Tutorial 5 - 2010 - Solution/Question 3

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Question 3 [1]

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1. . Write the Newton-Raphson iteration formula that you would use to solve this nonlinear equation.
2. . Apply 3 iterations of this formula, also starting from :math:`T = 380` K, and calculate the error tolerances.

Solution

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1. . The Newton-Raphson algorithm is given on slide 15-17 of the *Section C: Nonlinear Algebraic Equations* slide set.

To apply the Newton-Raphson method we must first calculate the derivative of our function:

.. math::

f'(T) = - 0.26 + 3.38\text{x}10^{-3}T - \frac{1.5\text{x}10^{5}}{T^{2}}

Therefore the Newton-Raphson iteration formula we use is the following:

.. math::

T^{(k+1)} = T^{(k)} - \frac{f(T^{(k)})}{f'(T^{(k)})} = T^{(k)} - \frac{- 24097 - 0.26T + 1.69\text{x}10^{-3}T^{2} + \frac{1.5\text{x}10^{5}}{T} - (-23505)}{- 0.26 + 3.38\text{x}10^{-3}T - \frac{1.5\text{x}10^{5}}{T^{2}}}

1. .

We start with :math:`T^{(0)} = 380 K`:

.. math::

$$\begin{array}{rl}\mathrm{\Delta }{H}_r^0\left(T^{(0)}\right)& =-24097-0.26\left(T^{(0)}\right)+1.69\text{x}{10}^{-3}{\left(T^{(0)}\right)}^2+\frac{1.5\text{x}{10}^5}{\left(T^{(0)}\right)}\\ \mathrm{\Delta }{H}_r^0\left(T^{(0)}\right)& =-24097-0.26\left(380\right)+1.69\text{x}{10}^{-3}{\left(380\right)}^2+\frac{1.5\text{x}{10}^5}{\left(380\right)}\\ \mathrm{\Delta }{H}_r^0\left(T^{(0)}\right)& =-23557.02715\end{array}$$

*ITERATION 1*

.. math::

$$\begin{array}{rl}{T}^{(1)}& =T^{(0)}-\frac{-24097-0.26T^{(0)}+1.69\text{x}{10}^{-3}(T^{(0)}{)}^2+\frac{1.5\text{x}{10}^5}{T^{(0)}}-(-23505)}{-0.26+3.38\text{x}{10}^{-3}{T}^{(0)}-\frac{1.5\text{x}{10}^5}{(T^{(0)}{)}^2}}\\ T^{(1)}& =(380)-\frac{-24097-0.26(380)+1.69\text{x}{10}^{-3}(380{)}^2+\frac{1.5\text{x}{10}^5}{(380)}-(-23505)}{-0.26+3.38\text{x}{10}^{-3}(380)-\frac{1.5\text{x}{10}^5}{(380{)}^2}}\\ T^{(1)}& =-3237.72940\\ & \\ \mathrm{\Delta }{H}_r^0\left(T^{(1)}\right)& =-24097-0.26\left(T^{(1)}\right)+1.69\text{x}{10}^{-3}{\left(T^{(1)}\right)}^2+\frac{1.5\text{x}{10}^5}{\left(T^{(1)}\right)}\\ \mathrm{\Delta }{H}_r^0\left(T^{(1)}\right)& =-24097-0.26(-3237.72940)+1.69\text{x}{10}^{-3}{(-3237.72940)}^2+\frac{1.5\text{x}{10}^5}{(-3237.72940)}\\ \mathrm{\Delta }{H}_r^0\left(T^{(1)}\right)& =-5585.4322\end{array}$$

.. math::

$$\begin{array}{rl}{ϵ}_{tol,x}^{(1)}& =\left|\frac{T^{(1)}-T^{(0)}}{T^{(0)}}\right|<{ϵ}_{tol}?\\ {ϵ}_{tol,x}^{(1)}& =\left|\frac{(-3237.72940)-\left(380\right)}{\left(380\right)}\right|<\left({10}^{-6}\right)?\\ {ϵ}_{tol,x}^{(1)}& =9.52<\left({10}^{-6}\right)?\to \text{No, therefore keep going}.\\ & \\ {ϵ}_{tol,f}^{(1)}& =\left|\frac{\mathrm{\Delta }{H}_r^0\left(T^{(1)}\right)-\mathrm{\Delta }{H}_r^0\left(T^{(0)}\right)}{\mathrm{\Delta }{H}_r^0\left(T^{(0)}\right)}\right|<{ϵ}_{tol}?\\ {ϵ}_{tol,f}^{(1)}& =\left|\frac{(-5585.4322)-(-23557.02715)}{(-23557.02715)}\right|<\left({10}^{-6}\right)?\\ {ϵ}_{tol,f}^{(1)}& =0.763<\left({10}^{-6}\right)?\to \text{No, therefore keep going}.\end{array}$$

