Question 11.AE.11: Figure E11.11.1 illustrates the sharing (increase) of the ad......
Figure E11.11.1 illustrates the sharing (increase) of the additional load among a shortterm storage plant (e.g., R_2=0.01 pu), a long-term-storage plant (e.g., R_3=10 pu), and a PV plant (e.g., R_1=10 pu) causing a frequency decrease. One obtains a stable frequency control when the short-term storage plant compensates the intermittent power output of the PV plant and the plant with spinning reserve (natural-gas fired plant) is replaced by a long-term storage plant. The long-term storage plant is connected all the time to the power system and serves therefore as frequency leader. The PV and the short-term storage plants may operate intermittently only.
In Figure 11.26 the spinning reserve plant can be replaced by a long-term storage plant (e.g., pump-hydro or compressed air facility), as shown in Figures E11.11.2a, b, c. The intermittently operating PV and WP plants are complemented by short-term storage plants (e.g., battery-fed inverter). During times of high power demand, the storage plants supply power to maintain the power balance between generation and the served loads. At low power demand, the renewable sources will supply the storage plants. The continuous load increase of conventional peak-power plants (which can provide additional load through spinning reserve) are replaced by putting short-term (located next to renewable plants) and long-term storage plants [1] online, reducing fossil-fuel generation and contributing to renewable portfolio standards. Renewable sources are operated at their peak-power point e.g., the slopes of the droop characteristic R_3\ and\ R_5 of Figure E11.11.2b are large while those of R_1\ and\ R_2 are relatively small to permit the increase of their output power upon demand. Thus the PV and WP plants cannot participate in frequency/load control. In Figure E11.11.2c all droop characteristics have a relatively small slope and all participating plants can output increased power and participate in frequency/load control.
In the block diagram of Figure E11.11.2a the PV plant, the governor, and prime mover represent the solar array and inverter/rectifier while in the short-term storage plant associated with the PV plant the governor and prime mover represent the storage device (e.g., battery, super-capacitor, flywheel) with inverter/rectifier. Similar considerations apply to the WP plant and its associated short-term storage plant.
Data for natural-gas fired plant (system #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 pu, the frequency dependent load change (ΔP_{L1}|_{frequ}) per angular frequency change (Δω_1), that is D_1= \frac{\left.\Delta P_{\mathrm{L} 1}\right|_{\text {frequ }}}{\Delta \omega_1}=0.8 \mathrm{pu}, \text { step-load change } \Delta P_{\mathrm{L} 1}(\mathrm{~s})=\frac{\Delta P_{\mathrm{L} 1}}{s} \mathrm{pu}=\frac{0.1}{s} \mathrm{pu}, angular momentum of gas turbine and generator set M_1=4.5, base apparent power S_{base}=500 MVA, governor time constant T_{G1}=0.3 s, valve charging time constant T_{CH1}=0.9 s, and load ref_1(s) =0.8 pu.
Data for long-term storage plant (system #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}=0.1 \mathrm{pu} (e.g., hydro-power 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 hydro turbine and generator set M_2=6, base apparent power S_{base}= 500 MVA, governor time constant T_{G2}=0.2 s, valve charging time constant T_{CH2}=0.2 s, and load ref_2(s) =0.5 pu.
Data for tie line: T=\frac{377}{X_{\text {tie }}} \text { with } X_{\mathrm{tie}}=0.2 \mathrm{pu}.
Data for PV plant (system #3): Angular frequency change (Δω_1) per change in inverter output power (ΔP_3), that is R_1=\frac{\Delta \omega_1}{\Delta P_3} \mathrm{pu}=0.3 \mathrm{pu}, governor time constant T_{G3}=0.1 s, equivalent valve time constant T_{CH3}=0.1 s, and load ref_3(s) =0.01 pu.
Data for short-term storage plant associated with PV plant (system #4): Angular frequency change (Δω_1) per change in generator output power (ΔP_4), that is R_4=\frac{\Delta \omega_1}{\Delta P_4} \mathrm{pu}=0.5 \mathrm{pu}, governor time constant T_{G4}=0.2 s, equivalent valve time constant T_{CH4}=0.1 s, and load ref_4(s) =0.01 pu.
Data for wind power (WP) plant (system #5): Angular frequency change (Δω_2) per change in generator output power (ΔP_5), that is R_5=\frac{\Delta \omega_1}{\Delta P_5} \mathrm{pu}=0.7 \mathrm{pu}, governor time constant T_{G5}=0.1 s, equivalent valve time constant T_{CH5}=0.1 s, and load ref_5(s) = 0.01 pu.
