Need some help on getting a small inverter for home loss of power

If you only want to power a few electronics, a low power blower fan, and maybe some LED lights consider one of these combined with a small generator to recharge it a time or two per day, depending on how many amps the fan draws.  Of course this will not run the microwave or any power tools

https://smile.amazon.com/PROGENY-Generator-Portable-Generators-Emergency/dp/B07S74MCQM/

or
https://smile.amazon.com/Enginstar-Portable-Generator-Cellphones-Electronics/dp/B07WQN41V9/

Say combined with something like
https://smile.amazon.com/Westinghouse-iGen2200-Portable-Inverter-Generator/dp/B01MUP6L1U

Run the fan over night on the battery, run the generator in the morning to power the fan, and recharge the battery for 2-3 hours, repeat.

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JackL said:

That first one is exactly what I had in mind

Many thanks,

Jack L

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Keep in mind the first one is not a "generator".  It is a battery inside of a box with an inverter to supply 120V. 

It still needs to be charged with a generator.

Jack, 

You will need to calculate your power requirements.

The fan if it indeed 5 watts then that will be a draw of 0.41 amps of 12V DC from your battery through the inverter.  If the fan runs constantly for 24 hours that is about 10amp hours (AH) of battery. 

However a small inverter, say 300-400watts, will use about 1 amp of 12V DC while being on and not supplying 120V AC.  That adds another 24 AH of battery power needed.

For every 60 watt light bulb that takes 5amps of 12V DC through the inverter.  If you use the light bulb for 3 hours that is another 15AH of battery power. 

To add this up:
10AH  Fan  (maybe less if the fan doesn't run constantly)
24AH  Inverter
15AH Each light bulb for 3 hours
49AH  Total for each day. 

If you have a single 100AH lead acid battery that will handle one days use if you only use one 60watt light bulb.  Each day you will need to run the generator for 3-4 hours to charge the battery. 

Note that it would be best if you had an inverter/charger or a stand alone 3 stage charger for the battery. 

The PROGENY 280W Portable Power Station (note I deleted the word "generator) will probably work for you.  However you will have to charge it every day.  Maybe for less time than if you tried charging a lead acid battery.  You might not find out for sure how long it will take to charge it until you buy the unit and test it.

Here are links to a website with great basic 12V battery info:

http://www.marxrv.com/12volt/12volt.htm
http://www.marxrv.com/12volt/12volta.htm

Note it says 5-6 hour charge time from the car charger option, if you go that way, though that would eat up a lot of gas idling your car to recharge the battery.

Actually, he said 5W and 0.34 A for his 120V fan! A fan that both circulates air (heat) and feeds the flame is much more likely to be 0.34A given that both figures cannot be correct).

Ernie

Ernie n Tara said:

Actually, he said 5W and 0.34 A for his 120V fan! A fan that both circulates air (heat) and feeds the flame is much more likely to be 0.34A given that both figures cannot be correct).
Ernie

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The fan just circulates warm air
Engraved on the motor is 120 volts  5W  0.34A

I'll double check on my spare motor and if I don't reply back, that's what it is, (at this old age, I make lots of mistakes!)

Jack L

A 5W fan would be along the lines of something used for a computer case or CPU cooler.  A "blower" running off of 120V seems more along the lines of a 40W unit at .34A.  A basic multimeter or Kill-a-watt unit would conclusively prove one way or the other.

Mark B.
Albuquerque, NM

JackL said:

Reading all the replies I think I will just settle for a small inverter with no battery and use the battery from my POP up trailer if necessary.

All I want it for is for up coming November, December, January and February and in the past the outages have been very few with the power turned on within a few hours.
I won't even need it then, if they come out with a vaccination sooner and we can leave here.

I guess there is no one out there with a small inverter that can or will respond

Thanks to all who tried to help, but that is not the kind of advice or help I was looking for.

Jack L

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Based on all the things you are trying to tell us here's my final thought.

1 - Pull the battery out of the pop up trailer. "Maybe" buy a second battery and wire them in parallel. That will get you about 100 amps usable.  But them on a trickle charger/battery tender (like $20) and leave them on trickle charge. Buy a 750W or so plug in inverter for another $40 or so. If you don't already have a normal battery charger buy one.

- Walmart everstart marine battery Group 27 - $85
- Trickle charger - $20
- Everstart 750W inverter - $50

$155 bucks.  Basically you want to maximize your batteries so I would add 2 more marine batteries to the bank for $170 and your total is like $325. (edit - Whoops! You are gonna need 4-6 shortie battery cables to wire it all together. A few more bucks)

If you need a battery charger buy the Walmart 15amp one for $60 and don't buy one of the batteries. If you use up the power in the battery bank you want to charge it as fast as you can hence the 15 amp charger.

You realize that this cannot be correct.

Engraved on the motor is 120 volts  5W  0.34A

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Power equals voltage times current (P=ExI)

40.8Watts = 120Volts * 0.34Amps

or

5 Watts = 120Volts * 0.041666Amps

This is what happens when you let the marketing department near the power specs LOL.

I have a 300 watt inverter connected to my coach battery. I'm either driving which keeps my coach battery charged, I'm plugged in or running the generator.

You can buy a 100 Ah deep cycle battery, small invertor and charger for about $150.

You can also get a small, light generator for about $400. This one will run your RV AC.
https://www.amazon.com/gp/product/B00YFT914I/ref=ox_sc_act_title_1?smid=A2LH8DUZ1Q6T63&psc=1

Power equals voltage times current (P=ExI)

40.8Watts = 120Volts * 0.34Amps

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That simple formula is not accurate for a motor, where inductance is a factor.  "Power factor" is added to the equation.

Power factor is just Cos (angle), it's the difference between the real power caused by the resistance and the apparent power caused by the inductive reactance PF = Real Power / Apparent Power

So a single phase motor just becomes P = E * I * Cos (angle) or P = E * I * PF

I think I've figured out what happened here...

A Typical PF for a no load motor is about 0.15. To make the specs, stamped on the motor work it would need to be 0.122549, so that's close. The specs on the motor were probably put there by the motor company before it became a fan.

The trouble is that as soon as you put the impeller on it and load it, the PF will jump to about 0.85 because now it's actually working.

I knew my formulas were simple but I was trying to point out that what was stamped on the motor was incorrect but I mistakenly assumed the specs were for the fan. He can run the motor by itself and draw only 5W but he can't run it as a fan and expect to only draw 5W, once it's asked to work it will draw closer to 35W.