Dear Amrizal

In normal modern units :

Power = Torque x radian speed

2 x pi x rpm

kW = kNm x ------------------

60

rpm of belt-driving drum

or

rpm of drive

gearbox ratio = rpm driving drum / rpm drive

etc.

But, may be you are referring to the belt conveyor design, which is much more complicated.

Then it is better to orentate yourself with literature, experts and manufacturers.

success ■

Dear Mr. Amrizal

When you are changing the capacity, you have to redesign the conveyor and provide the various items / components accordingly. Obviously, in such case, one tries to make a design in such a manner that there are minimum changes in existing items so that the item to be replaced are minimum. You have to follow the following steps while revising the capacity of conveyor :

1) I presume you are not changing the belt width. Possibly you are only increasing the speed to get the required capacity. However, if the speed becomes excessive to achieve the capacity for a particular belt width then you have to increase the belt width and use appropriate speed.

2) If the belt width is same, you need to examine / check the suitability of the existing idlers.

3) Now design the conveyor according to the revised parameters. There are well-established design procedures for designing the conveyor. This is a bg subject and cannot be narrated in this forum in few sentences or pages.

4) The conveyor design one has to first decide the tractive pull or sum of all resistances. The sum of resistances = tractive pull. The tractive pull multiplied by the drive pulley radius will give you the load torque. There is no short cut for deciding the load torque. This cannot be calculated from motor side.

5) Having decided the load torque, you can decide the required power at motor shaft, available motor size, gear ratio and so on.

6) I will suggest you to take the help of conveyor designers, who may be also available in your country. You will be expending in pennies for design charges, but ultimately you will be saving pounds in your procurement and project implementation.

Regards,

Ishwar G Mulani.

Author of Book : Engineering Science and Application Design for Belt Conveyors.

Author of Book : Belt Feeder Design and Hopper Bin Silo

Advisor / Consultant for Bulk Material Handling System & Issues.

Email : parimul@pn2.vsnl.net.in

Tel.: 0091 (0)20 25882916 ■

5252 = 33,000 / ( 2 * PI )

Hope this helps. ■

From my basic engineering training 30+ years ago ---

HP = 2 * PI * N (Rev/min) * T (ft lbf) / 33,000

(I don't think it's changed) ■

The definition of horsepower is:

HP = 33,000 lbs lifted one foot in one minute.

Some englishman invented it when he invented the measure of the foot, the bushel, the peck and so on. Probably when the steam engine began to replace the horse in rememberance.

PI is the circumferance of the circle when the diameter is unity or = 1, as circumference = diameter x PI. ■

Regarding the "Some Englishman" ......

"Horsepower is defined as work done over time. The exact definition of one horsepower is 33,000 lb.ft./minute. Put another way, if you were to lift 33,000 pounds one foot over a period of one minute, you would have been working at the rate of one horsepower. In this case, you'd have expended one horsepower-minute of energy.

Even more interesting is how the definition came to be. It was originated by James Watt, (1736-1819) the inventor of the steam engine and the man whose name has been immortalized by the definition of Watt as a unit of power. The next time you complain about the landlord using only 20 watt light bulbs in the hall, you are honouring the same man.

To help sell his steam engines, Watt needed a way of rating their capabilities. The engines were replacing horses, the usual source of industrial power of the day. The typical horse, attached to a mill that grinded corn or cut wood, walked a 24 foot diameter (about 75.4 feet circumference) circle. Watt calculated that the horse pulled with a force of 180 pounds, although how he came up with the figure is not known. Watt observed that a horse typically made 144 trips around the circle in an hour, or about 2.4 per minute. This meant that the horse travelled at a speed of 180.96 feet per minute. Watt rounded off the speed to 181 feet per minute and multiplied that by the 180 pounds of force the horse pulled (181 x 180) and came up with 32,580 ft.-lbs./minute. That was rounded off to 33,000 ft.-lbs./minute, the figure we use today. "

Clearly not just ANY Englishman. ■

WOW! I break my arm with a pat on the back to my memory. I am more amazed at the dept of your knowledge and your memory. Good work. ■

Isn't Google a wonderful thing :-)) ■

Well! I take it back although I still have a broken arm from retaining some fact of 40+ years ago.

To the original question of how to calculate the power required to upgrade the belt with the given specifications, more data is needed:

1. lift and any added geometry variances

2. idler spacing and its variances

3. idler diameter

4. idler trough angle

5. skirtboard length

6. when you say belt density we assume you mean bulk density

7. starting control: across-the-line, fluid coupling, voltage control..

6. belt weight per meter

You can get a reasonable idea of power and power change with the upgradewith these add facts.

However, the forum is for free advice not for free engineering. ■

".. the forum is for free advice not for free engineering."

A man after my own heart. ■

i m coming from basically electrical side

i used 11kw motor for lime shifting it is very small conveyor

i observed that

it takes near about 7 to 8 A with maximum load

as in electrical sytem if u r use over rated Motor so cu. losses and hysteriss losses come in the picture also consider motor eficiency.

if i used 7kw motor in this conveyor some of the enegy i will save.

can i do it on tachnical bassis?? ■

I assume you are using a 3 phase induction motor.

It's 11kW, but what is it's full load motor current? The supply voltage, that would be helpful, as would be the motor speed

Without that you stand no chance of even a rough estimate of what power is being produced from the motor at 7-8 amps. And without an indication of what power the conveyor is taking how can you decide on a smaller motor?

