Ok math people (conversion problem)

[cartman]

I'm having a heck of a time trying to figure something out. I know I have some needed information missing, but not sure how I should approach it.

If you have the "cubic feet per second" (cps) flow rate of a river, how do you convert this into miles per hour for the flow of the river? I'm guessing at a minimum you'd need the average width of the river channel, and the average water depth. Any ideas for a formula to use?

[RendeR]

The speed of the river is going to change depending on all those items you listed so the question is are you looking for the speed of the river at a certain point on the river or ar you looking for an average speed for the river from one point to another?

If the latter is the case simply ignor the cubic notation and take your data for Feet/sec and convert to miles/hour. This won't be precise of course but will give you a general idea of the velocity that you're looking for.

[oliegirl]

I must leave this thread before my head explodes! Gah...too much math!!!!

[JediKooter]

Answer: It's flowing faster than it looks.

[cartman]

Yeah, we are looking for the average speed between two points. I know the answer won't be exact, but a ballpark guesstimate will help.

I think I might have it figured out what you were saying, RendeR. If the rate is 200cfs, and the avg. width of the river channel is 50 feet, and the average depth is 2 feet, that means 2 feet of distance was measured by the sensor (50x2x2 is 200). So you are saying take the 2f/s and convert that to mph. Which would be 1.36 mph.

[Passacaglia]

I'm not sure how you can do the conversion -- a cubic foot is a measurement of volume, and a mile is a measurement of length.

[cartman]

Originally Posted by Passacaglia

I'm not sure how you can do the conversion -- a cubic foot is a measurement of volume, and a mile is a measurement of length.

True, but I think you can derive the length portion out of the cubic foot (l x w x h) to get the part needed, if you know the width and depth of the river.

[Passacaglia]

Originally Posted by cartman

True, but I think you can derive the length portion out of the cubic foot (l x w x h) to get the part needed, if you know the width and depth of the river.

Yes, that makes sense.

[gstelmack]

Yeah, just take a 1ftx1ftx1ft cube, your flow-rate = your speed.

[Toddzilla]

That is an invalid comparison.

1 cubic foot per second through a drinking straw would produce a much faster flow than 1 cubic foot per second through a sewer pipe which is faster still than 1 cubic foot per second down a river.

Pass is correct - one is a rate of volume, the other is a rate of speed, and you cannot compare the two.

[Toddzilla]

Originally Posted by cartman

True, but I think you can derive the length portion out of the cubic foot (l x w x h) to get the part needed, if you know the width and depth of the river.

No, because - as stated above - the length of the cubic foot is irrelevant in terms of cubic feet per second. One is a measure of volume, the other is of distance. Apples and oranges.

[cartman]

I understand what you are saying, but I guess what I'm missing is that there is a common component between the two measurements, one of the measure of feet (f/sec and f³/sec).

To use my assumptions from above on the 50 foot wide 2 foot deep river that has a 200cfm flow, the speed of the water would definitely be faster if another river, also with a 200cfm flow was 25 foot wide and 2 foot deep. To get 200cfm there, the distance portion would be 4 feet (25x2x4 =200), and the speed would be 2.72mph. If the river was 1 foot wide and 1 foot deep (the straw example) with a 200cfm flow, then 200 feet would be the distance measured, and the speed would be ~135 mph.

[sterlingice]

I seem to remember a ridiculous exercise like this from Physics one time and was wondering what one had to do with the other.

SI

[Daimyo]

If you know the area of a cross-section of the river (ie, the average depth and width) you should be able to convert between the two easily enough.

[flounder]

From the Army Water Measurement Manual.

4. Discharge-Area-Velocity Relationships

Flow rate or discharge, Q, is the volume of water in cubic feet passing a flow section per unit time, usually measured in cubic feet per second (ft3/s). The distance, dv, in feet that water will travel at a given velocity in a pipe of constant diameter is velocity, V, in feet per second (ft/s) multiplied by time, t, in seconds, or:

dv = V * t

The volume, Vo, in cubic feet passing from the upstream to the downstream ends of this distance is the distance, dv, in feet times area, A, in square feet of the flow section. Thus:

Vo = dv * A = A * V * t

To get the time rate of flow or discharge, Q, in cubic feet per second, divide the right and left sides of equation 2-8 by time, t, in seconds, resulting in:

Q = A * V

So the velocity V is Q / A. Or in other words, what Daimyo and cartman said.

[JeeberD]

Dude, just tie your beer tube to your tube and then you won't lose it no matter how fast the river is going. Geez...

[RPI-Fan]

Confirming what others said... V = Q/A (and A = HW for the longitudinal velocity).