24 AC to 32 DC. How?
How did he calculate the value of the capacitor in order to boost the voltage to 32 V DC from 24 V DC (after rectification)?


If the input is really 24VAC, after rectification the DC peak voltage at the capacitor will be given by:

$$V_{cap} = 14VAC \cdot \sqrt2 - 2\cdot (V_{DIODE})\approx 32VDC$$

Provided that the load connected to the capacitor does not draw too much power, down to the point where it would considerably discharge, the average output voltage would be slightly lower than 32VDC.

The capacitor value depends on the load, the allowed voltage drop between cycles, and the line frequency according to the following.

$$C\ge\dfrac{P_{max}\left(\dfrac{1}{2} + \dfrac{\arcsin\left(\dfrac{V_{min}}{V_{peak}}\right)}{\pi}\right)}{(V_{peak}^2-V_{min}^2)\cdot f_{line}}$$

Where: 

\$V_{min}\$ is the minimum allowed voltage during discharge;

\$V_{peak}\$ is the peak voltage across the capacitor (e.g. 32VDC);

\$f_{line}\$ is the frequency of the input AC voltage (e.g. 60Hz);


  I don't have an oscilloscope to see the waveform after rectification and filtration but, is there a better way to obtain 32 DC from 24 AC regarding the output waveform?


If by that you mean to measure the voltage, you can simply measure the RMS voltage across the capacitor. It should give something close to 32VDC. 

EDIT#1

The equation proposed above assumes that no input cycles are skipped.

EDIT#2

Just for clarification, here is the derivation. Assume thatthe voltage across the capacitor is the following:



The capacitor formula is given by:

$$\Delta I = C \dfrac{dV}{dt}$$

By replacing \$\Delta I\$ with \$\dfrac{P}{\Delta V}\$ and integrating the equation, the follwing is obtained:

$$P\cdot t = \dfrac{C}{2} \cdot \left(V_{peak}^2 - V_{min}^2\right)$$

where \$t=t_1 + t_2\$ in the above picture.

Since \$t_1\$ is a quarter of period, its radian equivalent is given by:

$$t_1 = \dfrac{2\pi}{4}\cdot \dfrac{1}{f}=\dfrac{\pi}{2\cdot f_{line}}$$

\$t_2\$ on the other hand can be calculated through the following way. It it is not so clear to see it, assume for the moment that \$V_{peak} = V_{min}\$ in the above image, it would mean that \$t_2=\dfrac{\pi}{2}(90°)\$. This in turn could be used to reverse calculate and check the amplitude ratio of both voltages according to:

$$\arcsin\left( \dfrac{V_{min}}{V_{peak}} \right)=\arcsin(1)=90°$$

That means that the following holds true

$$t_2 = \arcsin\left(\dfrac{V_{min}}{V_{peak}}\right)\cdot \dfrac{1}{f_{line}} $$

Rearranging everything you end up with the formula shown in the original answer.