From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 2642/CH6/EX6.4/Ex6_4.sce | 31 +++++++++++++++++++++++++++++++ 1 file changed, 31 insertions(+) create mode 100755 2642/CH6/EX6.4/Ex6_4.sce (limited to '2642/CH6/EX6.4') diff --git a/2642/CH6/EX6.4/Ex6_4.sce b/2642/CH6/EX6.4/Ex6_4.sce new file mode 100755 index 000000000..0cc5c81e7 --- /dev/null +++ b/2642/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES +// M.A.SALAM +// NAROSA PUBLISHING HOUSE +// SECOND EDITION + +// Chapter 6 : CONTROL AND STARTING OF A DC MOTORS +// Example : 6.4 + +clc;clear; // clears the console and command history + +// Given data +N = 1100 // speed of dc series motor in rpm +P = 4 // number of poles +I_a1 = 15 // supply current to dc series motor in A +V_t = 220 // supply voltage in V +R_a = 0.9 // series armature resistance in ohm +R_se1 = 0.6 // series field resistance in ohm +I_a2 = 25 // supply current to dc series motor in A +// phi2 = 0.8*phi1 +// phi1/phi2 = 1.25 + +// caclulations +E_b1 = V_t-I_a1*(R_a+R_se1) // back emf in V +R_se2 = 0.6/4 // value of resistance per path in ohm +E_b2 = V_t-I_a2*(R_a+R_se2) // back emf in V +N_2 = (E_b2/E_b1)*1.25*N // new speed in rpm + +// display the result +disp("Example 6.4 solution"); +printf(" \n New speed \n N_2 = %.1f rpm \n", N_2); + -- cgit