Centifugal Blower Motor

Asking the question again will not change the answer. And, this time you omitted the original question!

Let me try. NO, changing the impeller from steel to aluminum, titanium or any other lighter material will have no measurable effect on the power requirements of the motor.

This was all explained before.

Once again. NO.

If you want to lower the power requirement of the motor, trim the impeller.

thanks for the prompt help in form of guiding me .

but what about the difference in rated current(250 amps) and running current(150 amps).dont you think extra energy is being consumed resulting in extra energy cost.try to explain this point thanks

This depends on where and how you got these two current numbers. Your "rated" current number may be the combination of your running current and your starter current circuitry. Your rated current number may be the power distribution current rating number to feed this blower. Then again it maybe the rated current for the wiring. My point is that without knowing precisely where and how these current numbers come from, nobody can say if any energy is being wasted.

the specification plate fitted to the motor reads 132 kw. rated current amps- 250,voltage-415,rpm-985.

the motor is connected through soft starter.

the running current was noted through ammeter connected to the panel and also with the help of tong tester which was showing 150 amps.

can you explain what is meant by your "starter current circuitry".

Getting an AC motor to spin the rotor shaft from rest up to a minimum speed will often take much more current than the operating current. One of the problems an AC motor confronts in moving from rest comes from the complication that the powering voltage supply changes polarity many times a second. So before the rotor and mechanical load can get up to running speed in one direction on one polarity cycle, the voltage polarity will switch which pulls the rotor assembly back to where it started. Some motors to solve this problem have a speed sensitive or timed switch in them that adds a second set of stator windings that will add more poles to the stator windings (rarely), or permit the addition of a capacitor delayed current to prevent there ever being a reverse direction force on the rotor. Most important to this discussion, the current draw while the motor is starting will be considerably higher than one will expect from the loss of the back EMF voltage. I suspect that this might be the considerably higher, 250A, current rating your motor label shows.

Because of this high starting current many large motors have an additional soft-start circuit that limits the voltage applied to a staring motor and thus limit the current available during starting, too. The usually acceptable drawback of this is that it will take longer for the motor to get up to operational speed.

We don't have the manufacturer's fan curve showing power consumption of the fan at various speeds, so this is a bit of a guess. From your numbers, it appears that the fan is actually drawing about 104kva [150a × 400v × √3 ÷ 1000]. Because this is a light load for this motor, the PF is probably not very high; let's guess 0.75. Then 104 x 0.75 ≈ 78kw.

If the air flow is satisfactory, a smaller motor would probably work. It would not save much energy (kw), but it would operate at higher PF and thereby decrease the reactive current and the kva. If your utility charges for low PF, this would save you some money.

Alternatively, you could add PF correction capacitors to the existing motor, if this has not already been done. (The motor may have initially been oversized to allow for increasing the fan speed if needed.)

i have already asked the industry personal for the manufacturer manual so that i can trace out curve showing power consumption of the fan at various speeds.will also find out if the motor was initially over sized .will pass on the details as soon as i get hold of the required data

Power consumption in an AC motor is comprised of two main components: Load Current and Motor inefficiency as a percent of Load Current. Your load current is the same unless, as suggested, you change something such as by trimming the impellers. Assuming you have a relatively new motor built to the latest efficiency standards, you are probably somewhere in the neighborhood of >90% efficiency at full load, maybe 1-2% less at <75% load. So using your current only as an indicator of loading percentage (actually a very poor indicator)the net difference you may obtain in using a motor that is closer to the load use is only that difference, the 1-2%, between full load efficiency and partial load efficiency.

The cost of that is going to be lower accelerating torque and no ability to handle any load transients, as well as a severely limited ability for the motor to ride through any line voltage dips. In other words you have more shaft torque than you need right now with that 132kW motor, but torque diminishes by the square of the voltage drop. So if your line voltage drops just 5%, your shaft torque drops 10%. If you had 15% extra, no problem but if you had no margin of error because you used a 90kW motor, then a 5% VD causes more slip, higher current draw and if it doesn't cause an over load trip, it shortens your motor life significantly. This by the way is depended on that next lower size motor creating enough torque at all!

So weigh the potential 1-2% energy savings money against the cost of lost production due to the motor being tripped on overload or out of commission for a day or more while being rebuilt or replaced.

I think you are not satisfied with our answers in earlier posts. Let me try once again to convince you that nothing is required to be done.

