4. Proposed method of protection

Figures 4 and 6 shows a single-phase (internal and external) fault differential protection system. Obviously, a pair of current sensors by using RC surrounded the protected area. Since this was a natural trend, differential protection provided protection for system equipment wire carrying current from the RC called a pilot wire.

Figure 4. Differential relay currents at the time internal fault.

a constant turn density n. Second, the effects of adjacent conductors carrying a large current to the coil of the output signal should be minimal. The following formula

When the core has a constant cross-section S, μ<sup>o</sup> is the permeability of the free space, and the winding line is perpendicular to the midline m, with a constant density n. The output voltage is proportional to the measured rate of change of

Where M is the mutual inductance of the coil, also called the sensitivity of the RC. The CT iron core has a nonlinear characteristic and is therefore saturated when a high current or a direct current component is present in primary current. When CT is saturated (that is, the CT ratio error increases), which adversely affects the performance of the relay. Figure 2 displays the equivalent circuit of a current transformer.The current phase angle between the primary coil and the secondary voltage is almost 90° (due to the coil inductance Ls). Figure 3 shows the equivalent

Rogowski coils are linear and can be used in measurement applications. The oscillatory response of RC can be represented by voltage response and natural

dt :<sup>e</sup>

where ω<sup>n</sup> is the natural frequency, and ε is the damping coefficient. As a result,

�ξωnt sin ω<sup>n</sup>

ffiffiffiffiffiffiffiffiffiffiffiffi <sup>1</sup> � <sup>ξ</sup><sup>2</sup> � � q

t, (3)

dt ¼ �<sup>M</sup> di tð Þ

used for protection purposes, review the essential characteristics.

RCs can replace conventional CTs for measurement and protection. IEEE Std C37.235TM-2007 [19] provides guidance on the application of RC sensors when

dt ¼ �<sup>M</sup> di tð Þ

v tðÞ¼�μonS di tð Þ

M ¼ μ<sup>o</sup> � n � S (1)

dt (2)

defines mutual inductance M:

Micro-Grids - Applications, Operation, Control and Protection

current, as shown in Eq. (2) [18]:

frequency, as described below:

Current transformer equivalent circuit.

Rogowski coil equivalent circuits.

VoðÞ¼� <sup>t</sup> <sup>M</sup> di tð Þ

circuit of RCs.

Figure 2.

Figure 3.

116

In case of no fault, current input protection unit Ip is to be at all times the same to the current going out of the protected zone. Considering current in transformer A, the pilot wire was carrying current as follows:

$$\mathbf{I\_{AS}} = \mathbf{a\_{A}}\mathbf{I\_{p}} - \mathbf{I\_{Ae}}\mathbf{I\_{AS}} = \mathbf{a\_{A}}\mathbf{I\_{p}} - \mathbf{I\_{Ae}} \tag{4}$$

where aA is the conversion ratio for RC A and IAe is the secondary excitation current for RC A. Similarly, for current in transformer B, the equation is as follows:

$$\mathbf{I\_{BS}} = \mathbf{a\_B}\mathbf{I\_p} - \mathbf{I\_{Be}} \tag{5}$$

By considering the appropriate characteristic impedance and propagation constant, we can apply the above conversion method to the zero, positive and negative sequence components of the current. In this paper, the "0," "1," and "2" subscripts

> Um0 Zc0

Um1 Zc1

Um2 Zc2

Un0 Zc0

Un1 Zc1

Un2 Zc2

> 3 7 5

> 3 7 5

2 6 4

2 6 4

Imk0 Imk1 Imk2

Ink0 Ink1 Ink2

Uf0 ¼ Un0ch γ<sup>0</sup> ð Þ� lnf In0Zc0sh γ<sup>0</sup> ð Þ lnf (20) Uf1 ¼ Un1ch γ<sup>1</sup> ð Þ� lnf In1Zc1sh γ<sup>1</sup> ð Þ lnf (21) Uf2 ¼ Un2ch γ<sup>1</sup> ð Þ� lnf In2Zc1sh γ<sup>1</sup> ð Þ lnf (22)

