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Determine temperature distribution in the wall.

A composite wall consist of three materials, as shown in fig. Convection heat transfer takes place on inner surface of the wall with T=8000c and h=30W/m20c. The outer surface temperature T0=200c. Determine temperature distribution in the wall.

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1 Answer
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(i) Number of nodes and elements

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(ii) Element matrix equation

{[hcA000]+kAhe[1111]}{θ1θ2}={Q1Q2}

For element 1

hc = 30 W/m2; A = 1m2; K = 25; he = 0.3

KAhe=25×10.3=83.33

{[30000]+[83.3383.3383.3383.33]}{θ1θ2}={Q1Q2}

[113.3383.3383.3383.33]{θ1θ2}={Q1Q2}

For element 2

hc = 0; A = 1m2; K = 30; he = 0.2

KAhe=30×10.2=150

[150150150150]{θ2θ3}={Q2Q3}

For element 3

hc = 30 W/m2; A = 1m2; K = 70; he = 0.15

KAhe=70×10.15=466.67

{[00030]+[466.67466.67466.67466.67]}{θ3θ4}={Q3Q4}

[466.67466.67466.67466.67]{θ3θ4}={Q3Q4}

(iii) Global matrix equation

[113.3383.330083.33233.3315000150616.67466.6700466.67496.67]{θ1θ2θ3θ4}={Q1Q2Q3Q4}

Imposing boundary condition,

θ1=800Cθ4=20Cθ2=θ3=0(Forbalancing)

[113.3383.330083.33233.3315000150616.67466.6700466.67496.67]{800θ2θ320}={Q100Q4}

Now,

(113.33 x 800) - 83.33 θ2 = θ1 ...(1)

(-83.33 x 800) + 233.33 θ2 - 150 θ3 = 0 ...(2)

(-150 θ2 + 616.67 θ3 - 466.67 x 20) = 0 ...(3)

(-466.67 θ3 + 496.67 x 20) = θ4 ...(4)

By solving equation (2) and (3), we get

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