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Design a transportable capacitorby stunner5000pt
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#1
Oct3105, 11:31 PM

P: 1,440

You have been assigned to design a transportable capacitor taht can store 250 kj of energy. You select parallel plate capactor type with a diaelectric
What is the minimum capacitor volume that is achievable using a diaelectric with strenght k? now capacitor voluem is Area times distance between plates? now the energy in this capactor is given by [tex] 250 = U = \frac{1}{2} \frac{\kappa \epsilon_{0} A}{x} \Delta V^2 [/tex] now multiplying both sides by d to make Ax = Vol [tex] U = \frac{\kappa \epsilon_{0} Vol}{2x^2} \Delta V^2 [/tex] now how would igo about minimizing the colume... would i differentiate by x? And set taht equal to zero? Please help! A capacitor has 250 kJ of energy wit ha volume of 0.087 m^3. Assuming the same dielectric used in teh above question what is the diaelectric constant? Now using the smae expressiong from the aboe question how would i figure out distance between the plates?? Have a look at my diagram for this one A capacitor has square plates each of side a, making an angle theta with each other. Show taht for small angles theta the capacitance is given by [tex] C = \frac{\epsilon_{0} a^2}{d} (1 \frac{a \theta}{2d} [/tex] Hint: The capacitor may be divided into differential strips that are effectively in parallel) now for the SQUARE plates [tex] C = \int_{d}^{d+ a \sin{\theta}} \frac{\epsilon_{0} a^2}{x} dx [/tex] is the integral setup correctly? It turns into [tex] C = \epsilon_{0} a^2 \ln{(1 + \frac{a \theta}{d})} = \epsilon_{0} a^2 (\frac{a \theta}{d}  \frac{1}{2} \frac{a^2 \theta^2}{d^2} [/tex] this doesnt look like it's heading in the right direction... Please help! Thank you in advance! 


#2
Nov105, 07:47 PM

P: 147

For energy storage the thickness of the dielectric is important because it determines the voltage and thus the stored energy. It also unfortunately, contributes to the volume.
I have some 80 volt capacitors and about 20 of them would make a Farad which also would be about a cubic foot and twenty pounds. This gives at one half C times V squared 3200 jules per cubic foot. I have about a factor of 70 to go which means 70 cubic feet. That is about three cubic yards which would fill the inside of a small car. I think a quarter megajoule is way too small to be useful. I figured out the integral too. The integrand is dx /(1+(x/d)sin theta) and the limits are 0 to a. The outside factor is epsilon a/d. 


#3
Nov105, 08:54 PM

P: 1,440

i dont quite understand what you are saying in the first two lines of ur post...
true the dielectric does contirbute to the volume... but let the dielectric be inserted directly into the capacitor such that there is no air between the two capaitor plates? Also why is the integrand that? 


#4
Nov105, 09:05 PM

P: 92

Design a transportable capacitor
here's what happens with different dielectric strength materials...
Material* Dielectric strength (kV/inch) =========================================== Vacuum  20 Air  20 to 75 Porcelain  40 to 200 Paraffin Wax  200 to 300 Transformer Oil  400 Bakelite  300 to 550 Rubber  450 to 700 Shellac  900 Paper  1250 Teflon  1500 Glass  2000 to 3000 Mica  5000 from http://www.allaboutcircuits.com/vol_1/chpt_12/8.html 


#5
Nov105, 09:12 PM

P: 1,440

ok ok but the question is asking to find the min imum capacitor volume with a dielectric that can store 250 kJ of energy. The choices for the dielectric are vaccuum
air polystyrene paper transformer oil pyrex mica porcelain siliconm water at 298K or 293.15K, titonia ceramic or strontium titanate But ignoring all theses choices How would i minimize the volume using equations only?? [tex] U = \frac{1}{2} CV^2 [/tex] is this not the energy stored in a capactiro with a dielectric? now the capacitance is given by by [tex] \frac{\kappa \epsilon_{0} A}{d} [/tex] but how would these two be related when trying to minimize the volume? 


#6
Nov105, 09:14 PM

P: 92

Dielectric Constant The ratio of the capacitance (AC voltage) of electrodes with the insulating material between them to the capacitance of the same electrodes with a vacuum or dry air in between. It's a measure of how good a material works to separate the plates in a capacitor. Dielectric Strength (V/mil) The voltage difference (DC) between two electrodes at which electrical breakdown occurs. It's an indication of how effective an "insulator" is. 


#7
Nov105, 09:20 PM

P: 92

Dielectric Constant Reference Guide Dielectric Constant (k) is a number relating the ability of a material to carry alternating current to the ability of vacuum to carry alternating current. The capacitance created by the presence of the material is directly related to the Dielectric Constant of the material. Knowing the Dielectric Constant (k) of a material is needed to properly design and apply instruments such as level controls using radar, RF admittance, or capacitance technologies. There are also analytical reasons to know the (k) of a material. and this is a great page, too...http://www.audienceav.com/on_capaci...c_material.htm adding a dielectric between the plates of a capacitor can increase the capacitance [Farads/square inch], but can lower the voltage capability before the applied electric field breaks down across the plates. 


#8
Nov105, 09:21 PM

P: 1,440

i already know waht these things are...
I wnt to know how to minimize the volume of this capacitor. Also note that i do not have to PHYSICALLY DESGIN this capacitor. It is only a problem for an HAND WRITTEN assignment 


#9
Nov205, 04:36 PM

P: 1,440

can anyone help with the first and second ones
know that [tex] Volume = \frac{250* 10^3}{\kappa \epsilon_{0} E^2} [/tex] but what is applied Electric field E?? Is it hte breakdown voltage for the dielectric? 


#10
Nov205, 09:43 PM

P: 147

I know of no straightforward way to formulate an equation for the capacitor energy density that can be maximized through differentiation. The dielectric constant is the capacitence parameter such that the greater K is the bigger the capacitence is. Thus good dielectric strength and a large dielectric constant are both needed.
In the integral the distance between the plates is d + x sin theta where x starts at 0 and goes to a. But that form is not of the form 1+ c so it is necessary to divide by d and pull it out of the integral. Try integrating dx/(1+x sin theta /d). It works out nicely. 


#11
Feb1011, 07:51 PM

P: 2

hahah wow this is a few years late, but i'll try my best to answer this anyways...
for the first part, you know that energy density: u = (electric potential energy)/(volume) or u = U/Volume energy density is also: u = (1/2)(kappa)(epsilon)(E^2) so set those equal to eachother: U/Volume = (1/2)(kappa)(epsilon)(E^2) for E, you use the dielectric strength of the dielectric which i believe if you are using Strontium titanate it is 8 x 10^6 (V/m) use this b/c it is the max. E that the dielectric can take, so it makes sense you would use the max. E to find the min. Volume. Rearranging the equation we get: Volume = 2U/((kappa)(epsilon)(E^2))=2.85 (m^3) (considering you used Strontium titanate as dielectric). For the second part, using your given volume, the same dielectic strength, and the same equation we already solved for, plug in the numbers and you should get about k=1.01 x 10^4 hope this helped >_< 


#12
Feb1111, 07:50 AM

Mentor
P: 11,614

Read the problem statement carefully. You are not given the dielectric constant but rather the dielectric strength! k is the dielectric strength.



#13
Feb1411, 07:28 PM

P: 2




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