Calculating maximum flow through cooling pipe

[THX-1138]

I plan to make a small induction heater using copper tubing for the work coil, and run water through the tubing for cooling purposes.

 

Assuming water as the coolant fluid, an inside diameter of [math]d[/math], a maximum height differential of [math]h[/math]" (going through a cooling block), and a total tubing length of [math]l[/math], how can I calculate the maximum volume of water and the necessary PSI of the pump?

 

There'll either be a sharp 180⁰ bend where the tubing doubles back, or else the 180⁰ turnaround will be made more gently by transitioning to a nonconductive pipe -- depends upon how much of an impact the sharp bend would have on the maximum flow.

 

I assume there'll be different answers depending upon whether it's an open system (the return dumps elsewhere than the intake) or a closed one (integrated tank/pump unit).

 

I never did any work with fluid engineering, so this is totally out of my area.

 

Thanks!

 

[EdEarl]

Your problem is H_in flow H_out

 

Pipe geometry and insulation determine heat and resistance losses. If you insulate the pipe and pump, you can ignore heat losses until after you build the system, unless you will have very long runs of pipe. Similarly, you can ignore resistance of water flowing in the pipe for short runs of pipe.

 

Flow of water is Volume per sec (V/s).

Water carries 1 calorie / cm³ for each degree centigrade.

Efficiency and operating temperature of your heat exchangers will determine how much heat you can transfer into the system and out of the system. Unless the water in the system is pressurized, the boiling point of water determines one temperature parameter.

 

You need to provide additional information to get a numeric answer.

 

I hope this helps you to make some equations. Of course, someone may give you more of the answer, but that will not be as much fun for you.

[THX-1138]

For the first pass, let's ignore the heat-exchanger aspect of the application.

 

I got here because I was looking at pumps, and noticed that one I rather liked in all other respects had fittings larger (6mm) than the I.D. of my proposed tubing (3mm). The pump was rated 4-7 litres/minute, so I wondered whether the tubing could support that volume of flow. Then I tried to find out, ran into my own wall of ignorance, and generalised the question a bit to ask it here.

 

Thanks!

[EdEarl]

It is best to use the same size pipe as fittings on the pump. Smaller will tend to reduce flow rate. Larger is more expensive. If you use a smaller pipe, use enough parallel pipes to make up the same cross sectional area (or a bit larger) as the fittings on the pump.

[Enthalpy]

Water combines uneasily with electricity. Even when water doesn't short the current path, one always gets corrosion from the voltage drop. De-ionized water leaks little electricity but corrosion introduces ions quickly.

 

The pressure drop is not trivial at all. It can be estimated from experimental curves, taking all bends into account, plus some magic like the tube smoothness, the inlet shape... This will require to invest some learning time (knowledge reusable in the future). Usable books are scarce, because most ones tell only the scientific-looking tensors and differential equations which lead to nothing, but very few ones give the ugly empirical equations and curves that are poorly justified and take pages and pages.

 

If you read German (I haven't seen it translated), one perfect book is:

Technische Fluidmechanik, from Herbert Sigloch

 

[EdEarl]

Here is a graph to find "Pressure Loss (psi/ft) of Water Due to Friction in Copper Tubes".

 

http://www.engineeringtoolbox.com/pressure-loss-copper-pipes-d_930.html

 

Home hot water systems use a sacrificial anode to minimize electrolysis problems.

 

 

There are a number of other things to consider, for example will your system be exposed to freezing temperatures. If so, you will need to use polypropylene glycol anti-freeze.

 

Since water expands and contracts with temperature you need a pressure equalization tank with some air in it.

Edited June 28, 2013 by EdEarl

[THX-1138]

It is best to use the same size pipe as fittings on the pump.

That's fairly straightforward. However, the pump's fittings can handle 7 L/m at full bore, and have no expectation of getting that kind of flow through the system. However, the lower speed of the pump is 4 L/m, which might go. But I'd rather calculate it out ahead of time -- at least roughly -- than find out empirically.

Water combines uneasily with electricity. Even when water doesn't short the current path

Cooling induction heaters by pumping coolant through the work coil itself is very common. Replacing the coolant or the coil itself is a simple matter, though. I'll tackle that when it becomes needed.

There are a number of other things to consider, for example will your system be exposed to freezing temperatures. If so, you will need to use polypropylene glycol anti-freeze.

No, freezing temperatures are not an issue.

 

I lack knowledge of the proper terminology, and ISTM we're wandering rather far afield from my original basic question. So let me simplify it further.

 

Assume a 60" straight length of hardware-store 0.1875" I.D. copper tubing, parallel to the ground. What PSI would I need to apply at the source to drive a flow of 4 L/m through it?

[EdEarl]

http://www.engineeringtoolbox.com/pressure-loss-copper-pipes-d_930.html

 

According to the above chart, 1/4" tubing looses about 0.15 psi/ft at 1 gal/min flow.

1 gal/m is about 4 L/m

Your tubing is 5 ft. Thus, 5 ft * 0.15 psi/ft = 0.75 psi loss in 1/4" tubing. Since your tubing is 3/16, which is smaller, the loss will be greater.

 

Double check please, I haven't done it before.

 

From looking at the chart, I would guess that loss in a 3/16 pipe would be 2-3 times greater than a 1/4 pipe.

[Enthalpy]

Engineeringtoolbox is always an excellent address...

 

The head loss estimate is for a straight tube, but THX-1138 told about bends, so losses will exceed it. By how much, depends on many details.

 

If someone has made comparison trials by himself with a sacrificial anode, he shall please tell us! My corrosion trials showed instead no significant effect by corrosion couples.

 

I know that water is used to cool electric wires and coils... But expect it to be difficult.