18.5. Power flow study

Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

18.4.7. Systematic fault analysis using Z_bus matrix

For large networks, this method is applicable. Consider the network shown in Figure 18.32. Assume a three-phase fault at bus k has occurred through fault impedance Z_f.

Prefault-bus voltages are obtained from load-flow solution and are a column vector.

$V_{bus} (0) = [\begin{array}{c} V_{1} (0) \\ ⋮ \\ V_{k} (0) \\ ⋮ \\ V_{n} (0) \end{array}]$ $V_{bus} (0) = [\begin{array}{c} V_{1} (0) \\ ⋮ \\ V_{k} (0) \\ ⋮ \\ V_{n} (0) \end{array}]$

Short circuit currents are larger than steady-state values and hence can be neglected. Represent the bus load by a constant impedance evaluated at the prefault bus voltage, that is,

$Z_{i L} = \frac{{|V_{i} (0)|}^{2}}{S_{L}^{2}} .$ $Z_{i L} = \frac{{|V_{i} (0)|}^{2}}{S_{L}^{2}} .$

The change in network voltage caused by fault in impedance Z_f, is equivalent to those caused by the added voltage V_k(0) with all other sources short circuited. The bus voltage changes caused by the fault in this circuit are represented by:

$\underline{∆} V_{bus} (0) = [\begin{array}{c} \underline{∆} V_{1} (0) \\ ⋮ \\ \underline{∆} V_{k} (0) \\ ⋮ \\ \underline{∆} V_{n} (0) \end{array}] .$ $\underline{∆} V_{bus} (0) = [\begin{array}{c} \underline{∆} V_{1} (0) \\ ⋮ \\ \underline{∆} V_{k} (0) \\ ⋮ \\ \underline{∆} V_{n} (0) \end{array}] .$

Using Thévenin’s theorem, bus voltages during the fault are obtained by superposition of the prefault bus voltages and changes in bus voltages are given by,

$V_{bus} (F) = V_{bus} (0) + \underline{∆} V_{bus}$ $V_{bus} (F) = V_{bus} (0) + \underline{∆} V_{bus}$

It is known that I_bus = Y_busV_bus.

Current entering every bus is zero except at the faulty bus. As the current leaves the faulty bus, it is taken as negative current entering bus k:

$\begin{array}{l} [\begin{array}{c} 0 \\ ⋮ \\ - I_{k} (F) \\ ⋮ \\ 0 \end{array}] = [\begin{array}{c} Y_{11} & \dots & Y_{1 k} & Y_{1 n} \\ ⋮ & ⋮ & ⋮ \\ Y_{k 1} & \dots & Y_{k k} & \dots & Y_{k n} \\ ⋮ & ⋮ & ⋮ \\ Y_{n 1} & \dots & Y_{n 1} & \dots & Y_{n n} \end{array}] [\begin{array}{c} \underline{∆} V_{1} \\ ⋮ \\ \underline{∆} V_{k} \\ ⋮ \\ \underline{∆} V_{n} \end{array}] \\ I_{bus} (F) = Y_{bus} ∆ V_{bus} . \end{array}$ $\begin{array}{l} [\begin{array}{c} 0 \\ ⋮ \\ - I_{k} (F) \\ ⋮ \\ 0 \end{array}] = [\begin{array}{c} Y_{11} & \dots & Y_{1 k} & Y_{1 n} \\ ⋮ & ⋮ & ⋮ \\ Y_{k 1} & \dots & Y_{k k} & \dots & Y_{k n} \\ ⋮ & ⋮ & ⋮ \\ Y_{n 1} & \dots & Y_{n 1} & \dots & Y_{n n} \end{array}] [\begin{array}{c} \underline{∆} V_{1} \\ ⋮ \\ \underline{∆} V_{k} \\ ⋮ \\ \underline{∆} V_{n} \end{array}] \\ I_{bus} (F) = Y_{bus} ∆ V_{bus} . \end{array}$

Solving for ∆V_bus,

$\begin{array}{l} ∆ V_{bus} = Z_{bus} I_{bus} (F) \\ therefore, V_{bus} (F) = V_{bus} (0) + Z_{bus} I_{bus} (F) \\ for fault in bus k, the bus voltage, V_{k} (F) = V_{k} (0) - Z_{k k} I_{k} (F) and fault current is \\ I_{k} (F) = \frac{V_{k} (0)}{Z_{k k} + Z_{f}} . \end{array}$ $\begin{array}{l} ∆ V_{bus} = Z_{bus} I_{bus} (F) \\ therefore, V_{bus} (F) = V_{bus} (0) + Z_{bus} I_{bus} (F) \\ for fault in bus k, the bus voltage, V_{k} (F) = V_{k} (0) - Z_{k k} I_{k} (F) and fault current is \\ I_{k} (F) = \frac{V_{k} (0)}{Z_{k k} + Z_{f}} . \end{array}$

18.5. Power flow study

The load flow analysis is required to be performed to decide upon addition/removal of a substation, distribution transformers, tap changers, reactive power controlling devices, etc. From the conservation of energy, real power supplied by the source is equal to the sum of real powers absorbed by the load and real losses in the system. Reactive power must also be balanced between the sum of leading and sum of lagging reactive power-producing elements. The total complex power delivered to the loads in parallel is the sum of the complex powers delivered to each,

$\begin{array}{l} 0 = \sum P_{gen} - \sum P_{loads} - \sum P_{losses} \\ 0 = \sum Q_{leading} + \sum Q_{caps} - \sum Q_{lagging} - \sum Q_{ind} \\ 0 = \sum S_{gen} - \sum S_{loads} - \sum S_{losses} . \end{array}$ $\begin{array}{l} 0 = \sum P_{gen} - \sum P_{loads} - \sum P_{losses} \\ 0 = \sum Q_{leading} + \sum Q_{caps} - \sum Q_{lagging} - \sum Q_{ind} \\ 0 = \sum S_{gen} - \sum S_{loads} - \sum S_{losses} . \end{array}$

Complex power injected into the ith bus is S_i = P_i + jQ_i = V_iI_i^* and i = 1, 2, …, n where, V_i is the voltage at the ith bus and I_i is the source current injected into the bus,

Hence

S_{i} = P_{i} - j Q_{i} = V_{i}^{*} I_{i} = V_{i}^{*} \sum_{k = 1}^{n} Y_{i k} V_{k} .

$S_{i} = P_{i} - j Q_{i} = V_{i}^{*} I_{i} = V_{i}^{*} \sum_{k = 1}^{n} Y_{i k} V_{k} .$

Therefore,

P_{i} = real [V_{i}^{*} \sum_{k = 1}^{n} Y_{i k} V_{k}] .

$P_{i} = real [V_{i}^{*} \sum_{k = 1}^{n} Y_{i k} V_{k}] .$

$Q_{i} = - i_{m} [V_{i}^{*} \sum_{k = 1}^{n} Y_{i k} V_{k}]$ $Q_{i} = - i_{m} [V_{i}^{*} \sum_{k = 1}^{n} Y_{i k} V_{k}]$

In polar form

V_{i} = |V_{i}| e^{j δ i} and Y_{i} = |Y_{i k}| e^{j q_{i k}}

$V_{i} = |V_{i}| e^{j δ i} and Y_{i} = |Y_{i k}| e^{j q_{i k}}$

$\begin{array}{l} P_{i} = |V_{i}| \sum_{k = 1}^{n} |Y_{i k}| |V_{k}| \cos (θ_{i k} - δ_{i} + δ_{k}) \\ Q_{i} = - |V_{i}| \sum_{k = 1}^{n} |Y_{i k}| |V_{k}| \sin (θ_{i k} - δ_{i} + δ_{k}) . \end{array}$ $\begin{array}{l} P_{i} = |V_{i}| \sum_{k = 1}^{n} |Y_{i k}| |V_{k}| \cos (θ_{i k} - δ_{i} + δ_{k}) \\ Q_{i} = - |V_{i}| \sum_{k = 1}^{n} |Y_{i k}| |V_{k}| \sin (θ_{i k} - δ_{i} + δ_{k}) . \end{array}$

The P and Q equations are known as static load flow equations. To solve the static load flow equations, no explicit solutions are possible with the variables: P_i, Q_i, |V_i|, δ_I, as it involves trigonometry and only iterative solutions.

