Ayyıldız2 | 2008 TR Yapısı • 1-99 Orta Emek Destan • Oto Avsız • 10 Temmuz 21:00 HEMEN TIKLA!
Dc Load Flow Method
Because The Resistances Of Transmission Lines Are Rather Small Compared With Reactances Of Those Lines ,for Certain Types Of Operations (e.g., Fault Studies ) The Resistances Are Neglected To Simplify The Solution.
Letâs Consider The Nominal ï° Model Of A Medium Line. The Circuit Parameters Are :
A=1+ [ (1/2)yz ]
B=z+ [ (1/4)yz 2 ]
C=y
D=1+ [ (1/2)yz ]
Neglecting Line Resistance And Shunt Admittance , We Can Represent The Line With Its Inductive Reactance Only, As Shown In Figure 1.
Figure 1. Simplified Transmission Line Model
The Transmission (abcd) Parameters For This Representation Are:
A=d=1 0
B= Jxik
C= 0
And The Power Flows Through The Line ,
Ps = ï²vs Vr Cos(90 + ï¤ï©ï ï¯ï x = (vs Vr Sinï¤ï©ï ï¯ X
Pr = ï²ï vs Vr Cos(90 + ï¤) / X= (vs Vr Sinï¤ï©ï ï¯ï x (1)
For ï¤ï << 1 Radian, Sinï¤ ï¤ï®hence, In General, The Line Flows Become
Pik=(vi Vk ï¤ikï©ï ï¯ï xik (2)
Where
ï ï ï ï ï ï¤ikï ï½ï ï¤i ï²ï¤k (3)
Since All The Bus Voltages Of A Power System Are Around 1 Pu, Then Let
Vi = Vk =1 Pu
Bik =ï²1 / Xik (4)
Then Equation (3) Becomes
Pik = ï²bik ï¤ik = (ï¤i ï²ï¤k) / Xik (5)
Now, The Bus Power At Any Bus Is The Sum Of The Power Flows In The Lines Connected To That Bus. .hence ,
Pi = Pik = (ï²bik ï¤ik) I = 1, 2, ...., N (6)
Or In Matrix Form ,
P1 B11 B12 ... B1n ï¤1
P2 B21 B22 ... B2n ï¤ï²
. = . . ... . . (7)
. . . ... . .
Pn Bn1 Bn2 ... B Nn ï ï¤ N
Which Can Be Abbreviated As
[p] = [ ï¤ï ] (8)
Where
Bik = ï²1 / Xik (9)
And
Bii = (ï²bik) (10)
The Matrix Is The Imaginary Part Of Ybus . The Solution For [ï¤ïï is
[ï¤] = â1 [p] (11)
Matrix Is An (n â1) ï³ï (nï²1) Matrix Dimensionally For An N-bus System. The Diagonal And Off â Diagonal Elements Of The Matrix Can Be Found By Adding The Series Susceptances Of The Branches Connected To Bus I And By Setting Them Equal To The Negated Series Susceptance Of Branch Ik , Respectively.
Until Now , We Have Kept The System Ground As The Reference Bus. However Since We Have Dropped All Shunt Brunches In Simplifying Things , We Have Lost Our Reference. This Means That The Matrix Of Equation (7) Is Obtained By Equation (9) And (10) Will Be A Singular Matrix. Hence , ï²1 Of Equation (11) Does Not In Fact Exist.
Because The Resistances Of Transmission Lines Are Rather Small Compared With Reactances Of Those Lines ,for Certain Types Of Operations (e.g., Fault Studies ) The Resistances Are Neglected To Simplify The Solution.
Letâs Consider The Nominal ï° Model Of A Medium Line. The Circuit Parameters Are :
A=1+ [ (1/2)yz ]
B=z+ [ (1/4)yz 2 ]
C=y
D=1+ [ (1/2)yz ]
Neglecting Line Resistance And Shunt Admittance , We Can Represent The Line With Its Inductive Reactance Only, As Shown In Figure 1.
Figure 1. Simplified Transmission Line Model
The Transmission (abcd) Parameters For This Representation Are:
A=d=1 0
B= Jxik
C= 0
And The Power Flows Through The Line ,
Ps = ï²vs Vr Cos(90 + ï¤ï©ï ï¯ï x = (vs Vr Sinï¤ï©ï ï¯ X
Pr = ï²ï vs Vr Cos(90 + ï¤) / X= (vs Vr Sinï¤ï©ï ï¯ï x (1)
For ï¤ï << 1 Radian, Sinï¤ ï¤ï®hence, In General, The Line Flows Become
Pik=(vi Vk ï¤ikï©ï ï¯ï xik (2)
Where
ï ï ï ï ï ï¤ikï ï½ï ï¤i ï²ï¤k (3)
Since All The Bus Voltages Of A Power System Are Around 1 Pu, Then Let
Vi = Vk =1 Pu
Bik =ï²1 / Xik (4)
Then Equation (3) Becomes
Pik = ï²bik ï¤ik = (ï¤i ï²ï¤k) / Xik (5)
Now, The Bus Power At Any Bus Is The Sum Of The Power Flows In The Lines Connected To That Bus. .hence ,
Pi = Pik = (ï²bik ï¤ik) I = 1, 2, ...., N (6)
Or In Matrix Form ,
P1 B11 B12 ... B1n ï¤1
P2 B21 B22 ... B2n ï¤ï²
. = . . ... . . (7)
. . . ... . .
Pn Bn1 Bn2 ... B Nn ï ï¤ N
Which Can Be Abbreviated As
[p] = [ ï¤ï ] (8)
Where
Bik = ï²1 / Xik (9)
And
Bii = (ï²bik) (10)
The Matrix Is The Imaginary Part Of Ybus . The Solution For [ï¤ïï is
[ï¤] = â1 [p] (11)
Matrix Is An (n â1) ï³ï (nï²1) Matrix Dimensionally For An N-bus System. The Diagonal And Off â Diagonal Elements Of The Matrix Can Be Found By Adding The Series Susceptances Of The Branches Connected To Bus I And By Setting Them Equal To The Negated Series Susceptance Of Branch Ik , Respectively.
Until Now , We Have Kept The System Ground As The Reference Bus. However Since We Have Dropped All Shunt Brunches In Simplifying Things , We Have Lost Our Reference. This Means That The Matrix Of Equation (7) Is Obtained By Equation (9) And (10) Will Be A Singular Matrix. Hence , ï²1 Of Equation (11) Does Not In Fact Exist.
