4 Engineering Example 2

4.1 Pressure in an ideal multi-component mixture

Introduction

An ideal multi-component mixture consists of

1.  $n$ -pentane (5%)

2.  $n$ -hextane (15%)

3.  $n$ -heptane (50%)

4.  $n$ -octane (30%)

In general, the total pressure, $P$ (Pa) of an ideal four-component mixture is related to the boiling point, $T$ (K) through the formula:

$P=x_1{p}_1^{\ast }+x_2{p}_2^{\ast }+x_3{p}_3^{\ast }+x_4{p}_4^{\ast }$

where, for component $i$ , the mole fraction is $x_i$ and the vapour pressure is $p_i^{\ast }$ , given by the formula:

$p_i^{\ast }=\text{exp}\left\{A_i-\frac{B_i}{\left(T+C_i\right)}\right\}i=1,2,3,4$

Here $p_i^{\ast }$ is in mm Hg (1 mm Hg $=133.32$ Pa), $T$ is the absolute temperature (K) and the constants $A_i,B_i$ and $C_i$ are given in the table below.

Problem 1

For the liquid compositions, $x_i$ given in the table above, plot a graph of the total pressure, $P$ (Pa) against temperature (K) over the range $250$ to $500$ K.

Solution

$p_i^{\ast }=\text{exp}\left(A_i-\frac{B_i}{T+C_i}\right)$ , expressed in millimetres of mercury, and so it is 133.32 times that in pascals. Therefore, expressed in pascals, we have

$P=133.32{\sum }_{i=1}^4{x}_i\text{exp}\left(A_i-\frac{B_i}{T+C_i}\right)$

Plotting this from $T=250$ to $500$ gives the following graph

Figure 15

Problem 2

Using the Newton-Raphson method, solve the equations to find the boiling points at total pressures of 1, 2, 5 and 10 bars. Show the sequence of iterations and perform sufficient calculations for convergence to three significant figures. Display these solutions on the graph of the total pressure, $P$ (Pa) against temperature $T$ (K).

Solution

We wish to find $T$ when $P=1,2,5$ and $10$ bars, that is, $10^5,2\times 10^5,5\times 10^5$ and $10\times 10^5$ Pa.

Reading crude approximations to $T$ from the graph gives a starting point for the Newton-Raphson process. We see that for $10^5,2\times 10^5,5\times 10^5$ and $10\times 10^5$ Pa, temperature $T$ is roughly $365,375,460$ and $485$ degrees K, respectively, so we shall use these values as the start of the iteration.

In this case it is easy to calculate the derivative of $P$ with respect to $T$ exactly, rather than numerically, giving

$P^{\prime }\left(T\right)=133.32{\sum }_{i=1}^4{x}_i\text{exp}\left(A_i-\frac{B_i}{T+C_i}\right)\times \left(\frac{B_i}{{\left(T+C_i\right)}^2}\right)$

Therefore to solve the equation $P\left(T\right)=y$ , we set $T_0$ to be the starting value above and use the iteration

$T_{n+1}=T_n-\frac{P\left(T_n\right)-y}{P^{\prime }\left(T_n\right)}$

For $y=100000$ this gives the iterations

We conclude that, to three significant figures $T=363^{\circ }$ K when $P=100000$ Pa.

For $y=200000$ this gives the iterations

We conclude that, to three significant figures $T=389^{\circ }$ K when $P=200000$ Pa.

For $y=500000$ this gives the iterations

We conclude that, to three significant figures $T=430^{\circ }$ K when $P=500000$ Pa.

For $y=1000000$ this gives the iterations

We conclude that, to three significant figures $T=469^{\circ }$ K when $P=1000000$ Pa.

4.2 An approximate Newton-Raphson method

The Newton-Raphson method is an excellent way of approximating zeros of a function, but it requires you to know the derivative of $f$ . Sometimes it is undesirable, or simply impossible, to work out the derivative of a function and here we show a way of getting around this.

We approximate the derivative of $f$ . From Section 31.3 we know that

$f^{\prime }\left(x\right)\approx \frac{f\left(x+h\right)-f\left(x\right)}{h}$

is a one-sided (or forward) approximation to $f^{\prime }$ and another one, using a central difference, is

$f^{\prime }\left(x\right)\approx \frac{f\left(x+h\right)-f\left(x-h\right)}{2h}.$

The advantage of the forward difference is that only one extra $f$ -value has to be computed. If $f$ is especially complicated then this can be a considerable saving when compared with the central difference which requires two extra evaluations of $f$ . The central difference does have the advantage, as we saw when we looked at truncation errors, of being a more accurate approximation to $f^{\prime }$ .

The spreadsheet program Microsoft Excel has a built in “solver" command which can use Newton’s method. (It may be necessary to use the “Add in" feature of Excel to access the solver.) In reality Excel has no way of working out the derivative of the function and must approximate it. Excel gives you the option of using a forward or central difference to estimate $f^{\prime }$ .

We now reconsider the problem we met in Examples 24 to 26.

Example 27

We know that the single positive zero of $f\left(x\right)=xtanh\left(\frac{1}{2}x\right)-1$   lies between $1.5$ and $2$ . Use the Newton-Raphson method, with an approximation to $f^{\prime }$ , to approximate the zero of $f$ .

