Calculating Demand Load of Microwave

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Vankirk Electric SR. PM

Member
Location
Winder, Ga
Team:

I've got a microwave nameplate that states, "1750 watts input, 1000 watts output".

Would I size my demand load based on the input power?

I'm assuming that this drop in output is the power efficiency loss?

Thanks for the help.
 
junkhound

junkhound

Senior Member
Location
Renton, WA
Occupation
EE, power electronics specialty
Vankirk Electric SR. PM said:
Team:

I've got a microwave nameplate that states, "1750 watts input, 1000 watts output".

Would I size my demand load based on the input power?

I'm assuming that this drop in output is the power efficiency loss?

Thanks for the help.
Click to expand...

FWIW, one of the standard methods for measuring output power is to heat a liter of water(in low mass, low specific heat container) for a given period of time and calculate thermal power into the water based on temperature rise.
 
Besoeker

Besoeker

Senior Member
Location
UK
Vankirk Electric SR. PM said:
Team:

I've got a microwave nameplate that states, "1750 watts input, 1000 watts output".

Would I size my demand load based on the input power?

I'm assuming that this drop in output is the power efficiency loss?

Thanks for the help.
Click to expand...

That's also what I would assume but that 750W would have to somewhere - probably as heat into the air I imagine.
 
G

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
171215-2018 EST

Vankirk Electric SR. PM:

Quite obviously it is input power that you use.

On the efficiency of a microwave I ran an experiment with my GE microwave in comparison with a hot pot about 6 years ago.

The hot pot has very tight thermal coupling between the heating element and the water. This means almost all the input electrical energy is used to heat the water.

In the microwave there is considerable electronics. The major loss is in the magnetron itself, and coupling its output energy to the water.

For 1 quart of water for a temperature change of 125 F it took 0.08 kWh of input electrical energy to the hot pot. The theoretical value for no loss is 0.076 kWh. The microwave required 0.18 kWh to do the same job.

.
 
V

Vankirk Electric SR. PM

Member
Location
Winder, Ga
gar said:
171215-2018 EST

Vankirk Electric SR. PM:

Quite obviously it is input power that you use.

On the efficiency of a microwave I ran an experiment with my GE microwave in comparison with a hot pot about 6 years ago.

The hot pot has very tight thermal coupling between the heating element and the water. This means almost all the input electrical energy is used to heat the water.

In the microwave there is considerable electronics. The major loss is in the magnetron itself, and coupling its output energy to the water.

For 1 quart of water for a temperature change of 125 F it took 0.08 kWh of input electrical energy to the hot pot. The theoretical value for no loss is 0.076 kWh. The microwave required 0.18 kWh to do the same job.

.
Click to expand...

I think that was what caught me off guard at first... I was shocked at the amount of power that is lost by the time the power is applied to heat the food/liquid. Not a very good Power Factor!
 
G

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
171215-2340 EST

That ratio does not imply a bad power factor. Input power factor could be 1.

It does indicate poor efficiency between input and output.

On the other hand I could have close to 100% efficiency, but a poor power factor. For example, a high quality capacitor in series with a resistor. Or parallel the capacitor and resistor and it may be more obvious.

.
 
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V

Vankirk Electric SR. PM

Member
Location
Winder, Ga
gar said:
171215-2340 EST

That ratio does not imply a bad power factor. Input power factor could be 1.

It does indicate poor efficiency between input and output.

On the other hand I could have close to 100% efficiency, but a poor power factor. For example, a high quality capacitor in series with a resistor. Or parallel the capacitor and resistor and it may be more obvious.

.
Click to expand...

I thought I was mildly competent when it comes to electrical theory... now you've really lost me!
 
G

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
171216-1259 EST

The definition of power factor is:

PF = Real power / apparent power = measured power / Vrms*Irms

Average Real power = the average of instantaneous voltage times instantaneous current over some defined averaging time period.

Vrms is an averaged voltage measurement over some time period. It turns out that this relates to the heating effect of that voltage when applied to a pure resistance load. It is calculated by taking the instantaneous value of voltage, squaring this, and integrating the squared value over some time period, dividing by that time period (getting the average), and taking the square root of that averaged value.

RMS means square a value (S), take the mean value (average) (M), and take the square root (R).

There is no correlation between an RMS voltage and RMS current unless these are directly associated with a resistance.

.
 
Besoeker

Besoeker

Senior Member
Location
UK
gar said:
171216-1259 EST

The definition of power factor is:

PF = Real power / apparent power = measured power / Vrms*Irms(R).
.
Click to expand...
Yes. But in this case we are given only real power for input and output. PF doesn't come into it.
 
G

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
171216-1436 EST

Besoeker:

Absolutely. I think that in some manner Vankirk Electric SR. PM was misinterpreting Power Factor as having something to do with power efficiency at the load end.

Poor power factor can can reduce overall efficiency in a total system, but possibly more important is its effect on utilization of power system equipment.

.
 
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