Do flow resistance in parallel add like parallel resistances?

[MoreGooder]

I was reading this article:

Maximizing Flow Rate (http://www.procooling.com/articles/html/maximizing_flow_rates_with_h2o.php)

and started to wonder about my upcoming revision to my water cooling rig. My current plan is to connect two 1/2" res outlets into a very large 1" ID "T" fitting, then connect 1" ID tubing from the T to my MAG 3 Pump like shown in this thread:
Linque (http://www.ocforums.com/showthread.php?s=&threadid=274198)

Unfortuately, the 1"-1"-1" T fitting is so huge that it is going to be next to impossible to cram all of that tubing into my new external box without using 90° elbows at both entrances of the T fitting. According to the Darcy equation, K = f(L/D) where f = friction factor, L = Length and D = diameter. the K coefficient for a 90° miter Elbow is 60f for a 1/2" to 1/2" fitting.

Can one draw an analogy between flow restriction, and therefore the K coefficient, to electical resistance? If so, would my two 90° elbows feading the two T inlets add like two resistances in parallel, K1K2/(K1 + K2)?

[Xymurgy]

I make the parallel analogy to aerodynamics. Makes much more sense to me. Basically the faster a car goes (water flows) the more air resistance/friction (head loss). So it makes more fuel sense to run two cars at 40 mph for a distance than to run one car at 80 mph because they are less affected by resistance.

Not sure if this answers your question, but two 90 elbows in parallel will be affected as two resistances in parallel.

Something else to keep in mind, is that I'd assume the resistance from a 1" 90 elbow to be rather negligable. Rule of thumb is to not use elbows, but this mainly applies to people using 1/2 or 3/8 ID tubing. You didn't say you were using 1" elbows, but with the setup you're thinking of, it would be possible (and decrease resistance at the same time).

[UberBlue]

I've always tought of water-cooling loops as electrical circuits. In series add, parrallel divide.

I *think* the above applies to K coefficients also. So yes, your two 90°'s feeding a T would be K1K2/(K1 + K2)?

[MoreGooder]

Here's a pic of the 1" 1" 1" T fitting with the 1" to 1/2" 90° Elbows installed. The large tubing shown is 1" ID and will slip over the inlet of the MAG3 pump.

[MoreGooder]

Bump.....

Opinions on the pic I posted above? Good idea? Stupid? I'm just about ready to do final assembly on my new external box, so any opinions would be appreciated.

[pauldenton]

Originally posted by MoreGooder
Bump.....

Opinions on the pic I posted above? Good idea? Stupid? I'm just about ready to do final assembly on my new external box, so any opinions would be appreciated.

hmm - the size differential looks a lot bigger than i would have expected... are you sure the elbows aren't 1/2" OD?

the elbows look awfully restrictive - is there no way to get a smoother setup? maybe given a bit more distance? a 90 degree curve is a lot less restrictive than an elbow iirc

[pauldenton]

just a thought but i take it you couldn't use 1" elbows, then reducers...

[MoreGooder]

Yeah, the sizes are a bit decieving in the picture. The 90° elbows are 1" NPT X 1/2" Barb, which fit properly into 1/2" ID Clearflex tubing. The monsterous tubing is 1" ID Tygon.

I also have fittings that are 1" NPT X 1/2" Barb straight (no 90°). Problem with them is that they will require significant space inside my external rig dedicated to tubing loops long enough to avoid kinks. Secondly, all of that tubing will interrupt airflow through my radiator because there will need to be air flow into the external box, around all that mess of tubing, and then through the radiator to finally exhaust outside of the box.

I think I've convinced myself that the 90's are necessary, but darn it I wanted to avoid 90's all together. Then again, I don't expect to see a substantial decrease in temps anyway through all of this effort. My goal is to get something that is more cosmetically pleasing and less exposed.

[MoreGooder]

Originally posted by pauldenton
just a thought but i take it you couldn't use 1" elbows, then reducers...


Great suggestion. However, there isn't enough room. The T fitting already consumes more space than I had visualized.

[pauldenton]

Originally posted by MoreGooder
Yeah, the sizes are a bit decieving in the picture. The 90� elbows are 1" NPT X 1/2" Barb, which fit properly into 1/2" ID Clearflex tubing. The monsterous tubing is 1" ID Tygon.

I also have fittings that are 1" NPT X 1/2" Barb straight (no 90�). Problem with them is that they will require significant space inside my external rig dedicated to tubing loops long enough to avoid kinks. Secondly, all of that tubing will interrupt airflow through my radiator because there will need to be air flow into the external box, around all that mess of tubing, and then through the radiator to finally exhaust outside of the box.

I think I've convinced myself that the 90's are necessary, but darn it I wanted to avoid 90's all together. Then again, I don't expect to see a substantial decrease in temps anyway through all of this effort. My goal is to get something that is more cosmetically pleasing and less exposed.

hmm - presumably the 1" tubing is on a 1" barb X 1" NPT.... i take it doing away with all the other hardware by dritting a new hole in the res tapped for 1" NPT isn't possible.... (or even using a different res...)