Solution

There is no requirement for $f^{\prime }$ this time, but the nature of this method is such that we will resort to a computer straight away. Let us choose $h=0.1$  in our approximations to the derivative.
Using the one-sided difference to approximate $f^{\prime }\left(x\right)$ we obtain this sequence of results from the spreadsheet program:

$\begin{array}{ccccc}\hfill n\hfill & \hfill x_n\hfill & \hfill f\left(x_n\right)\hfill & \hfill \frac{f\left(x+h\right)-f\left(x\right)}{h}\hfill & \hfill x_{n+1}\hfill \\ ̲& ̲& ̲& ̲& ̲\\ \hfill 0\hfill & \hfill 1.75\hfill & \hfill 0.231835\hfill & \hfill 1.154355\hfill & \hfill 1.549165\hfill \\ \hfill 1\hfill & \hfill 1.549165\hfill & \hfill 0.006316\hfill & \hfill 1.110860\hfill & \hfill 1.543479\hfill \\ \hfill 2\hfill & \hfill 1.543479\hfill & \hfill 8.16E-05\hfill & \hfill 1.109359\hfill & \hfill 1.543406\hfill \\ \hfill 3\hfill & \hfill 1.543406\hfill & \hfill 1.01E-06\hfill & \hfill 1.109339\hfill & \hfill 1.543405\hfill \\ \hfill 4\hfill & \hfill 1.543405\hfill & \hfill 1.24E-08\hfill & \hfill 1.109339\hfill & \hfill 1.543405\hfill \\ \hfill 5\hfill & \hfill 1.543405\hfill & \hfill 1.53E-10\hfill & \hfill 1.109339\hfill & \hfill 1.543405\hfill \\ \hfill 6\hfill & \hfill 1.543405\hfill & \hfill 1.89E-12\hfill & \hfill 1.109339\hfill & \hfill 1.543405\hfill \\ \hfill 7\hfill & \hfill 1.543405\hfill & \hfill 2.31E-14\hfill & \hfill 1.109339\hfill & \hfill 1.543405\hfill \\ \hfill 8\hfill & \hfill 1.543405\hfill & \hfill 0\hfill & \hfill 1.109339\hfill & \hfill 1.543405\hfill \end{array}$

And using the (more accurate) central difference gives

$\begin{array}{ccccc}\hfill n\hfill & \hfill x_n\hfill & \hfill f\left(x_n\right)\hfill & \hfill \frac{f\left(x+h\right)-f\left(x-h\right)}{2h}\hfill & \hfill x_{n+1}\hfill \\ ̲& ̲& ̲& ̲& ̲\\ \hfill 0\hfill & \hfill 1.75\hfill & \hfill 0.231835\hfill & \hfill 1.144649\hfill & \hfill 1.547462\hfill \\ \hfill 1\hfill & \hfill 1.547462\hfill & \hfill 0.004448\hfill & \hfill 1.095994\hfill & \hfill 1.543404\hfill \\ \hfill 2\hfill & \hfill 1.543404\hfill & \hfill -1E-06\hfill & \hfill 1.094818\hfill & \hfill 1.543405\hfill \\ \hfill 3\hfill & \hfill 1.543405\hfill & \hfill 7.95E-10\hfill & \hfill 1.094819\hfill & \hfill 1.543405\hfill \\ \hfill 4\hfill & \hfill 1.543405\hfill & \hfill -6.1E-13\hfill & \hfill 1.094819\hfill & \hfill 1.543405\hfill \\ \hfill 5\hfill & \hfill 1.543405\hfill & \hfill 0\hfill & \hfill 1.094819\hfill & \hfill 1.543405\hfill \end{array}$

We see that each of these approaches leads to the same value (1.543405) that we found with the Newton-Raphson method.

Task!

Use a spreadsheet to recompute the approximations shown in Example 24, for the following values of $h$ :

$h=0.001,0.00001,0.000001.$

Answer

Exercises

1. It is given that the function

   $f\left(x\right)=x^3+2x+8$

   has a single negative zero.

   1. Find two integers $a$ and $b$ which bracket the zero of $f$ .
   2. Perform one iteration of the bisection method so as to halve the size of the bracketing interval.
2. Consider a simple electronic circuit with an input voltage of 2.0 V, a resistor of resistance 1000 $\Omega$ and a diode. It can be shown that the voltage across the diode can be found as the single positive zero of

   $f\left(x\right)=1\times 10^{-14}exp\left(\frac{x}{0.026}\right)-\frac{2-x}{1000}.$

   Use one iteration of the Newton-Raphson method, and an initial value of $x_0=0.75$ to show that

   $x_1=0.724983$

   and then work out a second iteration.

3. It is often necessary to find the zeros of polynomials as part of an analysis of transfer functions. The function

   $f\left(x\right)=x^3+5x-4$

   has a single zero near $x_0=1$ . Use this value of $x_0$ in an implementation of the Newton-Raphson method performing two iterations. (Work to at least 6 decimal place accuracy.)

4. The smallest positive zero of

   $f\left(x\right)=xtan\left(x\right)+1$

   is a measure of how quickly certain evanescent water waves decay, and its value, $x_0$ , is near $3$ . Use the forward difference

   $\frac{f\left(3.01\right)-f\left(3\right)}{0.01}$

   to estimate $f^{\prime }\left(3\right)$ and use this value in an approximate version of the Newton-Raphson method to derive one improvement on $x_0$ .

Answer