# Microwave oven efficiency

#### Charles Bagwell

I am working at retrofitting a microwave oven to produce steam for my 3 HP steam engine/generator. I have read estimates of 50-70% efficiency, but I am trying to verify those figures. Can anyone point me in the direction of calculating the efficiency myself?

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#### russ_watters

Mentor
I am working at retrofitting a microwave oven to produce steam for my 3 HP steam engine/generator. I have read estimates of 50-70% efficiency, but I am trying to verify those figures. Can anyone point me in the direction of calculating the efficiency myself?
Wow, your project is extremely dangerous in multiple ways, and I strongly recommend you don't attempt it. And be aware, PF will not assist with the execution of a dangerous project.

However, measuring the efficiency of a microwave oven is an interesting, easy and stand-alone exercise.

1. Plug the microwave into a Kill-a-Watt.
2. Measure the temperature of a liter of room temp water.
3. Zap it for 2 minutes, stir, measure again.
4. Calculate!

#### anorlunda

Mentor
Gold Member
It is always good to search the PF archives before posting. I found this.

https://duckduckgo.com/?q=microwave+oven+efficiency&ia=web

3HP is about 2.2 kW. At 50% effeicncy you'll need a 4.4 KW or bigger microwave oven. I've never seen one so big.

Microwaves aside, you are building a boiler. Homemade boilers can and do explode and can kill people.
What experience do you have with boilers?
What pressure do you need? Do you have a steam turbine?
What is the location of the boiler with respect to people nearby?
Can you test your generator by some means other than steam?

I ask because we don't allow dangerous topics on PF. So you need to satisfy us that your project is safe.

#### Charles Bagwell

Wow, your project is extremely dangerous in multiple ways, and I strongly recommend you don't attempt it. And be aware, PF will not assist with the execution of a dangerous project.

However, measuring the efficiency of a microwave oven is an interesting, easy and stand-alone exercise.

1. Plug the microwave into a Kill-a-Watt.
2. Measure the temperature of a liter of room temp water.
3. Zap it for 2 minutes, stir, measure again.
4. Calculate!
Thank you for the feedback.
It is always good to search the PF archives before posting. I found this.

https://duckduckgo.com/?q=microwave+oven+efficiency&ia=web

3HP is about 2.2 kW. At 50% effeicncy you'll need a 4.4 KW or bigger microwave oven. I've never seen one so big.

Microwaves aside, you are building a boiler. Homemade boilers can and do explode and can kill people.
What experience do you have with boilers?
What pressure do you need? Do you have a steam turbine?
What is the location of the boiler with respect to people nearby?
Can you test your generator by some means other than steam?

I ask because we don't allow dangerous topics on PF. So you need to satisfy us that your project is safe.
Thank you for the interesting links. I will study them.

And, thank you for your concern. I am a licensed 1st class Stationary and Refrigeration Engineer with unlimited horsepower rating with the City of Detroit. I have been operating Boilers since my Navy days. Most recently 4, 800 MW Super-critical Boiler/Turbine units, operating at 3,600 PSI and 1000 deg. F super-heated steam, for Detroit Edison's Monroe Power Plant in Monroe, Michigan. I have since retired and building my own boiler with appropriate safety measures, such as a relief valve and high pressure cut-out. I will be operating at 100 PSI, slightly super-heated for moisture control, to run my 3 HP double acting D slide valve steam engine, powering a 2 kw electric generator. The main purpose of my research is to develop a heater to heat my home. All the BTUs developed by the system will be contained inside my basement work shop and ducted into my HVAC system.

The way I see it, all the power coming from the wall socket, 1300 watts, is converted to heat (I2R) in the magnetron tube and its electrical components, cooling fan, light, electronics. 1300 watts X 3.413 BTU/watt = 4436.9 BTU/hr entering the room. Now, on top of those BTU, there are an additional amount of BTU, coming from the vibrating water molecules, which heats the food. Assuming a 70% efficiency rating, That amount is (4436.9 X .7 ) = 3105.83 + 4436.9 = 7542.73 Total BTU/hr entering the room, or a COP of 1.7. Don't forget, that without the two permanent ring magnets, on either end of the tube, there would be no microwaves. So, that 70% is being produced by the magnets at the speed of light, in my opinion. In other words, without the magnets all you have is a 1300 watt common resistance heater.

I suspect the COP is really higher than 1.7, because, unaccounted for, is the heat generated in the side walls of the oven and other components.

