# phasors and power calcs

Discussion in 'Electronic Basics' started by [email protected], May 25, 2005.

1. ### Guest

Hello,

I am studying an introductory electronics course and have been given the
following problem:

A small factory has the following loads operating from the 240Vrms 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

2. ### Guest

Jan,

I can give you a little help along your way, but since you've already
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)) = A cos(W * t + Phi)
k=1

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
"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) = cos(Omega) + j * sin (Omega)

Real{ e^(Omega * j) } = cos(Omega)
Imaginary{ e^(Omega * j) } = 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=1
= Sum of A(k) Re{ e^(W(0) * t + Phi(k)) * j }
= Sum of A(k) Re{ e^(W(0) * t) * j + e^(Phi(k)) * j }
= 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) = 50 Hz
V = 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‡ sine wave is approximately .707*peak voltage."
[http://www.bcae1.com/voltages.htm]

V = I*R [Ohms law]

Thus,

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

a) Forty 50 watts incandescent lamps with unity power factor

Power = Watts = V * I

40 * 50 W = 240/0.707 Peak Voltage * cos(2*pi*50*t + 0) * I,
I = (40 * 50 W) / (240 / 0.707) Peak Current * cos(2*pi*50*t + 0)
= (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 = 240/0.707 Peak Voltage * cos(2*pi*50*t + 2*pi*0.9) * I
I = (35 * 40 W) / (240 / 0.707) Peak Current * cos(2*pi*50*t +
2*pi*0.9)
= (35 * 40 W) / (240 / 0.707) Peak Current * Re { e^(2*pi*50*t +
2*pi*0.9)*j }
= (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
0.65.

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

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

3. ### Dan HollandsGuest

Draw each of the three phasors

Break each down to a real and a reactive component

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

www.QuickScoreRace.com

4. ### Terry GivenGuest

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

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

5. ### John PopelishGuest

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
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
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.

6. ### Ed-Guest

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

: Hello,
:
: I am studying an introductory electronics course and have been given
the
: following problem:
:
: A small factory has the following loads operating from the 240Vrms
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.

8. ### Don KellyGuest

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.

9. ### David SauerGuest

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.