Chapter 7, Problem 4P

To determine

The missing values.

Explanation of Solution

The given values are shown in below table:

Alternator (A)		Load 1 (L1)		Load 2 (L2)		Load 2 (L2)
${\text{E}}_{\text{P}\left(\text{A}\right)}$		${\text{E}}_{\text{P}\left(\text{L1}\right)}$		${\text{E}}_{\text{P}\left(\text{L2}\right)}$		${\text{E}}_{\text{P}\left(\text{L3}\right)}$
${\text{I}}_{\text{P}\left(\text{A}\right)}$		${\text{I}}_{\text{P}\left(\text{L1}\right)}$		${\text{I}}_{\text{P}\left(\text{L2}\right)}$		${\text{I}}_{\text{P}\left(\text{L3}\right)}$
${\text{E}}_{\text{L}\left(\text{A}\right)}$	480	${\text{E}}_{\text{L}\left(\text{L1}\right)}$		${\text{E}}_{\text{L}\left(\text{L2}\right)}$		${\text{E}}_{\text{L}\left(\text{L3}\right)}$
${\text{I}}_{\text{L}\left(\text{A}\right)}$		${\text{I}}_{\text{L}\left(\text{L1}\right)}$		${\text{I}}_{\text{L}\left(\text{L2}\right)}$		${\text{I}}_{\text{L}\left(\text{L3}\right)}$
		${\text{R}}_{\left(\text{Phase}\right)}$	$12\Omega$	${\text{X}}_{\text{L}\left(\text{Phase}\right)}$	$16\Omega$	${\text{X}}_{\text{C}\left(\text{Phase}\right)}$	$10\Omega$
		P		${\text{VARs}}_{\text{L}}$		${\text{VARs}}_{\text{C}}$

Refer to the circuit shown in Figure 7-21, all the both loads L1, L2, and L3 are connected to the alternator. Thus, the line voltages are equal.

  ${\text{E}}_{\text{L}\left(\text{A}\right)}={\text{E}}_{\text{L}\left(\text{L1}\right)}={\text{E}}_{\text{L}\left(\text{L2}\right)}={\text{E}}_{\text{L}\left(\text{L3}\right)}=480$

Consider load 3:

Here, the load 3 is made by 3 capacitors and uses wye connection. In wye connection, the line voltage is equal to $\sqrt{3}$ times phase voltage.

Thus,

  $\begin{array}{c}{\text{E}}_{\text{L}\left(\text{L3}\right)}=\sqrt{3}{\text{E}}_{\text{P}\left(\text{L3}\right)}\\ {\text{E}}_{\text{P}\left(\text{L3}\right)}=\frac{{\text{E}}_{\text{L}\left(\text{L3}\right)}}{\sqrt{3}}\\ =\frac{480}{\sqrt{3}}\\ =277.13\end{array}$

Calculate the phase current of the load 3$\left({\text{I}}_{\text{P}\left(\text{L3}\right)}\right)$.

  $\begin{array}{c}{\text{I}}_{\text{P}\left(\text{L3}\right)}=\frac{{\text{E}}_{\text{P}\left(\text{L3}\right)}}{{\text{X}}_{\text{C}\left(\text{phase}\right)}}\\ =\frac{277.13}{10\Omega }\\ =27.71\end{array}$

In wye connection, the phase current $\left({\text{I}}_{\text{P}}\right)$ and line current $\left({\text{I}}_{\text{L}}\right)$ are equal.

Thus,

  $\begin{array}{l}{\text{I}}_{\text{L}\left(\text{L3}\right)}={\text{I}}_{\text{P}\left(\text{L3}\right)}\\ {\text{I}}_{\text{L}\left(\text{L3}\right)}=27.71\end{array}$

Here, the load 3 is pure capacitive, the voltage and current are 90° out of phase with each other. Thus, the power factor becomes zero.

Calculate the reactive power of load 3 $\left({\text{VARs}}_{\text{C}}\right)$.

