I chose the topic of heat transfer because I find it really intriguing to learn about. I wanted to look further into how heat from two substances reacts with one another through another material that was placed between them. I will be looking at how to calculate the rate of heat transfer in a one dimensional space. This means that I will only be focusing on two temperatures, one hot and one cold, and a medium of which the heat will pass through.
External factors such as other temperatures and time will not be used as they are for three dimensional spaces. Heat is a type of energy that transfers between two pieces of matter that have different temperatures. There are three ways heat can be transferred. The first way is through radiation and the second is convection. The third way is through conduction which is when an object or material conducts the heat from one substance through itself and to another.
This is the method I will be focusing on. As stated in the second law of thermodynamics, heat flows from the matter or objects with the higher temperature to the one containing lower temperature and is not possible from cold to hot. This will continue until both objects have reached a thermal equilibrium. At this point, one object does not contain a higher temperature than the other, so the heat transfer ends. The rate at which the heat is transferred depends on the composition of the material that separates the two temperatures. For example, the rate at which heat flows from hot water to cold water through a copper cup will be different than if the cup is porcelain.
The rate that the heat energy is transferred is directly proportionate to the rate at which the temperature changes. Also, since the lower temperature is gaining the same amount of heat that the higher temperature is losing, the graphs of these two should be negative reciprocals of each other. The higher temperature ‘A’ will generally have a negative slope whereas the lower temperature ‘B’ will have a positive slope. When both A and B have the same temperature, as mentioned before, they have reached a thermal equilibrium and therefore will have a slope of zero as neither one is gaining or losing heat. This chart represents the increase and decrease in temperature due to the loss and gain of heat.
It does not represent any set data and is not a completely accurate diagram. The heat that the hot water loses is gained by the cold water. This continues until the hot water is the same temperature as the cold water. At this point, neither one is warmer nor colder than the other, so the transfer stops, resulting in a gradient of zero. To start off, I have two variables to work with. The first variable is the difference in temperature of the two objects; in this case, it is water.
The second variable is the composition of the material that is separating the two different temperatures. I have considered this variable because it is directly involved with the heat transfer as the heat energy is passing from one object, through the material, and into the second object. As mentioned before, heat flows from the region with the higher temperature to the region with the lower temperature. Let G represent the rate at which the heat is transferred through conduction. This should be equivalent to the temperature gradient of dT/dx , where T(x) is the temperature and x is the distance travelled in the same direction and the heat is flowing.
G=dT/dxHowever, with the current equation, there is no variable for the area of which the heat is transferring through. It only states that the gradient of dT/dx is the rate at which heat is transferred. It implies that the heat transfer between two substances will be the same if the area is 5cm2 and if it is 5km2. Area is another variable that affects the rate at which heat transfers. The larger the area of a medium, the more heat is transferred because there is more surface area that is conducting the heat from one substance and into another. An example of this is a large window in a building compared to a smaller one.
Rooms with large windows tend to be colder because more heat has been lost from there. G=A dT/dxdT/dx is the temperature gradient in the direction which is normal to the area. This equation includes the area (A) of the surface that the heat is passing through, but it does not factor in the conductivity of the material placed between the two substances. Let the variable c represent the thermal conductivity of the material involved in the transfer. G=K A dT/dxIn addition, the second law of thermodynamics states that heat energy must flow from a warmer region to a colder one. A negative sign must be placed on the right side of the above equation since the heat is being transferred in the direction of increasing x values will result in a positive quantity.
Conduction heat flow?T/?x is negative if the value of x increases while the temperature decreasesG=-k A dT/dxThe equation, G=-k A dT/dx , can be altered to form another equation. G=-k A dT/dxG/A=(-k A dT/dx)/AG/A=-k dT/dxG/A=?_T1^T2??-k dT/dx?G/A=-?_T1^T2??k dT/dx?T1 represents the hotter temperature. T2 represents the colder temperature. The area (A) is expressed in squared metres (m2).
The temperature (K) is expressed in Kelvin. X is expressed in metres. The rate of heat flow is expressed in watts (W). The thermal conductivity (k) is measured as watts per metre per kelvin. Material Thermal ConductivityW/m KCopper 399Aluminum 237Carbon Steel, 1% C 43Glass 0. 81Plastics 0.
2-0. 3Water 0. 6Ethylene Glycol 0. 26Engine Oil 0. 15Freon (Liquid) 0.
07Hydrogen 0. 18Air 0. 026Thermal conductivity chart of different materialsThermal resistance is when a material resists the heat from flowing. The equation for thermal resistance is_R= L/AkL stands for thicknessA stands for areak stands for thermal conductivityG=?T/(L/Ak)G=Ak/L(T1-T2)These equations are measured in k/W which is equivalent to °C/W. Using these equations, I can find the thermal resistance and the rate of heat transfer through a medium.
Take for example, a large aluminum slate (k = 237 W/m K) of which the dimensions are 1 metre (m) in height, 0. 5 metres (m) as the width, and a 0. 5 centimetre (cm) thickness (depth) where the exterior temperature is 25°C and the interior temperature is 30°C. Thermal resistance_R= L/AkR= (0.
005 m)/(1 m ×0. 5 m ×237 W/m K)R= 4. 2194×?10?^(-5) k/WRate of heat loss_G=Ak/L(T1-T2)G=((T1-T2))/RG=((30-25)°C )/(4. 2194 ×?10?^(-10 ) k/W)G=(5°C)/(4.
2194 ×?10?^(-10 ) k/W)G=118,500 WThis graph shows the change is temperature as a result of heat transfer over time. One temperature starts off much warmer than the other. As the process of heat transfer commences, the hot temperature loses heat and gets cooler. At the same time, the cold temperature gains the lost heat and becomes warmer. When both have reached a thermal equilibrium, there is no more heat to transfer and both remain the same temperature.
Heat transfer that takes place in a room is similar to the examples I have given, except that it occurs in three dimensional spaces. This means that other factors are required to make the equations work and come up with an appropriate answer. From these equations, other ones can be created to solve many other problems on heat transfer. Anywhere that temperature exists, there is some sort of heat transfer happening. By predicting, analysing, and testing any equation in thermodynamics, we can learn how to conserve heat and energy for when we actually do need it.
BibliographyKreith, Frank, and Mark S. Bohn. Principles of Heat Transfer, 6th ed. New York: Brooks/Cole, 2001Massachusetts Institute of Technology. “16.
4 Thermal Resistance Circuits. ”, http://web. mit. edu/16.
unified/www/FALL/thermodynamics/notes/node118. htmlThe Physics Classroom. “Rates of Heat Transfer.”, http://www.physicsclassroom.com/class/thermalP/u18l1f.cfm