# How can I calculate the flow rate necessary for a given cross sectional area, angle, and length I'm making a toy for my kids and this problem came up. I have a channel on a slight angle (angle is between ground and length of channel) and I'm pouring water into it. I want to know how quickly I need to pour water in to make it flow continuously at a given height.

So the water is (meant to be) .5m wide, .2m high, and 1m long, angled at 5 degrees. How fast is that water going to fall out, in other words, how fast do I have to pour it in?

I'm particularly looking for the technique to do this, not just an answer.

The part I find very confusing is that the water at the top of the box will accelerate a little, and have a low velocity, while the water at the bottom of the box will have accelerated for a while, and have a higher velocity. But intuition tells me the water will stay cohesive, for lack of a better word. Can someone explain what's going on here and how this can be calculated?

• The situation is completely unclear. Is the box full? What do you mean flow continuously? – Ron Maimon Jul 7 '12 at 6:02
• Also you need to state the angle of inclination more clearly. 5 degrees with respect to what? – upapilot Jul 7 '12 at 8:27
• @RonMaimon - I appreciate your help, but 'completely unclear?' I stated the volume of water and cross sectional area of water (well, indirectly, with all dimensions), the point is gravity is going to remove this water at a certain rate and I need to replenish it at the same rate. I showed this question to an engineer and he said 'open channel flow rate, mannings equation'. – Henry Jul 7 '12 at 22:11
• @upapilot 5 degrees with respect to the surface of the Earth. – Henry Jul 7 '12 at 22:11
• – Bernhard Jul 10 '12 at 5:41

I was told by an engineer to use Manning's equation for open channel flow rate, as described here http://en.wikipedia.org/wiki/Manning_formula

Manning coefficient for some common materials: http://www.engineeringtoolbox.com/mannings-roughness-d_799.html - in my case it was acrylic sheet so 0.009 worked fine

Combining with discharge as stated in the wikipedia article means you can avoid calculating the velocity if you don't need it.

Q = cubic meters per second

A = .2 * .5 (cross sectional area in meters squared)

Rh = A / P, P is the wetted perimeter in this case .2 + .2 + .5

S = 0.09 (tan(5 degrees))

k = m^1/3/s

n = 0.009

So 0.08m^3/s, or in liters, 80 liters/second

Hmm, not sure if that is actually correct, but it's the right approach, and if correct, tells me I need to decrease the angle and decrease the depth in order to achieve a flow rate that I can find a cheap pump for!

• How is this an answer? The amount that flows out depends on how much flows in--- if you put more in, more comes out. – Ron Maimon Jul 8 '12 at 7:08
• @RonMaimon well, if you put more in, the cross sectional area increases, and if you put less in, it decreases. So if you put more in, yes more comes out, but it's no longer the same open channel situation (eg, you are overflowing the channel). This gives me the 'correct' amount to put in and take out (they are equal) to achieve this particular channel configuration. Does that make sense? – Henry Jul 8 '12 at 16:59
• What are you talking about!! It is impossible to understand you. I am voting to close and -1 on this answer. – Ron Maimon Jul 8 '12 at 17:59
• What is the "channel"? Why does the cross sectional area increase? You are always overflowing the channel when stuff comes out. Please draw a picture or link a movie--- I don't think you are describing something real. – Ron Maimon Jul 8 '12 at 18:02
• Please refer to this article for a drawing and some discussion: tiny.cc/9pr7gw In the context of my question and answer, imagine the channel is 3y tall, and I am aiming for a cross sectional area of water y tall. In other words I am controlling the flow rate in order to achieve a given area and hydraulic radius. To determine the correct flow rate, I use the manning equation and solve for Q, giving me a value to search out a pump. Also, please relax :) – Henry Jul 10 '12 at 4:53

There is a frictional force between the water flowing in the channel and the sides of the channel. This force slows the water down. If you wish to have the height of the water stay constant through the channel you will have to balance this force with that of gravity. This frictional force will depend on the velocity of the water; the faster the water the higher the frictional force.

To estimate the frictional force, look for equations that give the decrease in "head" in water due to flowing through the pipe. That is, if you apply a certain pressure to a long open pipe, you will get a certain flow. If you make the pipe longer, the flow will decrease. Since the pressure on the exit end of the pipe is zero, you can compute how much work was done on the water (in moving it from a pressure P to a pressure of 0). The work done on the water this way is equal to the frictional work = force x distance, and so you can calculate the friction.

Since you can adjust the gravitational force by changing the angle of tilt of the channel, you can rest assured that for any (reasonable) velocity, you can balance the frictional force with the gravitational force. By "reasonable" I mean that you can't make the water fall faster than gravity with no friction. And you channel needs to have a region towards the top where either (1) the water is slower and so the channel is higher, or (2) the water has an initial non-zero velocity because, for example, a pressure.

I think this should be enough to get you started on this problem. If you have further questions, put them in the comments and someone will probably work on it further.