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clc,clear
printf('Example 3.8\n\n')
V=500
Io=5 //no load current
R_a=0.5,R_sh=250//resistance of armature and field circuits
I=100 //current at unknown efficiency
P_in_NL=V*Io //no load input
I_sh=V/R_sh
Iao=Io-I_sh
arm_cu_loss_no_load=R_a*Iao^2 //No load armature copper loss
constant_losses= P_in_NL- arm_cu_loss_no_load
I_a=I-I_sh
arm_cu_loss= R_a*I_a^2 //New armature copper loss
Total_loss=arm_cu_loss + constant_losses
P_in=V*I
efficiency=(P_in-Total_loss)*100/P_in //required efficiency
printf('Efficiency is %.3f percent when motor takes %.0f A current',efficiency,I)
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