18.5.1. Load flow equations and methods of solution

Load flow solution is a solution of the network under steady-state condition subject to certain inequality constraints, under which the system operates. These constraints can be in the form of load nodal voltages, reactive power generation of the generators, the tap settings of a tap changing under load transformer, etc. To solve the load flow equation, the buses are classified as in Table 18.1:

$0 = \sum P_{gen} - \sum P_{loads} - \sum P_{losses}$ $0 = \sum P_{gen} - \sum P_{loads} - \sum P_{losses}$

Table 18.1

Bus classification

Bus type	Quantities specified	To be obtained
Load bus	P, Q	\|V\|, δ
Generator bus or voltage controlled bus	P, \|V\|	Q, δ
Slack bus/swing or reference bus	\|V\|, δ	P, Q

But loss remains unknown until the load flow equation is solved. Hence, one of the generator buses is made to take up additional real and reactive power to supply transmission loss and is named as swing bus. It is known that I_bus = [Y_bus]V_bus, where V and Y are obtained from the prefault-bus voltage and bus-admittance matrix, as discussed in the previous topics.

where

and

Y_{i_{p}} = |Y_{i_{p}}| γ_{i_{p}}

$Y_{i_{p}} = |Y_{i_{p}}| γ_{i_{p}}$

Complex power injected into the ith bus is,

The static load-flow equation can be rewritten as,

$\begin{array}{l} Real part = P_{i} = |V_{i}| \sum_{p = 1}^{n} |V_{p}| |Y_{i_{p}}| \cos (δ_{i} - γ_{i_{p}} - δ_{p}) \\ Imaginary part = Q_{i} = |V_{i}| \sum_{p = 1}^{n} |V_{p}| |Y_{i_{p}}| \sin (δ_{i} - γ_{i_{p}} - δ_{p}) . \end{array}$ $\begin{array}{l} Real part = P_{i} = |V_{i}| \sum_{p = 1}^{n} |V_{p}| |Y_{i_{p}}| \cos (δ_{i} - γ_{i_{p}} - δ_{p}) \\ Imaginary part = Q_{i} = |V_{i}| \sum_{p = 1}^{n} |V_{p}| |Y_{i_{p}}| \sin (δ_{i} - γ_{i_{p}} - δ_{p}) . \end{array}$

18.5.2. Solution of load flow equation

$\begin{array}{l} X_{i}^{k} = f_{i} (X_{1}^{k}, X_{2}^{k}, \dots X_{i - 1}^{k}, X_{i}^{k - 1}, X_{i + 1}^{k - 1}, X_{n}^{k - 1}) \\ ∆ X_{i} = X_{i}^{k} - X_{i}^{k - 1} \\ X_{i}^{k} = X_{i}^{k - 1} + a ∆ X_{i} . \end{array}$ $\begin{array}{l} X_{i}^{k} = f_{i} (X_{1}^{k}, X_{2}^{k}, \dots X_{i - 1}^{k}, X_{i}^{k - 1}, X_{i + 1}^{k - 1}, X_{n}^{k - 1}) \\ ∆ X_{i} = X_{i}^{k} - X_{i}^{k - 1} \\ X_{i}^{k} = X_{i}^{k - 1} + a ∆ X_{i} . \end{array}$

Its advantages are,

1. Simplicity of the technique.

2. Small computer memory requirement.

3. Less computation time per iteration.

Its disadvantages are,

1. Slow rate of convergence, therefore large number of iterations.

2. Increase in number of iterations directly with increase in number of buses.

3. Effect on convergence due to choice of slack bus.

Hence, Gauss–Seidel (G–S) method is used only for systems having a small number of buses.

18.5.2.1. Using Gauss-Seidel (G-S) method when PV buses are absent

Out of n buses, 1 is the slack bus, so n – 1 PQ buses.

$\begin{array}{l} I_{i} = Y_{i_{i}} V_{i} \sum_{p = 1}^{n} Y_{i_{p}} V_{p} \\ V_{i} = \frac{1}{Y_{i_{i}}} [I_{i} - \sum_{p = 1}^{n} Y_{i_{p}} V_{p}] . \end{array}$ $\begin{array}{l} I_{i} = Y_{i_{i}} V_{i} \sum_{p = 1}^{n} Y_{i_{p}} V_{p} \\ V_{i} = \frac{1}{Y_{i_{i}}} [I_{i} - \sum_{p = 1}^{n} Y_{i_{p}} V_{p}] . \end{array}$

Similarly,

$\begin{array}{l} S_{i}^{*} = P_{i} - j Q_{i} = V_{i}^{*} I_{i} \\ I_{i} = \frac{P_{i} - j Q_{i}}{V_{i}^{*}} \\ V_{i} = \frac{1}{Y_{i_{i}}} [\frac{P_{i} - j Q_{i}}{V_{i}^{*}} - \sum_{p = 1}^{n} Y_{i_{p}} V_{p}] i = 2,3, \dots n, since 1 is a slack bus. \end{array}$ $\begin{array}{l} S_{i}^{*} = P_{i} - j Q_{i} = V_{i}^{*} I_{i} \\ I_{i} = \frac{P_{i} - j Q_{i}}{V_{i}^{*}} \\ V_{i} = \frac{1}{Y_{i_{i}}} [\frac{P_{i} - j Q_{i}}{V_{i}^{*}} - \sum_{p = 1}^{n} Y_{i_{p}} V_{p}] i = 2,3, \dots n, since 1 is a slack bus. \end{array}$

Solve for V_2, V₃, …, V_n,

$\begin{array}{l} K_{i} = \frac{P_{i} - j Q_{i}}{Y_{i_{i}}} L_{i_{p}} = \frac{Y_{i_{p}}}{Y_{i_{i}}} i = 2,3, \dots n p = 1,2, \dots n p \neq n \\ V_{i}^{k + 1} = \frac{K_{i}}{{(V_{i}^{k})}^{*}} - \sum_{p = 1}^{n} L_{i_{p}} V_{p}^{k + 1} - \sum_{p = i + 1}^{n} L_{i_{p}} V_{p}^{k} i = 2,3, \dots n . \end{array}$ $\begin{array}{l} K_{i} = \frac{P_{i} - j Q_{i}}{Y_{i_{i}}} L_{i_{p}} = \frac{Y_{i_{p}}}{Y_{i_{i}}} i = 2,3, \dots n p = 1,2, \dots n p \neq n \\ V_{i}^{k + 1} = \frac{K_{i}}{{(V_{i}^{k})}^{*}} - \sum_{p = 1}^{n} L_{i_{p}} V_{p}^{k + 1} - \sum_{p = i + 1}^{n} L_{i_{p}} V_{p}^{k} i = 2,3, \dots n . \end{array}$

Problem 18.13

The following is the system data for a load flow solutions. The line admittances are,

Bus code	Admittance
1–2	2–j8.0
1–3	1–j4.0
2–3	0.666–j2.664
2–4	1–j4.0
3–4	2–j8.0

The schedule of active and reactive powers,

Bus code	P	Q	V	Remarks
1	–	–	1.06	Slack
2	0.5	0.2	1 + j0	PQ
3	0.4	0.3	1 + j0	PQ
4	0.3	0.1	1 + j0	PQ

Determine the voltages at the end of first iteration using G–S method. Take

\propto = 1.6

$\propto = 1.6$

$Y_{bus} = [\begin{array}{c} 3 - j 12 & - 2 + j 8 & - 1 + j 4 & 0 \\ - 2 + j 8 & 3 .666 - j 4 .664 & - 0 .666 + j 2 .664 & - 1 + j 4 \\ - 1 + j 4 & - 0 .666 + j 2 .664 & 3 .666 - j 14 .664 & - 2 + j 8 \\ 0 & - 1 + j 4 & - 2 + j 8 & 3 - j 12 \end{array}] .$ $Y_{bus} = [\begin{array}{c} 3 - j 12 & - 2 + j 8 & - 1 + j 4 & 0 \\ - 2 + j 8 & 3 .666 - j 4 .664 & - 0 .666 + j 2 .664 & - 1 + j 4 \\ - 1 + j 4 & - 0 .666 + j 2 .664 & 3 .666 - j 14 .664 & - 2 + j 8 \\ 0 & - 1 + j 4 & - 2 + j 8 & 3 - j 12 \end{array}] .$

Power for load buses is negative, generation buses is positive,

$\begin{array}{l} V_{2}^{1} & = \frac{1}{Y_{22}} [\frac{P_{i} - J Q_{i}}{Y_{i i}} - Y_{21} V_{1}^{0} - Y_{23} V_{3}^{0} - Y_{24} V_{4}^{0}] \\ = \frac{1}{3.666 - j 14.664} [\frac{- 0.5 + j 0.2}{1 - j 0} - 1.06 (- 2 + j 8) - 1 (0.666 - j 2.664) - 1 (- 1 + j 4)] \\ = 1.01187 - j 0.02888 \\ V'_{2 acc} & = (1 + j 0) + 1.6 (1.01187 - j 0.02888 - 1 - j 0) \\ V'_{3} & = \frac{1}{Y_{33}} [\frac{P_{3} - J Q_{3}}{V_{3}^{*}} - Y_{31} V_{1} - Y_{32} V'_{2} - Y_{34} V_{4}^{0}] \\ = \frac{1}{3.666 - j 14.664} [\begin{array}{l} \frac{- 0.4 + j 0.3}{1 - j 0} - (- 1 + j 4) 1.06 - (- 0.666 + j 2.664) \\ (1.01899 - j 0.046208) - (- 2 + j 8) (1 + j 0) \end{array}] \\ = 0.994119 - j 0.029248 \\ V'_{3 acc} & = 0.99059 - j 0.0467968 \\ V'_{4} & = \frac{1}{Y_{44}} [\frac{P_{4} - J Q_{4}}{V_{4}^{*}} - Y_{41} V_{1} - Y_{42} V'_{2} - Y_{43} V'_{3}] \\ = 0.9716032 - j 0.064684 \\ V'_{4 acc} & = 0.954565 - j 0.1034944 \end{array}$ $\begin{array}{l} V_{2}^{1} & = \frac{1}{Y_{22}} [\frac{P_{i} - J Q_{i}}{Y_{i i}} - Y_{21} V_{1}^{0} - Y_{23} V_{3}^{0} - Y_{24} V_{4}^{0}] \\ = \frac{1}{3.666 - j 14.664} [\frac{- 0.5 + j 0.2}{1 - j 0} - 1.06 (- 2 + j 8) - 1 (0.666 - j 2.664) - 1 (- 1 + j 4)] \\ = 1.01187 - j 0.02888 \\ V'_{2 acc} & = (1 + j 0) + 1.6 (1.01187 - j 0.02888 - 1 - j 0) \\ V'_{3} & = \frac{1}{Y_{33}} [\frac{P_{3} - J Q_{3}}{V_{3}^{*}} - Y_{31} V_{1} - Y_{32} V'_{2} - Y_{34} V_{4}^{0}] \\ = \frac{1}{3.666 - j 14.664} [\begin{array}{l} \frac{- 0.4 + j 0.3}{1 - j 0} - (- 1 + j 4) 1.06 - (- 0.666 + j 2.664) \\ (1.01899 - j 0.046208) - (- 2 + j 8) (1 + j 0) \end{array}] \\ = 0.994119 - j 0.029248 \\ V'_{3 acc} & = 0.99059 - j 0.0467968 \\ V'_{4} & = \frac{1}{Y_{44}} [\frac{P_{4} - J Q_{4}}{V_{4}^{*}} - Y_{41} V_{1} - Y_{42} V'_{2} - Y_{43} V'_{3}] \\ = 0.9716032 - j 0.064684 \\ V'_{4 acc} & = 0.954565 - j 0.1034944 \end{array}$

