# hw4 - MEGR 324 – HW#4 Chapter 5 Problem 36 A hot surface...

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Unformatted text preview: MEGR 324 – HW #4 Chapter 5, Problem 36. A hot surface at 100 ° C is to be cooled by attaching 3-cm-long, 0.25-cm-diameter aluminum pin fins ( k = 237 W/m × ° C) with a center-to-center distance of 0.6 cm. The temperature of the surrounding medium is 30 ° C, and the combined heat transfer coefficient on the surfaces is 35 W/m 2 × ° C. Assuming steady one- dimensional heat transfer along the fin and taking the nodal spacing to be 0.5 cm, determine ( a ) the finite difference formulation of this problem, ( b ) the nodal temperatures along the fin by solving these equations, ( c ) the rate of heat transfer from a single fin, and ( d ) the rate of heat transfer from a 1-m 3 1-m section of the plate. Figure P5.36 Chapter 5, Solution 36 One side of a hot vertical plate is to be cooled by attaching aluminum pin fins. The finite difference formulation of the problem for all nodes is to be obtained, and the nodal temperatures, the rate of heat transfer from a single fin and from the entire surface of the plate are to be determined. Assumptions 1 Heat transfer along the fin is given to be steady and one-dimensional. 2 The thermal conductivity is constant. 3 Combined convection and radiation heat transfer coefficient is constant and uniform. Properties The thermal conductivity is given to be k = 237 W/m × °C. Analysis ( a ) The nodal spacing is given to be ∆ x =0.5 cm. Then the number of nodes M becomes 1 MEGR 324 – HW #4 7 1 cm 5 . cm 3 1 = + = + ∆ = x L M The base temperature at node 0 is given to be T = 100 ° C. This problem involves 6 unknown nodal temperatures, and thus we need to have 6 equations to determine them uniquely. Nodes 1, 2, 3, 4, and 5 are interior nodes, and thus for them we can use the general finite difference relation expressed as ) )( ( 1 1 =- ∆ + ∆- + ∆- ∞ +- m m m m m T T x p h x T T kA x T T kA → ) )( / ( 2 2 1 1 =- ∆ + +- ∞ +- m m m m T T kA x p h T T T The finite difference equation for node 6 at the fin tip is obtained by applying an energy balance on the half volume element about that node. Then, m = 1: ) )( / ( 2 1 2 2 1 =- ∆ + +- ∞ T T kA x p h T T T m = 2: ) )( / ( 2 2 2 3 2 1 =- ∆ + +- ∞ T T kA x p h T T T m = 3: ) )( / ( 2 3 2 4 3 2 =- ∆ + +- ∞ T T kA x p h T T T m = 4: ) )( / ( 2 4 2 5 4 3 =- ∆ + +- ∞ T T kA x p h T T T m = 5: ) )( / ( 2 5 2 6 5 4 =- ∆ + +- ∞ T T kA x p h T T T Node 6: ) )( 2 / ( 6 6 5 =- + ∆ + ∆- ∞ T T A x p h x T T kA where C W/m 35 C, 100 C, 30 C, W/m 237 m, 005 . 2 ° ⋅ = ° = ° = ° ⋅ = = ∆ ∞ h T T k x and m 00785 . ) m 0025 . ( m 10 0.0491 cm 0491 . /4 cm) 25 . ( 4 / 2-4 2 2 2 = = = × = = = = π π π π D p D A ( b ) The nodal temperatures under steady conditions are determined by solving the 6 equations above simultaneously with an equation solver to be T 1 =97.9 ° C, T 2 =96.1 ° C, T 3 =94.7 ° C, T 4 =93.8 ° C, T 5 =93.1 ° C, T 6 =92.9 ° C ( c ) The rate of heat transfer from a single fin is simply the sum of the heat transfer from...
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hw4 - MEGR 324 – HW#4 Chapter 5 Problem 36 A hot surface...

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