the per unit system

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  • #10194
    admin
    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
    admin
    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
    admin
    Keymaster

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

    #11415
    admin
    Keymaster

    Thank you for your replies and help. Also I managed to find some useful papers on the subject on the internet and I think I understand  the P.U system now but still I have to apply it to a real life case to determine the short circuit current in a system.

    However I would like to call on your help again and ask if anyone can tell me the imp/Km of standard 11kV, 33kV,66kV and 132kV. overhead transmission lines. Thank you. 

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