Data for short-term storage plant associated with WP plant (system #6): Angular frequency change (Δω_2) per change in generator output power (ΔP_6), that is R_6=\frac{\Delta \omega_2}{\Delta P_6} \mathrm{pu}=0.5 \mathrm{pu}, governor time constant T_{G6}=0.2 s, equivalent valve time constant T_{CH6}=0.1 s, and load ref_6(s) = 0.01 pu.
a) List the ordinary differential equations and the algebraic equations of the block diagram of Figure E11.11.2a.
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} \text { pu, load } \operatorname{ref}_2(s)=\frac{0.5}{s} \text { pu, load } \operatorname{ref}_3(s)=\frac{0.01}{s} \mathrm{pu}, load \operatorname{ref}_4(s)=\frac{0.01}{s} \mathrm{pu}, \text { load } \operatorname{ref}_5(s)=\frac{0.01}{s} \mathrm{pu}, \text { load } \operatorname{ref}_6(s)=\frac{0.01}{s} \mathrm{pu}, and run the program with a zero step-load changes ΔP_{L1}=0, ΔP_{L2}=0 for 200 s. Save the steady-state values for all variables at 200 s. Plot the calculated angular frequency response.
c) Initialize the parameters with the steady-state values as obtained in Part b). After 300 s impose step-load change \Delta P_{l1}(s)=\frac{\Delta P_{L1}}{s}pu=\frac{0.1}{s}pu, and after 400 s impose \Delta P_{L2}(s)=\frac{\Delta P_{L2}}{s}pu=\frac{-0.1}{s}pu. Thereafter, for load ref_3(s),\ load\ ref_4 (s),\ load\ ref_3(s),
DPstorage_4(s), DPstorage_6(s) and load ref_4 (s):
Lr3[t_]:=If [t<600, 0, 0.06];
Lr4[t_]:=If [t<600.1, 0,- 0.6];
Lr3[t_]:=If [t<1200,If[t<1120,If[t<1000,If[t<940,If[t<910,0,0.03],0.09],0.05],
0.03],0.0];
Lr5[t_]:=If [t<1200,If[t<1120,If[t<1000,If[t<940,If[t<910,0,0.3],0.9],0.5],0.3],
0.0];
DPstorage4[t_]:=If[t<1200.2,If[t<1120.2,If[t<1000.2,If[t<940.2,If
[t<910.2,0,0.15],0.45],0.25],0.15],0.0];
DPstorage6[t_]:=If[t<1200.2,If[t<1120.2,If[t<1000.2,If[t<940.2,If
[t<910.2,0,0.15],0.45],0.25],0.15],0.0];
Lr4[t_]:=If [t<700, 0,- 0.01]; Plot the given WP plant load reference (Lr3[t]) and calculated the transient response Δω(t) for a total of 1500 s.
d) Initialize the parameters with the steady-state values as obtained in Part b). After 300 s impose step-load change \Delta P_{L1}(s)=\frac{\Delta P_{L1}}{s} pu=\frac{0.1}{s} pu, and after 400 s impose \Delta P_{L2}(s)=\frac {\Delta P_{L2}}{s}pu=\frac{-0.1}{s}pu. Thereafter:
Lr3[t_]:=If [t<600, 0, 0.6];
Lr4[t_]:=If [t<600.1, 0,- 0.6];
Lr3[t_]:=If [t<1200,If[t<1120,If[t<1000,If[t<940,If[t<910,0,0.03],0.09],0.05],
0.03],0.0];
Lr5[t_]:=If [t<1200,If[t<1120,If[t<1000,If[t<940,If[t<910,0,0.3],0.9],0.5],
0.3],0.0];
DPstorage4[t_]:=If[t<1200.2,If[t<1120.2,If[t<1000.2,If[t<940.2,If
[t<910.2,0,0.01],0.01],0.01],0,0.01],0.0];
DPstorage6[t_]:=If[t<1200.2,If[t<1120.2,If[t<1000.2,If[t<940.2,If
[t<910.2,0,0.30],0.90],0.50],0.30],0.0];
Lr4[t_]:=If [t<700, 0,- 0.3]; calculate and plot the transient response Δω(t) for a total of 1500 s.
