Question about Canon NP 6050 Copier

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Posted on Jan 02, 2017

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There is quite a lot of "that depends" in the answer.

That depends on how long you need to run the pump, whether it to be run in short bursts with long gaps or long bursts with short gaps or run continuously. That depends on whether any battery charging is carried out while in use, solar for instance.

That depends on whether the 1/4 hp is the input of the motor or the output. Some motors are rated one way and some the other. If the motor output is 1/4 hp, the input has to be more due to the relative inefficiency.

If the motor input is the equivalent of 1/4 hp the output will be less and it might be your motor might not be able to do the work adequately or for long enough.

A 1/4 hp input motor will consume 1/4 of 746 watts

A 1/4 hp output motor will consume 1/4 of 746 watts X the motor inefficiency, say, 130% which is a consumption of almost 250 watts.

That depends on the efficiency of your inverter.

Inverters aren't particularly efficient though great leaps forward have been made in this over early models, the inefficiency can't be ignored.

If the inverter efficiency was 100% the 250 watts the motor requires will require a current flow from the battery of about 21 amps but allow a notional inefficiency of the inverter similar to that of the motor that becomes around 27 amps.

There can be no definite answers until those questions are answered and it is best to provide wide safety margins to allow for falling battery capacities as they age. It is also best to assume the rated battery capacity is half of the actual due to the efficiency quickly falling off after the terminal voltage has fallen to half charge.

These understandably are rough figures. A fully charged battery of 100 AH (amp hour), considering the desirability of not allowing the battery to fall below half charge, will power the pump continuously for around two hours. Running the motor in bursts can extend the time considerably.

The batteries will need to be traction or deep cycle types rather than vehicle batteries but even such batteries will suffer a shortened life if the recharging is not carefully managed - frequently topping up the charge level of a partially discharged battery is infinitely more desirable than a less frequent complete recharge.

I hope this helps...

That depends on how long you need to run the pump, whether it to be run in short bursts with long gaps or long bursts with short gaps or run continuously. That depends on whether any battery charging is carried out while in use, solar for instance.

That depends on whether the 1/4 hp is the input of the motor or the output. Some motors are rated one way and some the other. If the motor output is 1/4 hp, the input has to be more due to the relative inefficiency.

If the motor input is the equivalent of 1/4 hp the output will be less and it might be your motor might not be able to do the work adequately or for long enough.

A 1/4 hp input motor will consume 1/4 of 746 watts

A 1/4 hp output motor will consume 1/4 of 746 watts X the motor inefficiency, say, 130% which is a consumption of almost 250 watts.

That depends on the efficiency of your inverter.

Inverters aren't particularly efficient though great leaps forward have been made in this over early models, the inefficiency can't be ignored.

If the inverter efficiency was 100% the 250 watts the motor requires will require a current flow from the battery of about 21 amps but allow a notional inefficiency of the inverter similar to that of the motor that becomes around 27 amps.

There can be no definite answers until those questions are answered and it is best to provide wide safety margins to allow for falling battery capacities as they age. It is also best to assume the rated battery capacity is half of the actual due to the efficiency quickly falling off after the terminal voltage has fallen to half charge.

These understandably are rough figures. A fully charged battery of 100 AH (amp hour), considering the desirability of not allowing the battery to fall below half charge, will power the pump continuously for around two hours. Running the motor in bursts can extend the time considerably.

The batteries will need to be traction or deep cycle types rather than vehicle batteries but even such batteries will suffer a shortened life if the recharging is not carefully managed - frequently topping up the charge level of a partially discharged battery is infinitely more desirable than a less frequent complete recharge.

I hope this helps...

Mar 28, 2018 | HP Electronics - Others

The correct size of inverter is an easy calculation and important to ensure long life from the inverter and the life of the appliances getting there power from the inverter. To calculate you take the volts and multiply by the amps which equals the wattage. Every appliance has a tag which must state the volts and amps. Locate the tag to find the information. Here is an example.

