3
$\begingroup$

I measured the heat absorbed by water in a microwave, and it was about 80% of the max. power of its specifications.

But when the amount of water decreases too much (100 g or less) the efficiency decreased sharply.

I wonder where the EM energy goes in that case, (or when there is no thermal load inside). It is true that the internal walls of the device are heated, but not too much.

I believe that the current is the same (at the same power set), no matter how much food is inside. Maybe the power factor changes according to the thermal load, and we are not charged more for heating several times half cup of coffee, instead of one liter one time?

$\endgroup$
2
  • 1
    $\begingroup$ Interesting. How did you do the experiments? I'd like to reproduce them at home. In particular, how are you comparing the heat absorbed by the water with the power of the microwave? Are you just multiplying it by the time in the microwave? I'd be very surprised if the microwave changed anything depending on what was put it in, but then I've been very surprised about things before! :) $\endgroup$
    – Philip
    Aug 16, 2020 at 23:19
  • 1
    $\begingroup$ @Philip My multimeter has also a thermocouple. I weighted different volumes of water in a kitchen scale, and measured the $\Delta T$ after 1 minute of heating. After calculating the heat transfered in Joules, I divided by $60$ to get $W = J/s$. And compared with the supplier specification. $\endgroup$ Aug 17, 2020 at 0:32

4 Answers 4

1
$\begingroup$

According to a couple of answers at https://electronics.stackexchange.com/questions/206206/is-a-microwaves-output-power-proportional-to-the-mass-of-its-contents/361645 , where they conducted experiments, the power draw (practically) does not depend on the amount of food in the microwave oven. As for the efficiency, some energy is absorbed in the walls and other parts of the oven.

$\endgroup$
1
$\begingroup$

If the microwave oven is operated without a dissipative load in the cavity, the microwave energy in the cavity increases until it builds to a level that begins affecting the operation of the magnetron itself, with the potential of destroying the magnetron in the process. This is why the microwave manufacturers assert you must not operate the oven without a load in it.

It is possible that your microwave detects the no-load condition and cuts back the power output of the tube so it doesn't get damaged. Experiments will answer this question.

The magnetron puts out the same amount of RF energy regardless of how much food is in its cavity (unless it has a no-load detect feature). This means that if you double the amount of food in it, you double the time it takes to heat it all up, and if you cut the food load in half, you cut the cooking time in half too.

$\endgroup$
0
$\begingroup$

The magnetron in most common home microwave ovens only has one power rating. Different power settings are accomplished by turning the magnetron on and off at different intervals. See: https://en.wikipedia.org/wiki/Microwave_oven#Control_panel

$\endgroup$
0
$\begingroup$

Microwaves are, unsurprisingly, waves and act accordingly.

When waves are created, they travel away from the source until they are either absorbed or reflected. This is known as forward power. A closed cavity reflector such as a metal oven absorbs very little, but reflects them around until they return to the source. This is known as reflected power.

At some point the wave power in the closed box must be absorbed, causing local heating. If the oven is empty, most of the power will be re-absorbed by the emitting circuitry, causing it to overheat. Depending on the technology the oven will either turn itself off briefly, turn itself down, or break down.

Food is the obvious absorber, or load. The maximum power theorem states that the maximum amount of power will be absorbed when the source and load impedances are equal, or matched. The oven will have a design limit on the amount of food it can cook and will (or should) be designed to match the impedances at this point.

Any different amount of food will cause some power to be reflected and the source must be capable of absorbing that reflected power. This is as true for excessive amounts of food as it is for small amounts. In practice food is not closely-characterised electromagnetically and there will almost always be 20% or more reflected power to deal with.

As well as the power absorbed by the food, the total power emitted includes any reflected component.

The power actually consumed also depends on inefficiencies such as the internal resistance of the circuit. The "power" charged for by the supplier will also depend on the power factor of the circuit in operation, as suppliers actually charge for current not power; this is true for all electrical appliances.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.