# capacitor for starting surge

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## Comments

2,490✭✭✭It looks like you have 3 different types of Battery capacity, thats a big issue for any voltage system.

17,615✭✭It also appears your batteries are "laddered" meaning they won't be sharing current very well at all.

Have a look at this page: http://www.smartgauge.co.uk/batt_con.html

What you've got is method #1, which isn't good. What you want is any of the others, preferably method #3.

8,380✭✭✭✭✭When I need to do this, I use a automotive 12V taillight bulb, the bulb limits the amps as the cap charges (and when discharging it) and you will be amazed how long the bulb glows at charge and discharge. After it's charged up through the bulb, you can remove the bulb and connect the cap to the DC Buss. And use a bulb to discharge, because your screwdriver will get all messed up from arc welding across the terminals, and it chews up the terminal and internals of the cap. But you are not going the cap route anyway.

Battery bank - Well, it's a prime example of how NOT to wire batteries. Looks like the charger and load feed off the big battery on the left ? That leaves the other batteries at the ends of longish jumpers, and the fractional ohm resistance greatly reduces their contribution to the system.

Inverter - ack ! it's a chapter of how it's not supposed to be done. Fortunately, going to a 48V pure sine (even a 24V pure sine) inverter will lessen the amps on the system, but all your series batteries, need to be the same size (golf cart batteries would work OK)

Charger - I take it you are using a generator to charge this, only when the batteries are so low it won't start the pump? Any plans for solar, or a auto-start module for the generator ?

Well, you won't find better advice about this stuff on planet earth, except for here.

|| 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✭✭✭✭@BB Good Stuff, I'll see if I can follow it.

- In following your post - I want to make it clear that the 6.7 amps of running current was done at the pressure switch bypass, on the output side of the inverter.

FYI the 220v wires were removed from the pressure switch and connected together to create a long "extension cord" to the well pump. The pressure switch is wired to a relay that turns the 220v inverter on and off. This gets me around having to have the inverter on 24x7 for a few minutes of run time.

Also note that the inverter claims to operate 220v at 50Hz

@Blackcherry04 - Noted, and thanks. But this has been an Ad-Hoc system now for about two years. At present we just throw anything handy in the mix and hope for the best. That's why I am currently planning out how to do it right. initially i was considering 12v deep cycle with capacitors to take up the initial motor surge.

In reviewing the system tonight I found out that the "good" batteries I was given to work with in the beginning were the old T105's from the house bank. So it seems I was trying to pull these loads from a collection of four year old golf cart batteries. oddly enough, it didn't work. So lets disregard that part of the discussion.

OK, Granted that MSW is not as efficient as pure sine; on the other hand I can pick up a 3k surge to 6K 48v Inverter/charger for under $300. Now, the existing 12v version has performed well, except for losing an argument with lightning, for a couple of years. Ultimately it's not an expensive gamble.

next, Well pump stats.

Franklyn 230v three wire 1/2 HP, around 80' down and about a hundred yards from the house.

Initial starting surge; around 4500 watts. (credit to cariboocoot)

Running wattage; around 1500 watts. (Ditto)

Running time per cycle around 8 minutes. This I'd like to up to 12 minutes by increasing the shut off pressure of the well pump.

That brings up the first question; am I better off running a larger number of lower psi pump cycles or do I get a better energy use from running higher PSI cut off cycles - and fewer cycles? Currently I think the pump kicks in around 20psi and out at 55psi

@BB, I have two RMS meters and one non RMS. - Could you point me in the right direction for info on how to understand the data I get. I'm not yet grasping the Vector approach. I get watts, volts and amps and am now trying to grasp the implications of PF and THD in the MSW.

next, In dealing with MSW, @50hz and the Franklyn pump, this is clearly not optimal. But it works.

If the PF of the well pump is off because of the MSW and 50hz and the inverter is not as efficient as the manufacture claims (Gasp) - why not just guesstimate the inefficiencies and oversize the batteries accordingly?

In order to get a decent estimate of energy used I'd need: Volt, Amps, (easy enough) but then I would also need Voltage-Current Phase angle and the PF of the motor.

Guess which part of this I am having trouble with?

Fortunately I have running time and an approximation of the wattage used. Since I'd like to increase the running time to twelve minutes, then I'd like to plan my batteries for a 48v system that will provide 48 minutes of running time with no less then a 60% SOC at the end of the four cycles.

Running wattage is @1500 x 48 = 72000 watt minutes. /60 = 1200 or 1.2KwH

At this point how do I best factor in the four @4500 watt starting surges?

