Parallelling capacitors

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Electron_Sam78

Senior Member
Location
Palm Bay, FL
Tomorrow I will have to parallel some capacitors together temporarily until we get the right replacement in. I haven't done a whole lot of work with capacitors over the years so I'd like to run this by the members here. We have a 65uf, 240VAC cap. (this is a factory cap. on a commercial gate opener motor, 1/2 HP, 120V). I have spare motor run caps. with ratings of 3uf, 370VAC + a 30uf, 440VAC + a 32uf, 400VAC. All parallelled I would have a 65 uf circuit, with a maximum safe operating voltage of 370 VAC correct? Would this setup work or do all the parallelled caps need to be around the same capacitance? Also, from what I've researched the voltage rating only determines maximum safe operating voltage. You can go lower but not higher and circuit voltage doesn't change capacitance. Is that correct?
 
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kwired

Electron manager
Location
NE Nebraska
Tomorrow I will have to parallel some capacitors together temporarily until we get the right replacement in. I haven't done a whole lot of work with capacitors over the years so I'd like to run this by the members here. We have a 65uf, 240VAC cap. (this is a factory cap. on a commercial gate opener motor, 1/2 HP, 120V). I have spare motor run caps. with ratings of 3uf, 370VAC + a 30uf, 440VAC + a 32uf, 400VAC. All parallelled I would have a 65 uf circuit, with a maximum safe operating voltage of 370 VAC correct? Would this setup work or do all the parallelled caps need to be around the same capacitance? Also, from what I've researched the voltage rating only determines maximum safe operating voltage. You can go lower but not higher and circuit voltage doesn't change capacitance. Is that correct?

Everything you said is correct to my knowledge, caps in parallel are additive no matter what their values are. If you had a 3 and a 30 it is kind of like building one with a little more substance to it to give it a rating of 33.

Is this a start or run capacitor? I can't ever recall that small of a motor with that large of a run capacitor, and at same time never that low of capacitance for start capacitors. Most PSC motors are for low torque applications like fans, but I guess this could be a PSC motor with higher torque.
 

Phil Corso

Senior Member
Electron_Sam...

1) You will have to calculate the "equivalent" capacitance on a common voltage basis!

2) Also, adding to Kwired's comments, the 'start-cap' is usually an electrolytic-type, while the 'run-cap' is polypropylene or oil-filled!

Regards, Phil
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
I understand that capacitors with different voltages are not an issue if the lowest voltage unit is the proper voltage. In other words, the voltage will only be as high as the voltage of the lowest unit. Is that not true?
 

Electron_Sam78

Senior Member
Location
Palm Bay, FL
Electron_Sam...

1) You will have to calculate the "equivalent" capacitance on a common voltage basis!

2) Also, adding to Kwired's comments, the 'start-cap' is usually an electrolytic-type, while the 'run-cap' is polypropylene or oil-filled!

Regards, Phil

1) I thought I was calculating the equivalent capacitance by applying the formula C=C1+C2+C3 to get 65uf?

I'm not sure if this is a start or run cap. I assumed it was a run cap based on the ratings though 240 VAC is unusual for a run cap
 

Dennis Alwon

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Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
1) I thought I was calculating the equivalent capacitance by applying the formula C=C1+C2+C3 to get 65uf?

I'm not sure if this is a start or run cap. I assumed it was a run cap based on the ratings though 240 VAC is unusual for a run cap

As long as the capacitors are paralleled then you add them. Capacitors are opposite resistors so in series they are 1/C +1/C +....
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
120616-0812

The lowest voltage rating capacitor in a group of paralleled capacitors determines the maximum voltage that should be applied to the parallel combination. The maximum voltage applied to a capacitor should be moderately below its rating. In other words you should operate with some safety factor.

When a capacitor is operated in a series circuit with an inductor, then you have a series resonant circuit at some frequency. At resonance the voltage across both the capacitor and inductor may be very much larger than the applied voltage to the series circuit. As you move off of resonance the voltage drops. In a motor with a run capacitor the capacitor voltage may be substantially higher than supply voltage. This might be 1.5 times line voltage, possibly higher.

A good discussion on capacitor motors is in "Alternating-Current Machinery", by Bailey and Gault.

The sum of the individual capacitance vales determines the parallel capacitance.

It would be desirable that all the capacitors be polypropylene. This is a low leakage high quality dielectric material. For a starting capacitor it is normal to use an electrolytic capacitor to reduce size and cost, but the losses are too high to use for a running capacitor. I think for a temporary measure don't bother with the 3 ufd.