*ITERATION 2*

.. math::

$$\begin{array}{rl}{T}^{(2)}& =T^{(1)}-\frac{-24097-0.26T^{(1)}+1.69\text{x}{10}^{-3}(T^{(1)}{)}^2+\frac{1.5\text{x}{10}^5}{T^{(1)}}-(-23505)}{-0.26+3.38\text{x}{10}^{-3}{T}^{(1)}-\frac{1.5\text{x}{10}^5}{(T^{(1)}{)}^2}}\\ T^{(2)}& =(-3237.72940)-\frac{-24097-0.26(-3237.72940)+1.69\text{x}{10}^{-3}(-3237.72940{)}^2+\frac{1.5\text{x}{10}^5}{(-3237.72940)}-(-23505)}{-0.26+3.38\text{x}{10}^{-3}(-3237.72940)-\frac{1.5\text{x}{10}^5}{(-3237.72940{)}^2}}\\ T^{(2)}& =-1640.311\\ & \\ \mathrm{\Delta }{H}_r^0\left(T^{(2)}\right)& =-24097-0.26\left(T^{(2)}\right)+1.69\text{x}{10}^{-3}{\left(T^{(2)}\right)}^2+\frac{1.5\text{x}{10}^5}{\left(T^{(2)}\right)}\\ \mathrm{\Delta }{H}_r^0\left(T^{(2)}\right)& =-24097-0.26(-1640.311)+1.69\text{x}{10}^{-3}{(-1640.311)}^2+\frac{1.5\text{x}{10}^5}{(-1640.311)}\\ \mathrm{\Delta }{H}_r^0\left(T^{(2)}\right)& =-19214.81711\end{array}$$

.. math::

$$\begin{array}{rl}{ϵ}_{tol,x}^{(2)}& =\left|\frac{T^{(2)}-T^{(1)}}{T^{(1)}}\right|<{ϵ}_{tol}?\\ {ϵ}_{tol,x}^{(2)}& =\left|\frac{(-1640.311)-(-3237.72940)}{(-3237.72940)}\right|<\left({10}^{-6}\right)?\\ {ϵ}_{tol,x}^{(2)}& =0.493<\left({10}^{-6}\right)?\to \text{No, therefore keep going}.\\ & \\ {ϵ}_{tol,f}^{(2)}& =\left|\frac{\mathrm{\Delta }{H}_r^0\left(T^{(2)}\right)-\mathrm{\Delta }{H}_r^0\left(T^{(1)}\right)}{\mathrm{\Delta }{H}_r^0\left(T^{(1)}\right)}\right|<{ϵ}_{tol}?\\ {ϵ}_{tol,f}^{(2)}& =\left|\frac{(-19214.81711)-(-5585.4322)}{(-5585.4322)}\right|<\left({10}^{-6}\right)?\\ {ϵ}_{tol,f}^{(1)}& =2.44<\left({10}^{-6}\right)?\to \text{No, therefore keep going}.\end{array}$$

*ITERATION 3*

.. math::

$$\begin{array}{rl}{T}^{(3)}& =T^{(2)}-\frac{-24097-0.26T^{(2)}+1.69\text{x}{10}^{-3}(T^{(2)}{)}^2+\frac{1.5\text{x}{10}^5}{T^{(2)}}-(-23505)}{-0.26+3.38\text{x}{10}^{-3}{T}^{(2)}-\frac{1.5\text{x}{10}^5}{(T^{(2)}{)}^2}}\\ T^{(3)}& =(-1640.311)-\frac{-24097-0.26(-1640.311)+1.69\text{x}{10}^{-3}(-1640.311{)}^2+\frac{1.5\text{x}{10}^5}{(-1640.311)}-(-23505)}{-0.26+3.38\text{x}{10}^{-3}(-1640.311)-\frac{1.5\text{x}{10}^5}{(-1640.311{)}^2}}\\ T^{(3)}& =-908.1982\\ & \\ \mathrm{\Delta }{H}_r^0\left(T^{(3)}\right)& =-24097-0.26\left(T^{(3)}\right)+1.69\text{x}{10}^{-3}{\left(T^{(3)}\right)}^2+\frac{1.5\text{x}{10}^5}{\left(T^{(3)}\right)}\\ \mathrm{\Delta }{H}_r^0\left(T^{(3)}\right)& =-24097-0.26(-908.1982)+1.69\text{x}{10}^{-3}{(-908.1982)}^2+\frac{1.5\text{x}{10}^5}{(-908.1982)}\\ \mathrm{\Delta }{H}_r^0\left(T^{(3)}\right)& =-22632.0781\end{array}$$

.. math::

$$\begin{array}{rl}{ϵ}_{tol,x}^{(3)}& =\left|\frac{T^{(3)}-T^{(2)}}{T^{(2)}}\right|<{ϵ}_{tol}?\\ {ϵ}_{tol,x}^{(3)}& =\left|\frac{(-908.1982)-(-1640.311)}{(-1640.311)}\right|<\left({10}^{-6}\right)?\\ {ϵ}_{tol,x}^{(3)}& =0.45<\left({10}^{-6}\right)?\to \text{No, therefore keep going}.\\ & \\ {ϵ}_{tol,f}^{(3)}& =\left|\frac{\mathrm{\Delta }{H}_r^0\left(T^{(3)}\right)-\mathrm{\Delta }{H}_r^0\left(T^{(2)}\right)}{\mathrm{\Delta }{H}_r^0\left(T^{(2)}\right)}\right|<{ϵ}_{tol}?\\ {ϵ}_{tol,f}^{(3)}& =\left|\frac{(-22632.0781)-(-19214.81711)}{(-19214.81711)}\right|<\left({10}^{-6}\right)?\\ {ϵ}_{tol,f}^{(3)}& =0.178<\left({10}^{-6}\right)?\to \text{No, therefore keep going}.\end{array}$$

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