  $\begin{array}{c}{\text{VARs}}_{\text{C}}=\sqrt{3}\times {\text{E}}_{\text{L}\left(\text{L3}\right)}\times {\text{I}}_{\text{L}\left(\text{L3}\right)}\\ =\sqrt{3}\times 480\times 27.713\\ =23040.16\end{array}$

Consider load 2:

Here, the load 2 is made by 3 inductors and uses delta connection. In delta connection, the line voltage is equal to phase voltage.

  $\begin{array}{l}{\text{E}}_{\text{P}\left(\text{L2}\right)}={\text{E}}_{\text{L}\left(\text{L2}\right)}\\ {\text{E}}_{\text{P}\left(\text{L2}\right)}=480\end{array}$

Calculate the phase current of the load 2$\left({\text{I}}_{\text{P}\left(\text{L2}\right)}\right)$.

  $\begin{array}{c}{\text{I}}_{\text{P}\left(\text{L2}\right)}=\frac{{\text{E}}_{\text{P}\left(\text{L2}\right)}}{{\text{X}}_{\text{L}\left(\text{phase}\right)}}\\ =\frac{480}{16\Omega }\\ =30\end{array}$

Calculate the line current of the load 2 $\left({\text{I}}_{\text{L}\left(\text{L2}\right)}\right)$.

  $\begin{array}{c}{\text{I}}_{\text{L}\left(\text{L2}\right)}=\sqrt{3}\times {\text{I}}_{\text{P}\left(\text{L2}\right)}\\ =\sqrt{3}\times 30\\ =51.9615\\ \approx 51.96\end{array}$

Calculate the inductive power of load 2 $\left({\text{VARs}}_{\text{L}}\right)$.

  $\begin{array}{c}{\text{VARs}}_{\text{L}}=\sqrt{3}\times {\text{E}}_{\text{L}\left(\text{L2}\right)}\times {\text{I}}_{\text{L}\left(\text{L2}\right)}\\ =\sqrt{3}\times 480\times 51.96\\ =43198.73\end{array}$

Consider load 1:

Here, the load 1 is made by 3 resistors and uses wye connection. In wye connection, the line voltage is equal to $\sqrt{3}$ times phase voltage.

Thus,

  $\begin{array}{c}{\text{E}}_{\text{L}\left(\text{L1}\right)}=\sqrt{3}{\text{E}}_{\text{P}\left(\text{L1}\right)}\\ {\text{E}}_{\text{P}\left(\text{L1}\right)}=\frac{{\text{E}}_{\text{L}\left(\text{L1}\right)}}{\sqrt{3}}\\ =\frac{480}{\sqrt{3}}\\ =277.13\end{array}$

Calculate the phase current of the load 1$\left({\text{I}}_{\text{P}\left(\text{L1}\right)}\right)$.

  $\begin{array}{c}{\text{I}}_{\text{P}\left(\text{L1}\right)}=\frac{{\text{E}}_{\text{P}\left(\text{L1}\right)}}{\text{R}}\\ =\frac{277.13}{12}\\ =23.0942\end{array}$

Here, the load 1 is made by 3 resistors and uses wye connection. In wye connection, the line current is equal to phase current.

  $\begin{array}{l}{\text{I}}_{\text{L}\left(\text{L1}\right)}={\text{I}}_{\text{P}\left(\text{L1}\right)}\\ {\text{I}}_{\text{L}\left(\text{L1}\right)}=23.0942\end{array}$

Calculate the resistive power of load 1$\left(\text{P}\right)$.

  $\begin{array}{c}\text{P}=\sqrt{3}\times {\text{E}}_{\text{L}\left(\text{L1}\right)}\times {\text{I}}_{\text{L}\left(\text{L1}\right)}\\ =\sqrt{3}\times 480\times 23.0942\\ =19200.13\end{array}$

Consider the alternator:

Since, the alternator is connected to the loads such as resistive $\left(\text{R}\right)$, inductive $\left(\text{L}\right)$, and capacitive $\left(\text{C}\right)$.

