# How would I work this continuity problem?

I am trying to design a system for a project and I need some input on if this is right or not.

I have $10 \frac{kg}{s}$ of fluid flowing through a pipe. It breaks up into $10$ 1 inch diameter pipes. I am giving a picture below. Does this mean that the feed pipe, the one supplying the $10 \frac{kg}{s}$, should be just 10 inches in diameter instead of 1 inch?

In case your wondering, the fluid is ethylene glycol, I dont think that matters

I assume your objective is to minimize resistance to flow, in which case the dominating influence would be the length of the 1-inch pipes and the shape of their entry points where they enter the plenum.

If you want the fluid velocity at the entrance to the plenum to be the same as it is in the 1-inch pipes, then the plenum should have the same cross-sectional area as the small pipes, and a diameter of sqrt(10) = 3.16 inches would be sufficient.

However, you could use a smaller diameter plenum, with a higher fluid velocity, without introducing too much drag since its ratio of volume to surface area is larger. On the other hand, a larger plenum would have less drag, at the cost of its containing a larger amount of fluid.

• I am designing a piping network that goes under an ice rink for a project. So I am not really sure if the fluid velocities should be the same. But I know that the flow rate of each smaller pipe is $\frac{1}{10}$ of the flow rate in the big feed pipe Nov 17, 2012 at 20:49
• @Greg: Sounds like a fun project. Since the 1" pipes will be pretty long, the resistance in each pipe will be pretty high compared to anything in the plenum. I wouldn't expect the plenum to have to be larger that 3" diameter. A good plumber might give you better advice, though. Nov 18, 2012 at 0:36
• Thanks. It's a cool project, if I did save it for the last two weeks. I still have to design the vapor compression refrigeration cyce. It's for my Energy Systems design course. Yes the pipes will be pretty long but they will be winding under the ice rink. Maybe not 1" but between 1" and 2". Do you think I should keep the fluid velocities in the larger pipe and the smaller pipes the same? Nov 18, 2012 at 2:48
• @Greg: I would liken it to a resistor network. Each pipe under the ice is a resistor (voltage = pressure drop) in parallel with the others. The big pipe and whatever completes the circuit is in series with those. As long as the big pipe diameter is large enough to provide minimal pressure drop, it should not matter. Check out pipe flow calculator. Nov 18, 2012 at 17:13

To conserve momentum of the fluid, it's the cross sectional area that should be conserved. The area is in a square relationship to the diameter, so simply multiplying the diameter of the small pipes by 10 won't be correct.

Here's how I figure the calculation should go:

$A_{in} = A_{out}$

$\pi(D_{in}/2)^2 = \pi(D_{out}/2)^2 * 10$

$D_{in}/2 = \sqrt{(D_{out}/2)^2 * 10}$

$D_{in} = \sqrt{10} D_{out}$

• How could the fluid conserve momentum while changing directions, and why would we want to do that? Nov 17, 2012 at 8:27
• @MarkEichenlaub You're right, I mis-stated that. I meant to argue that the density is equal and hence the cross-section, but I see now that the fluid could be denser but slower at the one end ... Nov 17, 2012 at 9:01
• @MarkEichenlaub I think the crux is that the cross-sectional areas should be equal, but I'm having a hard time justifying that. Nov 17, 2012 at 9:05
• @dbaseman: it depends what is controlling the fluid flow. If the pump delivers fluid at a controlled rate you just need to make sure the pressure drop in the pipes isn't too great. There's not necessarily any need for the pipe area before and after to be the same. Nov 17, 2012 at 10:16
• A pump will be controlling the fluid flow so there will be a constant flow rate. I am essentially designing an ice rink cooling system and I am trying to design the piping network that will be under the rink. Will the fluid velocities be the same between the smaller pipe and the bigger feed pipe or would that be a design choice? Nov 17, 2012 at 20:47