THE POWER SYSTEM-AN OVERVIEW AND MODELLING

UNIT-I
THE POWER SYSTEM-AN OVERVIEW AND MODELLING
Explain the modeling of generator, load, transmission line and transformer for power flow, short circuit and stability studies.
Refer - modeling chart
Draw the per unit reactance diagram for the power systems shown below. Neglect resistance and use a base of 100MVA, 220KV in 50 ohms line. The ratings of the generator, motor and transformers are







G: 40MVA, 25KV, X’’ = 20%
M: 50MVA, 11KV, X’’ = 30%
T1: 40MVA, 33 Y/ 220Y KV, X = 15%
T2: 30MVA, 11 Δ / 220Y KV, X = 15%
Load: 11KV, 50MW+j68 MVAR

Solution:
Base MVA, MVAb, new = 100MVA
Base KV, KVb, new = 220KV

Reactance of the transmission line:
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)
Actual reactance =50Ω
Base impedance= (KV_(b,new) )^2/MVA_(b,new) = 〖220〗^2/100=484Ω
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)=50/484 =0.1033 p.u

Reactance of the Transformer T1:
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.15 p.u.
MVA_(b,old)=40 ; MVA_(b,new)=100
KV_(b,old)=33 ; KV_(b,new)=?
Base KV on LT side of transformer T_1=Base KV on HT side×(LT voltage rating)/(HT voltage rating)
Base KV on LT side of transformer T_1=220×33/220=33KV
KV_(b,new)= 33 KV
X_(p.u,new)=0.15×(33/33)^2×(100/40)=0.375 p.u
Reactance of the Generator G:
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.2 p.u.
MVA_(b,old)=40 ; MVA_(b,new)=100
KV_(b,old)=25 ; KV_(b,new)= 33
X_(p.u,new)=0.2×(25/33)^2×(100/40)=0.287 p.u
Reactance of the Transformer T2:
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.15 p.u.
MVA_(b,old)=30 ; MVA_(b,new)=100
KV_(b,old)=11 ; KV_(b,new)=?
Base KV on LT side of transformer T_2=Base KV on HT side×(LT voltage rating)/(HT voltage rating)
Base KV on LT side of transformer T_1=220×11/220=11KV
KV_(b,new)= 11 KV
X_(p.u,new)=0.15×(11/11)^2×(100/30)=0.5 p.u
Reactance of the Motor M:
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.3 p.u.
MVA_(b,old)=50 ; MVA_(b,new)=100
KV_(b,old)=11 ; KV_(b,new)= 11
X_(p.u,new)=0.3×(11/11)^2×(100/50)=0.6 p.u

Reactance Diagram
















Draw the reactance diagram using a base of 50MVA and 13.8KV on generatorG1















G1: 20MVA, 13.8KV, X’’=20% ; G2: 30MVA, 18.0KV, X’’=20%
G3: 30MVA, 20.0KV, X’’=20% ; T1: 25MVA, 220/13.8 KV, X =10%
T2:3Single phase unit each rated 10MVA, 127/18 KV, X =10%
T3: 35MVA, 220/22 KV, X =10%

Solution:
Base MVA, MVAb, new = 50MVA
Base KV, KVb, new = 13.8KV
Reactance of the Generator G1:
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.2 p.u.
MVA_(b,old)=20 ; MVA_(b,new)=50
KV_(b,old)=13.8 ; KV_(b,new)= 13.8
X_(p.u,new)=0.2×(13.8/13.8)^2×(50/20)=j0.5 p.u
Reactance of the Transformer T1 :( Primary side)
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.1 p.u.
MVA_(b,old)=25 ; MVA_(b,new)=50
KV_(b,old)=13.8 ; KV_(b,new)= 13.8
X_(p.u,new)=0.1×(13.8/13.8)^2×(50/25)=j0.2 p.u


Reactance of the transmission line j 80Ω:
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)
Actual reactance =80Ω
Base KV on HT side of transformer T_1=Base KV on LT side×(HT voltage rating)/(LT voltage rating)
Base KV on HT side of transformer T_1=13.8×220/13.8=220KV
KV_(b,new)= 220 KV
Base impedance= (KV_(b,new) )^2/MVA_(b,new) = 〖220〗^2/50=968Ω
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)=80/968 =j0.0826 p.u

Reactance of the transmission line j 100Ω:
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)
Actual reactance =100Ω
Base KV on HT side of transformer T_1=Base KV on LT side×(HT voltage rating)/(LT voltage rating)
Base KV on HT side of transformer T_1=13.8×220/13.8=220KV
KV_(b,new)= 220 KV
Base impedance= (KV_(b,new) )^2/MVA_(b,new) = 〖220〗^2/50=968Ω
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)=100/968 =j0.1033 p.u

