Intelliflow Flow Rate and Run Time

TheOne

LifeTime Supporter
Mar 28, 2007
167
Houston, TX
Ok so I am wanting to take advantage of my Intelliflow pump's variable speeds and have a question regarding the flow rate of my pump.

Here is the flow rate chart for my 4x160 pump.



I have a 15250 gallon pool. My question is if I understand the flow rates correctly. For now assume I want to have 2 turns of day for a total of 30,500 gallons going through the filter daily. The 30500 gallons divided by 1440 available minutes in a day give me 21.18 GPM needed for 2 turns. Based on the flow chart above wouldnt I be able to achieve that by running my pump at <750 rpm?

Also the speed of the pump is adjustable between 400 - 3450 RPM for any program. When I want the pool cleaner (Legend) running I can increase the RPMs if needed.
 

TheOne

LifeTime Supporter
Mar 28, 2007
167
Houston, TX
Ahh yes that would be too easy. I have seen caclulations for how to get TDH and it made my head hurt. I suppose it will be trial and error then to find the right RPM.
 

JohnT

Admin
Mod Squad
TFP Expert
Apr 4, 2007
9,624
SW Indiana
JimW said:
Is there an "good" estimate for normal pools to calculate the head loss? I have the same pump and wondered how to set it for optimal efficiency.
There are many variables. Plumbing run length, pipe size, fittings, elevation changes, flow rate, filter type etc., etc. Only way to know for sure is to get a vacuum gauge and pressure gauge to measure your suction and pressure head. Not an easy calculation. At 50 gpm, 100ft of 1.5" pipe adds about 10ft of head. Every elbow adds another 7.5ft. The filter can't add more than 7ft (I think) when it's clean according to industry standards.
 

TheOne

LifeTime Supporter
Mar 28, 2007
167
Houston, TX
Not sure if this will help or not. This is from the pump sizing guide from the Hayward site which is available here. They give an example as well.

Friction Loss

Everything that the water must pass through within the recirculating system –
plumbing and equipment – creates resistance, or Head Loss. The sum of all
the resistance is called Total Dynamic Head, and is measured in Feet of Head.

Often, we are unable to determine the total amount of pipe and fittings in an
existing installation … it’s underground. Therefore, what follows is a simplified
“rule-of-thumbâ€
 

chem geek

TFP Expert
LifeTime Supporter
Mar 28, 2007
12,083
San Rafael, CA USA
There are two parts to efficiency. One is whether you are on the efficient part of the pump curve and that only makes a difference in terms of "wear" on the pump if you are way off the efficiency point. It also affects actual power efficiency, but there are other fixed electrical losses. So, the only real way to know what speed is truly best for efficiency is to compare your water flow rate vs. your electrical power consumption (measured at your electrical meter, for example). Unfortunately, it is hard to know your flow rate though knowing your RPM you could look that up on the pump curve if you know your head. If you have a pressure meter on your filter, then you know your head on output and just need to estimate head on suction which isn't that hard to do (the other post gives a decent way to estimate that).

I did some curve fitting and calculations to come up with the following formula that relates Head in feet to RPM and flow rate in GPM:

Head = (RPM/350)^2 - (GPM^2)/470

The output power vs. input power is approximately 50% at the most efficient point. However, for the very low speeds below about 1500 or 1000 RPM, the efficiency drops even at the best point due to fixed electrical losses. For example, at 690 RPM and 20 GPM with 3 feet of head the efficiency is only about 12%. Generally speaking, the lowest RPM will be more efficient, but at some point below 1500 or 1000 RPM you will start to lose efficiency. This post on the Pool Forum shows a chart that gives the same sort of data as in your curves PLUS the electrical power consumption as well.
 
