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Potential energy of suspended mass expressed in kWh

CircutScoper

Mar 29, 2022
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Well then, if friction causes heat, how long does it take a monkey to screw a bucket of water (10 litres) to boiling point.

Turns out, that experiment (or one very similar to it) was actually done and the results published in 1798.

Heat is a Form of Motion: An Experiment in Boring Cannon
Benjamin Thompson (Count Rumford)
Philosophical Transactions (vol. 88), 1798

http://spiff.rit.edu/classes/phys314/refs/dbhs.wvusd.k12.ca.us/Chem-History/Rumford-1798.html

To convert Thompson's results to your units, I'll rely on well documented physical facts (citations provided)...

1. The well-known thermodynamic temperature reference point: Cold enough to freeze the balls off a brass monkey.
https://www.phrases.org.uk/meanings/cold-enough-to-freeze-the-balls-off-a-brass-monkey.html
and...
2. 10 liters = 2.64172 gallons.

https://www.thecalculatorsite.com/conversions/liquidvolume/liters-to-gallons-(us).php

Therefore, while in Thompson's historic experiment the orifice being drilled was the bore of a brass cannon (instead of a bucket), and the tool used was a dull bit instead of a monkey's -- um -- tool, the similarity of the materials involved and the almost identical volumes of water (i.e., 2.64 vs 2.25 gallons) clearly allows direct transfer of the quantitative duration of screwing intervals to achieve the boiling point.

Viz: "At two hours twenty minutes it was 200 °F; and at two hours thirty minutes it actually boiled!"

So there you have it: 2.5hr. Ain't science grand?
 
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davenn

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I do think you are correct about kilo abbreviation. kWHr not KWHr

that is correct ... lower case k ... I took the liberty of editing one of your previous posts :)
 

davenn

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So you are expressing the m in PE = mgh in SI units not in traditional units where the mass base unit is the gram and not the kg. Hence, 1000 x 9.8 x 3 = 29,400.

no, it is mgh = mass in kg and you plug in what ever the mass is that you are dealing with :)
so if the mass in 1kg. 10kg etc
if you have 500 grams, then that is 0.5kg

so .... 0.5 x 9.8 x h (whatever that is in your experiment)
 

Bluejets

Oct 5, 2014
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Turns out, that experiment (or one very similar to it) was actually done and the results published in 1798.

Which more than likely turns out to be the obvious answer.

Same time as it takes to shove 500g butter up an elephant's butt with a red hot needle.:)
 

HANKMARS

Jul 28, 2019
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Here is the skinny. Those that prefer to work in SI units know to change an actual real mass of 1000g to 1g in their working formula. What has not been mentioned by those same people is that the results of their calculations need an adjustment of adding 3 to the exponent of their result. Probably much in the same spirit that I use the term Amp instead of Ampere and see no reason to explain why. A measured mass of 500g becomes .5g when using SI units. So when I am dealing with a Mg, it is plugged into the formula as 1000g. And the final (nearly final) result is what is offered up as the correct result. Kind of a rude thing to do to a genetically flawed orangutan.
 
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Harald Kapp

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A measured mass of 500g becomes .5g when using SI units.
That is nonsense. 500 g are 500 g. In SI units. Or 0.5 kg.
So when I am dealing with a Mg, it is plugged into the formula as 1000g.
Again nonsense and not related to either SI or imperial units.
1 M is the prefix for Mega which means 10^6.
So 1 Mg = 1000 kg = 1 000 000 g

Just because 1 kg is the base unit doesn't mean that the prefixes like k, M, m etc. lose or change their meaning. 1 kg = 1000 g because k ist the prefix for kilo meaning 1000.
 

davenn

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Here is the skinny. Those that prefer to work in SI units know to change an actual real mass of 1000g to 1g in their working formula. What has not been mentioned by those same people is that the results of their calculations need an adjustment of adding 3 to the exponent of their result. Probably much in the same spirit that I use the term Amp instead of Ampere and see no reason to explain why. A measured mass of 500g becomes .5g when using SI units. So when I am dealing with a Mg, it is plugged into the formula as 1000g. And the final (nearly final) result is what is offered up as the correct result. Kind of a rude thing to do to a genetically flawed orangutan.

seems you didnt read my previous 2 posts ;) ;)
 

HANKMARS

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That is nonsense. 500 g are 500 g. In SI units. Or 0.5 kg.

Again nonsense and not related to either SI or imperial units.
1 M is the prefix for Mega which means 10^6.
So 1 Mg = 1000 kg = 1 000 000 g

Just because 1 kg is the base unit doesn't mean that the prefixes like k, M, m etc. lose or change their meaning. 1 kg = 1000 g because k ist the prefix for kilo meaning 1000.
You get no argument from me. Tell it to these guys.

SI EXAMPLE 1.1.png
 

HANKMARS

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no, it is mgh = mass in kg and you plug in what ever the mass is that you are dealing with :)
so if the mass in 1kg. 10kg etc
if you have 500 grams, then that is 0.5kg

so .... 0.5 x 9.8 x h (whatever that is in your experiment)
I think your opinion makes sense but the physics' department has a limitless vault of details and bs with which to dazzle and baffle you. Please read post #49.
 

HANKMARS

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Lost? I was thrown somewhere between Narnia and a bad acid trip when I received the following reply to my question regarding potential energy of a one tonne mass suspended 3 meters above the Earth's surface.