- Gas power required for your blower is about 73 kW (as per the formula given by a contributer in earlier case). It is based on static pressure at discharge. You have not given the total pressure, so pressure due to velocity head is missing. Roughly adding 5% for it, gas power required for total pressure is about 77 kW. Efficiency of such fans are in the order of 70 to 80% and motor efficiency around 95%. Assuming the overall efficiency as 70%, the shaft power requirement will be 110 kW. Motor rating should be 110% of shaft power requirement, i.e. 121 kW. So selecting next higher standard size 132 kW for your blower is very much correct. In fact, some customors demand 120% of shaft power, perticularly when soft start is not provided.

- Trimming the impeller is not the solution, as suggested by someone to reduce power. It will reduce the capacity/static pressure required.

- Reason for low power consumption may be due to lower flow rate or pressure or incorrect measurement of flow rate or pressure or current or combination of it.

- If your blower is really very efficient, say 90%(hypothetical) and you are really consuming less power, then also it is not advisable to change the motor. Because, reduction in power consumption with smaller motor will be nigligible and break even time for replacement cost will be much longer. Moreover, with slightly oversized motor you need not worry about plant outage due to overload trips.

dear pritam.thanks for the detail answer.hope you will not mind to clear what you mean by Gas power.you mean the electric power.regarding velocity head pressure will find out from the industry personal. the air volume 4000 mcube/m and static pressure 1100pa is mentioned on the specification plate fixed to the blower.

An aluminum fan wouldn't save anything on running kw, but it could help with starting acceleration. And wouldn't a VFD be better than just a soft starter for ramping the fan up to speed? If any of the paint booths have supply air dampers that could be closed when not in use, the fan could then be run slower, which would save energy. Mr. Zia might thus have some useful options to consider.

the process is such that they cant close any one of the supply air dampers.

also soft starter are used.

instead of mentioning soft starter they term it "Inverter drive".

any difference between soft starter and Inverter drive .or simply they have written inverter since first A.C is converted to D.C and than D.C to A.C in inverter.

any difference between soft starter and Inverter drive .or simply they have written inverter since first A.C is converted to D.C and than D.C to A.C in inverter.

Giggle. You really have no idea what these words mean? Giggle. Now it could even be possible that you actually have a variable frequency drive running a synchronous motor. I'm afraid this software package has failed the Turin test for me.

Have Fun, Goodbye

hi,

what is the BKW/ bhp of the fan ?

Motor KW depends on the same.

The B.KW is 114. Motor rating 132 KW ,6 pole , 415 V , 250 A.

Air flow 4000 cube m/min

static pressure 1100 pa

revolutions/ m 530

air temp. 20 C

air density 1.198 kg/cube m

Fan efficiency 58 %

1. Dear Mr. Salahuddin, you are slowly opening up your cards. You should have given all these information in the begining itself, my contribution at post-7 would have been more meaningful. First I want to answer your questions at post-8.

a) Gas power is not an electric power. Gas power is the ideal power required to bring the air from the condition at inlet to the outlet of the fan. It excludes all losses like inlet leakage, blade profile, fictional, outlet losses etc.

b) What is velocity head pressure? This you can calculate yourself using the formula

Pvh (Pa) = 0.5 x density (kg/cu.m) x v**2

where v = outlet velocity (m/s). For your case, if I assume cross-section area of outlet pipe/duct as A = 4 sq.m, v = Q/A = 1000 m/min OR 16.67 m/s.

So, Pvh = 0.5 x 1.198 x (16.67)**2 = 166 Pa (Please note it is assuming A = 4 sq.m). You should calculate correct value using correct cross section area.

2. Now, comming to your present post, the fan efficiency 58% appears to be too low. Please check whether it is overall efficiency including motor and belt drive.

You have given B.kW 114. After considering fan efficiency as 58%, gas power required is 114 x 0.58 = 66 kW only, which is too less considering the required air flow rate and static pressure. Therefore, somewhere something is wrong Mr. Salahuddin.

Considering B.kW = 114 also motor rating 132 kW is OK.

At this point of time I want to ask you whether you have the actual measurement for air flow rate and static pressure? And whether they are matching with the book value.

we are carrying out electrical energy audit of the automobile industry.they dont have any solid information.i have been repeatedly asking them to provide us the manual.they have promised to provide us the manual soon ,only than i will be able to pass on the correct data.we have gathered the data from the staff working in the industry and was posted on CR4

as soon as i have the actual data in hand i will pass on the same to you.in case the thread is closed for further questions i hope you will not mind if i send the mail on your email.as so many questions are coming up one off them is that we are still not sure whether they have soft starter or vfd.we have to find out also whether motor is induction or synchronous .total static pressure etc.i am sure all my gentlemen engineers wiil try to solve my problem thanks

Sorry Engineer the efficiency is not fan efficiency buy total efficiency 58 %