Calculating the Constant for Measuring Flow Rates in Open Channels

  • Thread starter Thread starter Frustra8ed
  • Start date Start date
  • Tags Tags
    Constant
Click For Summary

Homework Help Overview

The problem involves calculating a constant for measuring flow rates in open channels using a weir. The original poster presents an equation for discharge rate and seeks to understand how to convert it for different units.

Discussion Character

  • Exploratory, Assumption checking, Unit conversion

Approaches and Questions Raised

  • Participants discuss the importance of unit conversions between gallons per hour and cubic feet per second, as well as inches to feet. There is a focus on setting up the equation correctly with the appropriate conversions.

Discussion Status

Some participants have provided guidance on how to approach the unit conversion and have shared their own calculations. There is a recognition of differing results, with one participant noting a consistent discrepancy in their calculations.

Contextual Notes

The original poster expresses uncertainty about the derivation of the constant and mentions a specific answer they have encountered. The discussion includes verification of calculations and the exploration of unit relationships.

Frustra8ed
Messages
4
Reaction score
0
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:
 
Physics news on Phys.org
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:
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 56 ·
2
Replies
56
Views
4K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K