I have a harbor freight 12v diesel transfer pump that I would like to run off 120v ac when at the shop. Is there a converter that can do this directly without a battery? I have run it off my battery charger but the 10 amp charge under powers it and the 50 amp over powers it. What amp of converter would match the power of hooking directly to.the battery as this pump is designed to do.

Thanks

For the limited use, could you use the small electronic rechargeable battery jumpers?,

Those jumpers have limits on how long you can draw juice with out them getting hot. It is not that long. But how long you would crank a car to start it, then they require a cool down period that is quite a bit longer then the use.

Yes. I'm assuming it is 12vDC. You can buy a power supply which essentially is a 120vAc to 12vDC converter. It will have terminals where you can make the + and - connections. We used to get them from radio shack when they were around. Search "12v power supply" online.

I'm not a power specialist. That being said, there is an equation that comes to mind.

The first is: Power = Voltage X Current

In the USA, the 120 V AC is RMS, not peak, so it can be directly compared to a DC voltage (without needing to convert to RMS by dividing by root 2).

SO, assuming you want both configurations to draw the same power:

Voltage1 X Current1 = Power = Voltage2 X Current 2

Do you know how much power (Watts) your device uses? You know the DC (and AC) voltage, so if you know the power, you can calculate how much current (Amps) is being used. The AC configuration (for the same power use) is going to use less current than the DC configuration since the voltage is higher.

If this is your pump:

https://www.harborfreight.com/12-volt-diesel-transfer-pump-66784.html

The specifications say that it draws 13.5 A of current. So, at 12 V DC, the power usage is:

Power = 12 V X 13.5 A = 162 W

So you would need a power inverter capable of providing at least 162 W power continuous.

These are all back-of-the-napkin calculations.

What he needs is a rectifier. And it will be a small one. That is less than 1.5amps AC at residential voltage.
But that is only part of the story. A DC power supply is the unit that will include the rectifier along with other components to give smooth DC power.
Edit: with a motor, you also have inrush current at startup. I believe a good general rule is to have 150% of the stall current available from the power supply, but I would have to look that up for your particular motor type.
Second edit: You can determine the startup current by measuring the resistance of the windings. Use that measurement and the voltage to determine the stall/startup current, them oversize your DC supply to be sure you are covered.

Oh, you're right, since he's converting AC to DC. An inverter would go the other way (DC to AC). Brain fart.

The Canadian done answered your question. Thanks a lot american public schools

Been trying to get an old Fairlane Ford running and the resistor on the coil gave me fits for a couple days

Don't sleep on the Canadians.

Wattage rating using OHM's law can only be used with DC current accurately - with AC current your need to figure impedance into the equation. You don't really need to worry about sizing your power source to the load - the load will draw what it needs - having surplus power is better. In rush current on an electric load will always be at least 4 times the FLA (Full Load Amps). What's your old Ford Fairlane doing as far as the resistor?

You can always use the P = VI equation to calculate instantaneous power. There's no need to invoke Ohm's law for that, if you know the voltage and current.