1. ## Curious about parts of the velocity flow formula

I'm taking hydraulics this semester, and am trying to understand all parts of the formulas we use so I can hopefully actually grasp the concepts instead of simply crunching numbers. We went over the velocity flow formula the other day and the formulas we used were
V=12.14(square root of the pressure) when the pressure is known,
or V=8(square root of Head ) when the head pressure is known.

I can't find a correlation for either number in the above formulas. It seems like the number being multiplied is usually the weight of a 12" column of water i square inch at the base, or some other real world reference, and it seems like where the book is using a technical coefficient, it refers to the number in those terms. With these formulas, however, it simply dumps the numbers out and moves on. What exactly am I multiplying together here? Any help would be appreciated. We're using the Brady 3rd edition of FD Hydraulics.

Thanks!

2. Velocity Flow
V = 8 √H (when head is known)
V = velocity flow in feet per second
V = 12.14 √P (when pressure is known)

3. Phaedrus: I assume you are referring to Page 92 in Brady. Taking the equation of V = Sq.Rt. of 2 GH near the bottom of the page. The value of G is 32 ft/Sec. x Sec. (acceleration due to gravity) the dimensional analysis is then (Sq Rt. of 64 x H) but the dimensions are Ft. (feet of head) x Ft./ Sec. x Sec. When you actually take the square root the 64 becomes 8 and the dimensional analysis becomes ft./sec. while the H remains under the radical. It should be obvious that the dimensions for V are in feet per second, so the answer must also be in feet per second.
If we replace H with psi. we convert using o.435 psi per foot of head, but that value is in the divisor and under the radical...so take the sq. rt. of 0.435 (0.6595) and divide the 8 by that value and we get 12.13 for the multiplier of the square root of the pressure.

4. I won't tell you how many times I had to read what you wrote to get it all squared away in my skull, but I think that makes sense. Thanks for the reply!

5. Originally Posted by kuh shise
Phaedrus: I assume you are referring to Page 92 in Brady. Taking the equation of V = Sq.Rt. of 2 GH near the bottom of the page. The value of G is 32 ft/Sec. x Sec. (acceleration due to gravity) the dimensional analysis is then (Sq Rt. of 64 x H) but the dimensions are Ft. (feet of head) x Ft./ Sec. x Sec. When you actually take the square root the 64 becomes 8 and the dimensional analysis becomes ft./sec. while the H remains under the radical. It should be obvious that the dimensions for V are in feet per second, so the answer must also be in feet per second.
If we replace H with psi. we convert using o.435 psi per foot of head, but that value is in the divisor and under the radical...so take the sq. rt. of 0.435 (0.6595) and divide the 8 by that value and we get 12.13 for the multiplier of the square root of the pressure.
I had a math instructor in school who's favorite quote was "It's intuitivly obvious to the most casual observor". Yeah Right.

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