Motor selection for fan application

[srinivasan]

Vendor recommended 132kw motor for fan having following datas,

operating medium: gas
flow rate:22000 Nm3/hr
pressure : 1200 mmWC. (normal operation is 800 mm WC at 50% speed operation using VFD)
Full load speed:2975 rpm
full load torque: 353 Nm at 1.2kg/m2
GD^2 value =59.6kg.m2.

I used the below formula for calculating motor sizing but i have not done before this like calculation.
Power rating=(T*N)/5250 where T is in Lb ft and Speed is in RPM. Please give correct rating of the motor for above data

 

[Besoeker]

Vendor recommended 132kw motor for fan having following datas,

operating medium: gas
flow rate:22000 Nm3/hr
pressure : 1200 mmWC. (normal operation is 800 mm WC at 50% speed operation using VFD)
Full load speed:2975 rpm
full load torque: 353 Nm at 1.2kg/m2
GD^2 value =59.6kg.m2.

I used the below formula for calculating motor sizing but i have not done before this like calculation.
Power rating=(T*N)/5250 where T is in Lb ft and Speed is in RPM. Please give correct rating of the motor for above data

2975 rpm and 353Nm equates to 110kW.

Simply P=Tω
T in Nm is 353
ω in radians per second is 311.5
That gives you watts.

 

[Julius Right]

You can use also the known equation if you translate the IS unit in Imperial:
1Nm=0.7376 Lb.ft
353 Nm=0.7376*353=260.3728 Lb.ft
HP =Torque (Lb. - Ft.) x RPM/5250
HP =260.3728* 2975/5250=147.544
kw=HP*.746=147.544*.746=110.068 kw
See:
http://www.engineeringtoolbox.com/torque-d_991.html

 

[topgone]

Vendor recommended 132kw motor for fan having following datas,

operating medium: gas
flow rate:22000 Nm3/hr
pressure : 1200 mmWC. (normal operation is 800 mm WC at 50% speed operation using VFD)
Full load speed:2975 rpm
full load torque: 353 Nm at 1.2kg/m2
GD^2 value =59.6kg.m2.

I used the below formula for calculating motor sizing but i have not done before this like calculation.
Power rating=(T*N)/5250 where T is in Lb ft and Speed is in RPM. Please give correct rating of the motor for above data

2975 rpm and 353Nm equates to 110kW.

Simply P=Tω
T in Nm is 353
ω in radians per second is 311.5
That gives you watts.

If the vendor recommended 132kW, and the pump capacity calcs yielded 110kW, your vendor is contemplating on supplying you with an equipment with a total expected efficiency of 110/132 = 83%!

 

[Besoeker]

You can use also the known equation if you translate the IS unit in Imperial:
1Nm=0.7376 Lb.ft
353 Nm=0.7376*353=260.3728 Lb.ft
HP =Torque (Lb. - Ft.) x RPM/5250
HP =260.3728* 2975/5250=147.544
kw=HP*.746=147.544*.746=110.068 kw
See:
http://www.engineeringtoolbox.com/torque-d_991.html

Simpler without the conversions......
P=Tω

 

[Besoeker]

If the vendor recommended 132kW, and the pump capacity calcs yielded 110kW, your vendor is contemplating on supplying you with an equipment with a total expected efficiency of 110/132 = 83%!

83% of motor full load rating.
And a 132kW motor efficiency isn't going to be much lower at 83% of full load. I ran the calcs for a motor that I have data for. Efficiency went from 93.4% at full load to 93.3% at 83% load.

93.3% 93.4%

0.830 1.000

Then there is the question of tolerances.
The 2975rpm and 353Nm gives 109.97kW.
What if the torque was 354Nm? Maybe bearing friction is marginally more until bedded in or deteriorates during the life of the equipment?
That marginal (<0.3%) change takes it over 110kW.
The next standard size motor is 132kW.
I think the vendor made the prudent and conservative choice.