> Un0 Zc0

Un1 Zc1

3 7

Imk<sup>∅</sup> ¼ �Ink<sup>∅</sup> (19)

3 7

sh γ<sup>0</sup> ð Þ lmk (11)

sh γ<sup>1</sup> ð Þ lmk (12)

sh γ<sup>1</sup> ð Þ lmk (13)

sh γ<sup>0</sup> ð Þ lnk (14)

sh γ<sup>1</sup> ð Þ lnk (15)

sh γ<sup>1</sup> ð Þ lnk (16)

<sup>5</sup> (17)

<sup>5</sup> (18)

sh γ<sup>0</sup> ð Þ lnf (23)

sh γ<sup>1</sup> ð Þ lnf (24)

were described as zero, positive, and negative sequences, respectively.

Innovative Differential Protection Scheme for Microgrids Based on RC Current Sensor

DOI: http://dx.doi.org/10.5772/intechopen.85473

Imk0 ¼ Im0ch γ<sup>0</sup> ð Þ� lmk

Imk1 ¼ Im1ch γ<sup>1</sup> ð Þ� lmk

Imk2 ¼ Im2ch γ<sup>1</sup> ð Þ� lmk

Ink0 ¼ In0ch γ<sup>0</sup> ð Þ� lnk

Ink1 ¼ In1ch γ<sup>1</sup> ð Þ� lnk

Ink2 ¼ In2ch γ<sup>1</sup> ð Þ� lnk

Imka Imkb Imkc

2 6 4

Similarly, the relationship to current is as below:

2 6 4

the phase current should satisfy the following equation:

fault location of the current and voltage from the end n,

119

Inka Inkb Inkc

where lmk and lnk ¼ distance of point k from end m and n respectively:

2 6 4

2 6 4

where ∅ ¼ a, b, c shows the phase from where the current belong to.

If0 ¼ In0ch γ<sup>0</sup> ð Þ� lnf

If1 ¼ In1ch γ<sup>1</sup> ð Þ� lnf

<sup>a</sup> <sup>¼</sup> ej2π=<sup>3</sup>

11 1 1 a<sup>2</sup> a 1aa2

11 1 1 a<sup>2</sup> a 1aa2

If the fault is outside the protected area or there is no fault in the system, then

The following explanation makes it possible to understand the actual current and the position where the fault and the derivation of the current occur. To calculate the

where aB is the conversion ratio for RC B and IBe is the secondary excitation current for RC B.Considering the equal ratio, aA = aB = a, relay working current Iop is:

$$\mathbf{I}\_{\rm op} = \mathbf{I}\_{\rm Ae} - \mathbf{I}\_{\rm Be} \tag{6}$$

When there was a normal operation or the external fault of the system, the relay's working current Iop was very small, but never tends to zero. Nevertheless, within the time the internal fault was in the protected area, the current was no longer equal to the outgoing current. The operating current of the differential relay is nothing more than an increase of the input current as the same as the feed fault.

$$\mathbf{I}\_{\rm op} = \mathbf{a} (\mathbf{I}\_{\rm f1} + \mathbf{I}\_{\rm f2}) \mathbf{I}\_{\rm Ae} - \mathbf{I}\_{\rm Be} \tag{7}$$

The uniform distribution of the line is as shown in Figure 5. Where, co is a shunt capacitance (F/km), go the shunt leakage conductance (S/km), Io the series inductance (H/km) and ro is a series resistance (Ω/km). The distribution of current and voltage along the transmission line is given by the equations of the current and voltage display in the diagram at the ends of the line.

$$-\frac{\partial u}{\partial \mathbf{x}} = \mathbf{r}\_o \mathbf{i} + \mathbf{l}\_o \frac{\partial \dot{\mathbf{r}}}{\partial t} \tag{8}$$

$$-\frac{\partial \dot{i}}{\partial \mathbf{x}} = \mathbf{g}\_o \mathbf{u} + \mathbf{c}\_o \frac{\partial \dot{i}}{\partial t} \tag{9}$$

