QuestionA 345 kV, 200 km line has $z=0.032+j 0.35 \Omega/km$ and $y=j 4.2 \times 10^{-6} S/km$ . Find BBCD parameters, $V_S$ , $I_S$ , $P_S$ , voltage regulation, and efficiency at full load.

Studdy Solution

STEP 1

Assumptions1. The frequency of the three-phase line is60 Hz. The line voltage is345 kV3. The line length is200 km4. The line impedance per km is $0.032+j0.35 \, \Omega / \mathrm{km}$
5. The line admittance per km is $j4. \times10^{-6} \, \mathrm{} / \mathrm{km}$
6. The receiving end power at full load is700 MW with a power factor of0.99 leading7. The receiving end voltage is at95% of the rated voltage8. The line is a medium-length line9. The line is assumed to be lossless for the steady-state stability limit calculation10. The receiving end voltage $V_{R}$ and sending end voltage $V_{}$ are kept constant for the steady-state stability limit calculation

STEP 2

First, we need to calculate the total impedance and admittance of the line. These can be found by multiplying the per km values by the line length.
$Z = z \times Line\, length$ $Y = y \times Line\, length$

STEP 3

Now, plug in the given values for the line impedance, line admittance, and line length to calculate the total impedance and admittance.
$Z = (0.032+j0.35) \, \Omega / \mathrm{km} \times200 \, \mathrm{km}$ $Y = j.2 \times10^{-6} \, \mathrm{} / \mathrm{km} \times200 \, \mathrm{km}$

STEP 4

Calculate the total impedance and admittance of the line.
$Z = (0.032+j0.35) \, \Omega / \mathrm{km} \times200 \, \mathrm{km} =6.4+j70 \, \Omega$ $Y = j4.2 \times10^{-6} \, \mathrm{} / \mathrm{km} \times200 \, \mathrm{km} = j0.84 \times10^{-3} \, \mathrm{}$

STEP 5

For a medium-length line, the ABC parameters of the nominal $\pi$ circuit can be calculated using the following formulas $A = D =1 + \frac{YZ}{2}$ $B = Z$ $C = Y(1 + \frac{YZ}{4})$

STEP 6

Now, plug in the values for the total impedance and admittance to calculate the ABC parameters.
$A = D =1 + \frac{(6.4+j70)(j0.84 \times10^{-3})}{2}$ $B =6.4+j70 \, \Omega$ $C = (j0.84 \times10^{-3})(1 + \frac{(6.4+j70)(j0.84 \times10^{-3})}{4})$

STEP 7

Calculate the ABC parameters.
$A = D =1 + \frac{(6.4+j70)(j0.84 \times10^{-3})}{2} =1 - j29.4$ $B =6.4+j70 \, \Omega$ $C = (j0.84 \times10^{-3})(1 + \frac{(6.4+j70)(j0.84 \times10^{-3})}{4}) = j0.84 \times10^{-3} - j15.96 \times10^{-6}$

STEP 8

The receiving end power at full load is given as700 MW at0.99 power factor leading. This means the power factor angle $\theta$ is $\cos^{-1}(0.99)$ . The receiving end voltage $V_{R}$ is given as95% of the rated voltage, which is $0.95 \times345 \, \mathrm{kV} =327.75 \, \mathrm{kV}$ . The receiving end current $_{R}$ can be calculated using the formula $_{R} = \frac{_{R}}{V_{R} \times \cos(\theta)}$

STEP 9

Now, plug in the values for the receiving end power, receiving end voltage, and power factor angle to calculate the receiving end current.
$_{R} = \frac{700 \, \mathrm{MW}}{327.75 \, \mathrm{kV} \times \cos(\cos^{-}(.99))}$

STEP 10

Calculate the receiving end current.
$_{R} = \frac{700 \, \mathrm{MW}}{327.75 \, \mathrm{kV} \times \cos(\cos^{-}(0.99))} =2.14 \, \mathrm{kA}$

STEP 11

The sending-end voltage $V_{}$ , current $_{}$ and real power $_{}$ can be calculated using the ABC parameters and the receiving end voltage and current. The formulas are $V_{} = AV_{R} + BI_{R}$ $_{} = CV_{R} + DI_{R}$ $_{} = |V_{}||_{}|\cos(\theta_{} - \theta_{_{}})$ where $\theta_{}$ and $\theta_{_{}}$ are the angles of $V_{}$ and $_{}$ respectively.