Reactance of the Transformer T2:( Primary side)
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.1 p.u.
Y/Δ Connection; voltage rating: √3 ×127/18 KV= 220/18KV

MVA_(b,old)=3×10=30 ; MVA_(b,new)=50
KV_(b,old)=220 ; KV_(b,new)=220?
X_(p.u,new)=0.1×(220/220)^2×(50/30)=j0.1667 p.u


Reactance of the Generator G2:
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.2 p.u.
MVA_(b,old)=30 ; MVA_(b,new)=50
KV_(b,old)=18 ; KV_(b,new)= ?
Base KV on LT side of transformer T_2=Base KV on HT side×(LT voltage rating)/(HT voltage rating)
Base KV on LT side of transformer T_2=220×18/220=18KV
KV_(b,new)= 18KV
X_(p.u,new)=0.2×(13.8/13.8)^2×(50/20)=j0.5 p.u

X_(p.u,new)=0.2×(18/18)^2×(50/30)=j0.333 p.u

Reactance of the Transformer T3 :( Secondary side)
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.1 p.u.
MVA_(b,old)=35 ; MVA_(b,new)=50
KV_(b,old)=220 ; KV_(b,new)= 220
X_(p.u,new)=0.1×(220/220)^2×(50/35)=j0.1429 p.u

Reactance of the Generator G3:
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.2 p.u.
MVA_(b,old)=30 ; MVA_(b,new)=50
KV_(b,old)=20 ; KV_(b,new)= ?
Base KV on LT side of transformer T_3=Base KV on HT side×(LT voltage rating)/(HT voltage rating)
Base KV on LT side of transformer T_3=220×22/220=22KV
KV_(b,new)= 22KV
X_(p.u,new)=0.2×(20/22)^2×(50/30)=j0.2755 p.u

Reactance Diagram













A simple power system is shown in fig. Redraw this system where the per unit impedance of the components are represented on a common 5000 VA base and common system base voltage of 250V.






Solution:
Base MVA, MVAb, new = 5000 VA = 5MVA
Base KV, KVb, new = 250V = 0.25KV
Impedance of the Generator G1:
Z_(p.u,new)=Z_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
Z_(p.u,old)=0.2 p.u.
MVA_(b,old)=1 ; MVA_(b,new)=5
KV_(b,old)=0.25 ; KV_(b,new)= 0.25
Z_(p.u,new)=0.2×(0.25/0.25)^2×(5/1)=j1.0 p.u
Impedance of the Generator G2:
Z_(p.u,new)=Z_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
Z_(p.u,old)=0.3 p.u.
MVA_(b,old)=2 ; MVA_(b,new)=5
KV_(b,old)=0.25 ; KV_(b,new)= 0.25
Z_(p.u,new)=0.3×(0.25/0.25)^2×(5/2)=j0.75 p.u
Impedance of the Transformer T1: (Primary side)
Z_(p.u,new)=Z_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
Z_(p.u,old)=0.2 p.u.
MVA_(b,old)=4 ; MVA_(b,new)=5
KV_(b,old)=0.25 ; KV_(b,new)= 0.25
Z_(p.u,new)=0.2×(0.25/0.25)^2×(5/4)=j0.25 p.u
Impedance of the transmission line Z= 40+ j 100Ω:
p.u.impedance of the transmission line= (Actual impedance,Ω )/(Base impedance,Ω)
Actual impedance = (40+j150) Ω
Base V on HT side of transformer T_1=Base V on LT side×(HT voltage rating)/(LT voltage rating)
Base V on HT side of transformer T_1=250×800/250=800V
V_(b,new)= 800 V
Base impedance= (V_(b,new) )^2/VA_(b,new) = 〖800〗^2/5000=128Ω
p.u.impedance of the transmission line= (Actual impedance,Ω )/(Base impedance,Ω)=(40+j150)/128 =0.3125+j1.17 p.u
Impedance of the Transformer T2: (Primary side)
Z_(p.u,new)=Z_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
Z_(p.u,old)=0.06 p.u.
MVA_(b,old)=8 ; MVA_(b,new)=5
KV_(b,old)=1 ; KV_(b,new)= 0.8
Z_(p.u,new)=0.06×(1/0.8)^2×(5/8)=j0.0585 p.u
Impedance of the Load
Z_(p.u,new)=2500/5000=j0.5
Reactance Diagram





