Gene Earl

21-08-2003, 11:15 AM

Can anyone tell me how many watts a 4 cu. ft. 110 volt refrigerator uses?

Thanks,

Gene Earl

Thanks,

Gene Earl

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Gene Earl

21-08-2003, 11:15 AM

Can anyone tell me how many watts a 4 cu. ft. 110 volt refrigerator uses?

Thanks,

Gene Earl

Thanks,

Gene Earl

RogGoetsch

22-08-2003, 06:53 AM

Manufacturers rating plate should show amps at nominal 115v. Power in watts = amps X volts.

***Oops, bad answer! See DaBit, below! Only works for resistive loads.***

If the unit is a refrigerator without a freezer it will be more efficient: 2 amps or less. Refrig/freezer combo will run about 2.5 amps, *** but can't just multiply to get watts due to inductive load.***

***Oops, bad answer! See DaBit, below! Only works for resistive loads.***

If the unit is a refrigerator without a freezer it will be more efficient: 2 amps or less. Refrig/freezer combo will run about 2.5 amps, *** but can't just multiply to get watts due to inductive load.***

DaBit

22-08-2003, 08:48 AM

You cannot just multiply amps with volts to get Watt, unless the load is pure resistive (like a lightbulb)

Based on power consumption of the compressors I use, I would say that the device pulls about 150W.

Based on power consumption of the compressors I use, I would say that the device pulls about 150W.

herefishy

22-08-2003, 02:10 PM

DaBit, do you think that you are referencing the refrigerating capacity (output) of your compressors? You have losses in the process, and the input will be greater than the output (in any machine).

But of course, if you add the work done in the condenser (heat exchange to air), to the heat exchange in the evaporator, maybe then you would be near 280W?

In regard to the original question, determining the power consumption would be related to the usage (how many hours a day the machine runs.

But of course, if you add the work done in the condenser (heat exchange to air), to the heat exchange in the evaporator, maybe then you would be near 280W?

In regard to the original question, determining the power consumption would be related to the usage (how many hours a day the machine runs.

DaBit

22-08-2003, 02:25 PM

Originally posted by herefishy

[B]DaBit, do you think that you are referencing the refrigerating capacity (output) of your compressors? You have losses in the process, and the input will be greater than the output (in any machine).

No, I was referencing to the electical power consumed by the compressor, and I made the guess by referencing to the power draw of my compressors in my systems. After all: they are household freezer compressors. I was not referring to the work done by the compressor.

Simply multiplying volts by amps to get consumed power only works when the load is completely resistive. In reality, only heaters and incandescent lightbulbs fullfill this requirement.

As soon as capacitors or coils are used, you cannot calculate power by multiplying volts by amps.

To prove this: take any electrical tool with a motor in it. The type plate mentions consumed power, nominal voltage and amps. When ams and volts are multiplied, the result is not the power mentioned on the type plate.

About the fridge: the fridge is very well capable of removing 300W of heat with only 150W of imput power. At these relatively high evaporation temperatures, the COP is higher than 1.

[B]DaBit, do you think that you are referencing the refrigerating capacity (output) of your compressors? You have losses in the process, and the input will be greater than the output (in any machine).

No, I was referencing to the electical power consumed by the compressor, and I made the guess by referencing to the power draw of my compressors in my systems. After all: they are household freezer compressors. I was not referring to the work done by the compressor.

Simply multiplying volts by amps to get consumed power only works when the load is completely resistive. In reality, only heaters and incandescent lightbulbs fullfill this requirement.

As soon as capacitors or coils are used, you cannot calculate power by multiplying volts by amps.

To prove this: take any electrical tool with a motor in it. The type plate mentions consumed power, nominal voltage and amps. When ams and volts are multiplied, the result is not the power mentioned on the type plate.

About the fridge: the fridge is very well capable of removing 300W of heat with only 150W of imput power. At these relatively high evaporation temperatures, the COP is higher than 1.

RogGoetsch

27-08-2003, 06:19 AM

Originally posted by DaBit

You cannot just multiply amps with volts to get Watt, unless the load is pure resistive (like a lightbulb)

DOH! Of course you are right. I guess it's true what they say about a net loss of gray cells after the age of fifty! Guess I'll go pin my address to my shirt now, while I still remember it.

Rog

You cannot just multiply amps with volts to get Watt, unless the load is pure resistive (like a lightbulb)

DOH! Of course you are right. I guess it's true what they say about a net loss of gray cells after the age of fifty! Guess I'll go pin my address to my shirt now, while I still remember it.