18.5.2.2. Modification of G–S method when PV buses are present

Now, i = 1 slack bus, i = 2, 3, …, m PV bus, and i = m + 1, …., n PQ bus.

Conditions to be met by PV buses are:

1. |V_i| = |V_i|_specified for i = 2, 3,…, m

2. Q_i,min < Q_i < Q_i,max for i = 2, 3,…, m.

The second requirement can be violated if the specified bus voltage |V_i|_specified is either too high or too low. It is possible to control |V_i| only by controlling Q_i. Hence, if Q constraint is violated, treat it as a PQ bus with Q equal to maximum or minimum values.

Steps:

1. Calculate

Q_{i} = |V_{i}| \sum_{p = 1}^{n} |V_{p}| |Y_{i p}| \sin (δ_{i} - γ_{i p} - δ_{p})

$Q_{i} = |V_{i}| \sum_{p = 1}^{n} |V_{p}| |Y_{i p}| \sin (δ_{i} - γ_{i p} - δ_{p})$

2. For every iteration |V_i| must be set equal to |V_i|_specified.

Q_{i}^{k + 1} = {|V_{i}|}_{specified} \sum_{p = 1}^{i = 1} |V_{p}^{k + 1}| |Y_{i p}| \sin (δ_{i}^{k} - γ_{i p} - δ_{p}^{k + 1}) + {|V_{i}|}_{specified} \sum_{p = i}^{n} |V_{p}^{k}| |Y_{i p}| \sin (δ_{i}^{k} - γ_{i p} - δ_{p}^{k})

$Q_{i}^{k + 1} = {|V_{i}|}_{specified} \sum_{p = 1}^{i = 1} |V_{p}^{k + 1}| |Y_{i p}| \sin (δ_{i}^{k} - γ_{i p} - δ_{p}^{k + 1}) + {|V_{i}|}_{specified} \sum_{p = i}^{n} |V_{p}^{k}| |Y_{i p}| \sin (δ_{i}^{k} - γ_{i p} - δ_{p}^{k})$

4. Check for constraints, if violated, treat as PQ bus.

Problem 18.14

If in Problem 18.13, bus two is taken as a generator bus with |V₂| = 1.04 and reactive power constraint is 0.1 ≤ Q₂ ≤ 1.0, determine the voltages starting with a flat voltage profile and assuming the accelerating factor as 1.

Since bus two is a PV bus, Q is not specified. Hence, to find V₂′, Q₂ must be calculated, with

V₂ = 1.04 and the phase angle of voltage = 0.

$\begin{array}{l} P_{2} - j Q_{2} & = V_{2}^{*} \sum_{q = 1}^{4} Y_{2 q} V_{q} \\ = V_{2}^{*} [Y_{21} V_{1} + Y_{22} V_{2} + Y_{23} V_{3} + Y_{24} V_{4}] \\ Q_{2} & = |1.04| [Y_{21} V_{1} \sin (δ_{1} - γ_{21} - δ_{2}) + Y_{22} V_{2} \sin (δ_{2} - γ_{22} - δ_{2}) + Y_{23} V_{3} \sin (δ_{2} - γ_{23} - δ_{3}) \\ + Y_{24} V_{4} s i n (δ_{2} - γ_{24} - δ_{4})] \\ = 0.1108. \end{array}$ $\begin{array}{l} P_{2} - j Q_{2} & = V_{2}^{*} \sum_{q = 1}^{4} Y_{2 q} V_{q} \\ = V_{2}^{*} [Y_{21} V_{1} + Y_{22} V_{2} + Y_{23} V_{3} + Y_{24} V_{4}] \\ Q_{2} & = |1.04| [Y_{21} V_{1} \sin (δ_{1} - γ_{21} - δ_{2}) + Y_{22} V_{2} \sin (δ_{2} - γ_{22} - δ_{2}) + Y_{23} V_{3} \sin (δ_{2} - γ_{23} - δ_{3}) \\ + Y_{24} V_{4} s i n (δ_{2} - γ_{24} - δ_{4})] \\ = 0.1108. \end{array}$

Since Q₂ lies within limits, V₂ = V_2specified:

$\begin{array}{l} V_{2} = \frac{1}{Y_{22}} [\frac{0.5 - j 0 .1108}{1 .04 - j 0} - Y_{22} V_{1} - Y_{23} V_{3} - Y_{24} V_{4}] \\ V_{2}^{'} = 1 .0472846 + j 0 .0291476, δ = 1 .59 ° \\ V_{2} = 1 .04 = 1 .0395985 + j 0 .02891159 \\ V_{2acc}^{'} = negative \\ V_{3}^{'} = \frac{1}{Y_{33}} [\frac{R_{3} - j Q_{3}}{V_{3}} - Y_{31} V_{1} - Y_{32} V_{2}^{'} - Y_{34} V_{4}] \\ = 0 .9978866 - j 0 .015607057 \\ = 0 .998065 - j 0 .022336. \end{array}$ $\begin{array}{l} V_{2} = \frac{1}{Y_{22}} [\frac{0.5 - j 0 .1108}{1 .04 - j 0} - Y_{22} V_{1} - Y_{23} V_{3} - Y_{24} V_{4}] \\ V_{2}^{'} = 1 .0472846 + j 0 .0291476, δ = 1 .59 ° \\ V_{2} = 1 .04 = 1 .0395985 + j 0 .02891159 \\ V_{2acc}^{'} = negative \\ V_{3}^{'} = \frac{1}{Y_{33}} [\frac{R_{3} - j Q_{3}}{V_{3}} - Y_{31} V_{1} - Y_{32} V_{2}^{'} - Y_{34} V_{4}] \\ = 0 .9978866 - j 0 .015607057 \\ = 0 .998065 - j 0 .022336. \end{array}$

Problem 18.15

For the same problem, if the reactive power constraint on generator two is 0.2 ≤ Q₂ ≤ 1, solve the problem for voltages at the end of the first iteration.

Q₂ = 0.1108. Hence, violated. So assume as PQ bus with P₂ = 0.5, Q₂ = Q_2min = 0.2, and

V_{2}^{°}

$V_{2}^{°}$

= 1 + j0 and as usual procedure, but P₂ + jQ₂ is positive, though it is assumed to be a PQ bus.

$\begin{array}{l} V_{2}^{'} = \frac{1}{Y_{22}} [\frac{0.5 - j 0 .2}{1 - j 0} - Y_{22} V_{1} - Y_{23} V_{3}^{°} - Y_{24} V_{4}^{°}] \\ = 1 .098221 + j 0 .030105. \end{array}$ $\begin{array}{l} V_{2}^{'} = \frac{1}{Y_{22}} [\frac{0.5 - j 0 .2}{1 - j 0} - Y_{22} V_{1} - Y_{23} V_{3}^{°} - Y_{24} V_{4}^{°}] \\ = 1 .098221 + j 0 .030105. \end{array}$

Similarly

V_{3}^{'}

$V_{3}^{'}$

and

V_{4}^{'}

$V_{4}^{'}$

are calculated.

18.5.2.3. Newton–Raphson method

It is suitable for large systems.

Its advantages are:

1. Increased accuracy and surety of convergence.

2. Only about three iterations are required compared to more than 25 or so required by the G–S method.

3. Number of iterations is independent of system size.

4. This method is insensitive to factors like slack bus selection, regulating transformers, etc.

Its disadvantages are:

1. Solution technique is difficult.

2. More calculations are involved, hence more computation time/iteration.

3. Memory requirement is large.

This method can be used with both rectangular and polar coordinates. But the rectangular coordinate requires more number of equations as compared to the polar form. Hence, the polar form is preferred.

