Motor Generator

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mull982

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Hello

I am involved with a project at work where we are using a 1000 h.p. motor generator to convert AC to DC to power various motors on a dragline.

My question was if there was any kind of direct conversion number between the number of horsepower the motor generator was running at, and amount of DC output power. Does it depend on the equiptment producing the DC power or is their a direct correlation between the amount of DC power that can be converted from a given amount of AC power?

I would appreciate any information on this Motor Generator (AC to DC) topic since it is a fairly new one to me.

Thank you for your help.

Mull982
 
While I don't know the actual formulas....someone will chime in:

watts out = watts in - watts lost as heat
or
watts out =watts in x efficiency (%)

You'll need to know your efficiency of the unit to realize how much energy is being "lost" as heat.
 
You might want to do a tradeoff and cost analysis between using a rotary converter system which is what you describe, and a modern solid-state converter system.

If the dragline is yet to be built, you might want to consider using Variable Frequency Drives.

I once visited a massive coal mining shovel in Kansas that was powered by electricity. They dragged the wires along behind it, and motors on the machine powered hydraulic pumps which supplied hydraulic motors and cylinders to operate the shovel and move the rig.

Since you are new to this, I suggest visiting somewhere that you can see what has been used on other successful machines.

The coal digger was not successful in the end because it was just too big for the job. It is now a tourist attraction.
 
mhulbert said:
While I don't know the actual formulas....someone will chime in:

watts out = watts in - watts lost as heat
or
watts out =watts in x efficiency (%)

You'll need to know your efficiency of the unit to realize how much energy is being "lost" as heat.
Click to expand...

I don't see anything wrong with this statement.
Any motor power application can be split into two considerations: power required for the work to be performed, and losses. The power required by the work to be preformed, whether it is converted or not, will be the same because it is determined only by the load requirements. Once converted to DC and used by your downstream motors on the drag line, the power used becomes fixed regardless of the source. In an MG set however, you have loses in BOTH systems, the AC motor and the DC generator. The DC generator has copper losses, eddy-current losses, hysteresis losses and mechanical losses. The AC motor has all those plus additional eddy-current losses.

If all of your motors are DC, it may be daunting to replace them all with AC drives and motors, but you may still want to consider replacing that MG set if you are concerned for power consumption. Were you to switch to a solid state converter you would have only the switching losses of the converter, usually significantly less than the combined losses of the rotating machines, and of course considerably less maintenance!.
 
Thanks for all the help guys.

It makes sense what your saying about heat loss, and mechanical loss between the motor and generator but is DC horse power the same as AC horse power. Another words if i have two DC swing motors that are 10 h.p. each therefore requiring 20 h.p. from the DC generator, would that also correlate to a 20 h.p. draw from the AC motor (plus losses of course) or is DC horse power less than that of AC?

So I gues my overall question is if whatever h.p. or FLA is being drawn from the DC generator is the same as the h.p. and FLA that is being drawn from the AC motor (plus losses)

Thanks for all the help!

mull982
 
You are dealing with a thorny terminology issue here.

First you have the issue of power. 1 Horsepower is a unit of power. It is equal to 33000 foot-pounds per minute, or 0.74569987158227022 kW. It turns out that the electrical industry has come up with a different unit, also called the horsepower, which is equal to 0.746kW _exactly_. If you presume perfect efficiency, then:

mechanical load horsepower = DC electrical load horsepower = AC electrical supply horsepower

or mechanical load horsepower * 0.746 = AC electrical load kW

Since motors are on the order of 95% efficient, the above is a pretty good approximation.

The problem that you have is that the "horsepower rating" of a motor is _not_ the mechanical load horsepower. A horsepower rating is similar to the rating of a circuit; it is a measure of the power that the motor is capable of passing to a load. It is a rough number that includes things like the maximum power that the motor can pass on a continuous basis, the starting capability of the motor, the breakdown torque, etc.

A '10 Hp' motor will not necessarily provide 10 horsepower of mechanical power to the load, and will not necessarily consume 7460 watts of power (plus losses). Depending upon the load it might consume more or less power, and will quite likely have transient components (starting, stopping) that exceed 10 hp. Some of these transient components can be absorbed by the rotational inertia of your MG set, some will force the use of larger MG components.

So to move forward, you will need to figure out what your mechanical loads are, and if the '10 hp' motors are oversized or undersized for the load.

-Jon
 
Horsepower is all the same whether provided by an AC or DC motor. The thing that is uncertain is what the motor does and requires for power, relative to its rating. One HP is 550 ft-pounds per second whether it is produced by AC or DC motor, by a windmill, or water power, or slaves running in a squirrel cage.

Many of the old motors would at times draw more current than you would expect for the horsepower. To be safe, you need to determine the current that the DC motors will require under worst-case starting and load and be sure that the generator will deliver that current.

A motor expert will be able to help more, but my recollection is that torque is roughly proportional to the current. You will have to deliver enough torque to the generator to produce the starting current for any large DC motor. A big flywheel might help the AC motor keep that generator up to speed when the big DC motors start.
 
Thanks for the help guys everything seems a lot clearer now.

I do have one last question however. I'm trying to size mining cable to be used as our new feeder cable to the dragline. Does anybody know the ampacity of a 4/0 AWG 3/C 5000V MSHA mining cable. Is there a place that I could find a table showing the ampacities for 5000 V conductors of this type.

Thanks again for all the help.

mull982
 
mull982 said:
I'm trying to size mining cable to be used as our new feeder cable to the dragline. Does anybody know the ampacity of a 4/0 AWG 3/C 5000V MSHA mining cable. Is there a place that I could find a table showing the ampacities for 5000 V conductors of this type.

mull982
Click to expand...
Here are a couple of links to a supplier of the cable. Their table says 321 Amps. They should be able to help.

http://www.aiwc.com/catalogsection/miningcable.html

http://www.aiwc.com/catalogsection/catalogpdfs/pccatalog/AmerMusttypeSHD-GC.pdf
 
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