Simple way to measure thermal conductivity

[Maiapa]

hello!
I'm investigating on thermal conductivity of concrete blocks.
I've got an equation for calculating K, which is Fourier's law.
Q'=-k×A× ∆T/l

I have to make the experiment simple as possible as i can.
My plan is to heat one side of a block of concrete and measure the temperature difference then the put the all values needed in the equation.
I can get values of A(area), ∆T(temp difference) and l(length of the block).

but i really don't get what Q' is. i have no idea and nothing have clued me up yet.
could anyone teach me what it means and how to get the value of Q' for my experiment?
(assume that i can get whatever that is needed for the experiment)
Available apparatus from my school are Joulemeter, power supply and thermocouple etc

the really important thing is 'how to measure the value'

God will bless replier
thank you!

 

[Dadface]

hello!
I'm investigating on thermal conductivity of concrete blocks.
I've got an equation for calculating K, which is Fourier's law.
Q'=-k×A× ∆T/l

I have to make the experiment simple as possible as i can.
My plan is to heat one side of a block of concrete and measure the temperature difference then the put the all values needed in the equation.
I can get values of A(area), ∆T(temp difference) and l(length of the block).

but i really don't get what Q' is. i have no idea and nothing clued me up yet.
could anyone teach me what it means and how to the value of Q' for my experiment?
(assume that i can get whatever that is needed for the experiment)
Available apparatus from my school are Joulemeter, power supply and thermocouple etc

God will bless replier
thank you!

Hello Maiapa Q stands for the heat flow per second when the block reaches steady state.If you heat one side of the block the temperatures on both sides will rise and at steady state both will have reached their maximum value the heat lost per second becoming equal to the heat supplied per second.You can measure Q with your Joulemeter and a clock.I think you can get extra details if you google "Lees Disc"

 

[Maiapa]

Thanks for your reply.
I got one more question.
I'm going to heat one side of a block using a plate of copper (or other metals).
If electricity flows through the copper it will heat the concrete up.

When I'm doing this, how can i measure Q (heat flow per second).

Could you give me an example about how i should do?

 

[Dadface]

Thanks for your reply.
I got one more question.
I'm going to heat one side of a block using a plate of copper (or other metals).
If electricity flows through the copper it will heat the concrete up.

When I'm doing this, how can i measure Q (heat flow per second).

Could you give me an example about how i should do?

You need to get some sort of electrical hot plate and make sure you you have your experiment checked for safety.
1.Set it up as you described with the joule meter connected to the heating supply.
2.Switch on the heater and using your thermocouples measure the temperatures,both will be increasing.
3.Now you need to be patient and wait for steady state to be reached.Measure the temperatures every so often.Steady state is reached when both temperatures stop rising but there's likely to be some small fluctuations because of changes in the local conditions.For a thick slab this will take a long time.
4.At steady state take a reading from your joule meter and start a clock .Keep it running for several minutes record the Joules transferred and the time and find Q by division.

 

[Maiapa]

Thank you again!

But the thing is...
what is the real concept of joule? Is it electricity that i flow through the metal?
or.. Is it just heat?
If it is heat, how can i measure it?
and how to use joule meter?

hmm... it gets vaguer.

One more question, if you were me, what would you choose for heating method.
I think metals such as copper and iron produces really big power (e.g 12v->1300kW)
Could you suggest any feasible alternatives?

 

[Dadface]

Joules here refers to both the electrical energy fed to the heater and the heat energy it generates

 

[Dadface]

Thank you again!

But the thing is...
what is the real concept of joule? Is it electricity that i flow through the metal?
or.. Is it just heat?
If it is heat, how can i measure it?
and how to use joule meter?

hmm... it gets vaguer.

One more question, if you were me, what would you choose for heating method.
I think metals such as copper and iron produces really big power (e.g 12v->1300kW)
Could you suggest any feasible alternatives?

