# How to find the capacitance per phase?

Discussion in 'Homework Help' started by ADCapacitor, Oct 4, 2016.

Feb 3, 2016
38
0
Three similar coils,each of resistance 20ohm and inductance 0.07H are connected in star to a 415V,3-phase,50 Hz supply
A delta-connected capacitor bank connect to the system to improve the overall power factor to unity.
Find the capacitance per phase:

IL (Line curret)= IP(phase current)
IL=Vp/XL
IL=[415/√3]/ [20+2*pi*50*j0.07]
=8.06<-47.7 deg
power factor=cos-47.7
=0.673
P(Real power)=√3*VL*IL*cosθ
=3899W
S=3899/0.672
=5793 VA
For unity power factor,
cos θ=1
θ=90

When the apparent power times the sinθ, I get 0 for the reactive power.I will also get 0 for the capacitance?
How can I find the current going into capacitor ?Then , I can make use of V=RI to find the impedance?

2. ### MrAl Well-Known Member

Jun 17, 2014
2,206
433
Hi,

Real quick, maybe you can start by selecting a star connected capacitor bank then transform the values later.
If this does not help i'll try to help more later sometime unless someone else does first.
In any case you should be able to analyze the circuit using something like Nodal to check your final results.

Feb 3, 2016
38
0
Thank you
I try to draw the circuit diagram and transform the delta-connected capacitor bank to star connected
Then,the size of capacitor bank change into the one-third from the original ?
I1=IL+IC
I get IL from the above result
However, this question do not give any value of power?
How can I find the value of I1 and calculate the current IC?

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4. ### MrAl Well-Known Member

Jun 17, 2014
2,206
433
Hi again,

Do you understand complex math? Each source can be represented as a complex quantity and then you can use an analysis technique such as Nodal to find any other quantity you want to know such as Ic, I1, I2.
Because the two networks are symmetrical, you might be able to call each center node neutral and then use the line to neutral voltages to calculate the currents Ic and I2, and the sum would be I1. You should really try it both ways though.

LATER:
I verified that the symmetrical technique works too when all of the source voltages are the same and all components in all three phases are the same and connected the same way. Thus we can use the line to virtual neutral to calculate the currents and compute the value of one capacitor, which will then be the value for all three capacitors.
To transform to delta each capacitor value becomes 1/3 of the wye capacitor value (so yes you are right about the transformation of cap values).

It is still wise to do the analysis for the complete problem though, with all three sources and all three caps and all three loads. That's because when a problem that comes up that is not symmetrical, you'll still be able to handle it. For this example everything is the same for all three phases, but if you come across a network where say C1=1, C2=2, C3=3, L1=4, L2=5, L3=6, R1=1, R2=2, R3=3, you will not be able to use the symmetrical network trick.

What did you mean by 90 degrees for unity power factor? Unity power factor would be cos(0)=1 which requires zero degrees. Thus the angle between current and voltage is zero.

Last edited: Oct 6, 2016

Feb 3, 2016
38
0
Sorry,I can't follow you .....May be I am not clear about the concept of 3-ph power
I1=Ic+I2
I only know the value of I2 and there are two unknowns(Ic,I2) in this equation. How can I use nodal analysis to find Ic and I1?

When the power factor equal to 1, it means that sinθ=0. The reactive power become 0?(because Q=S*sinθ)

6. ### MrAl Well-Known Member

Jun 17, 2014
2,206
433
Hi,

I meant use Nodal to find all the quantities if you care to do that, which would be a good idea too. However, for the star version it's actually just a matter of doing each phase individually anyway assuming a virtual zero neutral (all symmetrical).

To solve the circuit as a single phase repeated three times, you have the inductor (and resistor) in parallel with the cap, that means i1=i2+iC. You know i2 because i2=V/(R+jwL) but you also know indirectly iC because iC=V/(1/(jwC)). If you put that together you can solve for C because:
i1=i2+iC=V/(R+jwL)+V/(1/(jwC))

or just:
i1=V/(R+jwL)+V/(1/(jwC))

So you see how this works?
This is an AC equation so we can either solve for the amplitude or the phase angle or both. Since we dont know C yet, we have to look for another way. We can note that since for a unity power factor we have:
cos(A)=1
and solving for the primary angle A we get:
A=0
and so we know the angle must be zero. Thus, we know that the angle of i1 is zero and so that means if we use the above equation:
i1=V/(R+jwL)+V/(1/(jwC))

and only look at the angle, we have:
angle(i1)=angle(V/(R+jwL)+V/(1/(jwC)))
or:
0=angle(V/(R+jwL)+V/(1/(jwC)))

Now we are down to only one unknown, the capacitor value C.

Try that and see if you can solve this.

Another way to look at it is if we have two complex currents a+bj and cj, we get zero phase angle when b=-c (ie when the currents add we get only a real part). This means that we can calculate the imaginary part of the inductor/resistor current and then try to match that with the negative of the imaginary part of the capacitor current. Or:
imagpart(iL)=-imagpart(iC)

If you care to try that it's actually easier. After that it's just a matter of transforming back into delta.

Keep in mind though that these ideas only work when the entire network is symmetrical, or you can do each phase individually as in a complete wye setup.

I've included a small diagram. The formula there is just to show the relationship between the angle and amplitudes.

MUCH LATER:
If you want to double check once you get the delta connected caps connected up, you can calculate the cap current from:
iC1=(Va-Vb)/(zC)
iC3=(Va-Vc)/(zC)
then:
iC=iC1+iC3
and you can add that to the current through the inductor/resistor caused by Va.
Va, Vb, Vc, are the three complex phase voltages here.
Check that the phase of the current from Va is zero or alternately that the imaginary part of the total current from Va is zero.

• ###### PowerFactorCorrection-1.gif
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Last edited: Oct 7, 2016

Feb 3, 2016
38
0
Thank you so much !!
I finally get the value of 2.631uF. And I use different method to check my answer and get the same result.

8. ### MrAl Well-Known Member

Jun 17, 2014
2,206
433
Hi,

Oh ok, that's very good, but are you sure you have the decimal point in the right position in your final answer for the cap value? Maybe double check to make sure you did not type it out wrong
You did say it was a 50Hz system right, and R=20 Ohms and L=0.07 Henries ?
You typed: 2.631uF, with the decimal point after the '2'.