# 170L cube - JBL Cristal Profi e1902 to much flow?

#### Furgan

##### Member
I have a 170L cube tank with the tetra tec 800, would a jbl Cristal Profi e1902 be too much, too powerful for the tank? Creating too much flow. Looking for a filter that I could use if I upgrade the tank size later!

tetra filter has started leaking at the hose connector and I don’t feel the flow is strong enough.

thanks

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#### The grumpy one

##### Member
Depends what fish you have and what flow they require.
It is quite fast, but not too fast for some types of fish.

My thoughts: Please do not read, it is a mental exercise for Sunday morning.
If you do read then as it is Sunday morning, I might have made more than one error and needs checking over.

The e1902 has a flow rate of 1900 L/h. That is measured by JBL with the canister empty and zero head. Once you put mechanical and/or biological media in and put the filter under the tank (giving you a head of at least the height of the tank plus the height of the canister filter), you can expect the flow rate to reduce to about 50%. This will give you a flow rate of 950 L/h. Now with your tank being 170 L, you will get a turnover of 5.6 tanks/h. So the turnover is about right, for your tank size; for the recommended 5 complete water turn over per hour. Or if you take the 1900 L/h then 10 times.

Velocity of water in pipe (m/s) = flow rate (L/s) divided by cross sectional area of pipe (m²)

Cross sectional area of pipe = Pi times the radius times the radius or πr²
Now the diameter of the pipe is 19 mm, giving a radius of 9.5 mm. Or 0.0095 m.
So the cross sectional area of the pipe = 3.141 x 0.0095² = 0.0002835 m²

Flow rate is 950 L/h or 15.83 L/m or 0.264 L/s

So: Velocity of water in pipe (m/s) = 0.264 L/s divided by 0.0002835 m² = 930.7 m/s.

I am guessing your tank dimensions to be 550 mm x 550 mm x 570 mm (H)
If we assume that the water flows round the tank.
Then if you think of the tank as a pipe. You have a pipe that is 275 mm x 570 mm
Giving a cross sectional area of 0.15675 m².

So recalculating velocity of water in tank.
V = Flow rate / csa.
V = 0.264/0.15675 = 1.68 m/s = 6.048 km/h = 3.76 miles per hour.

So the speed of the water on the tank will be 930.7 m/s at the out pipe then slow to 1.68 m/s as it spreads out round the tank.

But if the intake and outtake are at opposite ends of the tank then you would, effectively, have a pipe that is 550mm x 570mm CSA of 0.3135 m².
Then the velocity of the water in the tank would be 0.84 m/s.

But this is a simplified solution.

True formula:
F=(π(P1−P2)r⁴)/(8ηL) where η is a constant dependant on the viscosity, temperature and corners/obstructions in the pipe.

I told you not to read this ramble.

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