To answer your first question it is both.The electrical energy fed to the heater in Joules is equal to the heat energy output in Joules.The Joule meter is connected to the input of the heater.If it were me I would go for a lees disc arrangement but I am assuming that is not available so I would use a commercially made hot plate with a flat top so that the concrete slab could sit on it and make good thermal contact.The slab should entirely cover the heater and be as thin as possible.

 

[Maiapa]

Thank you again and again.
Now I think I'm getting closer and closer to the answer.

But still have some questions.

If I use a commercial hot plate how can i measure the power supplied.
Should I just put the value of watt from specification of the hot plate?
For example, it says AC220V, 60Hz and 1300W, then which should i put for value of Q'?

and what is the reason for waiting until the temperatures don't change? / what size do you recommend? (height)

 

[Dadface]

Thank you again and again.
Now I think I'm getting closer and closer to the answer.

But still have some questions.

If I use a commercial hot plate how can i measure the power supplied.
Should I just put the value of watt from specification of the hot plate?
For example, it says AC220V, 60Hz and 1300W, then which should i put for value of Q'?

and what is the reason for waiting until the temperatures don't change? / what size do you recommend? (height)

Hello Maiapa,
You can use the specifications from the hot plate for example 1300 watts means 1300 joules per second,this would be your value for Q.
When you heat your concrete initially most of the heat is used to raise the temperature and internal energy of the concrete and only a little bit of the heat passes through by conduction.The temperature rise is rapid at first and slows down as steady state is reached more heat passing through.At steady state the temperatures stop rising and all of the heat supplied per second passes through.
Ideally your concrete slab would be smooth to make good thermal contact with the heater(difficult to achieve with concrete),thin to get a good heat flow rate and of the same area as the heater area so that it sits snugly on top of it and less heat flows sideways rather than upwards.
Using the apparatus available to you you will not get brilliant results but that does not matter provided that you get an awareness of and evaluate the errors and perhaps, can make suggestion about how the experiment can be improved using more sophisticated equipment.

Maiapa, the main concern here is safety.I have seen a small concrete block fail under an increasing compressive load and fragments went flying round the room at high speed.I am not familiar with the properties of concrete or whether it is adviseable to subject it to a high temperatures.I tried a search but found nothing relevant.Please discuss safety issues with your supervisor.

 

[Maiapa]

Hello and Thank you.
I just bought a hot plate which shows how much power is supplied to produce heat.
My experiment will be easier as i can just take the value and put it in the equation.

Wiki dictionary defines the term 'steady state' as "a state when a recent observed behavior of a system continues". I'm curious about how it reaches the 'steady state' even if it's supplied heat continuously.

A point in question is that,, whether i should use insulator or not.
Because of the complicated concept of steady state (for me), i just don't get what i should do.

 

[Dadface]

Think of an ordinary electric fire.When you first switch it on the temperature rises but not indefinately.As the filament gets hotter the amount of heat lost to the surroundings increases.Eventually ,after about a second or so,the filament reaches a highest steady temperature this being when the heat energy lost to the surroundings per second becomes equal to the electrical energy fed to the filament per second.An ordinary filament lamp reaches its highest temperature (steady state) in about one millisecond.The time taken for your concrete depends, amongst other things, on its thickness but because it is a poor conductor I am guessing you will need at least thirty minutes.Please don't be put off by the concept of steady state because all it is is a state when the highest temperatures are reached.The side of your sample in contact with the hot plate will reach the temperature of the plate fairly quickly but the temperature of the cooler side will rise slowly.All you need to do is to measure these temperatures every so often until they level off.In fact it would be more interesting if you measured the temperatures at regular intervals and plotted graphs of temperature against time.If you decide to use a conductor you will still need to wait for steady state but this will be reached much more quickly.With a conductor, however,the sample will need to be long to get an accurately measurable temperature difference and ideally it will be lagged because the heat losses from sides will be appreciable.

