Question

A source supplies power to the following three loads connected in parallel: (1) a lighting load drawing 10 kW, (2) an induction motor drawing 10 kVA at 0.90 power factor lagging, and (3) a synchronous motor operating at 10 hp, 85% efficiency and 0.95 power factor leading (1 hp = 0.746 kW). Determine the real, reactive, and apparent power delivered by the source. Also, draw the source power triangle.

Short answer

The real, reactive and apparent power is obtained as, 27.78 kW, 1.474 kVAR and 27.82 kVA. Thr power triangle is drawn as,

Step-by-step solution

Describe the given information

The total power drawn by the parallel loads is sum of the individual powers

${\mathbf{S}}_{\mathbf{1}}\mathbf{+}{\mathbf{S}}_{\mathbf{2}}\mathbf{+}{\mathbf{S}}_{\mathbf{3}}$

It is given that three loads are connected in parallel. First load is motor drawing 10 kW. Thus the apparent power drawn by first load is $S_1=10+jkW$Second load is a synchronous motor drawing drawing 10 kVA at 0.90 power factor lagging. Thus, the apparent power drawn by second load is $S_2=10\angle {\cos }^{-1}0.9$.

Simplify $S_2=10\angle {\cos }^{-1}0.9$

$$\begin{gathered} S_2=10\angle {\cos }^{-1}0.9 \\ =10\left(\cos \left(\angle {\cos }^{-1}0.9\right)+j\sin \left(\angle {\cos }^{-1}0.9\right)\right) \\ =10\left(0.9+j0.4359\right) \\ =9+j4.359 \end{gathered}$$

Third load is a synchronous motor operating at 10 hp, 85% efficiency and 0.95 power factor leading .As thers is 0.746 kW in 1 hp. Thus, there is $10\times 0.746\mathrm{kW}$in 10 hp, that is equal to 7.46 kW. The efficiency is 85% and power facor is 0.95 , thus the real part obtained by dividing the operating power by efficiency and power factor.Thus, the apparent power drawn by third load is $S_3=\frac{10\times 0.746}{0.85\times 0.95}\angle {\cos }^{-1}0.95$

$$\begin{gathered} S_3=\frac{10\times 0.746}{0.85\times 0.95}\angle {\cos }^{-1}0.95 \\ =9.238\angle {\cos }^{-1}0.95 \\ =9.238\left(\cos \left(\angle {\cos }^{-1}0.95\right)+j\sin \left(\angle {\cos }^{-1}0.95\right)\right) \\ =9.238\left(0.95+j0.312\right) \\ =8.776-j2.885 \end{gathered}$$

obtain the total power

Substitute $8.776-j2.885$for $S_3,9+j4.359\mathrm{for}{\mathrm{S}}_2$and $10+j0\mathrm{for}{S}_3$into ${\mathrm{S}}_1+{\mathrm{S}}_2+{\mathrm{S}}_3$.

$$\begin{gathered} {\mathrm{S}}_1+{\mathrm{S}}_2+{\mathrm{S}}_3=8.776-\mathrm{j}2.885+9+\mathrm{j}4.359+10+\mathrm{j}0 \\ =27.78+\mathrm{j}1.474 \end{gathered}$$

Thus, the total apparent power drawn by the loads or delivered by the souce in the circuit is $27.78+\mathrm{j}1.474$.

Whereas for a pure capacitive load, the current into the load is in perpendicular with the load voltage, where current lead voltage in phase by $90°$.

Thus, the (a) is incorrect option.

Obtain the total real power

The total real power delivered by the source is the real part of the total power, which is equal to 27.78 kW.

Obtain the total reactive power

The total reactive power delivered by the source is the imaginary part of the total power, which is equal to 1.474 kVAR.

Obtain the total apparent power

The total apparent power delivered by the source is the magnitude of the total power, which is equal to $\left|27.78+j1.474\right|$.

Simplify $\left|27.78+j1.474\right|$.

$$\begin{gathered} \left|27.78+j1.474\right|=\sqrt{{\left(27.78\right)}^2+{\left(1.474\right)}^2} \\ =27.82\mathrm{kVA} \end{gathered}$$

Draw the source power triange

In the source triangle, the source or hypotenuse is the apparent power, 27.82 kVA , the base is the real power, 27.78 kW , and perpendicular is the reactive power, 1.474 kVAR.

Thus, the source power triangle is obtained as,