Calculating the Constant for Measuring Flow Rates in Open Channels

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The discussion focuses on calculating the constant for measuring flow rates in open channels using a weir. The original equation for a rectangular weir is given as Q=288.8 LH^(3/2), where Q is in gallons per hour, and L and H are in inches. To convert this equation for larger channels, the units must be changed to cubic feet per second, leading to a new constant of 5.35. Participants emphasize the importance of unit conversions, specifically from gallons per hour to cubic feet per second and inches to feet. The conversation highlights the need to understand the units of the original constant to derive the new one effectively.
Frustra8ed
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I'm having trouble a problem and I was hoping someone could show me some direction.

Q:
A weir is used to measure flow rates in open channels. For a rectangular weir the expression can be written as Q=288.8 LH^(3/2).
Where Q= discharge rate, in gal/h
L= Length of weir opening parallel to liquid, in inches
H= height of fluid above crest, in inches

As the channels become larger, the weir opening can be expressed as follows:
Q in cubic feet per second, with L and H in feet.

Determine the new constant.


The answer is 5.35, and I can't figure out how they derived it. I think it has to do with converting units, but I don't know.

Any help is appreciated.

Thanks,

Frustra8ed :confused:
 
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You're right, it's all about the units.

You have Q, which is initially in gal/hour, and you want to get it into ft^3/second. How would you do this?

You have L, which is initially in inches, and you want to get it into feet.

W is like L.

--J
 
You have the right idea. Set up the same equation but add the units conversions on the appropriate sides:

One side will be the conversion from gal/hr to ft^3/sec. The other side will be the unknown times the two conversions from ft to in. Don't forget that one of those connversions is to a power. Solve for the unknown.

That being said, I went through this problem 4 times and I keep getting the answer, except a factor of 10 off. I keep getting 53.9. Maybe someone else can verify that.
 
I'm getting 5.35.

The easier way to do this problem is to figure out the units of the constant 288.8, and then convert these units into the units of the 5.35 constant.

The 288.8 constant has units gal/hr/in^(5/2). The desired units are ft^.5/s.

--J
 
Thanks!
I was on the right track, just needed a boost. I appreciate the help.

Frustra8ed :approve:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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