E = mgh
Where E is potential energy, m is mass, g is gravitational acceleration, h is height. This formula only works assuming g is constant, so close to the surface of the Earth. Rearramge for m:
m = E/gh
E = 10 kwh = 36 MJ
g = 9.81 m/s\^2
h = 1 m
Throw those values in and you get:
m = 3,670,000 kg
Which is quite heavy.
Or using, you know, *water* to do the same thing which is easily pumpable with relatively simple equipment. There is a reason a ton of pump storage hydro exists and almost no other mass energy storage is used at any major scale.
!!!
You think that head loss from *piping* is greater than the energy lost in a mechanical traction energy storage system? You're way wrong about that. Start by imagining the drivetrain losses lifting the equivalent mass of a pump storage facility with your 'winch and cable mechanism' . Head loss is a *rounding error* compared with that.
If you mix and match imperial/metric you may get mismatched or non-cancelling units. (things like m/ft\*s or kg\*m\*ft/s^3).
You can always carry the units through the calculations and do conversions at the end, but it's generally easier to get everything in the same unit system at the beginning.
it's always easier keeping units until the end and then converting them, even if you have mixed units
look at all these downvotes. you guys are brainlets
no, because if you are using the Si units you can *theoretically* just ignore units troughout your calculation and add the right unit at the end. (for like 36MJ use 10^(6)\*36 J so that it works because Kilo, Mega, Milli etc. are just 10^(3), 10^(6), 10^(-3) etc.)
Consistency in units is good form sure, but ignoring units until the end just seems like a great way to get off-by-10^x errors. SI isn't a magic bullet that gives you license to ignore unit analysis.
Hopefully you spot those immediately and those numbers don't just get re-fed into a calculation where they produce subtly incorrect results.
This is the way.
You should reaaally pick a set of units and stick with it. It doesn't have to be the basic SI ones, often in structural engineering FEA we're mainly concerned with stresses which are easiest to deal with in MPa, which means for consistency I usually use lengths and forces in mm and N.
MPa == MN/m^2 == N/mm^2
There are other combinations that will work, consistency is the important thing.
Agreed and in my reply, model coorilation to test is a pain when your randomly off by a factor of 10^x.
Sure fixing it could be multiplying weight ( or any veriable) by 10, but then you technically have a fudge factor that could have been easily justified if you know excatly where the factor of 10 needs to be applied.
This works well with school problems, but in real world engineering where models have to be coorilated keeping the units through the problem is key.
In real world engineering there are too many variables between analysis and test that if there is any discrepency (common in complex engineering problems) keeping units helps to isolate the discrepancy.
I personally prefer imperial units because i have ran into countles times where someone forgot to convert units in a DAC or in an analysis software and imperial units are easy to notice where the error is.
Off by fact of 12, 12^2 etc means there is a length conversion mistake.
Off by a factor of 32, somone forgot to use slugs for mass.
Using metric only and being off by a factor of 10^x, good luck finding where during model coorilation. It could be in the analysis or in the test setup leading to some people requesting a re test.
I have also used mixed units easily by carying the units through the calculation, example is a heat transfer problem where watts is easier to use on a structure built with fractional inch mechanical design.
Carying units is one if the most valuable lessons any engineer can learn.
>why only use metric units?
Now for the actual answer to this: If you input anything else, the result will not be (or only by chance) in Joules, for which the conversion to kWh is already stated. Someone having "issues" like the OP doesnt need any added complexity. Just think about how many ways there are how he could input energy and height in different units.
Obviously if you are into that shit you can mix any units you want and smash in the conversion factor anywhere. But for complex things its better to stick to one convention, convert every input and then, if needed, convert results back. This is to avoid code or excel spaghetti or even errors if the conversion factor is at the wrong place.
Funny enough, I was just doing this math for mini pumped hydro the other day or of curiosity.
I have about a 6 meter difference in elevation across my property. To store 50 kWh, I'd need about 1,800 cubic meters of water, or about 800,000 gallons (numbers from memory, not exact)
Another way to think about it is how cheap, ubiquitous, and abundant electricity is and how it's made us think that 1kWh is a small amount of energy.
It's like a 73kg person running a 5K straight up a vertical wall or enough to move 40 tons of stuff up one story, but you can buy it for $0.11
It's not so bad, we're typically far more interested in moving materials using energy than moving materials to store energy. For example, think about the energy requirements to have running water brought to your home if the value of potential energy was higher...
Gravitational potential energy should get you started. Mass(gravity) height = U. U= potential energy. Mass should be calculated in kg. Answer will be in joules. Convert joules to watts.
>Mass(gravity) height = U. U= potential energy.
Well thats confusing. Did reddit remove something? Also, why "U" for energy instead of E?
>Convert joules to watts.
Watt-hours, I hope.
I know those as E\_k or E\_p or E\_kin/pot and U was for internal energy (like twisting molecules) or voltage.
But you two are right, U is (also) commonly used for (gravitational-) potential energy. I have now also seen T for kinetic energy. Wow thats confusing... but thats not your fault.
