torque speed and HP relation of three phase motor

[Muhammad Umer ansar]

I am a bit confused of The relation betweeen the parameters mentioned above.

According to the below formula, speed is inversely proportional to Torque as:
T=HPx9549/Rpm
But i have been taught in my college that speed is directly proportional to torque because to rotate an object at high speed we need more torque(POWER).
Coperation is needed.

 

[retirede]

Remember that torque and thus HP are dependent on the mechanical load.

Speed and torque would only be direct inverse proportion if the HP was constant - virtually never the case in real applications.

I suggest the you look at a a speed/torque curve for an induction motor and see if that helps.

 

[WasGSOHM]

9549 should be 5252?
You using kg or lbs?

You solved for T. Try also solving for HP and RPM.

 

[drcampbell]

Remember that a speed-torque curve depicts the maximum rated torque. Actual torque may be more or less, depending on what the motor's driving.

As an extreme case, consider a motor that's electrically energized and spinning, but the output shaft isn't connected to anything. Regardless of what the s-t curve says, actual net torque will be zero.

 

[drcampbell]

It'll make a lot more sense if you substitute
(33,000 / 2·π) (foot·pounds per minute) per horsepower
for "5252".

 

[WasGSOHM]

Another angle.
Basically the formula is T = HP/RPM.
The units of RPM is "things per minute or second),
HP is ft-lb per second
and
T is ft-lb.

The seconds cancel out and so the posted formula is "dimensionally correct."

And the odds are you speak punjabi.

 

[winnie]

As others have noted, your equation looks wrong because you don't state units and the scaling constant is not what we are used to. If torque is in pound feet, and Hp is power in US horsepower and RPM is rotations per minute, then the scaling factor is 5252.

But back to your deeper question: this equation makes torque the inverse of speed when previously you'd been taught that torque was proportional to speed.

This comes down to determining what is being held constant so that you can correctly apply the equation.

If you ask the question 'How much torque is 10 Hp at different speeds?' then by the way you have stated your question you are holding power constant. In this situation torque has to increase as speed decreases. You might encounter this situation if you were doing design work and had a particular power budget, knew the load torque requirements, and were selecting motor speed to fit the two.

On the other hand, you might be holding torque constant and asking what power was required to move at a particular speed. If torque is constant (for example lifting a load) then power is proportional to speed.

For many loads torque is proportional to speed, in which case power is proportional to speed squared. Still other common loads have torque proportional to speed squared in which case power is proportional to speed cubed.

Same base equation T=power/speed, but applied in different situations where different terms are constant or scale in different ways.

A completely analogous situation is the relationship between power, voltage, and current when you consider Ohm's Law. Power = volts * amps (DC case for simplicity) if you are asking 'how much current is needed to deliver some number of watts' then current goes down as voltage goes up. This question explicitly assumes constant power.

But if instead you are supplying power to a particular resistor, current goes up in proportion to applied voltage and power goes up as the square of the voltage. In this situation you are using the same equations but holding resistance constant.

And if you are supplying power to something like an incandescent lamp, you use the same equations but have to deal with the fact that even the lamp resistance is not constant.

Hope this helps.

Jon

 

[Carultch]

9549 should be 5252?
You using kg or lbs?

You solved for T. Try also solving for HP and RPM.

9549 is the numeric value of this conversion factor in the Metric system, using RPM, Newton-meters, and kilowatts as the units.
5252 is the US customary value of this conversion factor, using RPM, ft-lbs, and horsepower as the units.

9549 comes from 1000 W/kW * 60 sec/min / (2*pi radians/revolution)
5252 comes from 550 (ft-lbs/sec)/Horsepower * 60 sec/min / (2*pi radians/revolution)

The way to avoid this conversion factor altogether, is to use radians/second for rotation speed (omega), Newton-meters for torque (tau), and Watts for power (P). This is how a Physicist would use this formula (P = omega*tau), using radians as the natural angle unit, and consistent SI units that don't require conversion factors.