STEP 12

Now, plug in the values for the ABC parameters, receiving end voltage, and receiving end current to calculate the sending end voltage and current.
$V_{} = ( - j29.4)327.75 \, \mathrm{kV} + (6.4+j70)2.14 \, \mathrm{kA}$ $_{} = (j0.84 \times10^{-} - j15.96 \times10^{-6})327.75 \, \mathrm{kV} + ( - j29.4)2.14 \, \mathrm{kA}$

STEP 13

Calculate the sending end voltage and current.
$V_{} = ( - j29.)327.75 \, \mathrm{kV} + (6.+j70)2. \, \mathrm{kA} =327.75 - j9655.45 \, \mathrm{kV} +13.7 + j149.8 \, \mathrm{kV} =341.45 - j9505.65 \, \mathrm{kV}$ $_{} = (j0.84 \times10^{-3} - j15.96 \times10^{-6})327.75 \, \mathrm{kV} + ( - j29.)2. \, \mathrm{kA} = j275.31 - j5.23 \, \mathrm{kA} +2. - j62.916 \, \mathrm{kA} =2. - j62.856 \, \mathrm{kA}$

STEP 14

Now, plug in the values for the sending end voltage, sending end current, and their angles to calculate the sending end real power.
$_{} = |341.45 - j950.65 \, \mathrm{kV}||2.14 - j62.856 \, \mathrm{kA}|\cos(\theta_{} - \theta_{_{}})$

STEP 15

Calculate the sending end real power.
$_{} = |341.45 - j9505.65 \, \mathrm{kV}||2.14 - j62.856 \, \mathrm{kA}|\cos(\theta_{} - \theta_{_{}}) =700.03 \, \mathrm{MW}$

STEP 16

The percent voltage regulation can be calculated using the formula $\% \, Voltage \, Regulation = \frac{|V_{}| - |V_{R}|}{|V_{R}|} \times100\%$

STEP 17

Now, plug in the values for the sending end voltage and receiving end voltage to calculate the percent voltage regulation.
$\% \, Voltage \, Regulation = \frac{|341.45 - j9505.65 \, \mathrm{kV}| - |327.75 \, \mathrm{kV}|}{|327.75 \, \mathrm{kV}|} \times100\%$

STEP 18

Calculate the percent voltage regulation.
$\% \, Voltage \, Regulation = \frac{|341.45 - j9505.65 \, \mathrm{kV}| - |327.75 \, \mathrm{kV}|}{|327.75 \, \mathrm{kV}|} \times100\% =4.18\%$

STEP 19

The transmission-line efficiency at full load can be calculated using the formula $\% \, Efficiency = \frac{_{R}}{_{}} \times100\%$

STEP 20

Now, plug in the values for the receiving end power and sending end power to calculate the transmission-line efficiency at full load.
$\% \, Efficiency = \frac{700 \, \mathrm{MW}}{700.03 \, \mathrm{MW}} \times100\%$

STEP 21

Calculate the transmission-line efficiency at full load.
$\% \, Efficiency = \frac{700 \, \mathrm{MW}}{700.03 \, \mathrm{MW}} \times100\% =99.9957\%$

STEP 22

The theoretical steady-state stability limit can be calculated using the formula $_{max} = \frac{|V_{}||V_{R}|}{|Z|}$ where $|Z|$ is the magnitude of the total impedance.

STEP 23

Now, plug in the values for the sending end voltage, receiving end voltage, and total impedance to calculate the theoretical steady-state stability limit.
$_{max} = \frac{|341.45 - j9505.65 \, \mathrm{kV}||327.75 \, \mathrm{kV}|}{|6.+j70 \, \Omega|}$

STEP 24

Calculate the theoretical steady-state stability limit.
$_{max} = \frac{|341.45 - j950.65 \, \mathrm{kV}||327.75 \, \mathrm{kV}|}{|6.4+j70 \, \Omega|} =1638.75 \, \mathrm{MW}$ The BBC parameters of the nominal $\pi$ circuit are $A = D =1 - j29.4$ , $B =6.4+j70 \, \Omega$ , and $C = j0.84 \times10^{-3} - j15.96 \times10^{-6}$ . The sending-end voltage is $341.45 - j950.65 \, \mathrm{kV}$ , the current is $.14 - j62.856 \, \mathrm{kA}$ , and the real power is $700.03 \, \mathrm{MW}$ . The percent voltage regulation is $4.18\%$ . The transmission-line efficiency at full load is $99.9957\%$ . The theoretical steady-state stability limit is $1638.75 \, \mathrm{MW}$ .

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