diff options
Diffstat (limited to '1445/CH8/EX8.33/Ex8_33.sce')
| -rw-r--r-- | 1445/CH8/EX8.33/Ex8_33.sce | 27 |
1 files changed, 5 insertions, 22 deletions
diff --git a/1445/CH8/EX8.33/Ex8_33.sce b/1445/CH8/EX8.33/Ex8_33.sce index aaf8cdd40..9e16a1081 100644 --- a/1445/CH8/EX8.33/Ex8_33.sce +++ b/1445/CH8/EX8.33/Ex8_33.sce @@ -1,10 +1,10 @@ //CHAPTER 8- DIRECT CURRENT MACHINES //Example 33 +clc; disp("CHAPTER 8"); disp("EXAMPLE 33"); -//230 V DC series motor //VARIABLE INITIALIZATION v_t=230; //in Volts N1=1500; //in rpm @@ -15,34 +15,17 @@ r_se=0.2; //series field resistance in Ohms //SOLUTION //solution (a) -//for series motors, phi dir prop Ia -// therefore, Te dir prop Ia^2 -// at starting Eb=0 and Vt= Ia1.(r_a+r_se+r_ext) -//rearranging for r_ext, we get -// r_ext = (Vt-Ia1.(r_a+r_se))/ Ia1 -E_b=0; //back emf at starting -nr1=v_t-I_a1*(r_a+r_se); //value of numerator in the expression for r_ext +E_b=0; //at starting +nr1=v_t-I_a1*(r_a+r_se); //value of numerator r_ext=nr1/I_a1; -disp(sprintf("(a) At starting, the resistance that must be added is %.0f Ω",r_ext)); +disp(sprintf("(a) At starting, the resistance that must be added is %f Ω",r_ext)); //solution (b) -//Ia2=Ia1=20 A -//as phi dir prop Ia, we get -//Eb2/Eb1 = phi2.n2/ phi1. N1 = Ia2.N2/Ia1.N1 -//=> Eb2/Eb1=N2/N1 as Ia2=Ia1 (eq 1) I_a2=I_a1; N2=1000; ratio=N2/N1; -// now, we know that Eb1=Vt-Ia1.(r_a+r_se) and -// Eb2 = Vt - Ia2.(r_a+r_se+r_ext) -//substituting values of Eb1 and Eb2 in eq 1 above, we get -//n2/n1 = (Vt - Ia2.(r_a+r_se+r_ext))/ (Vt-Ia1.(r_a+r_se)) -//since ia1=Ia2 (rated torque) -//we get -//r_ext = (N2/N1).(v_t-I_a1*(r_a+r_se))/Ia2 -(v_t-I_a2*(r_a+r_se))/Ia2 -// nr2=v_t-I_a2*(r_a+r_se); r_ext=((ratio*nr1)-nr2)/(-I_a2); -disp(sprintf("(b) At 1000 rpm, the resistance that must be added is %.3f Ω",r_ext)); +disp(sprintf("(b) At 1000 rpm, the resistance that must be added is %f Ω",r_ext)); //END |