1. Aug 19, 2011

### yshtar

I'm trying to get the correct capacitor for my 2 water pumps. One is 125watt and the other is 300watt, and both run at 220v. I don't know if it's the start or run capacitor, btw.

I'm getting that since:

1F = 1C / V, and 1J = 1C / V, and 1W = 1J / second, that 1F = 1W.

So the capacitor for the 125 watt motor should be: 125F / 1V = x / 220V = 0.568F, to get the 125watt power.

Am I correct with this equation?
If not, what is the equation to get the correct capacitor for the motor?

2. Aug 19, 2011

### Gordianus

Do you need a starter or a reactive-power compensating capacitor?

3. Aug 20, 2011

### yshtar

I don't understand that, I'm afraid.
I have read that the start capacitor is to boost the speed and stop at a later time, but I haven't known about the other function of capacitor you mentioned. And I don't know how to determine the function of the capacitors on my water pumps. In fact, it's only a few days ago when I knew that a capacitor with different capacities would give different results.

4. Aug 20, 2011

### K^2

If the motor runs from AC, I'm guessing it's a compensating cap. In that case, you need to know the inductance of the motor's coils to know the capacitance you need.

Do you have manufacturer/model numbers for these motors? It might be something we can look up.

Edit: Actually, there is one more way. Can you measure the RMS current these motors draw when running at full power?

5. Aug 20, 2011

### FireStorm000

I think your units for Jules is wrong. Power is Current times Voltage, or W = A * V, and energy is W * s = A * V * s = C * V.

And I agree with K squared that the purpose is most likely to reduce the power factor of the motors.

6. Aug 20, 2011

### yshtar

Yes, they use AC power. 220 volts.
I'm not sure if I can' get the manufacturer's data on this. I"m in Indonesia and they don't publish such thing. Besides the 300 watt one is about 30 years old, and the 125watt is about 20 years old.
And I don't know how to measure the inductance, and the RMS. What is RMS and what do I need to do to measure both the RMS and inductance?

@firestorm000: "W = A * V, and energy is W * s = A * V * s =... C * V...."
So, 1J = 1C * V, not 1C / V ?

Last edited: Aug 20, 2011
7. Aug 20, 2011

### uart

No. CV^2 has the units of Joules.

Re the capacitors.

- Do the motors currently run or do they fail to start?

- Why do you need new capacitors. Can you see the exiting ones damaged?

BTW You're probably looking at capacitors in the "ballpark" of 2 to 20 uF, but it's not possible to say much more without more details.

8. Aug 20, 2011

### K^2

Impedance is what you really need, but there are indirect ways of measuring it.

RMS just means Root-Mean-Square. It's a way of measuring "average" current or voltage in an AC circuit. The 220V is RMS voltage, for example. Instantaneous voltage would oscillate between roughly ±300V. An amp-meter or multi-meter designed to measure AC current will typically give you RMS current. If the pump really is running 125watt at 220V it should take .57A of current. That's assuming that two are in phase, which is what you are trying to achieve. Lower value would suggest that you aren't running the thing at peak efficiency. If the difference is big, you can estimate the capacitor you'd need to compensate.

9. Aug 20, 2011

### Gordianus

After reading the thread again, my best guess is the OP asks for a starter capacitor. The optimum capacitance varies from one model to another, but a fair average choice would be 4.7 microfarads (non-polarized).

10. Aug 20, 2011

### yshtar

I assume you talked about the resistance of the wire, the Ohm...?

I don't know if it's start/run/compensate capacitor.Both the motors run, but not as fast as before. My guess is that the capacitors had an expiry date as they have the year on them..
We changed the capacitor several times before, and the last time about 2 years ago. Replacing the capacitor run the motor fast again but not as fast as before. I don't know the capacitance. I don't check it since I don't think it matters because it doesn't run as expected.

11. Aug 20, 2011

### uart

A compensation capacitor wont significantly effect the power (hence speed) unless there's something seriously wrong with your power supply, so it must be a start or run capacitor.

If it's a start capacitor then the wrong value will effect the starting torque, but assuming that it get's up to the cut-out speed, it shouldn't make any difference to the final running speed.

So I'd say it's one of two possibilities.

1. A capacitor start split phase motor that's not ever getting enough speed to switch (centripetal) out the start winding.

OR

2. A permanent capacitor split phase motor with the wrong size (or degraded) capacitor.

Can you tell us the markings on your existing capacitors. They should have a capacitance (uF) and a voltage rating (Vac) marked on them.

12. Aug 20, 2011

### yshtar

I don't know the capacitance. I haven't taken the pump out. It's in the well. I didn't look at the capacitance since I didn't know it was the important thing. I thought it was the voltage that was important, and it's 370V.

13. Aug 20, 2011

### uart

The voltage rating is very important for the survival and longevity of the capacitor, but it's the uF value that measures the actual capacitance.

14. Aug 20, 2011

### yshtar

Did you mean that the higher the voltage, the longer it can work?

15. Aug 20, 2011

### uart

To some extend yes. If the voltage rating is close to the actual line voltage then the capacitor would typically fail sooner than than if it were rated 30% higher than required. At some point however further increases in voltage rating wont make any difference.

16. Aug 21, 2011

### yshtar

I'll remember that. Thanks.:)

Btw, does put 2 capacitors in series add its capacitance? 20uF + 30uF = 50uF ? I'm thinking of doing it if I can't get a capacitor with big enough capacitance.

Last edited: Aug 21, 2011
17. Aug 21, 2011

### yshtar

I haven't got the 125w's capacitance, but the 300w's is 30uF, 220AC.

I just tried to calculate the 125 watts PFC with 0.8 power factor and the result is 4.9uF (a little over what Giordanus said, 4.7uF) , and considering that the 300 watts motor had 30uF, 4.9uF is way much smaller, so it's not about power factor correction, right?

18. Aug 21, 2011

### uart

Yeah it's either a "start" or a "run" capacitor.

The other thing to think about is that there could be other reasons why the motors are running slow.

1. There could be a problem with the centripetal switch in the case of a start capacitor.

2. There could be other mechanical issues (bearing - increased friction) that are increasing the load.

19. Aug 21, 2011

### yshtar

It's a run cap, I think, since I saw no arm at all.

I just got this and need help to understand it.

Start cap: C = 350000 * I/2p * f * U * cosφ where:
I — current;
f — frequency;
U — voltage;
2p — Take 2 big power factor, power factor Minor 4;
cosφ — power factor (0.4 ~ 0.8).

---What should be the value of "2p?" And the cosφ, should it be cos 0.4 to cos 0.8, or we just use 0.4 to 0.8?

Run cap: C = 120000 * I/2p * f * U * cosφ where:
I — current;
f — frequency;
U — voltage;
2p — take 2.4;
cosφ — power factor (0.4 ~ 0.8).

---Should the "2p" be "2.4" ? And the cosφ question is the same as above.

The results from my case: the run cap:

120000 * 0.568 / 2.4 * 50 * 220 * 0.8 = 249,920,000.
120000 * 0.568 / 2.4 * 50 * 220 * (cos 0.8) = 217,651,175.8916

I don't understand how it can be so huge.

Note:
It was on a Chinese cap manufacturer, and I think my english is much better. lol.
There was no more explanation on the page.