I'd like to know what the operating temperature of the magnetron tube body is with and without a water load.

I'd like to know what temperature the actual microwaves are. Is the tube endothermic or exothermic?

I have not seen these questions addressed in the limited research I've done so far. So in the near future I will be addressing these questions plus others that pop up as I proceed.

#### 256bits

Gold Member
all you have is a 1300 watt common resistance heater
And you won't do any better than that to heat water, actually less using a kitchen microwave.
An efficiency rating, which you quoted as 70%, means that 70% of the energy input from the wall socket will heat the water and the rest is used to heat the electronics, circuits and parts of the machine due to inefficiencies.

#### Asymptotic

The way I see it, all the power coming from the wall socket, 1300 watts, is converted to heat (I2R) in the magnetron tube and its electrical components, cooling fan, light, electronics. 1300 watts X 3.413 BTU/watt = 4436.9 BTU/hr entering the room. Now, on top of those BTU, there are an additional amount of BTU, coming from the vibrating water molecules, which heats the food. Assuming a 70% efficiency rating, That amount is (4436.9 X .7 ) = 3105.83 + 4436.9 = 7542.73 Total BTU/hr entering the room, or a COP of 1.7. Don't forget, that without the two permanent ring magnets, on either end of the tube, there would be no microwaves. So, that 70% is being produced by the magnets at the speed of light, in my opinion. In other words, without the magnets all you have is a 1300 watt common resistance heater.
This doesn't sound right.
1300 watts input at 70% efficiency is 910 watts of microwave radiation coupled into the load.
No matter how you slice it, all you have is 1300 watts to work with.

http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/mwoven.html

#### Charles Bagwell

And you won't do any better than that to heat water, actually less using a kitchen microwave.
An efficiency rating, which you quoted as 70%, means that 70% of the energy input from the wall socket will heat the water and the rest is used to heat the electronics, circuits and parts of the machine due to inefficiencies.
Thank you for your feedback. You may be right. But,
And you won't do any better than that to heat water, actually less using a kitchen microwave.
An efficiency rating, which you quoted as 70%, means that 70% of the energy input from the wall socket will heat the water and the rest is used to heat the electronics, circuits and parts of the machine due to inefficiencies.
And you won't do any better than that to heat water, actually less using a kitchen microwave.
An efficiency rating, which you quoted as 70%, means that 70% of the energy input from the wall socket will heat the water and the rest is used to heat the electronics, circuits and parts of the machine due to inefficiencies.
You may be right. But, I will still run my own tests to verify. I will furnish feedback as to my results.

#### russ_watters

Mentor
And, thank you for your concern. I am a licensed 1st class Stationary and Refrigeration Engineer with unlimited horsepower rating with the City of Detroit. I have been operating Boilers since my Navy days. Most recently 4, 800 MW Super-critical Boiler/Turbine units, operating at 3,600 PSI and 1000 deg. F super-heated steam, for Detroit Edison's Monroe Power Plant in Monroe, Michigan. I have since retired and building my own boiler with appropriate safety measures, such as a relief valve and high pressure cut-out. I will be operating at 100 PSI, slightly super-heated for moisture control, to run my 3 HP double acting D slide valve steam engine, powering a 2 kw electric generator.
This resume and demonstrated understand of the thermodynamics and safety involved does indeed reduce my level of concern. That said - ironically perhaps - the electrical engineers here were more worried about the boiler safety and the mechanical engineer (me) more worried about the microwave safety.
The way I see it, all the power coming from the wall socket, 1300 watts, is converted to heat (I2R) in the magnetron tube and its electrical components, cooling fan, light, electronics.
This doesn't sound right. Most of the power is converted to microwave radiation, not thermal energy, and certainly not via I2R. I2R is [a portion of] the loss, not the desired conversion. That said:
1300 watts X 3.413 BTU/watt = 4436.9 BTU/hr entering the room.
Yes, eventually that's true. The microwaves get absorbed by what is inside the microwave and then over time released into the room as heat.
Now, on top of those BTU, there are an additional amount of BTU, coming from the vibrating water molecules, which heats the food. Assuming a 70% efficiency rating, That amount is (4436.9 X .7 ) = 3105.83 + 4436.9 = 7542.73 Total BTU/hr entering the room, or a COP of 1.7.
No, that's already accounted for; you're double-counting the microwaves. A microwave oven does not have a COP of ~1.7, it has an efficiency of ~70%. Having a >1 COP would violate conservation of energy.

