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Energy Converted to Electricity

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ronandkryn
#1
Nov12-09, 11:08 PM
P: 15
Energy Converted to Electricity

A piston is ten feet in diameter and travels through a sleeve in one direction for a distance of exactly nine feet. The one-way trip travel time is six hours. It is then immediately pushed in the opposite direction for a return trip at the same distance and speed. In twenty-four hours the piston makes four trips. A constant force of 27,100 tons is exerted against the piston head at all times.

No consideration is given to weight, temperature, lubrication, drag, energy required to run the piston or to run the mechanical linkage or to run the electrical generator itself. Additionally, and no consideration is to be given as to the rated efficiency of the generator or any other factors. However, it would be nice to have a list of the important factors that have been omitted.

How much electricity will such a contraption manufacture within a 24 hour period? Except for the few parameters listed in the first paragraph, any assumptions can be made regarding "anything" that might make the computation [of electrical output] easier or more accurate. What is the approximate margin of error of such an estimate? 5%, 10%, 15%, 20%, 25%; or margin of error cannot be determined.
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stewartcs
#2
Nov13-09, 07:58 AM
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Quote Quote by ronandkryn View Post
Energy Converted to Electricity

A piston is ten feet in diameter and travels through a sleeve in one direction for a distance of exactly nine feet. The one-way trip travel time is six hours. It is then immediately pushed in the opposite direction for a return trip at the same distance and speed. In twenty-four hours the piston makes four trips. A constant force of 27,100 tons is exerted against the piston head at all times.

No consideration is given to weight, temperature, lubrication, drag, energy required to run the piston or to run the mechanical linkage or to run the electrical generator itself. Additionally, and no consideration is to be given as to the rated efficiency of the generator or any other factors. However, it would be nice to have a list of the important factors that have been omitted.

How much electricity will such a contraption manufacture within a 24 hour period? Except for the few parameters listed in the first paragraph, any assumptions can be made regarding "anything" that might make the computation [of electrical output] easier or more accurate. What is the approximate margin of error of such an estimate? 5%, 10%, 15%, 20%, 25%; or margin of error cannot be determined.
Define what you mean by "electricity". If you mean "power", then calculate the work done by the piston and divide it by the time. That will give you the power without any losses.

CS
Mech_Engineer
#3
Nov13-09, 08:45 AM
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Given a force and a distance that force is applied for, you can easily calculate work done (energy). I'm not sure a margin of error can be estimated with the information given unless you start making assumptions about the mechanical linkage and electrical generation method being used, but the total energy generation is around 730 kW-hr for the 24-hour period, and the power generation is 30 kW.

ronandkryn
#4
Nov13-09, 12:49 PM
P: 15
Energy Converted to Electricity

Thank you
Thank you
Thankyou
Ron
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Quote Quote by Mech_Engineer View Post
Given a force and a distance that force is applied for, you can easily calculate work done (energy). I'm not sure a margin of error can be estimated with the information given unless you start making assumptions about the mechanical linkage and electrical generation method being used, but the total energy generation is around 730 kW-hr for the 24-hour period, and the power generation is 30 kW.
ronandkryn
#5
Nov13-09, 03:03 PM
P: 15
Thank you again for such a quick response.

I just got back from meetings & had a chance to read through it.
I guess that maybe I am lost.
Basically, I am talking about a piston being moved at a constant rate of speed for twenty-four hours for a total distance of 36 feet; with a force against the piston head of
27,100 TONS. Are we talking about the same thing?


Quote Quote by Mech_Engineer View Post
Given a force and a distance that force is applied for, you can easily calculate work done (energy). I'm not sure a margin of error can be estimated with the information given unless you start making assumptions about the mechanical linkage and electrical generation method being used, but the total energy generation is around 730 kW-hr for the 24-hour period, and the power generation is 30 kW.
Mech_Engineer
#6
Nov13-09, 03:17 PM
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Work (energy) is force times distance, so to find the total amount of energy expended all you have to do is multiply the force times the distance (keep in mind you need the proper units). For example if you apply a force of 1N to a piston over a distance of 1m, that equates to 1J of energy expended.

