Chapter 30, Problem 1P

Repeat Example 30.1, but use the midpoint method to generate your solution.

To determine

To calculate: The solution of the one-dimensional heat conduction equation using midpoint method for a thing rod of length $10\text{ cm}$, and the rod possess following values:

$k^{\prime }=0.49\text{ cal/}\left(\text{s}\cdot \text{cm}\cdot °\text{C}\right)$, $\Delta x=2\text{ cm}$

Answer to Problem 1P

Solution: The desired result is as shown below.

Explanation of Solution

Given Information:

The expression of the temperature distribution of long, thin rod is,

$l=10\text{ cm},k\text{'}=0.49\text{ cal/}\left(\text{s}\cdot \text{cm}\cdot \text{°C}\right),△x=2\text{cm},\text{ and }△t=0.1\text{s};$

$\text{at }t=0,T_0=0,T\left(0\right)=100°C\text{ and }T\left(10\right)=50°C,$

$C=0.2174cal/\left(g\cdot °C\right),\rho =2.7g/c{m}^3,$

Calculation:

Calculate $k$

and $\lambda$

$\begin{array}{c}k=0.49/\left(2.7\cdot 0.2174\right)\\ =0.835{\text{ cm}}^{\text{2}}\text{/s and}\\ \lambda =\frac{0.835\left(0.1\right)}{{\left(2\right)}^2}\\ =0.020875\end{array}$

The expression of temperature,

$\frac{d{T}_i}{dt}=k\frac{T_{i-1}-2T_i+T_{i+1}}{\Delta x^2}$

Rewrite the above equation,

$\frac{T_i^{l+1}-T_i^l}{\Delta t}=k\frac{T_{i-1}-2T_i+T_{i+1}}{\Delta x^2}$

Rewrite the above equation,

$T_i^{l+1}=T_i^l+\lambda \left(T_{i-1}-2T_i+T_{i+1}\right)$

Substitute the value at t = 0.1 s for the node at x = 2 cm,

$\begin{array}{c}{T}_1^1=0+0.020875\left[0-2\left(0\right)+100\right]\\ =2.0875\end{array}$

The results at the other interior points are,

$x=4,6,\text{ and }8\text{ cm},$

Therefore,

$\begin{array}{c}{T}_2^1=0+0.020875\left[0-2\left(0\right)+0\right]\\ =0\end{array}$

Next,

$\begin{array}{c}{T}_3^1=0+0.020875\left[0-2\left(0\right)+0\right]\\ =0\end{array}$

And,

$\begin{array}{c}{T}_4^1=0+0.020875\left[50-2\left(0\right)+0\right]\\ =1.0438\end{array}$

The value at t = 0.2 s; the interior points are four are,

$\begin{array}{c}{T}_2^1=2.0875+0.020875\left[0-2\left(2.0875\right)+100\right]\\ =4.0878\end{array}$

Therefore,

$\begin{array}{c}{T}_2^2=0+0.020875\left[0-2\left(0\right)+2.0875\right]\\ =0.043577\end{array}$

Next,

$\begin{array}{c}{T}_3^2=0+0.020875\left[1.0438-2\left(0\right)+0\right]\\ =0.021788\end{array}$

And,

$\begin{array}{c}{T}_4^2=1.0438+0.020875\left[50-2\left(1.0438\right)+0\right]\\ =2.0439\end{array}$

The Midpoint method in this subpart is,

$y_{i+\frac{1}{2}}=y_i+f\left(x_i,y_i\right)\frac{h}{2}$

And,

$y_{i+1}=y_i+f\left(x_{i+\frac{1}{2}},y_{i+\frac{1}{2}}\right)h$

Use $i=0,$

predictor is calculating as follow:

$\begin{array}{c}{y}_{0+\frac{1}{2}}=0+\left(\frac{100-0+100}{4}\right)\frac{0.1}{2}\\ y_{\frac{1}{2}}=0\end{array}$

Slop-midpoint,

$\begin{array}{c}f\left(x_{\frac{1}{2}},y_{\frac{1}{2}}\right)=\frac{100-0+100}{4}\\ =0\end{array}$

Calculate corrector,

$\begin{array}{c}{y}_1=y_0+f\left(x_{\frac{1}{2}},y_{\frac{1}{2}}\right)h\\ =0+0\\ =0\end{array}$

Use Excel to create the table as follow,

Use excel to solve this problem.

Step 1 Open the excel-spreadsheet and then press Alt+F11.

Step 2 Then there is a window opened in which write the coding to find optimal solution is as below,

Option Explicit

Sub Explicit()

'define the variables which are integers

Dim i As Integer, j As Integer, Q As Integer, W As Integer

'define the variables which takes double values

Dim R(20) As Double, U(20) As Double, O(20) As Double, S(20, 20) As Double

Dim k As Double, E As Double, L As Double, B As Double, A As Double

Dim C As Double, t As Double, F As Double, h As Double, P As Double

Dim x As Double

'Initialize the value to the variable P

P = 0.000001

'Initialize the value to the variable L

L = 10

'Initialize the value to the variable W

W = 5

E = L / W

'Initialize the value to the variable k

k = 0.835

'Initialize the value to the variable R(0) and R(5)

R(0) = 100

R(5) = 50

'Initialize the value to the variable B

B = 0.1

'Initialize the value to the variable A

A = 12

'Initialize the value to the variable C

C = 3

'Initialize the value to the variable Q

Q = 0

O(Q) = t

'construct a loop that executes from 0 to W

For i = 0 To W

'moves from next i, if the condition is satisfied

S(i, Q) = R(i)

Next i

'the condition are checked

Do

'Do loop is initialised

F = t + C

If F > A Then F = A

h = B

Do

If t + h > F Then h = F - t

'Call a five parameters function

Call Derivs(R, U, W, E, k)

'construct a for loop from 1 to W-1

For j = 1 To W - 1

R(j) = R(j) + U(j) * h

Next j

t = t + h

'check the condition

If t >= F Then Exit Do

Loop

Q = Q + 1

O(Q) = t

'construct a loop

For j = 0 To W

S(j, Q) = R(j)

Next j

'check the condition

If t + P >= A Then Exit Do

Loop

Sheets("sheet1").Select

Range("a4:bb5000").ClearContents

Range("a4").Select

ActiveCell.Value = "time"

x = 0

'construct a loop for j variable

For j = 0 To W

ActiveCell.Offset(0, 1).Select

ActiveCell.Value = "x = " & x

x = x + E

Next j

Range("a5").Select

'construct a loop for i variable

For i = 0 To Q

ActiveCell.Value = O(i)

'construct a loop for j variable

For j = 0 To W

ActiveCell.Offset(0, 1).Select

ActiveCell.Value = S(j, i)

Next j

ActiveCell.Offset(1, -W - 1).Select

Next i

Range("a5").Select

End Sub

'call a function Derives which has five parameter

Sub Derivs(R, U, W, E, k)

Dim j As Integer

'construct a loop for j variable

For j = 1 To W - 1

U(j) = k * (R(j - 1) - 2 * R(j) + R(j + 1)) / E ^ 2

Next j

End Sub

Step 3 Now press F5, a new popup window will appear as shown below.

Step 4 Press run after selecting the program name, the desired result will be,

The flotation of the table is,