On Tue, 17 Apr 2007 10:33:28 -0700, you wrote: >On 4/17/07, Peter P. wrote: >> Vasile Surducan 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 ? Probably the cheap transformer saturating. However this doesn't necessarily mean that the magnetron only takes 1A of the line current when connected. > > > > > > >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 -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist