## Stump the Scientist: Force spinning a large generator

We’ve got another one for you!  A Stump the Scientist question for you comes from reader Chris, posted way back in December.  Sorry to keep you waiting Chris!  Hopefully your curiosity is finally quenched with this answer.  Feel free to submit more questions to @EdisonsDesk!

Question posted by Edison’s Desk reader Chris:

I would like to know, exactly, or at least very closely, how much force is required to spin a large generator. Something in the range of 50Mw. By force, I mean Newton’s, or horsepower, or foot pounds…something I can relate to!

If I wrapped a rope around the pulley or shaft and pulled on it, what would it take to make the generator turn? For the sake of argument, let’s say the gen is on its side and I wrap a wire rope around the pulley or shaft, then I hang weights on the rope to produce torque. How much weight to turn it? How much torque is required?

I got an answer of 154,000 Hp from someone else. I think 1,500 Hp would be closer, and maybe as little as 500Hp, but I’d really like to know for sure! I realize the pulley size, if it has one, changes things a little, as does the production of electricity.

The closer you get to operating at 100%, the harder it is to turn. I’d just like to be in the ballpark! Help a guy out will ya?

Answer from Chief Scientist Jim Bray:

The way to solve this problem is first to think about the power you want to generate (and, for general information, power is the time rate of production of energy). You have picked 50 MW (50 million watts). If I tell you that small utility generators are about 98% efficient (as they are, and 50 MW is fairly small), then you know you have to supply 50/0.98 = 51 MW of power to the generator to get the required output. You would like to know the force you have to apply, and there are several ways you might apply a force to produce this power.

You specifically mention wrapping a rope around a shaft. Power (P) equals force (F) times velocity/speed (V) that the force is acting (P=FV). If the shaft were a meter in diameter and the rope is wrapped around it, we must say how fast we want to pull the rope in order to find the force. Most utility generators run at 3600 rpm, or 60 revolutions per second. For a 1 meter diameter shaft, the surface speed is therefore π*(1 meter)*60/sec = 188 m/s approximately. If we divide this into the input power, we find that the needed force on the rope is 2.71×105 Newtons, approximately. (By the way, horsepower (Hp) is a power unit, and foot-pound is an energy unit, so neither one measures force.)

Comments

I’m guessing that last number is supposed to be 2.71E5? (Not 2.71 times 105?)

I’m still a little bit confused, but I understand that’s a result of the wording in my original question.

Let’s say I have a way to produce a very strong pulling action, and I can generate 20,000 pounds of pull, and can do so over a half mile.

My problem is…I’m trying to figure out if that is strong enough to turn a 50 MW generator.

We have the speed of rotation for the generator, 3600 RPM. And we know what it takes to turn it…2.71×105 Newtons. 284.55 Newton meters?

I’m not sure how to rectify what I have left.

I have 20,000 pounds of pull that travels 10 meters per second, so it will have to be geared to bring the speed up to 3600 at the generator.

I think half my problem is that I’m working with two systems here. I should probably find out what 20,000 pounds of pull really means! I think it should be in Newton’s for starters. But is that Newton meters? Is that different than Newton’s? Everything was pounds when I was a child…I find it very confusing.

Thanks for your time.

I’m a operator and am doing a project at work and need the some what of the same information. Unfortunately I don’t have any new information that can be of help but I am sure there is a formula you can use to change ft lbs of torque into newton meters. I plan on looking for this information and if I come across it if you would do the same that would be great?
Dave

**** if I come across it ill share the information, sorry about the mistake above

David,

Google has a pretty good calculator to convert from Newton meters to foot-pounds. Try http://www.google.com/search?q=135500+newton+meters+in+foot+pounds (approximately 100,000 foot-pounds).

Note however that Jim correctly computed 271,000 Newtons which is approximately 61,000 pounds-force (lbf). We must be very careful not to confuse force (Newtons or lbf) with torque (Nm or ft-lbs). They can be related but are only equal when the moment arm is the relative unit distance. In Jim’s case, the moment arm is 0.5 meter radius so 271,000 Newtons at 0.5 m gives us 135500 Nm.

Back to Chris’s modified question of a 20,000 pound weight on a pulley wrapped around the shaft (assuming 1 meter diameter generator shaft). If we start from the moment that the weight starts to drop, the torque produced at the generator shaft would be 20,000 lbf times 1.64 feet radius which is 32800 foot-pounds. This is approximately 1/3 of the needed 100,000 foot-pounds needed to generate 51MW at 3600 rpm.

For the sake of the thought experiment however let’s assume that the virtual generator, weight and pulley system we are dealing with is 100% efficient and has no losses such as copper resistance, windage, friction or iron magnetizing losses. Then we can leave the weights free fall for a long enough time (approximately 57 seconds and 53000 feet later at 1,260 mph) and continue to speed the fictitious generator up to 10,734 rpm (assuming it doesn’t fly apart at that speed). Once we reach that speed, we can start to pull 50MW out of the generator electrical circuit. The weights will continue to fall but they will stop accelerating downwards. However don’t get your hopes up about a perpetual motion machine and free energy. Eventually the weights are going to reach an end somewhere and all the energy we pulled out will have to be expended again pulling the weights back up. In fact this is similar to what we do with hydro generators where the weight is made up of falling water and the energy to pull the weight back up came from the sun evaporating water and creating rain. Most hydro generators use more weight of water and less velocity however.

If you are interested in similar power and energy issues, I would recommend that you read the Energy chapter of “Physics for Future Presidents” at http://muller.lbl.gov/teaching/physics10/PffP_textbook_F08/PffP-01-energy-F08.pdf

hey jim

here i am at diff situation from Chris, i got a pull of 500 kg on a shaft. how can i figure out the capacity and shaft diameter? just still i know that pull which i can provide.

Dev

here is my problem.

what size pipe and what amount of distance plus head? degree of decline would I need to produce 500 foot lbs of torque?

Lets asume 24 inch pipe and 1,000 feet in hieght and a natural occuring slope? which the average could be easily duplicated with machinery available, if this helps work the problem thru.

I then would assume that a 12 inch pipe and 2,000 feet with a similar decline would yeild a similar result? or are you going to spank me for such broad assumptions?

I want to produce 500 ft lbs of torque from a water pipe.

then my question is how much electricity can I produce with 400 ft lbs of torque or NM.. what ever works, I am an American so I think in ft lbs, but I can use google to swap the two (2).

Thank You for being such a GREAT! Sport and helping people.. it really is VERY! Nice of You!

My Best to You and Yours!

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