Chapter 20, Problem 6P

To determine

To calculate: The thermocline depth and the flux across the interface by the use of a cubic spline fit where $k=0.02\text{cal}/\text{s}\cdot \text{cm}\cdot \text{°C}$ and heat flux from the surface can be calculated from Fourier’s law as $J=-k\frac{dT}{dz}\text{and}\frac{d^{2}T}{d{z}^2}=0$.The table is given as,

Depth, m	0	0.5	1.0	1.5	2.0	2.5	3.0
Temperature, Celsius	70	68	55	22	13	11	10

The provided graph shows the relationship between depth and temperature as,

Answer to Problem 6P

Solution:

The value of thermocline depth and the flux across the interface is $-72.18°\text{C}/\text{m}\text{and}0.014436\text{cal}/{\text{cm}}^{\text{2}}\text{s}$ respectively.

Explanation of Solution

Given Information:

The table is given as,

Depth, m	0	0.5	1.0	1.5	2.0	2.5	3.0
Temperature, Celsius	70	68	55	22	13	11	10

The provided graph shows the relationship between depth and temperature as,

Calculation:

Consider the Fourier’s law,

$J=-k\frac{dT}{dz}$ …… (1)

The value of $k=0.02\text{cal}/\text{s}\cdot \text{cm}\cdot \text{°C}$ is given. Therefore, it is required to calculate $\frac{dT}{dz}$.

From the graph, this can be interpreted that curve has zero slope at $z=1.2\text{m}$.

Since, the cubic spline fit is required, so this problem can be solved by the Excel VBA(Visual Basic for applications). The steps are,

Step 1. Insert the data in excel as shown below,

Step 2. Press ALT+F11 and write the code as shown below,

OptionExplicit

SubSplines()

'nop and u is declared as integer type variable.

Dim u AsInteger,nopAsInteger

'arrayArr_x, Arr_y, fder, js, fder, sder and ms is declare as double type variable.

DimArr_x(100)AsDouble,Arr_y(100)AsDouble,jsAsDouble,msAsDouble

DimfderAsDouble,sderAsDouble

'rp is declared as variant.

DimrpAsVariant

'5th row is selected.

Range("a5").Select

'store the value of 5th row in nop.

nop=ActiveCell.Row

'last shell of excel is selected.

Selection.End(xlDown).Select

'store the value of last shell in nop.

nop=ActiveCell.Row-nop

'5th row is selected.

Range("a5").Select

'for loop is run uptonop.

For u =0Tonop

'store the value in array x.

Arr_x(u)=ActiveCell.Value

ActiveCell.Offset(0,1).Select

'store the value in array y.

Arr_y(u)=ActiveCell.Value

ActiveCell.Offset(1,-1).Select

'u value is increamented.

Next u

'5th row of column is selected.

Range("c5").Select

Range("c5:d1005").ClearContents

'for loop is run uptonop.

For u =0Tonop

'spline function is call.

CallSpline(Arr_x(),Arr_y(),nop,Arr_x(u),ms,fder,sder)

'first derivative value is stored.

ActiveCell.Value=fder

ActiveCell.Offset(0,1).Select

'second derivative value is stored.

ActiveCell.Value=sder

ActiveCell.Offset(1,-1).Select

'value is incremented by 1.

Next u

Do

'thismsg is asked from user.

rp=MsgBox("Do you want to interpolate?",vbYesNo)

'if condition is to check the value.

Ifrp=vbNoThenExitDo

js=InputBox("z = ")

'Spline function is called

CallSpline(Arr_x(),Arr_y(),nop,js,ms,fder,sder)

MsgBox"For z = "&js&Chr(13)&"T = "&ms&Chr(13)& _

"dT/dz = "&fder&Chr(13)&"d2T/dz2 = "&sder

'loop is ended.

Loop

'function is ended.

EndSub

'spline function is defined.

SubSpline(Arr_x,Arr_y,nop,js,ms,fder,sder)

'arraye,f,g,r and d2x is declared as double type

Dim e(100)AsDouble, f(100)AsDouble, g(100)AsDouble, r(100)AsDouble, d2x(100)AsDouble

'tridiag function is called.

CallTridiag(Arr_x,Arr_y,nop, e, f, g, r)

'decomp function is called.

CallDecomp(e(), f(), g(),nop-1)

'Substite function is called.

CallSubstit(e(), f(), g(), r(),nop-1, d2x())

'interpolation function is abbreviated as interpol.

CallInterpol(Arr_x,Arr_y,nop, d2x(),js,ms,fder,sder)

'function is ended.

EndSub

'Tridiag definition is given.

SubTridiag(Arr_x,Arr_y,nop, e, f, g, r)

'u is declare the variable as integer type.

Dim u AsInteger

'f(1),g(1) and r(1)value is calculated.

f(1)=2*(Arr_x(2)-Arr_x(0))

g(1)=Arr_x(2)-Arr_x(1)

r(1)=6/(Arr_x(2)-Arr_x(1))*(Arr_y(2)-Arr_y(1))

r(1)= r(1)+6/(Arr_x(1)-Arr_x(0))*(Arr_y(0)-Arr_y(1))

'for loop is run upto nop-2.

