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Three phase AC concepts and problems

  • Thread starter PainterGuy
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  • #1
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Hi

Problem:
Three loads are connected in parallel across a 12.47 kV three-phase supply.
Load 1: Inductive load, 60 kW and 660 kvar.
Load 2: Capacitive load, 240 kW at 0.8 power factor.
Load 3: Resistive load of 60 kW.
(a) Find the total complex power, power factor, and the supply current.
(b) A Y-connected capacitor bank is connected in parallel with the loads.
Find the total kvar and the capacitance per phase in μF to improve the overall
power factor to 0.8 lagging. What is the new line current?

Attempt at the solution:
Please have a look here, or the code below which might some formatting errors.

Three loads
are connected in parallel across a 12.47 kV three-phase supply.


Load 1: Inductive load, 60 kW and 660 kvar.


Load
2: Capacitive load, 240 kW at 0.8 power factor.


Load 3: Resistive load of 60 kW.


(a) Find the total complex
power, power factor, and the supply current.


(b) A Y-connected capacitor bank is connected in parallel with the loads.




Find the total kvar and the capacitance per phase in ##\mu##F to improve the overall


power factor to 0.8 lagging. What is
the new line current?


Load 1: ##\mathbf{S}_{l o a d 1} =60 k +j 660 k##


Load 2: Using Power Factor ## =\frac{P}{S} =\cos (\theta _{v} -\theta _{i}) \Longrightarrow S =\frac{P}{\cos (\theta _{v} -\theta _{i})} =\frac{240 k}{0.8} =300 k## ;


##S =\vert \mathbf{S}\vert =\sqrt[{2}]{P^{2} +\vert Q\vert ^{2}} \Longrightarrow \vert Q\vert =\sqrt[{2}]{S^{2} -P^{2}} =\sqrt[{2}]{300 k^{2} -240 k^{2}} =180 k##


##\mathbf{S}_{l o a d 2} =240 k -j 180 k##


Load 3: ##\mathbf{S}_{l o a d 3} =60 k +j 0##


##\mathbf{S}_{t o t a l} =\left (60 k +j 660 k\right ) +\left (240 k -j 180 k\right ) +60 k =360 k +j 480 k##


Using ##S =\vert \mathbf{S}\vert =\sqrt[{2}]{P^{2} +\vert Q\vert ^{2}}## \qquad and\qquad Power Factor ## =\frac{P}{S} =\cos (\theta _{v} -\theta _{i})##


##S =\vert \mathbf{S}\vert =\sqrt[{2}]{P^{2} +\vert Q\vert ^{2}}## ## =\sqrt[{2}]{360 k^{2} +\vert 480 k\vert ^{2}} =600 k##


##\frac{P}{S} =\cos (\theta _{v} -\theta _{i}) \Longrightarrow \cos (\theta _{v} -\theta _{i}) =\frac{360 k}{600 k} =0.6##


Assuming ##12.47## ##k V## as RMS line voltage of wye-connected source.


Using


##V_{L} =\sqrt[{2}]{3} V_{p}## where ##V_{L} =V_{a b} =V_{b c} =V_{b c}## and ##V_{p} =V_{a n} =V_{b n} =V_{c n}##


Also the line voltages lead their corresponding phase voltages by ##30 \mbox{{\ensuremath{{}^\circ}}}##, i.e.


##\mathbf{V}_{L} =\left (\sqrt[{2}]{3} V_{p}\right )\angle \left (\theta _{V_{p}} +30 \mbox{{\ensuremath{{}^\circ}}}\right )##


## \Longrightarrow V_{p} =V_{a n} =\frac{V_{L}}{\sqrt[{2}]{3}} =\frac{12.47 k}{\sqrt[{2}]{3}} =7.2## ##k V##


hence ##\mathbf{V}_{a n} =7.2 k \angle 0 \mbox{{\ensuremath{{}^\circ}}}##


##\mathbf{S}_{t o t a l} =360 k +j 480 k## which means total power supplied is ##360## ##k W## and ##480## ##k V A R##


by all three phases of the source therefore power supplied by one phase


of the source
is ##120 k +j 160 k##.


