ECE421/521 POWER SYSTEM ANALYSIS Homework #2 521 421 2.1a 10’ 15’ 2.1b 10’ 15’ 2.1c 5’ 10’ 2.2a 5’ 2.5’ 2.2b 5’ 2.5’ 2.2c 5’ 2.5’ 2.2d 10’ 2.5’ 2.3 10’ 20’ 2.4 10’ 20’ 2.5 10’ 20’ 2.1 Run the following code, %ECE421/521 Homework #2 2.1 Vm = input('Enter the value of Vm(v): '); thetav = input('Enter the value of thetav(degree): '); Z = input('Enter the value of Z(omega): '); gama = input('Enter the value of gama(degree): '); thetai = thetav - gama; theta = (thetav - thetai)*pi/180; Im = Vm/Z; wt = 0:.05:2*pi; v = Vm*cos(wt); i = Im*cos(wt + thetai*pi/180); p = v.*i; V = Vm/sqrt(2); I = Im/sqrt(2); P = V*I*cos(theta); Q = V*I*sin(theta); S = P + 1i*Q; pr = P*(1 + cos(2*(wt +thetav))); px = Q*sin(2*(wt + thetav)); PP = P*ones(1,length(wt)); xline = zeros(1, length(wt)); wt = 180/pi*wt; subplot(2,2,1),plot(wt,v,wt,i,wt,xline),grid title(['v(t)=Vm coswt, i(t)=Im cos(wt+',num2str(theta),')']) xlabel('wt degree') subplot(2,2,2),plot(wt, p, wt, xline),grid title('p(t)=v(t) i(t)'), xlabel('wt,degree') subplot(2,2,3),plot(wt, pr, wt, PP, wt, xline),grid title('pr(t)'),xlabel('wt, degree') subplot(2,2,4),plot(wt, px, wt, xline),grid title('px(t)'),xlabel('wt, degree'),subplot(111) For an inductive load, Z=1.25, gama=60 degree, input the following quantities, 2.6a 10’ 5’ 2.6b 10’ 5’ Total 100’ 100’+20’ We have the plots of 𝑖(𝑡), 𝑣(𝑡), 𝑝(𝑡), 𝑝𝑟 (𝑡), 𝑝𝑥 (𝑡), (1’) Also, in the same way, we can have plots of a capacitive load, Z = 2.0, gama = -30 degree, (1’) Finally we have figures of a resistive load Z = 2.5, gama = 0 degree, (1’) (a) From the 𝑝𝑟 (𝑡) and 𝑝𝑥 (𝑡) plots, the real and reactive power for each load can be estimated, Inductive load: 𝑃 = 2000𝑊, 𝑄 = 3400𝑣𝑎𝑟 (2’) Capacitive load: 𝑃 = 2100𝑊, 𝑄 = −1200𝑣𝑎𝑟 Resistive load: 𝑃 = 2000𝑊, 𝑄 = 0𝑣𝑎𝑟 (2’) (2’) The conclusion regarding the sign of reactive power for inductive and capacitive loads is, Inductive load: 𝑄 > 0 Capacitive load: 𝑄 < 0 (0.5’) (0.5’) (b) Inductive load: 𝐼= 𝑉 𝑍 = 100 ∠0° √2 1.25∠60° = 56.7∠ − 60° (1’) 𝑆1 = 𝑉𝐼 ∗ = 4000∠60° = 2000𝑊 + 𝑗3463𝑣𝑎𝑟 (2’) Capacitive load: 𝐼= 𝑉 𝑍 = 100 ∠0° √2 2∠−30° = 35.4∠30° (1’) 𝑆2 = 𝑉𝐼 ∗ = 2500∠ − 30° = 2165𝑊 − 𝑗1250𝑣𝑎𝑟 (2’) Resistive load: 𝐼= 𝑉 𝑍 = 100 ∠0° √2 2.5∠0° = 28.3∠0° (1’) 𝑆3 = 𝑉𝐼 ∗ = 2000∠0° = 2000𝑊 + 𝑗0𝑣𝑎𝑟 (2’) The results are close to the ones obtained from curves. (1’) (c) If the above loads are all connected across the same power supply, 𝑆 = 𝑆1 + 𝑆2 + 𝑆3 = 6165𝑊 + 𝑗2213𝑣𝑎𝑟 (5’) 2.2(a) We have, 