And yes, still thinking in Watts as opposed to Volt-Amps. Fortunately since I have, volts, amps and running times I don't need to estimate Volts*amps*pf to arrive at the reserve power required. . . . . . or do I?

@BB Power = V*A*PF = 240 VAC * 6.7 Amps * 0.67 PF = 1,077 Watts > This I do not understand. Since I measured 6.7 amps on one of two 110v legs going to the well pump, then why am I using the PF to decrease the Watt demand?

At this point, after I get another beer, I just need to do some more basic math.

1200 Watt Hours /12v = 100Ah worth of power.

Since I want the 100ah usage to represent no more than 40% of capacity. then I am looking at a 250ah battery. Since I rarely have fully charged batteries - drawing power from say 90% of charge to 60% of charge I should bump this up 10% or so.

However in looking at the trojan site the closest I can come to this is the J185H-AC, BCI group size 921 with a five hour rate of 185ah.

So, what did I do wrong and how do most other people work through this?

Thanks again for all the help.

80✭✭✭✭@cariboocoot - Found that site earlier, thanks for pointing it out though.

Actually all of the batteries should be connected with the #2 method, it might not be obvious in the photo. Also, the last two batteries are not connected. Didn't ask, don't know. Kinda why I am trying to come up with a plan to do this correctly, once.

Edit to add:

Crap. You're right. The guy that owns the place did it because he claims it works better like this. He has a buddy at a junk yard for free batteries. . . .

Anyway, I'll print out the page again and go over it tomorrow. Because if it really is working better like this then the other batteries are crap.

17,615✭✭More likely it is on at 25, off at 45: there's usually a 20 psi spread, with an adjustment of 10 psi or so on the switch.

As for dividing between # cycles and longer run time - pick run time if you can. Every turn-on is a hit on the batteries and the pump. So you should have as big a pressure tank as you can fit.

THD (Total Harmonic Distortion) is Voltage variance from the sine wave output. It looks like up and down spikes on what should be a smooth curve. As such the Volta/current relationship is off at these points.

PF (Power Factor) can be thought of as the difference between the current demand curve and the Voltage supply curve (it is a function of the varying resistance caused by induction). A motor with a PF of 0.6 may read 800 Watts on a Kill-A-Watt meter but in reality it is using 1333 Volt Amps.

The net effect of these two things is the same: more power consumed than what appears to be.

You don't want to be judging battery capacity by the 5 hour rate: it will be lower than the 20 hour rate we normally use (the higher the current draw the lower the capacity). The battery you mention is 225 Amp hours @ the 20 hour rate.

The big issue is the 'hit' the batteries will take to supply the start surge. One string of these would stand 90 Amps max, which is only 1kW on 12 Volts.

230 Volts does not have two 110 Volt legs: it simply has 230 Volts, the current goes out one side and in the other (so to speak). It is not like split phase 240 Volts, although that functions the same for 240 Volt loads. 6.7 Amps on 230 VAC is 1541 Watts.

NOTO BENE: Someone has been gabbing in my ear here so this is out of order and may have mistakes. You can never get people to shut up and leave you alone when you're doing something important.

8,380✭✭✭✭✭Larger accumulator tank, fewer run cycles. Toss the one with a bad membrane, it will just collect stagnant water and eventually rust through. Get the largest you can afford, you want to reduce cycle time, with a larger accumulator .

Higher PSI makes pump run harder, consumes more power. Starting the motor is hard on the motor & batteries.

|| 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✭✭✭✭@Cariboocoot, Thanks for the input. Don't mean to disagree but:

The inverter supplies two 110v legs and a ground/neutral. I double checked this tonight with the meter. This is not what I would call a single leg 220v like I see in Europe but a double leg 220 volt output like we get here.

As far as:

THD I can follow from the earlier posts on generators. But the implications of how it affects each class of equipment is what I am working through. Since I can't yet purchase a large enough Pure Sine inverter to handle all of the loads here I am trying to balance them between the 6K MSW and the 1K PSW.

PF, Good information. I use the Effergy system to track power usage here. I've played around with it some to compensate for the MSW but at best it's an approximation. If I understand this correctly will my RMS meter show correct amps being drawn through that circuit?

Batteries; so in sizing the batteries to the ones I mentioned, and switching to a 48v bank, four of the 921 size, then I can expect to provide a surge of 90 x 48 = 4320 watts? If so, this is still not enough. Especially if my clamp meter is not giving an accurate reading of the amps being drawn.

What kind of system do most people put in place to meet these demands?