Run capacitors are typically in the range of maybe 1/4 to 1/3 of the capacitance of a start capacitor.
 

Phil Corso

Senior Member
Electron_Sam, Kwired, and Dennis,
Like any other calculation involving electrical components with varying capacity and voltage ratings, including capacitors, then their impedances should be adjusted to a common base.
The Run- and Start-Winding of a single-phase motor are considered a series R-L-C circuit, which under certain conditions, could result in severe over-current due to circuit-resonance Thus, motor mfgs often issue a warnings regarding ?Capacitor? replacement. And, in many case voltage across the cap is well above the supply voltage. Hence 240, 370, 440 cap ratings in 120 Vac rated motors. And, NO! Never substitute a cap have a lower than specified voltage!
The calculation is straight forward, but it insures that substitution will not materially alter circuit parameters!
In closing, will the cap substitution proposed by Electron_Sam signicantly affect winding integrity? Or alter Running-Mode Torque? Probably not! And, if Electron_Sam's already installed his "fix? then my cautionary comments are unwarranted! On the other hand? !
Regards, Phil
 

a.bisnath

Senior Member
paralelling is fine

paralelling is fine

I have done it many times especially in fan or blower circuits one thing I must say never go higher in value ,I have seen this done by accident and it caused the winding on a pump to burn.A good book that explains it is Electrical Theory by Hughes ,It shows you lots of thing that capacitors can do from a theoretical point of view which I have actually done and used
 

iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
Occupation
EE (Field - as little design as possible)
... 1) You will have to calculate the "equivalent" capacitance on a common voltage basis! ...
I'm missing your point.

If you were discussing reactive kva then yes, the kva is adjusted by the ratio of the square of the voltages. But not the capacitance.

ice
 

Phil Corso

Senior Member
Ice...
When using a Base system, all reactances, capacitive as well as inductive, must be modified with a correction factor.

You mentioned one of two... square of voltage-ratio. The other is the ratio of their rated capacity, say VAcap (or kVAcap) to the chosen base VAbase (or kVAbase). Vcap is its rated voltage. And its rated currrent, Icap, is found from, Vcap/Zcap, thus VAcap = Vcap x Icap!

Once the reactance is modified, then an equivalent, Ceq, can be determinedfound.

Regards, Phil Corso
 

kwired

Electron manager
Location
NE Nebraska
Ice...
When using a Base system, all reactances, capacitive as well as inductive, must be modified with a correction factor.

You mentioned one of two... square of voltage-ratio. The other is the ratio of their rated capacity, say VAcap (or kVAcap) to the chosen base VAbase (or kVAbase). Vcap is its rated voltage. And its rated currrent, Icap, is found from, Vcap/Zcap, thus VAcap = Vcap x Icap!

Once the reactance is modified, then an equivalent, Ceq, can be determinedfound.

Regards, Phil Corso

Could not either of those words in the dictionary:D
 

iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
Occupation
EE (Field - as little design as possible)
Ice...
When using a Base system, all reactances, capacitive as well as inductive, must be modified with a correction factor.

You mentioned one of two... square of voltage-ratio. The other is the ratio of their rated capacity, say VAcap (or kVAcap) to the chosen base VAbase (or kVAbase). Vcap is its rated voltage. And its rated currrent, Icap, is found from, Vcap/Zcap, thus VAcap = Vcap x Icap!

Once the reactance is modified, then an equivalent, Ceq, can be determinedfound.Regards, Phil Corso

Phil -
From your previous posts you are plenty shart. However you really need to read this again.

A capacitance, say 25uf at 12.46KV is still 25uf at 7200V. At 12.46KV, 25uf is ~1.5MVARs. At 7200V, 25uf is ~500KVARS

The reactive loading changes, the capacitance does not.

And, I'm thinking you already know this.

ice
 

Phil Corso

Senior Member
Iceworm? I apologize for my late response to your question, but I was distracted trying to satisfy a member?s craving for minutia!
Addressing your reply # 15: Paraphrasing the original question, ?Is the performance of a single-phase motor affected if its RUN or START Cap is replaced with one having different parameters? My answer was, and still is, Yes!

For simplification assume Electron_Sam replaces the original 65uF/240V cap Co, with a new 65uF/ 420V cap Cn. Also assume capacitor losses are zero. Then:

1) Co?s rated-impedance Xo is 40.8Ω, and Cn?s rated-impedance, Xn is 40.8Ω.

