1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
|
// Electric Machinery and Transformers
// Irving L kosow
// Prentice Hall of India
// 2nd editiom
// Chapter 12: POWER,ENERGY,AND EFFICIENCY RELATIONS OF DC AND AC DYNAMOS
// Example 12-17
clear; clc; close; // Clear the work space and console.
// Given data
// code letter = J of SCIM (Ex.12-16)
// Calculations
// case a
// From Appendix A-3,Table 430-7(b),the starting kVA/hp (with rotor locked) is
// less than 7.99,which, when substituted in the following equation, yields a
// maximum starting current of :
// subscript u for I_s indicates upper limit of starting current
I_s_u = (7.99*(7.5*1000))/(sqrt(3)*220) ;
// case b
// The lower limit,code letter J,is 7.1 kVA/hp. Thus :
// subscript l for I_s indicates lower limit of starting current
I_s_l = (7.1*(7.5*1000))/(sqrt(3)*220) ;
// Display the results
disp("Example 12-17 Solution : ");
printf(" \n a: From Appendix A-3,Table 430-7(b),the starting kVA/hp ");
printf(" \n (with rotor locked) is less than 7.99,which, when substituted ");
printf(" \n in the following equation, yields a maximum starting current of :");
printf(" \n I_s = %.1f A \n",I_s_u);
printf(" \n b: The lower limit,code letter J,is 7.1 kVA/hp.\n Thus :");
printf(" \n I_s = %.1f A ",I_s_l );
|
