Pumping uphill

He could use a generator and black poly. 40 to 50 cents per foot or less for 100psi. Judging by the abandoned pot farms here, it lasts a LONG time on surface. Drain down would be easy and it could run back into his tank at the bottom anyway. PVC is only 20 cents a foot or less, if he can bury it.

Wally Hays said:

He could use a gasoline powered, self priming pump. A Google search finds several that will pump 120' of head at a pretty reasonable cost. The outlet is 1.25 which if he stays with that size will yield less friction loss and cost about the same for the pipe.View attachment 10986

Click to expand...

The Zoeller 320 that you reference has a shutoff head of 120', which is absorbed by rise, with no pipe or fittings calculated, so that one wouldnt work.

I was only posting a picture of a gas powered pump, not a specific model. Yes, he will need to match the curve to the conditions thanks. Sorry for the confusion

Nukeman, you'll have to do some serious calculations to change my mind that to run a 1/2 hp pump at say 53 psi to get a trickle 2500' away for 24 hrs and running the same 1/2 hp pump at 5 gpm for 1.66 hrs uses the same amount of electric; I do not believe it.

If we make any assumptions here we should assume the 120' elevation is throughout the whole 2500'. And the 43 psi friction loss + the 53 psi says he sets the switch at 90/110 and in 1.66 hrs of pumping he has 500 gals in the tank at the house. That assumes he allows the tank to go empty before he gets water and I'll bet he doesn't.

So pumping fewer gallons means you reduce the 1.66 hrs of run time; which saves electric. .Now if you want to pump fewer gals, you need to control the output of the pump which increases the TDH and you need more pressure and a flow control. Which means he spends more money.

He's much better off with say a 1/2 hp 10 gpm and fill the tank faster so he can shut off the pump and it's electric use sooner. And my money is on the OP agreeing with that.

NHMaster, no normal gas pump is going to do this job.

The formula for calculating work is

W= D*T where W=work ( or the load moved ) D= Distance and T = Time

Changing any one changes the other. If the load remains constant ( in our case 500 gallons @ 8.33 lbs/gal or 4165 lbs of water ) and the distance remains constant ( in our case 2500' ) the driving force ( electricity in our case ) remains the same. If it takes all day with the pump amp draw and friction loss at a minimum or an hour with the amp draw and friction loss at max the total amount of electricity used will be, has to be the. same or we violate the laws of physics. This is all right out of any high school physics book

The most basic way to show this is with affinity laws (for centrifugal style pumps). For this style of pump, the difference is even more dramatic. First off, the volume (gpm) is directly proportional to shaft speed:

Q1/Q2 =N1/N2

So, going from 1 GPM to 5 GPM with the same pump means 5x faster speed (makes sense).

The head goes as the square of the pump speed, so at 5 gpm, it will generate 25x (5^2) more head.

The power goes as the cube of the pump speed, so at 5 gpm, it will use 125x (5^3) more power compared to 1 gpm.

http://en.wikipedia.org/wiki/Affinity_laws

Say you move 500 gallons.

At 5gpm, this would take 100 minutes

At 1 gpm, this would take 500 minutes

Energy is power*time (kwh). So, at 1 gpm, it takes 5 times longer, but uses 125 times less power. So, pumping at 5 gpm, will use 25x the energy (kwh) to get the job done.


If this isn't clear, how about this:

WHP = Q*TDH / 3960

WHP = Power
Q = FLow
TDH = Total Dynamic Head (friction + elevation)

http://www.ag.ndsu.edu/pubs/ageng/irrigate/ae1057w.htm#Basic Pump

If we ignore friction, then by this equation, if Q increased by 5x, then power would also increase by 5x. Again, power*time = energy. 5x the power at 5x less time = same energy theroretically used. So if friction didn't exist (or was very small compared to elevation head), the best you could do it use the same electricity pumping at the higher rate. However, we know that TDH is not constant due to friction. We can see this impact by picking a pipe size and look at the loss per 100' at 1 gpm and 5 gpm.

From Charlotte pipe:

1in, sch 40 PVC

1 gpm = 0.24 psi (0.55') per 100' of pipe
5 gpm = 1.37 psi (3.17') per 100' of pipe

Assuming everything else is the same, we have:

TDH (1 gpm) = 120' + 0.55'*25 = 133.75'
TDH (5 gpm) = 120' + 3.17*25 = 199.25'

Ratio the power:

[WHP (5gpm) / WHP (1gpm) ] = (5gpm * 199.25) / (1 gpm * 133.75) = 996.25 / 133.75 = 7.45

So, it takes 7.45x more power to get 5x the flow in our case. Again, you are expending more energy pumping faster for shorter times. If we used the flow that I used assuming running for 24 hr, then the difference would be even greater.

