ch03_101-152 - PROBLEM 3.101 KNOWN: Dimensions of a plate...

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PROBLEM 3.101 KNOWN: Dimensions of a plate insulated on its bottom and thermally joined to heat sinks at its ends. Net heat flux at top surface. FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperature distribution and heat loss to heat sinks. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in x (W,L>>t), (3) Constant properties, (4) Uniform surface heat flux, (5) Adiabatic bottom, (6) Negligible contact resistance. ANALYSIS: (a) Applying conservation of energy to the differential control volume, q x + dq = q x +dx , where q x+dx = q x + (dq x /dx) dx and () o dq=q W dx . ′′ Hence, xo dq / dx q W=0. From Fourier’s law, x q k t W dT/dx. =− Hence, the differential equation for the temperature distribution is 2 o o 2 q dd T d T ktW q W=0 0. dx dx kt dx  −− + =   < (b) Integrating twice, the general solution is, 2 o 12 q T x x C x +C 2kt + and appropriate boundary conditions are T(0) = T o , and T(L) = T o . Hence, T o = C 2 , and 2 oo o1 2 1 qq L T L C L+C and C . 2kt 2kt + = Hence, the temperature distribution is ( ) 2 o o qL Tx x L x T. 2kt + < Applying Fourier’s law at x = 0, and at x = L, () ( ) x=0 x=0 W L L q 0 k Wt dT/dx) kWt x kt 2 2 x=L x=L W L L q L k Wt dT/dx) kWt x kt 2 2 =+ Hence the heat loss from the plates is q=2 q WL/2 q WL. = < COMMENTS: (1) Note signs associated with q(0) and q(L). (2) Note symmetry about x = L/2. Alternative boundary conditions are T(0) = T o and dT/dx) x=L/2 =0.
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PROBLEM 3.102 KNOWN: Dimensions and surface conditions of a plate thermally joined at its ends to heat sinks at different temperatures. FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperature distribution and an expression for the heat rate from the plate to the sinks, and (c) Compute and plot temperature distribution and heat rates corresponding to changes in different parameters. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x (W,L >> t), (3) Constant properties, (4) Uniform surface heat flux and convection coefficient, (5) Negligible contact resistance. ANALYSIS: (a) Applying conservation of energy to the differential control volume xo x d xc o n v qd qq d q + += + where () xd x x x q q dq dx dx + =+ ( ) conv dq h T T W dx =− Hence, ( ) ( ) x x q q W dx q dq dx dx h T T W dx ′′ +⋅ = + + x o dq hW T T q W dx +− = . Using Fourier’s law, x qk t W d T d x , 2 o 2 dT ktW hW T T q dx −+ = 2 o 2 q dT h TT 0 kt kt dx −− + = . < (b) Introducing T T θ ≡− , the differential equation becomes 2 o 2 q dh 0 kt kt dx = . This differential equation is of second order with constant coefficients and a source term. With 2 hk t λ and o Sqk t , it follows that the general solution is of the form xx 2 12 Ce C e S λλ θλ =++ .( 1 ) Appropriate boundary conditions are: oo LL (0) T T (L) T T θθ ∞∞ (2,3) Substituting the boundary conditions, Eqs. (2,3) into the general solution, Eq. (1), 002 o1 2 S 2 L1 2 S (4,5) To solve for C 2 , multiply Eq. (4) by -e + λ L and add the result to Eq. (5), L 2 L oL 2 eC e e S e 1 ++ + = + + + ( ) L2 L L L 2L o S e 1 e e +  
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This note was uploaded on 03/17/2008 for the course M E 410 taught by Professor Savasyavuzkurt during the Fall '07 term at Pennsylvania State University, University Park.

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ch03_101-152 - PROBLEM 3.101 KNOWN: Dimensions of a plate...

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