17,615✭✭Fine (220/110 was an old power standard in NA), but if the load is 220 Volts there is only 220 Volts to be considered and the current carried on one leg is the

samecurrent (not the same number X2) as on the other. And while we're at it, check the specs on that pump motor for Voltage and frequency; running it off spec for either will, once again, add to the power consumed.It has less of an ill effect on an induction motor than it will on many electronic components, but it still has an effect.

Good question. It isn't that the current or Voltage is wrong per se, but rather the curve function with an inductive load is off; the resistance varies creating greater/less demand not in sync with the Voltage curve of the sine wave. PF is a "load-based" problem and occurs on either MSW or pure sine. Mostly it is ignored on the grid because of the huge power available, except in the case of industrial applications where it is such a large factor that the utility has to measure and bill for it.

Two things: the Amp hour (and thus Watt hour) capacity and its relation to maximum current draw. So if you have 225 Amp hours @ 48 Volts you have a maximum of about 5.4 kW hours stored (not including all the little things that eat into it) @ 50% DOD. You'd want to limit this to 25% DOD ideally, or 2.5 kW hours. The bank could still supply a 90 Amp 'hit', but @ 48 Volts that is a 4320 Watt draw. It is easy to confuse Watts and Watt hours and not understand why you need to accommodate both.

I have my own rules-of-thumb regarding picking system Voltage and battery capacity. The first is to avoid 12 Volt systems unless there is a specific need for 12 VDC such as a mobile application. The second is to only shift up to 48 Volts if there is a large Watt hour need because 48 Volts needs some specific (expensive) equipment for circuit protection and switching. I would not want to supply a 4kW+ load from 24 Volts, however. Nor would I want to do it with the surge rating of, say, a 3kW or 4kW inverter.

If I were you, I'd fix up the water system first to make sure it is working as best possible. Then I'd replace those batteries with something decent to see if you can get by on the inverter you've got. Why? Because frankly the kind of inverter normally used to supply that amount of power is quite expensive. Like this: http://www.solar-electric.com/maenms4444wa.html

80✭✭✭✭@cariboocoot: Sorry to pester you but I am definitely missing something here in my understanding of current.

Fine (220/110 was an old power standard in NA), but if the load is 220 Volts there is only 220 Volts to be considered and the current carried on one leg is the same current (not the same number X2) as on the other.

6.7 amps at 110 volts per wire, times two wires? What's wrong with my thinking. If each wire is carrying around 737 Watts, then why does it matter how I end up with 1474 watts? 6.7 x 110 x 2 or 6.7 x 220? Current being electrical charge, Correct?

And while we're at it, check the specs on that pump motor for Voltage and frequency; running it off spec for either will, once again, add to the power consumed.

The fact that this is running out of spec and using more power, is pretty much a given.

Considering what I've spent here, the price of that inverter is not a deal breaker. But, like finally buying the Honda inverter generator I'd like to know how to do it right and why.

I do appreciate the help, thanks again.

29,518adminGood idea--Idling inverters can draw 6-20 watts or more... For something that is off more than it is on, that can be a huge waste of energy.

If you have other needs for power, eventually you may have to live with an "always on" or an inverter that has "search mode" (turns on ~once per second for a few cycles looking for >6+ watt AC loads--Then turns on 100%).

That is fine--We mostly work in AC Watts and then if 120/230/240 50/60 Hz is simply the right inverter for the job (Watts are--for the most part--voltage/frequency independent).

If you can get an 80Farad cap for the right price--Sure would not hurt to try. With the huge variablility in systems, components, existing vs new, size of bank account, etc.--Experimentation (well thought out and safely done) is certainly worth trying.

Which is why we use a lot of the "rules of thumb"--They have been found to give a pretty reliable system many years down the road (new battery bank working fine, but two years down the road and the system starts getting dicey--We try to avoid "on the edge designs". Leads to unhappy campers.

Yea--it is hard to ignore the $300 inverter that will power 80% of your loads without much in the way of problems--Vs a $3,000 inverter that will power 99% of your loads and may include many bells and whistles (AC battery charger, programmable settings, remote data access, etc.).

Note, if you increase the pressure, you will use more power to pump the water. If you do not "need" the increased pressure, may not be the best choice... And instead install another pressure tank.

If you are looking for better efficiency, you should look at the PM (Permanent Magnet pumps with VFD based controllers) such as Grundfos SQFlex and others. They are designed for very efficient operation (induction pumps use are about 20% less efficient). These are not cheap pumps--But some can be operated from battery, AC, and solar panels directly with the same pump setup (backup power, etc.). These are not cheap pumps (SQFlex are something like $2k each).

This is a pretty detailed explanation:

https://en.wikipedia.org/wiki/AC_power

But there is lots of math (AC Power "Is Math")...

For one way of looking at Power Factor-- PF = Cosine (voltage to current phase angle). The analogy of pulling straight on vs off to the side (straight on Cosine 0 = 1.0; Cos 60 = 0.5; 90 degrees off to the side is Cos 90 = 0.0)...

When you have non-sine waves, you can use PF sort of like "current efficiency"... For example many computer power supplies only draw current at the "peak" of the sine wave voltage wave forms (diode charging a high voltage capacitor). See this website for a good explanation:

http://www.data-acquisition.us/sss_6-0.html

A typical power supply may have a PF in the range of 0.67 to 0.80 where a "power factor corrected" power supply or load has a PF ~0.95 (PF=1.0 is "perfect", but because of other issues with circuit stability is not normally the end goal--0.95 is "good enough").

The conversion between 120 VAC RMS and 120 VDC equivalent is the Square Root of 2 for sine wave peak to DC voltage equivalent (it is the area under the sine wave curve squared):

120 VDC * srt(2) = 169.7 Volt Peak Sine Wave voltage equivalent power

A "cheap" Digital meter actually measures the sine wave peak voltage and divides by the Srt(2) to get the "DC" RMS equivalent voltage. However for each non-sine waveform (square, triangle, etc.) the conversion factor is different.

An RMS (root mean square) reading meter actually used digital signal processing to measure the area under an arbitrary waveform to get the "DC equivalent". So they are much more accurate when measuring MSW and other voltage/current wave forms (for amps/volts and computing VA and Watts).

We do--or Guess, or say VA=Watts--That makes a worst case power estimate (VA is always greater than, or equal to Watts--So making "errors" on the conservative side). Pumps age, Batteries age, etc... So this gives a bit more head room 5-10 years down the road.

If you use a good quality RMS reading meter (or even a non-RMS meter for measuring motors--and not computer power supplies) to measure the motor V&A and the DC input V&A to the inverter, you can "estimate" the Power Factor:

Power = Volts * Amps (DC) * Inverter Efficiency = Volts * Amps (RMS AC) * PF

Or:

VA (DC inverter) * 0.85 typical inverter eff * 1/(VA RMS AC) = PF (typically should be ~0.6 to 0.8 PF range for induction pump)

I like to use Current Clamp meters to measure DC and AC currents (cheap unit from Sears, as an example). Much safer, easier, and faster than using a DMM set in Current Mode wired into the AC wiring. And clamp meters can easily measure ~40-400 Amps -- Much more than the 10 amps typical for a DMM in current mode.

Note that batteries have design requirements. A typical flooded cell battery should be operated at a maximum of (roughly) C/8 to C/5 current rates (battery discharged to "dead" in 8 to 5 hours). You are looking at current rates in the C/1 or even less... You would need an AGM battery to come close to doing this.

And that is your dilemma--A much larger bank to manage the surge/run currents (and shallow discharging). Or an AGM (or other chemistry) that can support sustained high current outputs (C*4 or 15 minute discharge is possible with some AGM batteries--which cost 2x or more compared to flooded cell).

VA is always equal too, or large than Watts--So that is the "conservative" error (typically 20-30% error "too much power").

Because Watts is "real work" at the end of the day. VA is what the transformers, wiring, inverters, gensets "see" for I

^{2}R heating effects and other losses.Another analogy, if you "pedal just right" on a bicycle, you get very efficient power transfer from your legs to the wells. If you are "uncoordinated" on the pedals (pushing hard at the top/bottom of the stroke), you are out of phase with the pedal position and not getting maximum power transfer. PF is "coordination" factor -- 1.0 is perfectly coordinated, and 0.0 is perfectly uncoordinated.

Marc (Cariboocoot) has been helping you with the math...

-Bill

17,615✭✭Each wire is not carrying 6.7 Amps. One

circuitis carrying 6.7 Amps. Circuits are a loop: power out one wire and back in the other. So the current measured on one wire is the same power on the other one, just "coming back". The fact that there are two "hot" wires makes no difference. If this is a 220 Volt load the current goes out one hot and back in the other; it does not go out both hot wires at once and then disappear.Any time power specifications and load specifications don't line up more power is used and the system is less efficient. Too high/low Voltage, wrong frequency, distorted waveform all make for greater real power consumption. The pump motor can run slower, having to do more work (draw more current) and run hotter shortening its lifespan. You can well see the effect if the situation is exaggerate like trying to run a 240 Volt motor from 120 Volts; it tries to spin, can't get up to speed, heats up, shuts down. Although off frequency and the V-spikes of s distorted sine wave do not show up this easily, the effect is still there.