A simple (few buses) power system (Fig. E7.8.1) is used to illustrate in detail the Newton– Raphson load flow analysis. All impedances are in pu and the base apparent power is Sbase=1 kVA. Compute the fundamental admittance matrix for this system.

Question

A simple (few buses) power system (Fig. E7.8.1) is used to illustrate in detail the Newton– Raphson load flow analysis. All impedances are in pu and the base apparent power is [latex]S_{base}[/latex]=1 kVA. Compute the fundamental admittance matrix for this system.

Accepted Answer

For this example [latex]y_{14}=- ilde{y}_{14}, ext { where } ilde{y}_{14}[/latex] is the admittance between bus 1 and bus 4 and [latex]y_{14}[/latex] is an entry of the bus admittance matrix [latex]Y_{bus}[/latex]. Note the minus sign.

[latex]\begin{gathered} y_{14}=-\widetilde{y}_{14}=\frac{-1}{0.01+j 0.02}=\frac{-1}{0.02236 \angle 1.1071 \mathrm{rad}} \ =\frac{1 \angle \pm 3.1415 \mathrm{rad}}{0.02236 \angle 1.1071 \mathrm{rad}}=44.72 \mathrm{pu} \angle 2.034 \mathrm{rad} \ y_{12}=- ilde{y}_{12}=\frac{-1}{0.01+j 0.01}=\frac{-1}{0.01414 \angle 0.7854} \ =\frac{1 \angle \pm 3.1415 \mathrm{rad}}{0.01414 \angle 0.7854}=70.72 \mathrm{pu} \angle 2.356 \mathrm{rad} \ y_{11}=-\left\{ ilde{y}_{14}+ ilde{y}_{12} ight\}=-\{44.72 \cos (2.034)+70.72 \cos (2.356)+j[44.72 \sin (2.034) \ +70.72 \sin (2.356)]\}=-\{-19.98-49.997+j[40.008+50.02]\} \ =(69.98-j 90.03)=114.03 \mathrm{pu} \angle-0.910 \mathrm{rad} \end{gathered}[/latex].

[latex]y_{13}=-\widetilde{y}_{13}=0[/latex] (because there is no direct line between buses 1 and 3)

[latex]\begin{aligned} & y_{23}=- ilde{y}_{23}=\frac{-1}{0.02+j 0.08}=\frac{1 \angle \pm 3.1415 \mathrm{rad}}{0.082462 \angle 1.3258 \mathrm{rad}} \ & =12.127 \mathrm{pu} \angle 1.816 \mathrm{rad} \ & y_{24}=- ilde{y}_{42}=0 \ & y_{34}=- ilde{y}_{34}=\frac{-1}{0.01+j 0.02}=\frac{1 \angle \pm 3.1415 \mathrm{rad}}{0.02236 \angle 1.1071 \mathrm{rdd}} \ & =44.72 \mathrm{pu} \angle 2.034 \mathrm{rad} \ & y_{22}=-\left( ilde{y}_{12}+ ilde{y}_{23} ight)=-\{70.72 \cos (2.356)+ \ & 12.127 \cos (1.816)+j(70.72 \sin (2.356)+ \ & 12.127 \sin (1.816)]\}=-\{-49.997-2.994+ \ & j(50.102+11.764)\}=-\{-52.941+ \ & j 61.862\}=81.42 \mathrm{pu} \angle-0.863 \mathrm{rad} \ & y_{33}=-\left( ilde{y}_{23}+ ilde{y}_{34} ight)=-\{12.127 \cos (1.816)+ \ & 44.72 \cos (2.034)+j[12.127 \sin (1.8161)+ \ & 44.72 \sin (2.034)]\}=-\{-2.944-19.982+j(11.764+ \ & 4.008)\}=-\{-22.93+j 51.772\}=56.62 \mathrm{pu} \angle-1.154 \ & \end{aligned}\ \begin{aligned} & y_{44}=-\left( ilde{y}_{14}+ ilde{y}_{34} ight)=-\{44.72 \cos (2.034)+ \ & 44.72 \cos (2.034)+j(44.72 \sin (2.034)+ \ & 44.72 \sin (2.034)]\}=-\{-19.982-19.982+j(40.008+ \ & 40.008)\}=\{-39.964+j 80.016\}=89.44 \mathrm{pu} \angle-1.1076 \mathrm{rad} . \end{aligned}[/latex]

The bus admittance matrix is now

[latex]\bar{Y}_{ ext {bus }}=\left[\begin{array}{cccc} 114.03 \mathrm{pu} \angle-0.910 \mathrm{rad} & 70.72 \mathrm{pu} \angle 2.356 \mathrm{rad} & 0 & 44.72 \mathrm{pu} \angle 2.034 \mathrm{rad} \ 70.72 \mathrm{pu} \angle 2.356 \mathrm{rad} & 81.42 \mathrm{pu} \angle-0.863 \mathrm{rad} & 12.127 \mathrm{pu} \angle 1.816 \mathrm{rad} & 0 \ 0 & 12.127 \mathrm{pu} \angle 1.816 \mathrm{rad} & 56.62 \mathrm{pu} \angle-1.154 \mathrm{rad} & 44.72 \mathrm{pu} \angle 2.034 \mathrm{rad} \ 44.72 \mathrm{pu} \angle 2.034 \mathrm{rad} & 0 & 44.72 \mathrm{pu} \angle 2.034 \mathrm{rad} & 89.44 \mathrm{pu} \angle-1.1076 \mathrm{rad} \end{array} ight] .[/latex]

Note that [latex]\bar{Y}_{bus}[/latex] is singular; that is, [latex]\left(\bar{Y}_{ ext {bus }} ight)^{-1}[/latex] cannot be found, because there is no admittance connected to the ground bus. This is unimportant for the Newton-Raphson approach because for forming of the Jacobian the entries of [latex]\bar{Y}_{ ext {bus }} ext { are used and } \bar{Y}_{ ext {bus }}[/latex] is not inverted.

Question 7.AE.8: A simple (few buses) power system (Fig. E7.8.1) is used to i......

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