Thermodynamics Lab
Design Constraints:
· Composite insulation material must produce minimum heat loss, representing good insulating value.
· Composite insulation material must have overall uniform thickness less than or equal to one inch.
· Composite insulation material must have consistent internal composition.
· Individual insulation material(s) must be environmentally friendly.
· Individual insulation material(s) must be recyclable.
· Individual insulation material(s) must be economical.
· Composite insulation material dimensions must not exceed the overall dimensions of heat box apparatus top.
· The change in temperature inside the box is directly related to the heat absorbed or released by the air in the box.
· Composite insulation material must produce minimum heat loss, representing good insulating value.
· Composite insulation material must have overall uniform thickness less than or equal to one inch.
· Composite insulation material must have consistent internal composition.
· Individual insulation material(s) must be environmentally friendly.
· Individual insulation material(s) must be recyclable.
· Individual insulation material(s) must be economical.
· Composite insulation material dimensions must not exceed the overall dimensions of heat box apparatus top.
· The change in temperature inside the box is directly related to the heat absorbed or released by the air in the box.
Brainstorming When we were brainstorming for this project, we wanted to use notebook paper, tissues, aluminum foil, newspaper, and drywall. The challenge was deciding how to order them, because we had a few ideas. Our first idea was to have the aluminum foil on the bottom, the newspaper/tissues/notebook paper in the middle, and to finally have the drywall on top. The drywall is half an inch, and everything else we made an eighth of an inch thick. Our second idea was to have the drywall on the bottom, the aluminum foil/newspaper/tissues in the middle, and with the notebook paper on the top. The main difference between these two ideas was having the drywall on the top or the bottom. We decided that the rest of the materials were all paper, so it wouldn’t make a huge difference in what order we put them.
Final Results For our experiment, we simply heated a box with a light bulb. The box was made of Styrofoam, except for the top of the box, which was an acrylic-plastic material. We placed our insulation material above the plastic and poked a hole through our material. We placed a temperature sensor inside of the box, and one on top of the insulation material. The light bulb was inside of the box, and we waited 20 minutes after turning it on. After the 20 minutes was over, we turned the light bulb off. We calculated the energy gained by heating was 134.4022J, and the energy lost during cooling was 68.704J.
We did the same experiment for a different insulation material from another group. Their insulation material was made of only magazines. We heated the box for 20 minutes, but we couldn’t completely finish the cooling because class was over. So instead of 20 minutes, we cooled it for 16 minutes. This wasn’t a problem; we used 960 seconds instead of 1200 when we calculated the rate of the energy transfer (P). The energy gained by heating was 129.2494J, and the energy lost during cooling was 52.3868J. The net energy transfer (Q (net)) for our insulation material was 66.557J. If you do or don’t remember, work is measured in Newton-meters (which are the same as Joules). So roughly anywhere between 66.557Newtons X 1meter or 1Newton X 66.557meters is how much heat was transferred. For our next sample, the net energy transfer was 76.0038J. More energy was moved in our second sample, so what does this tell us? After we found the energy transfer for the two insulation materials, we decided to find the U/R-values for the two materials. The U-value is a material’s ability to conduct heat, and the R-value is a material’s ability to resist heat (the R-value is the reciprocal of the U-value). Materials with a higher R-value are better insulators because they don’t absorb heat. Instead, insulators act as a “shield” and don’t let heat pass through. The R-value of our insulation material was 54.585, and the other insulation material we tested had an R-value of 5.59. |
DocumentationOC=degrees Celcius
1.Inner Dimensions: l=.143m, w=.143m, h=.175m. 2. Volume of air: .003578m3. 3. Our insulation material is made of half an inch of drywall and an eighth of an inch of: aluminum foil, newspaper, tissues, and notebook paper. 4. Heat source light bulb wattage: P=? 5. Inside top acrylic panel initial surface temperature: 57.4OC 6. Inside top acrylic panel maximum temperature: 57.4OC 7. Inside top acrylic panel final surface temperature: 41.4OC 8. Top panel insulation material initial surface temperature: 26.1OC 9. Top panel insulation material maximum surface temperature: 26.7OC 10. Top panel insulation material final surface temperature: 25.9OC 11. Heating time (s): About 20 minutes. 12. Cooling time (s): 20 minutes. Constants: ρ=1.2kg/m3 (density of air) and Cp=1000J/kgOC (specific heat capacity of air). Reflection The R-value for our insulation material was significantly higher than the other group’s insulation material. Like I mentioned earlier, the R-value is a material’s ability to resist heat. Our material resisted the most heat, which means it performed the better job insulating the heat. The 2nd Law of Thermodynamics states that thermal energy flows from hot to cold. The 2nd Law of Thermodynamics can help you understand why some materials insulate heat better than others. Some materials, like stone, have a low R-value and can’t trap heat easily. When heat from a room in a stone house flows into the stone material, it gets absorbed. However, if the house was made like a modern day house, the heat would flow to the walls and wouldn’t get absorbed. The heat inside of the room would become trapped, keeping the temperature in the house warm.
We could’ve improved our insulation material by making the drywall a quarter of an inch thick, and placing it on the top and bottom of the insulation materials. This would’ve made it harder for the heat to escape from the box because it would have to gone through the drywall, then through the paper materials, and finally through the drywall again. Engineers need to understand thermodynamics today because we want more efficient ways of keeping ourselves warm; we want to stay warm while doing less work. Understanding the R-values of a material can help engineers satisfy society and help keep everybody happy. |