On 4/17/07, Peter P. <plpeter2006@yahoo.com> wrote:
> Vasile Surducan <piclist9 <at> gmail.com> writes:
>
> > The overall efficiency in a domestic microwave oven is very small.
> > Not because the magnetron but because the supplying transformer
> > (assuming the oven is not Panasonic and has no inverter) and the way
> > microwaves are absorbed in the food.
> > Need more details ?
>
> I strongly disagree. The efficiency is much higher than 50% (but does not reach
> the 90%+ possible with a magnetron acc. to books).
>
> One can calculate the efficiency of the oven as a first approximation using
> the water cup method. This is the standard way to test microwaves by techs,
> too. Here I describe it again:
>
> Measure mains voltage and put an AC ammeter in the mains circuit.


Then measure the sunked current with the magnetron connected and
"power" button on 100% (this is in fact the energy button).
The current drawn on 220V will be about 5.5A to 6A.
Then disconect the magnetron high voltage chatode and if you wish the filament.
(anyway tthe filament will source about 10A at 3V, this is
insignifiant compared with the rest).
Now measure the current with the same 100%  "power"
The current sourced will be between 4.5A and 5.5A
Do you know why ?
:)






This should
> be a clamp ammeter (do not use a shunt one unless it is spike-proof, the inrush
> can be huge).
>
> Take a cup of water,

Take a cup with alcohol and do the same...
What would you measure ?


weigh it (first empty cup then full, take the weight of
> the water alone), and measure its temperature. Put *another* cup in the oven
> and nuke 1 minute. Discard this quickly, put in the weighted cup quickly and
> again nuke for 1 minute. Remove the cup, stir quickly and take its temperature.
> Measure the oven current while it runs (better: use a real power meter).
>
> The heat that went into the water is about SPH = 4.2 Cal/deg C and gram. So:
>
>  P1 = Uac * Iac (or readout from real power meter)
>
> (take average Iac, P1 will read high due to the power factor of the
> transformer)
>
>  P2 = SPH * deltaT [deg. C] * (Weight [grams] / 60 [seconds])
>
> Efficiency is:
>
>  eta = P1 / P1
>
> Because of factors entering P1 this eta will read low wrt. the real value. To
> improve the reading look into power factor measurement of the transformer and
> so on. The measured output should be at least 400 Watts for a 750 Watt input
> oven(750 Watt is the nameplate rating). This is equivalent to 53% efficiency.
> Not so bad for a simple circuit involving a transformer, a diode, a capacitor
> and a fancy vacuum diode.

How much can you measure with a 750W resistor imersed in the same cup of water ?
:)

good luck with tests.


Vasile
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