Motor fan pulley sizing for HVAC

[Muhammad Hasan]

Hi Everyone!
I designed a motor fan system pulley system. So, basically, the motor is of 15 kw with a driver pulley of 7" dia. And through a v-belt I attached a backward curved fan with a 10" dia pulley on the shaft. My fan shaft power is 13.24 kw. This selection was given to me by the software.
Now the problem was that when I assembled this all, my current was coming out to be 48 amps. So, I had to reduce motor pulley size to 6" and amperage reduced to 24 amps. Still, my motor rating was at 22.3 amps. And when I reduced the dia, the rpm changed and hence the flow rate changed.
My motor is 15 kw 3 phase 380v motor. ABB 15 kw motor.

 

[CWatters]

If the fan draws 13.24kW at F_rpm and the motor is rated at 15kw at M_rpm then the pulley ratio must be selected to convert M_rpm to F_rpm.

If you set it up like that and it still draws too much current then something is wrong with the fan. It's drawing more than 13kw at the specified rpm.

You don't mention the rated rpm for either so I can't check the ratio.

 

[Muhammad Hasan]

If the fan draws 13.24kW at F_rpm and the motor is rated at 15kw at M_rpm then the pulley ratio must be selected to convert M_rpm to F_rpm.

If you set it up like that and it still draws too much current then something is wrong with the fan. It's drawing more than 13kw at the specified rpm.

You don't mention the rated rpm for either so I can't check the ratio.

Rated Motor RPM is 1466 and Fan RPM is 1014.
you mean i should change the fan??

 

[russ_watters]

My fan shaft power is 13.24 kw. This selection was given to me by the software.

Are you sure? Your motor is telling you the shaft power was 30 kW. What did you tell the selection software? What is the system configuration?

One common cause of higher than expected shaft power is lower than expected static pressure...

 

[Muhammad Hasan]

Are you sure? Your motor is telling you the shaft power was 30 kW. What did you tell the selection software? What is the system configuration?

One common cause of higher than expected shaft power is lower than expected static pressure...

what do you mean by 30 kw?

This is the selection that the software gave me. I calculated the pulley diameters using the blower speed RPM and Motor RPM(1440).

 

[russ_watters]

what do you mean by 30 kw?

48 amps at 380V and 90% efficiency is about 30 kW.

View attachment 233987

This is the selection that the software gave me. I calculated the pulley diameters using the blower speed RPM and Motor RPM(1440).

But where did you get the input data you told the selection software? Where did you get the airflow and static pressure?

 

[Muhammad Hasan]

48 amps at 380V and 90% efficiency is about 30 kW.

But where did you get the input data you told the selection software? Where did you get the airflow and static pressure?

A consultant gave me!

 

[Muhammad Hasan]

A consultant gave me!

just tell me something, i calculated sizes using simple RPM ratios. Is that calculation legit?? Or there some other process to calculate pulley dias?

 

[russ_watters]

A consultant gave me!

...Then it would be a good idea to go back to the consultant and make sure it was sized and selected properly. And/or if you can measure the pressure and airflow, check to make sure they match the design.

[edit] Also make sure it is installed properly. I've heard of contractors installing a fan and testing it before attaching the ductwork. You can't do that.

 

[russ_watters]

just tell me something, i calculated sizes using simple RPM ratios. Is that calculation legit??

Yes, your pulley sizing procedure was correct.

 

[CWatters]

+1

Rated Motor RPM is 1466 and Fan RPM is 1014.

That's a ratio of 1466/1014 = 1.45:1

10"/7" is a ratio of 1.43:1 which is close enough.

So yes the original pulley ratio should have worked fine.

I agree you need to speak to the consultant and ask why his fan appears to be drawing too much power. Perhaps a factor of 2 missing somewhere in his calculations?

 

[CWatters]

I might be wrong but I think have found part of the problem...

If the OUTPUT of the fan is 13.24kW and the efficiency is 65.8% then the INPUT power to the fan is..

= 13.24 * 100/65.8
= 20kW

That still does not explain why it is drawing 30kW but check with the consultant.

EDIT: What is the motor efficiency with a 33% overload (5kW/15kW=33%)? If the efficiency drops a lot then that might account for some of the other 10kW ?

 

[Dullard]

See post #9.

I believe that post #12 is incorrect - the 'software' picked a 15 KW motor - 'Efficiency' is accounted for in the 13.24KW. A rough calc of an ideal blower supports that notion.

As an experiment, you can restrict the inlet or outlet - you'll almost certainly see the current drop. You can determine the pressure difference across your fan with a piece of clear tubing and some water (hillbilly U-tube manometer). Those measurements and a fan curve (available from the manufacturer, if you don't already have one) will set you free. You are almost certainly (as Russ suggests) overflowing your blower.

 

[jrmichler]

Their fan curves are different from what I'm used to seeing:

If I'm correctly interpreting the fan curve, the fan shaft power will be 30 Kw at 1000 shaft RPM at about 45,000 m^3 per hour. That flow rate is larger than the fan is designed for. Also, the fan catalog specifies a 95% efficiency when designing belt drives, and also a 1.1 safety factor on the motor size.

Is the fan connected to the completed ductwork? Also @Dullard is correct about restricting inlet or outlet. And the clear tubing manometer he suggests, while it may be cheap appearing, is completely accurate. It is accurate enough to be used as a calibration standard for mechanical gauges.

 

[Dullard]

That curve shows what the data table claims:
approx 13 KW @ .95Kpa / 27K m3/Hr

 

[russ_watters]

I might be wrong but I think have found part of the problem...

View attachment 234007

If the OUTPUT of the fan is 13.24kW and the efficiency is 65.8% then the INPUT power to the fan is..

= 13.24 * 100/65.8
= 20kW

I don't know about their terminology, but a quick calc tells me the air power is 7.2 kW. 13.2 looks like motor drive shaft power.

 

[CWatters]

Yes I've just done the same and got 7.4kW for the air power. That gives a fan efficiency of 100 * 7.4/13.2 = 56%. So not sure how they get 65.9%.

 

[Muhammad Hasan]

I get what you all are saying but, what I don't understand is that why is the software selecting that Fan and Motor combination?

Furthermore, I am using TICA AHU software for designing! Any suggestion on a better software for designing AHUs?

I have also attached the Fan curve and selection done by the software! please have a look and help me out.
Thanks Everyone.

 

[russ_watters]

I get what you all are saying but, what I don't understand is that why is the software selecting that Fan and Motor combination?

The design engineer enters the required airflow, static pressure, altitude, and various desired configuration elements and the software lists fans and motors that will meet the specified requirements, and from that list the design engineer picks one.

 

[CWatters]

Perhaps one of the other contributors to this thread could comment on the data in the filter section in the TICA report. The air velocity in that section is listed as 0.15m/s where as the velocity out of the fan is 18m/s. Presumably the filter is much larger in diameter than the duct... From the flow rate and volume I calculated the duct has an area of about 0.4sqm (70cm Diameter). To drop the velocity to 0.15m/s wouldn't the filter have to be 40sqm? Is this actually lots of small filters around the building?

 

[Dullard]

I'm a bit puzzled by the "FanCurveReport." The 'design point' appears to require 925 RPM - I'm not sure where 1014 came from. The fan/system curve intercept will occur high/right of the plotted design point. The power gets high out there, based on the 'blank' curve in post # 14.