By reducing Eqs. (5) and (6) to their frequency domain,

$$
\begin{bmatrix} \mathbf{U\_m} \\ \mathbf{I\_m} \end{bmatrix} = \begin{bmatrix} \mathbf{ch}(\boldsymbol{\gamma}l\_{\mathbf{nm}}) & -\mathbf{Z\_c}\mathbf{sh}(\boldsymbol{\gamma}l\_{\mathbf{nm}}) \\ \mathbf{sh}(\boldsymbol{\gamma}l\_{\mathbf{nm}})/\mathbf{Z\_c} & -\mathbf{ch}(\boldsymbol{\gamma}l\_{\mathbf{nm}}) \end{bmatrix} \begin{bmatrix} \mathbf{U\_n} \\ \mathbf{I\_n} \end{bmatrix} \tag{10}
$$

where sh γl ð Þ nm and ch γl ð Þ¼ nm hyperbolic function, Zc ¼ characteristic impedance, γ ¼ the propagation constant and both are freqency dependent; lnm ¼ distance from end to end m.

Figure 5. Transmission line model with a fault located inside the two ends. Innovative Differential Protection Scheme for Microgrids Based on RC Current Sensor DOI: http://dx.doi.org/10.5772/intechopen.85473

By considering the appropriate characteristic impedance and propagation constant, we can apply the above conversion method to the zero, positive and negative sequence components of the current. In this paper, the "0," "1," and "2" subscripts were described as zero, positive, and negative sequences, respectively.

$$\mathbf{I}\_{\rm mk0} = \mathbf{I}\_{\rm m0} \mathbf{c} \mathbf{h}(\boldsymbol{\gamma}\_0 \mathbf{l}\_{\rm mk}) - \frac{\mathbf{U}\_{\rm m0}}{\mathbf{Z}\_{\rm c0}} \mathbf{s} \mathbf{h}(\boldsymbol{\gamma}\_0 \mathbf{l}\_{\rm mk}) \tag{11}$$

$$\mathbf{I\_{mk1}} = \mathbf{I\_{m1}} \mathbf{ch}(\boldsymbol{\gamma}\_1 \mathbf{l\_{mk}}) - \frac{\mathbf{U\_{m1}}}{\mathbf{Z\_{c1}}} \mathbf{sh}(\boldsymbol{\gamma}\_1 \mathbf{l\_{mk}}) \tag{12}$$

$$\mathbf{I}\_{\rm mk2} = \mathbf{I}\_{\rm m2} \mathbf{ch}(\chi\_1 \mathbf{l}\_{\rm mk}) - \frac{\mathbf{U}\_{\rm m2}}{\mathbf{Z}\_{\rm c2}} \mathbf{sh}(\chi\_1 \mathbf{l}\_{\rm mk}) \tag{13}$$

$$\mathbf{I}\_{\rm nk0} = \mathbf{I}\_{\rm n0} \mathbf{c} \mathbf{h}(\chi\_0 \mathbf{l}\_{\rm nk}) - \frac{\mathbf{U}\_{\rm n0}}{\mathbf{Z}\_{\rm c0}} \mathbf{s} \mathbf{h}(\chi\_0 \mathbf{l}\_{\rm nk}) \tag{14}$$

$$\mathbf{I}\_{\rm nk1} = \mathbf{I}\_{\rm n1} \mathbf{c} \mathbf{h}(\gamma\_1 \mathbf{l}\_{\rm nk}) - \frac{\mathbf{U}\_{\rm n1}}{\mathbf{Z}\_{\rm c1}} \mathbf{s} \mathbf{h}(\gamma\_1 \mathbf{l}\_{\rm nk}) \tag{15}$$

$$\mathbf{I}\_{\rm nk2} = \mathbf{I}\_{\rm n2} \mathbf{c} \mathbf{h}(\mathbf{y}\_1 \mathbf{l}\_{\rm nk}) - \frac{\mathbf{U}\_{\rm n2}}{\mathbf{Z}\_{\rm c2}} \mathbf{s} \mathbf{h}(\mathbf{y}\_1 \mathbf{l}\_{\rm nk}) \tag{16}$$