The single line diagram of a three phase power system is shown in fig. Select a common base of 100MVA and 13.8KV on the generator side. Draw per unit impedance diagram








G: 90MVA, 13.8KV, X=18% ; T1 :50MVA, 13.8/220KV, X=10%
T2:50MVA, 220/11KV, X=10% ; T3 :50MVA, 13.8/132KV, X=10%
T4:50MVA, 132/11KV, X=10% ; M : 80MVA, 10.45KV, X=20%
LOAD : 57MVA, 0.8 p.f lagging at 10.45 KV ; Line 1 = j 50Ω ; Line 2 = j 70Ω
Solution:
Base MVA, MVAb, new = 100MVA
Base KV, KVb, new = 13.8KV
Reactance of the Generator G1:
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.18 p.u.
MVA_(b,old)=90 ; MVA_(b,new)=100
KV_(b,old)=13.8 ; KV_(b,new)= 13.8
X_(p.u,new)=0.18×(13.8/13.8)^2×(100/90)=j0.2 p.u
Reactance of the Transformer T1 :( Primary side)
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.1 p.u.
MVA_(b,old)=50 ; MVA_(b,new)=100
KV_(b,old)=13.8 ; KV_(b,new)= 13.8
X_(p.u,new)=0.1×(13.8/13.8)^2×(100/50)=j0.2 p.u

Reactance of the transmission line j 50Ω:
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)
Actual reactance =50Ω
Base KV on HT side of transformer T_1=Base KV on LT side×(HT voltage rating)/(LT voltage rating)
Base KV on HT side of transformer T_1=13.8×220/13.8=220KV
KV_(b,new)= 220 KV
Base impedance= (KV_(b,new) )^2/MVA_(b,new) = 〖220〗^2/100=484Ω
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)=50/484 =j0.1033 p.u
Reactance of the Transformer T2 :( Primary side)
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.1 p.u.
MVA_(b,old)=50 ; MVA_(b,new)=100
KV_(b,old)=220 ; KV_(b,new)= 220
X_(p.u,new)=0.1×(220/220)^2×(100/50)=j0.2 p.u
Reactance of the Transformer T3 :( Primary side)
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.1 p.u.
MVA_(b,old)=50 ; MVA_(b,new)=100
KV_(b,old)=13.8 ; KV_(b,new)= 13.8
X_(p.u,new)=0.1×(13.8/13.8)^2×(100/50)=j0.2 p.u
Reactance of the transmission line j 70Ω:
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)
Actual reactance =70Ω
Base KV on HT side of transformer T_3=Base KV on LT side×(HT voltage rating)/(LT voltage rating)
Base KV on HT side of transformer T_3=13.8×132/13.8=132KV
KV_(b,new)= 132KV
Base impedance= (KV_(b,new) )^2/MVA_(b,new) = 〖132〗^2/100=174.24Ω
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)=70/174.24 =j0.4017 p.u
Reactance of the Transformer T4 :( Primary side)
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.1 p.u.
MVA_(b,old)=50 ; MVA_(b,new)=100
KV_(b,old)=132 ; KV_(b,new)= 132
X_(p.u,new)=0.1×(132/132)^2×(100/50)=j0.2 p.u
Reactance of the Motor M:
X_(p.u,new)=X_(p.u,old)×(KV_(b,old)/KV_(b,new) )^2×(MVA_(b,new)/MVA_(b,old) )
X_(p.u,old)=0.2 p.u.
MVA_(b,old)=80 ; MVA_(b,new)=100
KV_(b,old)=10.45 ; KV_(b,new)= ?
Base KV on LT side of transformer T_4=Base KV on HT side×(LT voltage rating)/(HT voltage rating)
Base KV on LT side of transformer T_4=132×11/132=11KV
KV_(b,new)= 11KV
X_(p.u,new)=0.2×(10.45/11)^2×(100/80)=j0.22565 p.u
The load is at 0.8 p.f lagging is given by
S_L (3ϕ)=57∠36.87˚
Load impedance is given by
Z_L=(V_(L-L) )^2/(S_L3ϕ^* )=〖10.45〗^2/(57∠36.87˚)=(1.532+j1.1495)Ω
Base impedance for the load is
Base impedance= (KV_(b,new) )^2/MVA_(b,new) = 〖11〗^2/100=1.21Ω
p.u.reactance of the transmission line= (Actual reactance,Ω )/(Base reactance,Ω)=(1.532+j1.1495)/1.21 =(1.266+j0.95) p.u
Reactance Diagram