G

Guest

JimW said:
Is there an "good" estimate for normal pools to calculate the head loss? I have the same pump and wondered how to set it for optimal efficiency.
I seem to remember reading somewhere that 50-60 feet of head is a good guesstimate for most pool installations unless things are nonstandard like lots of water features, spa, long distance from pool to equipment pad, etc. I will try and find the source if I can.
 

peterl1365

LifeTime Supporter
Mar 28, 2007
242
Murrieta, CA
waterbear said:
JimW said:
Is there an "good" estimate for normal pools to calculate the head loss? I have the same pump and wondered how to set it for optimal efficiency.
I seem to remember reading somewhere that 50-60 feet of head is a good guesstimate for most pool installations unless things are nonstandard like lots of water features, spa, long distance from pool to equipment pad, etc. I will try and find the source if I can.
Waterbear:

I've also seen that 50-60 ft estimate floating around the various forums. The only problem with that estimate is that it probably is only accurate for an average sized pump (1.5 hp, say) flowing at full speed. Since the TDH varies significantly with flow rate, I don't think that guesstimate would apply here.

Is there a way to estimate TDH by correlating it to filter pressure? For example, I also have an Intelliflo 4x160. When I run at low speed (around 800 rpm) my filter pressure is virtually zero. At 2100 rpm (which I use to run my Hayward Navigator), the filter pressure climbs to between 8 and 10 psi. At 2800 rpm (for surface skimming), the pressure rises to 15-20 psi.
 
G

Guest

Waterbear:

I've also seen that 50-60 ft estimate floating around the various forums. The only problem with that estimate is that it probably is only accurate for an average sized pump (1.5 hp, say) flowing at full speed. Since the TDH varies significantly with flow rate, I don't think that guesstimate would apply here.

Is there a way to estimate TDH by correlating it to filter pressure? For example, I also have an Intelliflo 4x160. When I run at low speed (around 800 rpm) my filter pressure is virtually zero. At 2100 rpm (which I use to run my Hayward Navigator), the filter pressure climbs to between 8 and 10 psi. At 2800 rpm (for surface skimming), the pressure rises to 15-20 psi.[/quote]

If you don't mind doing a little work there is a way to directly measure the head under various conditions. Plumb a vacuum gauge on the suction side of the pump where the pipe enters it and a pressure gauge on the pipe leaving the pump. The vacuum gauge will measure inches of mercury and each inch of mercury equals 1.13 feet of head. The pressure gauge measures pounds per square inch (PSI). Every 1 PSI equals 2.31 feet of head. If add the total suction side head and the total pressure head you will have the total dynamic head of the system.
 

chem geek

TFP Expert
LifeTime Supporter
Mar 28, 2007
12,083
San Rafael, CA USA
If your PSI pressure gauge were measuring correctly, then it should be measuring the pressure at the input to the filter so essentially should be measuring the output (pressure) head though you need to convert from PSI to feet of head by multiplying by 2.3 (these are just two ways of measuring pressure). That would just leave the input (suction) pressure that you would need to estimate. So the fact that you found nearly zero PSI on your pressure gauge seems a bit strange. At 800 RPM, you should be seeing around 4-5 feet of head or 1.7 - 2.2 PSI and around 20-30 GPM flow rate assuming your system curve is hitting somewhere near the efficiency point for this pump. Perhaps the 2 PSI looks like nearly zero on your gauge. As waterbear points out, an accurate vacuum gauge and pressure gauge will give you the precise numbers you want, but notice how relatively flat the pump curves are, especially at the lower RPM. It will be hard to know exactly what flow rate you have, even knowing RPM and Feet of Head. The formula I gave may be accurate, but your error in measuring low PSI or pressures will make the GPM estimate vary quite a lot.

If you measure some higher RPM and PSI points that are easier to measure and estimate the suction head, then you can calculate the GPM with the formula I gave and these multiple points (at different RPM) give you a system curve. You can extrapolate that curve and probably get a better estimate at what goes on at the lower RPM. As a rough estimate for the shape of most system curves, a doubling of GPM requires the Feet of Head to be 3.5 times higher. That is, the relationship is roughly

Head = constant * ( GPM^1.8 )

and at lower GPM of 30 and below the exponential factor is closer to 1.75 where this relationship is for the pressure loss found in pipes. The curve for the filter may be a bit different, but not usually by much since the physics is similar so long as the flow is turbulent which it generally is except for extremely low flow rates. The "constant" is a function of your pipe size and length as well as filter characteristics, but as I mentioned above you can just measure a few points at higher RPM to get a decent idea of the shape of the system curve and then see where it intersects with the pump curves. Or in formulas, you will have two simultaneous equations to solve at your known RPM.