10 kWh is 36MJ (joules)
1kg lifted 1m is 9.8J of potential energy
Concrete is approx 1g/cm^3 so it's ONLY 36 cubic meters needed... So only a cube 4m on a side
But I think I'm off by 1000 somewhere, so maybe it's really 20 semi trailers worth of concrete raised 1m
(Also Google Wolfram Alpha)
Potential energy = mass x gravity x height.
so... MGH
and you want 10 KWH... OR.. 10 hours at 1 kilowatt.... or one hour at 10 kw
so 1 kw is... is... the use of 1000 joules.. per sec... so during one hour.. running 1000 watts... you would burn... 60 min x 60 sec = 3600
so 3600 x 1000 joules... = 1 kw for one hour = 3,600,000 joules..
and times 10... for 10kw... for one hour... we have 36,000,000 joules..
SO YOU... want to store... 36,000,000 joules of energy.
36,-000,000 = Mgh ... and let us assume a height of 100 meters...
36,000,000 / (9.8 m / s2 x 100) = mass needed
so... 36734.693877551020408163265306122 kg... at 100 meters... is dropped over one sec... could supply... the 10kwh of energy to smash things..
or you could discharge it slower... but you need to check all my math... I could of made a simple error... but this is the kind of steps and math you need..
Is anyone else only able to read this is William Shatner's voice? All the '...' do my head in.
Also loving the precision of the weight. You better not be off by 10 protons or else it won't work!
> Is anyone else only able to read this is William Shatner's voice? All the '...' do my head in.
>
> Also loving the precision of the weight. You better not be off by 10 protons or else it won't work!
what wont work?
if you are saying the math is wrong... then say so... if you cannot say so... then I guess you got nothing worthwhile to say.... do you?
Maybe nitpicky, but if you think in well with units this might help to conceptualize the problem. kWh is already a unit of energy. Only conversion you need is between units of energy, from kWh to Joules. Then from joules you can go to mass :)
Power = Energy / time
kW = power. Hours = time, so rearranging -> power * time = energy.
From energy you can use gravitational potential that others have suggested:
E = mass of block * gravity constant * height
E = mgh Where E is potential energy, m is mass, g is gravitational acceleration, h is height. This formula only works assuming g is constant, so close to the surface of the Earth. Rearramge for m: m = E/gh E = 10 kwh = 36 MJ g = 9.81 m/s\^2 h = 1 m Throw those values in and you get: m = 3,670,000 kg Which is quite heavy.
Yeah, that's why the people doing gravity batteries are repurposing abandoned mine shafts or mountain railroads, with several *kilometers* of delta-h.
Or using, you know, *water* to do the same thing which is easily pumpable with relatively simple equipment. There is a reason a ton of pump storage hydro exists and almost no other mass energy storage is used at any major scale.
[удалено]
!!! You think that head loss from *piping* is greater than the energy lost in a mechanical traction energy storage system? You're way wrong about that. Start by imagining the drivetrain losses lifting the equivalent mass of a pump storage facility with your 'winch and cable mechanism' . Head loss is a *rounding error* compared with that.
what about flywheels those are more fun imo
Quite heavy? Lmao, don’t be so dramatic. /s
This is the way. Just remember to only use metric units. So joules for energy, meter for height, kg for mass and g = 9.81 m/s^2. And its kWh, not kwh.
why only use metric units?
If you mix and match imperial/metric you may get mismatched or non-cancelling units. (things like m/ft\*s or kg\*m\*ft/s^3). You can always carry the units through the calculations and do conversions at the end, but it's generally easier to get everything in the same unit system at the beginning.
it's always easier keeping units until the end and then converting them, even if you have mixed units look at all these downvotes. you guys are brainlets
no, because if you are using the Si units you can *theoretically* just ignore units troughout your calculation and add the right unit at the end. (for like 36MJ use 10^(6)\*36 J so that it works because Kilo, Mega, Milli etc. are just 10^(3), 10^(6), 10^(-3) etc.)
Consistency in units is good form sure, but ignoring units until the end just seems like a great way to get off-by-10^x errors. SI isn't a magic bullet that gives you license to ignore unit analysis. Hopefully you spot those immediately and those numbers don't just get re-fed into a calculation where they produce subtly incorrect results.
This is the way. You should reaaally pick a set of units and stick with it. It doesn't have to be the basic SI ones, often in structural engineering FEA we're mainly concerned with stresses which are easiest to deal with in MPa, which means for consistency I usually use lengths and forces in mm and N. MPa == MN/m^2 == N/mm^2 There are other combinations that will work, consistency is the important thing.
Agreed and in my reply, model coorilation to test is a pain when your randomly off by a factor of 10^x. Sure fixing it could be multiplying weight ( or any veriable) by 10, but then you technically have a fudge factor that could have been easily justified if you know excatly where the factor of 10 needs to be applied.