N–R method using rectangular coordinates is,

Changes in active and reactive power with bus as slack,

$\begin{array}{l} ∆ P_{i} = \sum_{p = 2}^{n} \frac{\partial p i}{\partial l_{p}} ∆ l_{p} + \sum_{p = 2}^{n} \frac{\partial p i}{\partial l_{p}} ∆ f_{p} \\ ∆ Q_{i} = \sum_{p = 2}^{n} \frac{\partial Q i}{\partial l_{p}} ∆ l_{p} + \sum_{p = 2}^{n} \frac{\partial Q i}{\partial f_{p}} ∆ f_{p} . \end{array}$ $\begin{array}{l} ∆ P_{i} = \sum_{p = 2}^{n} \frac{\partial p i}{\partial l_{p}} ∆ l_{p} + \sum_{p = 2}^{n} \frac{\partial p i}{\partial l_{p}} ∆ f_{p} \\ ∆ Q_{i} = \sum_{p = 2}^{n} \frac{\partial Q i}{\partial l_{p}} ∆ l_{p} + \sum_{p = 2}^{n} \frac{\partial Q i}{\partial f_{p}} ∆ f_{p} . \end{array}$

[\begin{array}{c} ∆ P \\ ∆ Q \end{array}] = [\begin{array}{c} j_{1} & j_{2} \\ j_{3} & j_{4} \end{array}] [\begin{array}{c} ∆ e \\ ∆ f \end{array}]

$[\begin{array}{c} ∆ P \\ ∆ Q \end{array}] = [\begin{array}{c} j_{1} & j_{2} \\ j_{3} & j_{4} \end{array}] [\begin{array}{c} ∆ e \\ ∆ f \end{array}]$

for PV bus |V_i|² = e_i² + f_i².

∆ V_{i}^{2} = \frac{\partial | V_{i} | 2}{\partial ρ i} ∆ ρ i + \frac{\partial | V_{i} | 2}{\partial f i} ∆ f i

$∆ V_{i}^{2} = \frac{\partial | V_{i} | 2}{\partial ρ i} ∆ ρ i + \frac{\partial | V_{i} | 2}{\partial f i} ∆ f i$

with PV bus (instead of Q_i and V_i).

Total number of equations = (n – 1)².

N–R method using polar coordinates is,

$P_{i} = f_{1} (δ, | V |), Q_{i} = f_{2} (δ, | V |) .$ $P_{i} = f_{1} (δ, | V |), Q_{i} = f_{2} (δ, | V |) .$

$∆ P_{i} = \sum_{p = 2}^{n} \frac{\partial p i}{\partial δ_{p}} \cdot ∆ δ_{p} + \sum_{p = 2}^{n} \frac{\partial p i}{\partial | V_{p} |} \cdot ∆ | V_{p} | .$ $∆ P_{i} = \sum_{p = 2}^{n} \frac{\partial p i}{\partial δ_{p}} \cdot ∆ δ_{p} + \sum_{p = 2}^{n} \frac{\partial p i}{\partial | V_{p} |} \cdot ∆ | V_{p} | .$

(18.1)

$∆ Q_{i} = \sum_{p = 2}^{n} \frac{\partial Q i}{\partial δ_{p}} \cdot ∆ δ_{p} + \sum_{p = 2}^{n} \frac{\partial Q i}{\partial | V_{p} |} \cdot ∆ | V_{p} | .$ $∆ Q_{i} = \sum_{p = 2}^{n} \frac{\partial Q i}{\partial δ_{p}} \cdot ∆ δ_{p} + \sum_{p = 2}^{n} \frac{\partial Q i}{\partial | V_{p} |} \cdot ∆ | V_{p} | .$

(18.2)

but for PV bus (18.2) does not exist.

The n bus system has 1 slack bus and g PV buses.

Total number of equations = (2n – 2 – g).

Therefore,

$\frac{\partial P_{i}}{\partial δ_{p}} - \frac{\partial Q_{i}}{\partial δ_{p}} = j (e_{i} - j f_{i}) (G_{i p} + j B_{i p}) (e_{p} + j f_{p})$ $\frac{\partial P_{i}}{\partial δ_{p}} - \frac{\partial Q_{i}}{\partial δ_{p}} = j (e_{i} - j f_{i}) (G_{i p} + j B_{i p}) (e_{p} + j f_{p})$

Replace ∆V_p by,

\frac{∆ |V_{p}|}{|V_{p}|}

$\frac{∆ |V_{p}|}{|V_{p}|}$

$\begin{array}{l} ∆ P_{i} = \sum_{p = 2}^{n} \frac{\partial p i}{\partial δ_{p}} \cdot ∆ δ_{p} + \sum_{p = 2}^{n} \frac{\partial p i}{\partial | V_{p} |} \cdot | V_{p} | \frac{∆ |V_{p}|}{|V_{p}|} \\ ∆ Q_{i} = \sum_{p = 2}^{n} \frac{\partial Q i}{\partial δ_{p}} \cdot ∆ δ_{p} + \sum_{p = 2}^{n} \frac{\partial Q i}{\partial | V_{p} |} \cdot | V_{p} | \frac{∆ |V_{p}|}{|V_{p}|} \\ [\begin{array}{c} ∆ P \\ ∆ Q \end{array}] = [\begin{array}{c} H & N \\ J & L \end{array}] [\begin{array}{c} ∆ δ \\ \frac{∆ |V|}{|V|} \end{array}] \\ H_{i p} = \frac{\partial p i}{\partial δ_{p}} N_{i p} = \frac{\partial p i}{\partial | V_{p} |} | V_{p} | J_{i p} = \frac{\partial Q i}{\partial δ_{p}} L_{i p} = \frac{\partial Q i}{\partial | V_{p} |} | V_{p} | \\ P_{i} - j Q_{i} = V_{i}^{*} \sum_{p = 1}^{n} Y_{i p} V_{p} ... (*) . \end{array}$ $\begin{array}{l} ∆ P_{i} = \sum_{p = 2}^{n} \frac{\partial p i}{\partial δ_{p}} \cdot ∆ δ_{p} + \sum_{p = 2}^{n} \frac{\partial p i}{\partial | V_{p} |} \cdot | V_{p} | \frac{∆ |V_{p}|}{|V_{p}|} \\ ∆ Q_{i} = \sum_{p = 2}^{n} \frac{\partial Q i}{\partial δ_{p}} \cdot ∆ δ_{p} + \sum_{p = 2}^{n} \frac{\partial Q i}{\partial | V_{p} |} \cdot | V_{p} | \frac{∆ |V_{p}|}{|V_{p}|} \\ [\begin{array}{c} ∆ P \\ ∆ Q \end{array}] = [\begin{array}{c} H & N \\ J & L \end{array}] [\begin{array}{c} ∆ δ \\ \frac{∆ |V|}{|V|} \end{array}] \\ H_{i p} = \frac{\partial p i}{\partial δ_{p}} N_{i p} = \frac{\partial p i}{\partial | V_{p} |} | V_{p} | J_{i p} = \frac{\partial Q i}{\partial δ_{p}} L_{i p} = \frac{\partial Q i}{\partial | V_{p} |} | V_{p} | \\ P_{i} - j Q_{i} = V_{i}^{*} \sum_{p = 1}^{n} Y_{i p} V_{p} ... (*) . \end{array}$

Hence, a Jacobian matrix can be formed from Y_bus:

$\begin{array}{l} Y = G + JB V_{i} = e_{i} + j f_{i} \\ a_{p} - j b_{p} = (G_{i p} + j B_{i p}) (e_{p} + j f_{p}) \\ \frac{\partial p_{i}}{\partial δ_{p}} = H_{i p} = a_{p} f_{i} - b_{p} e_{i} J_{i p} = - (a_{p} e_{i} + b_{p} f_{i}) \\ H_{i i} = - Q_{i} - {|V_{i}|}^{2} B_{i i} J_{i i} = P_{i} - {|V_{i}|}^{2} G_{i i} \\ N_{i p} = - J_{i p} and L_{i p} = H_{i p} \\ N_{i i} = P_{i} + {|V_{i}|}^{2} G_{i i} and L_{i i} = Q_{i} - {|V_{i}|}^{2} B_{i i} \\ P_{i} - j Q_{i} = |V_{i}| e^{- j δ_{1}} \sum_{p = 1}^{n} |Y_{i p}| e^{j γ L_{p}} |V_{p}| e^{j δ_{p}} \\ \frac{\partial p_{i}}{\partial δ_{p}} - j \frac{\partial Q_{i}}{\partial δ_{p}} = j (|V_{i}| e^{- j δ_{1}}) (|Y_{i p}| e^{j γ L_{p}}) (|V_{p}| e^{j δ_{p}}) \\ |V_{i}| e^{- j δ_{1}} = e_{i} - j f_{i} |Y_{i p}| e^{j γ L_{p}} = G_{i p} + j B_{i p} |V_{p}| e^{j δ_{p}} = e_{p} + j f_{p} . \end{array}$ $\begin{array}{l} Y = G + JB V_{i} = e_{i} + j f_{i} \\ a_{p} - j b_{p} = (G_{i p} + j B_{i p}) (e_{p} + j f_{p}) \\ \frac{\partial p_{i}}{\partial δ_{p}} = H_{i p} = a_{p} f_{i} - b_{p} e_{i} J_{i p} = - (a_{p} e_{i} + b_{p} f_{i}) \\ H_{i i} = - Q_{i} - {|V_{i}|}^{2} B_{i i} J_{i i} = P_{i} - {|V_{i}|}^{2} G_{i i} \\ N_{i p} = - J_{i p} and L_{i p} = H_{i p} \\ N_{i i} = P_{i} + {|V_{i}|}^{2} G_{i i} and L_{i i} = Q_{i} - {|V_{i}|}^{2} B_{i i} \\ P_{i} - j Q_{i} = |V_{i}| e^{- j δ_{1}} \sum_{p = 1}^{n} |Y_{i p}| e^{j γ L_{p}} |V_{p}| e^{j δ_{p}} \\ \frac{\partial p_{i}}{\partial δ_{p}} - j \frac{\partial Q_{i}}{\partial δ_{p}} = j (|V_{i}| e^{- j δ_{1}}) (|Y_{i p}| e^{j γ L_{p}}) (|V_{p}| e^{j δ_{p}}) \\ |V_{i}| e^{- j δ_{1}} = e_{i} - j f_{i} |Y_{i p}| e^{j γ L_{p}} = G_{i p} + j B_{i p} |V_{p}| e^{j δ_{p}} = e_{p} + j f_{p} . \end{array}$

