View Full Version : Calculating water pressure changes as pipe size changes

Patty O'Furniture

10-11-2007, 07:29 PM

Is it a straightforward linear relationship between the pressure and pipe diameter? In other words, if water comes into my home via a 1 inch pipe at 50 psi, will it be 25 psi if I use a reducing coupling to bring the pipe down to ½ inch? Does the length of the pipe have anything to do with it? What about the direction (vertical or horizontal) it runs?

Not linear at all. It actually depends on the radius to the 4th power. The equation looks something like this:

Q = (PiR^4P)/8Ln

Q is the flow rate, Pi is pi, R^4 is the radius of the pipe to the forth power, P is the pressure, L is the length of the pipe, and n is the viscosity of water. This is a simplified version, assuming laminar flow, no slip conditions, yadda yadda yadda.

And you have it backwards. A smaller pipe increases the pressure. Think of it this way: You have the same total amount of fluid moving in the pipe in the same amount of time across the same distance, but in a smaller maount of space to do it in. The only thing that can change to make the same amount of water move the same distance in the same time is the pressure. So, in your example, assuming everything else stays the same, then 50 psi with a 1 inch pipe will increase to 800 psi.

ETA: Oh, and as you can see, the length does play a factor. Increasing the total length of the pipe increases the pressure. If you wanted to have the same pressure at the same flow rate with that new 0.5 inch pipe, then the new pipe would have to be 0.0625, or 1/16, the length of the old one. And vertical vs horizontal does matter, as gravity comes into play in a vertical pipe.

Rhubarb

10-11-2007, 08:04 PM

First off, the static pressure will be roughly the same everywhere in the pipe, independent of the diameter. There will be changes in pressure due to differences in height (P=ρgh), and these are strictly linear. When the fluid is moving, it's a different story.

The pressure drop in the pipe for moving liquid due to diameter changes is a little trickier and not even remotely linear. The governing equations are the continuity equation, volume flow rate is constant = A1V1 = A2V2 and part of the Navier-Stokes equations, P=1/2 ρV2. So velocity is directly proportional to the pipe diameter, but pressure change is proportional to the square of the velocity.

To further compound the problem, you need to account for frictional losses which are a function of the turbulence, expressed by the Reynold's number, and the viscosity of the fluid, and the area and the pipe roughness and something else I've likely forgotten.

Here's the ubiquitous Wiki link (http://en.wikipedia.org/wiki/Navier-Stokes).

FYI, I am not a plumber, and if I were, I would not be your plumber, even if I were licensed to plumb in your state. This post does not constitute plumbous advice. Contact a licensed plumber in your jurisdiction for qualified plumbing advice.

boytyperanma

10-11-2007, 08:18 PM

For the purposes of you home, the water should be at the same pressure regardless of pipe size. Only height would make a difference.

Patty O'Furniture

10-11-2007, 08:35 PM

Q = (PiR^4P)/8Ln

That would have been my second guess. :)

And you have it backwards. A smaller pipe increases the pressure.

So, does pressure increase toward infinity as pipe size decreases to zero? Would this principle allow me to get water all the way up to the top of a sky scraper simply by reducing the pipe diameter at each floor? Granted by the time I get to the top I may only be able to run a drinking fountain...

How come I can easily stop water from flowing out of a 1/8" plastic tube (such as the one feeding the ice maker or water filter) with my thumb and minimal effort, but it's hard as hell to stop water rushing out of a half inch line? There seems to be more pressure from the latter than the former.

Santo Rugger

10-11-2007, 08:45 PM

Fluid flow, like any other physics phenomenon, obeys the law of conservation of energy. Bernouli's principle sums this up nicely for fluids, and can be expressed as:

v^2/2 + gh + P/p , where v is velocity, g is gravity, h is height, P is pressure, and p (rho) is density.

This equation must always equal the same value at any point along a pipe, after frictional losses are taken into account.

For this problem, assuming no losses (which would be an easy enough calculation) due to the pipe changing sizes, and knowing that h is a constant, we can calculate the change in velocity by assuming a unit velocity, traveling through a 1 inch diameter, which would have a cross sectional area of .785 square inches. A .5" pipe would have a cross sectional area of .196 inches squared. That means that, even though the diameter is halved, the effective area is one fourth of the original. This means that the resultant velocity would be quadrupled, so the pressure would drop in an amount to satisfy v^2/2 + P/p remaining a constant.