Suppose you spin a 1 meter radius wheel, with 1 Newton of tangential force, and you apply 1 Newton-meter of torque. This becomes 1 Joule of work, when you spin it through 1 radian of rotation (about 57 degrees), because that is the angle that corresponds to the application point traveling 1 meter of arc length. Newton-meters of torque become Joules of work, after applying a torque through 1 radian of angle change. Accomplish this task in 1 second, and that is 1 Watt of power.

 

[Julius Right]

The relation between torque and speed it is not constant. If you know the actual power delivered at the motor shaft and the speed the relation P=T*(2*π*rpm/60)
it is correct [P in W and T in Nm]. Actually, during induction motor start process there are 3 different torque: at start when rpm=0 Tstart=1-3 times rated torque [Tn]; at approximate 6-8% less than rated rpm it is the breakdown torque [maximum torque] and at 1-2% less then synchronous speed [close to no-load rpm] it will be the rated torque].
At a D.C. motor- it depends if the excitation is extern, shunt, series or compound- the torque dependence is different referring to speed.

 

[Muhammad Umer ansar]

9549 should be 5252?
You using kg or lbs?

You solved for T. Try also solving for HP and RPM.

i have used 9549 because i am calculating torque in Newton-Meter

 

[Muhammad Umer ansar]

As others have noted, your equation looks wrong because you don't state units and the scaling constant is not what we are used to. If torque is in pound feet, and Hp is power in US horsepower and RPM is rotations per minute, then the scaling factor is 5252.

But back to your deeper question: this equation makes torque the inverse of speed when previously you'd been taught that torque was proportional to speed.

This comes down to determining what is being held constant so that you can correctly apply the equation.

If you ask the question 'How much torque is 10 Hp at different speeds?' then by the way you have stated your question you are holding power constant. In this situation torque has to increase as speed decreases. You might encounter this situation if you were doing design work and had a particular power budget, knew the load torque requirements, and were selecting motor speed to fit the two.

On the other hand, you might be holding torque constant and asking what power was required to move at a particular speed. If torque is constant (for example lifting a load) then power is proportional to speed.

For many loads torque is proportional to speed, in which case power is proportional to speed squared. Still other common loads have torque proportional to speed squared in which case power is proportional to speed cubed.

Same base equation T=power/speed, but applied in different situations where different terms are constant or scale in different ways.

A completely analogous situation is the relationship between power, voltage, and current when you consider Ohm's Law. Power = volts * amps (DC case for simplicity) if you are asking 'how much current is needed to deliver some number of watts' then current goes down as voltage goes up. This question explicitly assumes constant power.

But if instead you are supplying power to a particular resistor, current goes up in proportion to applied voltage and power goes up as the square of the voltage. In this situation you are using the same equations but holding resistance constant.

And if you are supplying power to something like an incandescent lamp, you use the same equations but have to deal with the fact that even the lamp resistance is not constant.

Hope this helps.

Jon

Very Helpful it is jon!
Can you reckon or post the link of types of loads and which loads are following squared power or cubed power relationship ?

 

[drcampbell]

i have used 9549 because i am calculating torque in Newton-Meter

Then you should be expressing power in watts, not horses.

The relationship is Power = Torque x Speed.
Everything else is just converting incompatible units.

 

[Besoeker3]

Then you should be expressing power in watts, not horses.

The relationship is Power = Torque x Speed.
Everything else is just converting incompatible units.

I agree. And it makes it so much simpler. SI is the way to go!

 

[winnie]

Very Helpful it is jon!
Can you reckon or post the link of types of loads and which loads are following squared power or cubed power relationship ?

As I said, lifting (eg. a crane lifting a load) is roughly constant torque. This makes power proportional to speed.

Some type of pump loads have torque that scales linearly with speed. This makes power proportional to the square of speed.

Many pumps and fans have torque that scales as the square of speed. This makes power proportional to the cube of speed.

The energy in the wind scales as the cube of the wind speed.

Here is a useful link: https://www.bharathuniv.ac.in/colleges1/downloads/courseware_eee/Notes/sem7/SEM VII BEE703 EDC.pdf

-Jon

 

[Besoeker3]

As I said, lifting (eg. a crane lifting a load) is roughly constant torque. This makes power proportional to speed.

