From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 3472/CH17/EX17.8/Ex17_8.png | Bin 0 -> 9743 bytes 3472/CH17/EX17.8/Example17_8.sce | 58 +++++++++++++++++++++++++++++++++++++++ 2 files changed, 58 insertions(+) create mode 100644 3472/CH17/EX17.8/Ex17_8.png create mode 100644 3472/CH17/EX17.8/Example17_8.sce (limited to '3472/CH17/EX17.8') diff --git a/3472/CH17/EX17.8/Ex17_8.png b/3472/CH17/EX17.8/Ex17_8.png new file mode 100644 index 000000000..e82294481 Binary files /dev/null and b/3472/CH17/EX17.8/Ex17_8.png differ diff --git a/3472/CH17/EX17.8/Example17_8.sce b/3472/CH17/EX17.8/Example17_8.sce new file mode 100644 index 000000000..092fd7d61 --- /dev/null +++ b/3472/CH17/EX17.8/Example17_8.sce @@ -0,0 +1,58 @@ +// A Texbook on POWER SYSTEM ENGINEERING +// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar +// DHANPAT RAI & Co. +// SECOND EDITION + +// PART II : TRANSMISSION AND DISTRIBUTION +// CHAPTER 10: POWER SYSTEM STABILITY + +// EXAMPLE : 10.8 : +// Page number 273-275 +clear ; clc ; close ; // Clear the work space and console +funcprot(0) + +// Given data +V = 33.0*10**3 // Line voltage(V) +R = 6.0 // Resistance per phase(ohm) +X = 15.0 // Reactance per phase(ohm) + +// Calculations +V_S = V/3**0.5 // Sending end phase voltage(V) +V_R = V/3**0.5 // Receiving end phase voltage(V) +beta = atand(X/R) // β(°) +Z = (R**2+X**2)**0.5 // Impedance(ohm) +delta_0 = 0.0 // δ(°) +P_0 = (V_R/Z**2)*(V_S*Z*cosd((delta_0-beta))-V_R*R)/10**6 // Power received(MW/phase) +delta_1 = 30.0 // δ(°) +P_1 = (V_R/Z**2)*(V_S*Z*cosd((delta_1-beta))-V_R*R)/10**6 // Power received(MW/phase) +delta_2 = 60.0 // δ(°) +P_2 = (V_R/Z**2)*(V_S*Z*cosd((delta_2-beta))-V_R*R)/10**6 // Power received(MW/phase) +delta_3 = beta // δ(°) +P_3 = (V_R/Z**2)*(V_S*Z*cosd((delta_3-beta))-V_R*R)/10**6 // Power received(MW/phase) +delta_4 = 90.0 // δ(°) +P_4 = (V_R/Z**2)*(V_S*Z*cosd((delta_4-beta))-V_R*R)/10**6 // Power received(MW/phase) +delta_5 = 120.0 // δ(°) +P_5 = (V_R/Z**2)*(V_S*Z*cosd((delta_5-beta))-V_R*R)/10**6 // Power received(MW/phase) +delta_6 = (acosd(R/Z))+beta // δ(°) +P_6 = (V_R/Z**2)*(V_S*Z*cosd((delta_6-beta))-V_R*R)/10**6 // Power received(MW/phase) + + +delta = [delta_0,delta_1,delta_2,delta_3,delta_4,delta_5,delta_6] +P = [P_0,P_1,P_2,P_3,P_4,P_5,P_6] +a = gca() ; +a.thickness = 2 // sets thickness of plot +plot(delta,P,'ro-') +a.x_label.text = 'Electrical degree' // labels x-axis +a.y_label.text = 'Power in MW/phase' // labels y-axis +xtitle("Fig E10.7 . Power angle diagram") +xset('thickness',2) // sets thickness of axes +xstring(70,14.12,'P_max = 14.12 MW/phase(approximately)') +P_max = V_R/Z**2*(V_S*Z-V_R*R)/10**6 // Maximum power transmitted(MW/phase) +delta_equal = 0.0 // δ With no phase shift(°) +P_no_shift = (V_R/Z**2)*(V_S*Z*cosd((delta_equal-beta))-V_R*R)/10**6 // Power transmitted with no phase shift(MW/phase) + +// Results +disp("PART II - EXAMPLE : 10.8 : SOLUTION :-") +printf("\nPower angle diagram is plotted and is shown in the Figure 1") +printf("\nMaximum power the line is capable of transmitting, P_max = %.2f MW/phase", P_max) +printf("\nWith equal voltage at both ends power transmitted = %.f MW/phase", abs(P_no_shift)) -- cgit