Hopefully this makes it clear. Now for other advantage going smaller on the pump:

- lower powered pumps cost less money
- wire size may be able to be reduced at you are using 5x less amps in this example
- lower flows may allow for piping size to be reduced
- lower flow = less friction (wasted energy)

So you are best off getting the smallest pump that can handle the required flow and head.

I work with pumps that could fill 4 of these tanks in under 1 second, but he couldn't afford buying one of these pumps or getting a power source to one.

Westinghouse newest pumps for nuclear power plants can empty an Olympic pool in about 1.5 seconds. They claim that two of them could empty like Winnipasaukee in 4 days

In the real world of submersible pumps, you add stages and do not increase rotational speed in order to gain head potential , which makes much of those energy use calculations irrelevant. If you have ever bench [bucket] tested submersible pumps of varying stages and at various pressures [restricting output] and observe an ammeter, the difference in amp draw between 0 and max psi is shockingly small. Your theoreticals may be correct, but the variables of MFG, type of impellers, method of motor construction make that 7.45 vs. 5 unlikely in the real water world. I agree on carefully choosing the smallest pump running at its sweetest spot on the output curve for the desired GPM in line with the pipes diameter.

nukeman said:

The most basic way to show this is with affinity laws (for centrifugal style pumps). For this style of pump, the difference is even more dramatic. First off, the volume (gpm) is directly proportional to shaft speed:

Click to expand...

The rpm is 3450 for most if not all sizes of submersible 4" residential pumps.

nukeman said:

So, going from 1 GPM to 5 GPM with the same pump means 5x faster speed (makes sense).

Click to expand...

Not true, 3450 rpm for both. You add stages/impellers to increase output. And when you do, the amp draw hardly increases at all.

So the rest of your post concerning electrical usage is based on incorrect data/info but I can tell you that a 1/2 hp motor running for 24 hours will use lots more electric than if it only runs for say 2 hours.

So now we have graduated to arguing with the laws of physics

Nukeman, good luck with this one. I don't know how you could have explained the concept any clearer than you did.

[Quote Slusser] So the rest of your post concerning electrical usage is based on incorrect data/info but I can tell you that a 1/2 hp motor running for 24 hours will use lots more electric than if it only runs for say 2 hours.

Of course it is less, it's not doing an equivalent amount of work. We are talking about moving a finite amount of water gallons/lbs in a given period of time. To move anything at a faster rate consumes more energy. the number of pump stages has to remain the same for each.

Maybe this will help. Grab hold of two 8 x 8 x 16 two hole cement blocks and slowly walk them the length of a football field. Now take the same two blocks and run the length of the field as fast as you can.

1. Adding stages will increase head, but not flow (or you could add flow (and not head) by placing stages in parallel). The very best you could do would be to use 5x the power to get 5x the flow and that would be a wash in terms of energy used. So, 0.5 GPM for 10 hr, 1 GPM for 5 hr, 5 GPM for 1 hr, etc. would all use the same energy. However, this can only be true if friction is small in relation to everything else. Otherwise, you will always use more energy at a higher flow rate. In this case, the frictional losses at 5 GPM are roughly 50% of the total loss (friction is ~ equal to elevation head). Friction is a total loss. The energy used to overcome frictional forces only serves to increase the fluid temperature. Would you both agree with this? Why would pumping faster (for shorter time) make this better? Increasing the speed (as in my example) will gain head and flow (not one or the other).

2. Yes, these pumps may run at a constant speed, but that has no bearing. I was speaking in terms of centrifugal pumps in general. With a fixed design pump (say single stage) you increase/decrease flow by changing the pump speed or changing the head (restricting output). Say your pump does 10 GPM with this system. Now, you say "I now want 30 GPM". What are you going to do? You can either put in a bigger pump or drive the pump at a higher speed.

3. I am not saying to restrict flow or anything to lower the flow rate. I'm simply stating that the smallest pump to do the job (running for longer times) is better than a running a larger pump for less time.

