# the per unit system

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• #10194
Keymaster

can some one please suggest a book to learn the per unit calculation system as I find most of the papers and books on the subject ambigious in some parts therefore difficult to follow. Thank you.

#11385
Keymaster

Hello,

I dont know any book to suggest, but i can give you a simple example.

In a per unit system the values are refered to some values taken as bases of the system.

For example if we have many transformers conected in an energy distribution  system, we can take

the values of one of them to make de bases of the per unit system :

T1 ( we take this one)

Sn=2500 kVA

Un=15 kV

T2

Sn=630 kVA

Un=15 kV

T3

Sn=500 kVA

Un=15 kV

T……

Bases :

Ub=U(t1)n = 15 kV

Sb=S(t1)n = 2500 kVA

Ib=2500*10^3 /(1,73*15*10^3) = 96,33 A

– Now lets make some calculations in per unit system

For example  we Have :

S(T2) = (630 *10^3) / (2500*10^3) = 0,252 pu

In(T2) = 0,252 / 1 = 0,252 pu    ( 1the denominator  is because we have Un(T2)=15 kV=Ub= 1 pu)

S(T3) = (500 *10^3) / (2500*10^3) = 0,2 pu

In(T2) = 0,2 / 1 = 0,2 pu

So in pu :

Sn(T1) =1 pu  ,      Un(T1)= 1 pu ,  In(T1)= 1 pu

Sn(T2)=0,252 pu,  Un(T1)= 1 pu ,  In(T2)= 0,252 pu

Sn(T3)=0,2 pu,      Un(T1)= 1 pu ,  In(T3)= 0,2 pu

We can see very easily wich equipment is above or under our network reference, we dont need to do calculations with

Ampere, Volt, VA, etc, it is also very usefull in some complex calculations.

If we want the real values we simple multiply the pu values by the bases :

Sn(T1) =1 * 2500 kVA       = 2500 kVA  ,    Un(T1)= 1 * 15 kV = 15 kV ,  In(T1)= 1 * 96,33 A        = 96,33 A

Sn(T2)=0,252 * 2500 kVA = 630 kVA,        Un(T1)= 1 * 15 kV = 15 kV ,  In(T2)= 0,252 * 96,33 A = 24,275 A

Sn(T3)=0,2 * 2500 kVA     = 500 kVA,        Un(T1)= 1 * 15 kV = 15 kV ,  In(T3)= 0,2 * 96,33 A     = 19,266 A

#11409
Keymaster

el BEEMAN, Industrial Electric Power System  Talvez en bibliotecas , ediciones agotadas

#11415