cableguy
Has never sniffed a stink bug
I've been staring at these for about the last hour, wanted to start a thread on it to get some thoughts.
I was reworking the NCEES sample exam this afternoon, and bumped in to 115 again. I understand the integral and all that, but the dirty side of me wants something I can use more generically.
I then started looking at the Kaplan sample exam, which has several strange rectifier circuit questions.
I also looked at sources around the Internet, looking for that killer equation.
I came up with this:
Vaverage = (Vpeak / pi) * (cos a - cos d)
where a is the firing angle / start of the wave, and d is the zero crossing or end of the wave.
For example, for NCEES 115, this would be (1/pi) * (.707 + 1) = .543 (correct)
A complete half wave would be (1/pi) * (1+1) = .6366 (correct)
Now I start looking at the wonky Kaplan problems. I think they're doing them wrong.
On one problem (A34), they throw out the equation:
Vavg = (2/pi) Vpeak cos a
And on a three phase half wave rectifier problem (A36), they deus-ex-machina throw out the equation
Vavg = (3 sqrt(3)/2pi) Vpeak cos a
And I believe these formulas are incorrect.
I printed out this graphic
http://upload.wikimedia.org/wikipedia/en/6...ification_2.png
And drew vertical lines at the 30 degree marks. One of the Kaplan problems was to calculate the average voltage of a 3 phase half wave rectifier with a 120Vrms wave with a trigger that fires starting at 60 degrees and then triggers every 120 degrees after that.
I calculated 81 volts using my formula. They use the equation above and get 70.2 volts. I honestly have no idea where they got their equation.
Anyone have any insight on average values for rectifier pulses? Am I off my rocker?
Thanks!
I was reworking the NCEES sample exam this afternoon, and bumped in to 115 again. I understand the integral and all that, but the dirty side of me wants something I can use more generically.
I then started looking at the Kaplan sample exam, which has several strange rectifier circuit questions.
I also looked at sources around the Internet, looking for that killer equation.
I came up with this:
Vaverage = (Vpeak / pi) * (cos a - cos d)
where a is the firing angle / start of the wave, and d is the zero crossing or end of the wave.
For example, for NCEES 115, this would be (1/pi) * (.707 + 1) = .543 (correct)
A complete half wave would be (1/pi) * (1+1) = .6366 (correct)
Now I start looking at the wonky Kaplan problems. I think they're doing them wrong.
On one problem (A34), they throw out the equation:
Vavg = (2/pi) Vpeak cos a
And on a three phase half wave rectifier problem (A36), they deus-ex-machina throw out the equation
Vavg = (3 sqrt(3)/2pi) Vpeak cos a
And I believe these formulas are incorrect.
I printed out this graphic
http://upload.wikimedia.org/wikipedia/en/6...ification_2.png
And drew vertical lines at the 30 degree marks. One of the Kaplan problems was to calculate the average voltage of a 3 phase half wave rectifier with a 120Vrms wave with a trigger that fires starting at 60 degrees and then triggers every 120 degrees after that.
I calculated 81 volts using my formula. They use the equation above and get 70.2 volts. I honestly have no idea where they got their equation.
Anyone have any insight on average values for rectifier pulses? Am I off my rocker?
Thanks!
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