Rog

REL

27-08-2003, 06:41 AM

DaBit,

I disagree completly. Amps * Volts = watts. It does NOT change for an inductive load compared to a resistive load. The OUT put does change but the INPUT does not.

This person seems to be asking about the electrical useage of the appliance and amps * volts would be the most accurate method.

It does NOT mean that a 115 volt unit using 6.4 amps will be 1 horse power output but it certianly does mean the usage will be 736 watts.

I got the feeling from the question the person is inquiring about how much is it going to take to run this machine, not motor efficiencies and power factor.

I disagree completly. Amps * Volts = watts. It does NOT change for an inductive load compared to a resistive load. The OUT put does change but the INPUT does not.

This person seems to be asking about the electrical useage of the appliance and amps * volts would be the most accurate method.

It does NOT mean that a 115 volt unit using 6.4 amps will be 1 horse power output but it certianly does mean the usage will be 736 watts.

I got the feeling from the question the person is inquiring about how much is it going to take to run this machine, not motor efficiencies and power factor.

DaBit

27-08-2003, 09:22 AM

Originally posted by REL

[B]DaBit,

I disagree completly. Amps * Volts = watts. It does NOT change for an inductive load compared to a resistive load. The OUT put does change but the INPUT does not.

Thus in your opinion it doesn't matter for the amount of energy whether we have a resistor that pulls 1A @ 110VAC(=110W), or that we have a capacitor with enough reactance to pass through 1A @ 110VAC?

Amps * Volts is indeed Watts, but mathematically you must integrate that over the entire sine curve, and not take the average voltage and amperage. That is only allowed in a DC system, not in an AC system.

This person seems to be asking about the electrical useage of the appliance and amps * volts would be the most accurate method.

I still do not agree. Have a look at a random electrical tool. It will mention nominal voltage, nominal current draw, and wattage. These numbers do not match.

I got the feeling from the question the person is inquiring about how much is it going to take to run this machine, not motor efficiencies and power factor.

That's why I estimated power draw to be 150W. It is impossible to tell or estimate how many kWh/year the unit consumes without more information, so that 150W is the best guess I can make.

[B]DaBit,

I disagree completly. Amps * Volts = watts. It does NOT change for an inductive load compared to a resistive load. The OUT put does change but the INPUT does not.

Thus in your opinion it doesn't matter for the amount of energy whether we have a resistor that pulls 1A @ 110VAC(=110W), or that we have a capacitor with enough reactance to pass through 1A @ 110VAC?

Amps * Volts is indeed Watts, but mathematically you must integrate that over the entire sine curve, and not take the average voltage and amperage. That is only allowed in a DC system, not in an AC system.

This person seems to be asking about the electrical useage of the appliance and amps * volts would be the most accurate method.

I still do not agree. Have a look at a random electrical tool. It will mention nominal voltage, nominal current draw, and wattage. These numbers do not match.

I got the feeling from the question the person is inquiring about how much is it going to take to run this machine, not motor efficiencies and power factor.

That's why I estimated power draw to be 150W. It is impossible to tell or estimate how many kWh/year the unit consumes without more information, so that 150W is the best guess I can make.

REL

28-08-2003, 12:08 AM

Originally posted by DaBit

Amps * Volts is indeed Watts, but mathematically you must integrate that over the entire sine curve, and not take the average voltage and amperage. That is only allowed in a DC system, not in an AC system.

I stand corrected. THe volts RMS * the amps = watts.

It has nothing to do with the rating plates on power tools.

Your arguments about capacitance, reactance and inductance all affect the OUTPUT ratings. The input is directly proportional and does not change. iIf it is three phase you simply multiply the mean voltage times the square root of three ect ect. and ohms still applies.

Amps * Volts is indeed Watts, but mathematically you must integrate that over the entire sine curve, and not take the average voltage and amperage. That is only allowed in a DC system, not in an AC system.

I stand corrected. THe volts RMS * the amps = watts.

It has nothing to do with the rating plates on power tools.

Your arguments about capacitance, reactance and inductance all affect the OUTPUT ratings. The input is directly proportional and does not change. iIf it is three phase you simply multiply the mean voltage times the square root of three ect ect. and ohms still applies.

Servicefrigo

16-12-2004, 07:59 PM

If I understand nobody did't say nothing about cosinus fii

P=U X I X Cos fi

P=U X I X Cos fi

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