The bottom line, however, is that it sounds like what you are really trying to do is accurately estimate the flow rate so that you can optimize for one turnover of water per day. As you can see, that is harder to do with the IntelliFlow 4x160 since you only know RPM from the pump and not GPM while the full IntelliFlo with its flow meter gives you GPM directly.

[EDIT]

I'm going to use peterl365's numbers as an example to hopefully make things more clear. Remember that Feet of Head = 2.3 * PSI. At 2100 RPM the pressure head is, say, 9 PSI (say 21 feet of head). At 2800 RPM the pressure head is, say, 17 PSI (say, 39 feet of head). Let's first ignore the suction head and then estimate that later. From my formula for the Intelliflo curves we can solve for GPM as follows.

Head = (RPM/350)^2 - (GPM^2)/470
so GPM = sqrt( 470 * ( (RPM/350)^2 - Head ) )

At 2100 RPM, GPM = sqrt(470*((2100/350)^2 - 21)) = 84 GPM
At 2800 RPM, GPM = sqrt(470*((2800/350)^2 - 39)) = 108 GPM

Now let's estimate the suction head and I'll assume two separate 1.5" pipe lines for the skimmer and floor drains. There are many tables and formulas one can use and I am assuming Schedule 40 pipe. In that case, the following approximate formulas may be used:

For 1.5" (nominal; 1.610 inner diameter; 1.900 outer diameter) pipe: Head Loss per 100 feet = 0.011 * ( GPM^1.8 )
For 2" (nominal; 2.067 inner diameter; 2.375 outer diameter)) pipe: Head Loss per 100 feet = 0.0034 * ( GPM^1.8 )

So at 84/2 = 42 GPM, that's a suction head loss of around 9 feet per 100 feet and I assume 100 feet from the pump to the skimmer and floor drains. At 108/2 = 54 GPM, that's a suction head loss of around 15 feet of head per 100 feet. So including the suction head into our formulas above I get:

At 2100 RPM, GPM = sqrt(470*((2100/350)^2 - 21 - 9)) = 53 GPM
At 2800 RPM, GPM = sqrt(470*((2800/350)^2 - 39 - 15)) = 69 GPM

Doing another iteration, at 53/2 = 26.5 GPM, that's a suction head loss of around 4 feet per 100 feet. At 69/2 = 34.5 GPM, that's a suction head loss of around 7 feet per 100 feet. So,

At 2100 RPM, GPM = sqrt(470*((2100/350)^2 - 21 - 4)) = 72 GPM
At 2800 RPM, GPM = sqrt(470*((2800/350)^2 - 39 - 7)) = 92 GPM

I'll do one more iteration even though I made some approximate assumptions. At 72/2 = 36 GPM, that's a suction head of around 7 feet per 100 feet. At 92/2 = 46 GPM, that's a suction head of around 10 feet per 100 feet. So,

At 2100 RPM, GPM = sqrt(470*((2100/350)^2 - 21 - 7)) = 61 GPM
At 2800 RPM, GPM = sqrt(470*((2800/350)^2 - 39 - 10)) = 84 GPM

Now let's use the other formula to see if these two points are consistent with a system curve.

Head = constant * ( GPM^1.8 )
so constant = Head / ( GPM^1.8 )

At 2100 RPM, constant = (21+7) / ( 61^1.8 ) = 0.0171
At 2800 RPM, constant = (39+10) / ( 84^1.8 ) = 0.0168

so at least we have some consistency which is good. So let's say the "constant" is 0.017 and can now use the formula to calculate GPM as follows:

1) Head = (RPM/350)^2 - (GPM^2)/470
2) Head = 0.017 * ( GPM^1.8 )

0.017 * ( GPM^1.8 ) + (GPM^2)/470 = (RPM/350)^2

You can't easily solve for GPM in the above equation, but can iteratively estimate it for a given RPM. At an RPM of 750, the above formula gives a GPM of

0.017 * ( GPM^1.8 ) + (GPM^2)/470 = 4.59

so a GPM of 20 makes the above formula work. The Head is

Head = (750/350)^2 - (20^2)/470 = 3.74 feet = 1.6 PSI so barely measurable on a standard PSI pressure gauge on a filter.

