phasors and power calcs

Jan,

I can give you a little help along your way, but since you've already formulated an answer then my explanation can help you validate your work.

Expressing sinusoidal signals (such as AC RMS) with the same frequency (in this case 50Hz) as phasors is useful for consolidating by addition the signals into a signal formula. The key to phasors is that this can be done when the frequency is the same, but the amplitudes and phases differ. For example,

N Sum of A(k) cos(W(0) * t + Phi(k)) =3D A cos(W * t + Phi) k=3D1

Note that on the left hand of the equation, I have written in ASCII text as follows:

"W(0)" is short hand for writing W subscript 0 and represents the radian measure of angle, "Phi(k)" is short for Phi subscript k, "t" is used as the classic symbol representing time.

This equation states that the sum of N cosine signals of differing amplitudes and phase shifts, but with the same frequency, can always be reduced to a single cosine signal of the same frequency. [Excerpted from text "DSP First, A multimedia approach", by McClellan, et al. pp

31]. We know that this is only one half of the fun. Euler provided us with complete fun by showing us that we could express cosine signals as the real part of e^(W*t + Phi) * j. His brilliance helps us in one way because it saves us from having to write annoying trigonometric identities all over the place, and reduces the math to addition in the exponents. Thus,

e^(Omega * j) =3D cos(Omega) + j * sin (Omega)

Real{ e^(Omega * j) } =3D cos(Omega) Imaginary{ e^(Omega * j) } =3D sin(Omega)

Notes:

Omega is the English literal translation of the Greek symbol O, j represents the imaginary symbol representing square root of -1.

Hence,

N Sum of A(k) cos(W(0) * t + Phi(k)) k=3D1 =3D Sum of A(k) Re{ e^(W(0) * t + Phi(k)) * j } =3D Sum of A(k) Re{ e^(W(0) * t) * j + e^(Phi(k)) * j } =3D Re { e^(W(0) * t) * j } * Sum of A(k) * Re { e^(Phi(k)) * j }

So much for phasors. From your problem statement we have,

W(0) =3D 50 Hz V =3D 240 Vrms

We know that, "Peak Voltage: Peak voltage tells you how far the voltage swings, either positive or negative, from the point of reference... The RMS voltage of a pure=87 sine wave is approximately .707*peak voltage."

V =3D I*R [Ohms law]

Thus,

I =3D 240 Vrms/ 1 Ohm. =3D 240rms cos(2*pi*50*t + Phi) =3D 240/0.707 Peak Current * cos(2*pi*50*t + Phi)

a) Forty 50 watts incandescent lamps with unity power factor

Power =3D Watts =3D V * I

40 * 50 W =3D 240/0.707 Peak Voltage * cos(2*pi*50*t + 0) * I, I =3D (40 * 50 W) / (240 / 0.707) Peak Current * cos(2*pi*50*t + 0) =3D (40 * 50 W) / (240 / 0.707) * Re { e^(2*pi*50*t)*j }

b) Thirty-five 40 watts fluorescent lamps with a lagging power factor of 0.9

35 * 40 W =3D 240/0.707 Peak Voltage * cos(2*pi*50*t + 2*pi*0.9) * I I =3D (35 * 40 W) / (240 / 0.707) Peak Current * cos(2*pi*50*t + 2*pi*0.9) =3D (35 * 40 W) / (240 / 0.707) Peak Current * Re { e^(2*pi*50*t + 2*pi*0.9)*j } =3D (35 * 40 W) / (240 / 0.707) Peak Current * Re { e^(2*pi*50*t)*j * e^(2*pi*0.9)*j }

c) One 2400 watts air conditioning system with a lagging power factor of

0=2E65.

1 * 2400 W =3D 240/0.707 Peak Voltage * cos(2*pi*50*t + 2*pi*0.65) * I I =3D (1 * 2400 W) / (240 / 0.707) Peak Current * cos(2*pi*50*t +

2*pi*0.65) =3D ...