e) Initialize the parameters with the steady-state values as obtained in Part b). After 300 s impose step-load change \Delta P_{L2}(s)=\frac{\Delta P_{L1}}{s}pu=\frac{0.1}{s}pu. and after 400 s impose \Delta P_{L2}(s)=\frac{\Delta P_{L2}}{s}pu=\frac{-0.1}{s}pu. Thereafter:
Lr3[t_]:=If [t<600, 0, 0.6];
Lr4[t_]:=If [t<600.1, 0,- 0.6];
Lr3[t_]:=If [t<1200,If[t<1120,If[t<1000,If[t<940,If[t<910,0,0.03],0.09],0.05],
0.03],0.0];
Lr5[t_]:=If [t<1200,If[t<1120,If[t<1000,If[t<940,If[t<910,0,0.3],0.9],0.5],
0.3],0.0];
DPstorage4[t_]:=If[t<1205.2,If[t<1125.2,If[t<1005.2,If[t<950.2,If
[t<920.2,0,0.30],0.90],0.50],0.30],0.0];
DPstorage6[t_]:=If[t<1200.2,If[t<1120.2,If[t<1000.2,If[t<940.2,If
[t<910.2,0,0.01],0.01],0.01],0,0.01],0.0];
Lr4[t_]:=If [t<700, 0,- 0.3]; calculate and plot the transient response Δω(t) for a total of 1500 s.
Lr3[t_]:=If [t<600, 0, 0.6];
Lr4[t_]:=If [t<600.1, 0,- 0.6];
Lr3[t_]:=If [t<1200,If[t<1120,If[t<1000,If[t<940,If[t<910,0,0.03],0.09],0.05],
0.03],0.0];
Lr5[t_]:=If [t<1200,If[t<1120,If[t<1000,If[t<940,If[t<910,0,0.3],0.9],0.5],
0.3],0.0];
DPstorage4[t_]:=If[t<1205.2,If[t<1125.2,If[t<1005.2,If[t<950.2,If
[t<920.2,0,0.30],0.90],0.50],0.30],0.0];
DPstorage6[t_]:=If[t<1200.2,If[t<1120.2,If[t<1000.2,If[t<940.2,If
[t<910.2,0,0.01],0.01],0.01],0,0.01],0.0];
Lr4[t_]:=If [t<700, 0,- 0.3]; calculate and plot the transient response Δω(t) for a total of 1500 s.
f) Initialize the parameters with the steady-state values as obtained in Part b). After 300 s impose step-load change \Delta P_{L2}(s)=\frac{\Delta P_{L1}}{s}pu=\frac{0.1}{s}pu. and after 400 s impose \Delta P_{L2}(s)=\frac{\Delta P_{L2}}{s}pu=\frac{-0.1}{s}pu. Thereafter:
Lr3[t_]:=If [t<600, 0, 0.6];
Lr4[t_]:=If [t<600.1, 0,- 0.6];
Lr3[t_]:=If[t<1200,If[t<1120,If[t<1000,If[t<940,If
[t<910,0,0.03],0.09],0.05],0.03],0.0];
Lr5[t_]:=If [t<1200,If[t<1120,If[t<1000,If[t<940,If[t<910,0,0.3],0.9],0.5],0.3],
0.0];
DPstorage4[t_]:=If[t<1230,If[t<1150,If[t<1030,If[t<970,If
[t<940,0,0.030],0.090],0.050],0.030],0.0];
DPstorage6[t_]:=If[t<1200.2,If[t<1120.2,If[t<1000.2,If[t<940.2,If
[t<910.2,0,0.01],0.01],0.01],0,0.01],0.0];
Lr4[t_]:=If [t<700, 0,- 0.3]; calculate and plot the transient response Δω(t) for a total of 1500 s.
