2013美赛A题一等奖论文.pdf

Team 17999 2013 美赛 A 题一等奖论文 哈尔滨工业大学 杨宜蒙 李春柳 姜子木 注 此页非正式论文页 Contest 1 Introduction 1 2 General assumption for all models 1 3 What is the distribution of heat across the outer edge of a pan 1 3 1 Model establishment 1 3 1 1 Model Micro point model 1 3 1 2 Model Thermodynamics conduction model 3 3 2 Model solution and analysis 5 3 2 1 Model Micro point model 5 3 2 2 Model Thermodynamics conduction model 6 3 3 Sensitivity analysis 11 4 How to choose an optimal pan 14 4 1 Model assumption 14 4 2 Model requirement 14 4 3 Model establishment 14 4 4 Model solution and analysis 20 4 5 Sensitivity analysis 21 5 Superiority and weakness 22 6 Further research 22 7 Practical suggestions 23 8 References 23 9 ADS 24 10 Appendix 错误错误 未定义书签 未定义书签 Team 17999 Page 1 of 26 1 Introduction The brownie one of America s favorite baked treats was born in the U S A Even though it is a relatively recent entry to the food pantheon the recipe first appeared in print in Boston in the early 20th century there s no smoking gun And it s a sweet dessert that is like eating chocolate but not that sweet So it s easy to see that the brownie got its name from its dark brown color Ordinarily chocolate and butter is the main ingredient Thus we are easy to find the brownie pan is a tool to bake the brownie But the problem is when baking in a rectangular pan heat is concentrated in the 4 corners and the product gets overcooked at the corners This will make the food bakes unevenly which may lead to corners of baked paste case and middle section has not been baked While 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 respect to using the space in an oven Therefore we have proposed two models One is to show the distribution of heat across the outer edge of a pan for pans of different shapes The other is to select an optimal shape of the pan based on the length width ratio of the oven to make our performance optimization Figure 1 A kind of brownie pan used in daily life 2 General assumption for all models The oven is rectangular The heat distribute evenly in the oven that is the temperature is constant everywhere The baking pan is heated evenly The thickness of the pan is constant Without consideration of the influence caused by various material of the pan 3 What is the distribution of heat across the outer edge of a pan 3 1 Model establishment 3 1 1 Model Micro point model We build a micro point model to show the different distribution of heat in different position for different shapes of brownie pan As is mentioned in the 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 extent at the edges Intuitively we can make a conclusion that more heat distributes in the corners of the rectangle Team 17999 Page 2 of 26 and less heat on the edges Considering that the food in the oven gained more heat through the way of heat exchange by the pan we mainly take heat exchange into consideration neglecting heat convection and heat emission We select a particle in the space of the brownie pan let the distance from this point to the left wall to the right wall and to the bottom be X Y and Z Select one corner of the bakeware as the origin to establish a coordinate system in Figure 2 According to Law of heat conduction we can know that for the homogeneous medium with the thickness d and temperature difference between two sides T The heat goes through unit area in unit time is Where K is thermal conductivity coefficient In order to simplify the model we consider the particles in the same Z plane Thus for these particles the bottom surface offers equal heat QZ To one particle the heat of this point is Q We can get Q Figure 2 Establish a coordinate system in rectangular pan In a similar way we establish a coordinate system in a round brownie pan In order to simplify our model we still take the micro points at the same Z plane into consideration and is the heat Thus the heat here is Q K T 2 2 Team 17999 Page 3 of 26 Figure 3 Establish a coordinate system in round pan 3 1 2 Model Thermodynamics conduction model Assumption Without consideration of walls of the pan that is we only talk about the bottom of the pan in this model Assume the pan is rectangular or round In this model we no longer consider the heat conducting from baking pan to food but consider the heat exchange mode from the oven to baking pan in Figure 4 According to the assumptions and conclusions above we can know that each surface is thermally equivalent and thermal convection and thermal radiation are the two ways to conduct heat for a pan Thus it is easy to learn that there is a bigger heating area on the edges of the pan So the temperature at the edges is closer to the oven temperature Figure 4 The heat exchange mode of a pan in an oven Next we consider the application of thermodynamics in this model Basic knowledge of thermodynamics 1 shows that there are three basic ways of heat transfer heat exchange heat convection and heat emission The actual heat transfer is carried out in these three ways and most of the cases were carried out in two or three heat transfer modes simultaneously In terms of the relationship between temperature and the time course of the heat transfer process the heat transfer process can be divided into two types one is steady state heat transfer process which does not change with time and the other is metamorphosis of non steady heat transfer process over time Any process that the temperature of object does not change over time is called steady state heat transfer process otherwise it is called non steady state process Team 17999 Page 4 of 26 Thermal conductivity is a physical process linked to the random movement of microscopic particle such as atoms molecules and electrons All substances regardless of solid liquid or gas phase have the capacity to conduct heat but its value differs greatly Heat conduction is a composition of fundamental micro particle motion on