In physics, there is no free lunch. If you want more flow (means more work being done) that comes from somewhere. Frictional losses increase with increasing speed/flow. Your pumping faster would be like driving at 150 MPH instead of 50 MPH thinking that you'll save gas by driving for only 1/3 the time. It doesn't work that way.

I don't know how I can explain it any clearer. Again, if there was no friction, then 5gpm or 1gpm to move that 500 gallons would take the same total amount of energy (x number of pounds or water moved from one elevation to a higher elevation). With everything else being constant, the higher flow rate means more energy used because there is frciton in the real world [and the friction is significant at 5 gpm in this problem].

Actually larger pumps are a little more efficient than smaller pumps, so technically the faster you fill the tank the less energy you use. A 1 HP will pump 10 GPM at 300’ of head, yet it only takes about 80 HP to pump 1,000 GPM at the same head.

This only works if the pipe is large enough not to have friction loss at 1,000 GPM. Lowering the head required is the only way to reduce the HP required. In this case reducing the flow to 1 GPM causes less friction loss and only requires 133’ of head, compared to 199’ of head required at 5 GPM. Pumping at 133’ of head uses less energy than pumping at 199’ of head.

The problem is that it would be impossible to make or find a 1/25th HP pump that would efficiently produce 1 GPM at 133’ of head. We have to use a ½ HP pump that will produce 1 GPM at 400’ of head or 5 GPM at 300’ of head, and either slow it down or restrict it to do the job. Because head is lost by the square of the speed, it can only be slowed to 60% of full speed and still produce 1 GPM at 133’. Still this could get you down to 0.108 HP except for things like partial load efficiency of the motor and parasitic losses. A 1/2 HP motor will pull a minimum of about ¼ HP load, even without a pump attached. So you would be running 0.25 HP for 500 minutes to put 500 gallons in the tank.

Restricting the same pump to 5 GPM, would only be using about 0.35 HP, and would still easily produce 199’ of head. 0.35 HP for 100 minutes would use considerably less energy than using 0.25 HP for 500 minutes. In this case the faster you fill the tank and shut off the pump, the less energy is used.

Also I would suggest that if you do a pump test from zero flow to max flow, and you don’t see about 50% change in amperage, try a different brand of pump. Some pumps with floating impellers like a 10 GPM Sta-Rite will drop very little in amperage. Other pumps with fixed impeller designs, like a 25 GPM Grundfos, will easily drop 50% in amperage.

BTW my book shows 1” sch 40 couplings have the same friction loss as 3’ of pipe. I am sure belled ends would be less but, they would still have some loss.

I would be very interested to know what book that is because every single manufacturer of PVC that I know of does not even list couplings in their charts. Again all anyone has to do is to glue up a coupling and cut it lengthwise to see that there is no restriction at all there.

This whole pump conversation has about reached the point of maximum confusion with all sorts of numbers and formulas and such being thrown out there but the simple fact is that work = force times distance. when you change any one of the variables they all change. It is going to take the same amount of energy (not force) to move 500 gallons of water no matter how fast or slow you do it. Forget about pump curves and friction losses and TDH and anything else you can come up with. All those things are variables in the equation but they do not change the basic laws of physics.

They don't list them in the charts but I've seen x feet per coupling but I can't find where I have it.

Anyway, running this 2500' of PVC there are many couplers needed and I'll bet if they were all checked, we'd find a large number not together where there would be no friction loss because the pipe wouldn't be all the way in to the stop and there'd be plastic and cement sticking out inside about half of them due to the angle of the pipe being pushed into the coupler. And there may be some with the coupler being pushed on the pipe.

I think the confusion you mention has been cleared up, the longer you run the electric motor in this case, the more electric is used.

valveman: I agree with that. I'm talking about what the requirements should be. If you can't find a pump small enough to do the job, then you choose the next best thing. Pick the smallest pump that will do the job and weigh cost, quality, parts availability, etc. What I say still holds true:

pump 1: designed for 1 gpm at 133'
pump 2: designed for 5 gpm at 199'

Pump 1 will use less energy to do the job. Simple as that.

Gary started out saying that a huge pump and 2" poly would be needed for this job and that is simply not true. 1/2 hp is way more than enough. Remember also that we aren't limited to pumps designed for wells. Maybe a well pump would be best for the job, but maybe not.

This pump should do the job :

-125,000 GPM
- 330' head
- 9,000kW

Gary Slusser said:

They don't list them in the charts but I've seen x feet per coupling but I can't find where I have it.