a) Differential and algebraic equations
System # 1: \varepsilon_{11}=\operatorname{load~ref}_1-\frac{\Delta \omega_1}{R_1}, \quad \Delta P_{\text {valve } 1}+\tau_{G 1} \frac{d\left(\Delta P_{\text {valve }} 1\right)}{d t}=\varepsilon_{11},
\Delta P_{\text {mech } 11}+\tau_{\mathrm{CH} 1} \frac{d\left(\Delta P_{mech \_1}\right)}{d t}=\Delta P_{valve\_ 1}, \quad \varepsilon_{12}=\Delta P_{mech 11}-\Delta P_{\mathrm{L} 1}-\Delta P_{\text {tie }}+\Delta P_{mech\_ 3}+\Delta P_{mech\_ 4}-\Delta P_{storage\_4 }, \Delta \omega_1 D_1+M_1 \frac{d\left(\Delta \omega_1\right)}{d t}=\varepsilon_{12}Coupling (tie, transmission) network: \frac{1}{T}\frac{d(\Delta P_{\mathrm{ic}})}{d t}=\varepsilon_{3}, where \varepsilon _3=\Delta \omega _1-\Delta \omega _2
System # 2: \varepsilon_{22}=\text { load ref } f_2-\frac{\Delta \omega_2}{R_2}, \quad \Delta P_{\text {valve } \_2}+\tau_{\mathrm{G} 2} \frac{d\left(\Delta P_{\text {valve } \_2}\right)}{d t}=\varepsilon_{22},,
\begin{aligned} & \Delta P_{ mech \_2}+\tau_{\mathrm{CH} 2} \frac{d\left(\Delta P_{\text {mech } \_2}\right)}{d t}=\Delta P_{valve 2}, \quad \varepsilon_{21}=\Delta P_{mech \_2}-\Delta P_{\mathrm{L} 2}+\Delta P_{tie }+ \\ & \Delta P_{mech_5 }+\Delta P_{mech\_6 6}-\Delta P_{storage\_ 6}, \Delta \omega_2 D_2+M_2 \frac{d\left(\Delta \omega_2\right)}{d t}=\varepsilon_{21} \\ & System\ \# 3: \quad \varepsilon_{33}=\text { load } ref _3-\frac{\Delta \omega_1}{R_3}, \quad \Delta P_{valve _3 3}+\tau_{\mathrm{G} 3} \frac{d\left(\Delta P_{\text {valve }_3 3}\right)}{d t}=\varepsilon_{33}, \\ & \text { System } \# \quad 4: \quad \varepsilon_{44}=\text { load ref }{ }_4-\frac{\Delta \omega_2}{R_4}, \quad \Delta P_{\text {valve } 4}+\tau_{\mathrm{G} 4} \frac{d\left(\Delta P_{\text {valve } 4}\right)}{d t}=\varepsilon_{44}, \\ & \Delta P_{mech\_ 4}+\tau_{C H 4} \frac{d\left(\Delta P_{mech 44}\right)}{d t}=\Delta P_{valve\_ 4} \text {. } \\ & \end{aligned}.
System # 5: \varepsilon_{5}=\operatorname{load~ref}_5-\frac{\Delta \omega_2}{R_5}, \quad \Delta P_{\text {valve } 5}+\tau_{\mathrm{G} 5} \frac{d\left(\Delta P_{\text {valve } 5}\right)}{d t}=\varepsilon_{55},,
\Delta P_{mech\_ 5}+\tau_{\mathrm{CH} 5} \frac{d\left(\Delta P_{mech\_ 5}\right)}{d t}=\Delta P_{\text {valve } \_5}.
System # 6: \varepsilon_{6}=\text { load ref } \mathrm{r}_6-\frac{\Delta \omega_2}{R_6}, \quad \Delta P_{\text {valve } \_6}+\tau_{\mathrm{G} 6} \frac{d\left(\Delta P_{\text {valve } 6}\right)}{d t}=\varepsilon_{6}
\Delta P_{mech\_ 6}+\tau_{\mathrm{CH} 6} \frac{d\left(\Delta P_{\text {msch } 6}\right)}{d t}=\Delta P_{valve\_6 }The frequency variation of a power system should be within 59-61 Hz, that is, a frequency band of ±1.66 Hz. Table E11.11.1 lists the Mathematica program for Figure E11.11.2a on which the following figures are based. Figure E11.11.3a illustrates the establishment of steady-state conditions at time t=200 s for given systems parameters. Figure E11.11.3b illustrates the given WP plant load reference (Lr5[t]). Figure E11.11.3c shows the transient response Δω(t) if storage plants 4 and 6 absorb the additional energy generated by WP plant with 0.2 s delay, and Figure E11.11.3d depicts the transient response Δω(t) if only storage plant 6 absorbs the additional energy generated by WP plant with 0.2 s delay. If the change of the PW plant is absorbed by the short-term storage plant 4 associated with the PV plant then Figure E11.11.3e is obtained. Figure E11.11.3f shows the transient response Δω(t) if no storage plant absorbs the additional energy generated by WP plant. The influence of the transmission line, switching instants, and storage plants increasing Δω(t) can clearly be observed.
b) The results based on Mathematica are shown in Figures E11.11.3. The Mathematica input program is listed in Table E11.11.1.
