Question 11.AE.9: Angular frequency change Δω per change in generator output p......
Angular frequency change Δω per change in generator output power ΔP, that is R = \frac{\Delta \omega}{\Delta P} \mathrm{pu}=0.01 \mathrm{pu}, the frequency-dependent load change ΔP_L |_{frequ} per angular frequency change Δω, that is
D=\frac{\left.\Delta P_{\mathrm{L}}\right|_{\text {frequ }}}{\Delta \omega}=0.8 \mathrm{pu}, \text { step-load change } \Delta P_{\mathrm{L}}(s)=\frac{\Delta P_{\mathrm{L}}}{s} \mathrm{pu}=\frac{0.2}{s} \mathrm{pu}, angular momentum of steam turbine and generator set M=4.5, base apparent power S_{base} =500 MVA, governor time constant T_G=0.01 s, valve changing time constant T_{CH} =1.0 s, and load reference set point load(s)=1.0 pu:
a) Derive for Figure 11.27 Δω_{steady\ state} by applying the final value theorem. You may assume load reference set point load(s)¼1.0 pu, and \Delta P_{\mathrm{L}}(s)=\frac{\Delta P_{\mathrm{L}}}{s} \mathrm{pu}=\frac{0.2}{s} \mathrm{pu}. For the nominal frequency f*=60 Hz calculate the frequency f_{new} after the load change has taken place.
b) List the ordinary differential equations and the algebraic equations of the block diagram of Figure 11.27.
c) Use either Mathematica or Matlab to establish steady-state conditions by imposing a step function for load reference set point \operatorname{load}(s)=\frac{1}{s} \mathrm{pu} and run the program with a zero-step load change ΔP_L=0 (for 25 s) in order to establish the equilibrium condition without load step. After 25 s impose a step-load change of \Delta P_{\mathrm{L}}(\mathrm{s})=\frac{\Delta P_{\mathrm{L}}}{s} \mathrm{pu}=\frac{0.2}{s} \mathrm{pu} to find the transient response of Δω(t) for a total of 50 s and at 50 s impose a step-load change of \Delta P_L(s)=\frac{\Delta P_L}{s} p u=\frac{-0.2}{s} \mathrm{pu} to find the transient response of Δω(t) for a total of 75 s.
a) Define the transfer functions F_1=\frac{1}{1+s \tau_{\mathrm{G}}}, F_2=\frac{1}{1+s \tau_{\mathrm{CH}}}, F_3=\frac{1}{M s+D} \text {, and } F_4=\frac{1}{R}. From the block diagram one obtains \Delta \omega(s)=F_3 \cdot \Delta P_{\text {mech }}(s)-F_3 \cdot \Delta P_{\mathrm{L}}(s), \text { where } \Delta P_{\text {mech }}(s)=F_1 F_2\left\{\operatorname{load}(\mathrm{s})-F_4 \Delta \omega(s)\right\} \quad \text { resulting } \quad \text { in } \quad \Delta \omega(s)=\frac{F_1 F_2 F_3 \operatorname{load}(s)-F_3 \Delta P_{\mathrm{L}}(\mathrm{s})}{1+F_1 F_2 F_3 F_4} \text {. } Or
\Delta \omega(s)=\frac{\frac{\operatorname{load}(s)}{\left(1+s \tau_{\mathrm{G}}\right)\left(1+s \tau_{\mathrm{CH}}\right)(M s+D)}-\frac{\Delta P_{\mathrm{L}}(s)}{(M s+D)}}{\left\{1+\frac{1}{\left(1+s \tau_{\mathrm{G}}\right)\left(1+s \tau_{\mathrm{CH}}\right)(M s+D) R}\right\}} .At steady state load(s)=1.0 pu and \Delta P_{\mathrm{L}}(s)=\frac{\Delta P_{\mathrm{L}}}{s} \mathrm{pu}=\frac{0.2}{s} \mathrm{pu} one obtains with the finite-value theorem the droop characteristic in the frequency-load coordinate system \left.\Delta \omega\right|_{\text {steady state }}=\frac{-\Delta P_{\mathrm{L}}}{\left(D+\frac{1}{R}\right)}, \text { or for the given parameters }\left.\Delta \omega\right|_{\text {stedy state }}=\frac{-0.2}{\left(0.8+\frac{1}{0.01}\right)}=-1.98 \cdot 10^{-3} \mathrm{pu}=-0.198 \% \text { with } f_{\text {rated }}=f^*=60 \mathrm{~Hz} one obtains the new frequency f_{\text {new }}=f^*-0.00198^{\circ} 60 \mathrm{~Hz}=59.88 \mathrm{~Hz}.
b) The algebraic and differential equations of the block diagram are:
\begin{aligned} & \varepsilon_1=\left\{\text { load }-\frac{\Delta \omega}{R}\right\}, \Delta P_{\text {valve }}+\tau_{\mathrm{G}} \frac{d\left(\Delta P_{\text {valve }}\right)}{d t}=\varepsilon_1, \Delta P_{\text {mech }}+\tau_{\mathrm{CH}} \frac{d\left(\Delta P_{\text {mech }}\right)}{d t}=\Delta P_{\text {valve }} \\ & \varepsilon_2=\left\{\Delta P_{\text {mech }}-\Delta P_{\mathrm{L}}\right\}, \text { and } \Delta \omega D+M \frac{d(\Delta \omega)}{d t}=\varepsilon_2 \end{aligned}c) The results based on Mathematica are shown in Figure E11.9.1. The Mathematica input program is listed in Table E11.9.1.
Table E11.9.1 Mathematica input program for isochronous generation
\begin{array}{c|c}\hline R=0.01;&ic3=DPvalve[0]= =0;\\ d=0.8;&E1[t\_]:=Lr-Dw[t]/R;\\ M=4.5;&E2[t\_]:=DPmech[t]-DPL[t];\\ Tg=0.01;&eqn1=Dw’[t]= =(1/M)∗(E2[t]-d∗Dw[t]);\\ Tch=1;&eqn2=DPmech’[t]= =(1/Tch)∗(DPvalve[t]-DPmech[t]);\\ Lr=1;&eqn3=DPvalve’[t]= =(1/Tg)∗(E1[t]-DPvalve[t]);\\ DPL[t\_]:=If[t<50,If&sol=NDSolve[\left\{eqn1,eqn2,eqn3,ic1,ic2,ic3\right\}, \left\{Dw[t]\right., \\ [t<25,0,0.2],-0.2];& DPmech[t], DPvalve[t]\left.\right\},\left\{t,0,75\right\}, MaxSteps->100000];\\ ic1=Dw[0]= =0;& Plot[Dw[t]/.sol,\left\{t,0,75\right\},PlotRange->All,AxesLabel->\left\{"t\right.\\ ic2=DPmech[0]= =0;& [s]","Dw[t][pu]"\left.\right\}]\\ \hline \end{array}
