Greetings…
In the last six years I have designed and/or managed the design of civil improvements for three federal prisons and one state prison here in CA. ScottBortz is dealing with a real requirement, as strange as it may seem.
The Federal Bureau of Prisons currently requires that ALL pipes crossing under the double perimeter security fences be 8" or smaller. It used to be that water and sewer could be 12", but that changed just before my last prison project. I was told that they made the change was because a prisoner tried to get out through a 12" sewer!!! So now, nothing bigger than 8".
FYI, when I was younger I got inside a 20' stick of 18" pipe to check on a problem with a joint, and the forman on that job was so skinny he said he had once entered a 14" pipe to do the same.
Anyway, Scott, mroberts approach is a good one. Each of my three federal prisons had 48" SD pipes (well…1220 mm since this is a federal job!) in and out of the "pipe matrix" and I conservatively used 51x8" pipes. At FCI Herlong (the most recent) the 48" pipes have a slope of about 0.0008 and change, but we used 0.0025 for the 8" pipes in the pipe matrix.
On my first federal prison, which is the one where I did these calcs myself, I remember looking at this as a multi-barreled culvert problem with submerged inlet and outlet (a common condition in our flat country). Here's my approach:
* To compare the hydraulics of one large pipe to several small pipes, set up Mannings Equation for full flow in a culvert in terms of Q (i.e. Mannings Equation + entrance and exit losses)…I hope this is typed correctly…
Q = A * sqrt(dH/(((Kin+Kout)/2g)+((L*n^2)/((1.486^2)*(R^(4/3))))))
Q = flow (cfs)
A = pipe flow area (sf) = 0.25*pi*d^2
dH = head loss across the pipe matrix (feet)
Kin = entrance loss coefficient = 0.4
Kout = exit loss coefficient = 1.0
g = gravitational constant = 32.2 ft/s^2
L = pipe length (feet)
n = Mannings n
R = hydraulic radius (feet)
* Pick a head loss across the pipe matrix (it doesn't make much of a difference, I checked it for dH = 0.1' to 0.25' just because that's about what the guy who ran the hydraulic model got when he used a 48" pipe in lieu of the pipe matrix…and the answers were all pretty close)
* Calculate the flow through one large pipe based on the equation above.
* Calculate the flow through one small pipe based on the equation above, using the same head loss.
* Divide Q(big pipe) by Q(small pipe) to get the minimum number of pipes you should use, then add a few more.
One caveat though: if your big pipes are not flowing full, then this method cannot be applied as described herein. Open channel flow in the large pipe would then equate to full flow in the bottom pipes in the pipe matrix and possibly open channel flow in the top row of the pipe matrix. Even so, my method should be sound.
I determined the slope of the pipe martix (0.0025) by assuming the bottom row of pipes in the pipe matrix were half full and I wanted about 2 fps.
I hope this helps…or at least isn't too confusing.
Fred