We are reducing our Air Handling Unit air changes per hour to a set amount by changing the pulleys to slow down the fans. The motor will still be turning at 60 hz, but doing less work.

I’m wondering if motor current is linear between No Load Amps and Full Load Amps. Can I use a simple ratio to calculate motor amps? For instance, No Load = 0% and Full Load = 100%. If I reduce my work by 10% can I simply convert that 10% to amps? My ultimate goal is to convert the reduction to yearly electrical savings.

Is there a formula to calculate No Load Amps or will I need to remove the belts and run the motor to get it?
FLA is easy, it’s on the motor data tag.

Well, yes, kinda.
Trouble is, there is still a positive current at no-load, because there is never a real no-load unless the motor is off. The no-load current is a function of motor efficiency, and represents the energy necessary to turn the "unloaded" motor (and its internal fans, bearings, etc) plus the hysteresis losses in the magnetic components, plus I^2R losses in the coils, plus the slip losses if any.

So while full-load might be 100a, no-load might be 20a, and your current will vary over an 80a range from full-load to no-load. But even that will only approximate a linear function, because more current means more I^2R losses, a variable-speed drive on the motor will introduce harmonics which take power, and your slip angle may vary.

And in your particular application, fans are notoriously non-linear. Changing a fan's speed by 10% does not at all mean its power draw also changes by 10%. Fan manufacturers publish a fan curve from which you can compare air flow and fan speed to horsepower required.

Thanks Gary,
I’ve seen the fan curves. Since we are dealing with a built system, I’m trying to understand a way to relate the fan curves to less work at the existing motor.

1. So the fan curve will define hp required to deliver the new target air flow (HVAC guys already have this).
2. No VFD involved. The motor will still be turning at 60 hz.
3. I guess if I find the motor performance curve that shows hp to amps I’d have what I need.
4. I don’t know how closely the existing motors were sized. If I take the existing current reading and subtract the newly calculated hp/amp value I can calculate kWh savings from there.

Does that sound like the best way to tackle this problem?

As a matter of fact, the HVAC guys calculated the kWh savings from the fan curve values. My question originated because I wasn’t sure if using the larger motor (now oversized for the air volume they want) would keep their calculations accurate.

Attached is a spreadsheet to calculate amps vs load. You input the power factor at full load and the amps at full load. It has some assumptions listed (neglect leakage reactance and losses). Under these assumptions the real component of current varies directly proportional to the load and the reactive component of current is constant. The total current is of course square root of sum squares of real and reactive components. If that sounds complicated, remember all you have to do is fill in those green cells labeled FLA and PF (PF corresponding to full load conditions). If there is an efficiency listed on the nameplate, then nameplate data can be used to estimate power factor. If you tell us the horsepower and speed, we can guestimate power factor.

Also if your voltage deviates from nameplate voltage you would need to correct for that.

You can see from the curve comparing the yellow and blue curves that the simple approximation you suggested (current directly proportional to load) is fairly close if we are above 80% FLA, but way off if we are down at lower currents.

I'm not sure what the underlying problem is other than you want to estimate load. Another way to do that is to precisely measure speed (using tach or high resolution vibration). Then use linear relationship between slip and load using nameplate speed as full load point. There are also potential errors in this approach (sensitive to voltage, rotor temperature, wide tolerance for nameplate speed).

CurrentVsLoadR3.xls (18 KB, 41 downloads)

Thanks EPete,
I’m trying to calculate utility payback because we are currently supplying air changes in excess of our ISO requirements. Our AHU engineer calculated $150k/year in utility savings if we installed VFDs. I think the payback would be too long to install VFDs on a built system (about 14 supply/return fans) when considering all of the project costs- control systems upgrades, new equipment, electrical rerouting, air balancing, etc.

The discussion then shifted to slowing the fans down by simply changing the pulley ratio which has much lower project costs, but I wasn’t sure how to calculate the payback costs for that scenario. The AHU curves are more for calculating the proper motor size and are in hp. By changing the pulley ratio we have a fixed motor size (oversized at the lower ISO) and are working it less.

On your worksheet: I’m confused why the blue and pink curves would start at 0, wouldn’t they start at unloaded motor amps (~ 42 A in your worksheet). That’s where my question was about being linear- between 42 and 80 amps in this example.

Here’s the data for one example. The supply fan is 75 hp and the return fan 15 hp; 1775 is the typical RPM. In reviewing the air balance reports the design FLA on the 75 hp is 84 A, the actual measured current was 65 A. The EFF/PF was recorded at 95.4/88.0 respectively. The motor is rated at 460 VAC, our plant voltage is around 475 VAC. Lowering the fan speed should lower the motor current by an amount as yet undetermined. At 65 A we are loading the motor at 77%.


Here’s an interesting link: http://www.reliance.com/mtr/flaclcmn.htm
I don’t think I’ve ever thought about calculating savings using the same motor but reducing the work its doing. I just want to know project payback time to decide whether it’s worth it to proceed.

quote:

On your worksheet: I’m confused why the blue and pink curves would start at 0, wouldn’t they start at unloaded motor amps (~ 42 A in your worksheet). That’s where my question was about being linear- between 42 and 80 amps in this example.

Let's review the 4 curves:

Dark Blue = Reactive Current
pink/Magenta = "Real" Current = load component of current
Yellow = Total Current = sqrt(Reactive^2 + Real)^2
Light Blue = "Straight Line..." = % of full load.

Total current (yellow) is what you see on your ammeter. It starts at no-load amps, then increases from there. Near full load it approaches the light blue curve which is % full load. So in this range it is close.

I'll bet I threw you off the term "real" current. In this case "real" does not mean "actual" or "true" or "measured". It means the portion of the total curren that is associated with real (vs reactive) power. So the pink curve ("real" current) stars at 0 when the load is 0 because the no-load current is purely inductive (in this model with no losses).

I should talk more about the assumption neglecting leakage reactance. It does make the math a lot easier. But it results in a higher than actual prediction of no-load current. If you know no-load current it is better to start with that and use the same assumption (neglect leakage reactance and losses), then we can determine real component of current at full load as sqrt(FLA^2-NLA^2). Gives a better estimate acrross a wider range. In fact if you tell me your motor FLA and speed I can build that spreadhseet. The best estimate of all also accounts for leakage reactance.

The bottom line if you are very close to full load, then you can assume load is roughly proportional to amps. At lower loads this estimate is less and less accurate.

quote:

electricpete wrote;...If you know no-load current it is better to start with that and use the same assumption (neglect leakage reactance and losses), then we can determine real component of current at full load as sqrt(FLA^2-NLA^2). Gives a better estimate acrross a wider range. In fact if you tell me your motor FLA and speed I can build that spreadhseet.

Attach is revised spreadsheet that uses this approach. You have to enter full load amps and no-load amps. The four curves are the same but I clarified the names. The bottom line curve you are really interested in is yellow - it gives current vs load.

CurrentVsLoadR4.xls (18 KB, 24 downloads)

Another relationship that you might be able to use is that slip is roughly proportional to load. So the 100 HP motor that is rated at 1780 RPM is roughly delivering 50 HP if the slip is 10 RPM vs. 20, e. g., the motor is at 1790 RPM.

Many years ago there were a few companies that actually made instrumentation that displayed motor output in % based on this relationship. The drawback is being able to establish the slip that one can directly correlate to some relatively high load. Often the nameplate value is nominal and not truly representative of nameplate power.

John from PA