Explanation Given: The bottom of the domain is held by condensing steam at temperature, Te = 100°C Thermal conductivity isk = 15 W/m.K Convective heat transfer coefficient is h = 240 W/m 2 K. Calculation: Take nodes in the heat transfer systems which are shown below: Apply energy equation for node1, k ( Δ r 2 ) T 4 − T 1 r o Δ ρ + k ( r o Δ ρ 2 ) T 2 − T 1 Δ r = 0 Substitute all the given values as follows: 15 ( 10 × 10 − 3 2 ) × T 4 − T 1 40 × 10 − 3 × π 8 + 15 ( 40 × 10 − 3 × π 8 2 ) × T 2 − T 1 10 × 10 − 3 = 0 3.4674 T 1 − 2.4674 T 2 − T 4 = 0 ...... ( 1 ) Energy balance for the node 2, k ( r Δ ϕ 2 ) T 1 − T 2 Δ r + k ( Δ r ) T 5 − T 2 r Δ ϕ + k ( r Δ ϕ 2 ) T 3 − T 2 Δ r = 0 Substitute all the given values in above equation as follows: 15 ( 30 × 10 − 3 × π 8 2 ) T 1 − T 2 10 × 10 − 3 + 15 ( 10 × 10 − 3 ) T 5 − T 2 30 × 10 − 3 × π 8 + 15 ( 30 × 10 − 3 × π 8 2 ) T 3 − T 2 10 × 10 − 3 = 0 T 1 − 6.323 T 2 + T 3 + 4.323 T 5 = 0 ...... ( 2 ) Energy balance for node 3, k ( r i Δ ϕ 2 ) T 2 − T 3 Δ r + k ( Δ r 2 ) T 6 − T 3 r i Δ ϕ + h ( r i Δ ϕ 2 ) ( T ∞ − T 3 ) = 0 Substitute all the given values in above equation as follows: 15 ( 20 × 10 − 3 × π 8 2 ) T 2 − T 3 10 × 10 − 3 + 15 ( 10 × 10 − 3 2 ) T 6 − T 3 20 × 10 − 3 × π 8 + 240 ( 20 × 10 − 3 × π 8 2 ) ( 20 − T 3 ) = 0 T 2 − 2.7811 T 3 + 1.6211 T 6 + 3.2 = 0 ...... ( 3 ) Apply energy equation for node 4, k ( Δ r 2 ) T 1 − T 4 r o Δ ϕ + k ( Δ r 2 ) T 7 − T 4 r o Δ ϕ + k ( r o Δ ϕ ) T 5 − T 4 Δ r = 0 Substitute all the given values in above equation as follows: 15 ( 10 × 10 − 3 2 ) T 1 − T 4 40 × 10 − 3 × π 8 + 15 ( 10 × 10 − 3 2 ) T 7 − T 4 40 × 10 − 3 × π 8 + 15 ( 40 × 10 − 3 × π 8 ) T 5 − T 4 10 × 10 − 3 = 0 T 1 − 6.934 T 1 + 4.934 T 5 + T 7 = 0 ...... ( 4 ) Node 10 and 7 has same pattern in conduction as that of node 4. Thus, the equation must be same. Energy equation for node 7, T 4 − 6.934 T 7 + 4.934 T 8 + T 10 = 0 Energy equation for node 10, T 7 − 6.934 T 10 + 4.934 T 11 + T 13 = 0 Apply energy equation in node 5, k ( Δ r ) ( T 2 − T 5 ) r Δ ϕ + k ( Δ r ) ( T 8 − T 5 ) r Δ ϕ + k ( r Δ ϕ ) ( T 4 − T 5 ) Δ r + k ( r Δ ϕ ) ( T 6 − T 5 ) Δ r = 0 Substitute the given values in above equation as follows: [ 15 ( 10 × 10 − 3 ) ( T 2 − T 5 ) 30 × 10 − 3 × π 8 + 15 ( 10 × 10 − 3 ) ( T 8 − T 5 ) 30 × 10 − 3 × π 8 + 15 ( 30 × 10 − 3 × π 8 ) ( T 4 − T 5 ) 10 × 10 − 3 + 15 ( 30 × 10 − 3 × π 8 ) ( T 6 − T 5 ) 10 × 10 − 3 = 0 T 2 + 1.387 T 4 − 4.774 T 5 + 1.387 T 6 + T 8 = 0 ...... ( 7 ) Apply the energy equation for node 8, T 5 + 1.387 T 7 − 4.774 T 8 + 1.387 T 9 + T 11 = 0 Apply the energy equation for node 11, T 8 + 1.387 T 10 − 4.774 T 11 + 1.387 T 12 + T 14 = 0 Apply the energy equation for node 6, k ( Δ r 2 ) ( T 3 − T 6 ) r i Δ ϕ + k ( Δ r 2 ) ( T 9 − T 6 ) r i Δ ϕ + k ( r i Δ ϕ ) ( T 5 − T 6 ) Δ r + h ( r i Δ ϕ ) ( T ∞ − T 6 ) = 0 Substitute all the given values in above equation as follows: 15 ( 10 × 10 − 3 2 ) ( T 3 − T 6 ) 20 × 10 − 3 × π 8 + 15 ( 10 × 10 − 3 2 ) ( T 9 − T 6 ) 20 × 10 − 3 × π 8 + 15 ( 20 × 10 − 3 × π 8 ) ( T 5 − T 6 ) 10 × 10 − 3 + 240 ( 20 × 10 − 3 × π 8 ) ( 20 − T 6 ) = 0 T 3 + 1.233 T 5 − 3.4303 T 6 + T 9 + 3.947 = 0 ...... ( 10 ) Apply energy equation in node 9 and 12 as follows: T 6 + 1.233 T 8 − 3.4303 T 9 + T 12 + 3.947 = 0 ...... ( 11 ) T 9 + 1.233 T 11 − 3.4303 T 12 + T 15 + 3.947 = 0 ...... ( 12 ) Apply energy equation for node 13, k ( Δ x 2 ) T s − T 13 Δ y + k ( Δ r 2 ) T 10 − T 13 r o Δ ϕ + k ( Δ y 2 + r 0 Δ ρ 2 ) T 14 − T 13 Δ x = 0 Substitute all the values in the above equation as follows: 15 ( 10 × 10 − 3 2 ) T s − T 13 10 × 10 − 3 + 15 ( 10 × 10 − 3 2 ) T 10 − T 13 40 × 10 − 3 × π 8 + 15 ( 10 × 10 − 3 2 + 40 × 10 − 3 × π 8 2 ) T 14 − T 13 10 × 10 − 3 = 0 0

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ISBN: 9780470501962

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

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Introduction to Heat Transfer

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Solution Summary: The author calculates the energy balance for nodes in the heat transfer systems.

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