Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equivalent super Capacitor for a battery.

  1. Jun 2, 2014 #1
    I am working as a signalling engineer in Railways. This weekend i am going to give a presentation on super capacitors and their use in railway systems. Most importantly i am going to show how we can replace batteries with super capacitors (EDLC). The problem i am facing is i don't have any practical experience with capacitors and don't know how to show a relation between a battery and a corresponding equivalent capacitor. The type of battery we are using for driving the point machines (Google search "point machine railways") is of the type (Please refer to the attached image).

    Now what i need is an equivalent type of super capacitor which will be able to drive the same type of point machine. Guys if anyone of you who has an experience in this field please help me in finding the answer of the following questions:
    1. What will be the equivalent value of the super capacitor for the above mentioned battery?
    2. What will be the dimensions of such a super capacitor?
    3. How much will it weigh?
    4. Model or company's product which i will be able to use (links are most welcomed).

    All suggestions are welcome.

    Attached Files:

  2. jcsd
  3. Jun 2, 2014 #2
    It is about energy storage - research the amount of energy stored in a capacitor storage in a capacitor, then compare the ratings of the batteries to the capacitors. Next - cycle efficiency, how much of the energy placed into the storage device can you get back out ( a ratio) - the next item to review is long term storage - or loss of charge. All of this info should be available on line. Of course Cost, Size & Weight are all meaningful relative to these topics.
  4. Jun 2, 2014 #3


    User Avatar
    Science Advisor
    Gold Member

    Am I the only one who finds the above profound? Not saying that whatever it is that is being done or attempted will or won't work. I thought the only ones who pushed a product that they knew little about were car salesmen.
  5. Jun 2, 2014 #4
    No, you aren't the only one.

    Supercaps and batteries are pretty different. Its kind of insane to suggest flat out replacing one for the other without a lot of consideration for your requirements. Do you really need 25 A-hours? That is asking a huge amount from capacitors. Supercaps have a much lower energy density with a much high power density, but if you still need that same amount of total energy, you'll need 10x as much space with super caps. Or more.
  6. Jun 2, 2014 #5


    User Avatar
    Science Advisor
    Gold Member

    After I read the link I decided too that it is impractical to do in this case. That great big spark that you get when you short the terminals looks impressive but a capacitor is very seldom a good replacement for a battery.
  7. Jun 2, 2014 #6

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    25 AH = 90,000 coulombs

    if you're wanting say 12 volt drop (10%) , and C=Q/V that's 7500 farads ?

    China is making some 7500 farad caps for locomotives........
    http://www.gdleyin.com/en/NewsDetails.aspx?iNewsId=475 [Broken]

    it doesn't say what voltage. My locomotive experience was in 1960's, the old DC commutator machines. I understand newer ones run on AC.
    but i'd wager your roadforeman-locomotives has a trade magazine with a more detailed article on these locomotive scale super capacitors, or your company's engineering department...

    shoot this outfit an email inquiry ?
    Zhuzhou Electric Locomotive Co. http://www.csrgc.com.cn/g1733/s4222/t61706.aspx
    I'd think they would be pleased at an inquiry from a US railroad.

    old jim

    EDIT interesting GE short film here about graphene supercaps..
    Last edited by a moderator: May 6, 2017
  8. Jun 2, 2014 #7
  9. Jun 3, 2014 #8

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    hmmm 20 X 61kg = somewhat over a long ton....

    they're ~75 farads so it'd take a hundred of them to avoid the voltage converter

    i'm starting to appreciate the humble lead-acid battery with 10X the wh/kg.
  10. Jun 3, 2014 #9
    Thanks alot guys for replying...
    I am not saying that it is possible and we are going to do it...
    i am not an expert on this subject...but just consider this..
    if engineers are able to run a light metro train on super capacitor (http://www.railwaygazette.com/news/...upercapacitor-light-metro-train-unveiled.html)
    then why not a point motor...as you can see in the article the capacitor doesn't look like it weigh a ton..even then they are powering up a whole rolling stock...
    its is just an idea...and we are not going to use the capacitor as a power backup...the maximum time for which a point motor work is 10 sec at a time...after that the capacitor will again get the time to charge...so i dont think we will be needing that big capacitor...
    what i presented as an idea was we will directly take the supply...step it down with a transformer..then use a bridge rectifier for conversion and feeding the output to capacitor and use that capacitor drive the motor...
    the advantages we will get is..its cheaper and can work for up to 27 years without maintenance...and that a very important thing in our railway industry..cost and maintenance....
    when we are using batteries then first of all there is problem of replacement, maintenance and we are using condition monitoring circuits which is again very costly..
  11. Jun 3, 2014 #10
    @Jim: thanks for the reply..
    Will the calculations be same even i am using the capacitor for only a maximum time of 10 sec at a time..
    after that its will again get the time to recharge...i will try to find out the specifications of the motor we are using...will soon post it..
  12. Jun 3, 2014 #11
    Kind of agree here~
  13. Jun 3, 2014 #12
    yeah to some extent you are right..however neither i am trying to push a product nor m not selling anything...the only thing i wanted is get the fact whether it will work or not..that's why i came here...even if its possible then also i will not get anything out of it...its just if i am presenting some idea then i should have some practical insight of it...and i thought i will get some expert comments here..since the forum was created for the same...
    anyway thanks for the reply.. :)
    Last edited: Jun 3, 2014
  14. Jun 3, 2014 #13

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    We all took your battery specification as a starting point. All it gave us was the battery capacity.

