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I need help with the following problem:

Given symmetric three phase system (see attachment) of phase voltages with angular frequency ω=100rad/s, R=5ωL=100Ω. Find capacitance of capacitor C such that the power factor of three phase receiver has maximum value.

enter image description here

After transformation of Y capacitors to Δ (see attachment), it gives $$C_1=C/3.$$

enter image description here

Now we have a Δ connection of impedance Z which is a parallel of R,j5ωL and C1. Let $$\underline{Z_1}=R+j5\omega L.$$ From given data we can find that L=0.2H. This gives $$\underline{Z_1}=100(1+j)\Omega.$$ Now

$$\underline{Z}=\frac{\underline{Z_1}\cdot(-jX_{C_1})}{\underline{Z_1}+(-jX_{C_1})}=\frac{300(3+j(3−2⋅10^4C))}{2⋅10^8C^2−6⋅10^4C+9}\Omega $$

Now we have a three phase system with receiver in Δ connection (see attachment): enter image description here

After Δ to Y transformation (see attachment), we get new impedance: $$\underline{Z_2}=\frac{\underline{Z}}{3}=\frac{100(3+j(3−2⋅10^4C))}{2⋅10^8C^2−6⋅10^4C+9}\Omega.$$

enter image description here

Let

$$\underline{Z_3}=\underline{Z_2}+jX_L=\underline{Z_2}+j20=\frac{20(15+j8(3−2⋅10^4C+25⋅10^6C^2))}{2⋅10^8C^2−6⋅10^4C+9}\Omega.$$ Now we have a clean Y receiver connection with impedance Z3 (see attachment): enter image description here

Question: We are not given any values for voltage, current or power, so how to express power factor cosϕ without knowing any of those values?

EDIT: Power factor can be expressed by $$\cos\phi=\frac{\mathfrak{R}(\underline{S})}{\sqrt{P^2+Q^2}}$$ where $$\underline{S}$$ is complex apparent power, P is active and Q is reactive power. Power factor has a maximum value when reactive power tends to zero. Since we know only the impedance, we can look at the imaginary part of impedance Z3. If we introduce a function $$f(C)=\frac{160(25\cdot 10^6C^2-2\cdot 10^4C+3)}{2⋅10^8C^2−6⋅10^4C+9}$$, minimum value of f(C) is $$\frac{-40}{3}$$ at $$C=3\cdot 10^{-4}F.$$ So, maximum power factor is for C=0.0003F.

Question: Is this correct?

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  • \$\begingroup\$ Power factor is a unit-less ratio. So if you really want to start with voltages or whatever (TL/DR: I am pretty sure it is not needed), you can just introduce them yourself, as they should go away in the end anyway. \$\endgroup\$ – Eugene Sh. May 17 '16 at 13:40
  • \$\begingroup\$ @Eugene Sh. Is there an explicit formula for power factor without currents or voltages? \$\endgroup\$ – user300048 May 17 '16 at 13:42
  • \$\begingroup\$ \$pf = \frac {P} {S} = cos \theta \$ the phase angle. \$\endgroup\$ – StainlessSteelRat May 17 '16 at 15:30
  • \$\begingroup\$ @StainlessSteelRat Am I missing something? $$P=\sqrt{3}U_L I_L\cos\theta,S=\sqrt{3}U_L I_L$$ where $$U_L,I_L$$ are line voltage and current. How to evaluate the maximum power factor in this problem? \$\endgroup\$ – user300048 May 17 '16 at 16:55
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The correct answer is obtained by making the absolute value of the imaginary part of the total equivalent impedance as small as possible (explanation follows below). If possible, make it zero, as this will yield the maximum possible power factor, which is one.

In your case, assuming you did all the delta-wye and wye-delta conversions correctly (I didn't check that), then the correct answer is given by setting the imaginary part of \$Z_3\$ to zero, that is: $$ 25⋅10^6 C^2 − 2⋅10^4 C + 3 = 0 $$ which, as you mention in the comments to another reply, yields two valid solutions, \$C=6·10^{-4} F\$ and \$C=2·10^{-4} F\$. If this equation had no positive roots, then you would have to look for a minimum of its absolute value.

Explanation

When people talk about the power factor of a given device without any reference to the voltage and current applied to it, they are implicitly thinking about connecting it to an ideal voltage source \$V\$. The expression for the apparent power consumed by the device in this simple circuit is \$S=V I^*\$, and the real power \$P\$ is just the real part of \$S\$. The power factor is defined as: $$ \cos\phi = \frac{P}{|S|} $$ By Ohm's law, \$V=IZ\$, therefore \$S = V I^*= |I|^2 Z = |I|^2 R + j|I|^2 X\$. So the power factor can be written as: $$ \cos\phi = \frac{|I|^2 R}{|I|^2 \sqrt{R^2+X^2}} = \frac{R}{\sqrt{R^2+X^2}} $$ As you can see, neither the voltage nor the current appear in the final expression. Also, in order to maximize the power factor (the original question), it becomes obvious that you need to minimize \$|X|\$. To be more precise, what maximizes the power factor is the minimization of the ratio \$|X|/R\$, because if we divide both numerator and denominator by R, we obtain: $$ \cos\phi = \frac{R}{\sqrt{R^2+X^2}} = \frac{1}{\sqrt{1+(\frac{X}{R})^2}} $$

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Hints: -

You did the right thing and converted star C to delta C and now you have a parallel resonant circuit problem to solve. Yes, that sounds like radio-talk but getting PF to unity is the same as tuning a tuned circuit to resonance.

So, you are looking to solve the impedance of series R and L all in parallel with C - solve for a purely resistive value. It's about half a page of math.

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  • \$\begingroup\$ Your method is incomplete since there are inductors to the left of impedances of series R and L in parallel with C (see image 2). Could you look at my updated part of the question? \$\endgroup\$ – user300048 May 17 '16 at 18:43
  • \$\begingroup\$ You are quite correct but I'm only giving hints. \$\endgroup\$ – Andy aka May 17 '16 at 19:05
  • \$\begingroup\$ If we find a minimum of function f(C) we get that C=0.0003F, but if we solve f(C)=0 we get C=0.0002F or C=0.0006F (quadratic equation). So, which equation is correct to solve, f'(C)=0 or f(C)=0? \$\endgroup\$ – user300048 May 17 '16 at 19:25
  • \$\begingroup\$ I dunno - I've not looked deeply at your equations because you seemed to be making a mountain of it. I just wanted to hint that you needn't know voltage, current or power to find the answer. \$\endgroup\$ – Andy aka May 17 '16 at 20:11

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