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I have the following problem which require the calculation of Load L3 apparent power, real power and reactive power. Question1

I have done the following as the described in the figure trying to get 2 equations of 2 unknowns but got to a dead end

solution

how can i proceed with such problem with small information the global power factor is also giving but with no details on whether it is lagging or leading. i assumed it is lagging since L1 and L2 are inductive loads based on the description

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Intuitively, if all the load power factors are lagging, the overall power factor is lagging too.

$$% outer vertical array of arrays \begin{array}{c} % inner horizontal array of arrays \begin{array}{c|c} % inner array of minimum values \begin{align} S_1&=20\; \;\text{kVA}\\ V&=600 \angle{0}^{\circ} \\ I_1&=\frac{S}{V}=\frac{20k}{600}=33.3 \;\text{A}\\ \therefore {\overline{I}_{1}}&=33.3\angle{-36.87^{\circ}} \end{align} & % inner array of maximum values \overline{I}_{2}=\frac{V}{Z}=\frac{600}{15+30j}=8\sqrt{5}\angle{-63.43^{\circ}}\\ \end{array} \\ % inner array of delta values \hline \begin{matrix} &\overline{I}_{source}=\overline{I}=I\angle{-\arccos(0.707)}=I\angle{-45^{\circ}} \Longleftarrow\\ &\overline{I}_3=I_3\angle{-\arccos(0.6)}=I_3\angle{-53.13^{\circ}}\\ &{\overline{I}}={\overline{I}_{1}}+{\overline{I}_{2}}+{\overline{I}_{3}}\\ &\Im\{{\overline{I}_{1}}+{\overline{I}_{2}}\}=-35.98j\\ &\therefore I_3\sin(-53.13^{\circ})=-45^{\circ}-(-35.98^{\circ})=-4.02^{\circ}\\ &\text{So}\quad \overline{I}_{3}=5.03\angle{-53.13^{\circ}}\Longleftarrow\\ & \therefore I=54.94\angle{-45^{\circ}}\Longleftarrow\\ \end{matrix} \end{array}$$


$$\begin{array} &S_{L_3}={V}\times{I}_3=600\times5.03=3018\; \text{VA}\\ P_{L_3}=3018\times 0.6=1810.8 \;\text{W}\\ Q_{L_3}=\sqrt{3018^2-1810.8^2}=2414.4 \;\text{VAr}\\ P_{loss}=I^2R=54.94^2\times 0.3=905.5 \;\text{W}\\ V_i=\overline{V}+\overline{I}*\overline{Z}_{\text{line}}=600+(54.94\angle{-45^{\circ}})\cdot(0.3+j0.8)=643.0\angle1.73^{\circ} \end{array}$$

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You start by finding the impedance for each of the loads.

Load 1:

$$ I_{1}=\frac{20\times 10^{3}}{600}\angle -36.8 = 33.33A \angle -36.8 $$ $$ Z_{1}=\frac{V}{I_{1}}= \frac{600}{33.33\angle-36.8}=18\angle36.8$$

Load 2:

$$Z_{2}=R_{2}\left | \right |X_{2} $$

$$Z_{2}=((15)^{-1}+(j30)^{-1})^{-1}=13.42\angle 26.565$$

Load 3:

$$\cos(\tan^{-1}(\frac{X_{3}}{R_{3}}))=0.6$$ $$X_{3}=3.33R_{3}$$

Now consider all 3 loads in parallel and reduce it to a single impedance, you should get this in terms of one unknown variable.Then you know the overall power factor of the reduced load so solve it for one unknown variable R3 or X3.

From that point onwards you know the voltage drop across Z3 and hence I3, everything else derives itself as @Andyaka stated.

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Load 1 is a derivable impedance based on its VA, power factor and 600 volts applied. Load 2 has a fully defined impedance and Load 3 has a known power factor. This information along with the knowledge of what the overall power factor of the three loads are tells you enough to calculate the impedance of Load 3.

Everything else derives from this.

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  • \$\begingroup\$ Thankyou, but i Don't have S3, I3 and Z3 i will end up with 2 equations and 3 unknowns if i can get Stotal then i can get i total and then i3 since i got s1 s2 i1 i2 z1 z2 \$\endgroup\$ – chaosmind May 2 '15 at 14:47

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