where lmk and lnk ¼ distance of point k from end m and n respectively:

$$\mathbf{a} = \mathbf{e}^{j2\pi/3}$$

$$\begin{bmatrix} \mathbf{I}\_{\mathrm{mka}} \\ \mathbf{I}\_{\mathrm{mkb}} \\ \mathbf{I}\_{\mathrm{mkb}} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{1} & \mathbf{1} \\ \mathbf{1} & \mathbf{a}^2 & \mathbf{a} \\ \mathbf{1} & \mathbf{a} & \mathbf{a}^2 \end{bmatrix} \begin{bmatrix} \mathbf{I}\_{\mathrm{mkb}} \\ \mathbf{I}\_{\mathrm{mkb}} \\ \mathbf{I}\_{\mathrm{mkb}} \end{bmatrix} \tag{17}$$

Similarly, the relationship to current is as below:

$$
\begin{bmatrix}
\mathbf{I}\_{\rm nka} \\
\mathbf{I}\_{\rm nkb} \\
\mathbf{I}\_{\rm nkc}
\end{bmatrix} = \begin{bmatrix}
\mathbf{1} & \mathbf{1} & \mathbf{1} \\
\mathbf{1} & \mathbf{a}^2 & \mathbf{a} \\
\mathbf{1} & \mathbf{a} & \mathbf{a}^2
\end{bmatrix} \begin{bmatrix}
\mathbf{I}\_{\rm nk0} \\
\mathbf{I}\_{\rm nk1} \\
\mathbf{I}\_{\rm nk2}
\end{bmatrix} \tag{18}
$$

If the fault is outside the protected area or there is no fault in the system, then the phase current should satisfy the following equation:

$$\mathbf{I}\_{\text{mk}\mathcal{Q}} = -\mathbf{I}\_{nk\mathcal{Q}} \tag{19}$$

where ∅ ¼ a, b, c shows the phase from where the current belong to.

The following explanation makes it possible to understand the actual current and the position where the fault and the derivation of the current occur. To calculate the fault location of the current and voltage from the end n,

$$\mathbf{U\_{f0}} = \mathbf{U\_{n0}}\mathbf{ch}(\mathbf{y\_0l\_{nf}}) - \mathbf{I\_{n0}}Z\_{c0}\mathbf{sh}(\mathbf{y\_0l\_{nf}}) \tag{20}$$

$$\mathbf{U\_{f1} = U\_{n1}ch}(\mathbf{y\_1 l\_{nf}}) - \mathbf{I\_{n1}}\mathbf{Z\_{c1}sh}(\mathbf{y\_1 l\_{nf}}) \tag{21}$$

$$\mathbf{U\_{f2} = U\_{n2}ch}(\mathbf{y\_1 l\_{nf}}) - \mathbf{I\_{n2}}\mathbf{Z\_{c1}sh}(\mathbf{y\_1 l\_{nf}}) \tag{22}$$

$$\mathbf{I}\_{\rm f0} = \mathbf{I}\_{\rm n0} \mathbf{ch}(\chi\_0 \mathbf{l}\_{\rm nf}) - \frac{\mathbf{U}\_{\rm n0}}{\mathbf{Z}\_{\rm c0}} \mathbf{sh}(\chi\_0 \mathbf{l}\_{\rm nf}) \tag{23}$$

$$\mathbf{I}\_{\rm f1} = \mathbf{I}\_{\rm n1} \mathbf{ch}(\chi\_1 \mathbf{l}\_{\rm nf}) - \frac{\mathbf{U}\_{\rm n1}}{\mathbf{Z}\_{\rm c1}} \mathbf{sh}(\chi\_1 \mathbf{l}\_{\rm nf}) \tag{24}$$