Under Counter Refrigerator 1.3 amps (120 volts x 1.3 amps = 156 watts)

Microwave (120 volts x 10 amps = 1200 watts)

Alarm Clock Radio (120 volts x .03 amps = 3.6watts)

TV (120 volts x .6 amps = 72 watts)

You now add all the watts together, since the above list could likely be running at the same time. There is a combine wattage of 1431.6 watts. In this example you would need at least a 1500 watt inverter. However given some unknown variables personally I would get a 2500 watt inverter. Less stress on the inverter, will lengthen the life of the inverter. Had the total been 1,000 watts a 1,500 watt inverter would suffice.

You must also remember that your alternator on your vehicle must be able to keep up with the power demand of not only your vehicle needs but your inverter needs as well.

Under Counter Refrigerator 1.3 amps (120 volts x 1.3 amps = 156 watts)

Microwave (120 volts x 10 amps = 1200 watts)

Alarm Clock Radio (120 volts x .03 amps = 3.6watts)

TV (120 volts x .6 amps = 72 watts)

You now add all the watts together, since the above list could likely be running at the same time. There is a combine wattage of 1431.6 watts. In this example you would need at least a 1500 watt inverter. However given some unknown variables personally I would get a 2500 watt inverter. Less stress on the inverter, will lengthen the life of the inverter. Had the total been 1,000 watts a 1,500 watt inverter would suffice.

You must also remember that your alternator on your vehicle must be able to keep up with the power demand of not only your vehicle needs but your inverter needs as well.

on Apr 23, 2010 | Electronics - Others

Most power inverters require the vehicle to being running to operate, and make sure it's the correct wattage, if it's a small watt inverter it doesn't have enough power, as a coffee machine is considered higher wattage, hope that answers your question ?

Feb 21, 2015 | Magnum MM1524AE : 1500 Watt, 24V Inverter...

Hello, the answer is no. This inverter will work with a pump that below or about
400 watts power for operation. I notice it says that the inverter 400w/800w. The
800 watts is the surge required to get an electric motor to run. On a single
phase motor (120 AC) and it take 4 amperes to operator this motor. But to start
the it will 12 amperes to start it. This is one reason why on a lot of
electrical motors there will be a big tube or can, this is the starting
capacitor, it stores energy in form of current (amperes) to drop across the
windings of the motor when it starts. If some electric motors didn't have this
capacitor. The most likely scenario would be tripping a breaker because it
drawing to much current.

Now, for you problem. Example of Power and the unit of the formula is V (volt), R (resistance) I (current) P (power) Ohm's states that :

Now Power would be stated: P = V times(x) I. Therefore in your case, if the pump at normal operation is drawing 4 amps of current for operation pumping water. The power factor would be 120 V AC (house voltage) times (X) I (current) which 4 amps. So here is what you have: P = 120 X 4 equal= 480 watts.

Therefore, your inverter will not work because it operating at 80 watts more than the power inverter can deliver which is 400 Watts.

If you need to purchase a new power inverter, remember always take into account the Amperes required for operation. Just by the simple Ohm's Formula. Best of Luck. GB....stewbison

son

Now, for you problem. Example of Power and the unit of the formula is V (volt), R (resistance) I (current) P (power) Ohm's states that :

Now Power would be stated: P = V times(x) I. Therefore in your case, if the pump at normal operation is drawing 4 amps of current for operation pumping water. The power factor would be 120 V AC (house voltage) times (X) I (current) which 4 amps. So here is what you have: P = 120 X 4 equal= 480 watts.

Therefore, your inverter will not work because it operating at 80 watts more than the power inverter can deliver which is 400 Watts.

If you need to purchase a new power inverter, remember always take into account the Amperes required for operation. Just by the simple Ohm's Formula. Best of Luck. GB....stewbison

son

Aug 27, 2011 | Coleman Electronics - Others

on the model tag it gives you voltage and amps. multiply thes 2 numbers to get watts. to be more exact you need to check voltage at the source and take an amp reading while copier is running. Or if you have a watt meter that would be easiest.