Problem 18.16

Figure 18.33 shows a six-bus system. Assuming bus one is a slack bus, formulate the Jacobian matrix for this system.

Using the following conventions, the Jacobian matrix is formulated.

If bus i and bus m are both PQ buses, then all the Jacobian components are present.

If bus i is a PQ bus and bus m is a PV bus, then the Jacobian components present are H_im and J_im.

If bus i is a PV bus and bus m is a PQ bus, then the Jacobian components present are H_im and N_im.

If bus i and bus m are both PV buses, then the Jacobian component present is only H_im as ∆|V_m| = 0.

$\begin{array}{c} ∆ P_{2} \\ ∆ Q_{2} \\ ∆ P_{3} \\ ∆ P_{4} \\ ∆ Q_{4} \\ ∆ P_{5} \\ ∆ P_{6} \\ ∆ Q_{6} \end{array} = [\begin{array}{c} H_{22} & N_{22} & H_{23} & H_{24} & H_{24} \\ J_{22} & L_{22} & J_{23} & J_{24} & L_{24} \\ H_{23} & N_{23} & H_{33} & H_{35} & H_{36} & N_{36} \\ H_{42} & N_{42} & H_{44} & N_{44} & N_{45} \\ J_{42} & L_{42} & J_{44} & L_{44} & J_{45} \\ H_{53} & H_{54} & N_{54} & H_{55} & H_{56} & H_{56} \\ H_{63} & H_{65} & H_{66} & N_{66} \\ J_{63} & J_{65} & J_{66} & L_{66} \end{array}]$ $\begin{array}{c} ∆ P_{2} \\ ∆ Q_{2} \\ ∆ P_{3} \\ ∆ P_{4} \\ ∆ Q_{4} \\ ∆ P_{5} \\ ∆ P_{6} \\ ∆ Q_{6} \end{array} = [\begin{array}{c} H_{22} & N_{22} & H_{23} & H_{24} & H_{24} \\ J_{22} & L_{22} & J_{23} & J_{24} & L_{24} \\ H_{23} & N_{23} & H_{33} & H_{35} & H_{36} & N_{36} \\ H_{42} & N_{42} & H_{44} & N_{44} & N_{45} \\ J_{42} & L_{42} & J_{44} & L_{44} & J_{45} \\ H_{53} & H_{54} & N_{54} & H_{55} & H_{56} & H_{56} \\ H_{63} & H_{65} & H_{66} & N_{66} \\ J_{63} & J_{65} & J_{66} & L_{66} \end{array}]$

Problem 18.17

Figure 18.34 shows a five-bus system. Assuming bus one is a slack bus, formulate the Jacobian matrix for this system:

$[\begin{array}{c} H_{22} & N_{22} & H_{23} & H_{24} & H_{24} & 0 \\ J_{22} & L_{22} & J_{23} & J_{24} & L_{24} & 0 \\ H_{32} & N_{32} & H_{33} & 0 & 0 & H_{35} \\ H_{42} & N_{42} & 0 & H_{44} & N_{44} & H_{45} \\ J_{42} & L_{42} & 0 & J_{44} & L_{44} & J_{45} \\ 0 & 0 & H_{53} & H_{54} & N_{54} & H_{55} \end{array}]$ $[\begin{array}{c} H_{22} & N_{22} & H_{23} & H_{24} & H_{24} & 0 \\ J_{22} & L_{22} & J_{23} & J_{24} & L_{24} & 0 \\ H_{32} & N_{32} & H_{33} & 0 & 0 & H_{35} \\ H_{42} & N_{42} & 0 & H_{44} & N_{44} & H_{45} \\ J_{42} & L_{42} & 0 & J_{44} & L_{44} & J_{45} \\ 0 & 0 & H_{53} & H_{54} & N_{54} & H_{55} \end{array}]$

Figure 18.34 Five-bus system for Problem 18.17.

Problem 18.18

Determine the set of load flow equations at the end of first iteration using NR method. The load flow data for the given power system are given next. The voltage magnitude of bus 2 is to be maintained at 1.04 pu. The maximum and minimum reactive power limits of the generator at bus 2 are 0.35 and 0 pu, respectively.

Bus code	Assumed voltages	Generation		Load
Bus code	Assumed voltages	MW	MVAR	MW	MVAR
1	1.06 + j0	0	0	0	0
2	1 + j0	0.2	0	0	0
3	1 + j0	0	0	0.6	0.25

The admittance matrix is given as:

$Y_{bus} = [\begin{array}{c} 6.25 - j 18.75 & - 1.25 + j 3.75 & - 5 + j 15 \\ - 1.25 + j 3.75 & 2.916 - j 8.75 & - 1.666 + j 5 \\ - 5 + j 15 & - 1.666 + j 5 & 6.666 - j 20 \end{array}]$ $Y_{bus} = [\begin{array}{c} 6.25 - j 18.75 & - 1.25 + j 3.75 & - 5 + j 15 \\ - 1.25 + j 3.75 & 2.916 - j 8.75 & - 1.666 + j 5 \\ - 5 + j 15 & - 1.666 + j 5 & 6.666 - j 20 \end{array}]$

Solution

From the admittance matrix, it is observed that:

G₁₁ = 6.25	G₁₂ = −1.25	G₁₃ = −5	G₂₂ = 2.916	G₂₃ = −1.666	G₃₃ = 6.666
B₁₁ = 18.75	B₁₂ = −15	B₁₃ = −15	B₂₂ = 8.75	B₂₃ = −5	B₃₃ = 20

The voltage amplitude and phase angle of the bus are specified as:

e₁ = 1.06

e₂ = 1

e₃ = 1

f₁ = 0

f₂ = 0

f ₃ = 0

From the static load flow equations, active power in bus 2 can be calculated as follows. θ values are taken from the admittance matrix written in polar coordinates form.

$\begin{array}{l} P_{2} = |V_{2}| |V_{1}| |Y_{21}| cos (θ_{21} + δ_{1} - δ_{2}) + {|V_{2}|}^{2} |Y_{22}| cos (θ_{22}) + |V_{2}| |V_{3}| |Y_{23}| \\ cos (θ_{23} + δ_{3} - δ_{2}) = - 0.075 \end{array}$ $\begin{array}{l} P_{2} = |V_{2}| |V_{1}| |Y_{21}| cos (θ_{21} + δ_{1} - δ_{2}) + {|V_{2}|}^{2} |Y_{22}| cos (θ_{22}) + |V_{2}| |V_{3}| |Y_{23}| \\ cos (θ_{23} + δ_{3} - δ_{2}) = - 0.075 \end{array}$

Alternately, the active and reactive power can also be calculated as follows.

$\begin{array}{l} P_{i} - j Q_{i} & = {(e_{i} + j f_{i})}^{*} \sum_{p = 1}^{n} (G_{i p} - j B_{i p}) (e_{p} + j f_{p}) \\ = {(e_{i} - j f_{i})}^{*} \sum_{p = 1}^{n} (G_{i p} - j B_{i p}) (e_{p} + j f_{p}) \end{array}$ $\begin{array}{l} P_{i} - j Q_{i} & = {(e_{i} + j f_{i})}^{*} \sum_{p = 1}^{n} (G_{i p} - j B_{i p}) (e_{p} + j f_{p}) \\ = {(e_{i} - j f_{i})}^{*} \sum_{p = 1}^{n} (G_{i p} - j B_{i p}) (e_{p} + j f_{p}) \end{array}$

$P_{i} = \sum_{p = 1}^{n} e_{i} (e_{p} G_{i p} + f_{p} B_{i p}) + f_{i} (f_{p} G_{i p} - e_{p} B_{i p})$ $P_{i} = \sum_{p = 1}^{n} e_{i} (e_{p} G_{i p} + f_{p} B_{i p}) + f_{i} (f_{p} G_{i p} - e_{p} B_{i p})$

$Q_{i} = \sum_{p = 1}^{n} f_{i} (e_{p} G_{i p} + f_{p} B_{i p}) - e_{i} (f_{p} G_{i p} - e_{p} B_{i p})$ $Q_{i} = \sum_{p = 1}^{n} f_{i} (e_{p} G_{i p} + f_{p} B_{i p}) - e_{i} (f_{p} G_{i p} - e_{p} B_{i p})$

$\begin{array}{l} P_{2} & = e_{2} (e_{1} G_{21} + f_{1} B_{21}) + f_{2} (f_{1} G_{21} - e_{1} B_{21}) + e_{2} (e_{2} G_{22} + f_{2} B_{22}) + f_{2} (f_{2} G_{22} - e_{2} B_{22}) \\ + e_{2} (e_{3} G_{23} + f_{3} B_{23}) + f_{2} (f_{3} G_{23} - e_{3} B_{23}) \\ = - 0.075 \end{array}$ $\begin{array}{l} P_{2} & = e_{2} (e_{1} G_{21} + f_{1} B_{21}) + f_{2} (f_{1} G_{21} - e_{1} B_{21}) + e_{2} (e_{2} G_{22} + f_{2} B_{22}) + f_{2} (f_{2} G_{22} - e_{2} B_{22}) \\ + e_{2} (e_{3} G_{23} + f_{3} B_{23}) + f_{2} (f_{3} G_{23} - e_{3} B_{23}) \\ = - 0.075 \end{array}$

Similarly the active and reactive powers of bus 2 and 3 are calculated.

