EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 8220100254147
Author: Chapra
Publisher: MCG
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Chapter 31, Problem 12P
To determine

To calculate: The temperature distribution in a rod as shown in figure below with internal heat generation using the finite-element method and also derive the element nodal equations using Fourier heat conduction.

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 31, Problem 12P , additional homework tip  1

Expert Solution & Answer
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Answer to Problem 12P

Solution: The temperature distribution in a rod at various nodes is shown below,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 31, Problem 12P , additional homework tip  2

Explanation of Solution

Given Information:

To develop nodal equations for the temperature and their gradients at each of the six nodes. The element nodal equations is given as,

qk=kAdTx

With the heat conservation relationships is,

[qk+f(x)]=0

Here,

qk=heat flow(W)k=thermal conductivity(WmoC)A=cross-sectional area(m2)f(x)=heat source(Wcm)

The given rod is 50 cm long, the x-coordinate is positive to the right end and zero at the left end, kA=100 and 50 Wm/°C at x=0 and at x=50 respectively. The given data are,

kA=100,f(x)=3WcmT|x=0=100 oC, T|x=50=50 oC

Calculation:

The element 1 the equations for the node 1 and node 2 is,

Solve the equation for node 1,

qk+f(x)=0(kAdTdx|1+kA(1x2x1)(T2T1))+x1x2N1f(x)dx=0(kAdTdx|1+kA(1x2x1)(T2T1))+x1x2(x2xx2x1)f(x)dx=0

Substitute the value from given data,

((100)dTdx|1+(9510)(T2T1))+010(10x100)30dx=0((100)dTdx|1+9.5(T2T1))+3[10xx22]010=0

Change the temperature so the equation becomes,

9.5(T1T2)=(100)dTdx|1+150

Solve the equation for the node 2,

qk+f(x)=0(kAdTdx|2+kA(1x2x1)(T2T1))+x1x2N2f(x)dx=0(kAdTdx|2+kA(1x2x1)(T2T1))+x1x2(xx2x2x1)f(x)dx=0

Substitute the values from given data,

(90dTdx|1+(9510)(T2T1))+010(x0100)30dx=0(90dTdx|2+9.5(T2T1))+3[x22]010=0(90dTdx|2+9.5(T2T1))+3[1002]=0(90dTdx|2+9.5(T2T1))+150=0

So, the equation of node 2 is,

9.5(T2T1)=90dTdx|2+150

Similarly, the value for other nodes is calculated. For element 2 at node 2,

qk+f(x)=0(kAdTdx|2+kA(1x2x1)(T3T2))+x1x2N1f(x)dx=0(kAdTdx|2+kA(1x2x1)(T3T2))+x1x2(x2xx2x1)f(x)dx=0

Substitute the values from given data,

(90dTdx|2+(8510)(T3T2))+010(10x100)30dx=0((90)dTdx|2+8.5(T3T2))+3[10xx22]010=0(90dTdx|2+8.5(T3T2))+3[50]=0((90)dTdx|2+8.5(T3T2))+150=0

Change the state so the equation becomes,

8.5(T2T3)=90dTdx|2+150

Derive for node 3,

qk+f(x)=0(kAdTdx|3+kA(1x2x1)(T2T3))+x1x2N2f(x)dx=0(kAdTdx|3+kA(1x2x1)(T2T3))+x1x2(xx2x2x1)f(x)dx=0

Substitute the values from given data,

(80dTdx|3+(8510)(T2T3))+010(x0100)30dx=0(80dTdx|3+8.5(T2T3))+3[x22]010=0(80dTdx|3+8.5(T2T3))+3[1002]=0(80dTdx|3+8.5(T2T3))+150=0

So, the equation of node 3 is,

8.5(T2T3)=80dTdx|3+150

Derive the equation for the element 3,

Node 3

qk+f(x)=0(kAdTdx|3+kA(1x2x1)(T3T4))+x1x2N1f(x)dx=0(kAdTdx|3+kA(1x2x1)(T3T4))+x1x2(x2xx2x1)f(x)dx=0

