# Load Flow and Y Bus

on 24/2/2012 & Updated on Tuesday 1st of May 2018 at 12:34:13 PM### Formation of Bus Admittance Matrix (Y_{bus})

S_{1}, S

_{2}, S

_{3}are net complex power injections into bus 1, 2, 3 respectively y

_{12}, y

_{23}, y

_{13}are line admittances between lines 1-2, 2-3, 1-3 y

_{01sh}/2, y

_{02sh}/2, y

_{03sh}/2 are half-line charging admittance between lines 1-2, 1-3 and 2-3

The half-line charging admittances connected to the same bus are at same potential and thus can be combined into one

__Related pages__

If we apply KCL at bus 1, we have
Where, V_{1}, V_{2}, V_{3} are voltage values at bus 1, 2, 3 respectively
Where,

Similarly by applying KCL at buses 2 and 3 we can derive the values of I_{2} and I_{3}
Finally we have
In general for an n bus system

Some observations on Y_{BUS} matrix:

- Y
_{BUS}is a sparse matrix - Diagonal elements are dominating
- Off diagonal elements are symmetric
- The diagonal element of each node is the sum of the admittances connected to it
- The off diagonal element is negated admittance

### Development of Load Flow Equations

The net complex power injection at bus i is given by: Taking conjugate Substituting the value of I_{i}in equation (2) To derive the static load flow equation in polar form in equation (4) substitute On substitution of the above values equation (4) becomes In equation (5) on multiplication of the terms angles get added. Let’s denote for convenience Therefore equation (5) becomes Expansion of equation (6) into sine and cosine terms gives Equating real and imaginary parts we get Equations (7) and (8) are the static load flow equations in polar form. The above obtained equations are non-linear algebraic equations and can be solved using iterative numerical algorithms. Similarly to obtain load flow equations in rectangular form in equation (4) substitute On substituting above values into equation (4) and equating real and imaginary parts we get Equations (9) and (10) are static load flow equations in rectangular form.

Rating = 5 & Total votes = 2