Table E11.11.1 Mathematica input program for stability analysis of Figures E11.11.2, E11.11.3
| R1=0.01; |
DPvalve4b=(DPvalve4[t]/.sol1[[1]]/.t->200);
|
| d1=0.8; |
DPmech5b=(DPmech5[t]/.sol1[[1]]/.t->200);
|
| M1=4.5; |
DPvalve5b=(DPvalve5[t]/.sol1[[1]]/.t->200);
|
| Tg1=0.3; |
DPmech6b=(DPmech6[t]/.sol1[[1]]/.t->200);
|
| Tch1=0.9; |
DPvalve6b=(DPvalve6[t]/.sol1[[1]]/.t->200);
|
| Lr1=0.8; |
ic16=Dw1[200]=Dw1b;
|
| DPL1[t_]:¼If [t<210, 0, 0.1]; |
ic17=DPmech1[200]=DPmech1b;
|
| R2=0.1; |
ic18=DPvalve1[200]=DPvalve1b;
|
| d2=1.0; |
ic19=DPtie[200]=DPtie1b;
|
| M2=6; |
ic20=Dw2[200]=Dw2b;
|
| Tg2=0.2; |
ic21=DPmech2[200]=DPmech2b;
|
| Tch2=0.2; |
ic22=DPvalve2[200]=DPvalve2b;
|
| Lr2=0.5; |
ic23=DPmech3[200]=DPmech3b;
|
| DPL2[t_]:=If [t<400, 0, -0.1]; |
ic24=DPvalve3[200]=DPvalve3b;
|
| Xtie=0.2; |
ic25=DPmech4[200]=DPmech4b;
|
| Tie=377/Xtie; |
ic26=DPvalve4[200]=DPvalve4b;
|
| R3=0.3; |
ic27=DPmech5[200]=DPmech5b;
|
| Tg3=0.1; |
ic28=DPvalve5[200]=DPvalve5b;
|
| Tch3=0.1; |
ic29=DPmech6[200]=DPmech6b;
|
| Lr3[t_]:=If [t<600, 0, 0.06]; |
ic30=DPvalve6[200]=DPvalve6b;
|
| R4=0.5; | R1=0.01; |
| Tg4=0.2; | d1=0.8; |
| Tch4=0.1; | M1=4.5; |
| Lr4[t_]:=If [t<600.1, 0,- 0.6]; | Tg1=0.3; |
| R5=0.7; | Tch1=0.9; |
| Tg5=0.1; | Lr1=0.8; |
| Tch5=0.1; |
DPL1[t_]:=If [t<300, 0, 0.1];
|
| Lr5[t]:=If [t<600.2, 0, 0.3]; | R2=0.1; |
| R6=0.5; | d2=1.0; |
| Tg6=0.2; | M2=6; |
| Tch6=0.1; | Tg2=0.2; |
| Lr6=0.01; | Tch2=0.2; |
| ic1=Dw1[0]= =0; | Lr2=0.5; |
| ic2=DPmech1[0]= =0; |
DPL2[t_]:=If [t<400, 0, -0.1];
|
| ic3=DPvalve1[0]= =0; | Xtie=0.2; |
| ic4=DPtie[0]= =0; | Tie=377/Xtie; |
| ic5=Dw2[0]= =0; | R3=0.3; |
| ic6=DPmech2[0]= =0; | Tg3=0.1; |
| ic7=DPvalve2[0]= =0; | Tch3=0.1; |
| ic8=DPmech3[0]= =0; |
Lr3[t_]:=If [t<1200,If[t<1120,If[t<1000,If
|
| ic9=DPvalve3[0]= =0; | [t<940,If |
| ic10=DPmech4[0]= =0; |
[t<910,0,0.03],0.09],0.05],0.03],0.0];
|
| ic11=DPvalve4[0]= =0; |
DPstorage4[t_]:=If [t<1200.2,If[t<1120.2,If
|
| ic12=DPmech5[0]= =0; |
[t<1000.2,If[t<940.2,If
|
| ic13=DPvalve5[0]= =0; |
[t<910.2,0,0.15],0.45],0.25],0.15],0.0];
|
| ic14=DPmech6[0]= =0; |
