Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Chapter 27, Problem 29P

Repeat Prob. 27.28, but for the following heat source: f ( x ) = 0.12 x 3 2.4 x 2 + 12 x .

(a)

Expert Solution
Check Mark
To determine

To calculate: The temperature distribution by shooting method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=f(x)

, where heat source f(x)=0.12x32.4x2+12x and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 29P

Solution: The table of the solution of the boundary value problem is,

xTz040762179.0457.924261.05623.526276.5446.088246.59221.281020024

Explanation of Solution

Given:

A differential equation, d2Tdx2=f(x)

, where heat source f(x)=0.12x32.4x2+12x and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

Linear-interpolation formula:

f1(x)=f(x0)+f(x1)f(x0)x1x0(xx0)

Calculation:

Consider the following Poisson equation, d2Tdx2=f(x).

Since, f(x)=0.12x32.4x2+12x

. Then,

d2Tdx2=(0.12x32.4x2+12x)T(0)=40,T(10)=200

Now, change the above boundary value problem into equivalent initial-value problem. Then,

dTdx=zd2Tdx2=dzdx

But d2Tdx2=(0.12x32.4x2+12x).

Thus,

dTdx=zdzdx=(0.12x32.4x2+12x)

Use the shooting method in the above system of first order linear differential equation

Suppose, z(0)=1 and given that T(0)=40.

Then, the system of system of first order linear differential equation with initial condition is,

dTdx=zdzdx=(0.12x32.4x2+12x)T(0)=40,z(0)=1

Now, solve the above system of differential equation.

Thus,

dzdx=(0.12x32.4x2+12x)dz=(0.12x32.4x2+12x)dx

Integrate on both the sides of the above differential equation, to get

z(x)=(0.12x442.4x33+12x22)+C1=(0.03x40.8x3+6x2)+C1

Where C1 is constant of integration.

Now, use the initial condition, z(0)=1

. Thus,

z(0)=(0.03×040.8×03+6×02)+C11=0+C1C1=1

Put the value of C1 in z(x)=(0.03x40.8x3+6x2)+C1.

Thus,

z(x)=(0.03x40.8x3+6x2)1

Since, dTdx=z

. But z(x)=(0.03x40.8x3+6x2)1.

Therefore,

dTdx=(0.03x40.8x3+6x2)1dT=[ (0.03x40.8x3+6x2)1 ]dx

Integrate on both the sides of the above differential equation, to get

T(x)=[ (0.03x550.8x44+6x33)x ]+C2

Where C2

is constant of integration.

Use the initial condition, T(0)=40

. Then,

T(0)=[ (0.030550.8044+6033)0 ]+C240=0+C2C2=40

Put the value of C2 in T(x)

, then

T(x)=[ (0.03x550.8x44+6x33)x ]+40

Now, evaluate the above for T(10)

. Thus,

T(10)=[ (0.0310550.81044+61033)10 ]+40=(600+2000200010)+40=570

But the above of T(10) is much different from the boundary condition of T(10)=200.

Then, put another guess. Suppose z(0)=0.5.

Then, the system of system of first order linear differential equation with initial condition is,

dTdx=zdzdx=(0.12x32.4x2+12x)T(0)=40,z(0)=0.5

Now, solve the above system of differential equation. Then,

dzdx=(0.12x32.4x2+12x)dz=(0.12x32.4x2+12x)dx

Integrate on both the sides of the above differential equation

Thus,

z(x)=(0.12x442.4x33+12x22)+C3=(0.03x40.8x3+6x2)+C3

Where C3 is constant of integration.

Now, use the initial condition, z(0)=0.5.

Therefore,

z(0)=(0.03×040.8×03+6×02)+C30.5=0+C3C3=0.5

Put the value of C3 in z(x), then

z(x)=(0.03x40.8x3+6x2)0.5

Since, dTdx=z

. But z(x)=(0.03x40.8x3+6x2)0.5.

Thus,

dTdx=(0.03x40.8x3+6x2)0.5dT=[ (0.03x40.8x3+6x2)0.5 ]dx

Integrate on both the sides of the above differential equation. Then,

T(x)=[ (0.03x550.8x44+6x33)0.5x ]+C4

Where C4 is constant of integration.

Use the initial condition, T(0)=40

. Thus,

T(0)=[ (0.030550.8044+6033)0.5×0 ]+C440=0+C4C4=40

Put the value of C4 in T(x), then

T(x)=[ (0.03x550.8x44+6x33)0.5x ]+40

Now, evaluate the above for T(10)

. Thus,

T(10)=[ (0.0310550.81044+61033)0.5×10 ]+40=(600+200020005)+40=565

Since, the first guess value z(0)=1

corresponds to T(10)=570

and the second-guess value z(0)=0.5

corresponds to T(10)=565.

Now, use these values to compute the value of z(0)

that yields T(10)=200.

Then, by linear interpolation formula,

z(0)=1+0.5+1565(570)(200(570))=1+0.5×7705=1+77=76

Therefore, the right value of z(0)

which yields T(10)=200 is z(0)=76.

Then, the equivalent initial value problem corresponding to the boundary value problem is,

dTdx=zdzdx=(0.12x32.4x2+12x)T(0)=40,z(0)=76

Now, use the fourth order RK method with step size h=2.

