RPM to GPM estimates for VS-SVRS?

[mas985]

Those PSI numbers are consistent with the head loss estimates so it would seem to be pretty close to reality. The intent here is not to get exact numbers but to give the user an idea of what the operating points might be.

 

[Guest]

Remember, head loss drops in a given pipe as velocity drops due to less friction.

Scott

And I say:

'What is confusing to me (common sense wise) is that as wattage increases for the same RPM, head *decreases*. I always thought that longer runs with more resistance (more head) would make the pump work harder at any RPM level, thus increasing the wattage. And thus head should increase. Maybe you can correct my misunderstanding of basic fluid dynamics.'


So are you essentially saying that a pump needing to work harder (more watts) at the same RPM means the water is likely to be moving slower (even at that higher wattage) due to pipe limitations. And because of this, less friction and thus less head loss?

This isn't making sense to me. Take two pipes, 1" lines and 2" lines. I assume head loss at any RPM (or wattage for that matter) is greater in the 1" than the 2" pipe, as resistance is higher. At any RPM also, should wattage be higher in the 1" pipe versus the 2" pipe (as moving the water takes more -work- in 1" pipe)? I assume so. So didn't the greater head in the 1" pipe cause higher watt load than a 2" pipe at the same watt load? That is contradictory to the results of the spreadsheet, where higher wattages for the same RPMs yield *lower head*, not *higher head*. I thought head was higher if a total pipe run provides more resistance (whether it be from frictional, gravitational, or directional barriers/forces), so in a way head is a measure of aggregate pipe system resistance. Is this an incorrect understanding?

 

[Guest]

Those PSI numbers are consistent with the head loss estimates so it would seem to be pretty close to reality. The intent here is not to get exact numbers but to give the user an idea of what the operating points might be.

I'm just thrilled I'm getting closer to 30gpm versus 22-23gpm (which I thought I was getting) at 1400rpm. Allows me to cut a few hours/day off my filtering time.

 

[mas985]

What Scott was getting at is that head loss is related to the flow rate. In fact, close to the square of the flow rate. So for a given plumbing system, the higher the flow rate, the higher the head loss and visa versa as well.

 

[PoolGuyNJ]

I am the Induhvidual. I mistook the head for GPM. BRain fault on my part. Mas985's chart matches mine.

 

[Guest]

What Scott was getting at is that head loss is related to the flow rate. In fact, close to the square of the flow rate. So for a given plumbing system, the higher the flow rate, the higher the head loss and visa versa as well.

I wish to bring this back from the dead. I wanted to show that the results this spreadsheet spits out may not be accurate, and suggest the curve equations be fixed. Unfortunately, my physics knowledge is lacking to easily jump into those equations. But the chart here (produced from the equations in this spreadsheet) intuitively show that something is wrong with the calculations.. Most obviously, as more wattage is necessary to push water through at any fixed rpm (1400 in this case), head should be rising (as resistance is higher). Ie extreme example: Lets say we put oil instead of water in our pool pipes. It would take a lot more wattage to run the pump at 1400rpm, as the oil would provide substantially more resistance (due to viscosity differences) [I assume]. In other words, more work necessary to run the pump at X RPM means more watts. The spreadsheet spits out falling head instead. The rest of the derivations also don't make sense.

Please let me know what you think.

The results below contradict your statement. They show lower head loss at higher flow rates. Try the spreadsheet and you'll see. In fact, according to your point, the green line should by approx x^2 in the same direction as the red line. Here it is inverse falling.

BTW, the horizontal axis is both in feet of head, and gallons per minute. And the purple line (GPM/watt) never peaks, showing clearly something isn't right. It is read as at 300 watts for example, GPM/watt is approx .19.

 

[mas985]

I think you may be confusing the plumbing curve with the pump head curve. Think of a pump's head curve as potential head gain while the plumbing presents a head loss. A pump's head curve will generally fall off with increasing GPM because it cannot generate as much pressure with higher flow rates. So the maximum head of a pump is when the flow rate of the pump is 0 GPM (dead head) while the maximum flow rate of a pump is at very low head (run out). If the head loss of the plumbing decreases, then the flow rate produced by the pump increases or visa versa.

