Gravity-drain pipe-sizing problem.

[NadZ]

Hello all,

I am designing a drain system to handle a failure of a 5 to 11 gal/min 100 degF water line rupture. The trough is to be shallow (perhaps 6 in max) and must not overflow. There is no pump, just gravity to drain the 6 ft, vertical pipe with maybe one or two 45 deg bends and is smooth stainless or CuNi.

I'm new at fluids, and I would appreciate a sanity check. For drain pipes ranging from .5 in to 3 in I.D., I find Reynolds numbers ranging from 110000 down to 17000, and residency times from .3 sec to 12 (!) sec. At around 1 to 1.5 in I.D. the residency times are a realistic 1.4 sec, but as the diameter gets higher, the residency times rise to unbelievable numbers. I think my calcs do not include aeration, since I can't imagine a free-flowing pipe taking 12 seconds to get rid of a 6-foot column of water!

My primary reference for this was

http://www.pressure-drop-calculator.com

.

So what do you, my esteemed collegues, think? Will a 1.5 in by 6-ft pipe drain 11 gpm?

Thanks in advance, David

 

[pipesnpumps]

Your description is hard to follow. How is the drain line routed, vertically or horizontally?

If horizontally, a 2" pipe will flow 11.7 gpm at 50% fill ratio and a minimum slope of 0.0291 ft/ft.

This is from a plumbing book. A Reynolds number is too complex for plumbing, you need to bring it down a notch and get practical

You'll need a bell reducer (floor drain would be best) for the entrance, I'd guess at least a 6 x 2, or other size rated by the manufacturer for 11 gpm or more.

 

[Tremolo]

Starting with the simplest case, assume the drain pipe is open (i.e., no drain cover or grating) and ignore any entrance losses. Then the maximum drain velocity for a driving head equal to the trough depth (0.5 ft) is:

h = v^2/2g

where,

h = head, ft
v = velocity, fps
g = gravity 32.2 ft/s^2

Solved for v (ft/sec):

v = sqrt(2gh) = sqrt(2*32.2*0.5) = 5.67 fps

For a 0.5" drain diameter (Area = 0.00136 ft^2), the volumetric flow is 5.67 fps * 0.00136 ft^2 = .00771 ft^3/s = .00771 ft^3/s * 60 s/min *7.48 gal/ft^3 = 3.46 gpm.

This does not meet your design leak rate of 5 to 11 gpm, therefore the trough would overflow.

For the maximum drain size of 3" (Area = 0.0491 ft^2), the maximum drain rate is 125 gpm, which exceeds your design leak rate by more than a factor of 10. At the maximum drain size, you have substantial margin and the trough would not overflow.

The actual drain rate would be less than that shown because of entrance losses into the drain (especially if there is a grated drain cover). However, there is signficant margin at the maximum drain diameter, so entrance losses should not change the conclusion.

Once the water enters the 6' long pipe, I believe it will drain adequately despite the pipe friction losses. The limiting factor is the shallow driving head of the trough.

 

[NadZ]

Thanks guys,

my further calculations show that a little less than an inch diameter is the break-even point where the friction losses are canceled by gravity on the column, and the thing still flows at 11 gpm without any head. I assume no trough (just a small high-pressure pipe dumping into a larger pipe), and the entrance effects go away in a couple of inches.

I'll probably ask for 1.5 in to provide margin. I know you understand margin, Trem-the-nuc! Thanks for your take on trough, tho- gonna use that.

-David

 

[Tremolo]

1.5" sounds like a winner. Good luck.