The RK method for above system of first order linear differential equation with initial condition is,

Tn+1=Tn+16(k0+2k1+2k2+k3)zn+1=zn+16(l0+2l1+2l2+l3)

Where xn+1=xn+nh

And

k0=hf(xn,Tn,zn)k1=hf(xn+12h,Tn+12k0,zn+12l0)k2=hf(xn+12h,Tn+12k1,zn+12l1)k3=hf(xn+h,Tn+k2,zn+l2)

And

l0=hg(xn,Tn,zn)l1=hg(xn+12h,Tn+12k0,zn+12l0)l2=hg(xn+12h,Tn+12k1,zn+12l1)l3=hg(xn+h,Tn+k2,zn+l2)

Where f(xn,Tn,zn)=zn and g(xn,Tn,zn)=(0.12x32.4x2+12x).

Then, for n=0

x1=x0+0×2=0

And

k0=hf(x0,T0,z0)=2×z0=2×76=152l0=hg(x0,T0,z0)=2×[ (0.12x032.4x02+12x0) ]=0k1=hf(x0+12h,T0+12k0,z0+12l0)=2(z0+12l0)=2(76+12(0))=152l1=hg(x0+12h,T0+12k0,z0+12l0)=2[ 0.12(x0+12×2)32.4(x0+12×2)2+12(x0+12×2)]=19.44

Also,

k2=hf(x0+12h,T0+12k1,z0+12l1)=2(z0+12l1)=2(76+12(19.44))=132.56l2=hg(x0+12h,T0+12k1,z0+12l1)=2[ 0.12(x0+12×2)32.4(x0+12×2)2+12(x0+12×2)]=19.44k3=hf(x0+h,T0+k2,z0+l2)=2×(z0+l2)=2(7619.44)=113.12l3=hg(x0+h,T0+k2,z0+l2)=2×[ 0.12(x0+2)32.4(x0+h)2+12(x0+h) ]=30.72

Thus,

T1=T0+16(k0+2k1+2k2+k3)=40+16(152+304+265.12+113.12)=40+139.04=179.04

And

z1=z0+16(l0+2l1+2l2+l3)=76+16(038.8838.8830.72)=57.92

In the similar way, find the remaining Tn and zn

. Then,

T2=261.056 and z2=23.52

T3=276.544 and z3=6.08

T4=246.592 and z4=21.28

And

T5=200 and z5=24

Therefore, the table of the solution of the boundary value problem is

xTz040762179.0457.924261.05623.526276.5446.088246.59221.281020024

Hence, the graph of the temperature distribution is

(b)

Expert Solution
Check Mark
To determine

To calculate: The temperature distribution by finite difference method for a heated rod with a uniform heat source given by Poisson equation, d2Tdx2=f(x)

, where heat source f(x)=0.12x32.4x2+12x and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Answer to Problem 29P

Solution:

The table of the solution of boundary value problem is

xT0402184.1284266.8166280.3848247.87210200

Explanation of Solution

Given:

A differential equation, d2Tdx2=f(x)

, where heat source f(x)=0.12x32.4x2+12x and the boundary conditions, T(x=0)=40 and T(x=10)=200

, also Δx=2.

Formula used:

(1) The finite difference method is:

d2Tdx2=Ti+12Ti+Ti1Δx2

(2) The Gauss-Seidel iterative method is:

Ti(k)=1aii[ j=1i1(aijTj(k))j=i+1n(aijTj(k1))+bi ]

Calculation:

Consider the following Poisson equation, d2Tdx2=f(x).

Since, f(x)=0.12x32.4x2+12x.

Thus,

d2Tdx2=(0.12x32.4x2+12x)T(0)=40,T(10)=200

The finite difference method is given by,

d2Tdx2=Ti+12Ti+Ti1Δx2

Now, substitute the value of second order derivative in the boundary value problem.

Then, the boundary value problem becomes,

Ti+12Ti+Ti1Δx2=(0.12xi32.4xi2+12xi)Ti+12Ti+Ti1=Δx2(0.12xi32.4xi2+12xi)

Or

Ti1+2TiTi+1=Δx2(0.12xi32.4xi2+12xi)

Since, Δx=2

. Then,

Ti1+2TiTi+1=(2)2(0.12xi32.4xi2+12xi)Ti1+2TiTi+1=0.48xi39.6xi2+48xi

For the first node, i=1.

T11+2T1T1+1=0.48x139.6x12+48x1T0+2T1T1+1=0.48x139.6x12+48x140+2T1T1+1=0.48x139.6x12+48x12T1T2=101.44

For the second node, i=2.

T21+2T2T2+1=0.48x239.6x22+48x2T1+2T2T3=69.12

For the third node, i=3.

T31+2T3T3+1=0.48x339.6x32+48x3T2+2T3T4=46.08

For the fourth node, i=4.

T41+2T4T4+1=0.48x439.6x42+48x4T3+2T4200=0.48x439.6x42+48x4T3+2T4=215.36

Then, write the system of equations in matrix form

[ 2100121001210012 ]{ T1T2T3T4 }={ 101.4469.1246.08215.36 }

Since, the coefficient matrix is tridiagonal matrix, then use Gauss-Seidel iterative technique

The Gauss-Seidel iterative method is,

Ti(k)=1aii[ j=1i1(aijTj(k))j=i+1n(aijTj(k1))+bi ]

Now, evaluate Ti

by above Gauss-Seidel method

Then,

T1=184.128

T2=266.816

T3=280.384

And

T4=247.872

Then, the table of the solution of boundary value problem is

xT0402184.1284266.8166280.3848247.87210200

Therefore, the graph of temperature distribution is

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Chapter 27 Solutions

Numerical Methods for Engineers

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