PHYSICS FORUM CALC RESULTS 1.1.png
I rounded the value to 0.002 kWh which = 2Wh. Trying to visualize and understand the work required to raise a tonne 3 meters can be accomplished by providing a constant power flow of 2 watts for an hour, was starting to caramelize my onion. Couldn't see it. Converted value to Ws. 2Wh = 7,200Ws. Oh, 7,200 watts is a considerable amount of power, but raising a tonne 9ft in one second? Naw, still couldn't see it. I had to be missing something. Was this calc based on 0J point being the center of the Earth? Were forces such as the centripetal force that the Earth exerts on the sun in play here? I was thinking what Harold mentioned earlier, that even if the mass base unit is the kg, to be used per SI standards, that doesn't magically change 1,000,000g into 1000g, even if 1Mg is entered as 1000g, as is expected when using SI. I had to conclude that the person who provided this result was not proficient using SI units. A little later in another post reply, the definition of a kWh was stated as 1000 watts delivered in an hour. Not written as 1000 watts continuously delivered for an hour but 1000 watts in an hour. My confidence in this particular forum was waning. So, I have concluded that PE is derived using the formula PE = mgh. Plugging my values in it reads 1000kg X 9.8 X 3m = 29,400,000J. 29,400,000 / 3,600,000 = 8.17kWh. I'm going to run with that value. My goal is to recover those 8kWh for consumption. The energy transition from falling object to mechanical motion will exceed 90% efficiency, in theory, where as mechanical to electric energy can hopefully be achieved with 50% or above efficiency. If I can reach those values, 4kWh will meet my nominal daily power usage. That operates computer, security cameras, loudspeakers, lights. During summer months when running a/c unit, my usage approximately doubles. I appreciate your comments and assistance with my quandaries.
 

CircutScoper

Mar 29, 2022
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Lost? I was thrown somewhere between Narnia and a bad acid trip when I received the following reply to my question regarding potential energy of a one tonne mass suspended 3 meters above the Earth's surface.

View attachment 55359
I rounded the value to 0.002 kWh which = 2Wh. Trying to visualize and understand the work required to raise a tonne 3 meters can be accomplished by providing a constant power flow of 2 watts for an hour, was starting to caramelize my onion. Couldn't see it. Converted value to Ws. 2Wh = 7,200Ws. Oh, 7,200 watts is a considerable amount of power, but raising a tonne 9ft in one second? Naw, still couldn't see it. I had to be missing something. Was this calc based on 0J point being the center of the Earth? Were forces such as the centripetal force that the Earth exerts on the sun in play here? I was thinking what Harold mentioned earlier, that even if the mass base unit is the kg, to be used per SI standards, that doesn't magically change 1,000,000g into 1000g, even if 1Mg is entered as 1000g, as is expected when using SI. I had to conclude that the person who provided this result was not proficient using SI units. A little later in another post reply, the definition of a kWh was stated as 1000 watts delivered in an hour. Not written as 1000 watts continuously delivered for an hour but 1000 watts in an hour. My confidence in this particular forum was waning. So, I have concluded that PE is derived using the formula PE = mgh. Plugging my values in it reads 1000kg X 9.8 X 3m = 29,400,000J. 29,400,000 / 3,600,000 = 8.17kWh. I'm going to run with that value. My goal is to recover those 8kWh for consumption. The energy transition from falling object to mechanical motion will exceed 90% efficiency, in theory, where as mechanical to electric energy can hopefully be achieved with 50% or above efficiency. If I can reach those values, 4kWh will meet my nominal daily power usage. That operates computer, security cameras, loudspeakers, lights. During summer months when running a/c unit, my usage approximately doubles. I appreciate your comments and assistance with my quandaries.

The calculation might have gone smoother if you'd used 1000kg x 3m x 9.8N/kg = 29.4kJ = 0.00817kWhr.
 

HANKMARS

Jul 28, 2019
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Yeah, my buddy said maybe use ergs but I was unfamiliar with unit 10e-7 something.
So, if I have my calcs correct, I derive what seems to be a nonsensical conclusion. I am told it is because of unanticipated scaling problems. I'll look that phrase up later.
Given: AA battery; 1.2V, 2800mAh. Power = 1.2V X 2.8Ah = 3.36Wh. 3.36Wh = 0.00336kWh. 0.00336 X 3 = 0.01008kWh. This indicates that 3 AA batteries contain enough energy to lift a Mg 3 meters. A rather miraculous event but math is an exact science so there is no argument. Agreed? Or is there a grievous error in my calcs?
 

CircutScoper

Mar 29, 2022
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Yeah, my buddy said maybe use ergs but I was unfamiliar with unit 10e-7 something.
So, if I have my calcs correct, I derive what seems to be a nonsensical conclusion. I am told it is because of unanticipated scaling problems. I'll look that phrase up later.
Given: AA battery; 1.2V, 2800mAh. Power = 1.2V X 2.8Ah = 3.36Wh. 3.36Wh = 0.00336kWh. 0.00336 X 3 = 0.01008kWh. This indicates that 3 AA batteries contain enough energy to lift a Mg 3 meters. A rather miraculous event but math is an exact science so there is no argument. Agreed? Or is there a grievous error in my calcs?

No error that I can see.

Meanwhile, we have...

1 Joule = 10^7 ergs
1 bar = 10^5 Pascals
1 Tesla = 10^4 .Gauss

...which are at least as arbitrary, anti-mnemonic, and certifiably silly as anything you'd find in SAE.

Besides which, 12 has twice as many factors as 10.
 

Nanren888

Nov 8, 2015
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Here are the SI defenitions, from the posted PDF:
Interesting. We've just tossed our kg, so I thought that the new standard for kg had been accepted.

weights and measures have nothing to do with science. They're nothing but politics.
Well, maybe to do with trade between nations. But with that aim in mind, there is significant science in the definitions to make as many standards as possible based solely on fundamental constants that can be measured or defined; getting away from that rather arbitrary block of Pt and such.
 
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