In short, the magnitude of said drop would depend on the initial velocity, the initial pressure, and the density (which, for all intents and purposes, will remain constant)

The length of pipe will induce a friction loss, also known as "head loss", which is expressed as:

hL = f * L/D * v^2/2g , where hL is head loss, f is a friction factor, L is length, and D is Diameter.

Head loss is usually measured in feet (or meters), and for water, one foot of head is about half of a pound per square inch (.466, IIRC). So, for every vertical foot of travel, you'll be losing about half a psi.

As you can see, both the length and diameter are included in the above equation. The longer the pipe, the more molecules are going to be bumping against the sides, so there will be more friction. Similarly, the larger the pipe diameter, relatively fewer molecules will be bumping against the sides for the same amount of volume going through the pipe. Also note that frictional losses are proportional to the velocity squared. This means that the faster you go, the harder it becomes to go even faster.

Finally, the "f" value above, is a measure of many things. Most importantly is the "Reynolds number", which is a measure of how turbulent the flow is. A laminar flow can be observed by watching a lit cigarette inside with no wind. As the smoke convects up, it is a nice streamline. However, after it has traveled a few inches, it begins to swirl. These swirls are known as "eddy currents", and indicate the fluid mixing amongst itself. Surprisingly, in a pipe, a turbulent flow has less friction loss than laminar flow. In laminar flow, the fluid "sticks" to the side of the pipe, which slows it down more.

The "f" value also takes into account the roughness of the pipe. A PVC pipe is much more smooth than a cast iron pipe, and will therefore have less frictional losses.

Another thing that must be taken into account when designing a pipe system are elbows and valves. An equivalent length can be determined. For example, routing flow through a 90 degree elbow may be equivalent to having it flow through ten feet of pipe. These losses are known as "minor losses", while losses due to the friction on the pipe walls are known as "major losses".

Patty O'Furniture

10-11-2007, 08:51 PM

OK I guess I asked for that.

Why then, are large water appliances like softeners and water heaters (which usually operate inline with the 1" main) are rated for high pressures (up to 150 psi in the case of my water heater). Things like countertop filters and ice makers that are fed by ¼" or smaller lines are usually rated for lower pressures.

Santo Rugger

10-11-2007, 08:54 PM

That would have been my second guess. :)

So, does pressure increase toward infinity as pipe size decreases to zero? Would this principle allow me to get water all the way up to the top of a sky scraper simply by reducing the pipe diameter at each floor? Granted by the time I get to the top I may only be able to run a drinking fountain...

How come I can easily stop water from flowing out of a 1/8" plastic tube (such as the one feeding the ice maker or water filter) with my thumb and minimal effort, but it's hard as hell to stop water rushing out of a half inch line? There seems to be more pressure from the latter than the former.

Pressure will not increase toward infinity, because as the pipe gets smaller, frictional losses will become higher.

Water fountains, ice makers, etc. have mechanisms in them that drop the pressure, so what's coming out is not the same pressure that's available to them going in. Assuming the same velocity of water from an 1/8" and 1/2" line, it would take the same amount of force to stop each of the lines, but the smaller line would be easier to get your thumb around.

Santo Rugger

10-11-2007, 09:01 PM

OK I guess I asked for that.

Why then, are large water appliances like softeners and water heaters (which usually operate inline with the 1" main) are rated for high pressures (up to 150 psi in the case of my water heater). Things like countertop filters and ice makers that are fed by ¼" or smaller lines are usually rated for lower pressures.

It doesn't take much to create a pressure drop. A simple disk with a hole in it placed in line will cause quite a large pressure drop. Assuming the ideal relationship, which was done by several of us to simply the problem, does not take into account the frictional losses caused by the pipe reduction, which can be quite high.

A practical example of this is the little ring in your low flow shower head that severely lowers both the pressure and flowrate of your shower. :(

A.R. Cane

10-11-2007, 09:33 PM

For practical purposes, pressure is constant in a house plumbing system . I'd guess the OP is concerned about flow. When installing plumbing fixtures you size the lines according to flow rate. That way you don't get scalded in the shower when someone flushes.

Is that the kind of thing you're interested in?