DPstorage6[t_]:=If [t<1200.2,If[t<1120.2,If
|
| ic15=DPvalve6[0]= =0; |
[t<1000.2,If[t<940.2,If
|
| E11[t_]:=Lr1-Dw1[t]/R1; |
[t<910.2,0,0.15],0.45],0.25],0.15],0.0];
|
| E12[t_]:=DPmech1[t]-DPL1[t]-DPtie[t] | R4=0.5; |
| #ERROR! | Tg4=0.2; |
| E3[t_]:=Dw1[t]-Dw2[t]; | Tch4=0.1; |
| E22[t_]:=Lr2-Dw2[t]/R2; |
Lr4[t_]:=If [t<700, 0,- 0.01];
|
| E21[t_]:=DPmech2[t]-DPL2[t]+DPtie | R5=0.7; |
| [t]+DPmech5[t]+DPmech6[t]; | Tg5=0.1; |
| E33[t_]:=Lr3[t]-Dw1[t]/R3; | Tch5=0.1; |
| E44[t_]:=Lr4[t]-Dw1[t]/R4; |
Lr5[t]:=If [t<1200,If[t<1120,If[t<1000,If
|
| E55[t_]:=Lr5[t]-Dw2[t]/R5; |
[t<940,If[t<910,0,0.3],0.9],0.5],0.3],0.0];
|
| E66[t_]:=Lr6-Dw2[t]/R6; | R6=0.5; |
| eqn1=Dw1’[t]= =(1/M1)*(E12[t]- | Tg6=0.2; |
| d1*Dw1[t]); | Tch6=0.1; |
| eqn2=DPmech1’[t]= =(1/Tch1)* | Lr6=0.01; |
| (DPvalve1[t]-DPmech1[t]); |
E11[t_]:=Lr1-Dw1[t]/R1;
|
| eqn3=DPvalve1’[t]= =(1/Tg1)*(E11 |
E12[t_]:=DPmech1[t]-DPL1[t]-DPtie[t]
|
| [t]-DPvalve1[t]); | #ERROR! |
| eqn4=DPtie’[t]= =Tie*E3[t]; |
E3[t_]:=Dw1[t]-Dw2[t];
|
| eqn5=Dw2’[t]= =(1/M2)*(E21[t]- |
E22[t_]:=Lr2-Dw2[t]/R2;
|
| d2*Dw2[t]); |
E21[t_]:=DPmech2[t]-DPL2[t]+DPtie[t]
|
| eqn6=DPmech2’[t]= =(1/Tch2)* | #ERROR! |
| (DPvalve2[t]-DPmech2[t]); |
E33[t_]:=Lr3[t]-Dw1[t]/R3;
|
| eqn7=DPvalve2’[t]= =(1/Tg2)*(E22 |
E44[t_]:=Lr4[t]-Dw1[t]/R4;
|
| [t]-DPvalve2[t]); |
E55[t_]:=Lr5[t]-Dw2[t]/R5;
|
| eqn8=DPmech3’[t]= =(1/Tch3)* |
E66[t_]:=Lr6-Dw2[t]/R6;
|
| (DPvalve3 [t]-DPmech3[t]); |
eqn1=Dw1’[t]= =(1/M1)*(E12[t]-d1*Dw1
|
| eqn9=DPvalve3’[t]= =(1/Tg3)*(E33 | [t]); |
| [t]-DPvalve3[t]); |
eqn2¼DPmech1’[t]= =(1/Tch1)*(DPvalve1
|
| eqn10=DPmech4’[t]= =(1/Tch4)* | [t]-DPmech1[t]); |
| (DPvalve4[t]-DPmech4[t]); |
eqn3=DPvalve1’[t]= =(1/Tg1)*(E11[t]-
|
| eqn11=DPvalve4’[t]= =(1/Tg4)*(E44 | DPvalve1[t]); |
| [t]-DPvalve4[t]); |
eqn4=DPtie’[t]= =Tie*E3[t];
|
| eqn12=DPmech5’[t]= =(1/Tch5)* |
eqn5=Dw2’[t]= =(1/M2)*(E21[t]-d2*Dw2
|
| (DPvalve5 [t]-DPmech5[t]); | [t]); |
| eqn13=DPvalve5’[t]= =(1/Tg5)*(E55 |
eqn6=DPmech2’[t]= =(1/Tch2)*(DPvalve2
|
| [t]-DPvalve5[t]); | [t]-DPmech2[t]); |
| eqn14=DPmech6’[t]= =(1/Tch6)* |
eqn7=DPvalve2’[t]= =(1/Tg2)*(E22[t]-
|
