blob: 2f3d2ebc6384eae82bded187788e16f46a8df0fe (
plain)
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
|
// FUNDAMENTALS OF ELECTICAL MACHINES
// M.A.SALAM
// NAROSA PUBLISHING HOUSE
// SECOND EDITION
// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
// Example : 3.10
clc;clear; // clears the console and command history
// Given data
kVA = 12 // kVA ratingss of transformer
n = 0.97 // maximum efficiency at unity power factor
t_1 = 8 // time in hours
P_1 = 10 // load in kW
pf_1 = 0.8 // lagging power factor
t_2 = 10 // time in hours
P_2 = 15 // load in kW
pf_2 = 0.90 // leading power factor
t_3 = 6 // time in hours at no load
P_3 = 0 // load in kW
// caclulations
P_01 = kVA*1 // o/p power at full load and unity factor in kW
P_in1 = (P_01/n) // i/p power at full load
P_tloss = P_in1-P_01 // total loss in kW
P_cu = P_tloss/2 // copper loss at 12 kVA P_cu=P_i in kW
P_024 = P_1*t_1+P_2*t_2+P_3*t_3 // all day o.p power in kWh
P_i24 = 24*P_cu // iron loss for 24 hours in kWh
P_cu24 = P_cu*t_1*((P_1/pf_1)/P_01)^2+P_cu*t_2*((P_2/pf_2)/P_01)^2 // copper loss for 24 hours
P_in24 = P_024+P_i24+P_cu24 // all day i/p power in kWh
n_allday = (P_024/P_in24)*100 // all day efficiency
// display the result
disp("Example 3.10 solution");
printf(" \n All day efficiency \n n_allday = %.0f percent \n", n_allday);
|
