Electrical mystery

[Dean Abramson]

At the end of the season we use our pressure water pump and galley faucet to empty our freshwater tanks.

We have often done this while motoring to the boatyard where we haul out. But it will frequently trip the breaker.

We never trip the breaker with normal at-anchor use, even if we are taking long showers. And we just emptied all of the tanks this way at the dock with the motor off.

Can someone explain to me why it would trip the breaker on this circuit while the engine is running, but otherwise never? It is the only item on this circuit.

Dean

[Warren S]

With the engine running, most DC systems will exhibit 13 or 14 volts at the supply vs 12 or slightly less when there is no alternator spinning.

Perhaps the additional V in I=(V/R) is enough to source more current to trip the breaker.

Is the breaker sufficiently sized for the load?

This is a half baked theory, so feel free to shoot it down!

[Dean Abramson]

I had a similar thought. But then I thought:

Say I have a 120 watt pump. At 12 volts, it draws 10 amps, right. (120 divided by 12)

But at 14V, the math says it then draws 8.57 amps. LESS. (?)

Yes? No?

Or is my eletro-math wrong?

Confused,

Dean

[Warren S]

I'm going on the simple assumption that for a load capable of sinking variations in current, increasing the EMF or voltage potential will cause more current to flow.

That's it - I'm officially over my head! Will the EE's among us step in?

[mahalocd36]

I had a similar thought. But then I thought:

Say I have a 120 watt pump. At 12 volts, it draws 10 amps, right. (120 divided by 12)

But at 14V, the math says it then draws 8.57 amps. LESS. (?)

Yes? No?

Or is my eletro-math wrong?

Confused,

Dean

Hi Dean,

I guess it depends on what you think is "constant". I think the resistance through the pump is constant, ast least enough for this discussion. So, using the equation

I = V/R : as Voltage increases, current increases.

So, that's why you are blowing the breaker when the engine is running. More voltage, more current. I think Warren is right. Your breaker is just on the edge.

[Warren S]

The 120w rating does not assume a self-regulating load. I would think it is a non-regulating load being a coil of wire (motor winding). Therefore if you envision taking that 12v motor and deliberately apply 100vdc (read: "arbitrarily high voltage") across it, it will consume too much power that it is unable to dissipate and burn out. Of course in reality, the breaker will have tripped immediately. Why? The current component of P=IV will have increased proportionately. It's current flow that causes the heat. All the voltage in the world is fine until it starts to do work (current), then there is heat dissipation (expressed in watts?).

<Insert Standard Sea Hunt Disclaimer Here>

[mahalocd36]

Right, that is what I was getting at, power is not the constant here. P = V*I. It's a function of the voltage and the current. Usually power rating of a device is assuming a voltage, or given like (120W @12V)

-Melissa

[Warren S]

n/m

[river-rat]

An interesting discussion. A DC electric motor with a constant (shunt) field the speed will be approximately be proportional to the voltage and the current about proportional to the torque load.

With a higher voltage supply the motor will run faster and the load will be determined by the pump.

In the case of the tripping breaker, an ammeter would probably show the current overload that is causing the breaker to trip. Going to a slightly larger breaker is probably safe if the motor is running cool and the wiring is adequate. I am assuming that the voltage is within the motor manufacturers specs.

Burl Romick

[marv brinn]

Keep in mind that the surge current is often 40% more than the operating current.

and that the flux coming in can form a maxwell curve to impede the capacitive reactance Xc. this may produce anevoruim cells in the hopi tribe.........

see. the inductive cann recede too which provides matched imedance vs the coriolis effect..

[Dean Abramson]

Yes, that, and the breaker trips.

[Warren S]

What happens when you cross the equator into the southern hemisphere?

[John Danicic]

I am no electrical engineer but in my experience, I have found that if you constantly trip a breaker for what ever reason, that breaker becomes weaker and more likely to trip. For a mystery like this, try and replace the breaker and see if that cures it. The breaker could be broken or worn. If it's not that, look at the ground wire. Most electrical mystery's almost always turn out to have something to do with the ground.

Sail on

John Danicic
CD36 - Mariah - #124
Lake Superior - The Apostle Islands
CDSOA # 655

[Dean Abramson]

John, that sounds like a winner. It makes sense. I will replace the breaker to be safe. Finally something I understand!

Thanks,

Dean

[bottomscraper]

Dean,

John may be correct, it could be a bad breaker. I would not agree that just replacing the breaker is the correct first step. Before you take that step you should try to figure out what pump you have and what it should draw for current.

It's very likely that at some time since the boat was built the pump has been changed. The new pump may draw more current than the original. You may want to go to a bigger breaker assuming the wire can safely handle the additional current.

If you look at the various pump options from Shurflow and Jabsco you will see that there is a wide range of current ratings depending on which pump you buy.

If you just blindly replace the breaker with the same size you may find that next spring it behaves exactly the same, or might even trip without the engine running!

Since you only have the problem when the engine is running it sounds like the current that the pump is drawing is very close to the value that causes the breaker to trip. The small increase in voltage causes an increase in current. What we don't know is if the breaker is faulty (trips below it's rating), the pump is drawing more because it was designed that way (larger replacement pump) or the pump is faulty and is drawing too much current.

With an ammeter we could rule out the bad breaker.

Rich