are shown by solid line (with an arrow showing the direction) and by dotted line

Power injected by generator and voltage both are specified

equation that links the complex power with the voltage and current at the generator busbar \[S_{G}=P_{G}+jQ_{G}=V_{G}I^{*}\]

the current through the line can be calculated as \[I=\dfrac{P_{G}-jQ_{G}}{V_{G}^{*}}\]

Then, voltage at the load busbar \[\begin{aligned} V_{L} & =V_{G}-\left(R+jX\right)I\\ & =V_{G}-\left(R+jX\right)\left[\dfrac{P_{G}-jQ_{G}}{V_{G}^{*}}\right] \end{aligned}\]

By defining, \(V_{G}=V_{G}\angle\delta \Longrightarrow V_{G}^{*}=V_{G}\angle-\delta\)

Therefore: \[V_{L}=V_{G}\angle\delta-\left[\dfrac{RP_{G}+XQ_{G}}{V_{G}\angle-\delta}\right]-j\left[\dfrac{XP_{G}-RQ_{G}}{V_{G}\angle-\delta}\right]\]

Generator busbar volatge and power at the load busbar is known

The complex load power is

\[S_{L}=P_{L}+jQ_{L}=V_{L}I^{*}\]

Although the load power is known but load voltage is unknown, hence current cannot be determined

The generator power is \[S_{G}=P_{G}+jQ_{G}=V_{G}I^{*}\]

Again the current cannot be calculated, as the power at the generator is unknown

The load voltage does not have a closed form solution \[V_{L}=V_{G}-\left(R+jX\right)\left[\dfrac{P_{L}-jQ_{L}}{V_{L}^{*}}\right]\]

equation in relation to \(V_L\) is non-linear as it contains the product of the voltages at the load \[V_{L}V_{L}^{*}=V_{G}V_{L}^{*}-\left(R+jX\right)\left(P_{L}-jQ_{L}\right)\]

Solving the equation requires an iterative method

The solution can start with an initial value of \(V_L^{(0)}\), then find \(V_L^{*(0)}\)

A new value of \(V_L^{(1)}\) is calculated from the equation

The process is repeated for several iterations until the voltage of one iteration converges to the next iteration

Once \(V_L\) is obtained \(I\) can be calculated

Active losses = \(I^{2}R,\)

The active power output of the generator \(\Longrightarrow P_G = P_L + I^2R\)Reactive losses = \(I^2X\)

The reactive power production of the generator \(\Longrightarrow Q_G = Q_L +I^2X\)

Let the voltage at the \(i^{th}\) bus be denoted by \[V_{i}=\left|V_{i}\right|\angle\delta_{i}=\left|V_{i}\right|\left(cos\delta_{i}+jsin\delta_{i}\right)\]

Also let us define the self admittance at bus-\(i\) as \[Y_{ii}=\left|Y_{ii}\right|\angle\theta_{ii}=\left|Y_{ii}\right|\left(cos\theta_{ii}+jsin\theta_{ii}\right)=G_{ii}+jB_{ii}\]

Similarly the mutual admittance between the buses \(i\) and \(j\) can be written as \[Y_{ij}=\left|Y_{ij}\right|\angle\theta_{ij}=\left|Y_{ij}\right|\left(cos\theta_{ij}+jsin\theta_{ij}\right)=G_{ij}+jB_{ij}\]

Let the power system contains a total number of \(n\) buses. The current injected at bus-\(i\) is given as \[I_{i}=Y_{i1}V_{1}+Y_{i2}V_{2}+\cdots+Y_{in}V_{n}={\displaystyle \sum_{k=1}^{n}Y_{ik}V_{k}}\]

Note: assume current entering a bus to be \(+ve\) and that leaving the bus to be \(-ve\).

As a consequence the active power and reactive power entering a bus will also be assumed to be \(+ve\).

