Firing angle in rectifiers and inverters.

[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
For inductive load

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?

 

[ramox3]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

This is my attempt ;

V out = 5400/ 20 = 270
For inductive load

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?

 

[NascentOxygen]

Where is Fig 3?

 

[ramox3]

here's the figure :D

http://imageshack.us/photo/my-images/69/figure3t.png

 

[uart]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

This is my attempt ;

V out = 5400/ 20 = 270
For inductive load

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?

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.

 

[ramox3]

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.

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

 

[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 textbooks 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.

 

[berkeman]

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