Brake Horse Power for Fans

The input is volts multiplied by amps with a correction for power factor.

A 5hp motor doesn't always consume whatever it says on the label, for it depends upon the mechanical load attached to it. If it had nothing attached to it and were spinning, for example, it certainly wouldn't consume its maximum rated power from the supply at that time. Rest assured, though, that if the mechanical demand exceeded 5hp for any length of time by a significant margin, it will stop being a motor and turn into a rather spectacular convector heater instead. That is why there is an overload protection device in the motor starter circuit, the setting of which needs to correspond with the maximum rated current on the motor label plate.

Brake horsepower is the heating effect on the shaft and the surroundings as a result of applying a brake to that shaft. Its only connection with the motor rating is that the heating effect from the brake cannot exceed the rating of the motor without the convector heater end point as indicated above.

Thanks for the clarification.

I'm going to have to dog a bit deeper to find the amp draw and power factors, since I don't see them on the fan spec sheets in front of me.

I know how the input could be determined for an installed fan, but I am looking at design specifications and trying to determine fan efficiency without getting caught up in the misleading marketing hype.

You won't find power factor on the spec sheets, Boss. It's a feature of how the motor is behaving at any moment. You measure the current drawn using a tong-tester.

For design purposes, we can make assumptions for power factor, right?

http://www.engineeringtoolbox.com/power-factor-electrical-motor-d_654.html

We can't measure the PF right now - we're just doing a design exercise.

I think you are chasing your tail.

Are you talking about motor efficiency (electrical) or are you talking about fan efficiency (mechanical)?

Knowing the brake horse power of your motor won't tell you the efficiency of the fan.

If I were you, I'd be talking to anybody I could to learn about fan design and efficiencies of motors.

If you are, " just doing a design exercise" you are wasting your, and our, time with this.

Why not look on Search GlobalSpec for high efficiency fans, and get on with the exercise.

Thanks for the input. You suggest speaking with people about fan design, and that's what I am doing by posting a question here.

So if it's such a waste of time, why bother responding?

You must have a required duty i.e. flow and pressure rise. Even if the fan supplier doesn't publish efficiency, they should give fan input (shaft) power vs flow. Read off power at your selected flow. Then work out power into the air by formula in #3 or some other way (I use flow in m³/s x ΔP in kPa to get power in kW). Divide by the shaft power and you have fan efficiency. I'd expect something like 40 - 50%.

Motor efficiency for around 5 hp is probably 80 - 85%, I haven't got data to hand but it's easily obtainable.

Where did the 4.2 hp come from? As others have said, the motor can deliver power up to its rated output, but the actual output depends on the load it's driving, and if that's below rated load the electrical input power also falls, but by not quite as much as the motor efficiency falls with decreasing load.

No.

The concepts of brake horsepower and efficiency are not directly related.

However, there is an indirect relation in that if the brake horsepower of a load is less than the motor's rated horsepower, the efficiency will be slightly less (but not as low as the 80% in your example).

So to clarify, for better efficiency, we want the brake horsepower to be as close to the rated HP as possible (to maximize efficiency)?

Actually, for most squirrel cage induction motors, the peak efficiency is at about 75% load. The efficiency remains pretty good even down to 25% but the power factor drops a lot, as shown in the 2 tables below from the Baldor web-site.
It's more important (from an efficiency/power savings standpoint) to optimize the machine to the load at hand than to try to make up a bad machine/load match with a super efficient motor. That's one reason why VFDs have become so popular - slowing down the motor reduces the machine power requirements, even though the losses in the motor/drive package actually go up.

Baldor Premium Efficiency TEFC 1800 RPM, 5 HP

Baldor General Purpose Open 1800 RPM, 5 HP

Brake horse power is the actual horsepower a fan requires to move a certain amount of air because the load can vary due to air density and moisture content and other factors, including efficiency of motor..... BHP can be expressed as

P_BHP = q dp_inWG / μ 6356 (2)

where

P_BHP = Brake Horse Power (hp)

μ = efficiency

Link...

link....

Thanks for the equation.

I had seen the fan efficiency page from Engineering Tool box as part of my research, however, it seems that the sales reps for the fans don't know or won't tell what efficiencies their motors are. I do see some manufacturers mention the NEMA standards, but without enough info to give an accurate number. At the smaller HP ratings, I suppose a few % points won't matter much, so I could just use the lowest qualifying rating.

Yes, including a link to typical efficiencies under different loads..

http://www.itrc.org/reports/vfd/r06004.pdf

Dear Mr. sevetyseven.

The answer is NO. 5 H.P is the capacity, of the equipment and 4.2 H.P is the ABSORBED POWER. The absorbed Power multiplied by the EFFICIENCY is the NETT OUT PUT of the equipment.

DHAYANANDHAN.S

thanks to all for the comments - it has helped me to get a better understanding of the issues to consider.