Question

Figure 3.39 shows a oneline diagram of a system in which the three-phase generator is rated 300 MVA , 20kV with a sub transient reactance of 2.0 per unit and with its neutral grounded through a 0.4 V reactor. The transmission line is 64 km long with a series reactance of $\mathbf{0}\mathbf{.}\mathbf{5}\frac{\mathbf{V}}{\mathbf{km}}$. The three-phase transformer ${\mathbf{T}}_{\mathbf{1}}$is rated 350 MVA , 230/20kV with a leakage reactance of 0.1 per unit. Transformer ${\mathbf{T}}_{\mathbf{2}}$is composed of three single-phase transformers,each rated 100 MVA , 127/13.2 kV with a leakage reactance of 0.1 per unit. Two 13.2kV motors ${\mathbf{M}}_{\mathbf{1}}$and ${\mathbf{M}}_{\mathbf{2}}$with a sub transient reactance of 0.2 per unit for each motor represent the load. ${\mathbf{M}}_{\mathbf{1}}$has a rated input of 200 MVA with its neutral grounded through a 0.4 V current-limiting reactor. ${\mathbf{M}}_{\mathbf{2}}$has a rated input of 100 MVA with its neutral not connected to ground. Neglect phase shifts associated with the transformers. Choose the generator rating as base in the generator circuit and draw the positive-sequence reactance diagram showing all reactance in per unit.

Short answer

Draw the per-unit single line diagram below:

Step-by-step solution

Given data.

From the given data, the generator has ratings $S_{base}=300\mathrm{MVA}\mathrm{and}{X}_{d\left(G\right)}^{\text{'}\text{'}}=0.2\mathrm{pu}$.

Here, $S_{base}$is common base power of the system and $X_d^{\text{'}\text{'}}$is sub transient reactance of generator.

The transmission line length is $I=64\mathrm{km}$and the line reactance is $X_I=0.5\frac{\mathrm{\Omega }}{\mathrm{km}}$, therefore, the total reactance is $X_I=\left(0.5\times 64\right)\mathrm{\Omega }$

The transformer $T_1$ rating is role="math" localid="1655201747425" $X_{d\left(T_1\right)}^{\text{'}}=0.1\mathrm{pu},S_{T_2}=350\mathrm{MVA}\mathrm{and}{V}_k/V_I=230/20\mathrm{kV}$.

The transformer $T_2$ rating is $X_{d\left(T_2\right)}^{\text{'}}=0.1\mathrm{pu}\mathrm{and}{V}_m/V_n=127/13.2\mathrm{kV}$ .

The power and voltage ratings of motors $M_1\mathrm{and}{M}_2\mathrm{is}{X}_{d\left(M\right)}^{\text{'}\text{'}}=0.2\mathrm{pu}$,

$V_p=V_r=13.2\mathrm{kV},S_{M_1}=200\mathrm{MVA}\mathrm{and}{\mathrm{S}}_{{\mathrm{M}}_2}=100\mathrm{MVA}$.

Determine the formulas of new reactance, base impedance, per-unit impedance.

Write the formula of new reactance.

${\mathbf{X}}_{\mathbf{new}}\mathbf{=}{\mathbf{X}}_{\mathbf{old}}{\left(\frac{V_{\mathrm{old}}}{V_{\mathrm{new}}}\right)}^{\mathbf{2}}\left(\frac{S_{\mathrm{new}}}{S_{\mathrm{old}}}\right)$ …… (1)

Here, $X_{old}$is new reactance, $V_{old}$is old base voltage level, $V_{new}$is new base voltage level, $S_{new}$is new base power and $S_{old}$is old base power.

Write the formula of base impedance of transmission line.

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

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

Write the formula of per-unit impedance.

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

Here, $Z_{actual}$is actual impedance.

Determine the new reactance of motors and line.

