# Enegry math needed.



## Duncan (Dec 8, 2008)

Energy required
1/2 m x v2

3000lbs = 1363kg
65 mph = 104kph = 28.9m/sec
v2 = 835 x 1636 /2 = 683,000Joules

over 10 seconds = 68.3 kwatts

at 200v = 342 amps

Note
This is a constant power acceleration
For a constant acceleration rate the power will increase from zero to 136kw


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## lazzer408 (May 18, 2008)

How did you convert Joules to watts over time?


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## major (Apr 4, 2008)

kennybobby said:


> by definition, 1 Joule = 1 Watt / sec


Nope. 1 Watt = 1 Joule / second.

1 Joule = 1 Watt second.

3600 Joules = 1 Watt hour.

3,600,000 Joules = 1 kiloWatt hour.



lazzer408 said:


> What is the equation to figure out how much energy (in watts) is needed to propel a 3000lb to 65mph in 10 seconds?


Energy is measured in Watt hours, not Watts. Watts are units of power.


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## major (Apr 4, 2008)

Duncan said:


> 3000lbs = 1363kg
> 65 mph = 104kph = 28.9m/sec
> v2 = 835 x 1636 /2 = 683,000Joules


And you're busted by the math police  Transposed 1363 to 1636. It should be 569,195 Joules.


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## Duncan (Dec 8, 2008)

major said:


> And you're busted by the math police  Transposed 1363 to 1636. It should be 569,195 Joules.


Its a fair cop - I'll come quietly


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## lazzer408 (May 18, 2008)

I'm trying to explain capacitors to someone and how many farads it'll take to accelerate a car. The answer is currently "all of them".


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## major (Apr 4, 2008)

lazzer408 said:


> I'm trying to explain capacitors to someone and how many farads it'll take to accelerate a car. The answer is currently "all of them".


The two basic energy equations:

Kinetic Energy needed to put into the vehicle is KE = ½mv², (where m is mass & v is velocity).

Electrical Energy in the capacitor is E = ½CV², (where C is capacitance in Farads & V is voltage). 

Sound familiar? http://www.diyelectriccar.com/forums/showthread.php?p=381199#post381199


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## lazzer408 (May 18, 2008)

I hardly remember posting in that thread.

Mass is in grams right?
Velocity is in what? kms?

The equations never show the unit of measure. >.<


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## major (Apr 4, 2008)

lazzer408 said:


> I hardly remember posting in that thread.
> 
> Mass is in grams right?
> Velocity is in what? kms?
> ...


As long as you stick with an accepted system of units, you're o.k. Here use MKS, (meter, kilogram, second). So mass in kg, distance in m & time in s. Velocity is then m/s. As per post #2 by Dunc.


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## lazzer408 (May 18, 2008)

I swear. 5 years ago this would all make sense to me but lately it seems like I have a bad case of ADD. I must have a tumor or been drinking fluorinated water too long. Even MPH to m/sec became a daunting task tonight.

I'll try it again tomorrow.

What I'm trying to figure out is, if I started out with a fully charged 400v capacitor bank, how much capacitance (F) would be needed to get 3000lb to 65mph in 10sec to 50% discharge of 200v.


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## Hollie Maea (Dec 9, 2009)

lazzer408 said:


> Even MPH to m/sec became a daunting task tonight.


It's great: you can go to google and just type in "20 mph to meters per second" and it gives you the results in huge letters at the top of the page. You don't even have to think these days.


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## Duncan (Dec 8, 2008)

OK so energy required
= 569,195 Joules. (after Major spotted my dyslexia)

_Energy required
1/2 m x v2

3000lbs = 1363kg
65 mph = 104kph = 28.9m/sec
v2 = 835 x 1363 /2 = 569,000Joules_

Electrical Energy in the capacitor is E = ½CV², (where C is capacitance in Farads & V is voltage). 

_What I'm trying to figure out is, if I started out with a fully charged 400v capacitor bank, how much capacitance (F) would be needed to get 3000lb to 65mph in 10sec to 50% discharge of 200v. _

Energy is V squared 
So at 1/2 voltage we have 1/4 of the energy left

So to take 569 KiloJoules we would need to start with (569/3) x 4 = 758KJoules

= 1/2C x V squared
V= 400v 
So C = 2 X 758,000/(400 x 400) = 9.48 Farrads

The problem is Supercaps are limited to low voltages - about 2v
And when you stack capacitors to increase their voltage you DIVIDE their capacitance

So if you start with 10 Farrad capacitors and stack 200 together (to get 400v) 
The capacitance of the stack becomes 10/200 = 0.05 Farrads

To get 10 Farrads you would need not 200 Capacitors BUT 200 x 200 capacitors


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## lazzer408 (May 18, 2008)

"And when you stack capacitors to increase their voltage you DIVIDE their capacitance"

Why is that?


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## Tesseract (Sep 27, 2008)

lazzer408 said:


> "And when you stack capacitors to increase their voltage you DIVIDE their capacitance"
> 
> Why is that?


Capacitance per unit area between two plates is inversely proportional to the distance between them; when you wire two capacitors in series you effectively double the distance between the plates.


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## major (Apr 4, 2008)

Here is an ultracapacitor bank I did a number of years ago. 










It just happens to be approximately what you're talkin' bout. Close to 1 MJ useable energy when cycled from 200 to 400V. It consisted of 155 2600F, 2.6V cells in series. That figures to 16.8F. Each cell is about the size of a tall boy beer can.

Each cell can hold ½CV² or 8788J at 2.6V. If you have 2 cells each at 2.6V then there is twice the energy, 17576J total. If you connect those 2 cells in series, then you have 5.2V across the pair. Solve for C = 2E/V² = 2*17576/5.2² = 1300F. That is one half of the capacitance of a single cell.

With resistors, resistance adds when in series and divides when in parallel.

With capacitors, capacitance adds when in parallel and divides when in series.

That's just basic electrical circuit theory.


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## lazzer408 (May 18, 2008)

That's a sweet capacitor bank. What was the application?


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