For u =2Tonop-2

e(u)=Arr_x(u)-Arr_x(u -1)

f(u)=2*(Arr_x(u +1)-Arr_x(u -1))

g(u)=Arr_x(u +1)-Arr_x(u)

r(u)=6/(Arr_x(u +1)-Arr_x(u))*(Arr_y(u +1)-Arr_y(u))

r(u)= r(u)+6/(Arr_x(u)-Arr_x(u -1))*(Arr_y(u -1)-Arr_y(u))

'value is incremented by 1.

Next u

'e(nop-1)value is calculated.

e(nop-1)=Arr_x(nop-1)-Arr_x(nop-2)

'f(nop-1)value is calculated.

f(nop-1)=2*(Arr_x(nop)-Arr_x(nop-2))

'r(nop-1)value is calculated.

r(nop-1)=6/(Arr_x(nop)-Arr_x(nop-1))*(Arr_y(nop)-Arr_y(nop-1))

r(nop-1)= r(nop-1)+6/(Arr_x(nop-1)-Arr_x(nop-2))*(Arr_y(nop-2)-Arr_y(nop-1))

'function is ended.

EndSub

'Interpol is defined.

SubInterpol(Arr_x,Arr_y,nop, d2x,js,ms,fder,sder)

'u and flag is declared the variable as integer type

Dim u AsInteger, flag AsInteger

'variable C1,C2,C3 and C4 is declared as double type

Dim C1 AsDouble, C2 AsDouble, C3 AsDouble, C4 AsDouble

'variable T1,T2,T3 and T4 is declared as double type

Dim T1 AsDouble, T2 AsDouble, T3 AsDouble, T4 AsDouble

'flag and u is initialised as 0.

flag=0

u =1

Do

'if statement is to check the condition.

Ifjs>=Arr_x(u -1)Andjs<=Arr_x(u)Then

'C1 is calculated.

C1 =d2x(u -1)/6/(Arr_x(u)-Arr_x(u -1))

'C2 is calculated.

C2 =d2x(u)/6/(Arr_x(u)-Arr_x(u -1))

'C3 is calculated.

C3 =Arr_y(u -1)/(Arr_x(u)-Arr_x(u -1))- d2x(u -1)*(Arr_x(u)-Arr_x(u -1))/6

'C4 is calculated.

C4 =Arr_y(u)/(Arr_x(u)-Arr_x(u -1))- d2x(u)*(Arr_x(u)-Arr_x(u -1))/6

'T1 is calculated.

T1 = C1 *(Arr_x(u)-js)^3

'T2 is calculated.

T2 = C2 *(js-Arr_x(u -1))^3

'T3 is calculated.

T3 = C3 *(Arr_x(u)-js)

'T4 is calculated.

T4 = C4 *(js-Arr_x(u -1))

'ms is calculated.

ms= T1 + T2 + T3 + T4

'T1 is calculated.

T1 =-3* C1 *(Arr_x(u)-js)^2

'T2 is calculated.

T2 =3* C2 *(js-Arr_x(u -1))^2

'T3 is calculated.

T3 =-C3

'T4 is calculated.

T4 = C4

'fder value is calculated.

fder= T1 + T2 + T3 + T4

'T1 and T2 is calculated.

T1 =6* C1 *(Arr_x(u)-js)

T2 =6* C2 *(js-Arr_x(u -1))

'second derivative sder is calculated.

sder= T1 + T2

'flag is equal to 1

flag=1

Else

u = u +1

EndIf

'if statement is to check the calculated value.

If u =nop+1Or flag =1ThenExitDo

Loop

'flag is checked

If flag =0Then

'if condition is fulfil then output the msg.

MsgBox"outside range"

'loop is ended.

End

'end statement.

EndIf

'function is ended.

EndSub

'Decomp function definition is given.

SubDecomp(e, f, g,nop)

'k is declared as integer

Dim k AsInteger

'for loop is run uptonop.

For k =2Tonop

e(k)= e(k)/ f(k -1)

f(k)= f(k)- e(k)* g(k -1)

'k is increased by 1.

Next k

'function is ended.

EndSub

'substit definition is given.

SubSubstit(e, f, g, r,nop,Arr_x)

'k is declared as integer type.

Dim k AsInteger

'for loop is run uptonop.

For k =2Tonop

r(k)= r(k)- e(k)* r(k -1)

'k is increased.

Next k

'Arr_x(nop)value is calculated.

Arr_x(nop)= r(nop)/ f(nop)

'for loop is run uptoArr_x(k)

For k =nop-1To1Step-1

Arr_x(k)=(r(k)- g(k)*Arr_x(k +1))/ f(k)

' k is increased by 1

Next k

' function is ended.

EndSub

PrivateSubWorksheet_SelectionChange(ByVal Target As Range)

EndSub

Step 3. Press RUN then this dialog box appears.

Step 4. Enter the value of z.

Step 5. This output will appear.

Thus, the value of $\frac{dT}{dz}$ is $-72.18$. Substitute this value in equation (1),

$\begin{array}{c}J=-\frac{0.02\left(-72.18\right)}{100}\\ =0.014436\text{cal}/{\text{cm}}^{\text{2}}\text{s}\end{array}$

Hence, the value of thermocline depth and the flux across the interface is $-72.18°\text{C}/\text{m}\text{and}0.014436\text{cal}/{\text{cm}}^{\text{2}}\text{s}$ respectively.