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1、The perfect pan for oven The heat transfer in the oven includes heat conduction, heat radiation and heat convection. We use two-dimensional Fourier heat conduction equation ?u?t- (?2u?2+?2u?2) =f (? , ? , t) to make a research on distribution of heat for the pan. Heat source heats the pan by heat ra
2、diation. The pan interacts with air in the oven in the way of natural convection, so the pan realizes heat dissipation. We calculate heat radiation based on radiation ability of heat source and heating tube area. We use heat dissipation function to show the pans different parts loss of heat caused b
3、y natural convection. Both of them consist in heat source function f. The area of the pan is fixed at 0.085m2in this paper. When comparing temperatures at the edges of rectangular pans with different length to width ratios , we can get that the smaller is, the lower the temperature of the edges is.
4、But as long as it is still a rectangle, the amplitude of its drop wont be very big. When we make the pans with fixed area vary from square to round square to round, we find that the bigger the fillet radius is, the lower the temperature of its corners is and the extent of temperatures reducing is la
5、rge. We fix the bottom area of the oven and area of the pan. Through study, we find that round squares capacity for uniform distribution of heat is far higher than other shapes (except round). The larger the fillet radius of the round square is, the larger the pans waste of space is. But heat distri
6、bution is more uniform. We work out the optimal solution of pan s size under different weights p through optimizing the relationship between two conditions. Then we get several ovens width to length ratios of W/L by arranging the pans with the optimal size. 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - -
7、 - - - - - - - 名师精心整理 - - - - - - - 第 1 页，共 27 页 - - - - - - - - - Team # 22312 Page 1 of 26 I. IntroductionThe temperature of each point in the pan is different. For a rectangular pan, the corners have the highest temperature, so the food is easily overcooked. While the heat is distributed evenly o
8、ver the entire outer edge and the product is not overcooked at the edges in the round pan. To illustrate the model further, the following information is worth mentioning 1.1 Floor space of the pan The floor space of each pan is not the square itself necessarily. In this paper, there are 3 kinds of p
9、ans with different shapes, as rectangular pans, round pans and round rectangle pans. For rectangular pan, the floor space is the square itself, and the pans can connect closely without space. For round pans, the diagrammatic sketch of the floor space is as follows: shade stands for the round pan; sq
10、uare stands for the floor spaceFigure 1 Round pans have the largest floor space for a certain area. The space between each pan is larger than other two kinds of pans. The coefficient of utilization for the round pans is the lowest. For round rectangle pans, the diagrammatic sketch of the floor space
11、 is as follows: shade stands for the round rectangle pan; square stands for the floor space Figure 2名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 2 页，共 27 页 - - - - - - - - - Team # 22312 Page 2 of 26 If the area of the round rectangle is the same as the other two,
12、its floor space is between them. The coefficient of utilization of oven decreases with the radius of expansion. 1.2 Introduction of oven The oven is usually a cube, no matter it is used in home or for business. A width to length ratio for the oven is not a certain number. There are always two racks
13、in the oven, evenly spaced. There are one or more pans on each rack. To preserve heat for the oven, food is heated by radiation. Heating tube can be made of quartz or metal. The temperature of the tube can reach 800 high when the material is quartz. The heating tube is often in the top and bottom of
14、 the oven. Heating mode can be heating from top or heating from bottom, and maybe both1. 1.3 Two dimensional equation of conduction To research the heat distribution of pan, we draw into two dimensional equation2of conduction: ?u?t- (?2u?2+?2u?2) = f (? , ? , t) In this equation: u- temperature of t
15、he pan t- time from starting to heat x- the abscissa y-ordinate - thermal diffusivity f- heat source function The heat equation is a parabolic partial differential equation which describes the distribution of heat (or variation in temperature) in a given region over time. The heat equation is of fun
16、damental importance in diverse scientific fields. In mathematics, it is the prototypical parabolic partial differential equation. In probability theory, the heat equation is connected with the study of Brownian motion via the Fokker Planck equation. The diffusion equation, a more general version of
17、the heat equation, arises in connection with the study of chemical diffusion and other related processes. II. The Description of the Problem 2.1 The original problem When baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners (and to a lesser
18、 extent at the edges). In a round pan the heat is distributed evenly over the entire outer edge and the product is not overcooked at the edges. However, since most ovens are rectangular in shape using round pans is not efficient with 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 -
19、- - - - - - 第 3 页，共 27 页 - - - - - - - - - Team # 22312 Page 3 of 26 respect to using the space in an oven. Develop a model to show the distribution of heat across the outer edge of a pan for pans of different shapes -rectangular to circular and other shapes in between. Assume 1. A width to length r
20、atio of W/L for the oven which is rectangular in shape. 2. Each pan must have an area of A. 3. Initially two racks in the oven, evenly spaced. Develop a model that can be used to select the best type of pan (shape) under the following conditions: 1. Maximize number of pans that can fit in the oven (
21、N) 2. Maximize even distribution of heat (H) for the pan 3. Optimize a combination of conditions (1) and (2) where weights p and (1- p) are assigned to illustrate how the results vary with different values of W/L and p. In addition to your MCM formatted solution, prepare a one to two page advertisin
22、g sheet for the new Brownie Gourmet Magazine highlighting your design and results. 2.2 Problem analysis We analyze this problem from 3 aspects, showing as follows:2.2.1 Why the edge of the pan has the highest temperature? The form of heat transfer includes heat radiation, heat conduction and heat co
23、nvection. Energy of heat radiation comes from heat resource. The further from heat resource, the less energy it gets. Heat conduction happens in the interior of the pan, and heat transfers from part of high temperature to the part of low with the temperature contrast as its motivation. With the two
24、forms of heat transfer above, we find the result is that pan center has the high temperature and the boundary has the low. Depending on that, we can t explain why the product gets overcooked at the corners while at the edges not. We think that there is natural convection between pan and gas, because
25、 the temperature of pan is much higher than that of gas. The convection is connected with the contact area. The point in pan center has a larger contact range with air, so the energy loss from convection is more. While the point in the corners of arectangular pan has a narrow contact area, the energ
26、y loss is less than that in the inner part. Because that above, the energy in the pan center is more than that in corner, and the corners have higher temperature. 2.2.2 Analysis of heat distribution in pans in different shapes The shape of pan includes rectangle, round rectangle and round. When thes
27、e pans area is fixed, the rectangles with different shapes can be shown with different length to width ratios. Firstly, we study temperature (maximum temperature) in the corners of rectangles with different length to width ratios. Then we study how temperature in the corners changes when the pans va
28、ry from square to round square to round. After that, we select some rectangular pan名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 4 页，共 27 页 - - - - - - - - - Team # 22312 Page 4 of 26 and make it change from rectangle to round rectangle to study changes of temperatu
29、re in the corners. To calculate distribution of heat for the pan, we would use three main equations. The first one is the Fourier equation, namely heat conduction equation, the second one is the radiation transfer equation of heat source and the third one is the equation of heat dissipationthrough c
30、onvection. In the radiation transfer equation of heat source, we take heat source as a point. We get its radiating capacity through its absolute temperature, blackening and Stefan-Boltzmann law. We combine radiating capacity with surface area of quartz heating tube to get quantity of heat emitted by
31、 heat source per second, then we can get heat flux at each point of the pan. In the equation of heat loss through convection, Heat dissipating capacity is proportional to area of heat dissipation, its proportional coefficient can be found from the related material. Through the establishment of above
32、 three main equations, we can use pdetool in matlab to draw the figure about distribution of heat for the pan.2.2.3 How to determine the shape of the pan? To make heat distribution of the pan uniform, we must make it approach round. But under the circumstances of the pans fixed area, the closer the
33、pan approaches round, the larger its floor space is. In other words, the closer the pan approaches round, the lower the utilization rate of the oven is. More uniform distribution of heat for the pan is, the lower temperature in the corners of the pan is. Assuming the bottom area of the oven is fixed
