# Gardening Project: Flow Rate from pressure requirement

1. Jun 1, 2013

### mishima

Hi, I have a small gardening project and am trying to decide what kind of pump I need in order to maintain 50 psi (344.7 kPa) inside the line. The line is a 0.170 inch inner diameter vinyl tube with mister nozzles every few inches, for a total of 2 feet. This is a volume of about 9 cm^3.

I would like to know what flow rate, in GPH, I need since this seems to be the most common spec shopping around for pumps.

I tried using:

$$\Delta P = \frac{128\mu L Q}{\pi d^4}$$

With viscosity = 10^-3 Pa s, and solving for Q. This gave me a flow rate of .004829 m^3/s which seems to be about 4800 GPH.

I have a feeling that is way too high, and that I have misapplied this equation. I dont know much about fluids, can anyone help?

2. Jun 1, 2013

### etudiant

What is the performance of the individual mister nozzle and how many are there?
It seems to me the tube diameter and volume is irrelevant.
Without seeing your calculation, I wonder whether your huge flow rate, equivalent to an Olympic size pool every day, is calculated on an opening 9cm by 9cm.
In general, it is very much preferable to stick to metric measures for this kind of estimation. The English units are an abomination to work with.

3. Jun 1, 2013

### Staff: Mentor

The nozzles should have spec sheets that tell you flow rate and pressure.

4. Jun 1, 2013

### mishima

The only specification for the nozzles I have is that they are 1.5 GPH. There are 5 nozzles total.

For my calc, I used

d=0.004318 m
L=0.609 m
mu= 0.001 Pa s
P = 344700 Pa

So,

$$Q = \frac{\pi Pd^4}{128\mu L}$$
$$Q = \frac{3.14*344700*(0.004318)^4}{128*0.001*0.609}$$
$$Q = 0.0048 m^3 / s$$

I was also unsure of my value for μ, water with nutrients added at ~20 °C.

5. Jun 1, 2013

6. Jun 1, 2013

### mishima

Ok, I guess the way to do this is to use the pump manufacturers' "pump curves", which is a plot of pressure vs flow rate. A manufacturer typically plots all of its pumps on the same graph, and each pump has its own curve. You find a point for your required specs (50 psi and 7.5 gph in my case), and then use the pump associated with the curve above your point.

7. Jun 2, 2013

### SteamKing

Staff Emeritus
The flow rate you need is 5*1.5 GPH = 7.5 GPH, since you have five nozzles. The formula you were using is for laminar flow conditions.

This seems like a lot of work. Couldn't you just use a sprinkler can a couple of times a day?

8. Jun 2, 2013

### mishima

Right, that tutorial is for well pumps, and the pump curves there for demonstration. The pump curves I needed to look at were found on various manufacturers sites.

The flow rate isn't the only thing that matters, I'm using fogger nozzles which are like a fine spray and require a relatively high pressure inside the tubing. For example, a 15 dollar water pump that does 200 gph costing 14 dollars would superficially fulfill my flow rate needs, but in reality if I hooked that up to my nozzle series, the pressure would be insufficient for a mist (it would maybe just drip a little). There is a balance between flow rate and pressure that can only be found with the appropriate pump.

The purpose of the project is a classroom aeroponics demonstration my kids have been interested in (google "NASA aeroponics"). I was hoping to find a cheap way to get one going, since the nozzles and various tubing connections are on the order of a few cents each. The pump is the investment, and it seems after looking into it tonight I can NOT really get away without throwing down ~80 dollars on a water pump.

The alternative would be to use a hose outlet like normal houses have, which easily meet the psi req. The problem is you have to mix the water with nutrients for aeroponics to work, bringing me back to pumps (specifically, submersible, centrifugal). Maybe there is a way to inject nutrients into a hose stream I'm not thinking of.

Eventually I will need a adjustable recycling timer but as well, but that isn't too much. It might be fun to build the timing and control circuit in class.

But SteamKing, as mentioned I've never studied fluid mech in detail. Is it because the velocity of the flow is so great in my setup that a laminar flow assumption is ruled out? Thanks.

9. Jun 2, 2013

### SteamKing

Staff Emeritus
Laminar flow is applicable only for low Reynolds numbers (< 2000). The Reynolds number is calculated:

RE = V*D*rho/mu, where rho = 1000 kg/m^3, mu = 0.001 Pa-s, D = 0.004318 m and V= flow velocity in m/s

So, for the tubing,

RE = V*0.004318*1000/0.001 = 4318*V

This means that V must be less than about 0.5 m/s for laminar flow conditions to exist. Any higher flow velocity will become turbulent, and the flow formula from the OP will no longer apply. The Q at 0.5 m/s is equivalent to about 7 GPH, but the formula in the OP gives about 4500 GPH, so something has clearly gone wrong.

Without looking at it in detail, your system has 5 misters placed in only 24" of hose/tubing/whatever. In order for the misters to work properly, 50 psi must be available from the water supply, which means that by the time the fifth mister is reached, that 50 psi must be available, which also suggests that much higher pressure/flow must exist at the first mister. The laminar flow formula in the OP doesn't take this into account. It assumes that the flow velocity and Q are constant over the length of pipe.

10. Jun 2, 2013

### etudiant

Is it necessary to have the pump provide the pressure for your system?
You have a classroom environment, so presumably a school building.
Set up your nutrient reservoir on the second or third floor, lead a pipe down to your project and hydrostatic pressure does the work. All you need to do is to keep the reservoir upstairs filled.
Separately, a two foot long bed getting 7.5 gallons per hour sounds more like a bath than a plant growth demonstration. Are you sure you have all the parameters reconciled? It may be that the use of distinct measuring systems has introduced an error.

11. Jun 2, 2013

### Staff: Mentor

So this is really a pump selection issue.

At such low flow and high head, you may have trouble finding a suitable pump. One possibility would be to put two together in series to double the head.

Here's one that would probably do the job. Not cheap though. http://www.northerntool.com/shop/to...cm_pla=Google&cm_ite=https://www.google.com/&

I was actually thinking of doing some rain-barrel irrigation up a small hill, so I would need a similar sized pump. I'll look some more later this afternoon.

Last edited: Jun 2, 2013
12. Jun 3, 2013

### etudiant

There is a simple solution for using the faucet/hose option.
There are many systems that dispense fertilizer directly to the watering hose.
Here is one:
http://www.ezflofertigation.com/
but there are lots of others.
Given that this is a classroom demo, simplicity and low cost rule, plus pumps mean electricals, never a great idea around kids.

13. Jun 3, 2013

### sophiecentaur

Don't blame us for those units any more! We haven't used them in Science and Engineering for 50 years!! And if you must refer to them, give them their 'proper' name, which is Imperial Units. At least then we (the British) get a bit of acknowledgement for our past glories (sadly lacking, I fear, in some quarters).

It may be that they have that name in other parts of the World for the same reason that the 'French Disease', in England, was referred to as the 'English Disease', in France, in Napoleon's day.

14. Jun 3, 2013

### mishima

etudiant: Nice, I thought something like that probably existed. I will have to inspect the building outside my classroom further, but hopefully there is a tap. There is no water source inside my room either, sadly. I can at least get it going here at home and report my exploits. Thanks for taking a look at this.