Also, it takes more energy to generate flow rate than it does pressure or head. This is why the energy consumption of a pump will increase with increasing flow rate no matter if the increase in flow rate is due to an increase of RPM or a decrease in head loss. This can best be seen from the Intelliflo head curve and power curve from the manual shown here:

You can pick off a few points on the above curve to test the accuracy of the model but for most speeds it isn't too bad. But again, I am curve fitting to the CEC measured data vs the published data so it may not be an exact fit.

For example, the model produces an operating point of 3450 RPM, 2450 watts, 80 GPM and 80' of head which is pretty close to the above chart data point.

Another example at a lower speed is 2070 RPM, 550 watts, 40 GPM and 31' of head which is also fairly close to the above chart data point.

As for plumbing curves, the head loss increases with increasing flow rates which is opposite of the pump head curve. This can best be shown with a combination of the pump curve and plumbing curves shown here:

The plumbing curves are represented by the lines curving upwards to the right labeled A-G and again, the pump head curves, curving downwards to the right with varying HP. As you can see, they only intersect at a single point which is the operating point. Each pump, at a specific RPM, and each plumbing curve can have only one operating point.

I'm not sure if I answered all of your questions but hopefully the explanation helped a little bit. I also have a bit more information it the Hydraulics 101 sticky in my sig.

 

[Guest]

Looking at the 'VF Flow and Power Vs Flow Pump' curve, the most glaring and counterintuitive result is that at any RPM (ie 2070rpm), with increasing wattage comes increased flow. This is confusing me, as I imagine an impeller spinning an equal speed moving the same amount of gallons per minute no matter what, with wattage just increasing as plumbing head increases (or fluid viscosity increases) to move the water the same amount.

I read your hydraulic 101, but can you clarify on plumbing head vs pump head. I realize the pump head = total suction head+ total return head, but am a bit confused again.

 

[mas985]

What might help is using the analogy of an electric circuit.

Source voltage is similar to pump pressure or head gain. This is potential energy of the system.
Amps is similar to flow rate that circulates through the system. This is kinetic energy.
Resistance is similar to the friction or head loss in the pipes. This is the loss of energy through the system.

Much as the electrical energy is measured in watts and calculated as Volts x Amps, the pump energy can also be expressed in Watts as Head * GPM / 5.31. This is the amount of energy delivered to the water. The hydraulic HP delivered to the water can be expressed as Head * GPM / 3960. There is energy loss in both the motor and in the wet end so the motor input power required is much higher than the hydraulic power (~2x). Also, the hydraulic power increases with increasing GPM along the head curve which is why more input energy is required for the motor at higher GPM.

Conservation of energy must apply in hydraulics as well so all of the head losses in the plumbing must sum up to equal the head gain of the pump much like voltage drops across resistors in a circuit must add up to the source voltage. So to summarize, the pump provides head gain and the plumbing creates head loss and the net head for the system (suction port to return) is 0.

Also, the characteristics of the pump and the plumbing can be considered as two separate independent systems which is why they are shown as two separate curves. However, they do interact which is why there is only one operating point for a pump and a plumbing system together.

Think of the pump head curve as how the pump will perform for an infinite number of plumbing systems along the curve. Each point on a pump's head curve is for a different plumbing system. The left most side of the pump curve is for a plumbing system with maximum head loss so the flow rate is 0 GPM. The right most point of the head curve is where there is a minimum in head loss so the flow rate is a maximum.

You asked why the flow rates changes with head for a constant RPM. Inside the pump, there are losses. There is recirculation loss which is loss in flow rate due to the water traveling through the impeller and back into the inlet (never gets to the outlet). There is also friction loss inside the pump which reduces the exit pressure of the pump and thus the potential for flow rate. Both of these internal losses dictate how the pump head curve will look and the performance of the pump under varying conditions. So when there is a lot of pressure on the output of the pump, recirculation loss increases and the flow rate out of the pump decreases. When the exit pressure decreases, the recirculation losses decrease and the flow rate out of the pump increases. However, because of the friction loss inside the pump, there is a limit to how fast the water will move through the pump so the run out flow rate is fixed.