Bill Door

10-12-2007, 06:50 AM

The people who said the static pressure was all the same are correct, correcting for the loss of pressure due to static head, which is probably minimal in the average house. It takes 23 feet of elevation to lose 10 pounds per square inch pressure, and when I ran a well I set the pressure at the pump for 40 psi.

The frictional loss, usually measured in feet per hundred feet of pipe, is the important part. You have to measure a lot more than the overall length as well. Each fitting, elbow, bend, T-joint, whatever, has it's own equivalent feet measurement.

I don't have my Cameron Hydraulic Data (https://www55.ssldomain.com/fpdlit/Merchant2/merchant.mvc?) book with me, but I have a spreadsheet I put together, and for example, a flow of 10 gallons per minute in a 1-inch line has a linear flow rate of 3.7 feet per second and a head loss of 6.9 feet per hindred feet of pipe.

Double the size of the pipe to 2-inch, and the velocity decreases to 1.0 foot per second and the pressure drop to 0.3 feet per hundred feet. That's all assuming carbon steel schedule 40 pipe. If you've got any other type of pipe, that will be different too. A 2-inch pipe isn't necessarily 2-inches inside diameter, in schedule 40 it's 2.07 inches.

I don't have data down to the 1/4" line size you're talking about, because it's not that relevant in my business. My spreadsheet goes up to 60" diameter.

As far as why it's easier to block a small line than a large one, it's because the hole is bigger. A quarter inch hole at 40 psi exerts less than 2 pounds of pressure, because the cross sectional area is only 0.05 square inches or so. Expand the hole to one inch, and the cross sectional area grows to 0.79 inches, and the total pressure goes to 31 pounds.

GardenGuy

12-23-2012, 06:59 PM

I have a follow up question related to this subject. I apologize for the length but felt the need to give some background information as well as some things to consider with my question. Feel free to ask follow up questions if needed. If this is the wrong place to post this question I apologize in advance and tell me where I should post because this existing question seems close but not an exact match to my question.

Background:

Part 1:

I am working on a watering system in my raised beds at a community garden. I have been doing this for some time using gravity fed systems and PVC. This year we made changes that give each plot a water connection via PVC access to the water source. The change now gives me access to use timers as well as various types of automation. Watering my raised beds takes more time than one might think even with the changes I made using gravity systems. Given the recent changes I have been looking to limit that time even more now that I have water at my raised beds.

Part 2:

Most of my raised beds are 4' X 12'. I have 5 of them to be correct. I have one raised bed that is 4' X 29'. I have investigated various types of watering systems online and come across one that I am considering using given the various review by third parties. However, I do not want to spend the money to buy custom PVC parts from the vendor costing over $70/raised bed because a visit to my local hardware store I found out of the box parts that "appear to" work and do the same job and I love the DIY approach.

Part 3:

Clarifying the "appear to work" from above is now in order. The existing system, the one I like uses two chambers to water underground with jets of water. The first chamber is a big chamber that fills with water from the water source. The first chamber feeds into the second chamber that is much smaller. The water that goes into the second chamber due to the pressure change (increased) is basically turned into jets of water that is used to water underground.

Finally: The question.

I would like to do this myself however I am not sure about the water pressure I would create going from a 3/4" PVC down to a 1/2" PVC with two 90 degree turns. The idea would be that the water from the source runs from the front of bed (12') to the end of the bed and then two 90 degree turns would turn and downsize from 3/4" to 1/2" which would also run 12' back to the front of the bed using PVC pipe. In other words there would be two 12' pieces of PVC one 3/4" the other 1/2" that are 12' long with two 90 degree turns at the end of the raised bed running parallel to each other.

Things to consider with your answer:

1. The incoming pressure is 50psi. I could make changes to move it up some based on the settings of the pressure regulator but right now it is as 50psi.

2. I realize that many things could come into play answering this question most of which I forgot a long time ago after college. I am OK with estimates/best guesses even equations that are not exact nor take all variables into account. I am looking for a way to calculate ballpark numbers here within a range. An equation that takes into account length as it seems to apply would be useful since I have one odd ball bed that is 29' in length.

2. For me the bottom line here is the PVC for 1/2" that I can purchase has a max pressure of 600 psi, so ideally I would like to get the pressure around 500psi. Previous posts in this thread say going from 1" to 1/2" would be 800psi. I am going from 3/4" to 1/2" so I am not sure if we are talking about the same pressure given all the variables and is part of the reason for my question.