Using ##\mathbf{S} =\mathbf{V}_{r m s} \mathbf{I}_{r m s}^{ \ast }##


##\mathbf{S} =\mathbf{V}_{r m s} \mathbf{I}_{r m s}^{ \ast } \Longrightarrow \mathbf{I}_{r m s}^{ \ast } =\frac{\mathbf{S}}{\mathbf{V}_{r m s}} =\frac{120 k +j 160 k}{7.2 k \angle 0 \mbox{{\ensuremath{{}^\circ}}}} =27.78 \angle 53.13 \mbox{{\ensuremath{{}^\circ}}}##


## \Longrightarrow \mathbf{I}_{r m s} =27.78 \angle -53.13 \mbox{{\ensuremath{{}^\circ}}}##


(b)


Per phase power supplied by the source is ##120 k +j 160 k##


##\cos (\theta _{v} -\theta _{i}) =0.6##


## \Longrightarrow \theta _{1} =53.13 \mbox{{\ensuremath{{}^\circ}}}##


##\theta _{2} =\cos ^{ -1} (0.8) =36.9 \mbox{{\ensuremath{{}^\circ}}}##


##\theta _{1}## is the original power factor angle and ##\theta _{2}## is the desired angle. Likewise, ##Q_{1}##


is the original reactive power and ##Q_{2}## is the desired reactive power at ##\theta _{2}##.


##Q_{i n j e c t e d} =Q_{1} -Q_{2} =P (\tan \theta _{1} -\tan \theta _{2}) =120 k (\tan 53.13 \mbox{{\ensuremath{{}^\circ}}} -\tan 36.9 \mbox{{\ensuremath{{}^\circ}}}) =70## ##k V A R##


Original per phase power supplied by the source is ##120 k +j 160 k##


After addition of capacitor power supplied is


##\left (120 k +j 160 k\right ) +(0 -70 k) =120 k +j 90 k##


Using ##C =\frac{Q_{C}}{\omega V r m s^{2}}##


##C =\frac{Q_{C}}{\omega V r m s^{2}} =\frac{70 k}{2 \pi \times 60 \times 7.2 k^{2}} =3.6## ##\mu F##


Using formula \#1:


##R =\frac{V_{r m s}^{2}}{P} =\frac{7.2 k^{2}}{120 k} =432## \ \ \ ;
\ \ \ \ ##X =\frac{V_{r m s}^{2}}{Q} =\frac{7.2 k^{2}}{90 k} =576##


As power factor is lagging therefore ##\mathbf{Z} =432 +j 576##


##\mathbf{I}_{a} =\frac{\mathbf{V}_{a n}}{\mathbf{Z}} =\frac{7.2 k \angle 0 \mbox{{\ensuremath{{}^\circ}}}}{432 +j 576} =13.33 \angle -36.9 \mbox{{\ensuremath{{}^\circ}}}##


Using different formula \#2:


##V_{r m s} I_{r m s}## ##\angle \left (\theta _{v} -\theta _{i}\right ) =P +j Q##


##7.2 k I_{r m s}## ##\angle \left (0 -\theta _{i}\right ) =120 k +j 90 k##


## \Longrightarrow I_{r m s} \angle -\theta _{i} =20.83 \angle 36.87##


## \Longrightarrow \mathbf{I}_{a} =20.83 \angle -36.87##
My question:
Although I was able to find the line current in part (b) using formula #2 but am still curious to know that what went wrong with formula #1. I have used it a couple of times in the past to get the correct result. Where am I going wrong? I'm getting wrong Z which in turn leads to wrong value for Ia. I believe that the correct value for Z is 207.52+j207.39. Could you please help me? Thank you!
 

Answers and Replies

  • #2
689
33
Thank you for giving it a look. I have realized that it should have been S=Vrms^2/Z*, and it looks like that formula #1 is a wrong way get Z. Thank you.
 
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  • #3
scottdave
Science Advisor
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Thank you for giving it a look. I have realized that it should have been S=Vrms^2/Z*, and it looks like that formula #1 is a wrong way get Z. Thank you.
OK, good. I had started to look at it, but had not yet figured out enough to post a meaningful response.
 
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