Mar 23, 2011 | Kitchen Ranges

Your 8975 is rated at 120 Volts 11.6 amps and will consume approximately 1390 watts.

I can tell you that an inverter will eat your battery and put lots of strain on your charging system. I dont know what your application is, but I wouldn't rely on my car battery to run this device to defrost the window or heat the car.

Thanks for posting your question at FixYa.com

Please vote

thanks

a

I can tell you that an inverter will eat your battery and put lots of strain on your charging system. I dont know what your application is, but I wouldn't rely on my car battery to run this device to defrost the window or heat the car.

Thanks for posting your question at FixYa.com

Please vote

thanks

a

Dec 15, 2009 | Milwaukee Tool Milwaukee 8975 - 6 Dual...

What is it exactly that you want to accomplish? It looks like this unit is powered by the electrical system of your truck, but if it indeed can provide 2500 Watts of output power, its input demand from your truck electrical system will be in excess of 200 Amps!!! You'll need some pretty hefty wiring to support that type of demand, and I doubt if your alternator can put out that amount of current.

Are you sure you want/need to do this - and run it off your truck electrical system?

Are you sure you want/need to do this - and run it off your truck electrical system?

Sep 22, 2009 | Xantrex Technology DPS350KB 2500-Watt...

Ok, Heres the deal. A gel cell battery is 7.2 Ampere hrs, so to get 14,4 Ampere hours you would need two batteries connected in parallel. A F&P machine such as mine consumes around 60 Watts of power. Using Ohms law we get W= IxE or Watts = volts X amps[or amperes = watts / volts] so that a 60 Watt machine has a current draw of 60/110< =0.55A so that a 14.4 A/hr battery will run this machine for around 20hrs (allowing a residual charge of 25%)

Jun 23, 2009 | Duracell Standby Power Plug 400 Watt...

I have a background in electronics so i hope i can help. Now after reading this i have the feeling this item does not put out 2000 watts cause if that were the case it should draw somewhere near 147amps. This is because to find out amps you need to take the 2000 watts and divide it by 13.6 volts (assuming what the vehicle runs at) that gives you amps. watts/volts=amps.

Jan 13, 2009 | Atd Tools 2000-Watt Power Inverter # 5955

If you want to get more precise, figure out everything in terms of power (watts).

Basic electrical rule 1, 2 and 3:

voltage x current = power

or re-arranged:

current = power divided by voltage

or re-arranged:

voltage = power divided by current

For example, 12V X 2 amps = 24 watts.

or another example, 400 watts divided by 120 Volts = 3.33 amps

A 55W headlight that uses 12V would draw 55 /12 = 4.6 amps @ 12V

A 55 watt light bulb in a lamp at home would draw 55 / 120 = 0.46 amps @ 120V

As the previous post mentioned, inverters are not perfect when convertering 12V into 120V. If the converter consumes 1000W from the 12V battery, then a 90% effecient converter would generate 900W of 120V AC power best case. The other 100W is lost primarily as heat.

The other thing that gets tricky is that these ratings and the formula above are used for resistive loads, like light bulbs or hair dryers. Anything with a motor or transformer is considered an inductive load and can get much more tricky to calculate.

Consequently you need to give your self a safety margin when figuring out how big an inverter you need.

How does work in a practical sense?

Lets say you want an inverter for TV, DVD and Sat. Receiver. Look at the back of TV or in the manual. It should say how many watts it consumes. Lets say it is 400W. The DVD might be 100W and the Sat. receiver 50W - just as an example.

400 + 100 + 50 = 550 Watts. (just as an example)

You might think, well no problem, I'll use a 600 Watt inverter and have 50 watts left over. Depending on your inverter, that 600W might really be 600 x 90% effecient = 540 Watts of AC, less a 20% margin of error for the inductive transformers in the electronic of the TV, DVD and Sat. receiver 540 - 20% = 432 Watts.