By the way, the IntelliFlo chart that shows 750 RPM is wrong and is really 950 RPM. Pentair has other charts (see PDF page 53, manual page 47 of this link) for the full IntelliFlo (which is the same pump, but with extra electronics and a flow meter) that show 690 RPM and make clear that the 750 curve is incorrect and the pump formulas should be fairly consistent at all RPM since they are based on impeller shape more than anything else.

The thing to remember is that lower speeds have diminishing returns since the electrical efficiency drops at low RPM. It appears that the IntelliFlo pumps have a fixed electrical loss of around 80 Watts or so. Running the pump at very low RPM would theoretically have less frictional losses, but the fixed electrical losses make this less desirable. It would take another set of calculations to calculate the "sweet spot" which may very well be at flow rates below what it would take to achieve one turnover in 24 hours in which case the turnover requirement would take precedence.

I would estimate the IntelliFlo electrical power at or near the efficiency point at being roughly the following:

Input (electrical) Watts = (Output Watts) * 2 + 80 = (GPM * (Head In Feet) * 0.188165) * 2 + 80

[END-EDIT]

Richard
 

TheOne

LifeTime Supporter
Mar 28, 2007
167
Houston, TX
Installing a flow meter would probably be the best option then correct? That way I could gauge my turns more accurately. I suppose you would want the flow meter to be downstream of the filter but before the heater correct?
 

chem geek

TFP Expert
LifeTime Supporter
Mar 28, 2007
12,083
San Rafael, CA USA
Installing a flow meter would certainly make your settings incredibly simple, yes. So long as the flow meter were anywhere downstream from the pump and before the pipe splits into multiple pipes, I don't think the position matters (actually, it could even be before the pump, so long as it's before any split in the pipe). It's measuring flow, not pressure. But then, I'm not certain how flow meters work -- I would guess that they have some sort of spinning "vane" where the rate of spin determines the flow rate. Anyone else know about flow meters?
 

TheOne

LifeTime Supporter
Mar 28, 2007
167
Houston, TX
There are paddlewheel inline type and there are also clamp on type with a probe that goes into the water flow through a drilled hole. I know there are recommendations to not have any bends in the pipe for certain distances in front of or after the flow meter. I have some room in the pipe going to my salt cell to install one.

I know the inline type flow meters appear to be more expensive, of course you can get digital ones, etc.

Im not sure why they arent more common in installs other than maybe most builds just install a single speed pump and turn it on and let it go...
 

chem geek

TFP Expert
LifeTime Supporter
Mar 28, 2007
12,083
San Rafael, CA USA
You know, after all of my detailed calculations I now realize that a rough approximation works pretty well: the GPM is roughly proportional to the RPM. So if you measure or otherwise determine (via the one formula that relates GPM to RPM and Head) the GPM, then you can roughly calculate the GPM at lower RPM by simple proportion.

So, for 750 RPM we have:

61 * (750/2100) = 22
84 * (750/2800) = 23

and these are not far off from the 20 GPM that I calculated. So getting a flow meter is the sure-fire way of knowing your flow rate. Getting a suction gauge combined with your existing filter pressure gauge is the next best thing. Using the filter pressure gauge plus an estimate of the suction loss is the third option (and least accurate).
 

TheOne

LifeTime Supporter
Mar 28, 2007
167
Houston, TX
Thanks for the help. I only need approx 21 GPM to turn my pool once in 12 hours or a little over 10 GPM for one turn for 24 hours. My pump will go as low as 400 rpm. I have confirmed with Pentair that my filter doesnt have a minimum GPM flow rate requirement to filter properly.

I am working on tracking down a flow meter as they are kind of hard to find for some reason. I have found the phone number for a local distributor that sells a particular brand that should work. Their company Blue-White gave me a recommendation for flow meter for a pool/spa application. I have attached the data sheet for the model they recommended.

