Chapter 27, Problem 1P

A steady-state heat balance for a rod can be represented as

$\frac{d^{2}T}{d{x}^2}-0.15T=0T$

Obtain an analytical solution for a 10-m rod with $\left(0\right)=240\text{ and }T\left(10\right)=150$.

To determine

To calculate: An analytical solution for a 10-m rod with $T\left(0\right)=240\text{and}T\left(10\right)=150$, where a steady-state heat balance for a rod is represented as $\frac{d^{2}T}{d{x}^2}-0.15T=0$.

Answer to Problem 1P

Solution:

The solution is $T\left(x\right)=3.016944\times e^{\sqrt{0.15}x}+236.9831\times e^{-\sqrt{0.15}x}$.

Explanation of Solution

Given Information:

A steady-state heat balance for a rod is represented as $\frac{d^{2}T}{d{x}^2}-0.15T=0$ where the rod is 10-m long with $T\left(0\right)=240\text{and}T\left(10\right)=150$.

Calculation:

Consider the steady-state heat balance of rod given as $\frac{d^{2}T}{d{x}^2}-0.15T=0$.

Suppose $T=e^{\lambda x}$ be the solution of the above differential equation. Then,

$\begin{array}{l}{T}^{\prime }=\lambda e^{\lambda x}\\ T^{″}={\lambda }^2{e}^{\lambda x}\end{array}$ $$

Substitute the values of $T^{″}\text{and}T$ in the given differential equation.

$\begin{array}{c}\frac{d^2{e}^{\lambda x}}{d{x}^2}-0.15e^{\lambda x}=0\\ {\lambda }^2{e}^{\lambda x}-0.15e^{\lambda x}=0\\ e^{\lambda x}\left({\lambda }^2-0.15\right)=0\end{array}$

Since, $e^{\lambda x}\ne 0$. Then,

$\begin{array}{c}{\lambda }^2-0.15=0\\ {\lambda }^2=0.15\\ \lambda =\pm \sqrt{0.15}\end{array}$

Therefore, the general solution of the differential equation is,

$T\left(x\right)=A\times e^{\sqrt{0.15}x}+B\times e^{-\sqrt{0.15}x}$

The constant $A$ and $B$ can be evaluated by substituting each of the boundary conditions to generate two equations with two unknown constants.

Firstly, $T\left(0\right)=240$. Thus,

$T\left(0\right)=A\times e^{\sqrt{0.15}\times 0}+B\times e^{-\sqrt{0.15}\times 0}$

Solve further to get,

$240=A+B$.. .. .. (1)

The second value is $T\left(10\right)=150$. Thus,

$T\left(10\right)=A\times e^{\sqrt{0.15}\times 10}+B\times e^{-\sqrt{0.15}\times 10}$

Solve further to get,

$150=48.08563A+0.0207962B$.. .. .. (2)

Now to evaluate equations (1)and(2)

$\begin{array}{c}240=A+B\\ 150=48.08563A+0.0207962B\end{array}$

Multiply equation(1) by $48.0856$ on both sides then subtract equation (2) from equation(1).

Then,

$A=3.016944\text{and}B=236.9831$

Now put the value of $A$ and $B$ in general solution of the equation.

$T\left(x\right)=3.016944\times e^{\sqrt{0.15}x}+236.9831\times e^{-\sqrt{0.15}x}$

Substitute the values of $x=0,\dots ,10$

At $x=0$

$\begin{array}{c}T\left(x\right)=3.016944\times e^{\sqrt{0.15}\cdot 0}+236.9831\times e^{-\sqrt{0.15}\cdot 0}\\ =240\end{array}$

At $x=1$

$\begin{array}{c}T\left(1\right)=3.016944\times e^{\sqrt{0.15}\cdot 1}+236.9831\times e^{-\sqrt{0.15}\cdot 1}\\ =165.329\end{array}$

At $x=2$

$\begin{array}{c}T\left(2\right)=3.016944\times e^{\sqrt{0.15}\cdot 2}+236.9831\times e^{-\sqrt{0.15}\cdot 2}\\ =115.769\end{array}$

At $x=3$

$\begin{array}{c}T\left(3\right)=3.016944\times e^{\sqrt{0.15}\times 3}+236.9831\times e^{-\sqrt{0.15}\times 3}\\ =83.7924\end{array}$

At $x=4$

$\begin{array}{c}T\left(4\right)=3.016944\times e^{\sqrt{0.15}\cdot 4}+236.9831\times e^{-\sqrt{0.15}\cdot 4}\\ =64.5426\end{array}$

At $x=5$

$\begin{array}{c}T\left(5\right)=3.016944\times e^{\sqrt{0.15}\cdot 5}+236.9831\times e^{-\sqrt{0.15}\cdot 5}\\ =55.0957\end{array}$

At $x=6$

$\begin{array}{c}T\left(6\right)=3.016944\times e^{\sqrt{0.15}\cdot 6}+236.9831\times e^{-\sqrt{0.15}\cdot 6}\\ =54.0171\end{array}$

At $x=7$

$\begin{array}{c}T\left(7\right)=3.016944\times e^{\sqrt{0.15}\cdot 7}+236.9831\times e^{-\sqrt{0.15}\cdot 7}\\ =61.1428\end{array}$

At $x=8$

$\begin{array}{c}T\left(8\right)=3.016944\times e^{\sqrt{0.15}\cdot 8}+236.9831\times e^{-\sqrt{0.15}\cdot 8}\\ =77.5552\end{array}$

At $x=9$

$\begin{array}{c}T\left(9\right)=3.016944\times e^{\sqrt{0.15}\cdot 9}+236.9831\times e^{-\sqrt{0.15}\cdot 9}\\ =105.747\end{array}$

At $x=10$

$\begin{array}{c}T\left(10\right)=3.016944\times e^{\sqrt{0.15}\cdot 10}+236.9831\times e^{-\sqrt{0.15}\cdot 10}\\ =150\end{array}$

The values of $T$ generated corresponding to x are shown in the below table:

$\begin{array}{cc}x& T\\ 0& 240\\ 1& 165.329\\ 2& 115.729\\ 3& 83.7924\\ 4& 64.5426\\ 5& 55.0957\\ 6& 54.0171\\ 7& 61.1428\\ 8& 77.5552\\ 9& 105.747\\ 10& 150\end{array}$