Question for the math guys

  • We are currently upgrading MFK. thanks! -neo
The reason is because the pump box is an open system. The pump intake is one inch and is pump driven (high velocity). The vertical pipe is also one inch but is gravity driven (low velocity).

The difference in volume is made up by drawing in air through the top of the pump box. You could seal it (not very practical but do-able), you could remove the sponge filter and plumb it through, or start cutting holes :grinyes:

I don't remember the math, but I think the ball park result was four one inch holes or a single two incher. (The correct answer is velocity dependent.) Btw, the plastic strainer on the bottom of the vertical pipe puts restriction on the pipe and reduces it's flow.
 
Thanks, Hard plumb it is.
CHOMPERS;4159543; said:
The reason is because the pump box is an open system. The pump intake is one inch and is pump driven (high velocity). The vertical pipe is also one inch but is gravity driven (low velocity).

The difference in volume is made up by drawing in air through the top of the pump box. You could seal it (not very practical but do-able), you could remove the sponge filter and plumb it through, or start cutting holes :grinyes:
Click to expand...


I learned that the strainers reduce flow since I put them on, I just had a fish not to long ago do a header into my canister filter intake because I took the strainer off. Maybe I could do egg crate instead. But I have not noticed a lot of difference now that it is plumbed in
CHOMPERS;4159543; said:
I don't remember the math, but I think the ball park result was four one inch holes or a single two incher. (The correct answer is velocity dependent.) Btw, the plastic strainer on the bottom of the vertical pipe puts restriction on the pipe and reduces it's flow.
Click to expand...
 
Hard plumbed it is, and I figured out a couple little fixes also to make placing the filter on top easier.

Thanks for your help.

Bear
 
Another Newb question.

How much flow does a spray bar kill? Specific info not necessary (little/Lots) OK :)

I have a 27" spray bar with 3/8" holes drilled every 1.5"(ish). If I figured the head loss correct I am sitting right now without the spray bar at 690ish GPH. With 2 pumps that gives me apx. 6x turnover.

Are these spraybars gonna kill my flow?
 
By the time I type an answer with two fingers someone is going to give you a good answer. :)
 
Pi times the radius squared times the number of holes. If this is greater than or equal to the supply then you should see little or no loss.
 
dawnmarie;4190509; said:
Pi times the radius squared times the number of holes. If this is greater than or equal to the supply then you should see little or no loss.
Click to expand...

OK,

I really suck at math, but here are my figures;

Holes drilled are 3/8" or 18.75/100THs of an inch radius

Radius Squared * PI (37.5 * 3.1415 = 117.8)
Guessing 18 holes drilled (117.8 * 18 = 2120.5125)

Supply is around 690GPH.

So if I am doing the math right then I should see no loss at all from the spray bar.

Correct me if I am wrong.
 
The radius is .1875
r^2 is .035156
You multiplied r by two instead of squaring it.

.035156 times pi is .110447
.110447 times 18 is 1.988
 
Here's food for thought (pun intended)...

If you have six oranges and two apples,

and you divide the six oranges by two apples,

What do you get?
 
CHOMPERS;4190833; said:
The radius is .1875
r^2 is .035156
You multiplied r by two instead of squaring it.

.035156 times pi is .110447
.110447 times 18 is 1.988
Click to expand...


:flamed:

OK my head just exploded :ROFL:

please explain what the 1.988 means.
 
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