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?

 

[itsmoked]

You would take the instantaneous values, (V & I), as fast as you can multiply them together and integrate them all.

Better is to take a TrueRMS meter and measure it.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[zacky]

We can use the following formulas for the instanious current and voltage:
(Esin x /R) and (Esin x) ,multiply them together and ,make the integral over one period cycle and devide over the period cycle and that would be the wattmeter reading( the mean value of the product of current and voltage)
Now go to the secondary winding, multiply the rms value of the current and voltage ( transformer rating )and you will find it larger than the wattmeter reading though we assumed 100% efficiency.
Why that difference ?
Which one is the true power consumed in the resistor ?

 

[dpc]

Is this a real world problem?

 

[itsmoked]

I am not really understanding your question I think.

You are asking why the DC load power and the power on the low voltage side of the transformer are different? And specifically why the transformer side is higher?

If that is a the case. The power at the transformer has got to be higher because their are substantial heating losses at the rectifier. Also if you have a filter cap you have losses due the ESR of the capacitor. So - out with the 100%!

If this isn't what you are asking you must further clarify. Draw a picture perhaps.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[zacky]

This ciruit is very simple :
single phase transformer , its secondery is connected in series with a single diode and resistive load( simple half-wave rectifier)
The DC power Wdc = Idc x Vdc -- average values of the rectified voltage and current.

If you calculate the transformer's secondary VA rating using the rms values of the voltage and current there you will find it = 3.49Wdc

Now the wattmeter at the resistive load will caculate the power as the mean value of the product of the instanious current and voltage over one period - If you do this calculation mathematically you will find it =2.46Wdc

Why that difference between 3.49Wdc and 2.46Wdc ?




2.46

 

[dpc]

Because you are doing something wrong. Without a circuit diagram, including measurement points and your calculations, this is a pointless discussion. Energy is conserved, that is a given. Energy consumed is what it is, no matter how you arrive at the value.

If you are measuring at one point and calculating for another point, that's different.

 

[zacky]

Actually, I did two calculations:
one at the load and the other at the transformer secondary.
The two calculations give different results.
No measurments are taken- Just mathematical calculations and neglect any losses.

 

[JBinCA]

Zacky,

I think even on simple issues it is very difficult to convey specifics of engineering problems in pure text.

That being said, if I understand your situation, the answer is that one calculation is based on the transformer rating and the other is based upon the power consumed by the load.

You wrote: "If you calculate the transformer's secondary VA rating using the rms values of the voltage and current there you will find it = 3.49Wdc"

The XFMRs VA rating is the amount of apparent power that it can deliver without being overloaded. The XFMR does not always deliver this amount of power, but rather delivers the power demanded by the load. Loads that demand less than this amount of power can be served by the XFMR, loads that demand more will overload it.

It sounds like you're serving 2.46W worth of load with 3.49W worth of XFMR. No problem.

Another possible interpretation of your question would be that you know the circuit draws 3.49W (as calculated at the XFMR secondary), but you only calculate 2.46 watts of that power going to the resistor.

If this is the case, then you have ommitted the fact that the diode dissipates power as well. Diodes have characteristic voltage drops when conducting (Forward biased). 0.7V for Si based or 0.3V for Ge based diodes. I devise that has a voltage drop and is passing current is consuming or dissipating power.

I think one or the other of these explanations accounts for your discrepancy.

Regards.

 

[zacky]

JBinCA,
Thanks for the detailed explaination.
I still have one point is not clear.If the load is pure resistive,that means that the current and voltage are in phase and the power consumed by the load is real power,so if we calculate the power at the load should equal the power deliverd by the secondary winding-assume in all our calculations ideal diode and ideal transformer.
Is the product Vrms.Irms at the transformer's secondary does not the represent the real power delivered by it for that resistive load?

 

[jghrist]

If it's an ideal half-wave rectifier with no filtering and no losses, the power consumed would be 1/2 that of unrectified ac through the resistor. 0.5·E[sub]rms[/sub]²/R

 

[CJCPE]

The product Vrms X Irms at the transformer secondary is the voltamperes not the watts. The watts = Vrms X Irms X power factor. The power factor is less than 1 because one full cycle of secondary current is half a sine wave rather than a full sine wave. Mathematically, current consists of a sine wave at the power frequency plus harmonics. Only the RMS value of the fundamental current represents power delivered. The harmonic currents do not contribute to power delivered. This assumes an ideal transformer and an undistorted supply voltage.

 

[jghrist]

CJPE,
Are you saying that only the fundamental component of the current will cause the resistor to heat up?

 

[CJCPE]

jghrist:

When calculating the power from the AC voltage and the fourier coefficients of the AC current, only the instantaneous values of fundamental current times the voltage produces a non-zero result. That explains why voltamperes calculated from the AC RMS values is higher than the watts.

All of the current contributes to the power in the resistor, but I think that you need to go to the fourier components to make sense of the voltamperes vs. watts.

 

[dpc]

I guess I'm in a mood, because this thread is annoying me. You guys are arguing among yourselves, even though the OP provides absolutely no information except how simple it is.

Seems like either a homework problem or a troll.

 

[Marmite]

This is all getting very complicated for a simple transformer/diode/resistor circuit with talk of fourier coefficients and power factor. A resistor simply resists the flow of electrons at all periods of time. The waveform for the voltage across a resistor is always in phase with the current through it. Power factor is not applicable.
If you take the diode out of the circuit and imagine a 10V RMS AC source feeding a 2 Ohm resistor, a current of 5A RMS flows and dissipates 50W in the resistor. If you plug the diode back in the current will be halved to 2.5A RMS because you only have half a cycle, and the power dissipated in the resistor will be 12.5 W.
I'm still not sure what the issue is here, but the confusion may be arising because the RMS value of a sine wave is not the same as the average value. RMS=0.707(Peak)Average= 0.637 (Peak)
Regards
Marmite

 

[stevenal]

Basic rms addition:

5 amps rms on the first half cycle added to 0 amps rms on the second->

sqrt((5^2+0^2)/2half cycles)= 5/sqrt(2)=3.54 A rms on the full cycle.

Do the same for voltage and you get 10/sqrt(2) volts rms. Multiply them together and you get 25W average power dissipated. Now see jghrist's answer above.

I agree pf is not applicable (or unity) downstream from the diode where the load is resistive and linear. Upstream, the load looks non-linear and there will be less than unity pf due to distortion. Voltage is present during the second half cycle, but it does not contribute to power without the current. pf=25/(3.54*10)=.707

 

[LionelHutz]

To sum-up what the others have said in one line - You can not use the measured transformer voltage before the diode in a power calculation.

You get the same result with a SCR controller at part output voltage uf you measure voltage before an after and use that in a power calculation.

Keith hit it exactly in the first post. You would need a RMS power meter that does the rms calculation on instantaneous power measurements to measure the same power before and after. Many power meters don't do this, they calculate rms voltage and current and power factor and then do the math. I've seen a Fluke 43B ($4k meter or so??) display different power on the input and output of a SCR controller.

 

[Marmite]

Stevenal. Thanks for correcting my clumsy maths. My only excuse is it's a good few years since I was at college!
Regards
Marmite

 

[stevenal]

"You can not use the measured transformer voltage before the diode in a power calculation"

Sure you can. If you have true rms reading meters, and if the other assumptions (linear purely resistive load below a lossless diode) do not introduce too much error, just use the distortion pf I derived above in your calculation.