For your interest my motor catalogue gives

11 kW, 1470 rev/min, 380 V, 21.9 amps full load current ■

motor is 3-pahse 11kw its 1475 RPM & 415V

while it was running with full load it takes max current goes up to 8A

and also it o/L set at 10A

if this motor is overated due to cu losses and hysteriss losses come in the picture it comsumes some elecrcial anergy so i want to save this energy

can i put it 7.5kw 1470 rpm 415v

if i save energy then how much i save?

one factor is also that sytem voltage showing 400v

so voltage is goes down that it was also producess some losses it consumes some elecrical energy

so voltage level it 415 is really Needed rather than 400v is ok???

PLz guid. ■

Dear Pareshhjoshi,

Perform a cos(phi) or powerfactor measurement and you know exactly the consumed power under the measured conditions.

Then you can decide whether a smaller motor is possible.

However, before taking any actions also check the starting curves of the electric drive motor and the required starting torque for the conveyor.

"coming from basically electrical side" as you state, this cannot be too difficult?

success

Teus ■

Please tell me the motor full load current i.e. the current at 11 kW

It will be stamped on the motor nameplate along with the specified voltage and full load speed.

As Teus has said the power factor (cosine of the phase angle) is needed for a true power calculation. However it's probably easier to use a clip-on watt meter on each of the three phases. But this on the input side of the motor so an allowance still has to be made for motor efficiency.

As to the starting torque, take off the fan cowl, wrap some string/cord around the outside of the fan, pull the string with a spring balance and note the maximum value to start the conveyor moving. Multiply this by the fan radius and there's your starting torque.

Job done!

Next ■

James Watt's steam engine stands just outside Rutland Hall, my old Residency at Loughborough University. Wasn't he Scottish?

Remember that the Scots originated most of the things that embelish our modern society in one form or another. They are the most prolific per-capita inventors imaginable & it is a priveledge to live beneath them just to their south. (Now placing my tongue firmly against my cheek) Imagine what a wonderful world it could have been if they'd stayed sober for longer. ■

Originally Posted by louispanjang

Imagine what a wonderful world it could have been if they'd stayed sober for longer.

Maybe the world would be a worse place if if they were sober longer ■

in our elecrical system power factor is not big reason

because at MRSS HT side KVAR compansator is fixed

so any time inductive load is apply that time

this componont is stablised system power factor.

i not able to check this power factor at evry time because when all the motor is in off condition then it will show 0.2 PF

and when plant it running with fully load that time power factor will be gradually move up when motor is on respectively'

and it show 0.8 to 0.85 PF

so i asume this sytem pf is 0.8 stable

11 kw with 0.8pf full load current is 19A

and my sytem volage it 400v but on name plate i found 415v it may produce more power losses??

advise if voltage level will be at 415 to 425v it is able to reduce more losses in inductive circuit ?? ■

Dear Pareshhjoshi,

I was referring to the powerfactor of the electric motor. (Not to the powerfactor of a complete capacitor compensated powergrid)

From the measurement of the electric motor of the Voltage, Current and Powerfactor, the consumed power can be calculated under f.i. full mechanical load.

19 A for a 11 kW motor at 400 V is right.

If that motor is now consuming 8 A at full load, then it is not fully loaded and a smaller motor could also do the job. (If starting is not a problem for the smaller motor)

Normally, the energetic effieciency for an electric motor is at highest load and nominal voltage.

Another thing is that there is not much to win in improving such a small drive. (Considering the cost)

Success

teus ■

I think a crash course in the operation of three phase induction motors is necessary. Google would be a good start.

Alternatively, a visit from your motor suppliers technical representative to give an explanation.

It's too much to try and explain in a discussion forum.

Suffice it to say according to the Brook Crompton site

11 kW, 1470 rev/min motor (50 Hz)

FLC - 21.9A @ 380V, 20.9A @ 400V, 20.9A @ 415V

Efficiency - 91.5% full load, 91.5% @ 3/4 load, 90.5% @ 1/2 load

Power Factor - 0.83 @ full load, 0.79 @ 3/4 load, 0.67 @ 1/2 load ■

thanks you very much for u r guid line

my self Paresh i stated my job since two yr

so againg thanks to Dear disinger and jeus

Dear Jeus

in one big conveyor there a used 75 kw

and Feeder O/L setting set at to 65A

it current not gows below 65A

so i want to put it there 55 kw also

but first i want to do in small motor then i will be succes and found gud sign

then i will do in this motor

because i observe that one thing when i take a motor trial when it comes newly

whitout gear couple only motor

current takes half of the kw

means 75kw motor takes 40A. ■

Dear Pareshh joshi,

I second the advise of designer, to study the operation of three phase induction motors.

At the same time, it is necessary to study the characteristics of the driven installations.

The drive curves of motor and installation must cope with one another in order to work properly.

A special situation to consider is the starting sequence of the complete drive train.

Also keep in mind that the energy bill is mainly for kWh, and not for Amps.

After that, you can help us.

success

Teus ■

i have one question

if the motor is overrated so its starting torque also may be more

in belt we are done some joint

atlest one joint is there

so when belt is started with material/without material whatever it may be but

it starting torque is more due to motor is over rated

this starting torque can be damage the belt joint? ■