Making a smart electric kettle to heat water

[sygyzy]

I want to have a serious discussion about what the best design to heating water to specific temperatures. As a bonus we can also discuss maintaining that temperature. The reason this is important to me is I am a tea (and coffee but to a lesser degree) snob. I care about how many grams of tea I am brewing, what temp the water is and how long.

We will be focusing on electric kettles because they are convenient and heat up fast (efficient). They also don't require a stove, something I don't have access to at work. There are some variable temperature kettles on the market and I had the T-Fal one but found that it really did not heat up to a range of temps. Instead it was either 200 from the 0 position up to 50% then 212 the rest of the way.

I am not a physicists but this is how I imagine you could solve this problem.

1. Use a PID. Basically you know how much heat your element produces and you keep a thermometer inside the kettle. When it reaches that temp, you shut it off and sound the audible alarm.

2. This is more complicated and I am not sure how it'll work but here's my idea. The kettle has a known weight (empty) and since it must sit on the base/dock to heat the base can measure how much water is in there based on the additional weight. Water weighs about 1KG per litre so you can extrapolate that. Then you know how powerful your heating element is and you can figure how much long it needs to run to heat up that much water to a certain temp.

A kettle I am using holds 1.8 litres and has a 1500 watt element.

1.8 L = 1800 grams. A rise of 70F - 200F converted to C is 72.22 degrees.

1800 x 72.22 = 129,996 calories
129,996 x 4.184 = 543903.26 joules
543903.26 joules / 1500 Watts - 362 seconds
362 s / 60 seconds = 6.04 minutes

Since kettles specifically tell you to only put water in it, we are not concerned about the calories of other liquids such as juice or stock.

Is there a different way this can be done? Can anyone suggest a different kettle design?

 

[TuxDave]

Going along with #1:
Forget how much your heating element puts out (as long as you know that it exceeds the temp that you desire). Just let it go full blast and when the cup gets to the right temp, shut it off. When it gets too cold again, turn it back on. You'll have an error on the upper (element still tranferring heat to the water when off) and on the lower end (water cooling after the element turns off). I have no idea what that error bound is. Probably require a little bit of math.

#2 doesn't work since you will have heat loss and the heating element doesn't have a 100% transfer rate of energy (or does it? ummmm)

 

[sygyzy]

Hi Dave.

#1 - With a PID, that error would be eliminated. A PID should be able to even turn the element on and off in tiny spurts (0.5 s) to keep the temp pretty stable. I have a PID for sous vide and that's what it does. You can see the light go on and off for milliseconds at a time. If it's not using a microprocessor, you're right - you will have some error from getting the element up to speed as you turn it on from an off state.

#2 - You're right. No element is 100% efficient. I don't know much besides that. But it wouldn't be hard to figure out how efficient the particular element used is, right? So just adjust for that. If you know it's 90% efficient, could you just extend the time by 10%? I am not sure if that logic works.

 

[PsiStar]

buy a microwave oven with a thermal probe ... been around for 10+ years

 

[Born2bwire]

I have an air pot that has three temperature settings, 80, 90 and 98 which correspond to green, oolong and black/pu-er teas. For a practical solution, I would just look more closely at your options and you should find what you want. Even if there are limited temperature ranges, just get an electric thermometer with an alarm that can tell you when the water temp has dropped to your desired setting (should only take a few minutes).

To build one, it sounds like a very simple control theory problem. Just have a temperature probe and a means to control the element. Use control loop theory to adjust the element based on the temperature. Code this up into a say a PIC or some other microcontroller.

 

[Howard]

Scale will build up on the element and reduce its efficiency.

The only way I can see this is with a PID controller.

 

[CycloWizard]

I just got back from the UK, where I was working on a project that required me to design a small climate control chamber. I mention being in the UK because they use lots of electric "kettles" there for tea - just a 2L plastic pitcher which heats up the water pretty quickly. I am not a big tea person myself, so I'm not sure if it's actually temperature controlled, on/off, or even always on.

The system I actually built was designed to maintain body temperature. I achieved it using a simple on/off control algorithm. I modeled the system in considerable detail to arrive at the conclusion that this approach would work, and I can summarize them as follows:
1. The heating rate is small relative to the thermal mass of the system (or the temperature sampling rate is high relative to the system's thermal equilibration time at a given heating rate);
2. Heat transfer within the system occurs at a rate which is high relative to the heating rate (i.e. the system may be considered "well mixed") or, if not well mixed, at least you can estimate the thermal energy of the system based on measuring the temperature at one known point; and
3. Thermal losses are small relative to the heating rate (i.e. the system is "well insulated").