Calculate the total current supplied by the alternator to the RLC circuit.

  $\begin{array}{c}{\text{I}}_{\text{L}\left(\text{A}\right)}=\sqrt{{\left[{\text{I}}_{\text{L}\left(\text{L1}\right)}\right]}^2+{\left[{\text{I}}_{\text{L}\left(\text{L2}\right)}-{\text{I}}_{\text{L}\left(\text{L3}\right)}\right]}^2}\\ =\sqrt{{23.0942}^2+{\left(51.96-27.71\right)}^2}\\ =33.4873\\ \approx 33.49\end{array}$

Here, the alternator uses wye connection. In wye connection, the line current is equal to phase current.

  $\begin{array}{l}{\text{I}}_{\text{P}\left(\text{A}\right)}={\text{I}}_{\text{L}\left(\text{A}\right)}\\ {\text{I}}_{\text{P}\left(\text{A}\right)}=33.49\end{array}$

In wye connection, the line voltage is equal to $\sqrt{3}$ times phase voltage.

Thus,

  $\begin{array}{c}{\text{E}}_{\text{L}\left(\text{A}\right)}=\sqrt{3}{\text{E}}_{\text{P}\left(\text{A}\right)}\\ {\text{E}}_{\text{P}\left(\text{A}\right)}=\frac{{\text{E}}_{\text{L}\left(\text{A}\right)}}{\sqrt{3}}\\ =\frac{480}{\sqrt{3}}\\ =277.13\end{array}$

Calculate the apparent power of alternator $\left(\text{VA}\right)$.

  $\begin{array}{c}\text{VA}=\sqrt{3}\times {\text{E}}_{\text{L}\left(\text{A}\right)}\times {\text{I}}_{\text{L}\left(\text{A}\right)}\\ =\sqrt{3}\times 480\times 33.49\\ =27843.06\end{array}$

Thus, the all missing values are calculated and shown in below table:

Alternator (A)		Load 1 (L1)		Load 2 (L2)		Load 2 (L2)
${\text{E}}_{\text{P}\left(\text{A}\right)}$	277.13	${\text{E}}_{\text{P}\left(\text{L1}\right)}$	277.13	${\text{E}}_{\text{P}\left(\text{L2}\right)}$	480	${\text{E}}_{\text{P}\left(\text{L3}\right)}$	277.13
${\text{I}}_{\text{P}\left(\text{A}\right)}$	33.49	${\text{I}}_{\text{P}\left(\text{L1}\right)}$	23.09	${\text{I}}_{\text{P}\left(\text{L2}\right)}$	30	${\text{I}}_{\text{P}\left(\text{L3}\right)}$	27.71
${\text{E}}_{\text{L}\left(\text{A}\right)}$	480	${\text{E}}_{\text{L}\left(\text{L1}\right)}$	480	${\text{E}}_{\text{L}\left(\text{L2}\right)}$	480	${\text{E}}_{\text{L}\left(\text{L3}\right)}$	480
${\text{I}}_{\text{L}\left(\text{A}\right)}$	33.49	${\text{I}}_{\text{L}\left(\text{L1}\right)}$	23.09	${\text{I}}_{\text{L}\left(\text{L2}\right)}$	51.96	${\text{I}}_{\text{L}\left(\text{L3}\right)}$	27.71
VA	27843.06	${\text{R}}_{\left(\text{Phase}\right)}$	$12\Omega$	${\text{X}}_{\text{L}\left(\text{Phase}\right)}$	$16\Omega$	${\text{X}}_{\text{C}\left(\text{Phase}\right)}$	$10\Omega$
		P	33255.36	${\text{VARs}}_{\text{L}}$	43198.73	${\text{VARs}}_{\text{C}}$	23040.16