the basis of diffusion behavior which can also be understood as an energy transfer process that particles in higher energy level pass energy to the lower level It is a kind of internal energy exchange due to the temperature gradient of full contact between the two objects or between different parts of an object Heat conduction through the fixed plate is the most common and also the most simple heat conduction problem Our research in baking pan is just a kind of this issue baking pan heat transfer problems Here we can use Fourier s law to describe conductive heat flow or density of heat flow rate q quantitatively through tablet or q Where is heat through a given area per unit time q is heat transfer rate through unit area in unit time is heat conductivity coefficient the negative sign in the formula is used to indicate the direction of heat transfer is opposite to the direction of temperature increment Fourier s law establishes a numerical relationship between the object temperature field and density of heat flow rate therefore it is called the thermal conductivity of the heat flow rate equation In the case of the flat wall conducts heat steady state the temperature changes in one dimensional of X Integrate the Fourier s law we can get 1 2 Where is the difference in temperature between the bottom and the top surface of the tablet is the thickness of one dimensional tablet This equation demonstrates that is directly proportional to and is inversely proportional to According to Fourier s law and the equation above we can know that The greater is the smaller is and the lower temperature the pan can get in unit interval The smaller is the greater is and the higher temperature the pan can get in unit interval Thus the design of the thickness of the pan really helps a lot We have known that the thinner the pan is the higher temperature it can get in a certain period of time But considering that the pan is made up of aluminum diecasting it will be more difficult with poorer function to model the thinner pan So here we assume the thickness of the pan to be a fixed value Temperature gradient TEMPGRAD is the main reason that causes heat transfer Each mode Team 17999 Page 5 of 26 of heat transfer is closely related to the temperature of the object Temperature field is a kind of temperature distribution within all the points of the object in a transient Generally it is a function of the space coordinates and time Temperature field in the Cartesian coordinate system can be expressed as T f x y z t Where T is the temperature x y z are spatial Cartesian coordinates Temperature field has a non steady state temperature field and a steady state temperature field And the two fields both exist in heat conduction process of the pan While the temperature field can be divided into three categories one dimensional temperature field two dimensional temperature field and three dimensional temperature field Considering the actual situation it is a three dimensional temperature field for the pan Following analysis can be given by Fourier experimental laws In the infinitesimal period t the heat dQ flowing through an infinitesimal small area dS of an object along the normal direction is directly proportional to normal derivative of dS dQ f x y z The negative sign above indicates that the temperature always transmits from the high temperature side to the low side Figure 5 shows the relationship among the physical variables Figure 5 The analysis of the heat flow in micro perspective When the cross sectional area which is perpendicular to the direction of heat conduction gets larger the heat conducts faster and the temperature gets lower In terms of the central part of a baking pan cross sectional area perpendicular to the direction of heat conduction is the biggest heat conducts fastest and temperature is the lowest The edges of a pan reverse 3 2 Model solution and analysis 3 2 1 Model Micro point model Team 17999 Page 6 of 26 To show the heat distribution of rectangular and round pan in our model we use MATLAB programming to draw Figure 6 and Figure 7 to show the results Figure 6 The distribution of heat in rectangular pan From Figure 6 we can draw the following conclusions There exists a peak value of heat at the origin of coordinates which confirms that the rectangular angle gathers more heat indeed Rectangular edges have less heat compared with the distribution of heat on the corner Moreover the distribution of heat is more uniform on the edges The characteristics above prove that product in all 4 corners gets more easily overcooked but it may roast to a lesser extent on the edges in rectangular baking tray Figure 7 The distribution of heat of round pan From Figure 7 we can get the conclusions as follows Different micro points with the same distance to the origin have the same heat distribution The heat decreases as the distance between micro point and origin decreases The heat gets the lowest in the center That is the product will get less heat in the center when cooked in a round pan and this is also the reason why it is not easy to bake food in the center 3 2 2 Model Thermodynamics conduction model Team 17999 Page 7 of 26 Qualitative analysis is given above we are going to carry out quantitative analysis To simplify our model we only calculate two dimensional temperature field instead of three dimensional temperature field Isotherm The line connected by the points with the same temperature Temperature gradient TEMPGRAD represents temperature shift in the normal direction of isothermal level 0 grad t Where grad t is TEMPGRAD is temperature shift rate in the normal direction of isothermal level is unit vector in the normal direction of isothermal level pointing to the direction of temperature increasing are the partial derivatives of the temperature in the x y z direction are unit vectors in x y and z direction Basic laws of thermal conductivity