Etcetera, etcetera.... Now see if you can do the rest. You're Welcome,

-Beagle

Draw each of the three phasors

Break each down to a real and a reactive component

Add the real components

Add the reactive components

Draw the phasor from the the sum of real and reactive components

The length of the phasor gives the total current

The angle of the phasor gives the power factor

--

Dan Hollands

1120 S Creek Dr Webster NY 14580 585-872-2606 snipped-for-privacy@USSailing.net

guess the currents are all pretty sine waves, as is the voltage, and that the supply impedance is zero (no matter how much load you draw, the voltage never droops. not really true, but close enough :)

the incandescent lamps: =======================

50W/240V = 0.21Arms per lamp. there are 40 of them, so 8.33A = 0.21A*40

power factor = 1 so there is no reactive current (IOW its a resistive load)

The flouro's: =============

40W/240V = 0.17Arms per flouro. 35 of them, so 5.83A = 35*0.17A

power factor = 0.9 (lagging), so there *IS* reactive current (which is at 90 degrees to real current, so think right-angled triangles).

pf = cos(phi), phi = angle between real and actual current.

from SOH CAH TOA trigonometic identities, we know cos(phi) = Adjacent/Hypotenuse = 0.9.

so Hypotenuse = Adjacent/cos(phi) = 5.83A/0.9 = 6.48A

Pythagoras allows us to calculate the reactive current, as 6.48^2 = 5.83^2 + Ireact^2 Ireact = sqrt(6.48^2 - 5.83^2) = 2.83A

(you can check the cos(phi) comes out right)

The A/C: =============

2400W/240V = 10Arms

power factor = 0.65 (lagging), so there *IS* reactive current.

so Itotal = 10A/0.65 = 15.38A

Pythagoras allows us to calculate the reactive current, Ireact = sqrt(15.38^2 - 10^2) = 11.69A

(as you can see, the total current is a lot higher than the real current, so low PF loads kinda suck)

The result: ===========

to add vectors, add all the X and all the Y components separately.

Real current:

8.33A + 5.83A + 10A = 24.16Arms real

reactive current:

2.83A + 11.69A = 14.52Arms reactive (aka imaginary)

total current I = sqrt(24.16A + 14.52A) = 28.19Arms

power factor PF = 24.16A/28.19A = 0.86

Now, just to be a bastard, the question asks for PEAK current. which is of course sqrt(2)*RMS current, so Ipeak = 39.87A

HTH Cheers Terry

Break each of the individual load current phasors into a real (in phase with voltage) and imaginary (90 degrees shifted with respect to voltage) components. Add all the real components together as the real component of the total load phasor, and add all the imaginary components together as the total imaginary component of the total load phasor. Then put these two components start to end and draw the diagonal sum that represents their resultant. Of course, if you have a compass and protractor, you could just add the individual load phasors one to the next and get the same resultant. Breaking them into components just makes it easer to do the sums, graphically without those tools The divisions on the graph paper is the only tool needed.

next question: how can i get the electric meter to turn backwards by using some caps?

50Hz : supply : a) Forty 50 watts incandescent lamps with unity power factor : b) Thirty-five 40 watts fluorescent lamps with a lagging power factor of 0.9 : c) One 2400 watts air conditioning system with a lagging power factor of : 0.65. : : I am asked to draw the complete phasor diagram and determine the total : peak current drawn and overall power factor of the factory. : : I'm a little stuck :( I have some phasor diagrams put together which I : am a little unsure of, but I don't really know how to determine peak : current from this, can anyone shed some light or point me in the : direction of some material to explain this sort of problem? I've been : digging through a number of text books and am not really progressing. : : Thank you : : Jan

Dream on.

--
Former professional electron wrangler.

Michael A. Terrell
Central Florida

You can make it stop simply by turning off the main switch.

You can't.

Capacitors will change the power factor but not the real power(*time) which the meter measures. Changing the pf will result in less current and some decrease in losses. If you have demand metering, then the demand may be reduced.

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

Drawing is the best method, even if it is only a rough sketch, you can easily check that the answers are about the right values when you have finished.

And in an exam situation, even a correct drawing will normally get you part marks if you stuff up part of the equations.