100W Motor Lifting 20kg 50m in the Real World?

5 m maybe, depending on the gearing, not 50.

Yes; yes.

RE: Yes; Yes.

I presuming that is a disagreement with Lyn, and a yes answer to each of my questions.

I guess it drives home the real world value/cost of the 100w light bulb.

In theory, yes. However, a small gearmotor might be only 60% efficient (but not as bad as 10%).

Your calculations are correct but simplistic. To lift your mass will require some form of rigging between motor and mass. This rigging mass must also be lifted. The motor wattage rating can be output mechanical power or input electrical power. You will also have to provide energy to accelerate or more accurately move the mass at a velocity vertically. This kinetic energy to move the mass must also be stored or dissipated to put the mass at rest at the new elevation. The number you calculated is only the amount of energy stored in lifting a mass, it is not the amount of energy required to store this energy. Additional mechanical losses from bearing friction and rigging effects will also affect your results. I would be surprised if all of these additional power robbers will sum to 10% of the lifted mass energy but they must always be there.

"Your calculations are correct but simplistic."

Thanks for checking the figures.

I accept the charge that they were too simplistic, but to be honest, I was aware of the working losses, but I didn't want estimates to cloud the questioning of the 100w light bulb ON for 98 seconds, equalling 20kg being lifted 50m.

Allowing for losses, it is still pretty surprising, if we use 120w bulb as an example.

re: ramconsult....... I had been thinking about pumped storage as we see these all over when driving thru the Pyrenees, so it was a useful link.

I noted the statement: 1000 kilograms of water (1 cubic meter) at the top of a 100 meter tower has a potential energy of about 0.272 kW·h

Not much at all really.

So I checked out the linked page to Okinawa Yanbaru pumped storage.

They claim 30MW max output, from an effective drop of 136 m and maximum flow of 26 m³/s.

26,000Kg at 136m = 34652800 joules

I'm presuming it is just divided by 1sec with 4 652 800w lost

The final velocity calculates to 51.63 m/s

However, I don't believe this can be true for a liquid.
What does anybody think?

A column of water travelling 50m/s at the bottom, must be travelling at the same speed at the top.

hmm but with the syphon effect, perhaps the whole column does hit 50m/s (less resistance).

RE: gravel storage........ WOW!

This one had passed me by, but to be fair...... it's not lifting gravel between upper and lower hoppers (but it's still mind boggling).

Here's a link.

Did you miss the recent article where in England (rather UK) they are planning to use gravel as the energy storage medium, and using bucket conveyor system to rasie between lower and upper hopper bins? String along 100-1000 of these and the MWh of storage capacity really adds up to a fairly decent value. Low tech solution to high tech renewable' need for energy storage for levelling purposes indeed. Works in the most arid of places where solar energy is abundant, and some of the best wind areas as well.

Interesting idea. No loss of stored energy due to evaporation. Gravel will not level itself like any fluid will. I wonder how they distribute and retrieve the gravel?

They heat it using compressed gas, and uncompress the gas to the 2nd cell (freezing it) creating a potential difference.

bucket conveyor system similar to what would be used in a coal mine, or similar technology to a cable car system, only with many, many buckets on it, and a tilting system at each bin to dump on which ever is the active end (storage or generation).

First of all, indeed, the stored (potential) energy (which is equal to the work that you've done in order to lift the mass) is E=m_*g_*h=[20kg]_*[9,81m/sec²]_*[50m]=9810joules and, of course, it is equal to the energy consumed by a 100W lamp working for 98,1sec. It seems (to you) that you made a great effort in order to lift the mass. So, how can this be equal to a small lamp working for only 1,5 min? Well, have you ever tried to power a device by rotating a small magneto-generator using your hands??? I have tried this in army, when I had to rotate such a generator in order to give power to a ham-radio. It was not easy at all. (And the ham radio was probably consuming less than 100W.) So, try to power a 100W lamp using such a generator, using your hands, for 1,5min. I'm sure that you will, also, find out that the effort is great.

Yes you're right of course.
The example reversed shows just how much work is required to generate 100w.

It puts into context the energy that can be released by just a few ml of refined oil.

I did a calc of 1 000 000 kg (of say water) @ 10m = 98000000j / 86400sec = 1.134KW

To get the water back up in 24hrs would require l/min 1000000l/1440min = 694l/min

So I did a quick search on 1KW water pumps l/min and found this.

1.1 KW 600 l/min @2m head.
@11m it can only deliver 100 l/min.

Absolutely nowhere near the required 694 l/min. around 80% loss!
Perhaps there are better pumps for the job