Is horsepower horsepower? 2

[rharr]

We have a situation with a plastic processing extruder where we have changed the screw design for more output and the maximum speed is now limited to 65 rpm. The old screw would turn about 85 rpm.
The drive motor is a 300 Hp DC working through a 15.49:1 reducer. With 1750 rpm at the motor we would see 113 at the output of the reducer (the extruder)
The amp load tops out at the rated 470 amps and we only see 65 rpm at the screw. Working backward through the reducer, the motor is turning 1007 rpm at full load.
It has been suggested to change the ratio in the reducer to get more power from the motor to turn the screw faster.
Will this really work? Can we turn the screw faster with a larger reduction through the gear box?

 

[Compositepro]

A higher gear ratio will help. You are now maxing out the torque capability of the motor not the horsepower. Power is equal to torque times speed.

 

[rharr]

Thanks for the vote of assurance.

I guess I'm having trouble determining what the limiting condition is - Torque or horsepower?

I'm thinking about the power at the screw (output). It takes a certain torque to turn the screw this speed. This torque at this speed determines the horsepower out. If I want to turn the screw faster, I need more horsepower in?

Is it really this simple? The same torque in at a faster speed, because of the higher reduction, is more power. More power in results in more power out. More power out makes the screw turn faster.

Just because I provide full load amps to the motor, I'm not really getting full power? If you think about it, or should I say when I think about it, if I have 300 Hp at a given speed, I can back into the torque. (T=Hp/N) However, this can't really work, can it? Holding Hp constant would make torque sky rocket as speed approaches zero. There's my problem. What's the limiting factor? Torque or horsepower?

I asked for a performance curve like this from the motor's manufacturer a couple weeks ago and haven't seen anything yet. Given full load amps, what is the power (or torque) at various speeds? Is torque really directly proportional to amps?

 

[jraef]

HP = Speed (rpm) x Torque (ft. lbs.) / 5250

Mess with any side of that equation you like, but it always has to work out the same. But your MOTOR is designed for putting out a given torque at a given speed; or in other words, a Design Horsepower rating. There is usually a little fudge factor in there, but not enough to count on without testing.

Reduce the speed while keeping the torque the same and you are reducing the HP; no problem.

Increase the speed over the design while (proportionally) reducing the torque requirement; again no problem.

Increase the speed over the design rating while keeping the torque requirement the same; problem.

Demand more torque at a given speed than the motor was designed to deliver; problem.

In the last 2 cases you are attempting to get more HP out of a motor not designed to deliver it; you are over loading the motor. The motor is "dumb" and will attempt to do what you ask of it, but will pull more current to do that. More current = more heat. Eventually you will exceed the temperature rise that the motor is designed to handle.

 

[kevindurette]

http://kevinthenerd.googlepages.com/torque_vs_hp.html

If your amp load is topped out, and the voltage is fixed, your input power is topped out. That means the output power is also topped out. Gearing won't help. Gearing doesn't change power. (It only changes it by adding parasitic losses.)

Solution? Perhaps you can improve the cooling on the motor and over-load it. Just keep in mind that ohmic heating is proportional to the SQUARE of current but output power is only linearly proportional, which will result in diminishing returns for a given improvement in cooling efficiency. Check with EEs and make sure the motor is operating efficiently (with a good power factor and all that jazz).

You just might need a new motor.

 

[kevindurette]

Scratch that: a 300 hp motor is probably 3-phase and is already operating quite efficiently.

 

[davidbeach]

Actually all DC motors would be something like single phase.

 

[waross]

Good morning jraef;
When I first read the original post, I agreed with you.
After rereading the post several times, I am not so sure.

With 1750 rpm at the motor we would see 113 at the output of the reducer (the extruder)
The amp load tops out at the rated 470 amps and we only see 65 rpm at the screw. Working backward through the reducer, the motor is turning 1007 rpm at full load.

It seems that the motor is reaching it's rated torque/current limit before it reaches full speed. The motor is probably producing about (1007/1750)x300hp.
The drive is probably limiting the voltage to about (1007/1750)x rated voltage.
A gear ratio change should allow the motor to run faster and produce more hp. and more RPM on the output shaft of the reducer. The optimum ratio will depend on the speed/torque characteristics of the load.
Am I missing something here?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[kevindurette]

Is time-averaged current only dependent on load power? Or will the power just be traded between shaft output and ohmic losses? (In other words, the motor will be generating zero power when it's producing torque while stopped, but it'll produce plenty of heat while doing so.)

Sorry I'm just an ME, not an EE.

 

[LionelHutz]

You can think of most motors as torque producing devices. That is, they are capable of producing a certain amount of torque on the output shaft. The HP rating is derived from the rated speed.

This means the power output a motor is capable of producing goes up linearly with the speed. To describe it differently, at 0 speed you get 0hp and at rated speed you get rated power. You can create a graph by plotting and drawing a line between these 2 points. If you extend the graph and go to 2X rated speed you would be getting 2X rated power.

