Boost Your Power Calculations with Expert Help | Energy & Power Experimentation

[Don Bori]

Hello there! So confused here, I tried to do some calculations with my power generation experimention.

In terms of energy input, the energy output is lower.

But in terms of power input, the power output is way larger.

Thanks in advance! Cheers!

 

[anorlunda]

But in terms of power input, the power output is way larger.

Perhaps you said it wrong. You can not violate conservation of energy.

 

[Guineafowl]

Summary: What is the basis of feasible Power Generation, is it Power efficiency or Work efficiency?

In terms of energy input, the energy output is lower.

But in terms of power input, the power output is way larger.

Re-check (and post) your calculations - energy and power are two ways of looking at the same thing. One is a unit, the other a rate of delivery of that unit.
If you’re getting different results from those two different angles, and in doing so violating conservation of energy, work back to find your mistake.

 

[Fisherman199]

Please post these calculations. Conservation of energy will always hold.

As far as a basis for feasible power generation is concerned, there's much more to consider than just power input and received output. One must consider the type of load; current and projected markets; load growth; proximity of generation to fuel source, load, other generation, and transmission infrastructure; impact on grid stability; etc. What is the goal, here?

 

[Don Bori]

Thank you guys for your comments and suggestions, upon recalculations, I came up with this clearer query.
If a 1-liter water is elevated by 20 meters, it will have a Potential Energy of 200 Joules.
When I use a Water Pump to elevate that 1-liter water at 20 meters, it will require a Work of 200 Joules also. And if I pump the 1-liter water in 1 minute, it will require a Power of 0.33W.
Then when I use the mass of 1-liter water as load to a generator when going downwards in 1 minute, it will produce a Power of 3.33W. Which is 10x the input power of Water Pump.
If the generator allocates 0.33W for the Water Pump, then there is still 3W (90%) of Power being generated.
So how come that in terms of Energy Conservation it will be 0 and even negative in actual scenario due to losses, while the Power equation is significantly positive?

1-liter Water​
Volume, li.​	Volume, m3​	DensityWater, kg/m3​	Mass, kg​	Height, m.​	Gravity, m/s2​	Energy, kgm2/s2​	​	​	​	Work (Energy) = mass * gravity * height
1.00​	0.0010​	1,000.00​	1.00​	20.00​	10.00​	200.00​	​	​	​	Power = Work / Time
Water Pump​
Diameter, m​	Area, m2​	Volume, m3​	Volume, li.​	Stroke, m.​	Pressure, kg/ms2​	Force, kgm/s2​	Work, kgm2/s2​	Time, s​	Power, W​	Must pump 1 liter of water into the tank.
0.1596​	0.0200​	0.0010​	1.0000​	0.05​	20,000.00​	400.00​	20.00​	60.00​	0.33​	​
Generator​
Mass, kg​	Gravity, m/s2​	DiameterPinion, m​	Torque​	Circumference, m​	Height, m.​	Speed, rpm​	Speed, rps​	​	Power, W​	1 rev = 0.31 m
1.00​	10.00​	0.10​	0.50​	0.31​	20.00​	63.66​	1.06​	​	3.33​	​

Thank you and looking forward to my enlightenment! This is so confusing to me, I know my analysis must be wrong, but I don't know where. =)
Cheers!

 

[davenn]

When I use a Water Pump to elevate that 1-liter water at 20 meters, it will require a Work of 200 Joules also.

There's part of the problem ... it will require more than 200 Joules. You haven't accounted for losses
in the system ... the pump for example will not be 100% efficient, it may only be 70%.
Not knowing anything about your pump ( you haven't stated) it may require say 250 J to lift that 1l
of water to 20 metres ... it may be much more depending on the pump inefficiency.

Then when I use the mass of 1-liter water as load to a generator when going downwards in 1 minute, it will produce a Power of 3.33W. Which is 10x the input power of Water Pump.

If the generator allocates 0.33W for the Water Pump, then there is still 3W (90%) of Power being generated.

The moment you think you have more power out than in, then you need to re-evaluate what is happeningcheers
Dave

 

[Don Bori]

Hi Dave! Thanks to your inputs.

That's what mainly confuses me and I know it will never ever be 100% efficient for the pump. Considering that the pump is too inefficient that instead of only 0.33W input power, it uses 3W. But the power generated is 3.33W, so does it mean there's still 0.33W power that can be harnessed?

There's part of the problem ... it will require more than 200 Joules. You haven't accounted for losses
in the system ... the pump for example will not be 100% efficient, it may only be 70%.
Not knowing anything about your pump ( you haven't stated) it may require say 250 J to lift that 1l
of water to 20 metres ... it may be much more depending on the pump inefficiency.
The moment you think you have more power out than in, then you need to re-evaluate what is happeningDave

 

[davenn]

But the power generated is 3.33W, so does it mean there's still 0.33W power that can be harnessed?

Looking through your tables in post #5, I am not sure if you have what it needs to calc the power out ?
I don't see it standing out clearly, so I am interested in how you came to 3.33W generated ?