One more point is that water is considered an incompressible fluid so the density does not change much with pressure. However, water density does change with water temperature so the kinematic viscosity will change with temperature but not much with pressure. So viscosity only really plays a roll with differing water temperatures but still a small role.

If you want to learn more about pumps, head curves, plumbing and hydraulics in general, I would strongly suggest that you read through the articles that Joe Evens of Pentair put together on pumped101.com web site. It has everything you ever wanted to know about pumps and more. Plus he can probably explain things a lot better than I can. But I would also be happy to answer any more questions that I can.

 

[taekwondodo]

Lucky that I'm a double E (BSEE), or I would have been lost .

 

[Lershac]

This is confusing me, as I imagine an impeller spinning an equal speed moving the same amount of gallons per minute no matter what, with wattage just increasing as plumbing head increases (or fluid viscosity increases) to move the water the same amount.

That statement is a misunderstanding. That impeller does not move a set amount of water per rotation. It just imparts a pressure to the water to move in the desired direction. How much water moves as a result of that pressure or additional energy is a function of "head" or pressure exerted in the other direction.

A Piston pump would move a relatively "set" amount of water in each rotation, but an impeller pump does not. In Fact, given a tall enough pipe, an impeller can continue to rotate at the desired rate, and no water movement will occur.

 

[mas985]

To clarify a few points.

The fluid dynamics around the impeller are quite complex but as described in this reference and this reference, the impeller imparts kinetic energy to the water but it is the volute which converts the kinetic energy to pressure energy. According to Bernoulli's principle, when volume increases, velocity decreases but pressure increases and since the volute is larger in volume than the impeller, as water moves outward from the impeller into the volute, the velocity decreases but pressure increases. This transition can be large or small depending on where the operating point is on the head curve. Also, because the volute is asymmetrical, a diffusor is used to help equalize the loads on the impeller shaft and reduce turbulence to improve efficiency.

Because the impeller rotates at nearly a fixed RPM (+- a few percent), the water velocity at the edge of the impeller is nearly fixed and is actually quite high (75 ft/sec for a 5" impeller). However, the water velocity through the impeller is not fixed and depends upon operating point of the pump. The "extra" velocity of the water is accounted for by recirculation around the impeller. So some of the water comes up through the impeller inlet and out through the impeller exit. However, some water around the impeller exit is accelerated via eddies and recirculated back out. This is why the exit velocity of the impeller can remain nearly constant while the flow rate through the impeller changes. At the extreme of dead head, no flow comes out of the pump but there is still plenty of water circulating within the volute. It is just going in circles and providing no useful energy just pressure. However, there is a small amount of water which leaks out of the impeller ring back into the pump inlet but most of the energy is lost to recirculating within the volute.

Here are some great pictures from this reference which show both velocity and presure for various parts of the pump. The inner circle is the inlet, then you should be able to make out the impeller, diffuser and the outside is the volute.

First Velocity:

Then Pressure

And flow direction. Note the eddies around the diffuser.

 

[Guest]

Thanks a lot for the above posts. Questions for you: Does changing viscosity (fluid resistance) versus changing plumbing resistance have the same impact on the pump's power consumption?

I am starting to figure something out. If you have an increase of plumbing head in a system given a single RPM, less actual water will get to the pump's impeller. Thus the pump actually works LESS. This is why wattage falls as plumbing head increases (ie you hook a pool vacuum up to a line).

Now what if you increase the viscosity of the fluid, and make the pump move oil instead of water. What happens to the pump's wattage, given a fixed RPM, etc ?

Also unrelated question (actually more related to the thread topic). I've been measuring power consumption via a wrap around multimeter, and only reading current. I have accurate amp readings. Do I plug watts or VA into this spreadsheet? They aren't the same, assuming this pump has a power factor of something else than 1.0 (lets say .9).

So most importantly, what is the power factor on these intelliflo pumps?

 

[mas985]

All very good questions. I will try to answer them in-line.

Thanks a lot for the above posts. Questions for you: Does changing viscosity (fluid resistance) versus changing plumbing resistance have the same impact on the pump's power consumption?

