Apparent Power in an Center-Tap Rectifier

[MarcMTL]

Homework Statement

A highly inductive load is subjected to 10A and 150V via a center-tap recitifier powered by a monophased 240V supply.

See schematic below:

Homework Equations

I'm trying to calculate the apparent power at the primary and secondary windings.

For the secondary: S=Vse*Ise

The Attempt at a Solution

S = Vse*Ise where:

Vse = efficient voltage = (Vch * pi)/(2 * sqrt(2)) = 166.6v
Ise = ??

How do I find Ise? In a non-inductive load, Ise = Vse/R, however that doesn't apply in an inductive load.

I believe the key is that this is an 'highly inductive' load. Can anyone point me in the right direction?

Thanks,

Marc.

 

[gneill]

Do I_ch and V_ch represent RMS quantities for the rectified waveform?

 

[MarcMTL]

Do I_ch and V_ch represent RMS quantities for the rectified waveform?

No, they are both average qtys.

 

[gneill]

So presumably you could calculate their peak values given the nature of the full-wave rectified waveform (look up the conversion or do a bit of calculus). That peak will show up on the windings on either side of the secondary center tap. Since only one side or the other conducts current at any time the net effect is that the primary sinusoidal voltage will "think" that is creating a Vpk magnitude sinusoid in a single secondary coil, driving Ipk current. You should be able to tease out a transformer turns ratio from this...

 

[The Electrician]

Do I_ch and V_ch represent RMS quantities for the rectified waveform?

No, they are both average qtys.

See: http://en.wikipedia.org/wiki/Apparen...pparent_powers

about 75% of the way down the page under the heading "Basic calculations using real numbers".

Apparent power is defined in terms of RMS quantities. If you use average values, you won't be calculating apparent power.

 

[MarcMTL]

See: http://en.wikipedia.org/wiki/Apparen...pparent_powers

about 75% of the way down the page under the heading "Basic calculations using real numbers".

Apparent power is defined in terms of RMS quantities. If you use average values, you won't be calculating apparent power.

Exactly. What I'm not able to find is the rms value of Ich, as it's an inductive load.

 

[gneill]

Apparent power is $\sqrt{real^2 + reactive^2}$. Same goes for the voltage or current. If there's no real component, no worries. You're looking for apparent power!

If, as you say, Ich and Vch are averages for the rectified waveform, then you can convert to peak or rms by suitable maths. It seems to me that you need to determine the effective transformer turns ratio so so that you can convert the load values (average, peak, rms, or your choice) to values in the primary, where you can then determine the power there.

If you determine the peak value of the rectified waveform then you also have the peak of the sinewaves that are being induced in either side of the center tapped secondary. You can go RMS at this point, since you've got a full sine curve considering both conduction phases.

 

[The Electrician]

If, as you say, Ich and Vch are averages for the rectified waveform, then you can convert to peak or rms by suitable maths.

This assumes that the waveforms of Ich and Vch are rectified half-sine waves. This is more or less true for the voltage waveforms at various points in the circuit, but it isn't true for the current waveforms.

I built this circuit using a lower voltage transformer than the OP's problem shows, and the inductor I used is not perfect; I used a roll of magnet wire and the wire has some resistance. The inductor is about 10.5 mH with a resistance of 1.75 ohms. The measurements were made in the U.S. where the line frequency is 60 Hz. I used a variac in the primary and set the inductor current to 3 amps. This isn't exactly the OP's problem, but it demonstrates that assumptions about the inductor current must be carefully considered.

The first image shows the current in the primary of the transformer (orange), the voltage across the inductor (green) and the current in the inductor (purple).

The second image shows the same setup except the purple trace is now the current in one half of the secondary. The gray trace is the saved trace showing the inductor current from the first image.

This problem is a little bit trickier than it appears at first glance.

 

[gneill]

This problem is a little bit trickier than it appears at first glance.

We would be glad to hear of your suggestions for an analytical treatment of the scenario.

 

[The Electrician]

I don't see enough information given to come up with an analytical solution. The problem says "highly inductive load". What does this mean in numbers?

Consider; if the transformer is perfect (no leakage inductance, zero winding resistances), if the diodes are perfect, if the 240 VAC supply is perfect, and if the inductive load is perfect (no DC resistance in the winding, no core loss) then if any voltage at all is applied by this perfect circuit to the perfect inductor, the DC current in the inductor would be infinite.

