Complete Description of the Mathematical Formulae

Load Flow Problem

An alternating current in an electrical network satisfies Ohm's law with complex values. Thus, the complex values (X, V, R) are used in our program.

Ohm's Law for this simple 2-bus system is:

V - V0 = Z I

Where V is the voltage at the load bus, V0 is the voltage at the swing bus (Modulus voltage V₀, phase angle δ=0), and I is the complex current.

The simple solution for calculating the bus voltage is:

V = V0 + Z I

Since V0 and Z are known, if the current I consumed at the load bus is provided, the complex value of V is easily computed (The value of Z=0 is a short circuit and we are excluding it.)

In general, I is not known (it's not inexpensive to measure this in medium and high voltage systems), which could reduce the load flow problem to a linear one easily solved by matrix inversion. Loads are only known as complex power values; that is as P (real power) and Q (Reactive power).

The complex power S is given by the formula:

S = P + j Q

The relationship among these formulae is:

S = V I*

where I* is the complex conjugate of I

Therefore, Ohm's Law becomes:

V = V0 + Z S* / V*

where S* and V* are complex conjugates of S and V

The solution for V0≠1 is similar to the case where V0=1.

We have already defined the phase angle of the swing bus as δ =0, as the load flow calculations depend only on the phase angle difference between the two buses.

To illustrate this, we define the term H as:

H = Z S* = (R + jX)(P- jQ) = (RP + XQ) + j(XP-RQ)

Also,

VV* = V*(V0+ZS*/V*) = V*V0 + ZS*V*/V*

Since V0 = 1, then:

VV*=V*+ZS*

As ZS*=H,

VV* = V* + H

Substituting the complex values V = V_R + jV_I, where V_R = the real component of V and V_I = the imaginary component of V gives us:

VV* = (V_R + jV_I)(V_R - jV_I) = V_R² + V_I²

Also, as seen before, VV* = V* + H. Since V* = V_R - jV_I and H = H_R + jH_I, then:

V_R² + V_I² = V_R - jV_I + H_R + jH_I

Or:

V_R² + V_I² - V_R - H_R + j(V_I - H_I) = 0 + j0

Separating the real and imaginary parts gives us:

V_R² + V_I² - V_R - H_R = 0 and j(V_I - H_I) = j0

Therefore we have the imaginary part:

V_I = H_I

Substitution in the real part give us:

V_R² + H_I² = V_R + H_R

Therefore:

V_R² - V_R - (H_R - H_I²) = 0

And the real power part is:

V_R = ½ ±

Now we can conclude that:

• If H_R - H_I² < -1/4 (there is no physical solution.)

• If H_R - H_I² > -1/4 (the solution is double: one physical and one spurious.)

• If H_R - H_I² = -1/4 (both solutions coalesce and there is voltage collapse.)