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JasonLion said:

A typical 1.5 HP pump will draw somewhere between 1500 and 2000 watts, I will use 1500 watts in my calculations. 1500 watts for two hours a day is 3 kWh/day, or about 90 kWh/month. With electricity costing $0.11 per kWh, which is fairly low these days, that is about $10/month. Assuming a six month pool season and a 50% electric savings by switching to a variable speed pump, you could save around $30/year by switching.

For comparison, lets imagine that you were running the same pump for 12 hours a day and your electric rates were $0.45 per kWh. Then you would be spending about $245/month and could potentially save around $735/year.

As you can see, the savings are very dependent on the size of the pump, the number of hours you run the pump, and your electric rates. Larger pumps, longer run times, and higher electric rates all make it easier to justify the cost of replacing the pump.

For comparison, lets imagine that you were running the same pump for 12 hours a day and your electric rates were $0.45 per kWh. Then you would be spending about $245/month and could potentially save around $735/year.

As you can see, the savings are very dependent on the size of the pump, the number of hours you run the pump, and your electric rates. Larger pumps, longer run times, and higher electric rates all make it easier to justify the cost of replacing the pump.

JasonLion said:

The simplified rule of thumb is that an N HP motor draws N kWh per hour. So a 3/4 HP motor will draw 3/4 of a kWh each hour it is on. Real world numbers vary from that both up and down, but that is a good place to start. You can also use your electric meter to measure what the pump actually draws.

Brentr said:

The price difference between my pump and a variable speed pump is about $600. I guess it would take me 20 yrs to break even

mas985 said:

The service factor of the motor also needs to be taken into account so a N SFHP motor draws N kwh. For example, a 3/4 HP pump with a 1.65 service factor motor will actually draw about 1.2 kwh (1.2 SFHP). A 3/4 HP motor with only a 1.0 SF will draw 0.75 kwh.

Braking Horse Power

Total Horse Power

Yes, technically you should take SF into account. The more common formula is <listed HP> * <SF> * 0.746 * hours gives kWh. But even with that, it doesn't work for the current crop of variable speed pumps, two speed pumps need a correction factor for low speed, and high efficiency motors are a little off. Ultimately, there isn't any true standard of what a "HP" means in watts. Conventional usage is 746 watts == 1 HP, but that is not really true for all the situations that come up these days.

Since many people don't know their SF, and to keep the math really really simple, it is plausible to just use 1 kW for 1 HP. For high SF motors that will be off a bit, but is still usable as a rough estimate for the kinds of use discussed in this thread. I suppose my rough estimate formula could be a little higher, 1.2 kW for 1 HP, but 1 to 1 is so easy to work with and no estimate is really accurate. If you really want an accurate number it is better to measure the pump in actual use.

JasonLion said:

Ultimately, there isn't any true standard of what a "HP" means in watts.

1 BHP or SFHP = 745.699872 watts

Now the reason that the SFHP is a good approximation for consumed energy in kwatts for a motor operating at the service factor load is that the efficiency of an induction motor used in pump motors happens to be around 75%. Actually between 65% and 85% but 75% is a good average. So when you put it all together, the actual energy consumption of a motor is:

kWatts = SF * HP * 745.699872 / 1000 / Motor Efficiency

kWatts = SF * HP * 745.699872 / 1000 / 0.75 ~ SF * HP = SFHP

An approximation yes, but without any other data, it provides a fairly good upper limit.

then using an example which by some strange coincidence matches the numbers off the label on my pump... 3/4 HP, 1.5 SF, I get

kW = 1.5 * .75 * .746 / Efficiency

kW = .84 / Efficiency

Now it so happens the label also bears the inscription: KW: .55

which one would naively think means the motor draws .55 kilowatts. If we plug that in,

.55 = .84 / Efficiency, or Efficiency = 1.5

which makes me think I got lost somewhere.

--paulr

My old pool pump is more typical at 1 HP, 1.65 SF, 1.84 KW (on the faceplate, though I measured closer to 1700 Watts in actual use; perhaps the KW rating is more of a worst-case at the edge of normal operating limits).

I saved around 50% on my annual pool pump electricity costs going from around $1400 to $700 per year by moving to an Pentair Intelliflo, but my electricity rates are high at around 32 cents per KWh. If I didn't run my pump at higher speed needed for solar during part of the day, I'd save even more.

sjsoldo said:

This discussion confuses me some. For calculating power consumption, do you really need to know the "horsepower" of the pump?

Why not look up how many amps it uses and calculate kilowatts based on 110 vs. 220 volts, then multiply how many hours you run the sucker.

Is this incorrect????

Why not look up how many amps it uses and calculate kilowatts based on 110 vs. 220 volts, then multiply how many hours you run the sucker.

Is this incorrect????

sjsoldo said:

Why not look up how many amps it uses and calculate kilowatts based on 110 vs. 220 volts, then multiply how many hours you run the sucker.

Is this incorrect????

An electrical motor really has two "unrelated" definitions of HP going on, one measuring the amount of electrical energy going into the motor and the other measuring the amount of useful work coming out of the motor. For the traditional electrical motor design, these two numbers are equal. But for modern electrical motors they are not always equal. For example, the motors used in variable speed pumps have a noticeably higher inherent efficiency, they do more work for a given electrical input than a traditional motor would do. Likewise, high efficiency pumps do more work for a given electrical input than other pumps, though the difference is not as dramatic as it is for the current crop of variable speed pumps.

