Thermodynamics HW Solutions 346

# Thermodynamics HW Solutions 346 - o wall Q Q 427 1704 0846...

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Chapter 4 Transient Heat Conduction ( b ) The center of the top surface of the cylinder is still at the center of the long cylinder ( r = 0), but at the outer surface of the plane wall ( x = L ). Therefore, we first need to determine the dimensionless temperature at the surface of the wall. 857 . 0 ) 164 . 0 cos( ) 0050 . 1 ( ) / cos( ) , ( ) , ( ) 424 . 5 ( ) 164 . 0 ( 1 1 2 2 1 = = = = e L L e A T T T t x T t L i wall λ θ τλ Then the center temperature of the top surface of the cylinder becomes C 84.2 ° = ⎯→ = = × = × = ) , 0 , ( 494 . 0 20 150 20 ) , 0 , ( 494 . 0 577 . 0 857 . 0 ) , ( ) , 0 , ( , t L T t L T t L T T T t L T cyl o wall cylinder short i ( c ) We first need to determine the maximum heat can be transferred from the cylinder [ ] kJ 325 C ) 20 150 )( C kJ/kg. 389 . 0 )( kg 43 . 6 ( ) ( kg 43 . 6 m) 15 . 0 ( m) 04 . 0 ( ) kg/m 8530 ( max 2. 3 2 = ° ° = = = π = ρπ = ρ = T T mC Q L r V m i p o Then we determine the dimensionless heat transfer ratios for both geometries as 135 . 0 164 . 0 ) 164 . 0 sin( ) 869 . 0 ( 1 ) sin( 1 1 1 , max = = λ λ θ = wall
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Unformatted text preview: o wall Q Q 427 . 1704 . 0846 . ) 577 . ( 2 1 ) ( 2 1 1 1 1 , max = − = − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ J Q Q cyl o cyl The heat transfer ratio for the short cylinder is 504 . ) 135 . 1 )( 427 . ( 135 . 1 max max max max = − + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ wall plane cylinder long wall plane cylinder short Q Q Q Q Q Q Q Q Then the total heat transfer from the short cylinder during the first 15 minutes of cooling becomes kJ 164 = = = kJ) 325 )( 504 . ( 503 . max Q Q 4-57...
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