Firing angle in rectifiers and inverters.

1. Aug 16, 2011

ramox3

Hello everyone I am a newbie to electrical engineering, I have this problem I've been strugling with, no textbook seems to cover this problem..

A three-phase, 440-V, generator delivers 5400W of active power through two recti-fies, A and B, into a three-phase, 230-V, line as shown in Figure 3.
The dc current through the inductor is 20A. If power loss in both rectifiers are negligible, calculate:

(a). Firing angle of the rectifier A;
(b). Firing angle of the rectifier (inverter) B.

This is my attempt ;

V out = 5400/ 20 = 270

VOUT = 1.35 × VLINE × cos α

Where ‘α’ is the firing angle of the rectifier.

Therefore,

cos α = V / 1.35 x Vline

α = 29

how does this seem? and how is the inverter's firing angle different?

2. Aug 17, 2011

ramox3

1. The problem statement, all variables and given/known data

Hello everyone I am a newbie to electrical engineering, I have this problem I've been strugling with, no textbook seems to cover this problem..

A three-phase, 440-V, generator delivers 5400W of active power through two recti-fies, A and B, into a three-phase, 230-V, line as shown in Figure 3.
The dc current through the inductor is 20A. If power loss in both rectifiers are negligible, calculate:

(a). Firing angle of the rectifier A;
(b). Firing angle of the rectifier (inverter) B.

2. Relevant equations

3. The attempt at a solutionThis is my attempt ;

V out = 5400/ 20 = 270

VOUT = 1.35 × VLINE × cos α

Where ‘α’ is the firing angle of the rectifier.

Therefore,

cos α = V / 1.35 x Vline

α = 29

how does this seem? and how is the inverter's firing angle different?

3. Aug 18, 2011

Staff: Mentor

Where is Fig 3?

4. Aug 18, 2011

ramox3

5. Aug 18, 2011

uart

Seems ok. And you can use the same formula (with the relevant line voltage) for the inverter side.

BTW The constant 1.35 comes from 3 sqrt(2) / pi.

6. Aug 18, 2011

ramox3

but what is the relevent line voltage for the inverter? isn't it the same?

7. Aug 19, 2011

Mbert

Not sure this is the right place to post this (perhaps this is why no one answers, but I don't know I'm new to PF). Here's some help:

In fact, there are a lot of text books that cover this. For instance see "Power electronics", from Mohan. I can't see your Figure 3, but I assume you have a perfect voltage source connected to 2 full-bridge thyristor converters (1 rectifier, 1 inverter) and some load. There should also be an inductor on the DC bridge. To find the solution for rectifier A (source):

$$P=V_{DC}I_{DC}$$

where the voltage on the DC bus is given by:

$$V_{DC}=\frac{3\sqrt{2}}{\pi}V_{LL}\cos{\alpha}$$

where $V_{LL}$ is the line-line voltage of the AC side. The DC bus current is given in the problem, which is $I_{DC}$. Just isolate these for $\alpha$...

For rectifier (inverter) B, same procedure, but by using a negative power and use the 230V line-line voltage. Isolate again for $\alpha$... This angle should be between 90 and 180 degrees since this converter is operating in inverter mode.

M.

8. Aug 19, 2011

Staff: Mentor

(Moderator's note -- two threads merged...)