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Firing angle in rectifiers and inverters. 
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#1
Aug1611, 05:15 PM

P: 12

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 threephase, 440V, generator delivers 5400W of active power through two rectifies, A and B, into a threephase, 230V, 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? 


#2
Aug1711, 09:36 AM

P: 12

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 threephase, 440V, generator delivers 5400W of active power through two rectifies, A and B, into a threephase, 230V, 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 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? 


#3
Aug1811, 04:03 AM

HW Helper
Thanks
P: 5,166

Where is Fig 3?



#4
Aug1811, 09:48 AM

P: 12

Firing angle in rectifiers and inverters.



#5
Aug1811, 11:48 AM

Sci Advisor
P: 2,751

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


#6
Aug1811, 12:41 PM

P: 12




#7
Aug1911, 05:15 PM

P: 64

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 fullbridge 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): [tex]P=V_{DC}I_{DC}[/tex] where the voltage on the DC bus is given by: [tex]V_{DC}=\frac{3\sqrt{2}}{\pi}V_{LL}\cos{\alpha}[/tex] where [itex]V_{LL}[/itex] is the lineline voltage of the AC side. The DC bus current is given in the problem, which is [itex]I_{DC}[/itex]. Just isolate these for [itex]\alpha[/itex]... For rectifier (inverter) B, same procedure, but by using a negative power and use the 230V lineline voltage. Isolate again for [itex]\alpha[/itex]... This angle should be between 90 and 180 degrees since this converter is operating in inverter mode. M. 


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