Derek M. DeWitt
Brian Rapp
Kirsten Decoteau
Vince Selami
MA2051 Section A04
Project #1a Heat Loss Model
Part #1:
Problem Statement – For this part of the project we were asked to come up with an experiment to test the thermal conductivity for three different materials. With this data we will compare our experimental results to those suggested by the heat loss model:
T’(t) = -(Ak/cm)[T(t)-Tout], T(0)=Ti
We will also see if our data supports the model. For our experiment we have chosen three materials (aluminum, wood, glass) and we will analyze each material’s heat loss properties. A detailed description of our proposed experiment will follow.
Analysis– The first step we took was to find the surface area of our three objects. We chose a wooden cylinder, an aluminum soda can, and an ordinary kitchen glass. Since the heat disperses through the surface of the material, we chose water to fill our objects. We heated the water and then poured it in each object. We took an initial temperature reading of the water (T(0))and then began timing, taking readings after each two-minute interval. At this point we also took a recording of the ambient temperature (Tout). We recorded this data in order to compare our results to data suggested by the model. Our results are in the following table:
WATER TEMPERATURE [degrees Celsius] AT TIME (t) [minutes]
Material / T=0 / 2 / 4 / 6 / 8 / 10 / 12 / 14 / 16 / 18 / 20Glass / 73.0 / 70.0 / 68.0 / 64.5 / 62.0 / 61.0 / 60.0 / 58.5 / 57.5 / 57.0 / 55.5
Wood / 89.0 / 84.0 / 81.0 / 78.5 / 76.5 / 74.0 / 72.0 / 70.5 / 69.5 / 68.0 / 66.5
Aluminum / 85.0 / 82.5 / 81.0 / 79.0 / 77.0 / 75.5 / 73.0 / 71.5 / 70.0 / 69.0 / 67.5
At the time of our experiment room temperature was 23.5 degrees Celsius. This will be used as our Tout in the model at the beginning of the report.
It must be mentioned that there were most likely errors in our experiment. Some major causes of error would be that we had to approximate the surface area of our objects. We tried to choose uniformly shaped objects but there were still variations in the surface of the objects. We also did not have lids of the same material for our objects. The soda can had only a small hole but the wood container and the glass were open on the top. We used a paper cup to place over the top of these to minimize heat loss through the top, due to the heat rising. Another problem area could have been the bottom of the containers, which were often thicker than the rest, and depending of the thermal conductivity, this could affect heat loss. Also, some of the two-minute intervals were not exactly two minutes and this could affect the comparison results. The data that we measured is in the following table. The value of the thermal conductivity is, however, the accepted value for each material. We will compare our results for “k” next.
QUANTITATIVE DATA FOR MATERIALS
Material / Specific Heat (cal/g*K) / Thermal Conductivitycal/s*K*cm / Dimensions height/radius (cm) / Area (cm2) / Mass
(g)
Glass / 0.20 / 0.0025 / 14.8/3.4 / 346.4 / 351.3
Wood / 0.42 / 0.0003 / 11.8/2.6 / 214.0 / 194.6
Aluminum / 0.215 / 0.5903 / 12.6/3.3 / 329.7 / 16.2
Since one of the objectives of the project is to compare our results with the numbers suggested by the model, we also used the model to predict a value for the thermal conductivity for each material. Our complete calculations can be found in the Appendix in the back of the report but a sample calculation for glass is as follows:
Step #1 - Converting the variables from Celsius to Kelvin:
Ti = 73 + 273 = 346
Tf = 55.5 + 273 = 328.5
Tout = 23.5 + 273 = 296.5
Step#2 - Plug values into the model for heat loss:
T’(t) = -(Ak/cm)[T(t)-Tout], T(0)=Ti
Solving the above equation will yield:
T(t) = Tout + Ce-(Ak/cm)t
Solving for C we used the equation C= Tin - Tout which yields C = 346-296.5 = 49.5
This results in the following:
328.5 = 296.5 + 49.5e-(346.4k/(0.20*351.3)*20)
Solving k = 0.0044
**The values we used can be referenced from the tables above.
Conclusion - When we compared our experimental results and our calculated answers we found our values to be fairly close to the accepted values for 'k' for our materials, with the exception of aluminum.
Material / Accepted Value 'k' / Experimental 'k'Glass / 0.0025 / 0.0044
Wood / 0.0003 / 0.0080
Aluminum / 0.5903 / 0.0042
Part #2
Problem Statement - For this part, we have to come up with a model for the heat loss of and object with an internal heat source (furnace). Therefore we will use a balance equation using heat loss of a house plus heat added by a furnace. We are going to assume outside temperature remains constant and that our house is a typical ranch giving us a predetermined value for the quotient (Ak/cm) of 0.012 min-1. We will then illustrate our model by proposing numerical experiments. In other words, make up a problem involving a house with furnace and test different situations
Analysis – We can start with the basic heat loss equation. This model will work for a house as well as an object for our purposes. We are assuming Tout to be constant.
T’(t) = -(Ak/cm)[T(t)-Tout], T(0)=Ti
We will add in a heat source, denoted by S(t). S(t)/cm will be the heat output rate of the furnace. This will lead us to the following equation:
T’(t) = -(Ak/cm)[T(t)-Tout] + S(t)/cm