Dump Load

Discussion in 'General Questions' started by RickPR116, Oct 7, 2009.

  1. RickPR116

    RickPR116 WindyNation Engineer

    Question on dump load requirement.
    I have a couple Vitreous Enameled Resistors 2519, 6.3 ohm, 300 watt that I’m using for dump load. I’m running a 12 volt system (6-6v Trojan batteries) controlling the charge of my two Windynation 500 watt PMA with a Xantrex voltage regulator. Presently I have the two resistor connected in series for a 12 ohm + total, but I don’t know if this is Ok. Should I use only one of the resistors? Should I install a fuse between the charge controller and dump load?
  2. windyguru

    windyguru WindyNation Expert

    This is an easy answer but requires a long winded explanation:

    Just because a resistor is rated to "dump" 300 Watts, in no way means it will dump 300 Watts when you give the resistor 12 volts. A dump load or diversion load has to be chosen with a little bit more effort.

    Step 1. Let's learn the equation for calculating resistance for resistors in series and resistors in parallel.

    For resistors in parallel: 1/R(total) = 1/R1 + 1/R2 + 1/R3 + 1/R(n)

    So, let's say you use three of your 6.3 ohm resistors wired in parallel. What will the resistance be?

    1/R(total) = 1/R1 + 1/R2 + 1/R3 >>>>>> 1/R(total) = 1/6.3 + 1/6.3 + 1/6.3 >>>>> R(total) = 2.08 ohms

    For resistors in series: R(total) = R1 + R2 + R3 + R(n)

    So, let's say you use three of your 6.3 ohm resistors wired in series. What will the resistance be?

    R(total) = R1 + R2 + R3 >>>>>>>>>>> R(total) = 6.3 + 6.3 + 6.3 = 18.9 ohms

    Step 2. Now we have to calculate how many watts these resistors can dump or dissipate when used as dump loads on a 12 volt battery bank?

    First, we will calculate this for the three 6.3 ohm resistors wired in parallel.

    Power= P Volts= V Amps= I Resistance= R

    Power = volts x amps (P=VI) and voltage = amps x resistance (V=IR)

    We use V=IR to first calculate the current (amps) going through the three 6.3 ohm resistors wired in parallel:
    (I will assume that your charge controller is set to dump at 14.5 volts)

    V=IR >>>>> 14.5= I x 2.08 and I = 6.97 amps (where we are using the 2.08 ohms that we calculated above)

    Now we use P=VI to determine how many Watts these resistors will dump:

    P=VI >>>>>> P= (14.5 volts) x (6.97 amps) = 101 Watts


    Second, we will calculate this for the three 6.3 ohm resistors wired in series.

    V=IR >>>>>>> 14.5 volts = I x 18.9 ohms >>>>> I = 0.77 amps

    Now we use P=VI to determine how many Watts these resistors will dump wired in series:

    P=VI >>>>>> P= (14.5 volts) x (0.77 amps) = 11 Watts


    You need to be very careful. These resistors have a high resistance and can not dump much energy at 12 volts. Your dump or diversion load needs to be capable of dumping more power than your generators are putting in or you will destroy your expensive batteries!

    WindyNation has 300 Watt dump loads for 12 volts battery banks (0.73 ohms) and 24 volt battery banks (2.9 ohms). They will be available in the store in 1-2 days.
  3. RickPR116

    RickPR116 WindyNation Engineer

    Thanks for the information. It was an excellent lesson. Now base on your info. If I use a single WindyNation 300 watt dump load (0.73 ohm) for my 12V system it calculates to a 288 watt dump load. How many of these resistors I need for my system? Remember I’m using two WindyNation 500W PMA to charge my battery bank.
    Wind has been very calm and I have not been able to fully charge my battery and that is good until I fix this dump load problem.
  4. windyguru

    windyguru WindyNation Expert

    Four would be enough but I would go with five if you want to be very safe. And it never hurts to have a one or two extra resistors in case you want to increase the size of your system in the future with solar panels or another turbine.

    One thing you need to be careful about is your charge controller! If you have five of these resistors hooked up in parallel you will be dumping 100 amps! This is a lot of current! Make sure your your charge controller is rated to handle 100 amps or you will fry it. The relay will pretty much just melt if it is not rated to handle the current.

    Keep in mind that you will only be putting 1000 Watts (from two 500 Watt PMA's) on windy days. This will happen in about 27-30 mph wind and even higher winds will be necessary if your turbines are not mounted really high above neighboring obstacles (i.e. trees, houses, etc.).

    Just some more stuff you have to keep in mind! Building a solid system is not easy but once you have it up and running correctly, it is worth it.
  5. RickPR116

    RickPR116 WindyNation Engineer

    This is the email that I was drafting with more questions.

    I did the math last night with the your excellent information and I determine that I need 4 of the WindyNations .73 ohm 300 watt resistor for a total of .1825 ohm in parallel.
    Now this is going to dissipate close to the 1,200 watt that I need for my two 500 PMA plus the 20%.
    The new question is the total current. Base on calculation the current will be 79.4 Amp during full operation (If I ever get there). What this is going to do to my Xantrex C40 (40A Controller). Also do I need to increase my wire gage for the dump load circuit?

    Now, do I need to change the Controller, add a controller or can I use a series of relays?
  6. windyguru

    windyguru WindyNation Expert

    Once your charge controller switches to dump load mode it will instantaneously dump 79.4 amps because this is what the resistors will draw from the battery bank. It will melt the relays in your charge controller whether they are mechanical relays or solid state relays ... it won't matter as they will not be able to handle the current.

    You could add on more C40 charge controller (or a windynation 40 amp charge controller). You could have one dump 40 amps and the other dump 40 amps giving you 80 amps in total.