P₃ = −0.3

Q₂ = −0.225

Q₃ = −0.9

The ∆ values are the difference between specified values and the calculated values. Hence,

∆P₂ = 0.275

∆P₃ = −0.3

∆Q₂ = 0.225

∆Q₃ = 0.65

Q₂ violates the limits specified, therefore bus 2 is treated as load bus with Q_2spec = 0.

$\frac{\partial P_{p}}{\partial e_{p}} = 2 e_{p} G_{p p} + \sum_{\begin{array}{c} q = 1 \\ q \neq p \end{array}}^{n} (e_{p} G_{p q} + f_{q} B_{p q})$ $\frac{\partial P_{p}}{\partial e_{p}} = 2 e_{p} G_{p p} + \sum_{\begin{array}{c} q = 1 \\ q \neq p \end{array}}^{n} (e_{p} G_{p q} + f_{q} B_{p q})$

$\frac{\partial P_{2}}{\partial e_{2}} = 2 e_{2} G_{22} + \sum_{\begin{array}{c} q = 1 \\ q \neq 2 \end{array}}^{3} (e_{2} G_{2 q} + f_{q} B_{2 q})$ $\frac{\partial P_{2}}{\partial e_{2}} = 2 e_{2} G_{22} + \sum_{\begin{array}{c} q = 1 \\ q \neq 2 \end{array}}^{3} (e_{2} G_{2 q} + f_{q} B_{2 q})$

$\begin{array}{l} \frac{\partial P_{2}}{\partial e_{2}} = 2 e_{2} G_{22} + [e_{2} G_{21} + f_{2} B_{21} + e_{2} G_{23} + f_{3} B_{23} \\ = 2 x 1 x 2.916 + 1.06 (- 1.25) + 0 (- 3.75) + 1 (- 1.66) + 0 (- 5) = 2.848 \end{array}$ $\begin{array}{l} \frac{\partial P_{2}}{\partial e_{2}} = 2 e_{2} G_{22} + [e_{2} G_{21} + f_{2} B_{21} + e_{2} G_{23} + f_{3} B_{23} \\ = 2 x 1 x 2.916 + 1.06 (- 1.25) + 0 (- 3.75) + 1 (- 1.66) + 0 (- 5) = 2.848 \end{array}$

Similarly,

\frac{\partial P_{3}}{\partial e_{3}} = 6.367

$\frac{\partial P_{3}}{\partial e_{3}} = 6.367$

$\frac{\partial P_{p}}{\partial f_{p}} = 2 f_{p} G_{p p} + \sum_{\begin{array}{c} q = 1 \\ q \neq p \end{array}}^{n} (f_{q} G_{p q} + e_{q} B_{p q})$ $\frac{\partial P_{p}}{\partial f_{p}} = 2 f_{p} G_{p p} + \sum_{\begin{array}{c} q = 1 \\ q \neq p \end{array}}^{n} (f_{q} G_{p q} + e_{q} B_{p q})$

$\frac{\partial P_{2}}{\partial f_{2}} = 2 f_{2} G_{22} + [f_{1} G_{21} - e_{1} B_{21} + f_{3} G_{23} - e_{3} B_{23}] = 8.975$ $\frac{\partial P_{2}}{\partial f_{2}} = 2 f_{2} G_{22} + [f_{1} G_{21} - e_{1} B_{21} + f_{3} G_{23} - e_{3} B_{23}] = 8.975$

Similarly,

\frac{\partial P_{3}}{\partial f_{3}} = 20.9

$\frac{\partial P_{3}}{\partial f_{3}} = 20.9$

$\frac{\partial P_{p}}{\partial e_{q}} = e_{p} G_{p q} - f_{p} B_{p q}$ $\frac{\partial P_{p}}{\partial e_{q}} = e_{p} G_{p q} - f_{p} B_{p q}$

$\frac{\partial P_{2}}{\partial e_{3}} = - 1.666$ $\frac{\partial P_{2}}{\partial e_{3}} = - 1.666$

$\frac{\partial P_{3}}{\partial e_{2}} = - 1.666$ $\frac{\partial P_{3}}{\partial e_{2}} = - 1.666$

$\frac{\partial P_{p}}{\partial f_{q}} = e_{p} B_{p q} + f_{p} G_{p q}$ $\frac{\partial P_{p}}{\partial f_{q}} = e_{p} B_{p q} + f_{p} G_{p q}$

$\frac{\partial P_{2}}{\partial f_{3}} = - 5.0$ $\frac{\partial P_{2}}{\partial f_{3}} = - 5.0$

$\frac{\partial P_{3}}{\partial f_{2}} = - 5.0$ $\frac{\partial P_{3}}{\partial f_{2}} = - 5.0$

$\frac{\partial Q_{p}}{\partial e_{p}} = 2 e_{p} B_{p p} + \sum_{\begin{array}{c} q = 1 \\ q \neq p \end{array}}^{n} (f_{q} G_{p q} - e_{q} B_{p q})$ $\frac{\partial Q_{p}}{\partial e_{p}} = 2 e_{p} B_{p p} + \sum_{\begin{array}{c} q = 1 \\ q \neq p \end{array}}^{n} (f_{q} G_{p q} - e_{q} B_{p q})$

$\frac{\partial Q_{2}}{\partial e_{2}} = 8.525$ $\frac{\partial Q_{2}}{\partial e_{2}} = 8.525$

$\frac{\partial Q_{3}}{\partial e_{3}} = 19.1$ $\frac{\partial Q_{3}}{\partial e_{3}} = 19.1$

$\frac{\partial Q_{p}}{\partial f_{p}} = 2 f_{p} B_{p p} + \sum_{\begin{array}{c} q = 1 \\ q \neq p \end{array}}^{n} (e_{q} G_{p q} + f_{q} B_{p q})$ $\frac{\partial Q_{p}}{\partial f_{p}} = 2 f_{p} B_{p p} + \sum_{\begin{array}{c} q = 1 \\ q \neq p \end{array}}^{n} (e_{q} G_{p q} + f_{q} B_{p q})$

$\frac{\partial Q_{2}}{\partial f_{2}} = - 2.991$ $\frac{\partial Q_{2}}{\partial f_{2}} = - 2.991$

$\frac{\partial Q_{3}}{\partial f_{3}} = - 6.966$ $\frac{\partial Q_{3}}{\partial f_{3}} = - 6.966$

At the end of first iteration, the load flow equations are:

$[\begin{array}{c} 0.275 \\ - 0.3 \\ 0.225 \\ 0.65 \end{array}] = [\begin{array}{c} 2.846 & - 1.666 & 8.975 & - 5.0 \\ - 1.666 & 6.366 & - 5.0 & 20.90 \\ 8.525 & - 5.0 & - 2.991 & 1.666 \\ - 5.0 & 19.1 & 1.666 & - 6.966 \end{array}] [\begin{array}{c} ∆ e_{2} \\ ∆ e_{3} \\ ∆ f_{2} \\ ∆ f_{3} \end{array}]$ $[\begin{array}{c} 0.275 \\ - 0.3 \\ 0.225 \\ 0.65 \end{array}] = [\begin{array}{c} 2.846 & - 1.666 & 8.975 & - 5.0 \\ - 1.666 & 6.366 & - 5.0 & 20.90 \\ 8.525 & - 5.0 & - 2.991 & 1.666 \\ - 5.0 & 19.1 & 1.666 & - 6.966 \end{array}] [\begin{array}{c} ∆ e_{2} \\ ∆ e_{3} \\ ∆ f_{2} \\ ∆ f_{3} \end{array}]$

Problem 18.19

Consider the three-bus system. Each of the three lines has a series impedance of 0.02 + j0.08 pu and a total shunt admittance of j0.02 pu. The specified quantities at the buses are tabulated below.

