# How to make constant drip of water drops from container

I have a pot with water inside and I need to drip the water from it with a specific flow. I can control this flow changing the size of the hole at the bottom of the pot. Check image below.

The problem is that as time goes by, the height H of water column gets smaller and the flow of water gets smaller (of course). How can I overcome this problem?

I dont need a very precise solution to this problem. The thing is that I have a vacuum robot which has a mop that clears the floor. So I added a pot over it so the robot drips water in front of it and the mop goes over that drops of water and make the floor cleaner. It works perfectly however as time passes, the water drips too slow and the floor does not clean very well anymore.

So I was wondering if there is some type of thing/valve that I can easily build/buy that makes the flow constant, no matter the height?

note: my "hacked" solution to this problem was to use a very large pot with a small height, so as time goes by the H does not change much and so the flow keeps more steady, however even this causes too much variance in the water amount that drips.

You require a what is called a pressure regulator. Here is something which may work.

Drip irrigation systems need a carefully controlled water pressure source in order for each drip outlet to deliver its specified water flow rate, over a fairly great range of possible water source pressures.

This is done by inserting a pliable rubber plug into the inlet flow fitment to the drip system. This plug has a precisely-sized tiny hole going through its center. The plug is supported against the source pressure around its circumference but it deflects inward when the system is pressurized.

As it flexes, the hole in its center gets progressively pinched shut. By artfully selecting the stiffness of the rubber, the thickness of the plug, and the hole diameter, it is possible to hold the flow rate almost completely constant across a source pressure range of from about 10 psi to 80 psi- using a simple device which costs just a few pennies.

For your application, to start with you would need a soft rubber plug, fairly thin (about 1/4 inch), and of large diameter- say about an inch or two. Experiment will then yield the correct hole diameter, plug thickness, and diameter.

• My friend, I think what you said is exactly what I need! However I tried imagining the rubber you said but I couldnt understand it. I googled pressure regulator on images and couldnt find anything similar to what you said. Could you just point me to a image? Commented Feb 25, 2020 at 21:16
• try googling on drip irrigation regulator and see what you get! -NN Commented Feb 26, 2020 at 7:32
• My friend, I googled what you said and I can only find images of commercialized products, I cant find the inner workings of those. I tried adding DIY or HOW WORKS to the search and no luck! Commented Feb 26, 2020 at 13:23

You need a container shape where the height $$H$$ between the water level and dropping outlet is nearly independent of the current water volume.

Since you don't need a precise solution, I suggest to insert a long thin tube or hose between the dropping outlet and the pot.

• Yeap, I realized using a big height helps a lot because as the time goes by, the change in height inside the container is very small compared to the total height. Good solution however on my case I cant use a big height cause the robot needs to fit below beds, chairs... Commented Feb 26, 2020 at 13:38
• Intravenous bags face this issue. You might just place a standard weight of metal resting on the bag, or a spring. Commented Feb 26, 2020 at 15:33
• The weight is a good idea :) Commented Feb 26, 2020 at 18:53

From Torricelli’s law we have the flow rate $$v$$ with a water height of $$h$$ to be given by: $$v(t)=\sqrt{2gh(t)}$$ Now if we want to fix the volume flow rate to some constant $$\alpha$$ we will have the constraint equation: $$A(t)\sqrt{2gh(t)}=\alpha$$ $$\Rightarrow A(t)=\frac{\alpha}{\sqrt{2gh(t)}}$$ So if you can control the hole area by the above equation, you must get constant drip rate.

• Thanks man but my problem is with the water height, I cant change the area of the hole every single time to account for water column height being changed Commented Feb 25, 2020 at 21:32
• The statement “I can control this flow changing the size of the hole at the bottom of the pot.” in your question put me off then. Commented Feb 26, 2020 at 4:05
• Yeap, I can control the size of the hole once, but not everytime :) Commented Feb 26, 2020 at 13:38

What you need amounts to a "water clock". Basically, the water source fills a container that has an overflow tube, to ensure that the water height in the reservoir stays constant. A stopcock or pinhole at the bottom of the reservoir sets the drip rate. Of course you need an additional container to hold the overflowed water as it accumulates.

Here is perhaps a simpler alternative: water in a tall, closed, evacuated container with a small hole at the bottom will drain only until the height of the water in the container causes the pressure at the bottom to equal ambient air pressure. If the container is flexible like a plastic bag, it will keep draining but keep the same height until there is very little water left. That is the principle behind IV bags and their plumbing. You could throw together a crude version of that, or beg the setup from your doctor.

• Thank you my friend but I think what you said will not help with the problem I am facing! Cause I dont have a water source to fill my container. My container is filled initially and after that it's inserted into the robot and from that moment ahead it only gets emptier as water drips. Commented Feb 25, 2020 at 21:18

Worth exploring about a simple concept of mariott siphon , it's a simple vertical air tight tube, it's top end is open to the atmosphere and the bottom end always remain inside the container. It's bottom end should always remain submerged. This way the pressure at the outlet would always remain constant even if the water level changes.

The flow rate out of the hole is dependent on the pressure at the hole. Your current configuration results in pressure at the hole that is solely dependent on the height of liquid in the tank, which is given by $$P=\rho g h$$, where $$\rho = 1000 \frac{kg}{m^3}$$ for water. For a 1 meter initial water height in the tank, the starting pressure at the hole is in the range of 9810 Pa as measured by a gauge (aka gauge pressure), and that pressure drops to zero as the water level drops to zero, greatly influencing the flow rate out of the hole.

If you can put a small amount of air pressure in your tank, you can significantly decrease the amount that the pressure drops at the hole. One atmosphere of pressure is equal to 101,000 Pa, which is MUCH higher than the pressure that is produced solely by the water column in your tank. If you put a small amount of air pressure in your tank (a bicycle pump would do the trick), you will obtain a much more constant pressure at the hole in the tank as the liquid level drops. For example, an air pressure of of 0.4 atm (5.7 psi) will result in 80% of full flow when the tank runs dry relative to its starting flow if your liquid height starts at 1 meter and you have an air pump that can hold a constant pressure. If you intend to pump the tank up manually, you will have to design the tank such that there is more air volume than liquid volume and use the ideal gas law to determine how much the pressure drops when the tank runs dry. I don't know how your robot limits the dimensions of your tank, but there should be a way to design a pressurized system that gives you approximately what you need.