If these three conditions are met, which they probably are in your system, then an on/off temperature system would work well. It's quite easy and cheap to measure the temperature to a suitable degree of accuracy. The method I eventually settled on is simply to compare the temperature sensor output to a reference voltage using a comparator. If the temperature voltage is higher than the reference, shut off the heater; otherwise, keep heating. The reference voltage will depend on a few things, but it will probably be easiest to simply find it by trial and error, especially if the system is not well mixed. You can get an idea for what kind of heaters you need based on your approach #2 above, which is exactly the way I approached it.

 

[CycloWizard]

Hi Dave.

#1 - With a PID, that error would be eliminated. A PID should be able to even turn the element on and off in tiny spurts (0.5 s) to keep the temp pretty stable. I have a PID for sous vide and that's what it does. You can see the light go on and off for milliseconds at a time. If it's not using a microprocessor, you're right - you will have some error from getting the element up to speed as you turn it on from an off state.

This won't necessarily work well for your problem, as electrical heaters of this type have a fairly long dead time before producing sensible heat change in the surrounding fluid. This can lead to instabilities and other complications which are probably a lot more than you want to get involved with for a project like this.

#2 - You're right. No element is 100% efficient. I don't know much besides that. But it wouldn't be hard to figure out how efficient the particular element used is, right? So just adjust for that. If you know it's 90% efficient, could you just extend the time by 10%? I am not sure if that logic works.

Depending on where you're putting the element, thermal efficiency should be very close to 100%. If it's actually in the water, it would be 100%. If it's not, then it becomes a bit more complex.

 

[canis]

I want to have a serious discussion about what the best design to heating water to specific temperatures. As a bonus we can also discuss maintaining that temperature. The reason this is important to me is I am a tea (and coffee but to a lesser degree) snob. I care about how many grams of tea I am brewing, what temp the water is and how long.

We will be focusing on electric kettles because they are convenient and heat up fast (efficient). They also don't require a stove, something I don't have access to at work. There are some variable temperature kettles on the market and I had the T-Fal one but found that it really did not heat up to a range of temps. Instead it was either 200 from the 0 position up to 50% then 212 the rest of the way.

I am not a physicists but this is how I imagine you could solve this problem.

1. Use a PID. Basically you know how much heat your element produces and you keep a thermometer inside the kettle. When it reaches that temp, you shut it off and sound the audible alarm.

2. This is more complicated and I am not sure how it'll work but here's my idea. The kettle has a known weight (empty) and since it must sit on the base/dock to heat the base can measure how much water is in there based on the additional weight. Water weighs about 1KG per litre so you can extrapolate that. Then you know how powerful your heating element is and you can figure how much long it needs to run to heat up that much water to a certain temp.

A kettle I am using holds 1.8 litres and has a 1500 watt element.

1.8 L = 1800 grams. A rise of 70F - 200F converted to C is 72.22 degrees.

1800 x 72.22 = 129,996 calories
129,996 x 4.184 = 543903.26 joules
543903.26 joules / 1500 Watts - 362 seconds
362 s / 60 seconds = 6.04 minutes

Since kettles specifically tell you to only put water in it, we are not concerned about the calories of other liquids such as juice or stock.

Is there a different way this can be done? Can anyone suggest a different kettle design?

What you are describing is not a "smart" electric kettle. It is binary operation based on one input. Not worthy of a ucontroller. What exactly do you mean by PID?

Your calculation oversimplifies the physical model and is inaccurate. Your method #2 is complicated, but not complicated enough.

The easiest solution is to measure the time it takes for your desired temperature and use an egg timer. You need to take into account the temperature of the water put into the kettle.

Comments about heat production not being 100% efficient are incorrect. The correct statement is thermal conductivity is finite for real materials.

 

[CycloWizard]

What you are describing is not a "smart" electric kettle. It is binary operation based on one input. Not worthy of a ucontroller. What exactly do you mean by PID?

Your calculation oversimplifies the physical model and is inaccurate. Your method #2 is complicated, but not complicated enough.

His #1 neglects internal thermal resistances (i.e. transfer of heat through the water from the heater to the sensor), but can actually be shown to work quite well. You're right that this method doesn't require a PID, as it is simply an on/off system, whereas a PID would require an analog voltage to the heater, varying the heating rate according to the controller output. His #2 is actually the simplest possible method for determining the energy required to raise the temperature of x amount of water by y degrees and assumes negligible thermal losses from the system. This model will hold to a high degree of accuracy assuming the thermal efficiency of the heater is high (which it should be for an electric kettle with a submerged heating element) and where the actual losses in the system are limited by natural convection (which is almost certainly true in this case).

