# Increasing amperage in simple circuit

1. May 8, 2013

### Chark711

I am currently in an engineering project which is attempting to hook up some solar panels to run an irrigation pump. Right now, the pump that we have is a marine Bilge pump that says that at 12 Volts and 7 Amperes it can pump at around 3.3 Gallons Per Minute. We were provided with a standard zinc-lead car battery that produces 12 Volts. We hooked the pump and battery up, and measured its flow to be around 1.07 GPM, with the battery outputting around 1.84 A and 12.50 VDC.

So we know that the pump can take more amperage, but is at its limit in volts. How do we reach this fabled 3.3 GPM? If I = V/R, then we either need less resistance or more voltage, but both seem impossible. Are there other batteries with less internal resistance that are more capable of producing the amperage we need (without such pathetic life-spans that they are useless)? Does anybody know if perhaps marine batteries are more capable at producing amperage than auto ones?

2. May 8, 2013

### phinds

You do NOT want more voltage as that will likely burn out the motor. What you need is a 12 source that will provide more amps than your battery is providing.

3. May 8, 2013

### Staff: Mentor

Lead-zinc or lead-acid? Lead-zinc is not suitable for this sort of application; but a properly functioning and fully charged lead-acid car battery will have no trouble at all providing 7 amps (or several hundred amps - whatever power source you end up with, be sure you have a fuse or other overcurrent protection).

4. May 8, 2013

### Introyble

Yeah, agreed, you want more AH in the battery. But, don't just assume you going to reach your stated GPM. Don't we have to calculate the head loss and gains?

5. May 9, 2013

### 256bits

That is most likely where his "problem" lies.

He is assuming that the pump will give 3.3 gallons per minute at any head. A bilge pump would be rated for low head, and it is not mentioned at all in the question, ie the head at 3.3 gpm.

I suspect, but only guesing from limited information, he is operating on the lower part of the pump characteristic curve with higher head and lower flow rate, where the power needed would naturally be less than the 12 v and 7 amp.

Other reasons would be that the pump is worn, there is an obstruction at the inlet, or there is not enough positive suction head.

6. May 9, 2013

### pumila

This does not add up. Pump is rated at 7A at 12v to deliver 3.3 gallons per minute. It is at 12v and a bit and is drawing only 1.84A for 1.07 gallons so is clearly lightly loaded and nowhere near capacity.

So the reason for the reduced flow is not a blockage or impediment or high working head or current would be higher. Not low volts because we are at rated voltage.

So why is the pump so lightly loaded? Is there an air leak into the inlet? Is the impeller loose on the shaft? A poor seal or a misalignment?

7. May 9, 2013

### Staff: Mentor

No, I agree with 256bits: what I think is the problem here is misleading pump specs. Often, pump vendors will provide the maximum flow of the pump, which occurs if it isn't generating any head (pressure). If you attach piping and pump it uphill, the flow goes down and the energy input with it. There probably isn't anything that can be done here except to get a bigger pump.

Chark, do you have a pump performance curve? How much lift (head) do you have in your system? What is the pressure rating of the pump?

8. May 9, 2013

### Staff: Mentor

No. It may be counter-intuitive, but more resistance results in less power draw. The reason comes from the pump law: power draw is a linear function of pressure, but a square function of flow.

9. May 9, 2013

### Staff: Mentor

I strongly agree about "misleading pump specs" when flow rates are specified across a negligible pressure difference. But usually when the flow rate disappoints because the pressure differential is substantial, the energy input does not go down. Instead, the pump draws its max amperage and consumes its max power, but most of the energy is dissipated thrashing water around instead of pushing it uphill, so power and current remain high while efficiency suffers. (Note that if you completely stall an induction motor, the current through it goes up and the windings start to burn).

I'm just guessing without more information, but I'm still inclined to think that in this case OP is being let down by inadequate batteries. Lead-zinc is not suited for serious power delivery and 7 amps at 12v is 84 watts... Think holding a 75 or 100 watt light bulb in your hand for a while.

10. May 10, 2013

### Chark711

The pump is new and has not been used, so I highly doubt any damage or ware to the pump itself. The adapters are also new, and they are at least water-tight, so I do not know if there is going to be any air leaking through. However, our testing of the pump involves letting an intake hose leading down to a bucket of water. Is our testing faulty? The final use of this pump will involve the water source being placed above the pump, so will it magically work up to its displayed specs? Right now, for testing purposes, it is drawing up out of a bucket and then weakly bubbling out of the open side into the bucket. In the final use, it is going to be used for irrigation from stored rain water, so it is going to be fed by water bladders stored above the pump, and then output through a hose. Will the final setup provide a better flow right then?

So, I believe it is actually a Lead-acid battery. It just left with someone, so I can't confirm, but my instructor believes this to be the case.

The pump is a FLOJET Premium QUAD Series Pumps Quiet Quad. Additional label info says it has a 35 PSI pressure cutoff. The 3.3 GPM on the label is written as: "3.3 GPM [12.5 LPM] Max", so is this actually an absolute limit before damage, and not a gauge for actual performance?

Also, I have no idea what phinds means, I never said that we were going to give it more voltage. I said it was impossible. It is worrying enough putting 12.5 VDC through it.

11. May 10, 2013

### pumila

That's clearly 3.3 US gallons, not Imperial gallons. Battery is fine because it is holding its voltage. Pump is not delivering rated flow, but from current is not trying to do much.

Does sound from what you say that, given low power consumption, you are definitely having air or cavitation at the impeller. Also from what you say, no chance of air leaking in on inlet side.

Suspect air is either circulating in pump impeller or leaking back from outlet. You mention water bubbling out rather than rushing out. The pump may be designed to operate with a head of water on the inlet. If you cannot arrange that then clamp on an outlet hose and keep the end of it above the height of the impeller. What happens then to current and flow?