Bus	Real load demand (P_D)	Reactive load demand (Q_D)	Real power generation (P_G)	Reactive power generation	Voltage specified
1	2	1			1.04 + j0
2	0	0	0.5	1	(PQ)
3	1.5	0.6	0	?	V₃ = 1.04 (PV)

A controllable reactive power source is available at bus three with the constraint 0 ≤ Q_G3 ≤ 1.5 pu. Find the load flow solution using the N–R method; use a tolerance of 0.01 for power mismatch.

$\begin{array}{l} Y_{bus} & = [\begin{array}{c} 24.23 ∠ - 75.95 & 12.13 ∠ 104.04 & 12.13 ∠ 104.04 \\ 12.13 ∠ 104.04 & 24.13 ∠ - 75.95 & 12.13 ∠ 104.04 \\ 12.13 ∠ 104.04 & 12.13 ∠ 104.04 & 24.23 ∠ - 75.95 \end{array}] \\ P_{2} & = V_{2} V_{1} Y_{12} \cos (θ_{21} + δ_{21} - δ_{2}) + V_{2}^{2} Y_{22} \cos (θ_{22}) + V_{2} V_{3} Y_{23} \cos (θ_{23} + δ_{3} - δ_{2}) \\ P_{2} & = 0.063 pu, P_{3} = - 0.122 pu \\ Q_{2} & = 0.224 pu, Q_{3} = - 0.557 pu \\ J & = [\begin{array}{c} H_{22} & H_{23} & N_{23} \\ H_{32} & H_{33} & N_{33} \\ J_{32} & J_{33} & L_{33} \end{array}] \\ H_{22} & = - Q_{2} - {|V_{2}|}^{2} B_{22} = - 0.224 - {|1.03|}^{2} (- 17.18) = 18.002 \\ a_{3} + j b_{3} & = (G_{23} + j B_{23}) (a_{3} + j b_{3}) = - 2.035 + j 8.61 \\ H_{23} & = a_{3} f_{2} - b_{3} e_{2} = - 8.868 \\ J & = [\begin{array}{c} 18.002 & - 8.868 & - 2.035 \\ - 8.868 & 17.736 & 3.948 \\ 2.096 & - 4.192 & 16.623 \end{array}] \end{array}$ $\begin{array}{l} Y_{bus} & = [\begin{array}{c} 24.23 ∠ - 75.95 & 12.13 ∠ 104.04 & 12.13 ∠ 104.04 \\ 12.13 ∠ 104.04 & 24.13 ∠ - 75.95 & 12.13 ∠ 104.04 \\ 12.13 ∠ 104.04 & 12.13 ∠ 104.04 & 24.23 ∠ - 75.95 \end{array}] \\ P_{2} & = V_{2} V_{1} Y_{12} \cos (θ_{21} + δ_{21} - δ_{2}) + V_{2}^{2} Y_{22} \cos (θ_{22}) + V_{2} V_{3} Y_{23} \cos (θ_{23} + δ_{3} - δ_{2}) \\ P_{2} & = 0.063 pu, P_{3} = - 0.122 pu \\ Q_{2} & = 0.224 pu, Q_{3} = - 0.557 pu \\ J & = [\begin{array}{c} H_{22} & H_{23} & N_{23} \\ H_{32} & H_{33} & N_{33} \\ J_{32} & J_{33} & L_{33} \end{array}] \\ H_{22} & = - Q_{2} - {|V_{2}|}^{2} B_{22} = - 0.224 - {|1.03|}^{2} (- 17.18) = 18.002 \\ a_{3} + j b_{3} & = (G_{23} + j B_{23}) (a_{3} + j b_{3}) = - 2.035 + j 8.61 \\ H_{23} & = a_{3} f_{2} - b_{3} e_{2} = - 8.868 \\ J & = [\begin{array}{c} 18.002 & - 8.868 & - 2.035 \\ - 8.868 & 17.736 & 3.948 \\ 2.096 & - 4.192 & 16.623 \end{array}] \end{array}$

2. ∆P₂ = P_2spec – P_2calc = 1.5 – 0.063 = 1.437 pu

∆P₃ = –1.2 – (–0.122) = –1.078 pu

∆Q₃ = –0.5 – (–0.557) = 0.057 pu

$[\begin{array}{c} 1.437 \\ - 1.078 \\ 0.057 \end{array}] = [\begin{array}{c} 18.002 & - 8.868 & - 2.035 \\ - 8.868 & 17.736 & 3.948 \\ 2.096 & - 4.192 & 16.623 \end{array}] [\begin{array}{c} ∆ δ_{2} \\ ∆ δ_{2} \\ \frac{∆ |V_{3}|}{|V_{3}|} \end{array}]$ $[\begin{array}{c} 1.437 \\ - 1.078 \\ 0.057 \end{array}] = [\begin{array}{c} 18.002 & - 8.868 & - 2.035 \\ - 8.868 & 17.736 & 3.948 \\ 2.096 & - 4.192 & 16.623 \end{array}] [\begin{array}{c} ∆ δ_{2} \\ ∆ δ_{2} \\ \frac{∆ |V_{3}|}{|V_{3}|} \end{array}]$

18.5.2.4. Fast decoupled load flow method

Sparsity of Y_Bus and loose physical interactions between MW and MVAR flows are taken to make load-flow studies faster and more efficient. P→δ and Q→V are strong whereas P→V and Q→δ are weak. Therefore, MW–δ MVAR–V calculations are decoupled and hence N and J are neglected in the Jacobian matrix:

$\begin{array}{l} [∆ P] = [H] [∆ δ] \\ [∆ Q] = [L] [\frac{∆ |V|}{|V|}] \end{array} .$ $\begin{array}{l} [∆ P] = [H] [∆ δ] \\ [∆ Q] = [L] [\frac{∆ |V|}{|V|}] \end{array} .$

• Voltage angle corrected using P

• Voltage magnitude corrected using Q

$\begin{array}{l} |y_{i p}| e^{j y L_{p}} = G_{i p} + j B_{i p} \\ \frac{\partial p_{i}}{\partial δ_{p}} - j \frac{\partial Q_{i}}{\partial δ_{p}} = j |V_{i}| |V_{p}| e^{j (δ_{p} - δ_{i})} \cdot (G_{i p} + j B_{i p}) \end{array}$ $\begin{array}{l} |y_{i p}| e^{j y L_{p}} = G_{i p} + j B_{i p} \\ \frac{\partial p_{i}}{\partial δ_{p}} - j \frac{\partial Q_{i}}{\partial δ_{p}} = j |V_{i}| |V_{p}| e^{j (δ_{p} - δ_{i})} \cdot (G_{i p} + j B_{i p}) \end{array}$

But δ_p – δ_i is very small. Hence, e^{–j(δp–δi)} ≈ 1.

1. cos(δp – δi) ≈ 1 and sin(δp – δi) = (δp – δi).

Therefore,

$\frac{\partial p_{i}}{\partial δ_{p}} - j \frac{\partial Q_{i}}{\partial δ_{p}} = j |V_{i}| |V_{p}| (G_{i p} + j B_{i p})$ $\frac{\partial p_{i}}{\partial δ_{p}} - j \frac{\partial Q_{i}}{\partial δ_{p}} = j |V_{i}| |V_{p}| (G_{i p} + j B_{i p})$

Separating real and imaginary terms where I ≠ p:

\begin{array}{l} H_{i p} = \frac{\partial p_{i}}{\partial δ_{p}} = - |V_{i}| |V_{p}| |B_{i p}| but L = H \\ L_{i p} = H_{i p} = - |V_{i}| |V_{p}| |B_{i p}| \end{array}

$\begin{array}{l} H_{i p} = \frac{\partial p_{i}}{\partial δ_{p}} = - |V_{i}| |V_{p}| |B_{i p}| but L = H \\ L_{i p} = H_{i p} = - |V_{i}| |V_{p}| |B_{i p}| \end{array}$

In the expressions for L_ii and H_ii, Q_i is generally very small compared to |V_i|² B_ii:

$\begin{array}{l} Q_{i} ≪ {|V_{i}|}^{2} B_{i i} \\ H_{i i} = L_{i i} = - {|V_{i}|}^{2} B_{i i} \\ [∆ P] = [|V| |B'| |V|] [∆ δ] \\ [∆ Q] = [|V| |B ″| |V|] [\frac{∆ |V|}{|V|}] \end{array}$ $\begin{array}{l} Q_{i} ≪ {|V_{i}|}^{2} B_{i i} \\ H_{i i} = L_{i i} = - {|V_{i}|}^{2} B_{i i} \\ [∆ P] = [|V| |B'| |V|] [∆ δ] \\ [∆ Q] = [|V| |B ″| |V|] [\frac{∆ |V|}{|V|}] \end{array}$

B′ and B99 elements [–B_ip].

In the PV bus, Q is not specified and ∆|V| = 0.

Therefore such rows can be simply neglected.

The final algorithm is obtained by making the following approximations.

1. Omit from [B′] the representation of those network elements, which predominantly affect MVAR flow only and do not affect MW flow significantly, that is, shunt reactance, off nominal X^r.

2. Omit from [B99] the angle shifting effects of phase shifters.

3. Neglect the series resistance is calculating the elements of [B]. With these modifications:

$\begin{array}{l} [\frac{∆ P}{|V|}] = |B'| [∆ δ] \\ [\frac{∆ Q}{|V|}] = ||B ″|| [∆ |V|] . \end{array}$ $\begin{array}{l} [\frac{∆ P}{|V|}] = |B'| [∆ δ] \\ [\frac{∆ Q}{|V|}] = ||B ″|| [∆ |V|] . \end{array}$

B′ and B99 are real and sparse.