The fact that the current in the inductor is only 10 amps means that something in the circuit other than the inductance of the load has limited the current to that value.

If the inductance is very large, the current in the inductor will be nearly pure DC; the ripple in the inductor current (compare the purple trace in the scope captures) will become small. The ratio of the peak to RMS, the crest factor, of the inductor current will approach 1 as the inductor becomes large.

We would need to know the value of the resistances in the circuit which are limiting the inductor current, and the value of the inductance to come up with an analytical solution. At least, that's how it seems to me. I wonder if the problem image in the first post is all the information we have, or if the student is expected to make some assumptions we don't know about.

 

[The Electrician]

Thinking about the problem some more, if we can assume that the inductance of the load is very large, and the ratio of the inductor reactance to circuit resistance is high, the current in the inductor could be assumed to be nearly ripple free; pure DC. Its value would be 10 amps, average and RMS both.

The voltage across the inductor will be half-sine pulses. We can calculate the RMS value from knowledge of the sinusoidal shape and the average value of the current. This will give us the turns ratio if we assume the transformer parasitics and the diode losses are negligible.

The product of the inductor current and the RMS value of the inductor voltage will give us the inductor apparent power. The secondary apparent power will be half that amount for each half of the secondary.

The primary current will be square waves, with the value determined by the turns ratio of the transformer. The RMS value of a square wave of current is equal to the peak value. The RMS value of the primary voltage is given, and knowing the RMS value of the primary current, we can calculate the primary apparent power.

If the transformer and diode losses are negligible, we would expect the total secondary apparent power to be the same as the primary apparent power since the primary current is a square wave just like secondary currents (they have a DC offset as well).

This all depends on the assumption that the inductance of the load is so large as to make the inductor current essentially DC.

 

[The Electrician]

I dug out a really big inductor, about 150 mH--15 times as big as the one I used in post #8.

I've attached 2 more images of the same setup as in post #8 but with the larger inductor in place.

In the first image notice how the transformer primary current (orange) is nearly a square wave, and the inductor current is practically pure DC (purple trace). I could only apply a small voltage to the rectifier if I wanted the inductor current to be only 3 amps. The inductor voltage consists of half-sine waves whose negative going dips actually go negative due to the diode drop.

In the second image, we see that the secondary current is also a square wave with a DC offset.

 

[MarcMTL]

Thinking about the problem some more, if we can assume that the inductance of the load is very large, and the ratio of the inductor reactance to circuit resistance is high, the current in the inductor could be assumed to be nearly ripple free; pure DC. Its value would be 10 amps, average and RMS both.

This all depends on the assumption that the inductance of the load is so large as to make the inductor current essentially DC.

The Electrician got it right. Since the load is 'highly inductive', one can assume that the ripple voltage is negligible and then your average & RMS value would be the same. In this problem the solution is attainable by assuming the RMS value is aprox. equal to the average.

 

[MarcMTL]

Here's another one for you guys:

I found my transformer ratio to be: np=1.44*ns

So, 240Vrms/(1.44)=166.6Vrms on the secondary, which is correct as Vch = [ 2*Vst*sqrt(2) ]/2 gives us 150Vch, exactly as stated int he problem.

So apparent power on the primary would be: Sp=Vpe*Ipe = (240Vrms)*(1/1.44)*(10A) as Ipe is related to (ns/np)*Ise. Sp = 1666VA

For the secondary, based on valid formula, I obtain Se = 2*Vse*Ise = 2(166.6)(10) = 3332.2 VA. Apparently, the '2' comes from the fact that there are two secondaries in a center-tap. I'm having a hard time grasping why the 2 is put in play, as we are using only 1 set of the windings at a time.

We endup getting twice the amount of apparent on the secondary then on the primary.

I'm having a hard time grasping how the apparent on the secondary could be higher then the apparent on the primary. I would, at the very least, expect it to be the same value or less, but not higher.

Can anyone enlighten me?

 

[The Electrician]

I'm having a hard time grasping why the 2 is put in play, as we are using only 1 set of the windings at a time.

Look at the purple waveform in the second image of post #12. The current pulses occur only half the time, so you need to multiply by 1/2.

The VA out of one half of the secondary would be (166.6)(10) only if the current were there all the time, but, in fact, it's only there half the time, so you need another factor of 1/2.