Now, we will consider the case of a material which is initially at a temperature of T0 throughout. If heat is supplied at one surface at a constant rate while the temperature of the opposite surface is kept at a constant value, then the temperature difference between the surfaces will rise until a steady value is reached.
Below is an explanation of the experimental procedure that was followed:
- Observed and recorded the thermocouple voltages from the chart recorded and ensured that steady conditions had been reached
- Recorded the voltage and current being supplied to the heater, the heat supplied, Q, to each sample.
- Switched off the power supply, opened up the apparatus, and measured the area and thickness off the sample lightweight concrete slab tested. The thickness was measured in several places and an average of these measurements taken.
- Removed and weighed some of the concrete slab and obtained the bulk density
Calculations for thermal conductivity
Having re-arranged the equation earlier, we can now work out the thermal conductivity of lightweight concrete blockwork.
To find the thickness of the concrete blockwork (d), both samples were identical, therefore we measured one sample. We used a 30cm rule and got the result of 4cm. This measurement was then converted to meters due to the fact we was working out the thermal conductivity in “m” and got the answer 0.040m:
d = 0.040m
Firstly we measured the thickness of the lightweight concrete slab using a ruler which was 0.04m. Next, we needed to calculate the surface area which was the length multiplied by the width of one of the concrete slabs:
L x W = A
To find the surface area of the concrete blockwork (A), we simply measured the length and width, and multiplied the two measurements together:
0.5m x 0.5m = 0.25m A = 0.25m
This is the same surface area for both the samples as they were identical.
To work out the watts we used the formula (q) watts = (v) volts x (i) current. The answer will be divided by two because there are two lightweight concrete slabs.
The reading which we attained from the voltmeter was 61v and the reading which we got from the ammeter was 0.18i we simply put this into the equation which allowed us to obtain the overall number of watts.
Q = Volts x Amps 61 x 0.18 = 5.49w
2 2
To find out temperature difference (t₁ - t₂), we used the graph below which took the temperature in a point of time of both pieces of concrete blockwork.
The red line of the graph indicated 0.86 mv (millivolt) and the blue indicated 0.56 mv. We then multiplied these readings by 5 because 1 mv = 5°C.
0.56mv = 0.56 x 5 = 2.8°C
0.90mv = 0.90 x 5 = 4.5°C
We then needed to find the average temperature difference by finding the sum and dividing the answer by 2.
2.8 + 4.5 = 7.3
7.3 / 2 = 3.65°C
The calculated temperature difference is 3.65°C.
So gathering all the results together, we can put them into the equation and find out the thermal conductivity.
d x Q = λ → 0.040 x 5.49 = 0.241 W/mK
A (t₁ - t₂) 0.25 x 3.65
I have now found the thermal conductivity.
The moisture content of the material then needed to be determined which required that the material be dried and re-weighed. As this is a lengthy process, therefore we assumed that the moisture content (Mc) is 1% by weight.
Density λ = 0.241w/m°C at Mc = 1% by weight
Mc volume = Mc x λ of concrete λ of water
We weighed the concrete blockwork and it weighed 4.030kg. We then calculated the volume by multiplying the length, width and height and got 0.002964 m³. So:-
4.030kg = 13.59kg/m³
- m³
We also knew that the density of water is 1000kg m³. Now knowing the figures we needed to calculate the moisture content, we applied them to the equation.
1% x 13.59kg = 1.359%
1000kg/m³
From the table, we worked out moisture factor for 3 % is 1.6 and for 1.241% is 1.35.
Putting it into the equation we can work out the thermal conductivity at 3% using the table below:
The thermal conductivity at 3% by volume is 0.3060w/m°C and at 1.241% its 0.241w/m°C. It shows that there is a very small amount of difference between them. This indicates to me that the higher percentages in moisture content by volume, the higher the thermal conductivity the material has.
The graph below shows thermal conductivity against moisture content:
The moisture content of concrete is corrected to a standard moisture content of 3%. A reason for the choice of standard moisture content is when a coating manufacturer specifies maximum moisture content in a percentage; this method will allow an applicator to insure that the concrete is within tolerance.
Coating manufacturers have discovered effective methods of overcoming adhesion problems on damp concrete surfaces with the use of various penetrating primers. Most manufacturers will agree, however, that a dry surface will certainly improve adhesion. There is no question that excessive moisture in the concrete can create serious problems when it begins to migrate out of the concrete after the coating is applied. Coating application instructions often specify that the surface must be dry; sometimes they specify an allowable moisture content, expressed in a percentage.
What they fail to mention is how deep this dryness must be or if there is a maximum vapour transmission rate allowable. There is an increasing awareness in the industry of these parameters. The flooring industry has shown a great deal of concern over vapor transmission, generally agreeing that 3 lbs./1000 sq. ft./24 hours (1.4kg/90 sq. m/24 hours) Is an acceptable rate.
Evaluation
We have now calculated the thermal conductivity of concrete blockwork assuming the materials have a moisture content of 1% by weight. The concrete blockwork was required to be dried and reweighed numerous times to find an accurate moisture content; however, this is a lengthy process so we assume the moisture content is 1% by weight. Any errors in the calculation stage and the accuracy of the given answer would have been a issue. Therefore this is an uncontrollable variable in this experiment.
By carrying out all the necessary calculations, we arrived to the answer that the thermal conductivity of the concrete blockwork was 0.241W/mK. By doing further calculations, we also found the answer to the thermal conductivity of 3% moisture content by volume which was 0.3060W/mK.
The graph below indicates the thermal conductivity (W/mK) against density (kg/m³) for lightweight concrete blockwork:
Within this experiment, a few errors would be likely, which could alter our results. Some of these variables could be uncontrollable and some could just be lack of accuracy.
I have already explored one of the uncontrollable variables when working out the temperature difference (t1 – t2). The problem here was that there were two temperatures. There was the red line and the blue line each representing a sample of lightweight concrete blockwork. I think that the temperature difference was uncontrollable because of factors such as heating, lighting, the fact that hot air rises and cools and also the fact that students body temperature was adding to the overall temperature in the room. This could be a possible error, and is definitely an uncontrollable variable.
Another reason for the inaccuracy of results could have been human error in taking readings off the graph. The inaccuracy of these could cause the end result to be incorrect. This is not a big problem; however, increasing the accuracy would have given better results.
I think that the experiment was reasonably accurate and satisfactory results were achieved. If you look at the graph below, you can see the measurements for concrete are slightly off the mark, which could be due to the errors identified above. The two points recorded were in line with each other on the graph showing thermal conductivity against density, so I think the experiment was acceptable.
Other than this we could have used some other form of technique to find the temperature, but I think a thermometer would be less accurate than the temperature reader and it would break when slotted between two concrete blocks. In place of the ammeter I don’t think there is a more accurate way of measuring the current. In conclusion, I think that we worked out the thermal conductivity and on the graph it relates to my results, so this experiment was successful.