The complex power at bus-\(i\) is then given by

\[ \begin{aligned} P_{i}-jQ_{i} & =V_{i}^{*}I_{i}=V_{i}^{*}{\displaystyle \sum_{k=1}^{n}Y_{ik}V_{k}}\\ & =\left|V_{i}\right|\left(cos\delta_{i}-jsin\delta_{i}\right){\displaystyle \sum_{k=1}^{n}\left|Y_{ik}V_{k}\right|\left(cos\theta_{ik}+jsin\theta_{ik}\right)\left(cos\delta_{k}+jsin\delta_{k}\right)}\\ & ={\displaystyle \sum_{k=1}^{n}\left|Y_{ik}V_{i}V_{k}\right|\left(cos\delta_{i}-jsin\delta_{i}\right)\left(cos\theta_{ik}+jsin\theta_{ik}\right)\left(cos\delta_{k}+jsin\delta_{k}\right)} \end{aligned} \]

Note: \[\begin{aligned} & \left(cos\delta_{i}-jsin\delta_{i}\right)\left(cos\theta_{ik}+jsin\theta_{ik}\right)\left(cos\delta_{k}+jsin\delta_{k}\right)\\ = & \left(cos\delta_{i}-jsin\delta_{i}\right)\left[cos\left(\theta_{ik}+\delta_{k}\right)+jsin\left(\theta_{ik}+\delta_{k}\right)\right]\\ = & cos\left(\theta_{ik}+\delta_{k}-\delta_{i}\right)+jsin\left(\theta_{ik}+\delta_{k}-\delta_{i}\right) \end{aligned}\]

Finally, the real and reactive power can be expressed as

equation* \[\begin{aligned} P_{i} & ={\displaystyle \sum_{k=1}^{n}\left|Y_{ik}V_{i}V_{k}\right|cos\left(\theta_{ik}+\delta_{k}-\delta_{i}\right)}\\ Q_{i} & =-{\displaystyle \sum_{k=1}^{n}\left|Y_{ik}V_{i}V_{k}\right|sin\left(\theta_{ik}+\delta_{k}-\delta_{i}\right)} \end{aligned}\]

Let \[\begin{array}{cc} P_{Gi}: & \mbox{real power generated at bus-}i\\ Q_{Gi}: & \mbox{reactive power generated at bus-}i\\ P_{Li}: & \mbox{real power consumed at the }i^{th}~\mbox{bus}\\ Q_{Li}: & \mbox{reactive power consumed at the }i^{th}~\mbox{bus} \end{array}\]

Then the net real power injected in bus-\(i\) is \[P_{i,inj}=P_{Gi}-P_{Li}\]

Let the injected power calculated by the load flow program be \(P_{i,calc}\) and \(Q_{i,calc}\). Then the mismatch between the actual injected and calculated values is given by

equation* \[\begin{aligned} \varDelta P_{i} & =P_{i,inj}-P_{i,calc}=P_{Gi}-P_{Li}-P_{i,calc}\\ \varDelta Q_{i} & =Q_{i,inj}-Q_{i,cal}=Q_{Gi}-Q_{Li}-Q_{i,calc} \end{aligned}\]

The purpose of the load flow is to minimize the above two mismatches.

Real and reactive power can be calculated from the derived equations.

Since the magnitudes of all the voltages and their angles are not known a priori, an iterative procedure must be used to estimate the bus voltages and their angles in order to calculate the mismatches.

It is expected that mismatches \(\varDelta P_{i}\) and \(\varDelta Q_{i}\) reduce with each iteration and the load flow is said to have converged when the mismatches of all the buses become less than a very small number.

Consider a system with

2 generator and 3 load buses

bus-1 as the slack bus while taking bus-5 as the P-V bus

Buses 2, 3 and 4 are P-Q buses

Line \(Z\) & line charging \(Y\) is given and the \(Y_{bus}\) is evaluated.

\(|V|, \delta, P_G, Q_G, P_L, Q_L\) at each bus are given

some \(|V|\) and \(\delta\) are given in boldface letters indicating that these are initial data used for starting the load flow program

we do not need to calculate the quantities shown by dash (–), hence their initial estimates are not given

Also note that the slack bus does not contain any load while the P-V bus 5 has a local load and this is indicated in the load column

The basic power flow equations are nonlinear

In an \(n\)-bus power system: , \(n_p\): number of \(P-Q\) bus and \(n_q\): number of \(P-V\) bus

**Unnown quantities:**: \(2n_p\) (\(V\) and \(\delta\) of the P-Q buses) + \(n_g\) (\(\delta\) of the P-V buses), total quantities to be determined**Known quantities:**: \(2n_p\) (real and reactive power of \(P-Q\) bus) + \(2n_g\) (real power and voltage magnitude of \(P-V\) bus) + voltage magnitude and angle of slack busAlthough sufficient numbers of known quantities are available, but closed form equations cannot be formed, hence have to resort iterative methods to obtain the load flow solutions

At the beginning a set of values for the unknown quantities are chosen

These are then updated at each iteration

The process continues till errors between all the known and actual quantities reduce below a pre-specified value