In any case, the loss through the casing of the microwave would be hard to measure, but I would think it should be obvious that it's way less than 1300 watts. Just stand in front of an electric space heater for comparison.

#### Nik_2213

IMHO, heating water by microwave is a horribly inefficient system, wasting a third or more of the 'plug power' as low-grade heat.

A resistive element, an 'immersion heater', gives you 100% plug-power less 'edge effects' mitigated by lagging.

As I see it, you need a resistive heat source that robustly self-regulates, so will not 'run away' if water feed fails. I thought I had a reference for eg bitumen pipe heaters, but 'tis lost...

The only way to do better than 100% 'from the wall' would be a heat-pump, but I can't think of a compact, efficient cycle with an output temperature which would fit your needs. Nearest equivalent may be 'solar concentrator' tech. 'Ocean Thermal' systems, IIRC, run on a much lower temperature differential...

#### russ_watters

Mentor
Well I have a little time to kill before lunch, so I ran the test I described above. Spoiler alert: the US Department of Energy monitors and regulates the energy output of consumer devices, and my microwave's nameplate sticker lists its IEC-705 test results as 1480W input, 1000W output, or 68% efficiency. So let's see if I can match that....
• I have a food scale and used it to weigh a liter of water going into a Pyrex dish.
• Using both a meat thermometer reading in F and glass thermometer I get 15.0C (or 59.0F) for the water.
• Zap for 2 minutes.
• After zapping, I stir a little and get about 38C (100F). This reading had more uncertainty due to the uneven heating and stirring, but it should be within about half a degree C or 1F.
• My whole-house energy monitor was reading 535W +/-5W before and after the test and 2250W +/-5W during
Easy math:
2250W - 535W = 1715W Hmm....well that's way off.
1000g * 4.186J/gC * (38C-15C) / 120s = 802W Hmm...well that's even worse.

Then I remembered the Pyrex dish is not insignificant. Ok, well that's 728g. So let's add that in:
728g * 0.8J/gC * (38C-15C) / 120s = 112W Well that's a little better. Note: different sources give quite a bit of variation in the specific heat for Pyrex.

Results:
• Input: 1715W
• Output: 914W
• Efficiency: 53%
Obviously there are some potential sources of error here:
• Convective and evaporative heat loss. Over a short test these should be relatively small. I suppose I could wrap a towel around the dish and put a lid on it.... But I've actually calculated and monitored the temperature rise while heating up water for beer brewing because of how long it takes on a stove, and the temperature rise is shockingly linear right up to a few degrees below boiling. So I don't think that should be very significant.
• The other side of that coin though is perhaps the oven has a warm-up period and part of the heat heat rejection from a warm microwave should be considered in the output? This could be significant.
• On the input side, I'm not sure of the consistency of the waveform a microwave pulls. It's possible an irregular wave pulls lower energy while tricking my energy monitor. I could look up if my energy monitor can deal with that or maybe someone can hook theirs up to an oscilloscope....
• Otherwise, all I can think of is maybe the microwave somehow gets less efficient as it ages: It's 16 years old.
Now, a note on the heat rejection of the electronics. My microwave has a little inlet and fan in the back. The inlet was covered with dust, so I cleaned it. Near as I can tell the outlet of that fan is louvers under and in the microwave. It's certainly possible they are trying to use some of the heat rejection of the electronics to heat the food. Holding my hand around the perimeter of the microwave, I felt a small amount of airflow, but it was not noticeably warm. So the heat rejection - while the test indicates it should be significant - is too small and diffuse to easily measure. Again, that could change with a longer test. If my microwave is mostly 5kg of steel, that 800W should warm it up by about 20C per minute, so it should be at operating temperature in a couple of minutes and start impacting the water or be noticeable outside. Again, I could run the test again but I don't feel like it.

#### Charles Bagwell

This resume and demonstrated understand of the thermodynamics and safety involved does indeed reduce my level of concern. That said - ironically perhaps - the electrical engineers here were more worried about the boiler safety and the mechanical engineer (me) more worried about the microwave safety.

This doesn't sound right. Most of the power is converted to microwave radiation, not thermal energy, and certainly not via I2R. I2R is [a portion of] the loss, not the desired conversion. That said:

Yes, eventually that's true. The microwaves get absorbed by what is inside the microwave and then over time released into the room as heat.

