Question 11.AE.10: Data for generation set (steam turbine and generator) #1: An......
Data for generation set (steam turbine and generator) #1: Angular frequency change (Δω_1) per change in generator output power (ΔP_1), that is R_1=\frac{\Delta \omega_1}{\Delta P_1} \mathrm{pu}=0.01 \mathrm{pu} (e.g., coal-fired plant), the frequency-dependent load change (ΔP_{L1}|_{frequ}) per angular frequency change (Δω_1), that is D_1=\frac {ΔP_{L1}|_{frequ}}{Δω_1} = 0.8 pu, step-load change \Delta P_{\mathrm{L} 1}(s)=\frac{\Delta P_{\mathrm{L} 1}}{s} \mathrm{pu}=\frac{0.2}{s} \mathrm{pu}, angular momentum of steam turbine and generator set M_1=4.5, base apparent power S_{base}=500 MVA, governor time constant T_{G1}=0.01 s, valve charging time constant T_{CH1}=0.5 s, and load ref_1(s) =0.8 pu.
Data for generation set (steam turbine and generator) #2: Angular frequency change (Δω_2) per change in generator output power (ΔP_2), that is R_2=\frac{\Delta \omega_2}{\Delta P_2} \mathrm{pu}=0.02 \mathrm{pu} (e.g., coal-fired plant), the frequency-dependent load change (ΔP_{L2}|_{frequ}) per angular frequency change (Δω_2), that is D_2=\frac{\left.\Delta P_{\mathrm{L} 2}\right|_{\text {frequ }}}{\Delta \omega_2}=1.0 \mathrm{pu}, \text { step-load change } \Delta P_{\mathrm{L} 2}(s)=\frac{\Delta P_{\mathrm{L} 2}}{s} \mathrm{pu}=\frac{-0.2}{s} \mathrm{pu}, angular momentum of steam turbine and generator set M_2=6, base apparent power S_{base}=500 MVA, governor time constant T_{G2}=0.02 s, valve charging-time time constant T_{CH2}=0.75 s, and load ref_2(s)=0.8 pu.
Data for tie line: T=\frac{377}{X_{\mathrm{tie}}} \text { with } X_{\mathrm{tie}}=0.2 \mathrm{pu}.
a) List the ordinary differential equations and the algebraic equations of the block diagram of Figure 11.28.
b) Use either Mathematica or Matlab to establish steady-state conditions by imposing a step function for load \operatorname{ref}_1(s)=\frac{0.8}{s} \mathrm{pu}, \text { load } \operatorname{ref}_2(s)=\frac{0.8}{s} \mathrm{pu} and run the program with zero step-load changes ΔP_{L1}=0, ΔP_{L2}=0 (for 10 s) in order to establish the equilibrium condition. After 10 s impose step-load changes \Delta P_{\mathrm{L} 1}(s)=\frac{\Delta P_{\mathrm{L} 1}}{s} \mathrm{pu}=\frac{0.2}{s} \mathrm{pu} and after 30 s impose \Delta P_{\mathrm{L} 2}(s)=\frac{\Delta P_{\mathrm{L} 2}}{s} \mathrm{pu}=\frac{-0.2}{s} \mathrm{pu} to find the transient response Δω_1(t)=Δω_2(t)=Δω(t) for a total of 50 s. Repeat part b) for R_1=0.5 pu, (e.g., wind-power plant), and R_2=0.01 pu (e.g., coal-fired plant).
Differential and algebraic equations
System # 1 \varepsilon_{11}=\text { load ref }_1-\frac{\Delta \omega_1}{R_1}, \Delta P_{valve \_1}+\tau_{\mathrm{G} 1} \frac{d\left(\Delta P_{\text {valve } 1}\right)}{d t}=\varepsilon_{11}, \Delta P_{mech\_1 }+\tau_{\mathrm{CH} 1} \frac{d\left(\Delta P_{mech\_ 1}\right)}{d t}=\Delta P_{valve\_ 1}, \varepsilon_{21}=\Delta P_ {mech\_1}-\Delta P_{\mathrm{L} 1}-\Delta P_{\text {tie }}, \Delta \omega_1 D_1+M_1 \frac{d\left(\Delta \omega_1\right)}{d t}=\varepsilon_{21} .