4. There will be holes in the smaller PVC pipe that work as jets which release the pressure down the length of the pipe so I might be OK. I have drill bits that drill 1/32" holes so for each 12" section there would be two holes one on each side of the smaller PVC pipe. Total there would be 24 holes along the length of the PVC that work as pressure reducers.

5. An additional concern I have is there might not be enough pressure. An example would be I need to increase the incoming pressure to make this work or maybe make the holes smaller so that pressure along the pipe that is lost is much smaller.

I appreciate any feedback you have and again I apologize for the length but given the possible complexity of my question I wanted to give as much information in context as possible.

--Preston

GardenGuy

12-23-2012, 07:33 PM

I have a follow up question related to this subject. I apologize for the length but felt the need to give some background information as well as some things to consider with my question. Feel free to ask follow up questions if needed. If this is the wrong place to post this question I apologize in advance and tell me where I should post because this existing question seems close but not an exact match to my question.

Background:

Part 1:

I am working on a watering system in my raised beds at a community garden. I have been doing this for some time using gravity fed systems and PVC. This year we made changes that give each plot a water connection via PVC access to the water source. The change now gives me access to use timers as well as various types of automation. Watering my raised beds takes more time than one might think even with the changes I made using gravity systems. Given the recent changes I have been looking to limit that time even more now that I have water at my raised beds.

Part 2:

Most of my raised beds are 4' X 12'. I have 5 of them to be correct. I have one raised bed that is 4' X 29'. I have investigated various types of watering systems online and come across one that I am considering using given the various review by third parties. However, I do not want to spend the money to buy custom PVC parts from the vendor costing over $70/raised bed because a visit to my local hardware store I found out of the box parts that "appear to" work and do the same job and I love the DIY approach.

Part 3:

Clarifying the "appear to work" from above is now in order. The existing system, the one I like uses two chambers to water underground with jets of water. The first chamber is a big chamber that fills with water from the water source. The first chamber feeds into the second chamber that is much smaller. The water that goes into the second chamber due to the pressure change (increased) is basically turned into jets of water that is used to water underground.

Finally: The question.

I would like to do this myself however I am not sure about the water pressure I would create going from a 3/4" PVC down to a 1/2" PVC with two 90 degree turns. The idea would be that the water from the source runs from the front of bed (12') to the end of the bed and then two 90 degree turns would turn and downsize from 3/4" to 1/2" which would also run 12' back to the front of the bed using PVC pipe. In other words there would be two 12' pieces of PVC one 3/4" the other 1/2" that are 12' long with two 90 degree turns at the end of the raised bed running parallel to each other.

Things to consider with your answer:

1. The incoming pressure is 50psi. I could make changes to move it up some based on the settings of the pressure regulator but right now it is as 50psi.

2. I realize that many things could come into play answering this question most of which I forgot a long time ago after college. I am OK with estimates/best guesses even equations that are not exact nor take all variables into account. I am looking for a way to calculate ballpark numbers here within a range. An equation that takes into account length as it seems to apply would be useful since I have one odd ball bed that is 29' in length.

2. For me the bottom line here is the PVC for 1/2" that I can purchase has a max pressure of 600 psi, so ideally I would like to get the pressure around 500psi. Previous posts in this thread say going from 1" to 1/2" would be 800psi. I am going from 3/4" to 1/2" so I am not sure if we are talking about the same pressure given all the variables and is part of the reason for my question.

4. There will be holes in the smaller PVC pipe that work as jets which release the pressure down the length of the pipe so I might be OK. I have drill bits that drill 1/32" holes so for each 12" section there would be two holes one on each side of the smaller PVC pipe. Total there would be 24 holes along the length of the PVC that work as pressure reducers.

5. An additional concern I have is there might not be enough pressure. An example would be I need to increase the incoming pressure to make this work or maybe make the holes smaller so that pressure along the pipe that is lost is much smaller.

I appreciate any feedback you have and again I apologize for the length but given the possible complexity of my question I wanted to give as much information in context as possible.

--Preston

The other thing I should mention is that the watering system I like allows to connect beds together using the same water source. The system I investigated and like would allow me to connect several of my beds on the same water line. So, as an example in my situation their system would have enough pressure for 2 or 3 of my raised beds which are 4' X 12' on the same line with water pressure around 50psi. In other words, using standard water pressure there system would be enough pressure for 2 or 3 of my beds.

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