34、, the number of most pans which the oven can bear is equal to the quotient of the bottom area of the oven divided by floor space per pan. So in a certain weight P, we can get the best type of pan (shape) by optimizing the relationship between temperature in the corners and the number of most pans wh
35、ich the oven can bear. Then we get the ovens width to length ratio of W/L by arranging the pans with the optimal size.2.3 Practical problem parameterization u: temperature of each point in the oven; t: heating time; x: the abscissa values; y: the ordinate value; : thermal diffusivity; : degree of bl
36、ackness of heat resource; E: radiating capacity of heat resource; T: absolute temperature of heat resource; Tmax: Highest temperature of pans edge ; : the length to width ratio for a rectangular pan;R: radius for a round pan; L: length for the oven;名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - -
37、- - - 名师精心整理 - - - - - - - 第 5 页，共 27 页 - - - - - - - - - Team # 22312 Page 5 of 26 W: width for the oven; q: heat flux; k: coefficient of the convective heat transfer; Q: heat transfer rate; P: minimum distance from pan to the heat resource; Other definitions will be given in the specific models be
38、low 2.4 Assumption of all models 1. We assume the heat resource as a mass point, and it has the same radiation energy in all directions. 2. The absorbtivity of pan on the radiation energy is 100%. 3. The rate of heat dissipation is proportional to area of heat dissipation. 4. The area of heat dissip
39、ation changes in a linear fashion from the centre of the pan to its border.5. Each pan is just a two-dimensional surface and we do not care about its thickness. 6. Room temperature is 25 degrees Celsius. 7. The area of the pan is a certain number. III. Models Consider the pan center as origin, estab
40、lishing a coordinate system for pan as follows: Figure 3 3.1 Basic Model In order to explain main model better, the process of building following branch models needs to be explained specially , the explanation is as follows:3.1.1 Heat radiation model名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - -
41、 - - - 名师精心整理 - - - - - - - 第 6 页，共 27 页 - - - - - - - - - Team # 22312 Page 6 of 26 Figure 4 The proportional of energy received by B accounting for energy from A is dxdy4 (?2+?2+?2)The absolute temperature of heat resource A (the heating tube made of quartz) T= 773K when it works. The degree of bl
42、ackness for quartz =0.94. The area of quartz heating tube is 0.0088m2. Depending on Stefan-Boltzmann law3E=T4 (=5.67*10-8) we can get E=18827w/m2. The radiation energy per second is 0.0088E. At last, we can get the heat flux of any point of pan. =0.0088E4 (?2+?2+?2)Synthesizing the formulas above, w
43、e can get: =13.24 (?2+?2+?2)(1) 3.1.2 Heat convection model 3.1.2.1 Round pan heat dissipation area of round pan Figure 4 When the pan is round, the coordinate of any point in the pan is (x,y). When the point is in the centre of a circle, its area of heat dissipation is dxdy. When the point is in th
44、e boundaries of round, its area of heat dissipation is 12dxdy. According to equation of heat As shown in the figure, A is the heat resource and located at center of oven roof. B is any point of the pan, and its position is （x,y） ,its area is dxdy. We can get a global which takes AB connection as rad
45、ius. The spherical surface is namely the heat radiation surface, which area is 4(x2+y2+P2). P 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 7 页，共 27 页 - - - - - - - - - Team # 22312 Page 7 of 26 dissipation through convection dQ=kdxdy and assumption that the area of
46、 heat dissipation changes in a linear fashion along the radius4, we can get the pans equation of heat dissipation: q =k1-0.5(x2+y2)0.5/R (2) 3.1.2.2 Rectangular pan Heat dissipation area of rectangular pan(length: M width: N) Figure 5When the pan is rectangular, the coordinate of any point in the pa
47、n is (x,y). When the point is in the centre of a rectangle, its area of heat dissipation is largest, namely dxdy. When the point is in the center of the rectangular edges, its area of heat dissipation is 12dxdy. When the point is in rectangular vertices, its area of heat dissipation is minimum, name
48、ly 14dxdy. According to equation of heat dissipation dQ=kdxdy and assumption that the area of heat dissipation changes in a linear fashion from the centre of a rectangle to the center of the rectangular edges.The area of any other point in the pan can be regarded as the result of superposing two cor
49、responding points area in two lines. Heat dissipating capacity of any point is: =1-y/N1-x/M(3)3.2 Pan heat distribution Model 3.2.1 Heat distribution of rectangular pans Assume that the rectangles length is M and its width is N, the material of the pan is iron.For rectangular pans, we change its len
50、gth to width ratio, establishing a model to get the名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 8 页，共 27 页 - - - - - - - - - Team # 22312 Page 8 of 26 temperature of the corners (namely the highest temperature of the pan). We assume the area of pan is a certain num