| (DPvalve6[t]-DPmech6[t]); | DPvalve2[t]); |
| eqn15=DPvalve6’[t]= =(1/Tg6)*(E66 |
eqn8=DPmech3’[t]= =(1/Tch3)*(DPvalve3
|
| [t]-DPvalve6[t]); | [t]-DPmech3[t]); |
| sol1=NDSolve[{eqn1,eqn2,eqn3,eqn4, |
eqn9=DPvalve3’[t]= =(1/Tg3)*(E33[t]-
|
| eqn5,eqn6,eqn7,eqn8,eqn9,eqn10, | DPvalve3[t]); |
| eqn11,eqn12,eqn13,eqn14,eqn15,ic1, |
eqn10=DPmech4’[t]= =(1/Tch4)*(DPvalve4
|
| ic2,ic3,ic4,ic5,ic6,ic7,ic8,ic9,ic10,ic11, | [t]-DPmech4[t]); |
| ic12,ic13,ic14,ic15},{Dw1[t], |
eqn11=DPvalve4’[t]= =(1/Tg4)*(E44[t]-
|
| DPmech1[t],DPvalve1[t],DPtie[t],Dw2 | DPvalve4[t]); |
| [t],DPmech2[t],DPvalve2[t],DPmech3 |
eqn12=DPmech5’[t]= =(1/Tch5)*(DPvalve5
|
| [t],DPvalve3[t],DPmech4[t],DPvalve4 | [t]-DPmech5[t]); |
| [t],DPmech5[t],DPvalve5[t],DPmech6 |
eqn13=DPvalve5’[t]= =(1/Tg5)*(E55[t]-
|
| [t],DPvalve6[t]},{t,0,200},MaxSteps- | DPvalve5[t]); |
| >100000000]; |
eqn14=DPmech6’[t]= =(1/Tch6)*(DPvalve6
|
| Plot[Dw1[t]/.sol1[[1]],{t,0,200}, | [t]-DPmech6[t]); |
| PlotRange->All,AxesLabel->{“t |
eqn15=DPvalve6’[t]= =(1/Tg6)*(E66[t]-
|
| [s]”,”Dw1[t][pu]”}] | DPvalve6[t]); |
| Dw1b=(Dw1[t]/.sol1[[1]]/.t->200) |
sol2=NDSolve[{eqn1,eqn2,eqn3,eqn4,eqn5,
|
| DPmech1b=(DPmech1[t]/.sol1[[1]]/. |
eqn6,eqn7,eqn8,eqn9,eqn10,eqn11,eqn12,
|
| t->200); |
eqn13,eqn14,eqn15,ic1,ic2,ic3,ic4,ic5,ic6,ic7,
|
| DPvalve1b=(DPvalve1[t]/.sol1[[1]]/. |
ic8,ic9,ic10,ic11,ic12,ic13,ic14,ic15},{Dw1
|
| t->200); |
[t],DPmech1[t],DPvalve1[t],DPtie[t],Dw2[t],
|
| DPtie1b=(DPtie[t]/.sol1[[1]]/.t->200); |
DPmech2[t],DPvalve2[t],DPmech3[t],
|
| Dw2b=(Dw2[t]/.sol1[[1]]/.t->200) |
DPvalve3[t],DPmech4[t],DPvalve4[t],
|
| DPmech2b=(DPmech2[t]/.sol1[[1]]/. |
DPmech5[t],DPvalve5[t],DPmech6[t],
|
| t->200); |
DPvalve6[t]},{t,200,1500},MaxSteps->
|
| DPvalve2b=(DPvalve2[t]/.sol1[[1]]/. | 100000000]; |
| t->200); |
Plot[Evaluate[Lr5[t]],{t,200,1500},PlotRange-
|
| DPmech3b=(DPmech3[t]/.sol1[[1]]/. |
>All,AxesLabel->{“t (s)”,”Lr5[t] [pu]”}]
|
| t->200); |
Plot[Dw1[t]/.sol2[[1]],{t,200,1500},
|
| DPvalve3b=(DPvalve3[t]/.sol1[[1]]/. |
PlotRange->All,AxesLabel->{“t[s]”,”Dw1
|
| t->200); | [t][pu]”}] |
|
DPmech4b=(DPmech4[t]/.sol1[[1]]/.
|
|
| t->200); |
