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Module 3b FEA

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10 views9 pages

Module 3b FEA

Uploaded by

varhadivighnesh06
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FINITE ELEMENT

ANALYSIS
Dr. Richa Agrawal

MODULE 3 – 1D ANALYSIS – BAR ELEMENT


Heat Transfer
• GDE in Local Coordinate
𝑑 𝑑𝜃
𝑘𝐴 =0 (𝑑𝑥 = 𝑑𝑥)ҧ
𝑑 𝑥ҧ 𝑑 𝑥ҧ

where k – conductivity of material and 𝜃 is the temperature difference between the


temperature at any point within the bar and the ambient temperature. \

• Boundary Conditions
𝑑𝜃 𝑑𝜃
𝑘𝐴 = −𝑄 𝑒 1 𝑘𝐴 = 𝑄𝑒 2
𝑑 𝑥ҧ ҧ
𝑥=0
𝑑 𝑥ҧ ҧ 𝑒
𝑥=ℎ

• Element Matrix Equation

𝑘𝐴 1 −1 𝜃1 𝑄1 𝑒
=
ℎ𝑒 −1 1 𝜃2 𝑄2 𝑒
Heat Transfer – Problem 1
Using Finite Element Method, Determine the temperature
in the wall shown in the figure and calculate the heat flow
through the wall.
KA = 20 x 10-3 W/mm C
KB = 30 x 10-3 W/mm C
0C A B C 50C KC = 50 x 10-3 W/mm C

20mm 100mm 50mm

𝑘𝐴 1 −1 𝜃1 𝑄1 𝑒
=
ℎ𝑒 −1 1 𝜃2 𝑄2 𝑒
Heat Transfer – Problem 1
𝑘𝐴 1 −1 𝜃1 𝑄1 𝑒
=
0C A B C 50C ℎ𝑒 −1 1 𝜃2 𝑄2 𝑒
KA = 20 x 10-3 W/mm C
KB = 30 x 10-3 W/mm C
KC = 50 x 10-3 W/mm C

20mm 100mm 50mm

𝑘1 𝐴1 1 −1 20×10−3 ×1 1 −1 1 −1
𝐾1 = = =10−3
𝐿1 −1 1 20 −1 1 −1 1

𝑘2 𝐴2 1 −1 30×10−3 ×1 1 −1 0.3 −0.3


𝐾2 = = =10−3
𝐿2 −1 1 100 −1 1 −0.3 0.3

𝑘3 𝐴3 1 −1 50×10−3 ×1 1 −1 1 −1
𝐾3 = = =10−3
𝐿3 −1 1 50 −1 1 −1 1
Heat Transfer – Problem 1
1 −1 0 0
0C A B C 50C
−1 1.3 −0.3 0
𝐾= 10−3
0 −0.3 1.3 −1
0 0 −1 1
1 −1 0 0 𝜃1 𝑄1
20mm 100mm 50mm
−1 1.3 −0.3 0 𝜃2 𝑄
10−3 = 2
0 −0.3 1.3 −1 𝜃3 𝑄3
0 0 −1 1 𝜃4 𝑄4

𝜃1 =0, 𝜃4 = 50, 𝑄2 =𝑄3 = 0

𝜃2 = 9.37℃ , 𝜃3 = 40.61℃

𝑄4 = 9.39 × 10−3 𝑊

𝑄1 = −9.37 × 10−3 𝑊
Heat Transfer – Problem 2
A composite wall consists of three materials as shown. The
outer temperature T0 = 20 C . Convection heat transfer
takes place at the inner surface of the wall with T = 800 C
and h=30 W/m2 C . Determine temperature distribution in
the wall.
KA = 25 W/m C
A B C KB = 30 W/m C
T,h T0
KC = 70 W/m C

0.3m 0.2m 0.15m

𝑘𝐴 1 −1 𝜃1 𝑄1 𝑒
=
ℎ𝑒 −1 1 𝜃2 𝑄2 𝑒
Heat Transfer – Problem 2
KA = 25 W/m C
T,h A B C T0 KB = 30 W/m C
KC = 70 W/m C

0.3m 0.2m 0.15m

𝑘1 𝐴1 1 −1 25×1 1 −1 83.3 −83.3


𝐾1 = = =
𝐿1 −1 1 0.3 −1 1 −83.3 83.3

𝑘2 𝐴2 1 −1 30×1 1 −1 150 −150


𝐾2 = = =
𝐿2 −1 1 0.2 −1 1 −150 150

𝑘3 𝐴3 1 −1 70×1 1 −1 466.7 −466.7


𝐾3 = = =
𝐿3 −1 1 0.15 −1 1 −466.7 466.7
Heat Transfer – Problem 2
83.3 −83.3 0 0
T,h A B C T0 −83.3 233.3 −150 0
𝐾=
0 −150 616.7 −466.7
0 0 −466.7 466.7
0.2m 0.15m 83.3 −83.3 0 0 𝜃1 𝑄1
0.3m
−83.3 233.3 −150 0 𝜃2 𝑄
= 2
0 −150 616.7 −466.7 𝜃3 𝑄3
0 0 −466.7 466.7 𝜃4 𝑄4

𝜃4 = 20, 𝑄2 =𝑄3 = 0

𝑄1 = −ℎ 𝑇1 − 𝑇∞ = −30𝑇1 + 24000

𝜃1 = 319.8℃ 𝜃2 = 146.9℃ , 𝜃3 = 50.87℃


• Refer class notes for more types of ID Analysis

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