Repeat Example 5.7 using the dual-variable approach.

Question

Accepted Answer

By introducing the additional equation

[latex]P_{23} =V_{2} V_{3} Y_{23} \cos (\delta _{2}-\delta _{3}-\phi _{23} )-V^{2}_{2} Y_{23} \cos \phi _{23} =0.400 pu[/latex] (5.83)

to the equality constraints, an additional equation gets added to the set of g(x). Therefore an additional unknown must be added to the state vector x to yield a solvable set of equations (three equations in three unknowns). Either [latex]P_{G2} [/latex] or [latex]P_{G3} [/latex] can be chosen as the additional unknown. In this example, [latex]P_{G3} [/latex] will be chosen. The new system Jacobian becomes

[latex]\left[\frac{∂_{g } }{∂_{x } } ight] =\left[\begin{matrix}\frac{∂_{g_{1} } }{∂_{x_{1} } } & \frac{∂_{g_{1} } }{∂_{x_{2} } } &\frac{∂_{g_{1} } }{∂_{x_{3} } } \ \frac{∂_{g_{2} } }{∂_{x_{1} } } & \frac{∂_{g_{2} } }{∂_{x_{2} } } & \frac{∂_{g_{2} } }{∂_{x_{3} } } \ \frac{∂_{g_{3} } }{∂_{x_{1} } } & \frac{∂_{g_{3} } }{∂_{x_{2} } } & \frac{∂_{g_{3} } }{∂_{x_{3} } } \end{matrix} ight][/latex] (5.84)

where

[latex]\frac{∂_{g_{1} } }{∂_{x_{1} } }=V_{2}( V_{1} Y_{12}\sin (\delta _{2}-\delta _{1}-\phi _{21} )+V_{2} Y_{13}\sin(\delta _{2}-\delta _{3}-\phi _{23} ) )[/latex]

[latex]\frac{∂_{g_{1} } }{∂_{x_{2} } }=-V_{2} V_{3} Y_{32}\sin (\delta _{2}-\delta _{3}-\phi _{23} )[/latex]

[latex]\frac{∂_{g_{1} } }{∂_{x_{3} } }=0[/latex]

[latex]\frac{∂_{g_{2} } }{∂_{x_{1} } }=-V_{3} V_{2} Y_{23}\sin (\delta _{3}-\delta _{2}-\phi _{32} )[/latex]

[latex]\frac{∂_{g_{2} } }{∂_{x_{2} } }=V_{3} (V_{1} Y_{13}\sin (\delta _{3}-\delta _{1}-\phi _{31} )+V_{2} Y_{23}\sin(\delta _{3}-\delta _{2}-\phi _{32} )[/latex]

[latex]\frac{∂_{g_{2} } }{∂_{x_{3} } }=1[/latex]

[latex]\frac{∂_{g_{3} } }{∂_{x_{1} } }=-V_{2} V_{3} Y_{23}\sin (\delta _{2}-\delta _{3}-\phi _{23} )[/latex]

[latex]\frac{∂_{g_{3} } }{∂_{x_{2} } }=V_{2} V_{3} Y_{23}\sin (\delta _{2}-\delta _{3}-\phi _{23} )[/latex]

[latex]\frac{∂_{g_{3} } }{∂_{x_{3} } }=0[/latex]

and

[latex]\left[\frac{∂g}{∂u} ight] =\left[\begin{matrix} 1 \ 0 \ 0 \end{matrix} ight] ;\left[\frac{∂f}{∂u} ight]=\left[1+0.025P_{G2} ight][/latex]

Similar to Example 5.5, the chain rule is used to obtain [∂f/∂x]:

[latex]\left[\frac{∂f}{∂x} ight] =\left[\frac{∂C}{∂P_{G1} } ight]\left[\frac{∂P_{G1} }{∂x} ight] +\left[\frac{∂C}{∂P_{G3} } ight] \left[\frac{∂P_{G3} }{∂x} ight][/latex] (5.85)

[latex]=(1+0.125P_{G1} )\left[\begin{matrix} V_{1} V_{2} Y_{12}\sin (\delta _{1}-\delta _{2}-\phi _{12} ) \V_{1} V_{2} Y_{13}\sin (\delta _{1}-\delta _{3}-\phi _{13} ) \ 0 \end{matrix} ight][/latex]

[latex]+(1+0.050P_{G3} )\left[\begin{matrix} V_{3} V_{2} Y_{32}\sin (\delta _{3}-\delta _{2}-\phi _{32} ) \-V_{3}( V_{1} Y_{13}\sin (\delta _{3}-\delta _{1}-\phi _{31} )+V_{2} Y_{23}\sin(\delta _{3}-\delta _{2}-\phi _{32} )) \ 0 \end{matrix} ight][/latex] (5.86)

Substituting these partial derivatives into the expression for ∇C of Equation 5.54 yields the samegeneration scheduling as Example 5.7

[latex]∇C=\left[\frac{∂f}{∂u} ight]-\left[\frac{∂g}{∂u} ight]^T\left[\left[\frac{∂g}{∂x} ight]^T ight]^{-1}\left[\frac{∂f}{∂x} ight][/latex] (5.54)

Question 5.8: Repeat Example 5.7 using the dual-variable approach.

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