Question

Two identical synchronous machines, each rated 60 MVA and 15 kV with a sub transient reactance of 0.1 pu., are connected through a line of reactance on the base of the machine rating. One machine is acting as a synchronous generator, while the other is working as a motor drawing at 0.8 pf leading with a terminal voltage of 14.5 kV , when a symmetrical three phase fault occurs at the motor terminals. Determine the sub transient currents in the generator, the motor, and the fault by using the internal voltages of the machines. Choose a base of 60 MVA and 15 kV in the generator circuit.

Short answer

The subs transient current in generator, motor and the fault are $1593.21-j9963.33\mathrm{A},-1593.21-j23528.71\mathrm{A}$, and $-33492.04\mathrm{A}$respectively.

Step-by-step solution

Write the given data from the question:

The power rating of synchronous generator and motor,$S_{base}=60\mathrm{MVA}$

The voltage rating of synchronous generator and motor,$V_{base}=15\mathrm{kV}$

Line reactance,$X_L=0.1\mathrm{pu}$

Sub transient reactance,$X_d^{\text{'}\text{'}}=0.1\mathrm{pu}$

Termina voltage,$V_t=14.5\mathrm{kV}$

The motor load is$p=40\mathrm{MW}$ at

Load power factor $pf=0.8$leading.

The rated power and voltage in per unit is 1 pu .

Determine the formulas to calculate the sub transient currents in the generator, the motor, and the fault by using the internal voltages of the machines.

The equation to calculate base current is given as follows.

${\mathbf{I}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{base}}}$ …… (1)

The equation to calculate the fault voltage is given as follows.

${\mathbf{V}}_{\mathbf{t}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{t}}}{{\mathbf{V}}_{\mathbf{base}}}$ …… (2)

The equation to calculate the motor current is given as follows.

${\mathbf{I}}_{\mathbf{motor}}\mathbf{=}\frac{\mathbf{P}\mathbf{\angle }{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(\mathrm{pf}\right)}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{t}}\mathbf{cos}\left(f\right)}$ …… (3)

The equation to calculate the terminal voltage for generator is given as follows.

${\mathbf{V}}_{\mathbf{t}\mathbf{,}\mathbf{g}}\mathbf{=}{\mathbf{V}}_{\mathbf{f}}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{1}{\left(I_{\mathrm{motor}}\right)}_{\mathbf{pu}}$ …… (4)

Here, ${\mathbf{V}}_{\mathbf{t}\mathbf{,}\mathbf{g}}$is the generator terminal voltage and${\mathbf{I}}_{\mathbf{motor}}$is the motor current.

The equation to calculate the induced voltage for generator is given as follows.

${\mathbf{E}}_{\mathbf{g}}\mathbf{=}{\mathbf{V}}_{\mathbf{t}\mathbf{,}\mathbf{g}}\mathbf{+}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{1}{\left(I_{\mathrm{motor}}\right)}_{\mathbf{pu}}$ …… (5)

The equation to calculate the sub transient current for generator is given as follows.

${\mathbf{I}}_{\mathbf{g}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{g}}}{{\mathbf{X}}_{\mathbf{d}}^{\mathbf{\text{'}\text{'}}}\mathbf{+}{\mathbf{X}}_{\mathbf{L}}}$ …… (6)

The equation to calculate the terminal voltage for motor is given as follows.

${\mathbf{V}}_{\mathbf{t}\mathbf{,}\mathbf{g}}\mathbf{=}{\mathbf{V}}_{\mathbf{f}}$ …… (7)

The equation to calculate the induced voltage for motor is given as follows.

${\mathbf{E}}_{\mathbf{m}}\mathbf{=}{\mathbf{V}}_{\mathbf{t}\mathbf{,}\mathbf{m}}\mathbf{‐}\mathbf{j}\mathbf{0}\mathbf{.}\mathbf{1}{\left(I_{\mathrm{motor}}\right)}_{\mathbf{pu}}$ …… (8)

The equation to calculate the sub transient current for motor is given as follows.

${\mathbf{I}}_{\mathbf{m}}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{m}}}{{\mathbf{X}}_{\mathbf{d}}^{\mathbf{\text{'}\text{'}}}}$ …… (9)

The equation to calculate the sub transient current for three phase faults is given as follows.

${\mathbf{I}}_{\mathbf{f}}\mathbf{=}{\mathbf{I}}_{\mathbf{g}}\mathbf{+}{\mathbf{I}}_{\mathbf{m}}$ …… (10)

The equation to calculate the per unit quantity is given as follows.

$\mathbf{Per}\mathbf{}\mathbf{unit}\mathbf{=}\frac{\mathbf{Actual}\mathbf{}\mathbf{value}}{\mathbf{Base}\mathbf{}\mathbf{value}}$

Calculate the sub transient current in moto, generator and the fault by using the internal voltage.