Some type of pump loads have torque that scales linearly with speed. This makes power proportional to the square of speed.

Many pumps and fans have torque that scales as the square of speed. This makes power proportional to the cube of speed.

The energy in the wind scales as the cube of the wind speed.

Here is a useful link: https://www.bharathuniv.ac.in/colleges1/downloads/courseware_eee/Notes/sem7/SEM VII BEE703 EDC.pdf

-Jon

For the majority of industrial motors applications the speed is close to being fixed. So the the square of the speed isn't a proportionality.

 

[paulengr]

Very Helpful it is jon!
Can you reckon or post the link of types of loads and which loads are following squared power or cubed power relationship ?

Right off the top centrifugal loads (pumps and fans) follow the affinity laws...torque is proportional to the square of speed.

 

[paulengr]

For the majority of industrial motors applications the speed is close to being fixed. So the the square of the speed isn't a proportionality.

In machining, conveyors, lifting maybe. But with saws, cmpressora, fans, pumps, not true at all. And this is just a high level view not getting into time based effects. For instance while running most of a conveyor load is idler friction independent of load. During startup torque is linear with time as the belt stretches then about 140% of running torque as it accelerates with constant speed.

 

[Besoeker3]

In machining, conveyors, lifting maybe. But with saws, cmpressora, fans, pumps, not true at all. And this is just a high level view not getting into time based effects. For instance while running most of a conveyor load is idler friction independent of load. During startup torque is linear with time as the belt stretches then about 140% of running torque as it accelerates with constant speed.

Yes, but once running................

 

[paulengr]

Yes, but once running................

In a tire assembly machine with dozens of motors the biggest one is 20 HP. The 140% starting torque just means pay a little extra for 20 HP instead of 10 HP. We can simply ignore details.

In a 10,000 foot 36” wide belt conveyor requiring a few thousand HP wife 8 drive motors if you accelerate under full torque at startup you rip the belt apart. And 140% torque for 30-90 seconds is outside the capabilities of a NEMA type B motor. Simply doubling the size is very expensive. So once running these minor details don’t matter...

That’s just one example. Many customers think they can just put in any motor that will fit and fix everything with a VFD. Often they are surprised when it does little to nothing or worse, destroys motors. For instance with pump power proportional to the square of speed they put a VFD in thinking they can pump more only to find the “extra amps” vanish without any appreciable increase in flow or conversely adding a VFD won’t control a flow that is easily valve controlled.

As an example of a very common application, think of what happens to a city water system every morning when most people go to the bathroom. Or in the afternoons in summer when we hit peak cooling loads. Even within manufacturing plants compressed air and steam demands fluctuate wildly. You can do a lot with on/off control but variable speed is either more efficient or simply necessary, even once running...

 

[Besoeker3]

In a tire assembly machine with dozens of motors the biggest one is 20 HP. The 140% starting torque just means pay a little extra for 20 HP instead of 10 HP. We can simply ignore details.

In a 10,000 foot 36” wide belt conveyor requiring a few thousand HP wife 8 drive motors if you accelerate under full torque at startup you rip the belt apart. And 140% torque for 30-90 seconds is outside the capabilities of a NEMA type B motor. Simply doubling the size is very expensive. So once running these minor details don’t matter...

That’s just one example. Many customers think they can just put in any motor that will fit and fix everything with a VFD. Often they are surprised when it does little to nothing or worse, destroys motors. For instance with pump power proportional to the square of speed they put a VFD in thinking they can pump more only to find the “extra amps” vanish without any appreciable increase in flow or conversely adding a VFD won’t control a flow that is easily valve controlled.

As an example of a very common application, think of what happens to a city water system every morning when most people go to the bathroom. Or in the afternoons in summer when we hit peak cooling loads. Even within manufacturing plants compressed air and steam demands fluctuate wildly. You can do a lot with on/off control but variable speed is either more efficient or simply necessary, even once running...

Yes, I understand. And I do know a little bit about motors. My field is mostly variable speed drives, usually upwards of 100kW. But I was referencing to the usual fixed speed drives. Typically these are 60Hz with a SMALL slip angle. Around 1750 rpm. Or, in my case, 1450 rpm.