Thermodynamics HW Solutions 451

Thermodynamics HW Solutions 451 - W 2783 = + + + = + + + =...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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, 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 =118.8 ° C, T 2 =116.7 ° C, T 3 =103.4 ° C, T 4 =53.7 ° C, T 5 =254.4 ° C, T 6 =253.0 ° C, T 7 =235.2 ° C, T 8 =103.5 ° C, T 9 =263.7 ° C, T 10 =117.6 ° C ( c ) The rate of heat loss through a 1-m long section of the chimney is determined from
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: W 2783 = + + + = + + + = = = = C 263.7)/2]-(280 235.2)-(280 253.0)-(280 254.4)/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. 5-54...
View Full Document

Ask a homework question - tutors are online