If you have again just been relying on basic figures as you did for the pump, then you have also
forgotten that your generator is not 100% efficient ... it may only be 50% ...
Again you haven't given any info on this generator of yours. ... the details as you are starting to see
are very important, else all your calculations are based on false information

I suspect you will find that you generated power is substantially less than 3.33 Wcheers
Dave

 

[davenn]

Ohhh just noticed this...

If the generator allocates 0.33W for the Water Pump, then there is still 3W (90%) of Power being generated

That ISNT going to work ... the generator is NEVER going to produce enough energy to power the
pump let alone have something left over to power other things

That is another flaw in your system

there's no free lunch in physics

As Scotty from Star Trek (TOS) said on more than one occasion ...
"Ye canna break the laws of physics, Captain"

 

[Don Bori]

You're right and I believe also that there are flaws in my understanding of this concept that's why I want to figure out with the help of you and other experts.

But I prefer to based my understanding in mathematical figures rather than just telling myself it's WRONG because that's what I've been learned in school. I know it's wrong, but part of learning is to know the details.

I considered the formula for power generated, Power = 2*(3.1416)*Torque*Speed, so I came up with 3.33W.

For the Torque, I assumed to use a 0.1-meter pinion/sprocket and used the formula, Torque = Force*radius.

And for the Speed, since the elevation is 20 meters, I divided it with the circumference of the pinion/sprocket to get the revolutions and assumed that the 1-liter water reaches the bottom in 1 minute.

Ohhh just noticed this...
That ISNT going to work ... the generator is NEVER going to produce enough energy to power the
pump let alone have something left over to power other things

That is another flaw in your system

there's no free lunch in physics

As Scotty from Star Trek (TOS) said on more than one occasion ...
"Ye canna break the laws of physics, Captain"

 

[anorlunda]

You set off all our alarm bells by expressing the problem as you did. Perpetual motion and over-unity energy gain are forbidden topics on PF.

You have two methods of calculating.

1. force*distance/time
2. torque*speed

They disagree by a factor of 10. You could have swapped and used force*distance/time for the generator and torque*speed for the motor, and you would have been off by 10 with more motor power than generator power.

You could also have applied both methods to just pumping mode (or just generating mode), and seen that they disagree. No need to introduce the other mode at all.

I did not go through your numbers to locate the source of the error. But I do object to the way you phrased the questions, and especially the title. I'm going to change the title so something less confusing for future users who discover this thread via Google search.

Edit: I also changed the summary. There is no reason to include efficiency in this topic.

 

[Guineafowl]

And for the Speed, since the elevation is 20 meters, I divided it with the circumference of the pinion/sprocket to get the revolutions and assumed that the 1-liter water reaches the bottom in 1 minute.

You can’t do that, and I think there’s your problem. This assumes a kind of belt 20m long driving the pinion 20/0.3 times in a minute, ignoring the kinetic energy of the water.

I imagine the power calcs for a water turbine are quite complicated if you try to estimate torque and deceleration of the water and so on. But you don’t need to.

Water pumped up gains gravitational potential energy = mgh = 1x10x20 = 200 joules.
200 joules over 60 seconds = 3.33W. You got this wrong.

Water falling back down gains kinetic energy of 200J (less losses), which is available for generation.

Rates can get confusing. IN ONE SECOND, 1/60 litre of water falls 20m. Its energy mgh = 1/60x10x20 = 3.33J. So the trickling water converts gravitational potential energy to kinetic at a rate of 3.33J/s, or 3.33W.

Next stage is the generator, which converts kinetic to electrical energy. But you can see it can’t do that at a rate greater than the input. After all, it can’t create energy, merely convert from one form to another.

 

[Don Bori]

Finally, I got what I was looking for!

It made me easier to understand seeing some figures like this, thanks Guineafowl!

Thanks to all of you guys!

Good days ahead!

You can’t do that, and I think there’s your problem. This assumes a kind of belt 20m long driving the pinion 20/0.3 times in a minute, ignoring the kinetic energy of the water.

I imagine the power calcs for a water turbine are quite complicated if you try to estimate torque and deceleration of the water and so on. But you don’t need to.

Water pumped up gains gravitational potential energy = mgh = 1x10x20 = 200 joules.
200 joules over 60 seconds = 3.33W. You got this wrong.

Water falling back down gains kinetic energy of 200J (less losses), which is available for generation.

Rates can get confusing. IN ONE SECOND, 1/60 litre of water falls 20m. Its energy mgh = 1/60x10x20 = 3.33J. So the trickling water converts gravitational potential energy to kinetic at a rate of 3.33J/s, or 3.33W.

Next stage is the generator, which converts kinetic to electrical energy. But you can see it can’t do that at a rate greater than the input. After all, it can’t create energy, merely convert from one form to another.

 

[Guineafowl]

Finally, I got what I was looking for!

It made me easier to understand seeing some figures like this, thanks Guineafowl!

Thanks to all of you guys!

Good days ahead!

The value of a good physics teacher. Mine taught me that whenever a rate is involved, such as a bathtub filling and draining at particular rates, simplify it by considering what happens in each second.

Your answer will then automatically remain a rate calculation, while maintaining the simplicity of a quantity one.

 

[Guineafowl]

Here’s a useful problem, which encompasses many of the concepts above. Feel free to answer or not, but at least have a think:

If I throw a ball in the air, and catch it, is the average speed faster on the way up, or the way down?

 

[Don Bori]

Considering my understanding of projectile motion, it should be the same. Glad to be corrected if I'm wrong. =)

 

[Guineafowl]

Think it through in terms of energy. Yes, under utterly ideal conditions it could be the same, but not in this universe.

Ball leaves hand, it has X kinetic energy (KE) and 0 potential energy (PE).
During upward travel, it’s converting KE to PE.

At apex, it now has 0 KE and X PE. Less losses.

Downwards to hand, it converts PE to KE, but can only draw on whatever energy it had at apex.

Since KE is proportional to v[squared], you can use the energy conversions to say that only if the conversion from KE to PE and back again was perfect will the average speed be the same. Any losses from friction or air resistance, distortion of the ball etc., will ensure that the downward jorney MUST have a lower average speed.

The point of this is, energy conservation can cut through a lot of seemingly complex or hard-to-visualise situations.

 

[russ_watters]

When I use a Water Pump to elevate that 1-liter water at 20 meters, it will require a Work of 200 Joules also. And if I pump the 1-liter water in 1 minute, it will require a Power of 0.33W.

Then when I use the mass of 1-liter water as load to a generator when going downwards in 1 minute, it will produce a Power of 3.33W. Which is 10x the input power of Water Pump.

You did the math wrong in the first case. The power in both cases should be 3.33W. You didn't post the actual calculations, but in your table your pump pressure is low by that same factor of 10. Perhaps you forgot to multiply by g?

 

[Don Bori]

I used the formula of Pressure = mass x gravity x height

You did the math wrong in the first case. The power in both cases should be 3.33W. You didn't post the actual calculations, but in your table your pump pressure is low by that same factor of 10. Perhaps you forgot to multiply by g?

 

[russ_watters]

I used the formula of Pressure = mass x gravity x height

And the numbers associated with that are...?

 

[Don Bori]

Hi Russ! The formula for pressure instead is equal density x gravity x height (1,000 x 10 x 20) giving a pressure value of 200,000 kg/ms2.

Yes, in my calculations the Pump Input Power and Generator Output Power are the same if the Pumping Time is the same with the Generating Time. But what if the pumping is not urgent, the generating power will be higher than the Pumping power, therefore can we use or store that power in a battery?

And the numbers associated with that are...?

 

[Guineafowl]

Hi Russ! The formula for pressure instead is equal density x gravity x height (1,000 x 10 x 20) giving a pressure value of 200,000 kg/ms2.

Yes, in my calculations the Pump Input Power and Generator Output Power are the same if the Pumping Time is the same with the Generating Time. But what if the pumping is not urgent, the generating power will be higher than the Pumping power, therefore can we use or store that power in a battery?

Don Bori has given me some specifics of the above suggestion, to which I have replied as below. I’ve posted it here in the hope that a real physicist or engineer might chip into help clear up his confusion. I don’t believe this is a free energy scam, merely someone playing with the figures (which is a good thing) and coming up with surprising answers.

*****

Short answer? No. Energy out cannot exceed energy in. Look up the first law of thermodynamics.

Longer answer - In one second, your pump supplies 0.42 J of energy. The generator outputs 1.67 J. This is not possible continuously, as above.

If you ran the pump for 4 seconds with the tank outlet blocked, it would supply 4 x 0.42 = 1.68 J.
Shut the pump off, then open the outlet, and the generator would run for 1 second, outputting that 1.68 J.

See the 4 seconds/1second terms I’ve underlined? Also see the same 1.68 J at the end?

Energy (J) is a constant thing we can’t create or destroy. 1.68 J is what you have. All you’ve done with this new suggestion is stretch the time over which it is delivered, manipulating the power (W) or rate that it is delivered. But changing the time domain does not alter the energy one.

Your new set-up trickles energy in at a rate of 0.42W. You can either run this continuously to output 0.42W (less losses), or run it in batches, outputting four times the power but for 1/4 the time.

 

[russ_watters]

Yes, in my calculations the Pump Input Power and Generator Output Power are the same if the Pumping Time is the same with the Generating Time. But what if the pumping is not urgent, the generating power will be higher than the Pumping power, therefore can we use or store that power in a battery?

Sure, but I don't see why that would be a useful thing to do.

 

[Guineafowl]

the generating power will be higher than the Pumping power

The generator power would be limited by the pumping power - is that a better answer? A dripping tap will fill a pint glass eventually, and you can swallow the pint of water much quicker than the drip rate. Slow the drip rate down, and it will take longer to fill the glass. Your swallowing rate will be the same, but your overall water intake rate (cf generator power) cannot exceed that of the dripping tap supply (cf pumping power).

Sure, but I don't see why that would be a useful thing to do.

 

[anorlunda]

Pumping power, therefore can we use or store that power in a battery?

Did you notice my signature? Power is to Energy as Speed is to Distance

You can run to the store and walk back. So what? I would not call that a novel idea.

You can collect energy all day long with your solar panel, then use it in 5 minutes warming your sauna. So what?