Anyway, running this 2500' of PVC there are many couplers needed and I'll bet if they were all checked, we'd find a large number not together where there would be no friction loss because the pipe wouldn't be all the way in to the stop and there'd be plastic and cement sticking out inside about half of them due to the angle of the pipe being pushed into the coupler. And there may be some with the coupler being pushed on the pipe

Ok, so if you hack the couplings together then you get friction loss. good enough LOL

I think the confusion you mention has been cleared up, the longer you run the electric motor in this case, the more electric is used.

Click to expand...

But if you run is for a shorter period of time at more pressure it uses the same amount of electricity. Please check a high school physics book out of the local library and read a few chapters.

Nukeman, the westinghouse plant that makes those is right down the road. I've actually seen them being manufactured and assembled.

2” poly has the same friction loss at 5 GPM, as 1” PVC has at 1 GPM. Because larger pumps are more efficient than smaller pumps, pumping 5 GPM through 2” poly would use less energy than pumping 1 GPM through 1” PVC. This is splitting hairs because the difference would be minimal. It would never pay for the difference in price for the 2” poly.

Even with a jet pump or centrifugal a ½ HP is about as small as they make. If You could get a positive displacement pump that would do 1 GPM and stay around 60% efficient, you could get down to .05598 HP as in the following.

Horse Power Calculator

Flow (GPM) 1 Head (in feet) 133 Efficiency (%) 60 BHP .05598


Most pumps at 1 GPM will be very inefficient. If the efficiency is only 10% then the pump will use 1/3 HP as shown below.

Horse Power Calculator

Flow (GPM) 1 Head (in feet) 133 Efficiency (%) 10 BHP .33586


Running closer to the Best Efficiency Point for the pump available, you can pump 500% more water at 30% more head, while only using 20% more energy as shown below.

Horse Power Calculator

Flow (GPM) 5 Head (in feet) 199 Efficiency (%) 60 BHP .41877


0.419 HP running for 100 minutes will use less energy than running 0.336 HP for 500 minutes. Again, the quicker you fill the tank the less energy you use. The pump uses the least amount of energy when it is not running.

The Affinity Law assumes the two points on the curve that have the same efficiency. This is rarely the case as efficiency usually follows the flow rate. So you will usually get the wrong answer using the Affinity Law for a change in flow rate. This can be very confusing. Many people believe that reducing the speed of the pump will save energy. In reality reducing the pump speed will reduce the amp draw, but this actually increases the energy used per gallon produced. Here are a few related links.

http://drives.co.uk/features.asp?id=3

http://www.toshont.com/ag/vfdapplication/ag_24em_(application_considerations_for_vfds-part_2).pdf

http://www.powerqualityanddrives.com/payback_analysis_vfd/

http://www.lappusa.com/vfdarticle.html

http://files.engineering.com/downlo...-4665-b078-436661f6a439&file=20100212_VFD.ppt

Hunter Technical Manual 2002

You have a typo in your higher flow calc. Should be 5 gpm, 199' to get 0.42 bhp at 60%.

Anyway, you need to compare apples-to-apples. Maybe the problem is that these flow are below what are normal for pumps that you typically deal with. Just because this pump isn't good at 1 gpm, doesn't mean that you couldn't buy/build one that was efficient at 1 gpm. Let's instead move the numbers to what you may typically see to see the difference.

Pump 1: 10 gpm, 133'
Pump 2: 50 gpm, 199'

(Yes, I know the head values would change, but could be offset by using larger pipe). I get: 0.56 hp for pump 1, 4.19 hp for pump 2. Both assuming 60% efficiency.

0.56*5 = 2.8 compared to 4.19

Now to make actual comparisons, you would need to select a pump that was spec'd to each case and check the curve and then come back and compare the numbers. Pumping slower (with a pump designed for that purpose) is typically the same/better unless the pump is not correctly sized and you are killing the efficiency. Not to mention, a 3/4 HP pump would do the job in example 1, but you would probably have to go to a 5 HP pump in example 2. Difference means more $$ for the pump, larger cable, larger pipe, etc. More money to install and more cost to use. Doesn't sound good to me. You pick up some efficiency with the larger motor, but not enough to make up the difference.

If large flowrate for shorter time was best, then the pump that I posted would be great as it would only have to rotate about 4 times to fill the 500 gallon tank, eh?