The diagonal of a long triangular bar is well insulated, while sides of equivalent length are maintained at uniform temperatures T 75°C, and T, = 0°C. Establish a nodal network consisting of five nodes along each of the sides. For one of the nodes on the diagonal surface, define a suitable control volume and derive the corresponding finite-difference equation. Using this form for the diagonal nodes and appropriate equations for the interior nodes, find the temperature distribution for the bar. The temperature at nodes 2 and 5 are T₂ = 53.57°C, Ts = = 21.43°C. = Determine the temperature in 1, 3, 4, 6 nodes. Insulation Ta 2 3 Ax = Ay y↑ 4 5 6 T₁ = i T3= i Ть °C °C T4= i °C T6= i °C
The diagonal of a long triangular bar is well insulated, while sides of equivalent length are maintained at uniform temperatures T 75°C, and T, = 0°C. Establish a nodal network consisting of five nodes along each of the sides. For one of the nodes on the diagonal surface, define a suitable control volume and derive the corresponding finite-difference equation. Using this form for the diagonal nodes and appropriate equations for the interior nodes, find the temperature distribution for the bar. The temperature at nodes 2 and 5 are T₂ = 53.57°C, Ts = = 21.43°C. = Determine the temperature in 1, 3, 4, 6 nodes. Insulation Ta 2 3 Ax = Ay y↑ 4 5 6 T₁ = i T3= i Ть °C °C T4= i °C T6= i °C
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Transcribed Image Text:The diagonal of a long triangular bar is well insulated, while sides of equivalent length are maintained at uniform temperatures T
75°C, and T, = 0°C. Establish a nodal network consisting of five nodes along each of the sides. For one of the nodes on the diagonal
surface, define a suitable control volume and derive the corresponding finite-difference equation. Using this form for the diagonal
nodes and appropriate equations for the interior nodes, find the temperature distribution for the bar. The temperature at nodes 2 and
5 are T₂ = 53.57°C, Ts = = 21.43°C.
=
Determine the temperature in 1, 3, 4, 6 nodes.
Insulation
Ta
2
3
Ax = Ay
y↑
4
5
6
T₁ =
i
T3=
i
Ть
°C
°C
T4=
i
°C
T6=
i
°C
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