PROBLEM 4.43

KNOWN: Two-dimensional network with prescribed nodal temperatures and thermal conductivity of he material. t

FIND: Heat rate per unit length normal to page, q .′ SCHEMATIC:

Node Ti(°C) 1 120.55 2 120.64 3 121.29 4 123.89

5 134.57 6 150.49 7 147.14

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional heat transfer, (3) No internal

olumetric generation, (4) Constant properties. v ANALYSIS: Construct control volumes around the nodes on the surface maintained at the uniform temperature Ts and indicate the heat rates. The heat rate per unit length is a b c dq q q q q qe′ ′ ′ ′ ′ ′= + + + +

r in terms of conduction terms between nodes, o 1 2 3 4 5 7q q q q q q q′ ′ ′ ′ ′ ′ ′= + + + + + . Each of these rates can be written in terms of nodal temperatures and control volume dimensions using

ourier’s law, F

1 s 2 s 3 s 4 s

5 s 7 s

T T T T T T T Txq k k x k x k x2 y y y y

T T T Ty k x k .y 2 x

− − −Δ′ = ⋅ ⋅ + ⋅Δ ⋅ + ⋅Δ + ⋅ΔΔ Δ Δ− −Δ

+ ⋅Δ + ⋅ ⋅Δ Δ

−Δ

a nd since Δx =Δy,

( )( ) ( ) ( )( ) ( ) ( )( )

1 s 2 s 3 s4 s 5 s 7 s

q k[ 1/2 T T T T T T T T T T 1/ 2 T T ].′ = − + − + −+ − + − + −

S ubstituting numerical values, find

( )( ) ( ) ( )( ) ( ) ( )( )

q 50 W/m K[ 1/2 120.55 100 120.64 100 121.29 100 123.89 100 134.57 100 1/ 2 147.14 100 ]′ = ⋅ − + − + −+ − + − + −

q 6711 W/m.′ = <

COMMENTS: For nodes a through d, there is no heat transfer into the control volumes in the x-direction. Look carefully at the energy balance for node e, e 5 7q q q ,′ ′ ′= + and how are evaluated.

5 7q and q′ ′