The screw requires a certain amount of power to turn at a certain speed. Basically, the screw is doing work. If it turns faster it is doing more work.

So, you increase the gear ratio. Now, the motor would turn faster and have more available output power when it is turning the screw at 65rpm, meaning the motor should be able to turn the screw faster before it reaches rated current and rated load.

If you change the ratio enough you go over the rated speed of the motor there are other factors to consider. I'm not too familiar with DC motors, but I believe that to run the motor at a higher speed you need to have a higher supply voltage. The other factors are mechanical. Can the motor handle the centrifugal force? Will the brushes become a problem? Are the bearings OK?

 

[electrageek]

Hello rharr,

In a DC Motor, Torque is the product of Armature Current X Field Flux (field current times the number of turns in the field) time a constant. For simplity Torque is basically

T = Ia x If

A DC motor can product full torque at zero speed (a great asset)

Speed in a DC motor is another simple equation. (I know it's really Field Flux not field current but for our use its close enough.)

n = (Vcemf X K2)/If

So the speed equation tell us that speed is directly proportional to the armature voltage and inversely proportional to the field current. Weaken the field and the motor will run faster at the same armature voltage but because of the Torque equation it will produce less Torque at current limit which is where you are at.

Horsepower is:

HP = (T x n)/746

Now a gearbox it is a device that reduces speed and amplifies torque or the other way around, but it keeps horsepower constant. T x n going into a gearbox is the same amount as T x n coming out. As your Torque goes up the speed goes down and vis versa.

So we have a couple of questions:

1. What was the armature current running in the original configuration.(ie with the old gearbox)? 80%,90% If it was near 100% Ia then combined with 100% speed you have 100% horsepower being generated. Since the load is constant (Extruder) that means you would need more horsepower to run any faster. Remember HP stays constant through a gearbox. Only way to get more torque is to slow down not speed up. If you were well under full load then you should have been able to run a little faster.

I have found that most Extruders are usually loaded right up to max all the time.

2. What is the armature voltage of the motor at 1007 rpm it should be in direct proportion to the speed. If it is 500Vdc at 1750 then the armature volts should be 288Vdc if it is not then it means your field is set incorrectly and that that will limit your torque (remember T = Ia x If).

Or if you want you can cheat a little:

1. Most DC motor can run above their normal operating voltage by 10%. Let the voltage go to 550Vdc and you would gain 10% more torque. You might lose a little life just remember to keep the motor cool.

 

[rharr]

I have some of your answers, electrageek. Does this tell you any more?
The original screw turned 85 rpm (1317 at the motor) with motor amps recorded as 480. (The rated arm. amps from the motor's tag is 473.)
The new screw turns 65 rpm (1007 at the motor) with the motor amps recorded as 470.
I don't have the armature voltage. Sorry.

 

[electrageek]

I miss understood. I thought you had changed the gearbox ratio of the extruder to get more speed. You somehow changed the screw itself which is causing it to require more torque.

Since we know that your at 100% torque at 1007rpm which is 57.54% of Motor Speed, you can increase your gearbox ratio and not give up HP. We can go up to 1750rpm which would be approx. 40% more speed to give.

For example:
If you change the gearbox ratio by 20% from 15.49 to 18.58 you would have generated 20% more torque by letting the motor run up to 1208 rpm and still have 65rpm on the output. But having 20% more torque would also let the screw run faster so your speed would increase to say 1400rpm before you limit out on current again. The amount of torque that your screw requires as you go faster is the unknown in the equation.

But its always a good idea to check the armature voltage and make sure it agrees with the actual speed thus making sure you're at the nameplate field current.

 

[rharr]

Thank you all for your help. We would much rather increase the power to the screw than redesign the screw so it performs at a substandard rate.

 

[LionelHutz]

If you change the gearbox ratio by 20% from 15.49 to 18.58 you would have generated 20% more torque by letting the motor run up to 1208 rpm and still have 65rpm on the output. But having 20% more torque would also let the screw run faster so your speed would increase to say 1400rpm before you limit out on current again. The amount of torque that your screw requires as you go faster is the unknown in the equation.

I think we know what you're saying but it you change the gear box ratio by 20%, you increase the motor speed by 20% and you increase the available HP by 20%, all while still running at 65rpm on the screw. You do not change the available torque.

With that extra available HP you will be able to increase the speed of the screw by some amount before the screw is once again using the same HP as the motor is producing.

 

[electrageek]

LionelHutz,

You do increase the amount of torque available on the output of the new gearbox. But your right it's the HP that increasing on the input side of the gearbox and therefore on the output side as well. He can make better use of the motor if it can run all the way to it's maximum speed.

 

[LionelHutz]

You're right, you increase the HP on the motor side which in turn gives you more torque available on the output side.