𝑣(𝑡) = 200cos(377𝑡) Rearrange instantaneous power p(t), 𝑝(𝑡) = 800 + 1000 cos(754𝑡 − 36.87°) = 800 + 1000 cos(754𝑡) cos(36.87°) + 1000 sin(754𝑡) sin(36.87°) = sin(36.87°)sin(2 ∗ 377𝑡) 200 10 ∗ 2 √2 √ ∗ cos(36.87°) [1 + cos(2 ∗ 377𝑡)] + (1’) Compare it with the standard form of p(t), 𝑝(𝑡) = 𝑃[1 + cos(2𝜔𝑡)] + 𝑄𝑠𝑖𝑛(2𝜔𝑡) We have, P = 800W Q = 600var (1’) (1’) Thus the complex power supplied to the load is 200 10 ∗ √2 √2 ∗ S = P+ jQ = 800W + j600var (2’) (b) According to the rearranged p(t) equation, the instantaneous current is, 𝑖(𝑡) = 10 cos(337𝑡 − 36.87°) 𝐴 (3’) The rms value of it is 𝐼 = 5√2 𝐴 (2’) (c) The load impedance is, 𝑍= 𝑉 𝐼 200 = 10 √2 √2 ∠0° ∠−36.87° = 20∠36.87° 𝛺 (5’) (d) Run the following code, (2’) %ECE421/521 Homework#2 2.2 t = 0:.1*10^-3:16.67*10^-3; v = 200*cos(377*t); p = 800 + 1000*cos(754*t - 36.87*pi/180); i = p./v; subplot(3,1,1),plot(t,v),grid title('v(t)=200cos(377t)') xlabel('t/s') ylabel('v/V') subplot(3,1,2),plot(t,p),grid title('p(t)=800+1000cos(754t-36.87)') xlabel('t/s') ylabel('p/VA') subplot(3,1,3),plot(t,i),grid title('v(t)=10cos(337t-36.87)') xlabel('t/s') ylabel('i/A') we have the plots of v(t), p(t) and i(t), (2’) From the current plot, we can estimate, The peak amplitude is 10A (2’) The phase angle is approximately 𝜃 = 0.0015 ∗ 0.016 360° = 33.75° 1 1 (2’) The angular frequency is approximately 𝜔 = 2𝜋 𝑇 = 2𝜋 0.016 = 390 𝑟𝑎𝑑/𝑠 2.3 Given, P = 288kw cosθ = 0.8 𝑃 𝑄 = 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 = 288 ∗ 0.8 0.6 = 216𝑘𝑣𝑎𝑟 𝑆 = 𝑃 + 𝑗𝑄 = 288𝑘𝑊 + 𝑗216𝑘𝑣𝑎𝑟 𝑍= |𝑉|2 𝑆∗ 24002 (2’) (2’) = 288∗103 −𝑗216∗103 = 12.8 + 𝑗9.6 𝛺 (2’) (2’) R =12.8 Ω X = 9.6Ω (2’) (2’) 2.4 𝑅= |𝑉|2 𝑃 |𝑉||𝐼| = 24002 = 288∗103 = 20 𝛺 𝑃 𝑐𝑜𝑠𝜃 = 288 0.8 (3’) = 360 𝑘𝑉𝐴 (2’) 𝑄 = |𝑉||𝐼|𝑠𝑖𝑛𝜃 = 360 ∗ 0.6 = 216𝑘𝑣𝑎𝑟 𝑋= |𝑉|2 𝑄 24002 = 216∗103 = 26.7 𝛺 (2’) (3’) 2.5 3 𝑄 = 𝑃𝑡𝑎𝑛𝜃 = 400 ∗ 4 = 300𝑘𝑣𝑎𝑟 𝑆 = 400𝑘𝑊 + 𝑗300𝑘𝑣𝑎𝑟 𝑡𝑎𝑛𝜃1 = (2’) (2’) √1 − cos 2 𝜃1 √1 − 0.962 = = 0.29 𝑐𝑜𝑠𝜃1 0.96 𝑄1 = 𝑃1 𝑡𝑎𝑛𝜃1 = 120 ∗ 0.29 = 35𝑘𝑣𝑎𝑟 𝑆1 = 120𝑘𝑊 − 𝑗35𝑘𝑣𝑎𝑟 (2’) (2’) 𝑆2 = 𝑆 − 𝑆1 = (400 − 120)𝑘𝑊 + 𝑗(300 − (−35))𝑘𝑣𝑎𝑟 = 280𝑘𝑊 + 𝑗335𝑘𝑣𝑎𝑟 2.6(a) Given, |V||I|=30kVA Power factor 0.8 leading 𝑆 = |𝑉||𝐼|𝑐𝑜𝑠𝜃 + 𝑗|𝑉||𝐼|𝑠𝑖𝑛𝜃 = 30 ∗ 0.8 − 𝑗30 ∗ 0.6 = 24𝑘𝑊 − 𝑗18𝑘𝑣𝑎𝑟 𝑅= |𝑉|2 𝑃 12002 = 24∗103 = 60𝛺 (3’) (4’) (2’) 𝑋= |𝑉|2 𝑄 12002 = 18∗103 = 80𝛺 (3’) (b) 𝑉 𝑉 𝐼 = 𝑅 + −𝑗𝑋 = 1200 1200 + −𝑗80 60 = 20 + 𝑗15 𝐴 (3’) 𝑉𝑙𝑖𝑛𝑒 = 𝑍𝐼 = (20 + 𝑗15)(8.4 + 𝑗11.2) = 0 + 𝑗350 𝑉 (3’) 𝑉𝑖𝑛 = 𝑉𝑙𝑖𝑛𝑒 + 𝑉 = (0 + 𝑗350)(1200 + 𝑗0) = 1200 + 𝑗350 𝑉 (4’)