# Calculating Starting Torque

SevenToFive
I have been trying to figure out the starting torque needed to start rotating a pipe and valve in a weld fixture. The pipe is 304 stainless 24 inches outside diameter and has an inside diameter of 23.625 inches. The length of pipe varies but so far all of the lengths of pipe get a valve welded on, the valve itself weighs 12000lbs. From the center of the pipe diameter to top of the valve is 78 inches. The total weight of the weldment is 22000lbs but they want to be able to weld weldments that weigh 44000lbs.
The weldment is attached to the gearbox that has a 50:1 ratio and 7HP motor with a VFD, and is supported by pipe roller stands. I am just wondering how do I calculate the torque needed to start rotating a 22000lb object disregarding any losses due to the roller stands.
Any help is greatly appreciated.

You will first need to find the total Moment of Inertia of the combined pipe, valve, weld fixture and gearbox output shaft. The gearbox input shaft and the motor must be treated separately as they have a different angular velocity.
The moment of inertia is the sum of all the constituent masses multiplied by the square of their distance from the axis of rotation. That is easy for most parts but may be difficult for the valve.

If the valve is asymmetric and the axis of rotation is not vertical you will also need to allow for the starting position of the mass imbalance.

SevenToFive
Is there a quick and easy way to calculate the moment of inertia and come with with a conservative value for the starting torque needed to spin the piece of pipe and valve?
The customer tells us that the valve weighs 12000lbs total and 6000lbs at top of the valve with is 78 inches from the center of the 24 inch pipe. The total assembly weighs 22000lbs. Any help is greatly appreciated. Thanks to all who reply.

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Since the motor and gearbox are decided, let's look at this backwards. We need to know the maximum stall torque of the 7 HP motor when driven by the VFD with current limiting to prevent overheating of the motor durng the initial rotor-lock situation.

It will be necessary to run the VFD at some fixed initial frequency to maximise torque during startup.
That motor stall torque will be multiplied by the 50:1 ratio gearbox.
The acceleration of the weldment will be equal to the Motor_Stall_Torque * 50 / Moment_of_Inertia.
That will then determine how long it takes for the weldment to reach the normal rotational speed.

SevenToFive
SevenToFive
So for the valve I can just break it up into 3 components and add them together for the Moment of Inertia? Then add the moment of inertia for the length of pipe to that and then if I remember right us the equation Torque = (Moment of Inertia*Difference in Speed) / (308 * Time to accelerate) that should give me torque in ft-lbs if I remember correctly.

Torque = (Moment of Inertia*Difference in Speed) / (308 * Time to accelerate) that should give me torque in ft-lbs if I remember correctly.
But you do not know the difference in speed or time to accelerate so you must start with the characteristics of your motor stall torque when current limited in the initial rotor lock situation.
Since Power = Torque * RPM, the torque must be infinite, for a fixed power at zero speed. Starting torque must therefore be limited by safe motor current.

You are on the right track.
The kinetic energy finally stored in the rotating weldment, at the design speed of rotation, cannot be delivered to the weldment faster than 7 HP = 5.52 kW = 5.52 kJ per second. One problem is that a slow running motor on a VFD will not deliver the full rated 7 HP at lower speeds.

What time per revolution, or RPM, of the weldment is wanted ?
Is the motor an induction motor designed for three phase 400VAC at 60 Hz ?
How many poles, or what is the design RPM for the motor at what power frequency ?

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SevenToFive
SevenToFive
What time per revolution, or RPM, of the weldment is wanted ?
For the desired rpm, we would be going from 0-3rpm in about a 2 seconds.

Is the motor an induction motor designed for three phase 400VAC at 60 Hz ?
Yes a 3 phase 60Hz motor with an output speed of 1800rpm.

How many poles, or what is the design RPM for the motor at what power frequency ?
I was not told the number of poles.

Also would it make a difference how the valve is placed, for example if the valve actuator is pointed at 12 o'clock or 7o'clock and we are trying to turn it clockwise. I would need to calculate it at a position where the most starting torque would be required to start rotating the assembly.
Thank you for all of your help.

Gold Member
If you place the valve at any point beyond 12 o'clock in the clockwise position and turn the assembly clockwise the weight of the operator assembly will reduce the amount of motor starting torque by an amount proportional to its angle in degrees in the clockwise position form the 12 o'clock position up to a maximum at the 3 o'clock starting position. At the same time, the motor running torque must then be capable of delivering sufficient rotational torque to offset the operators' resisting torque through the 9 o'clock position as well.

The acceleration needed is from 0 to 3 RPM in about 2 seconds.
First work out what gearbox ratio is needed to run the motor at it's design speed and power.
1800 RPM / 3 RPM = 600:1 ratio. But you only have a 50:1 gearbox so the VFD will need to drive the motor at about 1/12 of the design speed. To regulate the motor speed the VFD will need to operate at 60 Hz / 12 = 5 Hz.

I don't think you have a chance of getting the motor to provide sufficient acceleration without changing to a slower motor and/or a greater reduction ratio. We can calculate acceleration later from the moment of inertia, but at this stage we need to find torque data for induction motors when driven at a low frequency by a VFD.

Commonly available motor winding choices are.
A two pole motor will rotate at 60 Hz * 60 sec / 2 poles = 1800 RPM.
A four pole motor will rotate at 60 Hz * 60 sec / 4 poles = 900 RPM.
A six pole motor will rotate at 60 Hz * 60 sec / 6 poles = 600 RPM.
Those common motor speeds will not give the 150 RPM speed required.

Is it possible to replace the motor with a 12:1 geared motor?
Or insert a 12:1 reduction stage between the existing motor and the 50:1 ratio gearbox?

On the subject of balance; if the motor is a simple induction motor and the rotating load is not balanced, then the motor slip may vary between about 2% and 10%, so the speed once running might vary throughout the rotation by maybe +/–4%.
It might be possible to balance the load but that will increase the total mass and inertia.

SevenToFive
Baluncore: Thanks for all of your help with this so far. JBA thanks you as well.
The output shaft of the motor goes to a 7" center distance gearbox with a 50:1 ratio, so when the motor sees all 60Hz the gearbox input is around 1750rpm and the output is 35rpm and 13011 in-lbs of torque. The output of the gearbox goes through another reducer with a 6:1 ratio and gives an output around 5.8rpm. It seems they are using the VFD to get to 2.5 or 3 rpm.

I attempted to calculate the moment of inertia;
Most of the dimensions for the valve are guesses since the piece in question is no longer in the shop.
The valve I=mass(OD^2 +ID^2)/2: with mass in slugs
body,mass 9600lbs = 298.37slugs
I=298.37 ( 26^2+24^2)/2
I=186780lb-sec2/in
Stem, mass 175lbs = 5.44slugs
I=5.44(8^2+6^2)/2
I=272lb-sec2/in
Valve Actuator mass 2225lbs =69.2slugs
I=69.2(20^2+2.5^2)/2
I=14056lb-sec2/in

Valve Total 201108lb-sec2/in

Pipe is 92.6lb per foot
2315lbs = 71.9 slugs
I=71.9(24^2+23.625^2)/2
I=40772lb-sec2/in

Weldment total = 241881lb-sec2/in

Alpha = (Final velocity*2pi/60)/Time
Alpha = (2.5*2pi/60)/3.5seconds

Torque =I*Alpha
T=241881 * 0.0748
T=18092.6 in-lbs

So the 18092 would be my required starting torque?
All of the VFD talk makes me wonder if you turn down the frequency, don't you also lose HP? Thanks everyone for the replies.