Question

Using base values of 20kVA and 115 volts in zone 3, rework Example 3.4.

Short answer

(a) The per-unit impedances are $0.1j\mathrm{pu},0.189j\mathrm{pu},0.07259j\mathrm{pu}$and $1.36+0.302j\mathrm{pu}$.

(b) The supply voltage is $0.9596\angle 0°\mathrm{pu}$.

(c) The single line-diagram is drawn below:

(d) The load current is$0.63\angle -26.01°\mathrm{pu}\mathrm{and}109.56\angle -26.01°\mathrm{A}$.

Step-by-step solution

Given data

The base power is 20kVA , and the base voltage at the load end is 115 V .

Determine the formulas of base impedance, per-unit impedance, per-unit voltage, per-unit current, base current and actual current

Write the formula of base impedance.

${\mathbf{Z}}_{\mathbf{base}\mathbf{}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{base}}^{\mathbf{2}}}{{\mathbf{S}}_{\mathbf{base}}}$ ……. (1)

Here, $S_{base}$is base power, and $V_{base}$is base voltage.

Write the formula of new reactance.

${\mathbf{X}}_{\mathbf{new}}\mathbf{=}{\mathbf{X}}_{\mathbf{old}}{\left(\frac{V_{\mathrm{old}}}{V_{\mathrm{new}}}\right)}^{\mathbf{2}}\left(\frac{S_{\mathrm{new}}}{S_{\mathrm{old}}}\right)$ ……. (2)

Here,$V_{old}$is the old base of voltage; $V_{new}$is the new base of voltage; $S_{old}$is the old power base; $S_{new}$is the new power base; $X_{old}$is the old reactance.

Write the formula of per-unit load impedance.

${\mathbf{Z}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{Z}}_{\mathbf{actual}}}{{\mathbf{Z}}_{\mathbf{base}}}$ ……. (3)

Here,$Z_{base}$is the base impedance, and$Z_{actual}$ is the actual impedance.

Write the formula of per-unit voltage.

${\mathbf{V}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{actual}}}{{\mathbf{V}}_{\mathbf{base}}}$ ……. (4)

Write the formula of the base current.

${\mathbf{I}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{{\mathbf{V}}_{\mathbf{base}}}$ ……. (5)

Write the formula of per-unit current.

${\mathbf{P}}_{\mathbf{pu}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{pu}}}{{\mathbf{X}}_{\mathbf{T}\mathbf{1}}\mathbf{+}{\mathbf{X}}_{\mathbf{T}\mathbf{2}}\mathbf{+}{\mathbf{X}}_{\mathbf{line}}\mathbf{+}{\mathbf{X}}_{\mathbf{pu}}}$ ……. (6)

Here, $X_{T1}$is the reactance of transformer 1; $X_{T2}$is the reactance of transformer 2; $X_{line}$is the reactance of the transmission line;$V_{pu}$ is the resupply voltage;$Z_{pu}$ is the impedance of the load.

Write the formula of the actual current.

${\mathbf{I}}_{\mathbf{actual}}\mathbf{=}{\mathbf{I}}_{\mathbf{pu}}{\mathbf{I}}_{\mathbf{base}}$ ……. (7)

Here, $I_{pu}$is per-unit current, and $I_{base}$is base current.

Determine the per-unit impedance

(a) Determine the baseload impedance.

Substitute $20\mathrm{kVA}\mathrm{for}{S}_{base}\mathrm{and}115\mathrm{V}\mathrm{for}{V}_{base}$in equation (1).

$$\begin{gathered} Z_{base}=\frac{{\left(115\mathrm{V}\right)}^3}{20\mathrm{kVA}} \\ =0.16612\mathrm{\Omega } \end{gathered}$$

Determine the per-unit load impedance.

Substitute $0.9+j0.2\mathrm{\Omega }\mathrm{for}{Z}_{actual}\mathrm{and}0.6612\mathrm{\Omega }\mathrm{for}{Z}_{base}$in equation (3).

$$\begin{gathered} Z_{pu}=\frac{0.9+\mathrm{j}0.2\mathrm{\Omega }}{0.6612\mathrm{\Omega }} \\ =1.36+j0.302\mathrm{pu} \end{gathered}$$

Determine the base impedance of the transmission line.

Substitute $20\mathrm{kVA}\mathrm{for}{S}_{base}\mathrm{and}460\mathrm{V}\mathrm{for}{V}_{base}$in equation (1).

$$\begin{gathered} Z_{base}=\frac{{\left(460\mathrm{V}\right)}^2}{20\mathrm{kVA}} \\ =10.58\mathrm{\Omega } \end{gathered}$$

Determine the per-unit transmission line impedance

Substitute $2\mathrm{\Omega }\mathrm{for}{Z}_{actual}\mathrm{and}10.58\mathrm{\Omega }\mathrm{for}{Z}_{base}$in equation (3).

$$\begin{gathered} Z_{pu}=\frac{2\mathrm{\Omega }}{10.58\mathrm{\Omega }} \\ =0.189\mathrm{pu} \end{gathered}$$

Determine the per-unit transformer 1 reactance.

Substitute $0.1\mathrm{pu}\mathrm{for}{X}_{old},20\mathrm{kVA}\mathrm{for}{S}_{new},30\mathrm{kVA}\mathrm{for}{S}_{old},480\mathrm{V}\mathrm{for}{V}_{old}$and $460\mathrm{V}\mathrm{for}{V}_{new}$in equation (2).

role="math" localid="1655281102522" $$\begin{gathered} X_{new}=\left(0.1\mathrm{pu}\right){\left(\frac{480\mathrm{V}}{460\mathrm{V}}\right)}^2\left(\frac{20\mathrm{kVA}}{30\mathrm{kVA}}\right) \\ =0.07259\mathrm{pu} \end{gathered}$$

Therefore, The per-unit impedances are$0.1j\mathrm{pu},0.189j\mathrm{pu},0.07259j\mathrm{pu}$ and $1.36+0.302j\mathrm{pu}$.

Determine the per-unit source voltage

(b) Determine the base voltage at the generator.

$$\begin{gathered} V_{base}=240\times \left(\frac{460}{480}\right) \\ =230\mathrm{V} \end{gathered}$$

Determine the per-unit supply voltage.

Substitute $230\mathrm{V}\mathrm{for}{V}_{base}\mathrm{and}220\mathrm{V}\mathrm{for}{V}_{actual}$in equation (4).

$$\begin{gathered} V_{pu}=\frac{220\mathrm{V}}{230\mathrm{V}} \\ =0.9565\mathrm{pu} \end{gathered}$$

Therefore, the supply voltage is$0.9596\angle 0°\mathrm{pu}$ .

Determine the single-line diagram

(c) Draw the single-line diagram.

Determine the load current

(d)

Determine the base current.

Substitute $20\mathrm{kVA}\mathrm{for}{S}_{base}\mathrm{and}115\mathrm{V}\mathrm{for}{V}_{base}$in equation (5).

$$\begin{gathered} I_{base}=\frac{20\mathrm{kVA}}{115\mathrm{V}} \\ =173.91\mathrm{A} \end{gathered}$$

Determine the per-unit load current.

Substitute $0.1j\mathrm{pu}\mathrm{for}{X}_{T2},0.189j\mathrm{pu}\mathrm{for}{X}_{line},0.07259j\mathrm{pu}\mathrm{for}{X}_{T1}$role="math" localid="1655282510949" $1.36+0.302j\mathrm{pu}$for$Z_{pu}\mathrm{and}0.9596\angle 0°\mathrm{pu}\mathrm{for}{V}_{pu}$in equation (6).

$$\begin{gathered} I_{pu}=\frac{0.9596\angle 0°pu}{0.1j\mathrm{pu}+0.189j\mathrm{pu}+0.07259j\mathrm{pu}+1.36+0.302j\mathrm{pu}} \\ =0.63\angle -26.01°\mathrm{pu} \end{gathered}$$

Determine the actual load current.

Substitute $0.63\angle -26.01°\mathrm{pu}\mathrm{for}{I}_{pu}\mathrm{and}173.91\mathrm{A}\mathrm{for}{I}_{base}$for and for in equation (7).

$$\begin{gathered} I_{actual}=\left(173.91\mathrm{A}\right)\left(0.63\angle -26.01°\right) \\ =109.56\angle -26.01°\mathrm{A} \end{gathered}$$

Therefore, the load current is $0.63\angle -26.01°\mathrm{pu}$and$109.56\angle -26.01°\mathrm{A}$ .