Pool pumps base their listed HP * SF number on the amount of work they can do, not on the electrical input required. Listed HP * SF gives brake HP, which is a measure of the work being done by the motor shaft. For the traditional motor designs, brake HP will equal the electrical HP. But for high efficiency motors, most variable speed motors, and other newer designs, the numbers will not be equal.

Using the listed amps and volts on the motor is another reasonable way to estimate electrical usage. Amps * volts * hours / 1000 = kWh. But again, this is just an estimate. Real world electrical usage will vary from that number, typically it will be a bit lower, though the actual number will vary from pool to pool. If you know the amps and volts number for your pump, this is a perfectly reasonable way to estimate.

Using an amp meter to measure your pool pump in operation is not such a good approach. You can use the power companies electrical meter, but not an amp meter. There is a good article at Wikipedia explaining the difference between real and apparent power. The difference is significant, but rather technical. Super briefly, inductive loads, like electrical motors, can shift the phase of the AC power, causing the real power measured by the electrical meter to be different from the apparent power measured by an amp meter.

All of which is getting way off the topic of estimating payback times for replacing your pump with a more efficient pump. For that purpose, even the simplest possible estimates are quite sufficient. That is why I gave my super simple formula to start with, and didn't try to explain any of the things I got into in this message. This is one of those situations where knowing a little is fine, knowing a medium amount can cause problems, and knowing a lot gets you back to where you were when you only knew a little.

.75 HP x 1.5 SF = 1.125 kW

13.0 amp x 115 V = 1.495 kW

The amp part of the label also says "max load" implying that the 1.495 is really a worst case, and maybe implying the lower number is probably a better one to use; and the label's "KW: .55" should just be ignored as it's half of even the lower estimate. At least until I can figure out how to measure kWh off my electric meter.

--paulr

PaulR said:

At least until I can figure out how to measure kWh off my electric meter.

PaulR said:

Okay, so in my case the rough rules of thumb give

.75 HP x 1.5 SF = 1.125 kW

.75 HP x 1.5 SF = 1.125 kW

My super simple rule gives 0.75 kW (within 11% of 0.84 kW).

I can't think of anything that is going to get it down to 0.55 (assuming this is an ordinary pool pump and not some fancy modern hi-tech thing).

horsepower (550 ft â€¢ lbf/s) watt (W) 7.456 999 E+02

Second, I really think that definitions are important when discussing pump and motors. The definitions below may have been sited before but I wanted to summarize everything in one place. These are terms often used to describe pumps and motors but are sometimes confusing.

Nameplate HP (NHP) - This is the HP rating on the motor nameplate but is pretty much meaningless without the service factor.

Nameplate KW = NHP * 0.7457 - This is the KW rating and is similar to the nameplate HP and is generally used outside the US. Note that this is not the input power to the motor only the rating for the output power of the motor.

Energy Horsepower (EHP) = watts / 745.7 = Volts * Amps * Power Factor / 745.7 - Energy input delivered to the motor

Brake Horsepower (BHP) = EHP * Motor Efficiency - Energy delivered by the motor shaft to the load. As will be explained below this is not the same as SFHP nor SF * NHP in general. BHP is a function of the load on the motor shaft and will change with Head, GPM and RPM.

Pumping HP (PHP) = BHP * Pumping Efficiency = Head (ft) * GPM / 3960 - Energy delivered to the water. Sometimes called water HP (WHP).

Motor Efficiency = BHP / EHP - I2R, magnetic and mechanical losses in the motor only.

Pumping Efficiency = PHP / BHP - Recirculation and internal friction losses in the wet end only.

Total Pump Efficiency = Motor Efficiency * Pumping Efficiency = PHP / EHP (note this is why total pump efficiency approaches 50%).

Service Factor - This is an overload rating for motors which states that the motor can be safely be operated over the NHP by the service factor for short periods of time. However, for pumps, this overload rating is typically used as the maximum load that a motor would need to deliver to the wet end. Because the load on a pump does not rapidly change over time, the service factor load is often used as the maximum design point for the pump.

Service Factor HP (SFHP) = NHP * Service Factor â€“ This is the maximum load that can be safely be driven by the motor. This is not the same as BHP although they are equal at a single point on the head curve, usually the best efficiency point (BEP). BHP is the energy delivered by the motor shaft while SFHP is the motor rating. A motor can be driven above the SFHP but will likely fail in a shorter period of time. For pumps, the motor is usually driven at BHPs that are less the SFHP.

Full Load Amps - This can mean several things depending on the motor manufacture. It is either the amps at the NHP or it can be the amps at the SFHP. I have seen it both ways so unfortunately, there is not a good standard for this one.

Service Factor Amps - The amp draw when the motor is loaded to the service factor. This is not as ambiguous as the above rating. Also, using this with voltage should also give you a pretty good approximation for maximum power draw. However, sometimes the motor is over dimensioned for the pump so it will not always be an accurate measure of input power. This is also true of SFHP.

There are others but I think that the above definitions are the most important.

For further reading on this subject, I strongly recommend a web site put together by Joe Evans, a PHD from Pentair: http://www.pumped101.com.