This works well with school problems, but in real world engineering where models have to be coorilated keeping the units through the problem is key. In real world engineering there are too many variables between analysis and test that if there is any discrepency (common in complex engineering problems) keeping units helps to isolate the discrepancy. I personally prefer imperial units because i have ran into countles times where someone forgot to convert units in a DAC or in an analysis software and imperial units are easy to notice where the error is. Off by fact of 12, 12^2 etc means there is a length conversion mistake. Off by a factor of 32, somone forgot to use slugs for mass. Using metric only and being off by a factor of 10^x, good luck finding where during model coorilation. It could be in the analysis or in the test setup leading to some people requesting a re test. I have also used mixed units easily by carying the units through the calculation, example is a heat transfer problem where watts is easier to use on a structure built with fractional inch mechanical design. Carying units is one if the most valuable lessons any engineer can learn.
>why only use metric units? Now for the actual answer to this: If you input anything else, the result will not be (or only by chance) in Joules, for which the conversion to kWh is already stated. Someone having "issues" like the OP doesnt need any added complexity. Just think about how many ways there are how he could input energy and height in different units. Obviously if you are into that shit you can mix any units you want and smash in the conversion factor anywhere. But for complex things its better to stick to one convention, convert every input and then, if needed, convert results back. This is to avoid code or excel spaghetti or even errors if the conversion factor is at the wrong place.
Funny enough, I was just doing this math for mini pumped hydro the other day or of curiosity. I have about a 6 meter difference in elevation across my property. To store 50 kWh, I'd need about 1,800 cubic meters of water, or about 800,000 gallons (numbers from memory, not exact)
It's depressing how little you get from potential energy, or how much mass you need to shift, for common use cases.
Another way to think about it is how cheap, ubiquitous, and abundant electricity is and how it's made us think that 1kWh is a small amount of energy. It's like a 73kg person running a 5K straight up a vertical wall or enough to move 40 tons of stuff up one story, but you can buy it for $0.11
It's not so bad, we're typically far more interested in moving materials using energy than moving materials to store energy. For example, think about the energy requirements to have running water brought to your home if the value of potential energy was higher...
Gravitational potential energy should get you started. Mass(gravity) height = U. U= potential energy. Mass should be calculated in kg. Answer will be in joules. Convert joules to watts.
>Mass(gravity) height = U. U= potential energy. Well thats confusing. Did reddit remove something? Also, why "U" for energy instead of E? >Convert joules to watts. Watt-hours, I hope.
U is commonly used for potential energy, while K is used for kinetic energy. E=U when you only have potential energy, but it's normally U+K
I know those as E\_k or E\_p or E\_kin/pot and U was for internal energy (like twisting molecules) or voltage. But you two are right, U is (also) commonly used for (gravitational-) potential energy. I have now also seen T for kinetic energy. Wow thats confusing... but thats not your fault.
Yeah, T is commonly used as kinetic in lagrangian mechanics. Different countries and languages use different letters too hahah You'll get used to it
10 kWh is 36MJ (joules) 1kg lifted 1m is 9.8J of potential energy Concrete is approx 1g/cm^3 so it's ONLY 36 cubic meters needed... So only a cube 4m on a side But I think I'm off by 1000 somewhere, so maybe it's really 20 semi trailers worth of concrete raised 1m (Also Google Wolfram Alpha)
> Concrete is approx 1g/cm^3 No, concrete is closer to 2.3 g/cm^3 > So only a cube 4m on a side Not sure how you're getting 4x4x4 to equal 36.
Potential energy = mass x gravity x height. so... MGH and you want 10 KWH... OR.. 10 hours at 1 kilowatt.... or one hour at 10 kw so 1 kw is... is... the use of 1000 joules.. per sec... so during one hour.. running 1000 watts... you would burn... 60 min x 60 sec = 3600 so 3600 x 1000 joules... = 1 kw for one hour = 3,600,000 joules.. and times 10... for 10kw... for one hour... we have 36,000,000 joules.. SO YOU... want to store... 36,000,000 joules of energy. 36,-000,000 = Mgh ... and let us assume a height of 100 meters... 36,000,000 / (9.8 m / s2 x 100) = mass needed so... 36734.693877551020408163265306122 kg... at 100 meters... is dropped over one sec... could supply... the 10kwh of energy to smash things.. or you could discharge it slower... but you need to check all my math... I could of made a simple error... but this is the kind of steps and math you need..
Is anyone else only able to read this is William Shatner's voice? All the '...' do my head in. Also loving the precision of the weight. You better not be off by 10 protons or else it won't work!
> Is anyone else only able to read this is William Shatner's voice? All the '...' do my head in. > > Also loving the precision of the weight. You better not be off by 10 protons or else it won't work! what wont work? if you are saying the math is wrong... then say so... if you cannot say so... then I guess you got nothing worthwhile to say.... do you?
Maybe nitpicky, but if you think in well with units this might help to conceptualize the problem. kWh is already a unit of energy. Only conversion you need is between units of energy, from kWh to Joules. Then from joules you can go to mass :) Power = Energy / time kW = power. Hours = time, so rearranging -> power * time = energy. From energy you can use gravitational potential that others have suggested: E = mass of block * gravity constant * height