    In addition to the switch specifics,

    How many times do you want the point switch to cycle before recharge is necessary?
    How long must it be able to sit unpowered and still be able to move the switch ?
    What happens if the switch is covered with mud or snow - does it take more power to cycle?
    How many point switches does the battery serve?
    Is there any other load on it , maybe communications ?

    Maybe that rail spec RT/E/C/11600 is enlightening ... we don't know.

    Tabulate your energy needs. That's the stating point.
  15. Jun 3, 2014 #14

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    Last edited by a moderator: May 6, 2017
  16. Jun 3, 2014 #15

    1. Generally the capacitor will get the time to recharge after every cycle i.e. as soon as point is moved and locked on the other end. One cycle takes nominally 2.8 secs but to be on a safer side we can consider up to 7 sec in case of failure as discussed in point 3.

    2. Most of the time the switch remain locked in one position or the other (we call is normal or reverse). But the position of the switch changes as per the requirement of the route set by the signaller. If the route required is normal and switch is in normal position then no movement is required.

    3. The switch areas are cleaned regularly for any mud or dirt, in case of snow or ice there are heaters available connected with external supply (no relation with the switch) for melting the snow. However even after that if the during the movement from one position to other some kind of obstruction comes in between then the it returns to its previous position and route is not set and the system will count it as a failure. So in all the power is more or less the same in all cases. It depends also upon the time. If in a certain time frame the point is not getting locked then it returns to previous position.

    4. For this scenario you can consider one switch per battery. Since practically in one case we keep 5 batteries connected together powering up to 4 switches.

    PS: we are using this kind of machines: http://www.signallingsolutions.com/hw.php [Broken]
    HW 2000 series...
    Last edited by a moderator: May 6, 2017
  17. Jun 3, 2014 #16

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    Re point 2: i'd guess the reason for having a battery in the first place is so the switch will move even if local power is lost for a while. And the system can alert somebody to come investigate the power loss.

    Your switch link doesn't say what is operating current, just it's a DC motor.

    So the question becomes: why was the battery sized 25 amp-hours ?
    Does it need to supply that much charge? Or is that how big the battery needs to be to delver a brief ten or twenty amp burst ?

    If the latter you could take advantage of the capacitor's high current capability and use fewer farads.
    If the former our numbers were about right.

    Remember that during a power failure the point switch has to rely on the battery long enough for some maintenance guy to come out and restore the local power.
    So find out how long that Maxwell unit can sit unpowered before its own inverter and fans deplete its stored energy.

    Lastly - the Signalling Solutions link says "Fitted with two externally controlled heaters the machines can operate efficiently in a wide range of ambient temperatures. " Does the battery have to power those heaters ?

    So - your first step is to define your power and energy needs, both peak power and total watt-seconds .
  18. Jun 10, 2014 #17

    Hi Jim,

    Sorry for late reply. Just wanted to inform you that that the batteries we are using are not really for backup. It is not expected that the main power supply will go off at any point of time. The main reason we are using the batteries is since the point switches draw a lot of power for a short time the cables used were very expensive. Now the length of those cables are shortened as they will go from battery to point machine (earlier it was from Function Supply Point (main supply point) to point machine thus increasing the cost as the distance was increased). So here in our case we don't need capacitor as a backup device, so but we need it just to run the motor for few seconds as explained in my earlier posts. The rating of the point motor we are using for calculating the over all load is 1100 VA. A supply of 120 V DC is required to run the point machine.

    The arrange should look more or less like this at the end

    supply (110 V) -->bridge rectifier-->super capactior arrangement-->dc-dc (for constant voltage)-->point machine motor.

    Just require the size and cost of capacitor which can achieve this operation. A link will be great.
  19. Jun 10, 2014 #18
    No Jim..the heaters are getting supply from some other source not from the batteries.
  20. Jun 10, 2014 #19

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    Are you familiar with basic electrical units
    the Coulomb , Ampere , Joule, Volt, Watt , Farad ?

    Farad = Coulombs per volt http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

    1100 VA at 110 volts is only 10 amps, which for 30 seconds is 3000 coulombs

    if you size for 10% drop , C = Q/[itex]\Delta[/itex]V = 3000/11 = 272 farads
    which sounds like about four of Willems units from post #7 .
    That'd run a 110 VDC switch well, like your HW1000 and HW2000 in the spec sheet - the supercap has plenty of current capacity
    If you use an AC switch and inverter be sure it'll handle starting current of your switch motor.

    Operating time of the switch is stated to be under 3 seconds

    using Coulombs = Amps X seconds and Farads = Coulombs/[itex]\Delta[/itex]Volts
    can you figure how many Farads you'd need to supply that with 10% drop ? How many of Willem's units would that take ?

    Probably there's a more elegant way to do it, maybe a stepup inverter that'd hold output voltage up as input falls allowing a smaller capacitor. But i'm for simplicity.
    Read a few of these guys' application notes

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook