# capacitor for starting surge

islandguy
Solar Expert Posts:

**80**✭✭✭✭
A quick search showed me people have tried these. What I didn't see was a discussion about capacitors on the 12 volt side.

At amazon the Absolute CAP8000 80 Farad Power Capacitor is listed with someone already writing in about using it for this purpose.

What's the downside of adding one or two of these to a battery bank to handle high starting surges?

Thanks

At amazon the Absolute CAP8000 80 Farad Power Capacitor is listed with someone already writing in about using it for this purpose.

What's the downside of adding one or two of these to a battery bank to handle high starting surges?

Thanks

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## Comments

17,615✭✭It will do nothing.

Converting DC to AC to power 120 VAC loads presupposes variation in Voltage on the input side.

A properly sized battery bank is all you need.

80✭✭✭✭Thanks cariboocoot.

The problem here is exactly a too small battery bank. To start the pump multiple times during the day takes a toll on the bank. I resorted to using a mix of car batteries and deep cycle because it was impossible for me to get enough deep cycle to handle the starting surge.

Also a charging rate of C20 for batteries mean that the cap will charge quicker.

If my reading is correct an 80 farad cap will hold 40,000 watts. Is this correct? Seems like a lot.

So, taking it as a given that I am not yet sizing the battery bank correctly will one of these safely and reliably add power? And how much?

The pump is a Franklyn 1/2 HP 3 wire 230v about 80 feet down and a good 100 yards from the house.

1,218✭✭✭✭Whoa up there. Youre running a half horse well pump off of car batterys?

Supercapacitors are just a rally expensive kind of 'battery'. A real battery will be cheaper for sure.

The problem is to run that kind of pump you need a significantly better spec of system that anything that involves car batterys. Maybe put some system specs in your signature area, and we can look at where the system needs more oomf.

http://zoneblue.org/cms/page.php?view=off-grid-solar

17,615✭✭The input on the inverter already has a big cap on it.

Caps don't create energy, they just store and release it.

A Farad does not equate directly to a Watt, as it is a unit of capacitance (change in Voltage over time as a function of applied current) not a unit of power.

1 Farad = 1 Coulomb / Volt (1 Coulomb = 1 Amp second)

So Voltage, time, and current all figure in to how the capacitor performs.

Adding more capacitance to the inverter's input is no guarantee of increased Voltage stability against current drain by the inverter.

Your pump draw in Watts is possibly 2kW + on start-up. You'd be better off to install a soft start on the pump motor and/or improve the battery bank. Putting the right cap on the pump motor start winding would probably be of more benefit.

8,661✭✭✭✭✭I too, have a 1/2 hp Franklin motor in my pump, on a 6KW inverter, driven by a 48V, 800ah battery bank. You can't have too much power to start the motor. Depending on your motor control box (some have a relay, some are electronic) you can modify the starting cap on non-electronic starters, to get a "softer start". And the 3 wire motor also adds some inherent soft start feature too.

And mixing deep and starting batteries, I don't expect any of them will last really long. If you rely on this pump for water, you may need to re-think your system, and re-wire to 48V, where a 2000W starting surge, would only pull around 45A from the battery bank, instead of 180A from a 12V bank. Yes, a new inverter will cost you, and new batteries will cost too. You can use 6V golf cart batteries (about 200ah for the small ones) wire 8 of them in series. Or even moving to 24V will be an improvement. You must also have good battery cables, with good, hydraulically crimped connectors on the ends, to handle the amps.

Well, if you discharge it all the way to 0V, it might, but you are only going to use the top 1/2 volt, which is not very much. The Farad caps often have more inductance in them and their wires, and cannot dump the power as fast as paper says it can go. Here's what a good "power" cap looks like, many flat, parallel ribbon connections with low inductance. Attachment not found.

http://www.futureelectronics.com/en/technologies/passives/capacitors/film-capacitors/Pages/1883602-700D10896-348.aspx?IM=0

|| Midnight Classic 200 | 10, Evergreen 200w in a 160VOC array ||

|| VEC1093 12V Charger | Maha C401 aa/aaa Charger | SureSine | Sunsaver MPPT 15A

solar: http://tinyurl.com/LMR-Solar

gen: http://tinyurl.com/LMR-Lister ,

80✭✭✭✭I thought that was exactly what these were supposed to do, add extra watts to keep voltage stable? After a few starts the inverter gives a low voltage alarm at the start surge. The idea is to eliminate that. The ones I am looking at are the car stereo variety at amazon. For around 100, it seems like an inexpensive experiment.

I forget the numbers, but, at 3x running voltage I think starting surge was around 4.5kw. Initially looked into a soft start but couldn't find one to use that wasn't to expensive.

In 'developing' the well bank system I found I couldn't get enough deep cycle batteries to handle the starting surge. The battery wiring is all professionally done, probably double or triple 0. The thin plate batteries was the only way to get that initial power.

17,615✭✭Capacitors can't add any Watts: they only borrow power and then release it.

I'd be concerned about a supposed 1/2 HP well pump drawing 4.5 kW. It possibly is a larger pump than you think or has a problem. At any rate as Mike said it takes a lot of power to start a pump. So the first thing you must check is that the inverter is capable of starting it. Apparently it is, as it will do it a few times and then you run into trouble. That means the batteries are simply low, and adding the cap is not going to make up for that.

If you found you "couldn't get enough deep cycle batteries to handle the starting surge" something is very wrong with the system design. It may be difficult to start a pump but it is not impossible. Most of us off-gridders do it.

Instead of trying to apply a band-aid, let's have a look at the whole system specifications and see how best to address the problem.

8,661✭✭✭✭✭1) how often does your pump start up ? Too many start cycles will wear out the pump, and heat up the inverter - it could be a hot inverter can't start the pump. I went with elevated tank storage to reduce nighttime pressure tank pumping, and start cycle losses.

2) what is recharging the batteries, maybe after several cycles, they are just too low.

3) just because a "professional" set up the battery cables, doesn't mean it was really done right. (we've see that before)

4) My "Grundfos 10 SO5-9 with 3 wire Franklin motor (1/2hp 240V 1ph # 214505 )" indicates 1,000 watts draw, on the inverter readout. simple math would indicate 1/2 hp = 400 watts, but that ignores motor losses, PF losses, and such, so 4.5KW starting surge is reasonable from what I see on my system.

5) you can limp along with a marginal system, but eventually, you will have to spend the $$

|| Midnight Classic 200 | 10, Evergreen 200w in a 160VOC array ||

|| VEC1093 12V Charger | Maha C401 aa/aaa Charger | SureSine | Sunsaver MPPT 15A

solar: http://tinyurl.com/LMR-Solar

gen: http://tinyurl.com/LMR-Lister ,

30,190adminBatteries use chemical process to keep battery at ~11.5 to 14.5 volts in most conditions. And the battery voltage only drops by 1-2 volts during heavy surge events and holds it there until the surge is gone (chemical stored power).

Capacitors only store/release power based on voltage change.... Energy = 1/2 * CV

^{2}So, you only have ~1-3 volts of swing on a properly working 12 volt battery bank limits the amount of storage energy that a capacitor can supply vs a "50 volt capacitor" discharging to 11.5 volts to run an AC inverter:

[(12.5 volts)= 12,00 Watt*Seconds = (* 1/3,600 hours/second^{2}- (11.5 volts)^{2}]3.33Watt*Hours of useful storage[(60 volts)= 1,733,875 Watt*Seconds = (* 1/3,600 hours/second^{2}- (11.5 volts)^{2}]481Watt*Hours of useful storageAdding capacitors to a battery bank is not usually very helpful. A 1,000 Farad capacitor (in electronics we usually work with capacitors 1/10,000 as large) would support a 500 watt surge load for 1 second (or a 250 watt load for 2 seconds, etc.). And if the battery was already heavily loaded, you have even less "usable" voltage swing for the capacitor).

-Bill

80✭✭✭✭Thanks for the math.

The Cap I was looking is an 80 Farad one. In the Amazon reviews 'some guy' claims he added it to his solar battery system and it works wonderfully. Presently there is an 8D and one or two other car batteries in this bank. Last year there was a mix of Golf cart and auto batteries. they have since been cycled out.

When the pump runs the voltage drops by a volt or two. Max. The issue is between charging when the voltage will be run down to say 11.7 or so. I haven't done a system check this year but last year the set-up would handle around four pump cycles. When the generator is running it is putting less than 150 watts back into the bank. As soon as the PM3-45 gets replaced that will be charging the bank.

If I follow the math the 80 Farad capacitor would .5 * 80 * (16v Squared) works out to .5 * (80x256) = 10.240 So I'm probably doing this part wrong.

Before I posted about this my concerns were:

- How much power the capacitor would add to the system and would it be enough to offset the starting surge.

- recharge time and efficiency, since the Cap would be recharging without the generator during the day what would it pull from the battery bank and at what rate.

-Where in the bank to wire it, at the head end, putting current in right next to the inverter's cable or at the tail end of the bank flowing through the battery cables.

-Why 16v. Most of the car audio Caps are at 16 volts?

I'll have more detailed information when I get back on site about what the present batteries are.

The Inverter is a Meind 12v to 220v 50Hz. Not optimal, but it works. Prior to setting this up a dedicated generator would have to be run to provide the 220v for the well pump. So at $225 this was a major step forward. As far as I know this model can only be found on ebay. http://www.ebay.com/itm/140917835054?ssPageName=STRK:MEWAX:IT&_trksid=p3984.m1438.l2649

Thanks again for all the help.

30,190adminThe "problem" with your math is that is the total energy storage in the capacitor (16 volts peak)... But you are not charging to 16 volts and discharging to near zero volts... You are running (say when the sun is down) around 11.5 to 12.7 volts at the battery bank. So, the amount of available energy available from the capacitor is the ~1.2 volt difference (pump is off, then pump starts).

And I probably messed up my math too... The correct equation is:

^{2}- (11.5V)^{2}] = 580 Watt*SecondsSo, an 80F capacitor with a 1.2 volt sag can carry about 580 Watts for 1 second (I will go back and fix my math/algebra ).

The amount of energy stored (vs battery bank) is almost zero... Converted to AH on a 12 volt battery bank:

The capacitor is only supplying a very short burst of current to support the battery bank. Over all, it will not affect your charging/discharging of your system.

Try placing the Capacitor at the AC inverter's DC input connectors (short heavy wires)--It will do the most good there... Note it is possible for long wire runs from the battery bank to the capacitor to create an LC (inductor capacitor) resonance circuit... Where the LC "rings like a bell"... So if you get any "weird behavior", try moving the capacitor to the battery bus.

Also, be really careful with this stuff... Putting a discharged capacitor on a battery bus can give you are really scary spark (pre-charging capacitors to 12 volt light, etc. can prevent the sparks--And similar, you want to discharge the cap after is is removed from the circuit). Big caps can be scary.

Regarding 16 volts maximum--That is usually the maximum voltage one would expect to see on a 12 volt battery bank during equalization on a cold day--I would not use any smaller voltage rating (and would even look for a bit higher for safety sake).

What was the size (watt rating for inverter) and battery bank (Amp*Hour at 20 hour rate)? And what size cabling (diameter and length) between Battery bank and AC inverter?

-Bill

17,615✭✭If your battery Voltage is dropping that low between charge cycles the bank is

veryundersized for the application. These batteries are being tortured and will not last long. Placing a cap of any size on it will not help.You need to calculate (since your pump is 230 VAC 50 Hz) your power consumption. As Mike said it can be lessened perhaps by a larger pressure tank which will reduce the number of pump starts (always a good idea). Then you need a battery bank sufficient to supply the load. You do not want your battery Voltage dropping below 12 Volts, which represents about 50% SOC at rest.

17,615✭✭Bill I'm glad you did that math - it gives me a headache.

Since I know it doesn't work I just cut to the chase and tell people not to bother.

It's similar to how people calculate a battery capacity as Amp hours * Volts and expect the total Watt hours to be available - it isn't.

80✭✭✭✭System specs will have to wait a bit longer. Tonight turned into Penance night for one of my clients. Employees at the hotel brought in laptops for me to 'fix' So mostly I'll be spending the evening uninstalling crapware.

As far as the battery bank is concerned it is very undersized. i think we are down to an 8D and a 4D maybe. The other batteries in the bank actually brought down the voltage. So yeah, we are beating this one to death. In the initial attempt there was a 3/4 HP well pump and 8 golf cart batteries. This set up never provided enough power to power the pump. So I bought a smaller pump. Even with the smaller pump these batteries were not provided enough initial surge. it was only when I kicked 4 old car batteries into the mix that it worked. So it ran ok for about nine months.

There are two water tanks at this location, the main holding tank, about 6' tall has a ruptured pressure membrane so the smaller tank is acting as the pressure reserve. How much water this holds, cycles, I have no idea. I do know it takes the pump about fifteen minutes to complete a cycle. Take that fact with the consideration that, this pump is around 5 GPM, running with 50hz and down about 80'. Also when I've been measuring run times the shower or something has been running.

Kinda like Dr. Frankenstein's lab, minus the body parts.

Next steps:

Take my clamp meter, RMS, and get some reading on Amps going through the well wiring.

Time the run cycle. I did adjust the Square D pressure switch to cut off before 60 PSI as the top end of the pump cycle was overloading the bank.

- The present batteries are not a consideration. They will need to be replaced in the next few months. The question is how to do it right, that's affordable.

At this point, it looks like my best option will be to buy the Meind 48V version of what I have and then maybe 4 AGM vmaxtanks or similar.

.

80✭✭✭✭@ BB - Very Helpful.

So, having discovered that a rock makes a bad hammer, I'm going to suggest using a bigger rock.

Once I work out some math on the system and wattage draw, Why not use two, or four, of these to mitigate that 3x starting surge? My understanding of thick plate deep cycle is that they are very bad at providing an initial surge vs the thin plate variety?

Or, once I do the math, am I going to find that if I am sizing my bank for say sixty minutes of pump run time, I'll need enough batteries that the high starting surge is built into the capacity?

Fun stuff, thanks again.

30,190adminMore or less--But the difference is not that great between "car" and "deep cycle" (something like 20% difference in surge current)...

I have been using C/2.5 for the maximum surge current--Probably pretty conservative, but see how it lines up against your pumping requirements.

C/5 is OK for some minutes or hour of pumping (at maximum battery draw). C/8 is the typical max continuous draw "until battery is dead" current level.

And C/20 is a good "average draw" (2-3 nights of average loads to 50% discharge during bad weather--something like ~3 hours of average draw per night for ~9-10 hours of use, then 10 hours left for 50% capacity) for a home.

Rules of Thumbs are made for smashing by rocks and hammers (ouch). Just starting points.

-Bill

80✭✭✭✭OK. So let's take the house bank. 10 Trojan T105. In 5 12v banks

At C/5 = 185. So let's call C/2.5 160ah

160ah x5 = 800ah. 800ah x 12v = 9600 watt hours.

So you are saying this bank can surge to 9,600 watts for starting?

17,615✭✭This has its own problems: five parallel batteries. Almost certainly current will not remain the same between all.

A Trojan T105 is 225 Amp hours. Five in parallel would be 1125 Amp hours. In theory you could pull 450 Amps by Bill's formula meaning 5400 Watts. Note the shift from Amp hours capacity to Amps draw and thus Watts draw.

I find in practical terms this tends to be more than you can expect on a 12 Volt system. More like C/5 or 225 Amps / 2700 Watts. This is danger territory for 12 Volt systems. For example you'd need some very good wires and excellent connections to be able to do this.

80✭✭✭✭I'll review this in the morning but I'm not following how you are getting 450 amps from 1125 Amp hours? OK, 450 x 2.5 = 1125. So what should I have divided 1125 by to get this rate? Also why am I using the C/20 rate?

The site has excellent wire and connections. A long time ago it was set up correctly.

Could you walk me through calculating surge amperage based on Ah capacity, and what C value to use again?

And thanks again, this is incredibly helpful

80✭✭✭✭@cariboocoot

D'uh. 2.5

Thanks again, I'm out for the night.

30,190adminThe rules of thumbs we use around here are based on the battery's 20 hour discharge rating (C/20 for 100 AH battery === 5 amps * 20 Hours = 100 AH)... But for actual capacity calculations, a C/20 = 100 AH rate may look like 80 AH at a C/5 rate.

If we were to use a C/10 capacity, we would probably "play" with the rules of thumb a bit (5% to 13% rate of charge becomes 8% to 18% rate of charge, etc.).

As Marc/Cariboocoot says, there is a limit to how much current you can reasonably move around at low voltages.

Generally, designing for around 100 amps nominal maximum current is a good point... 100 amps * 12 volts = 1,200 watts; 100 amps * 24 volts = 2,400 watts, etc..

For example, Xantrex makes a well regarded 12 volt 2,000 watt AC inverter:

2,000 watts * 1/0.85 inverter efficiency * 1/10.5 inverter cutoff = 224 Amps at 12 vdc

NEC requires a 1.25x safety factor:

224 amps * 1.25 NEC safety factor = 280 amp minimum rated fuse/breaker

A good AC inverter can supply upwards of 2x rated power for a few seconds to a few minutes. 560 Amps with a 1 volt drop (11.5 volt minimum battery voltage - 10.5 inverter cutoff voltage = 1.0 volt maximum wiring drop) would need a cable of (using generic voltage drop calculator):

15 feet (maximum one way run) of 4/O copper cable -> 1.0 voltage drop.

4/O cable costs around $18.50 per foot and the copper itself is almost 1/2" in diameter.

-Bill

80✭✭✭✭I spent some time on site so I'll post some measurements.

With my clamp meter on one leg of the well line I got a max starting surge of 18.98 amps. Call it 19.

19 x 120 = 2280 watts

2280 x 2 = 4560 watts starting surge for the 1/2 HP well pump.

Running amps maxed at 6.7

6.7 x 120 = 804 watts

804 x 2 = 1608 running wattage.

The inverter is a Meind 3000 watt that will surge to 6000 watts. Manufacturer claims.

17,615✭✭Uh, there's no multiplying the current by two: the Amps going through one wire are the same coming out the other; it's one circuit. Earlier posts indicate your pump is 230 VAC 50 Hz, so that would be 19 Amps * 230 Volts = 4,370 Watts.

Trying to supply that on nominal 12 Volts = 364 Amps (this doesn't include powering the inverter itself nor current variation for higher/lower Voltage).

I'm not familiar with that inverter, but if it is MSW type the demand is even worse. I would not trust the surge rating on any inverter for handling start loads, as those ratings can be optimistic and unrealistically short-term (2X the run rating for 10 seconds for example).

Running current of 6.7 Amps @ 230 Volts would be 1541 Watts. That's like 128 Amps on 12 Volts. Pretty hefty stuff.

I would not do this on a 12 Volt system. It would be hard on even a well-balanced 1000 Amp hour battery bank.

80✭✭✭✭Getting there.

Initially I was looking into adding capacitors. Now it looks like it would be better to go the 48 volt route.

Yes it is an MSW type.

17,615✭✭MSW presents another problem of about 20% more power used than is apparent because of the distortion in the waveform.

80✭✭✭✭@Cariboocoot. Thanks.

Just tested the Meind. With no load I was reading @ 127 volts per leg at 59 Hz. So this inverter operates with two 110v Legs at 50Hz. Well, it's supposed to.

Specs say 12v in and 220v out.

Inverter/charger function will accept 110 -240v for the battery charger function.

Took some pictures of the well bank and connections. Will post in a few.

80✭✭✭✭If I did this right there will be two pictures. One of the Well Battery bank and one of the type of connections and wire that is being used on site.

At the moment I am timing the pump cycle. then I think I will have the info I need to start planning out a 48 volt system.

Thanks again for the help.

30,190adminYou can skip this post if you are not "ready" for the whole issue of AC power, phase angles, power factor, and VA (Volt Amp) readings... But, if you are:

Remember that there is "Watts" and there is "VA" (volt*amp or "Volt Amp") and they are not really the same thing.

Power = Volts * Amps (true for DC and usually "close enough" for AC circuits).

But, in reality, for AC:

Power = Volts * Amps * Cosine (Voltage-Current Phase angle) = V*A*Power Factor (PF) = some other equations too...

Note that when you measure Volts and Amps on an AC circuit--Most people can only measure Volts and Amps as separate values. To measure Watts accurately, you need two things--First a meter that can measure Volts and Amps as "Vectors" (measure both at the same time) and needs an RMS (root mean square) ability to "calculate the area" under the Voltage and current wave forms (non-RMS meters are only "accurate" with pure sine waves on AC power... Square Waves and other "non-linear" wave forms have different conversion factors). The "Kill-a-Watt" meter is a cheap and simple to use meter that can measure both Volt and Current phase angles (probably does not do well on PF and non-linear wave forms).

So--When you measured your pump Voltage and Current, you got:

Running of 6.7 amps (forget RMS issue for this post). And, this is a ~240 VAC pump/circuit (?).

6.7 amps * 240 VAC = 1,608 VA

Note, this is not (necessarily) Watts... The VA number is good, because your wire rating and AC Inverter (and Generator) outputs "care" about VA--And are usually rated in maximum VA output. (when voltage and current are not in phase, the current must be higher to do the "same amount of work". Current "heats" the wiring and saturates the inductors/pole pieces in transformers/generators--So knowing the maximum VA is very important.

The other number we need to look at is Watts...

Power = Volts * Amps * Power Factor

Where PF is a "magic number" that ranges from 0.00 to 1.00 -- You can think of this as "efficiency" of the current and voltage wave forms. In the case of motors, this is sort of the difference of standing in front of a car and pulling on a rope (Cosine 0 degrees=PF=) 1.0 or 100% of your pulling force is moving the car forward.

If you were standing 60 degrees to the side of the car and pulling forward, only (Cosine 60 degrees=) 0.5 = 50% of your force is moving the car forward... So, either 1/2 of the force is wasted, or you have to pull 2x harder to move the care forward at the same speed vs standing in front. And the rope would have to be 2x stronger to support that 60 degree offset.

For an electric induction motor, they will run around 0.67 to near 1.00 PF and perhaps at less than 0.50 PF when starting.

Now--The DC side input to the inverter (and gallons per hour into the genset motor)--Watts is work--So, the energy from the battery or gallons per hour of fuel is converted into energy to pump water (or whatever). Here the PF is "your friend":

Power = V*A*PF = 240 VAC * 6.7 Amps * 0.67 PF = 1,077 Watts

So, the size of your battery bank and solar array (and $$ of fuel) is less than you would have expected. So, if you measure the AC and DC voltage/current on the input to your AC inverter (with pump and no other AC loads), you can estimate the power (Watts) needed to run the pump. And start the pump too.

There are many ways to design motors and compensate for "poor power factor"--So all I can guess is your motor is probably in the range of 0.67 to 0.95 PF. If you measure the AC inverter DC input V and A, as well as the AC output V and A, we can guess at the PF of the motor.

Or, you can guess that VA ~ Watts (VA is always equal to, or larger than Watts), and do the numbers--It will slightly oversize your battery bank and solar array--But that is not a huge problem (we are doing lots of estimates anyway).

Sorry for the bunch of words...

-Bill

2,356✭✭✭✭I like the pulling on the rope analog, that makes it much clearer to me now.

30,190adminThere is also the "non-linear" analogy too... If your average pull on the rope is 100 lbs, but instead you pull 400 lbs for 1/4 second and 0.0 lbs for 3/4 seconds--The average is still 100 lbs, but the peak pull for the rope is 400 lbs. and must be sized for that impulse load.

That is similar to when non-PF Corrected power supplies only pull current "at the peak" of the sine wave--Or have current surges at the leading edges of square waves. High peak current for short periods of time increase heating effects of I

^{2}R (double the current, 4x the heating).We can do "water analogies" too.

-Bill