Power Rating Calculation 4

[zacky]

A single phase transformer is feeding a half-wave rctifier with a resistive load across it.
What is the real value of the power cosumed in the resistor load( dc + ripples ),is it to :
- Multiply transformer's secondary rms values of the voltage and current- assume 100% efficiency.
-Or to mutiply the the intatanious values of the voltage and current at the load terminals and then thake the average over one cycle period- as the wattmeter does?

 

[jghrist]

I guess another way to look at it is to calculate the fundamental component of the current as suggested by CGCPE.

The fundamental component of half wave rectified sine wave current is 0.5·I·sin(2·[π]·f·t) where I is the peak current. The rms value would be I[sub]rms[/sub] = 0.5·I/[√]2. The fundamental component of the voltage is V·sin(2·[π]·f·t). The rms value would be V[sub]rms[/sub] = V/[√]2. The power is the product V[sub]rms[/sub]·I[sub]rms[/sub] or 0.5·V/[√]2·I/[√]2 = 0.25·V·I. Both the fundamental current and voltage have a zero phase angle so the displacement power factor is unity.

Using the example of 10 volts rms and a 2 ohm resistor, the peak voltage is 10·[√]2. The peak current is 10·[√]2/2. The power is 0.25·10·[√]2·10·[√]2/2 = 25W.

This is a very complicated method for this simple problem, where it is intuitively obvious that the power would be 1/2 that of unrectified current, but it would be useful in other cases.

 

[zacky]

stevenal, agreed with your calculation that the PF of this circuit is .707
so. at the transformer's secondary side the power delivered to the circuit will be Irms.Vrmsx(.707)
Now what will be the exact tranformer's rating:
Is it the product of Irms.Vrms
Or the real power it delivers Irms.Vrmsx(.707)

 

[LionelHutz]

stevenal, you are correct. What I should have said is to be careful using it and know how to properly correct for the diode in the circuit.

It is easy to back calculate the power factor when you do know the power already but it's not so easy to calculate the power facture to figure out the power. As i said, I've seen fairly expensive power meters wrongly measure power on the line side of SCR circuit.

I don't fully agree with the power factor arguement anyways. If you looked at the voltage and current on a scope you'd see an in-phase current for 1/2 of the voltage cycle and no current for the other 1/2 of the voltage cycle. Where is the phase shift? I could also argue that the only current that flows goes through a resistor. I'm in agreement with Marmite in that an ideal resistor does not have a power factor so power factor isn't applicable.

 

[CJCPE]

LionelHutz:

The power factor is not a displacement power factor that arises from a phase shift, but a distortion power factor that arises from the current not being sinusoidal over the complete cycle. With unity displacement power factor and a lower distortion power factor, the total power factor (sometimes called the "true" power factor is less than 1. It is not the resistor that causes the reduced power factor it is the diode.

zacky:

Transformers are rated in VA, not watts. The distorted waveform causes transformer heating is excess of what would be caused if the waveform was not distorted. At a minimum, the transformer needs to be sized for Vrms X Irms. You also need to consider that the transformer might be saturated due to the unidirectional secondary current. It would be a lot better to use a full wave rectifier.

 

[jghrist]

Stevenal,

In your 25 Jan post, you said I[sub]rms[/sub] was 5 for the first half cycle and V[sub]rms[/sub] was 10/?2. Shouldn't V[sub]rms[/sub] be 10 to get 5 amps rms through the 2 ohm resistor?

According to the McGraw-Hill Science and Technology Dictionary at

http://www.answers.com/topic/power-factor-1,

power factor (?pau?·?r ?fak·t?r)
(electricity) The ratio of the average (or active) power to the apparent power (root-mean-square voltage times rms current) of an alternating-current circuit. Abbreviated pf. Also known as phase factor.

Active power is 25W. I[sub]rms[/sub] = 5/?2 amps over the whole cycle. V[sub]rms[/sub] = 10. Apparent power is 10·5/?2 = 50/?2 = 35.4 VA. Power factor is 25/(50/?2) = 0.707.

The transformer would have to be rated for the apparent power of 35.4 VA.

 

[stevenal]

pf is defined as the ratio of real to apparent power. This reduces to the cosine of the angle between the current and voltage for purely sinusoidal quantities. That reduction is not valid in this case.

Transformers for non-linear loads also have a k factor rating which deals with the extra heating. I agree that full wave rectification would be better.

 

[GS3]

It seems to me the original question is quite simple and some people here are getting hung up on unnecessary details.

>>A single phase transformer is feeding a half-wave rctifier with a resistive load across it.

It can be transformer or just ac line. We assume a perfect AC input, an ideal rectifier and resistive load.

>> What is the real value of the power cosumed in the resistor load

It is one half what it would be without the diode. Without the diode it would be V^2/R and with the rectifier it is V^2/R/2. Rather obvious when you think about it.

>> Multiply transformer's secondary rms values of the voltage and current- assume 100% efficiency.

This is also correct. Basically the same thing.

>> Or to mutiply the the intatanious values of the voltage and current at the load terminals and then thake the average over one cycle period- as the wattmeter does?

This should also give the same result. Are you assuming they would give different results? Am I missing something?

 

[jghrist]

GS3,
You're not missing anything. There are multiple complicated ways to arrive at the same intuitively obvious answer.