Thermodynamics HW Solutions 449

# Thermodynamics HW Solutions 449 - W 3153 = + + + = + + + =...

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Chapter 5 Numerical Methods in Heat Conduction where l = 0.1 m, k = 1.4 W/m ⋅° C, h i = 75 W/m 2 ⋅° C, T i =280 ° C, h o = 18 W/m 2 ⋅° C, T 0 =15 ° C, T surr =250 K, ε = 0.9, and σ = 5.67 × 10 -8 W/m 2 .K 4 . This system of 10 equations with 10 unknowns constitutes the finite difference formulation of the problem. ( b ) The 10 nodal temperatures under steady conditions are determined by solving the 10 equations above simultaneously with an equation solver to be T 1 =94.5 ° C, T 2 =93.0 ° C, T 3 =82.1 ° C, T 4 =36.1 ° C, T 5 =250.6 ° C, T 6 =249.2 ° C, T 7 =229.7 ° C, T 8 =82.3 ° C, T 9 =261.5 ° C, T 10 =94.6 ° C ( c ) The rate of heat loss through a 1-m long section of the chimney is determined from
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Unformatted text preview: W 3153 = + + + = + + + = = = = C 261.5)/2]-(280 229.7)-(280 249.2)-(280 250.6)/2-m)[(280 1 m C)(0.1 W/m 75 ( 4 )] )( 2 / ( ) ( ) ( ) )( 2 / ( [ 4 ) ( 4 4 4 2 9 7 6 5 surface, surface inner element, chimney of fourth -one T T l h T T l h T T l h T T l h T T A h Q Q Q i i i i i i i i m m i m i & & & Discussion The rate of heat transfer can also be determined by calculating the heat loss from the outer surface by convection and radiation. 5-52...
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## This note was uploaded on 01/22/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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