電力系統(tǒng)仿真matpower潮流計(jì)算(參考版)
2024-10-31 18:33本頁(yè)面
【正文】 。s method power flow converged in 5 in seconds ======================= System Summary ======================= How many? How much? P(MW)Q(MVAr)Buses 5 Total Gen Capacity to Generators 2 Online Capacity to Committed Gens2Generation Loads 3 Load Fixed 3 Fixed Dispatchable0 of Shunts 0 Shunt(inj) Branches 5 Losses(I^2 * Z) Transformers 2 Branch Charging(inj)—— Interties 0 Total Intertie Flow Areas 1 Minimum MaximumVoltage Magnitude . bus 1 . bus 2 Voltage deg bus 1 deg bus 4 P Losses(I^2*R) MVAr line 21======================= Bus Data ======================= Bus Voltage Generation Load Mag(pu)Ang(deg)P(MW)Q(MVAr)P(MW)Q(MVAr)1 3 : ======================= Branch Data ======================= Brnch From To From Bus Injection To Bus Injection Loss(I^2 * Z) Bus Bus P(MW)Q(MVAr)P(MW)Q(MVAr)P(MW)Q(MVAr)1 2 1 2 3 1 3 3 4 3 5 2 : 最后向讀者介紹幾款常用的潮流計(jì)算軟件:。case2_539。]。3 5 0 0 0 0 0 0 1360 360。3 1 0 0 0 0 0 0 1360 360。]。%% generator data% bus PgQgQmaxQmin Vg mbase status PmaxPmin =[ 4 500 0 999909999 100 1 600 0。5 3 0 0 0 0 1 0 100 1 。3 1 370 130 0 0 1 1 0 100 1 。%% Bus data% bus_i type PdQdGsBs area VmVabaseKV zone VmaxVmin =[ 1 1 160 80 0 0 1 1 0 100 1 。239。讀者利用MATPOWER進(jìn)行潮流計(jì)算時(shí)。MATPOWER特點(diǎn)是簡(jiǎn)單、易懂且程序代碼公開，這為電力系統(tǒng)專業(yè)學(xué)生深入學(xué)習(xí)和理解掌握潮流計(jì)算中的難點(diǎn)（如節(jié)點(diǎn)導(dǎo)納矩陣、算法及迭代過(guò)程等）提供了一個(gè)開放、便捷的平臺(tái)。六 結(jié)論通過(guò)這個(gè)任務(wù)，自己在matlab編程，潮流計(jì)算，word文檔的編輯功能等方面均有提高，但也暴漏出一些問(wèn)題：理論知識(shí)儲(chǔ)備不足，對(duì)matlab的性能和特點(diǎn)還不能有一個(gè)全面的把握，對(duì)word軟件也不是很熟練，相信通過(guò)以后的學(xué)習(xí)能彌補(bǔ)這些不足，達(dá)到一個(gè)新的層次。牛頓拉夫遜法是求解非線性方程的迭代過(guò)程，其計(jì)算公式為DF=JDX，式中J為所求函數(shù)的雅可比矩陣；DX為需要求的修正值；DF為不平衡的列向量。S% 求節(jié)點(diǎn)注入電流I=Y*U39。Q(i)=U(i)*U(j)*(G(i,j)*sin(a(i)a(j))B(i,j)*cos(a(i)a(j)))+Q(i)。endk,U,a% 求節(jié)點(diǎn)注入功率i=5。U(i)=U(i)+oU(i)。oU(i)=x1(i+4)*U(i)。M,L]% J為雅克比矩陣x1=((inv(J))*x2)。L(i,i)=L(i,i)+2*(U(i))^2*B(i,i)。M(i,i)=M(i,i)U(i)*U(j)*(G(i,j)*cos(a(i)a(j))+B(i,j)*sin(a(i)a(j)))。endendend% 當(dāng)i=j時(shí)，求H,N,M,Lfor i=1:4for j=1:5if i~=j H(i,i)=H(i,i)+U(i)*U(j)*(G(i,j)*sin(a(i)a(j))B(i,j)*cos(a(i)a(j)))。L(i,j)=H(i,j)。% x2為不平衡量列向量% 求雅克比矩陣% 當(dāng)i~=j時(shí)，求H,N,M,Lfor i=1:4for j=1:4if i~=jH(i,j)=U(i)*U(j)*(G(i,j)*sin(a(i)a(j))B(i,j)*cos(a(i)a(j)))。oQ(i)=oQ(i)+Q(i)。oQ(i)=oQ(i)U(i)*U(j)*(G(i,j)*sin(a(i)a(j))B(i,j)*cos(a(i)a(j)))。oQ(i)=0。L(i,j)=0。N(i,j)=0。k=0。x1=ones(8,1)。U(5)=。Q=imag(S)。S(5)=0。S(3)=。% 輸入原始節(jié)點(diǎn)的給定注入功率S(1)=+。endY% Y為導(dǎo)納矩陣G=real(Y)。% 求節(jié)點(diǎn)導(dǎo)納矩陣中互導(dǎo)納for i=1:5for j=1:5if i~=jY(i,j)=y(i,j)。for i=1:5for j=i:5y(j,i)=y(i,j)。y(2,5)=1/(+)。y(1,5)=1/(+)。y(1,3)=1/(+)。% 輸入原始數(shù)據(jù)，求節(jié)點(diǎn)導(dǎo)納矩陣y(1,2)=1/(+)。由課本總結(jié)基本步驟如下：1
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