How do I calculate the average voltage of this wave?

Question

I need to calculate the average voltage of the wave below, the DC voltage is 120 V , the curve starts at 21.65 degrees and the peak voltage is 325.269 V. The solution is 221.3 V. This is from a uncontrolled full-wave bridge rectifier with RE load, the source is 230V rms at 50Hz, R=2 ohm and E=120V. I tried calculating the average voltage of the rectified sine wave and subtracting the average voltage of the portions before the curve starts.

Answer 1

Here's a more mathematical approach to your problem. I understand your graph as the flat parts are at 120V. We can divide the area of the half period into a DC part before and after the sine wave (green) when the sine voltage is higher than 120V (blue).

We start by finding the angles, \$\theta_1\$ by setting the sine voltage equal to the dc voltage and \$\theta_2\$ by symmetry:

\begin{align} V_{dc} = \sqrt{2} V_{RMS} * sin(\theta_1) \rightarrow \theta_1 &= sin^{-1} \left( \frac{V_{dc}}{\sqrt{2} V_{RMS}} \right) \\\\ \theta_2 &= \pi - \theta_1 \end{align}

Using \$V_{dc}=120V\$ and \$V_{ac}=230V\$, we get the values \$\theta_1 = 21.649^o \$ and \$\theta_2 = 158.35^o\$. We then calculate the average voltage from the basic formula:

\begin{align} V_{avg} &= \frac{1}{T_p} \int^{T_p} v(t) d \theta \\\\ &= \frac{1}{\pi} \left[ \int_0^{\theta_1} V_{dc} d \theta + \int_{\theta_1}^{\theta_2} \sqrt{2}V_{ac}sin(\theta) d \theta + \int_{\theta_2}^\pi V_{dc} d \theta \right] \\\\ &= \frac{1}{\pi} \left[ V_{dc} \theta \bigg\rvert_0^{\theta_1} - \sqrt{2}V_{ac}cos(\theta) \bigg\rvert_{\theta_1}^{\theta_2} + V_{dc} \theta \bigg\rvert_{\theta_2}^\pi \right] \\\\ &= \frac{1}{\pi} \bigg[ V_{dc} (\theta_1 + \pi - \theta_2) - \sqrt{2}V_{ac}\Bigr(cos(\theta_2) -cos(\theta_1)\Bigr) \bigg] \\\\ &= \underline{\underline{221.33V}} \end{align}

Answer 2

Vmax = 323 = 325Vp -2V for Vf=1V@ 150A rated diodes.
Vmin = 120V dc

Therefore Vpp = 323-120 = 203Vpp or 101.5Vp

Vacg = Vdc (min) + Vp = 120V + 101.5 = 221.5 ( close enuf for gov't work.)