Substitute the values from given data,

(80dTdx|3+(7510)(T3T4))+010(10x100)30dx=0(80dTdx|3+7.5(T3T4))+3[10xx22]010=0(80dTdx|3+7.5(T3T4))+3[50]=0((80)dTdx|3+7.5(T3T4))+150=0

Change the state so the equation becomes,

7.5(T3T4)=80dTdx|3+150

Derive for node 4,

qk+f(x)=0(kAdTdx|4+kA(1x2x1)(T3T4))+x1x2N2f(x)dx=0(kAdTdx|4+kA(1x2x1)(T3T4))+x1x2(xx2x2x1)f(x)dx=0

Substitute the values from given data,

(70dTdx|4+(7510)(T3T4))+010(x0100)30dx=0(70dTdx|4+7.5(T3T4))+3[x22]010=0(70dTdx|4+7.5(T3T4))+3[1002]=0(70dTdx|4+7.5(T3T4))+150=0

So, the equation of node 4 is,

7.5(T3T4)=70dTdx|4+150

Derive the equation for the element 4,

The equations for node 4 is,

qk+f(x)=0(kAdTdx|4+kA(1x2x1)(T5T4))+x1x2N1f(x)dx=0(kAdTdx|4+kA(1x2x1)(T5T4))+x1x2(x2xx2x1)f(x)dx=0

Substitute the values from given data,

(70dTdx|4+(6510)(T5T4))+010(10x100)30dx=0(70dTdx|4+6.5(T5T4))+3[10xx22]010=0(70dTdx|4+6.5(T5T4))+3[50]=0((70)dTdx|4+6.5(T5T4))+150=0

Change the state so the equation becomes,

6.5(T4T5)=70dTdx|4+150

Derive the equation for the node 5,

qk+f(x)=0(kAdTdx|5+kA(1x2x1)(T4T5))+x1x2N2f(x)dx=0(kAdTdx|5+kA(1x2x1)(T4T5))+x1x2(xx2x2x1)f(x)dx=0

Substitute the values from given data,

(60dTdx|5+(6510)(T4T5))+010(x0100)30dx=0(60dTdx|5+6.5(T4T5))+3[x22]010=0(60dTdx|5+6.5(T4T5))+3[1002]=0(60dTdx|5+6.5(T4T5))+150=0

So, the equation of node 5 is,

6.5(T3T4)=60dTdx|5+150

Derive the equation for the element 5,

For node 5,

qk+f(x)=0(kAdTdx|5+kA(1x2x1)(T6T5))+x1x2N1f(x)dx=0(kAdTdx|5+kA(1x2x1)(T6T5))+x1x2(x2xx2x1)f(x)dx=0

Substitute the values from given data,

(60dTdx|5+(5510)(T6T5))+010(10x100)30dx=0(60dTdx|4+5.5(T6T5))+3[10xx22]010=0(60dTdx|4+5.5(T6T5))+3[50]=0((60)dTdx|4+5.5(T6T5))+150=0

Change the state so the equation becomes,

5.5(T5T6)=60dTdx|5+150

Derive the equation for node 6,

qk+f(x)=0(kAdTdx|6+kA(1x2x1)(T5T6))+x1x2N2f(x)dx=0(kAdTdx|6+kA(1x2x1)(T5T6))+x1x2(xx2x2x1)f(x)dx=0

Substitute the values from given data,

(50dTdx|6+(5510)(T5T6))+010(x0100)30dx=0(50dTdx|5+5.5(T5T6))+3[x22]010=0(50dTdx|5+5.5(T5T6))+3[1002]=0(50dTdx|5+5.5(T5T6))+150=0

So, the equation of node 6 is,

5.5(T5T6)=50dTdx|5+150

The equation assembly is given as,

[9.59.59.5188.58.5167.57.5146.56.5125.55.55.5][T1T2T3T4T5T6]={100dTdx|1+15030030030030050dTdx|6+150}

Insert the boundary conditions then the matrix becomes,

[1009.5188.58.5167.57.5146.56.51205.550][dTdx|1T2T3T4T5dTdx|6]={8001250300300575125}

Use MATLAB to write code for solving the matrix,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 31, Problem 12P , additional homework tip  3

The desired output is,

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 31, Problem 12P , additional homework tip  4

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