Coupling (tie, transmission) network: \frac{1}{T} \frac{d\left(\Delta P_{\text {tie }}\right)}{d t}=\varepsilon_3, where ε_3 = Δω_1 – Δω_2.
System # 2: \varepsilon_{12}=\text { load ref }_2-\frac{\Delta \omega_2}{R_2}, \Delta P_{\text {valve }\_ 2}+\tau_{\mathrm{G} 2} \frac{d\left(\Delta P_{valve\_ 2}\right)}{d t}=\varepsilon_{12}, \Delta P_{mech\_ 2}+ \tau_{\mathrm{CH} 2} \frac{d\left(\Delta P_{mech \_2}\right)}{d t}=\Delta P_{\text {valve } \_2}, \quad \varepsilon_{22}=\Delta P_{mech\_2 }-\Delta P_{\mathrm{L} 2}+\Delta P_{\text {tie }}, \quad \Delta \omega_2 D_2+M_2 \frac{d\left(\Delta \omega_2\right)}{d t}=ε_{22}.
b) The results based on Mathematica are shown in Figures E11.10.1a, b. Figure E11.10.1a illustrates the occurrence of instability for similar (R_1=0.01 pu, R_2=0.02 pu) droop characteristics and Figure E11.10.1b stability for dissimilar (R_1=0.5 pu, R_2=0.01 pu) droop characteristics. The Mathematica input program is listed in Table E11.10.1.
Figures 11.28 and E11.10.1a, b lead to the conclusion that one power plant must be the so-called frequency leader having a small-sloped droop characteristic which can accommodate overloads. Note that Δω_1(t)=Δω_2(t)=Δω(t) are identical, which is due to the transmission line transfer function (T/s) acting as an integrator where at steady-state the perturbation at the receiving end of the line is zero.
| Table E11.10.1 Mathematica input program for two interconnected systems | |
| R1=0.01; d1=0.8; M1=4.5; Tg1=0.01; Tch1=0.5; Lr1=0.8; DPL1[t_]:=If [t<10, 0, 0.2]; R2=0.02; d2=1.0; M2=6; Tg2=0.02; Tch2=0.75; Lr2=0.8; DPL2[t_]:=If [t<30, 0, -0.2]; Xtie=0.2; Tie=377/Xtie; ic1=Dw1[0]= =0; ic2=DPmech1[0]= =0; ic3=DPvalve1[0]= =0; ic4=DPtie[0]= =0; ic5=Dw2[0]= =0; ic6=DPmech2[0]= =0; |
ic7=DPvalve2[0]= =0; E11[t_]:=Lr1-Dw1[t]/R1; E12[t_]:=DPmech1[t]-DPL1[t]-DPtie[t]; E3[t_]:=Dw1[t]-Dw2[t]; E22[t_]:=Lr2-Dw2[t]/R2; E21[t_]:=DPmech2[t]-DPL2[t]+DPtie[t]; eqn1=Dw1’[t]= =(1/M1)*(E12[t]-d1*Dw1[t]); eqn2=DPmech1’[t]= =(1/Tch1)*(DPvalve1[t]-DPmech1 [t]); eqn3=DPvalve1’[t]= =(1/Tg1)*(E11[t]-DPvalve1[t]); eqn4=DPtie’[t]= =Tie*E3[t]; eqn5=Dw2’[t]= =(1/M2)*(E21[t]-d2*Dw2[t]); eqn6=DPmech2’[t]= =(1/Tch2)*(DPvalve2[t]-DPmech2 [t]); eqn7=DPvalve2’[t]= =(1/Tg2)*(E22[t]-DPvalve2[t]); sol=NDSolve[{eqn1,eqn2,eqn3,eqn4,eqn5,eqn6,eqn7, ic1,ic2,ic3,ic4,ic5,ic6,ic7},{Dw1[t], DPmech1[t],DPvalve1[t],DPtie[t],Dw2[t], DPmech2[t],DPvalve2[t]},{t,0,50},MaxSteps->100000]; Plot[Dw1[t]/.sol,{t,0,50},PlotRange->All,AxesLabel ->{“t[s]”,”Dw1[t][pu]”}] |