For power, all you need to do is divide energy by time. So you take the energy expended on the piston, and divide it by the amount of time it took to expend that energy (make sure to watch your units). For example 1J/1s = 1W.
Bob S
#7
Nov13-09, 03:40 PM
P: 4,663
I got ~ 52 kW average for 8 nine-foot strokes in 24 hours. I would look at pushing water (viscousless?) through a small dia pipe that could run a hydraulic pump/motor at maybe 900-1200 RPM(?).
Bob S
stewartcs
#8
Nov13-09, 03:53 PM
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I got 30.6 kW (41.061 HP) for one stroke (i.e. 27100 tons x 9 feet / 6 hours).

30.6 kW times 4 strokes = 122.4 kW in a 24 hour period.

CS
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#9
Nov13-09, 04:49 PM
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Quote Quote by Bob S View Post
I got ~ 52 kW average for 8 nine-foot strokes in 24 hours.
8x 9-foot strokes equates to a total distance of 72 feet, and if they're done in 24 hours it equates to 61.2 kW. It looks like you're off somewhere... the OP posted that the cylinder completes a single revolution in 12 hours (6 per direction) and that equates to two full revolutions, or 4 trips of 9 feet (36ft total) in a 24 hour period.

Quote Quote by stewartcs View Post
I got 30.6 kW (41.061 HP) for one stroke (i.e. 27100 tons x 9 feet / 6 hours).
Yup...

Quote Quote by stewartcs View Post
30.6 kW times 4 strokes = 122.4 kW in a 24 hour period.

CS
Kilowatts (kW) are a measure of power, so multiplying the number by 4 would assume that there were 4 pistons working together. To calculate the total energy generated in a 24-hour period you just multiply the power by the number of hours, which equates to 734.8 kW-hr (kilowatt-hours is a measure of energy).
stewartcs
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Nov13-09, 06:27 PM
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Quote Quote by Mech_Engineer View Post
Kilowatts (kW) are a measure of power, so multiplying the number by 4 would assume that there were 4 pistons working together. To calculate the total energy generated in a 24-hour period you just multiply the power by the number of hours, which equates to 734.8 kW-hr (kilowatt-hours is a measure of energy).
Doh! Yep...brain fart on my part!

CS
ronandkryn
#11
Nov13-09, 10:42 PM
P: 15
Thank you again for everyone's thoughts.

But I really have a BRAIN FREEZE on this one.

Please allow me get a little more specific:
Assume that a certain unopened geothermal structure is estimated to be exerting a constant pressure of 27,100 TONS. In an effort to greatly simply the question and because of other very practical limitations; please assume that there is a single piston sleeve that contains a single piston head that moves horizontally. Also, for simplicity please ignore the mechanics or reality of having to manage that much pressure out of a single well bore. The length of the piston sleeve is A REQUIRED NINE FEET. After traveling for a period of six hours, the piston head reverses direction, and the new trip will take six hours. Assume that by necessity this is the configuration of the initial linkage into the geothermal structure. After this initial linkage any other linkage or enhancement of mechanics is fair game.

One Wind Turbine can produce 50 KW. I would have thought that the single geothermal connection such as the one described above, would produce more than an entire Wind Farm.

All comments are welcome.
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#12
Nov13-09, 11:08 PM
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Quote Quote by ronandkryn View Post
Please allow me get a little more specific:
Assume that a certain unopened geothermal structure is estimated to be exerting a constant pressure of 27,100 TONS. In an effort to greatly simply the question and because of other very practical limitations; please assume that there is a single piston sleeve that contains a single piston head that moves horizontally. Also, for simplicity please ignore the mechanics or reality of having to manage that much pressure out of a single well bore. The length of the piston sleeve is A REQUIRED NINE FEET. After traveling for a period of six hours, the piston head reverses direction, and the new trip will take six hours. Assume that by necessity this is the configuration of the initial linkage into the geothermal structure. After this initial linkage any other linkage or enhancement of mechanics is fair game.
You have described the same problem 3 times now. Sure 27,100 TONS sounds like a lot, but you're only moving 9 feet every 6 hours, which limits your power output. The answer is still 30kW.

