Does increasing pipe size increase power requirements? 5

Damp129,

I can understand your question, but first you need to understand some key points. Many people believe that centrifugal fixed speed pumps like your simple pool pump provide a fixed flow regardless of the pressure or head that exists. Pumps are actually chosen in conjunction with a system curve which is the head required at different flow rates and is a curve in the opposite form to the pump curve, i.e. starts low and gradually increase on a squared basis a friction is proportional to the square of the velocity/floodgate for a fixed diameter system. Where the two curves intersect is your flowrate.

When you change the system by increasing the diameter of the piping you change that system curve by reducing the required head for the same flow. So the sturm curve is a shallower curve which then intersects the pump curve further to the right. Although the head is reduced the flow increases more and power is proportional to head and flowratr. What your pump curve doesn't show is that the redirect of the pump changes as flow changes and pumps, at least bigger industrial pumps, are normally chosen to have their required duty at the highest efficiency, called the BEP, the best efficiency point. As you change the flow through the pump this efficiency can reduce therefore power goes up due to increased flow and decreased efficiency.

It can be a little difficult to get your head around, but a fixed speed electric mirror will take whatever power it needs to maintain that fixed speed.

So what are your options? On an industrial pump you could use a VFD, variable frequency drive, which reduces the speed of the pump and hence bring your flow back to the same flowrate for your lower friction system. Or you could reduce the diameter of the impellor which results in a similar curve, but has lower head for the same flow and hence the point where the pump curve intersects your system curve moves to the left and hence flow and power reduce. Or just buy a smaller pump.

So in summary, matching the right pump to the right system is not guesswork but can be calculated to get the flow you want for the least overall cost. If you then change one part of the system, then it has impacts on the design.

Hope this helps and sorry i can't upload a curve vs system as I'm on a tablet, but if you search pump and system curves either here our on google you will probably find what I'm going on about.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

Sorry, a few predictive text errors up there."redirect" should read as efficiency!

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

Simple answer, yes - there is no such thing as a free lunch. If you increase flow then it will require more power input.

Why you want to increase pipe size - just because you can or do you have a valid reason?

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

Artisi and "BigInch's Little Brother" are both on target.

Let an electrical guy like me try. I had to learn this stuff for modeling building energy consumption, but I don't have the type of understanding that the folks above do. An ME gave me this as an explanation:

Add system resistance with a smaller pipe and flow goes down = less power needed. Decrease system resistance and flow goes up = more power needed. Why? How? Why should I believe that?

A swimming pool pump will have isolation valves on it, you can go prove it to yourself with a clamp-on ammeter, or even better a wattmeter with clamp-on current transformers. With your meter in place and the pump running, close an isolation valve partway but don't shut it. Watch the current or power decrease. You added system resistance by restricting flow with the valve. It's intuitive that the flow decreased. You know that from your garden hose and spigot valve. Closing the valve partially is equivalent to using smaller pipe. They both increase head on the pump and reduce flow.

Or you can bypass the sand filter and disconnect the discharge from the pump so that it blows water like crazy. I bet the circuit breaker trips in a short time. Moving more water = doing more work.

Here's a not-totally-technically correct, but less complex explanation of what happens on the pump curve. Check out the curve in the link, which is typical. No matter where your operating point is, when you increase flow (heading right on the chart), your system curve is and new operating point move in the direction of crossing the horsepower lines in the more-power direction, even though your head decreases.


Best to you,

Goober Dave

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Although it doesn't always work, you can think of this as an electrical system. What you've done is increase the size of the cable, hence reducing its resistance (friction per unit flow). However your voltage (pressure)has remained much the same. Hence you will get more current (flow) along your cable (pipe). As we all remember from school, electrical power = V x Amps (Head x flow). Therefore unless you change the voltage (head) a bigger cable will take more power.

I know it's not exact and doesn't include the efficiency of the pump issue and the fact the head falls off a bit as flow increases, but I think it sometimes helps get over the principle??