2) Co?s rated-current Io is 5.9A, and Cn?s rated-current, In is 10.3A.

3) Co?s rated VAr(o) is (Vo)x(Io) or 1,410, while Cn?s rated VAr(n) is 2,850.

4) Putting Cn in the circuit having a ?working? voltage of 240 Volts, produces a VAr(n)? of 886, or about 31% of its capacity.

5) This corresponds to an adjusted Cn? = 0.31 x 65 or 21uF.


6) Impact on the RUN circuit will be to reduce power-factor!

7) Impact of the START circuit is more insidious because starting-current may be too low!

8) Only Electron_Sam can say whether or not it worked in his case!

Regards, Phil Cors
 

iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
Occupation
EE (Field - as little design as possible)
Phil -
You can't be serious. Generally, your responses show an excellent technical knowledge. I'm certain you already understand this. You really need to read your responses again - and do the math again.
Addressing your reply # 15: Paraphrasing the original question, “Is the performance of a single-phase motor affected if its RUN or START Cap is replaced with one having different parameters? My answer was, and still is, Yes! ...
And my answer is, 'It depends on which parameters are changed. If the only difference is the new cap has higher working voltage, then the motor performance is not affected.

...
1) Co’s rated-impedance Xo is 40.8Ω, and Cn’s rated-impedance, Xn is 40.8Ω
Okay X = 1/(jwC) = -j/(377 x 65 x 10^-6) = -j40.8. So both caps have the same impedance. And that impedance is independent of the working voltage. Good

...
2) Co’s rated-current Io is 5.9A, and Cn’s rated-current, In is 10.3A.
The current through a 65uf at 240V => 240/40.8 = 5.9A
The current through a 65uf at 420V => 420/40.8 = 10.3A
Alright those numbers are good

...
3) Co’s rated VAr(o) is (Vo)x(Io) or 1,410, while Cn’s rated VAr(n) is 2,850.
This is the first error in your calc/reasoning:
For Co rated vars, you used Co rated voltage, 240V, and Co rated current, 5.9A which gave you Co rated vars = 240 x 5.9 = 1416 Vars.
For Cn, you used 240V, and 10.3A to get Cn rated vars. That would be in error. Using your reasoning, Cn rated vars = 420 x 10.3 = 4326Vars. If you want the rated vars, you must use the rated voltage. You account for the circuit voltage later.

...
4) Putting Cn in the circuit having a ‘working’ voltage of 240 Volts, produces a VAr(n)’ of 886, or about 31% of its capacity.
You appear to have reduced Cn rated vars by the ratio of the squares of the circuit voltage to Cn rated voltage. (240^2)/(420^2) = .326 I get ~33%, but that is close enough to your answer of 31%. So that gives me (Cn vars in a 240V circuit) = 4326 x .33 = 1427 Vars.

Isn't that just amazingly close to the value for a 240V rated cap in a 240V circuit. Why I'll bet it is within the roundoff error.

...
5) This corresponds to an adjusted Cn’ = 0.31 x 65 or 21uF.
And that is just wrong.

Capacitance does not change with the circuit voltage.

Capacitive Impedance does not change with circuit voltage.

Reactive VARS change with circuit voltage.

But in this case the circuit voltage was 240v and remained 240V. Changing the capacitor did not change the circuit voltage. In the 240V circuit, Cn vars are the same as Co vars.

ice
 

Phil Corso

Senior Member
Ice...

I believe I'm talking apples and you, oranges. I am preparing a spread sheet to illustrate my point. Please be patient, I have a very old computer and it has a very slow keyboard!

In the interim, please carry out a very simple calculation. Consider a three-element series-circuit consisting of a generator, a current limiting reactor, and a capacitor. Their parameters are:

o Genenerator: 4,000 kVa, 4 kV, 9% ST reactance.

o Reactor: 2,000 kVAr, 4 kV, 6% reactance.

o Capacitor: Cap: 10,000 uF; 4 kV.

A) Calulate the current flow for a short-circuit at the end of the circuit.

B) Calculate current again, but change reactor and cap voltages to 3 & 6kV, respectively.

Regards, Phil
 

Phil Corso

Senior Member
iceworm, Mivey, and interested others...

I apologize for my presentation! It is classic FUBAR... oweing to a bit of hubris!

Hopefully, I will be able to reply on or about the July 10th (2012)!

Regards, Phil
 
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