Consider again the three machine system of Example 5.2 except that bus 3 has been converted to a generator bus with a voltage magnitude of 1.0pu. The cost functions of the generators are

$C_{1} :P_{1} +0.0625P^{2}_{1} \$/h$

$C_{2} :P_{2}+0.0125P^{2}_{2} \$/h$

$C_{} :P_{}+0.0250P^{2}_{3} \$/h$

Find the optimal generation scheduling of this system.

Question

Consider again the three machine system of Example 5.2 except that bus 3 has been converted to a generator bus with a voltage magnitude of 1.0pu. The cost functions of the generators are

[latex]C_{1} :P_{1} +0.0625P^{2}_{1} \$/h [/latex]

[latex]C_{2} :P_{2}+0.0125P^{2}_{2} \$/h [/latex]

[latex]C_{} :P_{}+0.0250P^{2}_{3} \$/h [/latex]

Find the optimal generation scheduling of this system.

Accepted Answer

Following the steepest descent procedure, the first step is to develop an expression for the gradient ∇C, where

[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)

where f is the sum of the generator costs:

[latex]f:c_{1} +C_{2} +C_{3} =P_{1}+0.0625P^{2}_{1} +P_{2} +0.0125P^{2}_{2}+P_{3} +0.0250P^{2}_{3}[/latex]

g is the set of load flow equations:

[latex]g_{1} :0=P_{2} -P_{L2} -V_{2} \sum\limits_{i \xlongequal[]{}1 }^{3}{}V_{i}Y_{2i} \cos (\delta _{2}-\delta _{i} -\phi _{2i} )[/latex]

[latex]g_{2} :0=P_{3} -P_{L3} -V_{3} \sum\limits_{i \xlongequal[]{}1 }^{3}{}V_{i}Y_{3i} \cos (\delta _{3}-\delta _{i} -\phi _{3i} )[/latex]

where [latex]P_{Li}[/latex] denotes the active power load at bus i, the set of inputs u is the set of independent generation settings:

[latex]u=\left[\begin{matrix} P_{2} \P_{3} \end{matrix} ight][/latex]

and x is the set of unknown states

[latex]x=\left[\begin{matrix} \delta _{2} \\delta _{3} \end{matrix} ight] [/latex]

The generator setting [latex]P_{1}[/latex] is not an input because it is the slack bus generation and cannot be independently set. From these designations, the various partial derivatives required for ∇C can be derived:

[latex]\left[\frac{∂g}{∂u} ight] =\left[\begin{matrix} 1 \ 1 \end{matrix} ight] [/latex] (5.55)

[latex]\left[\frac{∂g}{∂x} ight] =\left[\begin{matrix} \frac{∂_{g_{1} } }{∂_{\delta _{2} } } & \frac{∂_{g_{1} }}{∂_{\delta _{3} }} \ \frac{∂_{g_{2} } }{∂_{\delta _{2} } } &\frac{∂_{g_{2} } }{∂_{\delta _{3} } } \end{matrix} ight] [/latex] (5.56)

where

[latex]\frac{∂_{g_{1} } }{∂_{\delta _{2} } }=V_{2} (V_{1}Y_{12}\sin (\delta _{2}-\delta _{1}-\phi _{21} )+ V_{3}Y_{13}\sin (\delta _{2}-\delta _{3}-\phi _{23} ) ) [/latex] (5.57)

[latex]\frac{∂_{g_{1} } }{∂_{\delta _{3} } }=V_{2} V_{3}Y_{32}\sin (\delta _{2}-\delta _{3}-\phi _{23} )[/latex] (5.58)

[latex]\frac{∂_{g_{2} } }{∂_{\delta _{2} } }=V_{3} V_{2}Y_{32}\sin (\delta _{3}-\delta _{2}-\phi _{23} )[/latex] (5.59)

[latex]\frac{∂_{g_{2} } }{∂_{\delta _{3} } }=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] (5.60)

and

[latex]\left[\frac{∂f}{∂u} ight] =\left[\begin{matrix}1+0.025P_{2} \ 1+0.050P_{3} \end{matrix} ight][/latex] (5.61)

Finding the partial derivative [∂f/∂x] is slightly more difficult since the cost function is not written as a direct function of x. Recall, however, that [latex]P_{2}[/latex] is not an input, but is actually quantity that depends on x, i.e.,

[latex]P_{1} =V_{1} (V_{1}Y_{11}\cos (\delta _{1}-\delta _{1}-\phi _{11})[/latex]

[latex]+V_{2}Y_{12}\cos (\delta _{1}-\delta _{2}-\phi _{12})+V_{3}Y_{13}\cos (\delta _{1}-\delta _{3}-\phi _{13}))[/latex] (5.62)

Thus, using the chain rule,

[latex]\left[\frac{∂f}{∂x} ight] =\left[\frac{∂f}{∂P_{1} } ight] \left[\frac{∂P_{1} }{∂x} ight] [/latex] (5.63)

[latex]=(1+0.125P_{1} )\left[\begin{matrix} V_{1}V_{2} Y_{12} \sin (\delta _{1}-\delta _{2}-\phi _{12} ) \ V_{1}V_{3} Y_{13} \sin (\delta _{1}-\delta _{3}-\phi _{13} ) \end{matrix} ight] [/latex] (5.64)

If the initial values of [latex]P_{2}=0.56 pu[/latex] and [latex]P_{3}=0.28 pu[/latex] are used as inputs, then the power flow yields the following states: [latex]\left[\begin{matrix}\delta _{2} & \delta _{3} \end{matrix} ight] =\left[0.0286-0.0185 ight][/latex] in radians and [latex]P_{1}=0.1152[/latex]. Converting the generated powers to megawatt and substituting these values into the partial derivatives yields

[latex]\left[\frac{∂g}{∂u} ight] =\left[\begin{matrix}1 & 0\ 0 & 1 \end{matrix} ight] [/latex] (5.65)

[latex]\left[\frac{∂g}{∂x} ight] =\left[\begin{matrix}-13.3267 & 9.9366\ 9.8434 & -19.9219 \end{matrix} ight][/latex] (5.66)

[latex]\left[\frac{∂f}{∂u} ight] =\left[\begin{matrix}15.0000 \ 15.0000 \end{matrix} ight] [/latex] (5.67)

[latex]\left[\frac{∂f}{∂x} ight] =15.4018\left[\begin{matrix}-52.0136 \ 155.8040 \end{matrix} ight][/latex] (5.68)

which yields

[latex]∇C =\left[\begin{matrix}-0.3256 \ -0.4648 \end{matrix} ight][/latex] (5.69)

Thus, the new values for the input generation are

[latex]\left[\begin{matrix}P_{2} \ P_{3} \end{matrix} ight]=\left[\begin{matrix} 560 \ 280 \end{matrix} ight] -\gamma \left[\begin{matrix}-0.3256 \ -0.4648 \end{matrix} ight] [/latex] (5.70)

With γ = 1, the updated generation is [latex]P_{2} =560.3[/latex] and [latex]P_{3} =280.5 MW[/latex].

Proceeding with more iterations until the gradient is reduced to less than a user-defined value yields the final generation values for all of the generators:

[latex]\left[\begin{matrix}P_{1} \ P_{2} \ P_{3} \end{matrix} ight] =\left[\begin{matrix} 112.6 \ 560.0 \ 282.7 \end{matrix} ight] MW[/latex]

which yields a cost of $7664/MWh.

Question 5.5: Consider again the three machine system of Example 5.2 excep...

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