Question

Redo Example 11.12 with the assumption the generator is supplying 100 1 j10 MVA to the infinite bus.

Short answer

The active power current, the reactive power current and reactive voltage are $0.956\text{pu}$, $-1.614\text{pu}$and $1.29\text{pu}$respectively.

Step-by-step solution

Write the given data from the question.

The complex power,$S=100+j10\text{MVA}$

The base power .$S_{base}=100\text{MVA}$

The bus voltage.$V_{bus}=1\text{pu}$

The DFAGreactance.$X_{eq}=j0.8\text{pu}$

The equivalent reactance from example 11.3 is .$X=j0.22\text{pu}$

Determine the equation to calculate the active power current, reactive power current and reactive voltage.

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

$S_{pu}=\frac{S}{S_{base}}$...(1)

The equation to calculate the power factor is given as follows.

$pf=\frac{P}{\sqrt{P^2+Q^2}}$...(2)

The equation to calculate the current is given as follows., $I=\frac{P}{V\times pf}\angle co{s}^{-1}\left(pf\right)$…… (3)

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

$V_T=V_{bus}+IX$.....(4)

The equation to calculate the $I_{sorc}$is given as follows.

$I_{sorc}=I+\frac{V_T}{j{X}_{eq}}$....(5)

The equation to calculate reactive voltageis given as follows.

$E_q=-I_q{X}_{eq}$...(6)

Calculate the active power current, reactive power current and reactive voltage.

Calculate the per unit value of the power.

Substitute $100+j10\text{MVA}$for $S$and $100\text{MVA}$for $S_{base}$into equation (1).

$$\begin{gathered} S_{pu}=\frac{100+j10}{100} \\ {S}_{pu}=1+j0.1\text{pu} \end{gathered}$$

Calculate the power factor.

Substitute $1\text{pu}$for P and$0.1\text{pu}$for Q into equation (2).

$$\begin{gathered} pf=\frac{1}{\sqrt{1^2+{0.1}^2}} \\ pf=\frac{1}{\sqrt{1.01}} \\ pf=0.995 \end{gathered}$$

Calculate the current.

Substitute $1\text{pu}$for $V_{bus}$, $1\text{pu}$1for P , and $0.995$for $pf$into equation (3).

$$\begin{gathered} I=\frac{1}{1\times 0.995}\angle {\cos }^{-1}\left(0.995\right) \\ I=1.005\angle 5.73° \\ I=1-j0.1\text{pu} \end{gathered}$$

Calculate the terminal voltage.

Substitute 1pu for$V_{bus}$, $1-0.1\text{pu}$for l and $j0.22$for $X$into equation (4).

$$\begin{gathered} V_T=1+\left(1-j0.1\right)\left(j0.22\right) \\ {V}_T=1.022-j0.22 \\ {V}_T=1.045\angle 12.148°\text{pu} \end{gathered}$$

Calculate the value of $I_{sorc}$.

Substitute $1-0.1\text{pu}$for l ,$1.022-j0.22\text{pu}$for $V_T$and $j0.8$for $X_{eq}$into equation (5).

$$\begin{gathered} I_{sorc}=\left(1-j0.1\right)+\frac{1.022+j0.22}{j0.8} \\ {I}_{sorc}=1-j0.1+0.275-j1.275 \\ {I}_{sorc}=1.275-j1.375 \end{gathered}$$

The expression to calculate the active and reactive current.

$I_{sorc}=\left(I_p+I_q\right)\left(1\angle 12.148°\right)$

Here,$I_p$is the active power current and $I_q$is the reactive power current.

$\begin{array}{rcl}1.275-j1.375& =& \left(I_p+I_q\right)\left(1\angle 12.148°\right)\\ I_p+I_q& =& \frac{1.275-j1.375}{1\angle 12.148°}\\ I_p+I_q& =& 0.956-j1.614\\ & & \end{array}$

Compare both the sides,

$$\begin{gathered} I_p=0.956\text{pu} \\ {I}_q=-1.614\text{pu} \end{gathered}$$

Calculate the reactive voltage,

Substitute $-1.614\text{pu}$for $I_q$and for $X_{eq}$into equation (6).

$$\begin{gathered} E_q=-\left(1.614\right)\left(0.8\right) \\ {E}_q=1.29\text{pu} \end{gathered}$$

Hence, the active power current, the reactive power current and reactive voltage are$0.956\text{pu}$ ,$-1.614\text{pu}$ and$1.29\text{pu}$ respectively.