🚀 go-pugleaf

jan@janhurst.com wrote:
> 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

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