Quote Quote by ronandkryn View Post
One Wind Turbine can produce 50 KW. I would have thought that the single geothermal connection such as the one described above, would produce more than an entire Wind Farm.
Well all I can say is you thought wrong.
ronandkryn
#13
Nov14-09, 12:00 PM
P: 15
Quote Quote by Mech_Engineer View Post
You have described the same problem 3 times now. Sure 27,100 TONS sounds like a lot, but you're only moving 9 feet every 6 hours, which limits your power output. The answer is still 30kW.



Well all I can say is you thought wrong.
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That nails it down exactly in a concept that now makes sense to me.
THANK YOU TO EVERYONE that helped out in this.
EXIT
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reecem13
#14
Nov18-09, 08:11 PM
P: 3
Help me if i'm missing something here.


Horsepower = (Torque * RPM)/ 5252

The torque for for one 24 hour period would be 27100 tons * 2000 Lbs/ton * 36 feet = 1,951,200,000 ft*Lbs

The rate of 2 complete revolutions a day = 2 / (24 hours * 60 min/hour) = 0.001388889 RPM

So plugging those numbers into the original equation:

hp = (1,951,200,000 Ft*Lbs X 0.001388889 RPM) / 5252 = 515.99395 horsepower in one day

If one horsepower is equal to about 746 watts, then you get 384,931.49 watts
or about 385 Kilowatts.

Please explain how some of you are getting your answers. Thank you.
russ_watters
#15
Nov18-09, 08:31 PM
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reecem13, there is no circular motion described here. No torque. You are treating the linear motion as if it were circular and it isn't. Your 1.95 trillion ft-lb isn't torque, it's work.

w=f*d

Googling, I find that 1 kWh = 2.655 million ft-lb, so.....

1.95 trillion ft-lb / 2.655 million ft-lb/kWh= 735 kWh..... which is close enough to what Mech-E calculated in post #3.
reecem13
#16
Nov19-09, 11:37 AM
P: 3
Thank you very much. I guess since I read piston I immediately thought "car engine" and applied the formula like the machine was a regular gas engine since he didn't say how it was connected i assumed like a conventional gas engine with connecting rod and crank.
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#17
Nov19-09, 11:54 AM
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Quote Quote by reecem13 View Post
Thank you very much. I guess since I read piston I immediately thought "car engine" and applied the formula like the machine was a regular gas engine since he didn't say how it was connected i assumed like a conventional gas engine with connecting rod and crank.
You can think of it as a convetional piston engine such as a car (although the OP has defined the piston to be generating power in both directions), but the distance 36 feet is the STROKE of the piston, not the crankshaft's centerline to connecting rod radius. Therefore, since we are working with the piston stroke you multiply the distance by the force applied to get total work (energy) and divide it by the time it took to get power.

The way you're calculating torque is to say the crankshaft's centerline to connecting rod radius is 36ft (this would make the piston's stroke 72 ft, by the way), and that the torque is somehow constant although it would actually go to zero as the connecting rod came into alignment with the crankshaft (in accordance with a piston motion equation). So either way, you were misinterpreting how you could apply the calculation.
Mech_Engineer
#18
Nov19-09, 02:54 PM
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So if you wanted to caluculate the power being put out using torque and angular speed, it's a few more steps. The piston stroke is double the moment arm acting on the crankshaft, so given that the calculation is possible.

First, calculate the angular speed:
[tex]\omega=\frac{4\pi rad}{24 hr}[/tex]

Next, describe the torque on the crankshaft as a function of angle:
[tex]T(\theta)=Force*\frac{Stroke}{2}*sin(\theta)[/tex]

The energy per stroke is an integration of the torque function between 0 and [tex]\pi[/tex] (half a revolution):
[tex]E=\int T(\theta) d\theta[/tex]

To calculate the power output, calculate the average torque applied over one stoke (half revolution), the integral is from 0 to [tex]\pi[/tex]:
[tex]T_{avg}=\frac{1}{\pi-0}\int Torque(\theta) d\theta[/tex]

...and multiply the average torque by the angular speed. Voila! The same numbers as before!

I have attached a pdf of the solution, showing the same results.
Attached Files
File Type: pdf Mathcad - Piston angular speed and torque power.pdf (22.1 KB, 15 views)


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