18.5.2.5. Fixed slope decoupled Newton–Raphson

Load flow computations in a large system are,

1. sparsity: 97% sparse,

2. slope only diagonal and off diagonal + row, column matrix,

3. Gauss elimination, triangular factorization,

4. optimal ordering – least zero, most nonzero number, hence row to be eliminated is with least zero elements.

18.5.2.6. Load flow solution for microgrids

Load flow is the procedure used for obtaining the steady-state voltages of electric power systems at fundamental frequency. An efficient power flow solution looks for fast convergence, minimum usage of memory (computationally efficient), and a numerically robust solution for all the scenarios. Load flow studies on transmission networks are well developed using G–S and N–R methods and their decoupled versions. Because of the some of the following special features the distribution networks fall in the category of ill-conditioned power systems for these conventional load flow methods.

• Radial or weakly meshed networks.

• High R/X ratios.

• Multiphase, unbalanced operation.

• Unbalanced distributed load.

• Distributed generation.

A single-phase representation of three-phase system is used for power flow studies on a transmission system that is assumed as a balanced network in most cases. But the unbalanced loads, radial structure of the network, and untransposed conductors make the distribution system as an unbalanced system. Hence, three-phase power flow analysis needs to be used for distribution systems. The three-phase power flow analysis can be carried out in two different reference frames, namely, phase frame and sequence frame. Phase frame deals directly with unbalanced quantities and sequence frame deals with three separate positive, negative, and zero sequence systems to solve the unbalanced load flow conditions in the circuit. Load flow analysis in distributed generation is carried out using the forward and backward sweep method, compensation methods, implicit Z bus method, direct method (bus injection to branch current and branch current to bus voltage matrices method), or modified Newton methods.

18.6. Power system stability

The dynamics of power systems with the continuous load variations and varying power generation capacity impacts on the stability of the system. Power system stability is the ability of the system to bring back its operation to a steady-state condition within the minimum possible time if undergoing a transient or other disturbance in the system. In power plants, synchronous generators with different voltage ratings are connected to the bus terminals having the same frequency and phase sequence. For example, consider the case defined in the Problem 18.19. After careful analysis, it is found that synchronization depends upon load sharing based on the droop characteristics of the generator. The same is also true for microgrids.

Problem 18.20

Two generators rated 200 and 400 MW are operating in parallel. The droop characteristics shown in Figure 18.35, of their governors are 4 and 5%, respectively, from no load to full load. Assuming that the generators are operating at 50 Hz at no load, how would a load of 600 MW be shared between them? What will be the system frequency at this load?

As the generators run in parallel, they operate at the same frequency.

Let load on G1 (200 MW) = x MW and load on G2 (400 MW) = (600 – x) MW.

Reduction in frequency = ∆f,

$\begin{array}{l} \frac{∆ f}{x} = \frac{0.04 \times 50}{200} \\ \frac{∆ f}{600 - x} = \frac{0.05 \times 50}{400} . \end{array}$ $\begin{array}{l} \frac{∆ f}{x} = \frac{0.04 \times 50}{200} \\ \frac{∆ f}{600 - x} = \frac{0.05 \times 50}{400} . \end{array}$

Equating the ∆f and solving, x = 231 MW. Therefore the load on G1 is 231 MW (overloaded) and that of generator two is 369 MW (underloaded). The system frequency will be,

$System frequency = 50 - \frac{0.04 \times 50}{200} \times 231 = 47.69 Hz .$ $System frequency = 50 - \frac{0.04 \times 50}{200} \times 231 = 47.69 Hz .$

As the droop characteristics are different, G1 is overloaded and G2 is underloaded. If both the governors are 4% droop then they will share the load as 200 and 600 MW, respectively.

Apart from this, stable and quality power is required for ICT applications and therefore stability analysis is important. The steady-state power limit defines the maximum power permissible to flow through a part of the system when subjected to a fault or disturbances. Stability analysis is done under various types of disturbances,

1. Steady-state stability – ability of a system to maintain its stability following a small disturbance like normal load fluctuations, action of automatic voltage regulator, etc., where the variations are gradual and infinitely small power change.

2. Transient state stability – ability of the system to maintain its synchronism following a large disturbance like sudden addition or removal of a large load, switching operations, faults, or loss of excitation, which exists for a reasonably longer period of time.

Power system stability studies as shown in Figure 18.36 are classified based on the duration, such as, short-term or long-term disturbance, and the factors, such as, voltage, frequency, and rotor angle disturbing the stability of the power system.

Figure 18.36 Classification of power system stability.

Basic phenomena associated with rotor angle stability are,

1. imbalance between accelerating and decelerating generator torque,

2. temporary surplus energy stored in the rotating masses,

3. synchronizing torque limited by pullout torque,

4. loss of the synchronization.

Basic phenomena associated with voltage stability are,

1. increased reactive load reducing the voltage magnitude,

2. temporary load reduction,

3. reduction in transfer capability between areas,

4. if there is no solution to load flow, the voltage collapses.

Basic phenomena associated with frequency stability are,

1. connected load,

2. speed of the generators,

3. prime mover.

18.6.1. Power angle curve and the swing equation [6]

The steady-state power limit is defined by the equation,

$P = \frac{|E| |V|}{X}$ $P = \frac{|E| |V|}{X}$

where E is the generated voltage, V is the terminal voltage, and X is the transfer reactance.

The power P = P_m sin δ and the power angle curve are drawn as shown in Figure 18.37.

The dynamics of a generator depend on the inertia constant designated by the manufacturer and the kinetic energy developed while running.

$Kinetic energy = \frac{1}{2} M ω_{s}$ $Kinetic energy = \frac{1}{2} M ω_{s}$

where M is the moment of inertia in MJs/rad and ω_s is the rotor speed in rad/s.

The kinetic energy is also defined as:

$G H = Kinetic energy = \frac{1}{2} M ω_{s}$ $G H = Kinetic energy = \frac{1}{2} M ω_{s}$

where G is the machine rating in MVA_base and H is the inertia constant in MJ/MVA.

Considering the rotor angle (δ) torque, speed, flow of mechanical (P_m) and electrical (P_e) powers in a synchronous machine, the rotor dynamics are defined by the swing equation:

$M \frac{d^{2} δ}{d t^{2}} = P_{m} - P_{e}$ $M \frac{d^{2} δ}{d t^{2}} = P_{m} - P_{e}$

18.6.2. Solutions of swing equation [7]

For steady-state and transient stability analysis, swing equations can be solved using,

1. Equal area criterion method: in a power angle curve, the accelerating area should be equal to the decelerating area for synchronism to be achieved and stability to be regained following the disturbance.

2. Numerical solution of swing equation: modified Euler’s method.

18.6.3. Stability analysis in microgrids [2]

Apart from active and reactive power control using FACTS controllers, a microgrid’s overall stability depends on voltage control. With a large number of microsources connected, microgrids suffer from reactive power oscillations without proper voltage control resulting in circulating currents. This circulating control is controlled by using voltage-reactive power (V–Q) droop controllers. V–Q droop controllers will increase the voltage set point if the microsource reactive current becomes predominantly inductive and decreases the set point if the reactive currents are predominantly capacitive. The reactive power limits are set by the apparent power rating of an inverter and active power output of the microsource.

18.7. Summary

This chapter introduced basic methods and algorithms to study the performance of power systems. Based on the results, solutions may be provided to improve the reliability and security of a power-system network.

Problems

1. A 300 MVA, 20 kV, 3φ generator has a subtransient reactance of 20%. The generator supplies a number of motors over a 64-km transmission line having transformers at both ends. The motors are all rated at 13.2 kV and are represented by two equivalent motors. The rated inputs to the motors are 200 MVA and 100 MVA, respectively. Both have a reactance of 20%. The 3φ transformer T1 is rated 350 MVA, 230/20 kV with a leakage reactance of 10%. T2 is composed of three numbers of 1φ transformers each rated 127/13.2 kV, 100 MVA with a leakage reactance of 10.1. Reactance of the transmission line is 0.5 Ω/km. Draw the one-line diagram, impedance diagram, and reactance diagram.

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for 18.5. Power flow study

Create new playlist

Sign In

Sign Up

18.4.7. Systematic fault analysis using Zbus matrix

18.5. Power flow study

18.5.1. Load flow equations and methods of solution

18.5.2. Solution of load flow equation

18.5.2.1. Using Gauss-Seidel (G-S) method when PV buses are absent

18.5.2.2. Modification of G–S method when PV buses are present

18.5.2.3. Newton–Raphson method

18.5.2.4. Fast decoupled load flow method

18.5.2.5. Fixed slope decoupled Newton–Raphson

18.5.2.6. Load flow solution for microgrids

18.6. Power system stability

18.6.1. Power angle curve and the swing equation [6]

18.6.2. Solutions of swing equation [7]

18.6.3. Stability analysis in microgrids [2]

18.7. Summary

Table of Contents for
18.5. Power flow study

18.4.7. Systematic fault analysis using Z_bus matrix