The easiest solution is to measure the time it takes for your desired temperature and use an egg timer. You need to take into account the temperature of the water put into the kettle.

Yes, but what fun is that? This method is also susceptible to errors due to ambient air temperature differences and other disturbances.

Comments about heat production not being 100% efficient are incorrect. The correct statement is thermal conductivity is finite for real materials.

No, heating will not be 100% efficient in a number of cases. For example, if he is simply using a burner on a stove outside the body of the kettle, some of the heat will never enter the water. Thus, while all of the electrical energy is converted to heat energy, not all of that heat energy will actually find its way into the system. Any relevant measure of efficiency would involve computing the ratio of the input electrical energy to the heat input to the water he's trying to heat.

 

[Cogman]

maybe not a teapot, but I've been thinking of a water distillation design for some time.

Rather then just curving a tube out the top of the water distiller, have the tube go back into the distiller and make a few submerged loops. That should make the distiller efficient by transferring the energy from the steam to the boiling water. Make a few loops outside the pot to ensure the steam is cool and you have yourself some clean water via most energy sources at minimal energy cost.

 

[TuxDave]

No, heating will not be 100% efficient in a number of cases. For example, if he is simply using a burner on a stove outside the body of the kettle, some of the heat will never enter the water. Thus, while all of the electrical energy is converted to heat energy, not all of that heat energy will actually find its way into the system. Any relevant measure of efficiency would involve computing the ratio of the input electrical energy to the heat input to the water he's trying to heat.

/agree

Yeah that's what I originally thought when I posted the not 100% number. Then I started thinking about if it wasn't a stove top but more like a heating element immersed in water, and that's probably very close to 100%.

 

[CycloWizard]

/agree

Yeah that's what I originally thought when I posted the not 100% number. Then I started thinking about if it wasn't a stove top but more like a heating element immersed in water, and that's probably very close to 100%.

Yep. We need to campaign for an MS Paint add-on for the HT forum so we can have fun times interpreting high-tech schematics.

 

[canis]

His #2 is actually the simplest possible method for determining the energy required to raise the temperature of x amount of water by y degrees and assumes negligible thermal losses from the system. This model will hold to a high degree of accuracy assuming the thermal efficiency of the heater is high (which it should be for an electric kettle with a submerged heating element) and where the actual losses in the system are limited by natural convection (which is almost certainly true in this case).

The simplified model is inaccurate in determining the time required. This is true even if the model "assumes negligible thermal losses". The negligible thermal loss assumption is unrealistic as well. So the temperature will not be correctly deduced for a specific period of time according to the simplified model. Which is the point.

Yes, but what fun is that? This method is also susceptible to errors due to ambient air temperature differences and other disturbances.

True. Air temps do have an effect. But it doesn't mean the difference is large. People like to live in comfortable temperatures. There would be a large difference if someone was making tea outdoors in the arctic. Even if there was a temp difference, time can be measured with the different conditions in place. What "other disturbances" are you referring to?

No, heating will not be 100% efficient in a number of cases. For example, if he is simply using a burner on a stove outside the body of the kettle, some of the heat will never enter the water. Thus, while all of the electrical energy is converted to heat energy, not all of that heat energy will actually find its way into the system. Any relevant measure of efficiency would involve computing the ratio of the input electrical energy to the heat input to the water he's trying to heat.

Please note my original statement. Note the phrase "heat production":
Comments about heat production not being 100% efficient are incorrect. The correct statement is thermal conductivity is finite for real materials.

Your comment does not negate my original statement.

Using the word efficiency to describe how much heat is transferred into a material is layman's usage.

 

[CycloWizard]

The simplified model is inaccurate in determining the time required. This is true even if the model "assumes negligible thermal losses". The negligible thermal loss assumption is unrealistic as well. So the temperature will not be correctly deduced for a specific period of time according to the simplified model. Which is the point.

It is actually quite likely that the thermal losses in this system would be negligible. It is almost a certainty that the rate of heat transfer from the outer wall to the surrounding air will be rate-limiting (i.e. the rate of natural convection will be the rate of heat loss). This rate will on the order of tens of Watts, assuming the thing is poorly insulated. This is therefore small relative to the heating rate (1500 W).

True. Air temps do have an effect. But it doesn't mean the difference is large. People like to live in comfortable temperatures. There would be a large difference if someone was making tea outdoors in the arctic. Even if there was a temp difference, time can be measured with the different conditions in place. What "other disturbances" are you referring to?

Yes, one may empirically determine the conditions necessary for the perfect cup of tea under any set of circumstances one wishes, but it's a lot easier to simply build in a simple feedback system which will do the job without infinitely many experiments. That's the whole point of the exercise, if I'm not mistaken.