When pumping a different fluid there are two factors to take into account, the fluid's viscosity and specific gravity. A pump's head curve does not change with either viscosity or specific gravity and this is one of the reasons that the head is used instead of PSI. PSI changes with specific gravity but head doesn't. However, the plumbing curve is affected by viscosity so an increase in viscosity will increase the head loss in the plumbing and change the operating point of the pump thereby reducing the flow rate. The energy consumption is related to the hydraulic HP which in turn is related to specific gravity of the fluid so the energy demand from the impeller will change slightly with the type of fluid.


Now what if you increase the viscosity of the fluid, and make the pump move oil instead of water. What happens to the pump's wattage, given a fixed RPM, etc ?

Oil has a lower specific gravity (lighter) than water so for a given head and GPM, the power required to pump it is actually less than water. But because oil has a higher viscosity, the head loss in the plumbing will be higher for a given flow rate. Therefore with the same plumbing and pump, the oil will have higher head loss, lower flow rate and much lower energy consumption than water.


I am starting to figure something out. If you have an increase of plumbing head in a system given a single RPM, less actual water will get to the pump's impeller. Thus the pump actually works LESS. This is why wattage falls as plumbing head increases (ie you hook a pool vacuum up to a line).

That is pretty much correct although I might say that the flow rate through the pump is reduced rather than water is prevented from getting to the pump. Nothing prevents water from getting to the pump except for the water ahead of it. In other words, the flow rate.

Also with regards to lower energy use for lower flow rates, this is mainly true for pool pumps but not for all pumps in general. Some very high head pump designs, think water towers, have an inverted power curve. But this because the lift has to be so high. The energy consumption of a pump is related to the hydraulic HP which is the product of head AND GPM. So if head increases at rate higher than GPM decreases, then the pump will use more energy for lower flow rates but again, this doesn't apply for pool pumps.

Also unrelated question (actually more related). I've been measuring power consumption via a wrap around multimeter, and only reading current. I have accurate amp readings. Do I plug watts or VA into this spreadsheet? They aren't the same, assuming this pump has a power factor of something else than 1.0 (lets say .9).

So most importantly, what is the power factor on these intelliflo pumps?

The power factor should be very close to 1.0. This is because a VFD first converts the AC to DC and then runs the DC into gate arrays which create the PCM waveform. So the load presented to the AC line is basically real. However, the lines feeding the actual 3-phase motor may have a power factor but that will not translate back to the AC line waveforms because of the AC to DC conversion.

Because a VFD uses PCM, there are current pulses that end up on the AC supply line which might affect how the meter measures the current. If the VFD has sufficient filtering, this should not be a problem and you should still get a fairly accurate measurement. But the only way to tell for sure is to put a scope on the current probe.

 

[Guest]

Thanks!

 

[NYY]

Thanks for the spreadsheet Mas985. You've provided something that has eluded me since I put my pool in last August. I couldn't find it on the Pentaire website, and when I send a message to Pentaire tech support, they told me to install a flow meter. I appreciate it.

 

[mas985]

I'm glad you found it useful but keep in mind that the results are approximations to limited data that exists for the Intelliflo. The accuracy is better at the higher RPMs than at the lower but overall should be better than nothing.

 

[pooltech]

Can anyone explain how to use the calibrate tab on the Intelliflo V1.4 spreadsheet? It looks as if you can populate 9 cells values under the five headings (LS A, HS A, HS B, Curve A and Curve B). I can see that the results on the calc tab change if you change the values.

Thanks.

 

[mas985]

First, welcome to the forum.

The calibrate tab is for establishing the coeficients of the curve fit equation used in the other tabs. There are two separate equations, one for energy input and one for head. Currently it is populated with the CEC measurement data. Unless you have actual measured data, I wouldn't change those values.

Currently, the CEC takes two measurements, Curve A and Curve B. LS A represents low speed for Curve A and HS A is high speed for Curve A. Same for Curve B. If at some point I can collect enough data from VF users, then I may move to that data. The VF reports GPM, RPM and Watts so with enough measurements, a better fit to the energy equation can be made. However, there is only the published head curves to fit that equation.

 

[pooltech]

Thanks for the welcome!

Please forgive my ignorance, but what is "CEC" measurement data? I do have some actual data (both flow (spa drain and fill data) and Head data (pump suction and Filter return)), although I am not sure how accurate it is.