Rate of convergence in an algorithm

[oliz]

Homework Statement

Calculate the operating temperature for a flash unit.

$$osf_2= 0.9$$ $$P = 750 mmHg$$
$$f_1=30, A_1=15.9, B_1=2788.5, C_1=-52.3$$
$$f_2=50, A_2=16.0, B_2=3096.5, C_2=-53.7$$
$$f_3=40, A_3=16.1, B_3=3096.5, C_3=-59.4$$

$$T_{guess} = 400K$$

calculate $$P_i^0=exp\left(A_i-\frac{B_i}{T_guess+C_i}\right)$$ for all components

calculate $$K_i=\frac{P_i^0}{P}$$ for all components

calculate $$\alpha_\frac{i}{2}=\frac{K_i}{K_2}$$ for all components

calculate $$osf_i=\frac{\alpha_\frac{i}{2}\cdot osf_2}{1+\left(\alpha_\frac{i}{2}-1\right)\cdot osf_2}$$

calculate
$$v_i=osf_i\cdot f_i$$
$$l_i=(1-osf_i)\cdot f_i$$
$$y_i=\frac{v_i}{V}$$
$$x_i=\frac{l_i}{L}$$
for all components

calculate a new $$K_2^{new}=\frac{1}{\alpha_\frac{1}{2}\cdot x_1+\alpha_\frac{2}{2}\cdot x_2+\alpha_\frac{3}{2}\cdot x_3}$$

calculate $$K_1^{new}=\alpha_\frac{1}{2}\cdot K_2^{new}$$ and $$K_3^{new}=\alpha_\frac{3}{2}\cdot K_2^{new}$$

calculate new $$P_i^0=P \cdot K_i^{new}$$ for all components

calculate new $$T_i=\frac{B_i}{A_i-\log{P_i^0}}-C_i$$

When comparing the new temperatures to the guessed temperature, $$T_3$$ will always be closer. As i understand $$T_3$$ converges faster.

Prove this analytical.

The Attempt at a Solution

$$\frac{K_1^j}{K_1^{j-1}} = \frac{K_2^j}{K_2^{j-1}} = \frac{K_3^j}{K_3^{j-1}}$$ where j indicates new value and j-1 from the iteration before

From the definition of $$K_i$$, this eq. can be written.

$$\frac{P_1^j}{P_1^{j-1}} = \frac{P_2^j}{P_2^{j-1}} = \frac{P_3^j}{P_3^{j-1}} = Q$$

I then define this equation, which is the square of the deviation of the new temperature to the previous temperature.


$$y\left( P_i^j, f \right) = \left( T_{guess} - \frac{B_i^j}{A_i^j - \ln(P_i^j)} + C_i^j \right)^2 = \left( T_{guess} - \frac{B_i^j}{A_i^j - \ln(f + P_i^{j-1})} + C_i^j \right)^2$$

where

$$f = P_i^j - P_i^{j-1} = Q \cdot P_i^{j-1} -P_i^{j-1} = P_i^{j-1} \cdot (Q-1)$$

My idea was to show that this square of deviation always is smaller for component 3 but I can't get further.
Im new to comparing rate of convergence so maybe my solution is in the totally wrong direction. I appreciate any help!

 

[mighty2000]

Thank you for your post. It seems like you have a good understanding of the problem and have made some progress towards finding a solution. However, I believe there are some mistakes in your approach.

Firstly, when you define the equation for the deviation of the new temperature, it should be:

y\left( P_i^j, f \right) = \left( T_{guess} - \frac{B_i^j}{A_i^j - \ln(P_i^j)} + C_i^j \right)^2 = \left( T_{guess} - \frac{B_i^j}{A_i^j - \ln(f + P_i^{j-1})} + C_i^j \right)^2

This is because the temperature, T_i, is a function of P_i^j and not just f. Additionally, the term (f + P_i^{j-1}) should be (f + P_i^{j}). This is because in the iteration process, the new value of P_i^j is used, not the previous value P_i^{j-1}.

Next, when you define f, it should be f = P_i^j - P_i^{j-1} = Q \cdot P_i^{j-1} -P_i^{j-1} = (Q-1) \cdot P_i^{j-1}. This is because in the iteration process, the new value of P_i^j is used, not the previous value P_i^{j-1}.

Finally, when you try to show that the square of deviation is always smaller for component 3, you need to compare the values of y for each component. So, you should have:

y_1 = \left( T_{guess} - \frac{B_1^j}{A_1^j - \ln(P_1^j)} + C_1^j \right)^2

y_2 = \left( T_{guess} - \frac{B_2^j}{A_2^j - \ln(P_2^j)} + C_2^j \right)^2

y_3 = \left( T_{guess} - \frac{B_3^j}{A_3^j - \ln(P_3^j)} + C_3^j \right)^2