 

[canis]

It is actually quite likely that the thermal losses in this system would be negligible. It is almost a certainty that the rate of heat transfer from the outer wall to the surrounding air will be rate-limiting (i.e. the rate of natural convection will be the rate of heat loss). This rate will on the order of tens of Watts, assuming the thing is poorly insulated. This is therefore small relative to the heating rate (1500 W).

The electric kettles I saw were made from thin plastic. Latent heat of vaporization is also present. Heat loss with respect to time is non linear as well.

Yes, one may empirically determine the conditions necessary for the perfect cup of tea under any set of circumstances one wishes, but it's a lot easier to simply build in a simple feedback system which will do the job without infinitely many experiments. That's the whole point of the exercise, if I'm not mistaken.

Indoor temperatures are typically at a limited range of temperatures. I would call it a stretch to say infinite number of experiments are required.

 

[CycloWizard]

The electric kettles I saw were made from thin plastic. Latent heat of vaporization is also present. Heat loss with respect to time is non linear as well.

Latent heat loss will be negligible assuming the kettle is full. Heat loss with respect to time will be a nonlinear increasing function which will be maximized at maximum temperature. This maximum loss will be limited by the natural convection losses rather than conduction through the plastic walls as you will clearly see if you do the math yourself, even if you assume that the thermal conductivity of the plastic is infinite.

Indoor temperatures are typically at a limited range of temperatures. I would call it a stretch to say infinite number of experiments are required.

It's not infinite, but it's not an efficient approach, either. The most efficient way would be to employ a response surface method in which the heating time is measured as a function of the room temperature and initial water temperature. Or, you could just throw in a $2 circuit and do it all automatically without worrying about how to do the experiments.

 

[canis]

Latent heat loss will be negligible assuming the kettle is full.

Why will vaporization heat loss be negligible if the kettle is full as opposed to being partially filled?

It's not infinite

True.

IMHO people should not be fooling around with mains power unless they know damn well what they are doing.

 

[CycloWizard]

Why will vaporization heat loss be negligible if the kettle is full as opposed to being partially filled?

Because the amount of water vaporized will be virtually nill. The electric kettles I've seen are closed systems - there is no vent for the vapor to leave the system.

True.

IMHO people should not be fooling around with mains power unless they know damn well what they are doing.

Meh. Worse thing you'll do is start a small fire. Or kill yourself via electrocution.

 

[Born2bwire]

Because the amount of water vaporized will be virtually nill. The electric kettles I've seen are closed systems - there is no vent for the vapor to leave the system.

The true cheap kettles have a constant small vent with the spout. But every air pot that I have seen, which would be the closest to a closed system, has what I believe to be a pressure sensitive vent. When the water is under a full boil for a time, the pot will vent the vapor for what I assume is only to release some of the pressure in the pot. At other times though, the vent appears to be closed. As for insulation, you can even get air pots that have vacuum insulation like in a thermos.

 

[CycloWizard]

The true cheap kettles have a constant small vent with the spout. But every air pot that I have seen, which would be the closest to a closed system, has what I believe to be a pressure sensitive vent. When the water is under a full boil for a time, the pot will vent the vapor for what I assume is only to release some of the pressure in the pot. At other times though, the vent appears to be closed. As for insulation, you can even get air pots that have vacuum insulation like in a thermos.

I'm not familiar with tea brewing I suppose, but I had assumed the water wouldn't be boiling, so the pressure would be fairly low (with an upper limit of atmospheric pressure + the vapor pressure of water at the setpoint temperature).

 

[canis]

Because the amount of water vaporized will be virtually nill. The electric kettles I've seen are closed systems - there is no vent for the vapor to leave the system.

Where did you see such electric kettles? If such a kettle existed, I would be shocked at the utter stupidity and dangerousness of the design.

 

[canis]

The true cheap kettles have a constant small vent with the spout. But every air pot that I have seen, which would be the closest to a closed system, has what I believe to be a pressure sensitive vent. When the water is under a full boil for a time, the pot will vent the vapor for what I assume is only to release some of the pressure in the pot. At other times though, the vent appears to be closed. As for insulation, you can even get air pots that have vacuum insulation like in a thermos.

I don't drink coffee or tea (I find the smell, taste and caffeine unpleasant) so I was unfamiliar with the meaning of air pot. It would be complicated to have a vacuum insulated chamber and a heater heating the chamber contents at the same time. Have you seen a air pot with both?

 

[DominionSeraph]

http://www.sears.com/shc/s/p_10153_1...0070921x00003a

 

[canis]

..