Fourier s law q gradt In the Cartesian coordinate system density of heat flow rate vector can be expressed as For isotropic material is the same in various directions Thus Where The general form of heat conduction differential equation in the Cartesian coordinate system can be expressed as 2 c However this equation is too complex We continue to simplify it and obtain the following two dimensional heat conduction equation Team 17999 Page 8 of 26 2 2 2 2 Where k is determined by the thermal conductivity of the material density and heat capacity u u t x y represents the temperature Take advantage of this two dimensional heat conduction equation we can get the heat conduction simplified equation with regard to the rectangular pan u u t x y 2 2 2 2 0 0 1 1 1 Enter this equation MATLAB we can get Figure 8 Figure 8 Temperature distribution of the rectangular pan The temperature of four corners is relatively high In terms of the round pan two dimensional heat conduction equation can be expressed as follows u u t x y 2 2 2 2 0 0 2 2 1 1 In this equation the second boundary condition 2 2 1 1 can t be identified in MATLAB So convert the Cartesian coordinate system into a polar coordinate system Then the Team 17999 Page 9 of 26 equation can be expressed as x cos sin 2 cos 2 2 sin 2 0 0 2 1 1 Enter this equation MATLAB we can get Figure 9 Figure 9 Temperature distribution of the round pan Apart from the heat conductance of the material we also need to know the temperature field of the object when calculating the conductive heat flow through the object Therefore the analysis of the temperature field is necessary We have simplified the model to two dimensional Create a polar coordinate system at any point of the model In the case of none limitation around 360 of the space This point can conduct heat outwardly in a range of 180 that is the case in which thermal conductivity of the cross sectional area is the largest For round pan the central part is in accord with the case mentioned above 360 space unconstrained Figure 10 The point conducts heat outwardly in a range of 180 However the edges can conduct heat in a range of 180 and the other side in a range of 90 Team 17999 Page 10 of 26 Figure 11 The point conducts heat at the edges Figure 12 shows the temperature field of round pan distributes evenly in rings Figure 12 The temperature field is evenly distributed For a rectangular pan the thermal conductivity in the four corners is constrained by space From a corner to the adjacent edge the range to conduct heat outwardly increases gradually from 90 to 180 It is similar to 360 space unconstrained for the other points Figure 13 The change of heat conduction from point to point Thus the four corners of the rectangular pan conduct heat slower so temperature is higher at the corners Figure 14 shows the non uniform distribution of temperature field Figure 14 The temperature field isn t evenly distributed Team 17999 Page 11 of 26 3 3 Sensitivity analysis In order to correctly measure every point s temperature we divide the rectangular pan with grid method 1 We can see it in Figure 15 Figure 15 Grid method in rectangular pan Figure 16 shows the computational domain and L R B U are dimensionless length of boundary And Table 1 shows characteristic ratio grid number grid spacing Figure 16 Computational domain Eigenvalue X direction Y direction Dimensionless length l width Ny Nx height Grid number Nx Ny Grid spacing 1 Nx 1 Ny Table 1 eigenvalue of computational domain We assume the grid spacing in X and Y directions are the same so the finite difference equation of grid point i j is 4 1 1 1 1 2 0 After making the grid discrete in order to Improve the precision of the calculation we assume the space between wall and first grid line is half of grid spacing 2 so the difference scheme of two derivative at i 1 and are 2T 2 1 2 1 2 1 2 Team 17999 Page 12 of 26 2T 2 2 1 2 2 Hereinto subscript j is ignored and we use subscript L R to express left and right bound ary in the process of formula derivation So we get 0 1 According to formula above we derive the finite difference equation of grid line 1 2 1 1 1 1 1 2 2 0 We exchange the equation into matrix AT B We assume the heating temperature is 160 degrees and heating time is 30 minutes So we can get the temperature of every point 2 as Figure 17 Figure 17 Temperature of every point in rectangular pan Therefore the temperature difference at edge is 157 9281 157 9294 157 9321 157 9243 157 1921 157 1984 157 1988 157 1922 4 0 7331 And the temperature difference in the middle is 157 9281 157 9294 157 9321 157 9243 4 156 1964 1 7321 We can divide round pan with grid method as Figure 18 Figure 18 Grid method in round pan We also the heating temperature is 160 degrees and heating time is 30 minutes So we can get the temperature as Figure 19 Figure 19 Temperature of every point in rectangular pan Team 17999 Page 13 of 26 And we can get the temperature at edges is approximately a straight line as Figure 20 Figure 20 Temperature at edges in round pan The variance of temperature at edges is S 2 622 3 10 6 The temperature difference between inner and outer is 157 6272 157 6243 157 6271 157 6264 157 6274 157 6269 157 6271 157 6247 157 6246 157 6263 157 6223 157 6236 12 157 1019 0 5237 When heating time is constant and heating temperature changes we measure the changes of our results in rectangular pan We can see the changes in Table 2 Heating temperature Temperature difference at edge Temperature difference in the middle 161 0 7346 1 7345 162 0 7421 1 7369 163 0 7469 1 7401 164 0 7511 1 7424 165 0 7542 1 7449 Table 2 Sensitivity analysis in rectangular pan Conclusions Temperature difference grows as the increasing of heating temperature The temperature are the highest at corners in rectangular pan no matter how heating temperature changes The temperature are the lowest in the middle in rectangular pan no matter