The only difference for continuous-duty motors is the overload protection.Yeah, I was thinking of a conveyor motor that would be running indefinitely. Continuous duty is defined as "Operation at a substantially constant load for an indefinitely long time". Whereas, continuous load is max current for 3 hours or more. So the motor would be continuous duty, but would it be considered "max current"?
You replied "no" to my post. Please cite references if my post was incorrect.No.
The motor draws according to the load on it. That is, a 10-hp motor draws pretty much the same running amps as a 1-hp motor, when there's only a 1/2hp. load applied.
Yes, there are differences in the starting amps and how quickly the motor can respond to new loads.
I disagree with that. Both only deliver 1/2 HP to the load, but they will not have same characteristics that result in same input current. The mass of the rotor on the 10Hp motor will likely require more input power just to keep it at speed than the rotor on the 1 hp motor for a place to start with differences.That is, a 10-hp motor draws pretty much the same running amps as a 1-hp motor, when there's only a 1/2hp. load applied.
The larger motor will draw less current given the same load. Kind of like a car with a big motor putts up the hill while the guy in the little car has to put the pedal down.I disagree with that. Both only deliver 1/2 HP to the load, but they will not have same characteristics that result in same input current. The mass of the rotor on the 10Hp motor will likely require more input power just to keep it at speed than the rotor on the 1 hp motor for a place to start with differences.
And the car with the big motor used more fuel than the little one to accomplish the same task. It is not that simple.The larger motor will draw less current given the same load. Kind of like a car with a big motor putts up the hill while the guy in the little car has to put the pedal down.
And the car with the big motor used more fuel than the little one to accomplish the same task. It is not that simple.
The larger motor will not draw less power. In the example given, a load requiring 1/2 hp is driven. Both the 10 hp motor and the 1 hp motor will deliver the needed 1/2 hp to the load. That is the only thing that is the same in both situations. The 1 hp motor likely does so more efficiently, as both motors not only need the 1/2 hp of input energy that gets transferred to the load, they also need whatever energy necessary to spin its own rotor, overcome its own friction, and other losses. Because of physical differences, those other losses are going to be higher in the 10 hp motor than in the 1 hp motor, meaning it will take more input energy to do the same work with the 10 hp motor.
A 10 hp motor with that small of a load on it probably has a terrible power factor, which is not a direct loss but it does contribute to what is needed for input.
You ever clamped an meter on a lead supplying a (running)motor with no connected load?The current draw on a motor is load dependent. You can see this on a fan with adjustable pulleys. Lower the rpm of the fan and the motor current will drop. There are other factors of course, mass and inertia etc. Starting current will be higher for the larger motor. Perhaps the comparison of a 1/2 to a 10hp is not a good one. There's a curve on a chart there somewhere but that is beyond me. But given the same load, the smaller motor will have to work harder to move it. So, to run the same load, if the smaller motor uses more current then it is also using more power.
I do understand your point and you are right it isn't so simple.
For a single motor, you multiply by 125% regardless. If it is run continuously, you do not compound that multiplying by 125% twice. As for continuous-duty motors, see 430.32Smart $, we may be having a problem communicating here ....
The NEC has no provisions for "contunuous duty motors.' The NEC does not recognize the term. There are motors, period. Then there are loads, which may be continuous or not.
The NEC defines a continuous load as one that runs for three hours or more. The load can be a light bulb, or an 'ordinary' motor. In this situation - where the load is present for mor than three hours - you apply the 125% multiplier.
This has nothing to do with 'cintinuous duty' motors. A 'continuous duty' motor is one that says 'continuous duty' on the nameplate. This means that the manufacturer has optomised the design for long run times. That's all that means. The motor need not be actually run for long periods. You wire this motor as you would an ordinary motor - using the 125% multiplier only if the motor will actually be run for more than three hours at a time.
That multiplier has nothing to do with the motor; it is there for the wire.
He did not say that he said "The electrical power input that supplies the mechanical power output is exactly the same for both motors". Which is what I have also been saying. It means the same amount of power is used to drive the load in either case, but there is additional power use in both cases but it is not the same in both.gar,
You just explained all of the losses in the 10hp motor compared to the 1hp motor, and then said the input power is exactly the same?
Wouldn't the output be the same and the 10hp motor would draw more current compared to the 1hp motor?
You ever clamped an meter on a lead supplying a (running)motor with no connected load?
Input energy is always higher than output energy, until someone designs a motor that is 100% efficient.
Most motors are much more efficient at/near full load current than they are at only 5% load (like the 10 hp driving a 1/2 hp load would be).
I do understand what you are saying, but you have to remember it takes a certain amount of energy to spin the 10 hp motor whether there is a load connected to it or not. Same with the 1 hp motor, but that energy is likely less than for the 10 hp. The additional 1/2 hp load is simply added to the "no load" input energy.
Thanks for the info. I understand what you are saying here.121211-1037 EST
The load on the motor is 1/2 HP or 746/2 = 373 W. That is exactly the same for both motors.
Both motors run at essentially the same speed. The large motor has larger bearings, and more force load on the bearings than the smaller motor. Thus, more frictional loss from the larger motor bearings.
The larger motor has more rotor surface area and likely a larger diameter, thus much higher windage loss than the smaller motor (this is also friction) .
The magnetic core in the larger motor is larger, thus more core losses (again essentially friction).
All of these factors tell you that the larger motor will require more power input than the smaller motor for the same output mechanical load.
The electrical power input that supplies the mechanical power output is exactly the same for both motors.
You need to study conservation of energy, and techniques of using energy and power in the analysis of problems.
The smaller motor will be running neared its full load capability than the large motor. Thus, at a more efficient operating point. Really that is what the above loss analysis tells you.
If you could design it with zero friction, you may have some advantage. That advantage would mostly be a reserve of energy that is ready to use should the source energy not be able to respond, or respond quick enough to a surge in the load. All the stored energy in the inertia still had to be input at some time, but with friction it is constantly being taken away therefore it constantly needs replaced.Thanks for the info. I understand what you are saying here.
I wonder how much the larger motor would gain from inertia, with all that weight spinning being hard to stop. Wouldn't there be a point where the mass would have an advantage?