Now I just need to verify if I have 2" or 2.5" piping as that changes which flow meter is required.
 

chem geek

TFP Expert
LifeTime Supporter
Mar 28, 2007
12,083
San Rafael, CA USA
You most likely are using what is known as "Schedule 40" pipe so you can look at this link to measure your pipe's outer diameter and figure out what size pipe you have. 2" pipe has an outer diameter of 2.375" while 2.5" pipe has an outer diameter of 2.875" (1.5" pipe has an outer diameter of 1.900").
 

TheOne

LifeTime Supporter
Mar 28, 2007
167
Houston, TX
For the attached Intelliflo power chart, is the chart showing that the 690 RPM rate is the most efficient? I'm not clear on how to interpret the chart relating to the intersection of the red and black lines...
 

chem geek

TFP Expert
LifeTime Supporter
Mar 28, 2007
12,083
San Rafael, CA USA
The chart is really two separate charts put into one so it is confusing. The red lines and the black lines are totally different so intersection means nothing. The red lines show the power consumption (in killowatts) compared to flow rate (in GPM). The black lines show the pressure head (in feet of water) vs. the flow rate.

Basically, the slower speeds (RPM) are more energy efficient at a faster rate than the way that RPM and GPM vary. Remember that I said that at least approximately, RPM and GPM vary directly together so that if you cut the RPM in half you roughly cut the GPM in half. Well, for power consumption, the very rough rule is that cutting the RPM in half cuts the power consumption by one-eighth. So lower RPM is almost always more efficient in terms of converting electricity into a flow rate (the efficiency in terms of output power is about the same but output power is the product of flow rate with pressure).

The point I was making was that at some very low RPM this savings goes away because there is a fixed electrical cost even at 0 GPM and that appears to be around 80 Watts. So at very low RPM you don't get the same sort of savings as at higher RPM and in fact at some point you stop getting savings. Let's just make some very simple assumption to see how this might work using the following very rough formulas:

GPM = 0.03 * RPM
Electrical Input Watts = 80 + (RPM/250)^3

You want to achieve one turnover which is a fixed number of gallons (G) so the total power consumption for one turnover is:

Energy Per Turnover (KWh-per-1000 gallons) = Kilowatts / GPM / 60 = ( 80 + (RPM/250)^3 ) / (1.8 * RPM)

You want to find a minimum for this function which clearly goes down with lower RPM for a while, but at some point starts to go back up again. Though we could take the derivative to find this out, it's instructive to just look at a table of values to see how this roughly changes:
Code:
RPM     KWh-per-1000gal     GPM     Turnover-Time-per-1000gal (Hours)     Max. Size (gallons) for 24 hours
3450          0.44          104                 0.16                                    150,000
2000          0.16           60                 0.28                                     86,000
1500          0.11           45                 0.37                                     65,000

1200          0.09           36                 0.46                                     52,000
1000          0.08           30                 0.56                                     43,000
 750          0.08           23                 0.74                                     32,000
 600          0.09           18                 0.93                                     26,000

 500          0.10           15                 1.11                                     22,000
 250          0.18            8                 2.22                                     11,000
So you can see that the minimal amount of energy is consumed around 750-1000 RPM though 600-1200 RPM is still very efficient. However, if you have an extremely large pool, then you will be limited by the 24 hour turnover requirement. However, the sweet spot for best energy efficiency will handle pools up to around 50,000 gallons in size which is far larger than most residential pools. So running at 750-1000 RPM is best and would turnover a 15,000 pool in around 8-11 hours.

This is an interesting analysis that I was going to have to do anyway for my own pool and it shows that using extremely slow pump speeds to achieve 24 hour turnover is NOT the smartest thing to do for optimum energy usage.

The only way to really confirm all of this is to take your actual pump in your real pool and measure its energy usage at different RPM and compare against the total energy consumed by looking at the GPM flow rate and the size of your pool for one turnover. However, I'll bet that my equation assumptions aren't that far off. It would be interesting to find out, though. Be sure and keep us posted (if you do get a flow meter since that's the only way you'll know how long you truly need to run the pump for one turnover).