[Joe Camel]

do the 1" 90s take up too much room "side to side"? if it could fit...what if you put the T on the bottom of the case with the 90's (1" if they fit) at a 45 angle with the 2 x 1/2" tube diagonal up to the rez along the sides of the box (for air flow)

[pauldenton]

Originally posted by MoreGooder



Great suggestion. However, there isn't enough room. The T fitting already consumes more space than I had visualized.

what about having the 1" fitting on one of the arms of the "T" instead of the post (if you see what i mean) - then you could have a straight 1/2" fitting on the other arm (as there would be no need to change direction) and one elbow on the post (assuming you couldn't then get away with a less restrictive fitting....)

[MoreGooder]

Originally posted by pauldenton


hmm - presumably the 1" tubing is on a 1" barb X 1" NPT.... i take it doing away with all the other hardware by dritting a new hole in the res tapped for 1" NPT isn't possible.... (or even using a different res...)

You are correct. The middle of the T is 1" NPT with a 1" ID barb. The tube will slip over the outside of the MAG 3 inlet.

The res I will be using can be seen in this thread:
Link (http://www.ocforums.com/showthread.php?s=&threadid=274198)
(Scoll down a bit)

The res is totally sealed and still under warranty, so I don't want to mod it.

Besides, for 1" NPT thread, there wouldn't be enough threads to hold it in place due to the fact that the material to be threaded is only 1/4" thick. Plus, a 1" NPT tap would be too pricy to just do one hole.

Thanks for the suggestion however.

[MoreGooder]

Originally posted by Joe Camel
do the 1" 90s take up too much room "side to side"? if it could fit...what if you put the T on the bottom of the case with the 90's (1" if they fit) at a 45 angle with the 2 x 1/2" tube diagonal up to the rez along the sides of the box (for air flow)

Yes, that might work. I'll attempt it during install. I've played with the tubing while connected to the straight reducers to know that considerable space is required to prevent kinking. But, I'll give this idea a try.

Thanks!

[MoreGooder]

Originally posted by pauldenton


what about having the 1" fitting on one of the arms of the "T" instead of the post (if you see what i mean) - then you could have a straight 1/2" fitting on the other arm (as there would be no need to change direction) and one elbow on the post (assuming you couldn't then get away with a less restrictive fitting....)

I think I understand what you're saying. That would also be a possibility. I worry about having enough room lengthwise in the case though. I'll give it a try.

[cpufan]

Originally posted by MoreGooder
According to the Darcy equation, K = f(L/D) where f = friction factor, L = Length and D = diameter. the K coefficient for a 90° miter Elbow is 60f for a 1/2" to 1/2" fitting.

The Darcy-Weisbach equation, from the middle of the 19th century, conveniently expresses the frictional loss in in a length of pipe , in terms of the velocity head. You have left the velocity head (v^2/2g) out above. The equation should read h=f(L/D)V^2/g, where h = the headloss in the tube due to friction. Darcy did not address losses in fittings.

I think the confusion rises from the referenced article, which relates losses in fittings to the Darcy-Weisbach friction factor, f. This is perfectly valid, but not really necessary for your analisys.

What you really need to know is that the loss in fittings is proportional to the velocity head (v^2/2g).

Let's assume a loss coefficient (k) of 0.9 (a reasonable assumption).

Let's further assume a flowrate of 2.1 gallons per minute.

Now the velocities. Since the 90 degree fittings are actually about 3/8" diameter, the velocity will be approximatly 3 feet per second (flowrate/area) if the flow is split between two. The Velocity will be approximately 6 feet persecond through just one (actually lower due to the resultant higher headloss and subsequent lower flow rate).

Now we are able to calculate the velocity head for each of the two cases, which is 0.14 feet for the two-elbow case, and 0.58 feet for the one-elbow case.

And finally, the losses. Head loss in the fittings = the loss coefficient times the velocity head x the number of fittings. For the two-elbow case, the total headloss = 0.25 feet. The headloss for the one-elbow case is 0.52 feet.

So, for what it is worth, you are clearly on the right track by using a scheme that has a lower velocity of flow.

[MoreGooder]

Thanks cpufan!

Velocity:

1 Gallon of water = .1337 cubic feet.
2.1Gallons = .28077 cubic feet

Area of the fitting opening: 3/8" = .03125 ft.
.03125 ft^2*pi = .00307 sqare ft

Flow rate:
.28077 cubic feet/minute X 1 minute/60 seconds = .00468 cubic ft / second.

Flow rate/area =
.00468 /.00307 = 1.52 ft/ sec. That's for just one fitting.

How did you come up with 6 ft / sec through one fitting?

[cpufan]

Hi Mr. Gooder

You have used the diameter in your area calculation, but should have used the radius of the flow cross-section.

Let's compare the losses in one 3/4" fitting to that of two 3/8" fittings, just for fun. The velocity would be 1.52 feet per second(again making the over-simplified assumption that flow rate remains 2.1 Gallons per minute). So, the loss in the single 3/4" elbow would be only 0.03 feet, which is considerably lower than the loss in two 3/8" fittings. Further, the frictional loss in the tubing would be much less. One should also note that there will be other losses in the proposed configuration. The inlet and outlet losses in the big Tee should be considered as well.

I have experimented with 1/2" copper elbows with clearflex stretched onto them. The copper elbows are considerably less restrictive to flow than the mitered nylon fittings in your photo.

Good luck.