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

If you completely block the pipe, the pump will produce as much head as it can, the very left of the curve. For a small pipe, head will be less and it will move to the right. For a larger pipe, it will move further right. For open discharge (no pipe at all, just spraying to atmosphere) it will be on the far right of the curve. Draw a vertical line that intersects the head curve, read off the values for power and flow (and anything else that might show up on a pump curve.)

Because the power curve is continually rising, anytime you move the vertical line to the right (by reducing the pipe restriction, which lowers head at the pump outlet, and increases the flow) your power will increase. Not all centrifugal pump curves are like that, but most of them are, especially small ones.

Increase flow and power is not necessarily a bad thing as it also means that you do not have to operate the pump as often; another thing that I don't see in the graph is the pump efficiency curve which has a lot to do with amount of current drawn out.

pool pumps generally run continuously, so higher flow rate results in a cleaner pool at increased cost for the electricity. operating to the right side of the efficiency curve will result in even further increase in cost.

Wow, thanks for the detailed replies everyone! I am convinced if not from reasoning, then from sheer opposition to my opinion! I would like to know the true reason that explains why less resistance to work requires more power because it goes against pretty much everything I know... However, while typing this the reply I thought of something that if true, would definitely make sense to me.

My theory (true or not, I am sure this is somebody else's theory too!):
If water is moving through the pump at low rate, the inner impeller (which requires less torque from the motor) is actually contributing to the movement of water. At high flow rates, the inner impeller does not move fast enough to help push the water and could even hinder the movement of water at this rate. Only the outer parts of the impeller are actually moving fast enough to keep up with a high flow rate and this outer part requires more torque to turn.

Is my theory correct? If so that makes it so much easier for me to understand!

NO,

Read the posts carefully.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

I have read the posts and appreciate them. They refer to system and pump performance curves. What is not explained is "why" more power is required at high flow rate. There has also been no analogy where resistance to work has been removed but removal of that resistances REQUIRED more power.

Speed is constant, flow rate varies. Even at low pressure, you can't move water without energy.

The major concept you are missing is "water horsepower." Google for a definition, it should make things click in your head.

In very a simple exercise without getting too hung up on specifics take a look at your pump curve supplied, if head reduces flow increases, therefore

Power = flow x head / hydraulic efficiency.

so if flow is 1 and head is 2 and efficiency is 50%, then power = 4
increase pipe size so that flow is 1.5 (increase) and head is 1.5 (reduction) and efficiency is 0.5, then power = 4.5



It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

All good points for your education & edification DanP, and of course now I feel like throwing in my 2 pesos.

When getting wrapped up in system curves, friction losses, curve intersections and the like, trying to understand centrifugals; it helps me to remember that the pump doesn't know anything that happens beyond its flanges. If you can just think about the data at the flanges and that all the crap going on downstream only has its effect on the pump at the discharge flange, it may be easier to "get it".

HP = flow X pressure
If you increase the flow more than you lower the pressure, then HP increases.

DubMac, I tried to think about what was going on at the flanges and thought my idea that the outer flanges had to do the majority of the work at higher flow rates would explain the higher power requirement. Apparently it makes no difference.

Ok, i thought my electricity analogy would help, but maybe need to go back to basics.

In one sense you are correct, less resistance will mean less power, but only if the flow rate doesn't change. If you could turn your pump down, e.g. plug it into 110 v instead of 220, you might pump the same amount of water for less energy. However with a fixed speed centrifugal pump less resistance just means the pump can pump more water, hence more power.

If you had a fixed speed piston pump then this would pump the same amount of water for less energy. But you don't. If you can't grasp that a centrifugal pump works differently to a piston type pump then I don't think anyone here can help you much more than we already have.

I really hope this makes sense.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

voltage stays constant => flow increases => motor turns faster => more amps flow through the wires => therefore more $$$$ required to pay your bill => pool is cleaner

Cvg,

Sorry to be pedantic, but motor does not turn faster. It's a fixed speed pump. I understand it's not easy to get it, but that's the way it is.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way