Substitute 0.2pu for $X_{d\left(M\right)}^{\text{'}\text{'}}=X_{old},300\mathrm{MVA}\mathrm{for}{S}_{base}=S_{new},200\mathrm{MVA}$for $S_{M_1}=S_{old},13.2\mathrm{kV}\mathrm{for}{V}_{old}\mathrm{and}13.8\mathrm{kV}\mathrm{for}{V}_{new}$ in equation (1).

$$\begin{gathered} X_{new\left(M_1\right)}=\left(0.2\mathrm{pu}\right){\left(\frac{13.2\mathrm{kV}}{13.8\mathrm{kV}}\right)}^2\left(\frac{300\mathrm{MVA}}{200\mathrm{MVA}}\right) \\ =0.2745\mathrm{pu} \end{gathered}$$

Substitute $0.2\mathrm{pu}\mathrm{for}{X}_{d\left(M\right)}^{\text{'}\text{'}}=X_{old},300\mathrm{MVA}\mathrm{for}{S}_{base}=S_{new},100\mathrm{MVA}$for $S_{M_1}=S_{old},13.2\mathrm{kV}\mathrm{for}{V}_{old}\mathrm{and}13.8\mathrm{kV}\mathrm{for}{V}_{new}$in equation (1).

$$\begin{gathered} X_{new\left(M_1\right)}=\left(0.2\mathrm{pu}\right){\left(\frac{13.2\mathrm{kV}}{13.8\mathrm{kV}}\right)}^2\left(\frac{300\mathrm{MVA}}{100\mathrm{MVA}}\right) \\ =0.549\mathrm{pu} \end{gathered}$$

Substitute $300\mathrm{MVA}\mathrm{for}{S}_{base}\mathrm{and}230\mathrm{kV}\mathrm{for}{V}_{base}$in equation (2).

$$\begin{gathered} Z_{base}\mathit{=}\frac{{\left(230\mathrm{kV}\right)}^2}{300\mathrm{MVA}} \\ =176.3\mathrm{\Omega } \end{gathered}$$

Substitute $176.3\mathrm{\Omega }\mathrm{for}{Z}_{base}\mathrm{and}\left(0.5\times 64\right)\mathrm{\Omega }\mathrm{for}{Z}_{new}$in equation (3).

role="math" localid="1655203309250" $$\begin{gathered} Z_{p.u.}=\frac{\left(0.5\times 64\right)\mathrm{\Omega }}{176.3\mathrm{\Omega }} \\ =0.1815\mathrm{pu} \end{gathered}$$

Therefore, the motor and transmission line reactance is $X_{new\left(M_1\right)}=0.2745\mathrm{pu},$

$X_{new\left(M_2\right)}=0.549\mathrm{pu}\mathrm{and}{Z}_{p.u.\left(line\right)}=0.1815\mathrm{pu}$.

Determine the new reactance of transformers.

Substitute $0.1\mathrm{pu}\mathrm{for}{X}_{a\left(T_1\right)}^{\text{'}}=X_{old},300\mathrm{MVA}\mathrm{for}{S}_{base}=S_{new},350\hspace{0.17em}\mathrm{MVA}\mathrm{for}$$S_{T_1}=S_{old},230\mathrm{kV}\mathrm{for}{V}_{old}\mathrm{and}230\mathrm{kV}\mathrm{for}{V}_{new}$and for in equation (1).

$$\begin{gathered} X_{new\left(T_1\right)}=\left(0.1\mathrm{pu}\right){\left(\frac{230\mathrm{kV}}{230\mathrm{kV}}\right)}^2\left(\frac{300\mathrm{MVA}}{350\hspace{0.17em}\mathrm{MVA}}\right) \\ =0.0857\mathrm{pu} \end{gathered}$$

Substitute$0.1\mathrm{pu}\mathrm{for}{X}_{a\left(T_2\right)}^{\text{'}}=X_{old},300\mathrm{MVA}\mathrm{for}{S}_{base}=S_{new},350\hspace{0.17em}\mathrm{MVA}\mathrm{for}$$S_{T_2}=S_{old},13.2\mathrm{kV}\mathrm{for}{V}_{old}\mathrm{and}13.8\mathrm{kV}\mathrm{for}{V}_{new}$and for in equation (1).

$$\begin{gathered} X_{new\left(T_2\right)}=\left(0.1\mathrm{pu}\right){\left(\frac{13.2\mathrm{kV}}{13.8\mathrm{kV}}\right)}^2\left(\frac{300\mathrm{MVA}}{300\hspace{0.17em}\mathrm{MVA}}\right) \\ =0.0915\mathrm{pu} \end{gathered}$$

Draw the per-unit single line diagram below: