# Initial Torque

1. Jul 26, 2009

### sunovya

Sir,
I stuck with a project. It is a bearing supported 5 Meter long cylinder with 1.5 meter diameter and weighs approximately 8 tons. At first stage I need to rotate this cylinder from 0-200 RPM within 1 minute and keep that RPM for 10 minutes and again from 200-1200 within 1 minute and keep this RPM for 30-45 Minutes, then bring down to 0 RPM, using a speed controlled(ATB) A/c Motor with suitable reduction gear system.

Kindly advice how to calculate the Initial torque to find out the correct capacity motor for this project

Thanks

Sunovya

2. Jul 26, 2009

### FredGarvin

What is the basic kinematic relationship between torque and angular acceleration?

3. Jul 26, 2009

### Bob S

Here are properties of the cylinder
Mass = M =7300 Kg
z = length = 8 meters
Moment of inertia = I = MR2 = 4106 kg m2 (hollow cylinder)
w = radians/sec = 2 pi RPM/60
Energy E = (1/2)I w2 = (1/2) M R2w2
Angular momentum is J = I w (kilogram-m2/sec)
Torque L = I dw/dt
(kilogram-m2/sec2 = Newton-meters)
If the cylinder is accelerated from zero to 200 RPM in 60 sec, this is equivalent to 0.35 radians per sec2.
The starting torque is 4106 x 0.35 = 1437 Nm

Last edited: Jul 26, 2009
4. Aug 2, 2009

### Turv

Bob,

You seem like a really clever guy, i wonder if you can help me find an easier formula for fan starting times regarding motors.

At current we use a really long winded formula.

i'll try to explain the best i can.

First of all we have to find the starting time and loss coefficients that we have to use a chart. if we have say 15 Kw absorbed power we may use say 18.5 Kw motor so firstly we do this.

15000 Watts / ( 18500 x 1.7 ) = 0.48 we then look at the starting time and loss coefficent graph which would be 1.24.

1.7 is what we estimate for starting torque vs normal torque.

so starting torque will be estimated (18500 x 1.7) / (2 x pi x revs per sec (24) ) = 208 Nm

we then need to know the rotor inertia and impeller inertia, so if the rotor inertia was 0.12 kg m^2 and the impellor inertia was 30 kg m^2 we use the following calculation.

lets say the impeller is 12 revs per second.

1.24 x (2x pi x 24) / 208 Nm x ( 0.12 + 30 x (12 / 24 ) ^2 ) = 6.85 secs starting time.

Last edited: Aug 2, 2009
5. Aug 2, 2009

### Bob S

I will try to answer your question, although I do not understand the source of all of your numbers. First, at full speed (1440 RPM), your motor can supply

18,500 x 60/(2 pi RPM) = 122 N-m
15,000 x 60/(2 pi RPM) = 99 N-m

The total torque to accelerate your inertia at dw/dt is (using full RPM of impeller is half the rotor RPM)

L = I dw/dt = 0.12 dw/dt + 30 x 0.5 dw/dt = 15.12 dw/dt = 208 N-m

So dw/dt = 208/15.12 = 13.8 radians/sec2

1440 RPM = 151 radians per sec.

So time to accelerate to full RPM at no load and 208 N-m torque is 151/13.8 = 10.9 seconds.

[Added text] I did not include your loss coefficient of 1.24. If I include this factor, the accelerating time to full RPM at no load would be 1.24 x 10.9 sec = 13.9 sec.

I hope this helps.

Last edited: Aug 2, 2009
6. Aug 3, 2009

### Turv

Bob,

Thanks for your reply, just a quick question what is L = I dw/dt ? i take it I= inertia? can you explain the rest please.

cheers Paul.

7. Aug 3, 2009

### Bob S

Hi Paul-
w (actually lower case omega) is commonly used as the angular speed in radians per second, so for example
w = 2 pi RPM/60.
Then dw/dt = angular acceleration in radians per sec2.
Angular momentum is J = I w
Then torque L = dJ/dt = I dw/dt
These are analogs of
p = linear momentum = m v
Force = dp/dt = m dv/dt

Bob S

8. Aug 3, 2009

### Turv

Bob,

I'm having a practice, can you confirm if i'm going right.

motor = 1440 rpm - 22 Kw, starting torque = (22000 x 1.7) / (2 x pi x24 ) = 248 Nm

Rotor inertia = 0.14 , Impellor inertia = 16, Impellor speed = 1600 rpm = 26.66 rps

L=I dw/dt = 0.14 dw/dt + 16 x 1.111 dw/dt = 17.916 dw/dt = 248 Nm

so dw/dt = 248/17.916 = 13.84 radians/sec^2

1440 rpm = 24 rps = 24 x 6.283 = 150.79 radians per sec

Acceleration time therefore equals 150.79/13.84 = 10.89 secs.

I dont think using your way you need the loss coefficient?

9. Aug 3, 2009

### sunovya

Dear Mr Bob,

Thanks you very much for the reply. It was very useful to me

Thanks & Regards

Sunovya