Explanation Draw a schematic diagram to illustrate the node 0 as a control volume. Write the expression of energy balance for control volume at node 0. E i n − E o u t = E s t Here, E i n is the energy entering the control volume, E o u t is the energy exiting from the control volume and E s t is the energy stored in the control volume. There is no energy leaving the control volume therefore, E o u t is 0. Substitute 0 for E o u t in above expression. E i n = E s t ....... (1) Write the expression for energy entering the control volume. E i n = q 1 A + q 1 B + q 3 A + q 3 B + q 2 + q 4 Here, q 1 A , q 1 B , q 3 A , q 3 B , q 2 and q 4 are the heat energies entering from different nodes. Write the expression for energy stored in control volume. E s t = E s t A + E s t B Substitute q 1 A + q 1 B + q 3 A + q 3 B + q 2 + q 4 for E i n and E s t A + E s t B for E s t in equation (1). q 1 A + q 1 B + q 3 A + q 3 B + q 2 + q 4 = E s t A + E s t B ....... (2) Write the expression for heat transfer through conduction. q = k A d T L ....... (3) Here, q is the rate of heat transfer, k is thermal conductivity, A is the area, d T is change in temperature and L is thickness of material. Calculate heat q 1 A by considering the unit depth of material. Substitute q 1 A for q , k A for k , Δ y 2 ( 1 ) for A , Δ x for L and T 1 − T 0 for d T in equation (3). q 1 A = k ( Δ y 2 ) ( T 1 − T 0 ) Δ x Substitute Δ x for Δ y in above expression. q 1 A = k A ( Δ x 2 ) ( T 1 − T 0 ) Δ x = 1 2 k A ( T 1 − T 0 ) Similarly calculate heat q 1 B by considering the unit depth of material as follows: q 1 B = 1 2 k B ( T 1 − T 0 ) Similarly calculate heat q 3 A by considering the unit depth of material as follows: q 3 A = 1 2 k A ( T 3 − T 0 ) Similarly calculate heat q 3 B by considering the unit depth of material as follows: q 3 B = 1 2 k B ( T 3 − T 0 ) Similarly calculate heat q 2 by considering the unit depth of material as follows: q 2 = k A Δ x Δ y ( T 2 − T 0 ) = k A ( T 2 − T 0 ) Similarly calculate heat q 4 by considering the unit depth of material as follows: q 4 = k B Δ x Δ y ( T 4 − T 0 ) = k B ( T 4 − T 0 ) Write the expression for energy stored in material A. E s t A = ρ A Δ x Δ y 2 c A ( T 0 p + 1 − T 0 p Δ t ) = ρ A c A ( Δ x ) 2 ( T 0 p + 1 − T 0 p 2 Δ t ) Here, ρ A is the density of material A, c A is the specific heat of material A, T 0 p + 1 is the temperature of node 0 with respect to time p + 1 and T 0 p is the temperature of node 0 with respect to time p . Write the expression for energy stored in material B. E s t B = ρ B Δ x Δ y 2 c B ( T 0 p + 1 − T 0 p Δ t ) = ρ B c B ( Δ x ) 2 ( T 0 p + 1 − T 0 p 2 Δ t ) Here, ρ B is the density of material B and c B is the specific heat of material B

7th Edition

ISBN: 9780470917855

Author: Bergman, Theodore L./

Publisher: John Wiley & Sons Inc

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Chapter 5, Problem 5.115P

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The transient 2-dimensional finite difference equation for temperature at node 0 located at the boundary at the contact of two materials

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Chapter 5, Problem 5.114P

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The transient 2-dimensional finite difference equation for temperature at node 0 located at the boundary at the contact of two materials

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