Question

The single-line diagram of a simple power system is shown in Figure 9.23 with per unit values. Determine the fault current at bus 2 for a three-phase fault. Ignore the effect of phase shift.

Short answer

The fault current at bus 2 is $5.115\angle -84.36°\text{\hspace{0.17em}pu}$.

Step-by-step solution

Write the given data from the question.

The pre fault value of the voltage, ${\mathrm{V}}_3=0.98\angle -10°\text{\hspace{0.17em}pu}$

The zero sequence ${\mathrm{X}}_{\mathrm{line}0}$, positive sequence ${\mathrm{X}}_{\mathrm{line}1}$and negative sequence ${\mathrm{X}}_{\mathrm{line}2}$reactance’s of the line are $0.1,0.1$ and $0.15\text{\hspace{0.17em}pu}$respectively.

The zero sequence $X_{T0}$, positive sequence $X_{T1}$and negative sequence $X_{T2}$reactance’s of the transformer are $0.1,0.1$ and $0.1\text{\hspace{0.17em}pu}$respectively.

The zero sequence $X_{G0}$, positive sequence $X_{G1}$and negative sequence $X_{G2}$reactance’s of the generator are $0.1,0.1$ and $0.05\text{\hspace{0.17em}pu}$ respectively.

The load of the system,

$\begin{array}{l}S=P+jQ\\ S=1+j0\\ S=1\text{\hspace{0.17em}pu}\end{array}$

Determine the equation to calculate the fault current at bus 2.

The equation to calculate the load current is given as,

${\mathbf{I}}_{\mathbf{load}}^{\mathbf{*}}\mathbf{=}\frac{\mathbf{S}}{{\mathbf{V}}_{\mathbf{3}}}$ …… (1)

Here${\mathbf{I}}_{\mathbf{load}}$is the load current.

The equation to calculate the supply voltage is given as,

$\mathbf{E}\mathbf{=}{\mathbf{V}}_{\mathbf{1}}\mathbf{+}{\mathbf{ZI}}_{\mathbf{loag}}$ …… (2)

Calculate the fault current at bus 2.

Draw the per unit positive sequence network of the system.

Calculate the load current.

Substitute $1\text{\hspace{0.17em}pu}$for $\mathrm{S}$and $0.98\angle -10°\text{\hspace{0.17em}pu}$for $V_3$ into equation (1).

$\begin{array}{l}{I}_{load}^{*}=\frac{1}{0.98\angle -10°}\\ I_{load}^{*}=1.02\angle 10°\\ I_{load}=1.02\angle -10°\text{\hspace{0.17em}pu}\end{array}$

Calculate the supply voltage from the positive sequence network.

Substitute $1\angle 0°\text{\hspace{0.17em}pu}$ for $V_1$, $j0.1$ for $Z$ and $1.02\angle -10°\text{\hspace{0.17em}pu}$for $I_{load}$ into equation (2).

$\begin{array}{l}E=1\angle 0°+j0.1(1.02\angle -10°)\\ E=1+0.102\angle 80°\\ E=1.022\angle 5.63°\text{\hspace{0.17em}pu}\end{array}$

Calculate the fault current from the positive sequence network.

$\begin{array}{l}{I}_{fault}=\frac{1.022\angle 5.63°}{j0.1+j0.1}\\ I_{fault}=\frac{1.022\angle 5.63°}{j0.2}\\ I_{fault}=\frac{1.022\angle 5.63°}{0.2\angle 90°}\\ I_{fault}=5.115\angle -84.36°\text{\hspace{0.17em}pu}\end{array}$

Hence the fault current at bus 2 is$5.115\angle -84.36°\text{\hspace{0.17em}pu}$ .