Question

2.A n+400-k VA, 2300+n- 230+n-v, 60-hz transformer supplies a 0.8 lagging power-factor load whose KVA Is adjustable from no load to 120 percent rated KVA. The percent resistance, percent reactance, and percent core loss are m+1.56, m+3.16, and m+0.42 respectively. For 2 KVA increments of load, tabulate and plot the efficiency of the transformer from no load to 120 percent rated load. Use Matlab

n=20

m=.2

Answer 1

clc
close all
clear all

n = 20;

m = 0.2;

full_load_va = (400 + n)*1000;

primary_voltage = 2300 + n;

secondary_voltage = 230 + n;

pu_resistance = (m + 1.56)/100;

pu_reactance = (m + 3.16)/100;

frequency = 60;

pu_core_loss = (m + 0.42)/100;

power_factor = 0.8;

full_load_power = full_load_va*power_factor;

full_load_var = full_load_va*(sqrt(1 - (power_factor)^2));

base_VA = full_load_va;

base_voltage = primary_voltage;

base_current = base_VA/base_voltage;

actual_resistance = pu_resistance * ((base_voltage)^2)/base_VA;

actual_reactance = pu_reactance * ((base_voltage)^2)/base_VA;

actual_core_loss = pu_core_loss * full_load_power;

maximum_VA = 1.2*full_load_va;

load_VA = 0:2000:maximum_VA;
% efficiency = [];
% efficiency(1) = 0

for i = 1:length(load_VA)
  
primary_current(i) = load_VA(i)/primary_voltage;

efficiency(i) = (load_VA(i)* power_factor) / ((load_VA(i)* power_factor)+ actual_core_loss + ((primary_current(i))^2*actual_resistance))*100;

end

figure(1)
plot((load_VA/1000), efficiency)

title('Efficiency vs Load')
xlabel('No Load KVA <= KVA <= Maximum KVA')
ylabel('Efficiency')
grid on

Efficiency vs Load 100 90 80 70 Efficiency 60 50 40 30 20 10 0 600 100 200 300 400 500 No Load KVA <= KVA <= Maximum KVA

Efficiency vs Load 98 96 94 Efficiency 92 90 88 86 50 500 100 150 200 250 300 350 400 450 No Load KVA <= KVA <= Maximum KVA