 

[Naty1]

Maiapa...experiments are supposed to provide a check on theroy...if you don't study the theory before you try the experiment you'll have a real hard time learning anything.

What you are trying to do takes some theoretical understanding as well as some degree of practical application/experimental/measurement capability...I'd suggest you do some further study on the theory and measurement techniques before you try to design an experiment...

good luck

 

[Dadface]

Hello maiapa,
Pyrex glass,in the form of a large beaker may be a better choice than concrete.It should be safe smoother than concrete and it would be easy to fix a thermocouple to its base.Ideally the sides would be cut off but I imagine this is impractical for you so keep it as it is. The extra error due to heat conducting up the sides of the glass is likely to be small compared to the other errors.As I mentioned before,ideally the sample would have the same area as the hot plate, but again I imagine you will find it difficult to achieve this so get the best fit you can and later on estimate the errors for example what fraction of the 1300 watts do you estimate is actually conducted through the glass?To measure k accurately requires quite sophisticated equipment which you don't have access to but you should be able to get a rough value.As from tomorrow I am away for a few days so good luck with it and enjoy it.I would be interested to hear about any progress you make.
I have just read the comment made by Naty 1 and I am in complete agreement with it.It is really important that you get a good understanding of the theory.

 

[Maiapa]

Hello

Thank you for your advice..
I'm trying to get it done right away!
I've learned lots of things by preparing this experiment for my extended essay (IB program)
Actually, I'm little bit warring about the fact that IB syllabus doesn't include the concept of thermal conductivity and Fourier's law. Is that okay for me to write an essay about a topic that is not included in syllabus?

Um.. I completely understood what 'steady state' means.
All I ask is whether i should use insulator around the slab (on top as well) or not.
If I do so, i think the temperature on the top of the slab will be increasing continuously without stop because i prevent the heat loss through convection.

Could you suggest me what I should do?

I got to go to sleep... it's 3AM here...OMG
(I'm studying in Shanghai, China)

 

[Dadface]

No extra insulator is needed and the temperature will not rise continually,it will reach a highest value and level off.Think about my suggestion for using a pyrex beaker instead of your slab.If you have time perhaps you could use both.I'm not familiar with the IB syllabus but I imagine that students are encouraged to challenge themselves and go beyond the syllabus for tasks such as extended essays and courseworks.I think you will enjoy it .Best wishes.

 

[Maiapa]

Hello,

I thought about your suggestion but i think you maybe misunderstand my topic.
The topic is 'variety thermal conductivities according to different densities'. So I'm using 2 blocks of concrete with different densities.

I don't understand why I don't need to use insulator. If i use then the experiment will give more accurate value. Is that related to steady state? What is the exact reason not to use insulator?

All the best

 

[Dadface]

Hello,

I thought about your suggestion but i think you maybe misunderstand my topic.
The topic is 'variety thermal conductivities according to different densities'. So I'm using 2 blocks of concrete with different densities.

I don't understand why I don't need to use insulator. If i use then the experiment will give more accurate value. Is that related to steady state? What is the exact reason not to use insulator?

All the best

Just a quick reply maiapa because I'm getting ready for a trip.Insulation around the sides of the concrete would be an improvement but only slight because the sample is thin with a large area.Insulation at the top would reduce the heat flow rate and make it more difficult to estimate Q.If it were me I would try it without insulation and then with insulation.The broader your investigation the better.Good luck.

 

[Ron D]

The conduction equation that you are using requires heat flow in one dimension, in this case in the direction from the hot surface to the cold surface. The sides of the block must be well insulated to ensure that very little of the input energy is lost from side surfaces. This is especially important if you are relying on a meter (such as a Joulemeter) that is measuring toal supplied energy. I would not trust the hot plate label rating to be very accurate. Also, the thermocouple on top of the block (cold surface) may read slightly lower than it should (a source of experiment error) because of heat loos by convection to the surroundind air. This is more of a factor if the sample is concrete than it would be if the sample was a good conductor, such as a metal.