No, that's already accounted for; you're double-counting the microwaves. A microwave oven does not have a COP of ~1.7, it has an efficiency of ~70%. Having a >1 COP would violate conservation of energy.

In any case, the loss through the casing of the microwave would be hard to measure, but I would think it should be obvious that it's way less than 1300 watts. Just stand in front of an electric space heater for comparison.
Russ, I can’t tell you how enjoyable it is to be in a discussion on my favorite subject: Energy. Though, I will have to take exception to your statement “Having a >1 COP would violate conservation of energy”. I would agree with you if we were discussing a closed system. But the microwave oven is an open system, and not governed by that law, in my opinion.

Let me relate a little story concerning COP: Forty five years ago, when I was studying for my steam and refrigeration licenses, I was amazed to find that Freon air conditioners, AKA, heat pumps, were routinely operating with COP>1. I began building experimental heat pumps. I tapped into the Freon side and, through nozzles, was running gear pumps, little turbines and scroll compressors, backwards. They didn’t develop a lot of power, but it was fun and I learned a lot. I learned that the High COP was not because of the low expansive properties of the Freon 22, that I was using. The secret was in the phase change, from liquid to vapor, in the evaporator. Each pound of Freon would suck in 90 BTU in the low side and deposit them in the high side condenser. I was doing this with only 18 BTU/pound of electrical power running the scroll compressor. It was magic.

I was so impressed that 30 years ago I built and installed a closed loop, three ton geothermal heat pump system with three, one hundred foot deep, 4” bore holes, with four hundred feet of one inch pipe in each. I push 8.5 GPM through the system. The Delta T is 15 degrees. So, 8.5 GPM X 8 lb/gal X 15 deg. X 60 min/hr
= 61,200 Btu/hr brought into the home from the ground mass (Solar Energy). The Electrical energy required from the grid is: 240 V AC X 11 amps = 2,640 watts/hr.

Converted to Btu: 2,640 X 3.413 BTU/watt = 9.010.32, which, in heating mode, is added to the heat from the ground and = 70,210.32 output. So, COP = 70,210.32/9,010.32 = 7.75 COP. That is very good! It saves me a bunch of money every year on heating and cooling costs.

Anyway, I see the microwave oven system as an analog to the Geo system. It seems to me that the microwaves produced by the magnetron vacuum tube acts as a catalyst to release BTUs from the vibrating water molecules (The Aether). The microwaves do not heat other substances that do not contain water. What I am saying is the heat that cooks the food comes from the water molecules through friction rather than being directly heated by the microwave radiation, in my opinion. So, those BTUs are added to the room as well as the BTUs from the grid. It still seems to me that the 1.7 COP estimate is in the ballpark, but probably low. Now, my job is to figure out how low. Suggestions welcomed. Sorry for the long post.

#### russ_watters

Mentor
Though, I will have to take exception to your statement “Having a >1 COP would violate conservation of energy”. I would agree with you if we were discussing a closed system. But the microwave oven is an open system, and not governed by that law, in my opinion.
Conservation of energy always applies. What can sometimes be difficult in an open system is identifying the inputs and outputs in order to balance the energy equation.
... I was amazed to find that Freon air conditioners, AKA, heat pumps, were routinely operating with COP>1. The secret was in the phase change....It was magic.
This is a very common thing people are surprised by and very common that people get the "why" wrong. Saying the secret is in the phase change is at best incomplete (you can make an air conditioner or heat pump without a phase change, with compressed air) and misses the point of what an air conditioner or heat pump does. The reason an air conditioner or heat pump can have a COP greater than 1 is that the electrical input isn't the only input. The electrical input just facilitates heat transfer between two heat reservoirs. It's not unlike carrying a bucket of hot water or ice from one place to another: the energy required to move the heat is partly decoupled from the amount of heat being carried.
Anyway, I see the microwave oven system as an analog to the Geo system. It seems to me that the microwaves produced by the magnetron vacuum tube acts as a catalyst to release BTUs from the vibrating water molecules (The Aether).
There's no mystery to how microwave ovens work and no need to try to invent your own theory:
https://en.wikipedia.org/wiki/Dielectric_heating

Experimentation is generally fine, but there's safety risks here. I think it would be more productive if you took some online courses and learned thermodynamics. In either case, this thread has turned into a trifecta of things we don't allow discussion of here:
1. Perpetual motion machines.
2. Personal/non-mainstream theories.
3. Dangerous activities.

So I'm locking the thread. Good luck to you.

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