Heat Chap05-084 - Chapter 5 Numerical Methods in Heat...

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Chapter 5 Numerical Methods in Heat Conduction 5-84 A uranium plate initially at a uniform temperature is subjected to insulation on one side and convection on the other. The transient finite difference formulation of this problem is to be obtained, and the nodal temperatures after 5 min and under steady conditions are to be determined. Assumptions 1 Heat transfer is one-dimensional since the plate is large relative to its thickness. 2 Thermal conductivity is constant. 3 Radiation heat transfer is negligible. Properties The conductivity and diffusivity are given to be k = 28 W/m °C and α = × - 12 5 10 6 . m / s 2 . Analysis The nodal spacing is given to be x = 0.02 m. Then the number of nodes becomes 1 / + = x L M = 0.08/0.02+1 = 5. This problem involves 5 unknown nodal temperatures, and thus we need to have 5 equations. Node 0 is on insulated boundary, and thus we can treat it as an interior note by using the mirror image concept. Nodes 1, 2, and 3 are interior nodes, and thus for them we can use the general explicit finite difference relation expressed as τ i m i m i m i m i m i m T T k x g T T T - = + + - + + - 1 2 1 1 2 k x g T T T T i m i m i m i m i m 2 1 1 1 ) 2 1 ( ) ( + - + + = + - + The finite difference equation for node 4 on the right surface subjected to convection is obtained by applying an energy balance on the half volume element about node 4 and taking the direction of all heat transfers to be towards the node under consideration: t T T C x x g x T T k T T h k x g T T T T k x g T T T T k x g T T T T k x g T T T T i i i i i i i i i i i i i i i i i i i i i - = + - + - + - + + = + - + + = + - + + = + - + + = + + + + + 4 1 4 0 4 3 4 2 0 3 4 2 1 3 2 0 2 3 1 1 2 2 0 1 2 0 1 1 2 0 0 1 1 1 0 2 2 ) ( : n) (convectio 4 Node ) 2 1 ( ) ( : (interior) 3 Node ) 2 1 ( ) ( : (interior) 2 Node ) 2 1 ( ) ( : (interior) 1 Node ) 2 1 ( ) ( : ) (insulated 0 Node ρ or k x g T k x h T T k x h T i i i 2 0 3 4 1 4 ) ( 2 2 2 2 1 + + + - - = + where C 20 C, W/m 35 C, W/m 28 , W/m 10 m, 02 . 0 2 3 6 0 ° = ° = ° = = = T h k g x , and 6 10 5 . 12 - × = m 2 /s. The upper limit of the time step t is determined from the stability criteria that requires all primary coefficients to be greater than or equal to zero. The coefficient of i T 4 is smaller in this case, and thus the stability criteria for this problem can be expressed as 1 2 2 0 1 2 1 2 1 2 - - + + h x k h x k t x h x k ( / ) ( / ) since = t x / 2 . Substituting the given quantities, the maximum allowable the time step becomes s 6 . 15 C)] W/m. 28 /( m) 02 . 0 )( C . W/m 35 ( 1 /s)[ m 10 5 . 12 ( 2 ) m 02 . 0 ( 2 2 6 2 = ° ° + × - t Therefore, any time step less than 15.5 s can be used to solve this problem. For convenience, let us choose the time step to be t = 15 s. Then the mesh Fourier number becomes 5-75 h, T x Insulated 0 1 2 3 4 g
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Chapter 5 Numerical Methods in Heat Conduction τ α = = × = - t x 2 6 2 12 5 10 002 046875 ( .
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This note was uploaded on 08/24/2011 for the course ENGR 3150 taught by Professor Engel during the Spring '11 term at Georgia Southern University .

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Heat Chap05-084 - Chapter 5 Numerical Methods in Heat...

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