    The other option is you could possibly switch out the relay in the C40 for a bigger relay that can handle the current. Or you might be able to have the existing relay in your C40 drive a bigger relay (i.e. a 100 amp relay). I will ask my friend who is more skilled in electronics if and how one would go about doing this.
  7. myocardia

    myocardia WindyNation Engineer

    Okay, I posted this last night, but this software doesn't have editing capabilities (a bad thing for someone with a bad memory!).

    This isn't correct. If the resistors drew ~80 amperes from your batteries, how would your batteries ever charge? They wouldn't. The resistors absorb power that would have went to the batteries, but will never make it there since it will be absorbed by the resistors, which are there to absorb power that is output by the turbine. In other words, the resistors (along with the dump load controller) are there to protect the batteries, not to drain the batteries. :lol: The dump load controller has one input, and two outputs. Once the battery voltage reaches the point that it is set to not let the battery voltage go above, it switches to the other output, the one that has the resistors attached, and they absorb whatever the turbine (in this instance, two turbines) happen to be outputting, from .1A all the way up to a minimum of 80A, assuming you have four of them. To phrase it another way, a diversion controller is merely an automatic two-way switch that is voltage controlled.

    Now, as far as two PMA's that can output ~40A each destroying a charge controller/dump load controller that's only built to be able to handle up to 40A total, yes, they could easily do that. There's really only one thing to do in this instance, and that's buy another C40 for $125: http://cgi.ebay.com/Xantrex-C40-Cha...emQQptZLH_DefaultDomain_0?hash=item3a547d02f2. You need one charge controller (used in this instance as a dump load controller) for each turbine, and both will require a minimum of two of these resistors: http://www.windynation.com/shop/index.php?act=viewProd&productId=23. Both C40's will be able to charge the same battery bank, of course. Now, if you were worried about damaging your batteries should you ever get high winds for a sustained period time, you can see how attaching 3 of these 300 watt resistors to the diversion side of each of your C40's will protect your batteries even better than only having two resistors on each, and will be very cheap insurance for your batteries, and couldn't possibly damage your C40, unless each turbine happened to actually be providing more than the 50 amperes that each C40 is guaranteed to be able to handle (40A x the 125% they're guaranteed to be able to withstand = 50A). In other words, only the turbines/generators (or solar panels, if you had any) can damage the diversion controller, not the resistors.

    This is what I forgot last night: All of the above is true, with one caveat. A diversion controller doesn't work exactly like a simple switch. A simple switch would just let your turbine/turbines put all of the voltage and amperage it wanted into your diversion load after your batteries were full. Because that could easily start fires if the diversion load wasn't large enough, it works more like the overflow in a bathtub. So, once the batteries are full, it starts "filling up" the diversion load. As long as there is never more output that the diversion load can handle, everything works just fine. But as soon as there is more than the diversion load can handle (as soon as it's "full", IOW), it starts continually switching back and forth between the two outputs, trying to find one that will accept more, effectively overcharging both. There are two ways around this happening, of course. You can have a bit of extra resistor capacity, or you can be at home, ready to start using extra battery power. The extra two resistors is of course my recommendation. By the way, one last addition. You do have your C40 hooked up as a diversion controller, not as charge controller, don't you?

    BTW, now that I've given you some info, I want some in return. How are you liking your Windy Nation turbines? What mounts are you using? Have any pictures?
  8. PierDrop

    PierDrop WindyNation Engineer

    There is on thing I would like to add here. There are two ways to operate a dump load controller:

    1. Direct connect: The turbine or solar panel is connected directly to the battery bank. This means that the generators are not wired through the charge controller. The charge controller is also wired directly to the battery bank. Finally, the dump loads are connected directly to the battery bank but wired through the relay on the charge controller.

    The charge controller senses the voltage of the battery bank and decides whether to turn the dump loads "on" or "off". The relay is essentially a switch and the charge controller decides when to flip the switch.

    For example, the voltage of the 12 volt battery bank hits 14 volts. You have set your charge controller to start dumping at 14 volts. Therefore, the charge controller flips the switch (the relay) and this allows energy to flow from your batteries to your dump load. In this case, if you have 80 amps of dump load resistors, you will dump 80 amps directly from your battery bank. You will continue to dump until the voltage of your battery bank drops to the voltage that you set for it to stop dumping (~13.5 volts).

    2. The other method is as myocardia describes and in this case the generators are directly diverted to the dump load once the battery is charged. In this case, as myocardia says, you will only dump the energy that your generators are producing.
  9. myocardia

    myocardia WindyNation Engineer

    PierDrop, thanks for bringing that up. I looked up the Xantrex controllers, and was surprised to find out that they operate as you described, and not as I had described. I had assumed that since they are a more expensive controller, they would operate the way that I described. Oh well, live and learn, huh? I do want to mention one other possibility that the OP may or may not be interested in. He could always buy a 24V inverter instead of another C40, and run his battery bank @ 24V. One C40 @ 24V can handle exactly the same amount of power as two of them @ 12V, and it comes with two added benefits: lower wiring losses, and slightly better power output from his turbines.<---(that's not always true, but is usually) He would need to buy the 24V resistors, though. Of course, he would only need to buy half as many of them, which is always a plus.
  10. myocardia

    myocardia WindyNation Engineer

    Hmm, not sure what I was smoking when I typed that last sentence above. The dump load resistors are rated for 300 watts each. Since 300 watts @ 24V is half as many amperes as 300 watts @ 12V, having 1200 watts total of the 24V resistors would still require four of them. They would only total 40A instead of 80A, but since they're rated @ 10A each, and not 20A like the 12V, the total amount required wouldn't change. Sorry for the confusion. :oops:

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