. You get a speed if you divide that by the area of the pipe and the time it takes to enter, Speed = Volume/(Area x Time). In this case, its one meter per second. If your end piece has a smaller area, say half a meter squared, then it would be traveling at 2 meters per second.

It can be tricky to get the length of the curved segment right, should you use the inner or outer length? Well, if its very close to a circular segment then you can take the average of the two. If it is some other type of arc, you cant quite average it like that.

5. Jun 11, 2010

### AlRo

Thanks - that's really useful. It's cleared up quite some confusion for me.

The reason for these questions - a friend has some contacts in Indonesia, trying to work out how much power they might get from a flow rate, for local village projects. They have measured the flow rate at 500 - 600 litres a second, in a channel that varies between 1 - 3 m wide.

So would the best way to calculate the possible power be to use this formula?

Max available power = 0.5 x Cp x p x A x v^3
where
Cp = max available theoretical efficiency (I think) = 0.592
p = density (1000 Kg / m^3)
A = area (m^2)
v = velocity (m/s)

Some questions on the values...

1) Should I use the likely efficiency of whatever kit they find, say 30%, not 59.2%?
2) Do I leave p = 1000?
3) If we take an average channel width of 2m and depth of 1m then the area is 2m^2
4) If the flow rate is 500 litres, that is 0.5 m^3 (ignoring annual variations in the flow rate), I assume that means the speed is volume / (area x time) = 0.5 / (2 x 1) = 0.25 m/s ?
I know I've used speed and velocity to mean the same thing, but am thinking as it's in a straight line the vector aspect doesn't matter.

Slotting those figures into the equation gives a really lousy total of

0.5 x 0.3 x 1000 x 2 x (0.25 x 0.25 x 0.25) = 4.6875 W which seems way too low

So where have I gone wrong?

Is that the total per second, so I need to multiply by 3600 (to give an hourly total) which gives 16875 W or about 16.9 kWh? As that seems too high...

As you can see, I'm more than a little confused about the application of the formula. We'd really appreciate any help you can provide.

Last edited: Jun 11, 2010