Calculate the base current,

Substitute 60 MVA for $S_{base}$and 15 kV for $V_{base}$into equation (1).

$$\begin{gathered} I_{base}=\frac{60\times {10}^6}{\sqrt{3}\times 10\times {10}^3} \\ {I}_{base}=2309.5\mathrm{A} \end{gathered}$$

The pre fault circuit is shown below.

The post fault circuit is shown below.

Calculate the fault current.

Substitute 14.5 kV for$V_t$and 15 kV for $V_{base}$into equation (2).

$$\begin{gathered} V_f=\frac{14.5}{15} \\ {V}_f=0.967\angle 0° \end{gathered}$$

Calculate the motor current.

Substitute 40 MW for $P$, 0.8 for$pf$, 14.5 kV for $V_t$and 0.8 for $\cos \left(\varphi \right)$into equation (3).

$$\begin{gathered} I_{motor}=\frac{40\angle {\cos }^{-1}\left(0.8\right)}{\sqrt{3}\times 14.5\times 0.8} \\ {I}_{motor}=1.991\angle 26.87°\mathrm{kA} \end{gathered}$$

Calculate the motor current in per unit.

$$\begin{gathered} {\left(I_{motor}\right)}_{pu}=\frac{1.991\times {10}^3\angle 36.87°}{2309.5} \\ {\left(I_{motor}\right)}_{pu}=0.862\angle 36.87° \\ {\left(I_{motor}\right)}_{pu}=0.69+j0.52\mathrm{pu} \end{gathered}$$

Calculate the terminal voltage for generator.

Substitute 0.967 pu for $V_f$, $0.69+j0.52\mathrm{pu}$for${\left(I_{motor}\right)}_{pu}$into equation (4).

$$\begin{gathered} V_{t,g}=0.967+j0.1\left(0.69+j0.52\right) \\ {V}_{t,g}=0.967+j0.069-0.052 \\ {V}_{t,g}=0.915+j0.069pu \end{gathered}$$

Calculate the induced voltage for generator.

Substitute $0.915+j0.069\mathrm{pu}$for $V_{t,g}$, $0.69+j0.52\mathrm{pu}$for${\left(I_{motor}\right)}_{pu}$into equation (5).

$$\begin{gathered} E_g=0.915+j0.069+j0.1\left(0.69+j0.52\mathrm{pu}\right) \\ {E}_g=0.915+j0.069+j0.069-0.52\mathrm{pu} \\ {E}_g=0.863+j0.138\mathrm{pu} \end{gathered}$$

Calculate the sub transient generator current.

Substitute $0.863+j0.138\mathrm{pu}$for localid="1655889139996" $E_g,j0.1$for$X_d^{\text{'}}$and $j0.1$for$X_L$into equation (6).

$$\begin{gathered} I_g=\frac{0.863+j0.138}{j0.1+j0.1} \\ {I}_g=\frac{0.863+j0.138}{j0.2} \\ {I}_g=0.69-j4.315\mathrm{pu} \end{gathered}$$

Calculate the sub transient generator current in per unit value.

$$\begin{gathered} I_g=\left(0.69-j4.315\right)\times 2309.5 \\ {I}_g=1593.21-j9963.33\mathrm{A} \end{gathered}$$

Calculate the terminal voltage for motor.

Substitute 0.967 pu for $V_f$into equation (7).

$V_{t,m}=0.967\mathrm{pu}$

Calculate the induced voltage for motor.

Substitute 0.967 pu for $V_f,0.69+j0.52\mathrm{pu}$for${\left(I_{motor}\right)}_{pu}$into equation (8).

$$\begin{gathered} E_m=0.967-j0.1\left(0.69+j0.52\mathrm{pu}\right) \\ {E}_m=0.967-j0.069+0.52 \\ {E}_m=1.019+j0.069\mathrm{pu} \end{gathered}$$

Calculate the sub transient motor current.

Substitute $1.019+j0.069\mathrm{pu}$for $E_m$and $j0.1\mathrm{for}{X}_d^{\text{'}\text{'}}$for into equation (9).

$$\begin{gathered} I_m=\frac{1.019+j0.069\mathrm{pu}}{j0.1} \\ {I}_m=-0.69-j10.19\mathrm{pu} \end{gathered}$$

Calculate the sub transient motor current in per unit value.

$$\begin{gathered} I_m\left(-0.69-j10.19\right)\times 2309.5 \\ {I}_m=-1593.21-j23528.71\mathrm{A} \end{gathered}$$

Calculate the sub transient current for three phase faults.

Substitute $1593.21-j9963.33\mathrm{A}$for $I_g$, and $-1593.21-j23528.71\mathrm{A}$for $I_m$into equation (10).

localid="1655889009222" $$\begin{gathered} I_f=1593.21-j9963.33-1593.21-j23528.71 \\ {I}_f=-33492.04\mathrm{A} \end{gathered}$$

Hence, the subs transient current in generator, motor and the fault are $1593.21-j9963.33\mathrm{A},-1593.21-j23528.71\mathrm{A}\mathrm{and}-33492.04\mathrm{A}$respectively.