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# Class Note for CHEE 502 with Professor Saez at UA

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This 8 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 14 views.

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Date Created: 02/06/15

Radiative heating of a circular metallic plate A thin circular metallic plate is exposed to the sun from which it receives a constant and uniform heat flux CI energy per unit time and surface area The plate releases thermal energy to the surroundings by convection see figure The bottom of the plate is insulated and the edge is kept at ambient temperature The thickness averaged temperature of the plate obeys a thermal energy balance which leads to the PDE r T Ta 1 3T kaaT CI h 3r 8 8 where p and Cp are the density and heat capacity of the metal respectively 5 is the plate thickness 8ltltR h is the convective heat transfer coefficient Ta is the ambient temperature and k is the thermal conductivity of the plate incident solar radiation I COHVeCtiOH 105563 hT39Ta insulated bottom The plate is initially at ambient temperature TTa t0 2 and the edge of the plate is kept at ambient temperature at all times TTa rR 3 Additionally a symmetry boundary condition can be applied aT 0 0 4 3r r First we will make the equations dimensionless using the following definitions 2 2 81 XZQR BhR kt I39 I R Ta 5kTa k8 pCpRZ n It can be shown that the dimensionless version of the problem is n8a Bo D 6 81170 7 88 70 0 8 an n 81n1 9 Before solving transient transport problems such as this it is convenient to explore what happens at steady state which should be achieved as t gtoo At steady state the temperature only depends on radial position ie 88 and the following problem is obtained lina Bo 1o 10 Tl C171 cm The boundary conditions are E 0 n0 11 Ch 8 1 n1 12 The ODE 10 can be written as follows iin B ocB 13 Tl cm This nonihomogeneous ODE can be solved by superposition of a homogeneous solution and a particular solution 8 8h 8p 14 By inspection of equation 13 we can see that the particular solution can be a constant X 1 p B 15 The problem that the homogeneous solution must satisfy is obtained by substituting equation 14 into equations 10 to 12 to obtain using equation 15 1 d d8 Tlih B h 0 16 T Chi dTl 0 n0 17 dn 8h nl 18 Equation 16 can be rewritten as n2 if nBn2 h 0 19 Let tmB 20 Equation 19 becomes tZdZGh tzo o 21 dtZ dt This is the modified Bessel39s equation with V0 notes equation 153 Its solution is 8h A10 t BKO t 22 or 8h A10mEBK0mB 23 From equations 159 and 161 of the notes with V0 we obtain 11o 24 00 K1t 25 dt Therefore we have ABllnd BJEK1nJB 26 Since K10 gtoo to satisfy the boundary condition 1 7 we need B0 which leads to 8h Me ThB 27 From boundary condition 18 we get L HOME and therefore the homogeneous solution is 28 8h 2 3M 29 B 10 J3 and from equations 14 15 and 29 the final solution is 81 9 pm 30 B 10MB H The figure shows dimensionless temperature profiles for 510 Since this is only the steady state solution let us change notation and state g owB 855 1Bl log6 31 Let us go back to the original transient problem equations 6 to Since one boundary condition is non homogeneous equation 9 we will use superposition to obtain a problem that can be solved by separation of variables In this type of problem it is convenient to use the decomposition 8km Gss T1 11W 32 Next we substitute this equation into equations 6 to and use equations 16 to 18 to simplify terms involving the steady state solution The result is 145 135 i 13 i 125 i 12 7 115 i 117 105 Ewan a Buli BuBu u1 Ssn l70 al0y1quotl0 3n u0 111 We will solve this problem by separation of variables Let uTynFTGn and substitute into equation 33 to get GdFF d dG 7 77 7 FG d1 ndn ndn B Divide by FG and separate 33 34 35 36 37 38 ldF 1 d dG 77 77 7 7t2 39 F d1 B nG dn n J This leads to the following two problems CL Bkzdt 3 F17 Aexp Bx2c 40 i E 2 chin dnjk nG 0 41 The two boundary conditions 35 and 36 are separable and lead to E 0 n0 42 Ch 00 n1 43 The ODE 41 can be rewritten as 2 nziddnc n XZnZG0 44 This is Bessel39s equation with V0 lts solution is G BJO m CYO m 45 Since dJot dYot T J10 Tt Y1t 46 From equation 45 and properties of the Bessel functions we have rmlm mum 47 Since Y10 gt oo the boundary condition at 110 implies C0 The boundary condition at 111 then yields the eigenvalue condition J 0 k 0 48 The eigenvalues are then all the roots of Jo The eigenfunctions are Gn BJ0 Mn 49 and the solution of the problem is um ianio xnngtexp Bxgtri so nl To find the coefficients we use the initial condition 34 combined with equation 31 to get 06 lookE B1 WE Elamme 51 and apply the orthogonality condition 1 mo xmn o xnmdn 0 man 52 0 To do this we multiply equation 51 by T and then integrate T from 0 to 1 This leads to l 1 mo kmn o Th3km Tl 0 kmmdrl 0 oc E 10 J6 am j 1 53 mg kmmdn 0 Letting X7tmn we have 1 1 km mo xmmdn F ijo xdx 54 0 m 0 Now we use equation 177 of the course notes xJoxdxdxJ1 55 So that equation 54 leads to l jnJoommdn 3xmh un llim 56 0 m m From equation 187 of the notes we get directly 1 1 mg xmmdn 752 57 0 The following identity can be found in tables of properties of Bessel functions ij0 8X10bXdX lm bXJ0 ax a1010XJ18Xl 58 Consequently we can state that l we mug 11mm LEh ngtJoxmngtmeoJBnJ1xmml 59 0 Tl Mn B We can see that only the upper limit will contribute to the righthand side 1 6LBhJBgtJoxmgtmeoJBgtJ1xmgtl 60gt 1 Mo kmn o Th3km 0 The first term inside the brackets is zero by Virtue of the eigenvalue condition 48 so we get km10BJ1lm Mn 6 61 1 In 0 kmmlo ThEMT 0 Substituting equations 56 57 and 61 into equation 53 leads to after manipulations 20 zxmilamwm 6 62 aIn The final solution is E 10 E m 1 ex M 63 o 1B1 IONS ngknhan mmlM 11 pl 6 or

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