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Super Capacitor Bank Output

Did pump get up to speed? It could have been turning slowly, drawing excessive current (because it couldn't get enough all at once to reach full speed.) In this case maybe something boosting available power for a short time as you're considering might help.

If it was up to speed, then either inverter or battery couldn't supply the operating power.

Review running amps of motor and rating of inverter & battery.
Starting pump with grid or generator could let you measure, using a clamp ammeter with inrush.
 
Did pump get up to speed? It could have been turning slowly, drawing excessive current (because it couldn't get enough all at once to reach full speed.) In this case maybe something boosting available power for a short time as you're considering might help.

Something, like more super caps!
 
An extra 250W won't do anything for you.
But if delivered in 0.5 seconds, an extra 5kW is probably what you need.

30F, drained from 28V to 24V would deliver 3120 Joules or Watt-seconds, according to my math.
So 6kW for 1/2 second. That could be useful.
I just found the pump motor specs online. The locked rotor amps (in-rush) is 20.5A for .5 seconds, about 4.7kW so this super capacitor bank may do the trick.
 
Of course precharge capacitors through resistor before paralleling with battery.

Total energy in capacitor is 1/2 C V^2, but the energy available to inverter is difference between that energy when V = initial battery voltage minus that energy when V = low voltage disconnect of inverter.

Capacitor has a series impedance (resistance), and battery does too. That determines current sharing when inverter draws voltage lower, and voltage vs. current curves determine how much current is available.

Lithium battery can probably supply all the current inverter needs to start any motor. BMS maybe not. A system that bypassed BMS FETs and had BMS wired to remote on/off of inverter, if it provides suitable delay, could work.

What is surge of pump? If 40A 240V, about 10kW, from 24V battery that's about 400A. A 2C surge for one second from 200Ah battery? That would seem something it could handle. Maybe just get a BMS with specs to cover that.



An extra 250W won't do anything for you.
But if delivered in 0.5 seconds, an extra 5kW is probably what you need.



30F, drained from 28V to 24V would deliver 3120 Joules or Watt-seconds, according to my math.
So 6kW for 1/2 second. That could be useful.

I just found the pump motor specs online. The locked rotor amps (in-rush) is 20.5A for .5 seconds, about 4.7kW so this super capacitor bank may do the trick

only 20.5A? Can you give the link to the spec sheet?
I found my pump motor part # in the pump owner's manual, and a search led me to this.
 

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I don't see 20.5 under "locked rotor amps"

20.5A x 240V = 4920W

If your inverter is 5kW, or can surge to 5kW for a second, then it should be able to start the pump.
Given enough current. Is your battery up to that, or is that why you think supercap is needed?
From 24V battery, about 200A, maybe 220A or so depending on inefficiency and how low voltage drops.
 
I don't see 20.5 under "locked rotor amps"

20.5A x 240V = 4920W

If your inverter is 5kW, or can surge to 5kW for a second, then it should be able to start the pump.
Given enough current. Is your battery up to that, or is that why you think supercap is needed?
From 24V battery, about 200A, maybe 220A or so depending on inefficiency and how low voltage drops.
I've been told that I might need 25A at the inverter MOSFETs, so that would be maybe 200A input. What I'm planning is using a 24V lead acid starting battery bank. WalMart group size 26 batteries (~$60 each) combined to make a 24V bank with massive cold cranking amps.
 
I've been told that I might need 25A at the inverter MOSFETs, so that would be maybe 200A input. What I'm planning is using a 24V lead acid starting battery bank. WalMart group size 26 batteries (~$60 each) combined to make a 24V bank with massive cold cranking amps.
I thought you were using super caps? Did the plan change to lead acid starting batteries?
 
I thought you were using super caps? Did the plan change to lead acid starting batteries?
I may try both ways. My lead-acid amps may be limited by 2 AWG cables. According to the charts, that diameter is meant for 120A, although the length is just 3' total. A 220A surge through a short 2 AWG cable is reasonable?
 
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A short surge in copper wire is not a problem. The issue would be if it has enough time to overheat, which takes a long time.
Just calculate resistance and IR drop, which will reduce voltage at inverter.

A similar application of high current in undersized wire is ground wire. From the following chart, a 4 awg ground wire would be sufficient for 2/0 current-carrying wires (200A), stays cool long enough to trip breaker. 2 awg ground is shown for 4/0 up to 350 kcmil (which is rated for 350A.)



 
A short surge in copper wire is not a problem. The issue would be if it has enough time to overheat, which takes a long time.
Just calculate resistance and IR drop, which will reduce voltage at inverter.

A similar application of high current in undersized wire is ground wire. From the following chart, a 4 awg ground wire would be sufficient for 2/0 current-carrying wires (200A), stays cool long enough to trip breaker. 2 awg ground is shown for 4/0 up to 350 kcmil (which is rated for 350A.)



Good, thanks. BTW here is my exact pump motor electrical spec. The second one in the chart. LRA is 20.5, and running is about 12A.
 

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20.5 LRA would be remarkably low for 12A running; I would expect 60 LRA (5x running).
The table says 5.7A full load, 6.3A service factor.

240V / 5 ohm main windings would be 48A
240V / 12 ohm starting windings would be 20A (but if in series with a capacitor, that would be reduced by capacitor impedance.)

0.37 kW / 230V = 1.6A, much less than 5.7A
Could service factor = 1.6 mean that much excess "reactive" current from inductance?
 
20.5 LRA would be remarkably low for 12A running; I would expect 60 LRA (5x running).
The table says 5.7A full load, 6.3A service factor.

240V / 5 ohm main windings would be 48A
240V / 12 ohm starting windings would be 20A (but if in series with a capacitor, that would be reduced by capacitor impedance.)

0.37 kW / 230V = 1.6A, much less than 5.7A
Could service factor = 1.6 mean that much excess "reactive" current from inductance?

The listed winding resistance for my model is higher than bigger motors, but the LRA seems proportional to all the rest.
 
How about trying your inverter with the pump, and tell us what happens?

Could be like video linked in this post:

 
How about trying your inverter with the pump, and tell us what happens?

Could be like video linked in this post:


I will eventually. Right now, I'm building the lead-acid battery bank, and could finish that today. The next step is to build the capacitor bank, and test charge it with my scc, then discharge it with my inverter and get a feeling for the output. With the well, I need to improve the 240V plug connection to the inverter, test a slightly higher amp pump starting capacitor, then string temporary 12 gauge wire 50 feet from the inverter shed to the well.
 
SO what was the outcome? looking for a way to keep our sump pump from tripping the overload when it comes on at the same time as the fridge compressor.....there is a 22+ amp surge for just a moment....
 
I have two systems for my house. The smaller one has a supercap in it to allow greater current than the battery specs for motor startups. This battery is connected to a large reverse cycle AC, a bore pump, and a desalination unit. It can run the rest of the house if required.

The cost if the supercap was a fraction of the cost of sizing the battery to remain below the 0.5C discharge i was aiming for.

The supercaps perform as expected.

My opinion on why supercaps don’t get the airtime they deserve is the majority of people haven’t experienced the long term gains that understressing your battery delivers.
 

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Have you measured voltage with/without caps, battery current with/without, and capacitor current?
With Lithium, would be interesting to try that both fully charged into/past the knee, and in the flat region.
I figure voltage has to sag to get power from capacitor, but super-capacitor doesn't need to draw down nearly as far as electrolytic decoupling capacitors would. If reducing battery current peaks is your goal, designed-in resistance might help.
 
Lots of talk of coping with the surge, but have you considered just using a Soft Starter? It's what they're designed to do.

Hi I would want a soft starter for a 120v sump pump also
 
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