Let's say I have a vertical pipe, 1 inch in diameter, 12 inches tall and sealed at the bottom. I drill three holes, horizontally, at 1 inch, 2 inches and 3 inches starting from the bottom. I fill the pipe with water. I would like each hole to spout water the same distance away from the pipe. What size do i drill each hole, and more importantly what formula is used in determining the hole size?
The pressure drop across an orifice will increase as the square of the flow rate. Conversely, the flow rate will be the square root of the pressure.
The pressure will not vary due to the hole size in this case, since it is fixed by the height of the water column. The hole size at each point will therefore not affect the pressure available.
Changing the hole size will change the amount of water, but I don't expect it to change the pressure, or thus the exit velocity. The distance away from the pipe is a function of the velocity, and the local gravity, which is also hard to control with todays technology.
You could cheat, and have an electrostatic deflector for each stream...
Or do something funky with the nozzle design, that might "adjust" the flow velocity after it leaves the orifice.
OK, Someone show me where I am wrong about this.
It can't happen under gravity feed. The hole size has nothing to do with velocity, only volume. So long as the holes are large enough to not be affected by surface tension, the velocity of the stream from each hole will be proportional to the pressure, which, of course, will be less, the higher the hole.
LLoyd
Sure, but then what you're discussing is "nozzles", not plain holes.
LLoyd
But it will. The smaller the hole, the higher the friction/pressure-loss (Reynolds) and the less the pressure at the *outer* end of the hole. Pressure at the *inner* end of the hole will vary with height, no doubt.
Best thing is to make some jets and screw them in, finding the right diameter (and jet-length) by experimenting. Of course a "long hole" (read tube) increases pressure drop. So for cheating, put some wounded tube in front of the hole.
Nick
On Thu, 05 Apr 2007 21:39:54 -0400, with neither quill nor qualm, snipped-for-privacy@here.now quickly quoth:
Using RCM for your homeworks solutions, youngandgrouchy?
What if you used a lateral thinking approach? drill all holes of equal size. Attach a bent piece of pipe from the inside of each of the lower two holes which ends up at the level of the upper hole. You should now have equal pressure at all 3 holes and equal size water spouts. Or am I wrong?? Klaus
Wouldn't you still have a column of water inside your tube that is causing the pressure to rise for the lower holes?
What about the flow friction in the "bent pipes" ? Maybe could be compensated for. ?? ...lew...
Wrong. The column height of the pipe is still water column height, and will cause the pressure at the hole to be the same as it would if there were only a hole.
You will change the performance of the system in one way, though. Now, all three "holes" will run out of water at approximately the same time. (little delay to empty the pipes, but quicker than if the whole tank had to drain down).
LLoyd
Exit velocity is more a fn of pressure (head) than diameter -- but what you can do is drill the three holes oversize and solder or braze in orifices of same diameter but different lengths -- little rods with holes in them. Longer passages will present more drag, thus equalizing the pressures at the exits.
Look in Marks' Handbook for suitable formulae. Get to know Mr. Reynolds.
I think one thing we all missed on the OP's subject is that the three holes are only an inch apart, and near the bottom of the 12" column. There's not going to be much difference in the pressure, and the three streams will likely merge and "heal" from surface tension only a small distance away from the column.
LLoyd
Its one of those things that might work, but for the wrong reason.
Dave
Of course, whatever he does will be affected by the pressure drop as the water level in the tube decreases. Adjusting by means of jet size and length will be good for only a very narrow pressure range.
John Martin
As others have stated, for dry water there is no way to do it.
If your water is wet (no, this is NOT a joke, it's about viscosity and Reynolds number), there is a pressure drop in each aperture that can be tuned to alter the outflow speed of each spout. If you want 'the same distance' at THE BOTTOM, you will want the velocities to result in parabolas that are ballistic trajectories with horizontal initial velocities V3, V2, V1 and the resulting equation is
V3*T3 = V2*T2=V1*T1
where T3= k*sqrt(3), T2= k*sqrt(2), T1= k*sqrt(1)
For dry water, the flow from one hole doesn't interfere with the pressure on the others, but your water has to be wet, so use narrow holes. For dry water,
V3**2/2 = g*(12-3) V2**2/2 = g*(12-2) V1**2/2 = g*(12-1)
where g is the gravity constant (about 380 in/sec**2)
but your wet water is going to be rather slower because of viscous drag in the orifices. How much slower, is a VAST problem to put into a formula; semiempirical solutions exist.
On second thought, this is perfectly doable. Make each hole .004" diameter. The water will seep out, and drop straight down.
John Martin
You will somehow have to reduce the flow rate on the bottom two holes. One way would be to drill the bottom two holes to accomodate a tube that would be inserted in the hole with the id the same size as the top hole. The length of the tube would, by friction, reduce the flow rate to the same as the top hole. You could crip the tube to get the same results. The tube itself would be inside the vertical pipe.
John
hahaha, nope, but I imagine it looks like that. I'm just an old man playing in my shop, expermenting with vortex tube design in quest of a design that will allow co2 extraction efficently. Book learning vs shop experience...I'll take shop experience any day. There is enough on RCM to rebuilt the USA after the fall.
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