Blower

Think about how the energy is used in a centrifugal blower. How does the weight of a rotating element effect energy use?

Lowering the weight of a car can save energy because you frequently accelerate the load. Most blowers operate at a steady rate so the weight of rotating assemblies does not cause energy use for speed changes.

Of coarse I am generalizing because there may be other considerations you have not shared with us. Also there may be some gains in a material, and thus weight, change due to the lower friction loads on the bearings (small) or a different material may make a more efficient impeller design possible. If that is the case go to various blower manufactures and compare there fan curves based and cost.

Also do you understand that even though you motor is rated for 132 KW it may be actually consuming less power depending on the mechanical load requirements? You need to measure the power consumed with a probe and compare it to rated full load. An increase in efficiency does not necessarily require going to a smaller motor. Dropping the mechanical load probably lowers the electrical power usage without a motor change. We would have to know details to be sure.

To add to this post, induction motors are said to be most effective at about 75 % of their load - so, if the OP's 132kW motor now produces 95 kW, and he changed it with a 100 kW one, he would be losing more energy.

thanks for the detail explanation.the rated current of the motor is 250amps at 415 volts.the running current is 150amps at 390-400 volts

There's something wrong there, if that's 400 m³/minute. Fluid power works out to only 7.3 kW. If it were 4000 m³/min (or 11000 Pa ~ 1100 mm water gauge) it's 73 kW, and with typical blower efficiency 132 kW motor is about right.

Cheers.........Codey

thanks for the correction.yes it is 4000mcube/min and not 400.

hope you will not mind explaining in detail how you have calculated 73kw and when the requirement is 73kw than how come 132kw is okay with reference to typical blower efficency.

can you let me have detail regarding typical blower efficency.please dont mind my question as i am learner on havc side thanks

Fluid power kW = flow (m³/s) x pressure rise (kPa).

Typical efficiency 60% gives blower shaft power 122 kW.

I thought motors were most efficient at about full load. In theory if actual load is say 105 kW, it might be a little more efficient if changed to a 110 kW motor, but marginal, and in any case the full-load efficiency of a 132 kW is likely to be a bit higher than a 110 kW (efficiency tends to improve with increasing size). In practice it wouldn't be worth doing it, it would take years to recover the cost, if ever.

Cheers........Codey

thanks code master for the reply but i will still remind you here again that i am a learner and you have to help me out.

fluid power kw= flow(mcube/s) *pressure rise(kpa).

my question is for flow it is mcube/s or mcube/m.

also how to calculate 60% efficiency to be 122kw is there some data for this calculation if so can you please guide me from where i can find this data.

how to calculate pressure rise.

i once again request you not to ignore my questions as i am learner.thanks

You need to work in a consistent set of units. For SI units, basic quantities are metre, kg, second. So flow has to be in m³/s. Force unit is newton. Pressure unit is pascal, Pa = newton/m². Multiplying them gives power in watt. I took a short cut by putting pressure rise in kPa, which gives power in kW (usually more practical).

60% efficiency is my assumed figure, typical for a blower of this size. It's not calculated, but if you were testing the blower you could measure the flow and pressure rise, to get fluid power. Measure blower shaft power, or much simpler, get it by measuring motor electrical power and motor efficiency (from supplier's data). Fluid power/shaft power = efficiency.

You stated pressure, 1100 Pa in original post. I assume suction is at atmospheric pressure so this is pressure rise.

Cheers.........Codey

thanks codemaster for the detail answer.would like to clear one point that you have calculated blower motor to be 122kw where as the installed motor capacity is 132kw .you think 132kw is okay or the industry should have gone for 122kw

There's no such thing as a 122 kW motor, you have to select the next bigger (at least) above the actual power required. I can't remember the standard ratings, I think there's a 90, 110 and 132 kW, but you can easily find data on the web.

132 kW rating is perfectly reasonable for 122 kW absorbed, which in any case is only my estimate. Some specs call for a 10% margin over absorbed power.

Cheers......Codey

dear codemaster thank you once again for the answer.one more point to be cleared.the existing installed motor132kw has rated current 250amps/415 v.where as the running load is 150amps/390-400v and the starting method is soft starter.my question is that when running load is 150amps against the rated current 250amps is it feasible to replace the existing 132kw motor with lower motor rating to cope with 150amps/or less running current.rest data remains same i.e. air flow ,static pressure.hope you will not mind this question as i am learner on havc side and i have to solve this problem for the automobile industry thanks

With those figures the actual motor output power is about 78 kW, so a 90 kW motor should be OK. But are you sure this is the maximum power? Might the blower need to operate at a different point on its curve, where it could draw more power?

But if you do change it, I'd be surprised if the current drawn is significantly reduced. If you have a 90 kW motor to hand, and you have a use for the 132 kW, maybe worth doing, but I wouldn't bother otherwise.

Cheers..........Codey

Dear sir,

changing the material of your fan will have a less effect on the load of motor. It affect on the starting torque, where the heavier fan reuires more start up load. If the fan is properly balanced and the bearing are good it take approximately the same load at continuous operation. Use a reducer or pulley or series of pulley on your drive.

Error in vital information like flow rate (which decides kW) was surprising. It apears that it is a belt driven blower. If your requirement is always around 4000 m3/min then nothing is to be changed. But, if you have a vide variation in required flow rate then a VVFD may help to reduce power consumption considerably to get less payback period. For installing VVFD existing motor is to be derated by about 30%.

yes i have accepted my mistake reference flow rate.

the blower is belt driven through 5 belts and mode of starting is through Soft starter .cant find out the soft starter is connected 1) in line or 2)inside delta connection as no drawing are available .can you guide me how to trace the connection. thanks

also the requirement is 4000 mcube/min as the humidified air is used in booth painting area of automobile engineering

It is a total distraction from the main question & should be posted at Electrical Section. Also you have not replied correctly to my answer at post-8.

thanks for guiding me to the correct direction i.e.i should post my question on electrical side.

the requirement of 4000mcube/min is always required as the humidified air is used for different booths in paint shop of automobile industry.it is belt driven centrifugal backward blade blower driven by 5 belts.i hope i have answered your question properly now try to solve my question thanks