Now you can see your 600 Watt inverter isn't big enough to do the job.

If we really need 550 watts of AC, add 10% to make up the effiency loss, then add a safety margin for inductive loads.

550 + 10% = 605 + 20% = 726 Watts.

Sounds more like an 800W inverter fits the job.

What does that mean in terms of wiring the 12V batteries to the inverter?

from the formula above:

current = power divided by voltage

In our example, we have an 800W inverter that runs on 12V

The current would thererfore be:

current = power divided by voltage

current = 800 watts divided by 12V

current = 66 amps.

That is important info because you can not use light gauge wire to carry 66 amps worth of 12V to the inverter nor could you use a 20A fuse to protect your inverter.

Now that's a lot of science for a guy who just wants to run a toaster on an inverter right?

800W / 120V = 6.66 amps

Using garryp's ratio 11:1, 6.66 x 11 = 73 amps.

That is a good ratio with a good safety margin.

This is all just MHO and should not taken as solid technical advise. In other words, don't blame me if you blow yourself up.

Basic electrical rule 1, 2 and 3:

voltage x current = power

or re-arranged:

current = power divided by voltage

or re-arranged:

voltage = power divided by current

For example, 12V X 2 amps = 24 watts.

or another example, 400 watts divided by 120 Volts = 3.33 amps

A 55W headlight that uses 12V would draw 55 /12 = 4.6 amps @ 12V

A 55 watt light bulb in a lamp at home would draw 55 / 120 = 0.46 amps @ 120V

As the previous post mentioned, inverters are not perfect when convertering 12V into 120V. If the converter consumes 1000W from the 12V battery, then a 90% effecient converter would generate 900W of 120V AC power best case. The other 100W is lost primarily as heat.

The other thing that gets tricky is that these ratings and the formula above are used for resistive loads, like light bulbs or hair dryers. Anything with a motor or transformer is considered an inductive load and can get much more tricky to calculate.

Consequently you need to give your self a safety margin when figuring out how big an inverter you need.

How does work in a practical sense?

Lets say you want an inverter for TV, DVD and Sat. Receiver. Look at the back of TV or in the manual. It should say how many watts it consumes. Lets say it is 400W. The DVD might be 100W and the Sat. receiver 50W - just as an example.

400 + 100 + 50 = 550 Watts. (just as an example)

You might think, well no problem, I'll use a 600 Watt inverter and have 50 watts left over. Depending on your inverter, that 600W might really be 600 x 90% effecient = 540 Watts of AC, less a 20% margin of error for the inductive transformers in the electronic of the TV, DVD and Sat. receiver 540 - 20% = 432 Watts.

Now you can see your 600 Watt inverter isn't big enough to do the job.

If we really need 550 watts of AC, add 10% to make up the effiency loss, then add a safety margin for inductive loads.

550 + 10% = 605 + 20% = 726 Watts.

Sounds more like an 800W inverter fits the job.

What does that mean in terms of wiring the 12V batteries to the inverter?

from the formula above:

current = power divided by voltage

In our example, we have an 800W inverter that runs on 12V

The current would thererfore be:

current = power divided by voltage

current = 800 watts divided by 12V

current = 66 amps.

That is important info because you can not use light gauge wire to carry 66 amps worth of 12V to the inverter nor could you use a 20A fuse to protect your inverter.

Now that's a lot of science for a guy who just wants to run a toaster on an inverter right?

800W / 120V = 6.66 amps

Using garryp's ratio 11:1, 6.66 x 11 = 73 amps.

That is a good ratio with a good safety margin.

This is all just MHO and should not taken as solid technical advise. In other words, don't blame me if you blow yourself up.

Nov 26, 2008 | Coleman 5640B807 Compact Refrigerator

Jul 11, 2015 | Canon NP 6050 Copier

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