In case of no fault, current input protection unit Ip is to be at all times the same to the current going out of the protected zone. Considering current in transformer

where aA is the conversion ratio for RC A and IAe is the secondary excitation current for RC A. Similarly, for current in transformer B, the equation is as follows:

where aB is the conversion ratio for RC B and IBe is the secondary excitation current for RC B.Considering the equal ratio, aA = aB = a, relay working current Iop is:

When there was a normal operation or the external fault of the system, the relay's working current Iop was very small, but never tends to zero. Nevertheless, within the time the internal fault was in the protected area, the current was no longer equal to the outgoing current. The operating current of the differential relay is nothing more than an increase of the input current as the same as the feed fault.

The uniform distribution of the line is as shown in Figure 5. Where, co is a shunt capacitance (F/km), go the shunt leakage conductance (S/km), Io the series inductance (H/km) and ro is a series resistance (Ω/km). The distribution of current and voltage along the transmission line is given by the equations of the current and

<sup>∂</sup><sup>x</sup> <sup>¼</sup> roi <sup>þ</sup> lo

<sup>∂</sup><sup>x</sup> <sup>¼</sup> gou <sup>þ</sup> co

<sup>¼</sup> ch <sup>γ</sup><sup>l</sup> ðÞ� nm Zcsh <sup>γ</sup><sup>l</sup> ð Þ nm sh γl ð Þ nm =Zc �ch γl ð Þ nm

where sh γl ð Þ nm and ch γl ð Þ¼ nm hyperbolic function, Zc ¼ characteristic impedance, γ ¼ the propagation constant and both are freqency dependent;

Un

∂i

∂i

IAS ¼ aAIp � IAeIAS ¼ aAIp � IAe (4)

IBS ¼ aBIp � IBe (5)

Iop ¼ IAe � IBe (6)

Iop ¼ a Ið Þ f1 þ If2 IAe � IBe (7)

<sup>∂</sup><sup>t</sup> (8)

<sup>∂</sup><sup>t</sup> (9)

(10)

In 

A, the pilot wire was carrying current as follows:

Micro-Grids - Applications, Operation, Control and Protection

voltage display in the diagram at the ends of the line.

Um Im 

Transmission line model with a fault located inside the two ends.

lnm ¼ distance from end to end m.

Figure 5.

118

� ∂u

� ∂i

By reducing Eqs. (5) and (6) to their frequency domain,

$$\mathbf{I\_{f2} = I\_{n2}ch}(\gamma\_1 \mathbf{l\_{nf}}) - \frac{\mathbf{U\_{n2}}}{Z\_{c2}}\mathbf{sh}(\gamma\_1 \mathbf{l\_{nf}}) \tag{25}$$

The differential current is the actual measure of relay operation. In addition, the brake current is one of the avoided currents of the mal-tripping differential relay. The effect of this current has been in breaking the relay.

$$\mathbf{I\_B} = |\mathbf{I\_{mk1}} + \mathbf{I\_{nk1}}|\tag{26}$$

5. SCADA systems

DOI: http://dx.doi.org/10.5772/intechopen.85473

orders to equipment.

distributed areas away from the focal point.

3. The software package used in the system.

the following:

Figure 7.

121

It is considered as a means of monitoring and control of power plants also used in renewable power plants and these systems transfer data to the heart of the system which is a master computer and receives orders from many remote terminals; Figure 7 illustrates the structure of the SCADA system, the SCADA system includes

Innovative Differential Protection Scheme for Microgrids Based on RC Current Sensor

1. RTUs or PLC which transfers information to the central unit and transfers

2. Radio system or satellites to secure communication between central units and

The operation of electricity distribution networks is monitored and controlled by supervisory control and data acquisition (SCADA) systems. SCADA systems are linked through various communications networks such as microwave and optical fiber networks to ensure the functioning of the system. They are used to connect transmission substations with the major generators to facilitate an integrated system. The operation of the methods of communication in the networks of energy is produced by lines along the system with advanced optical networks. The loss of these communication cables is possible and could make the protection and monitoring of the network more complex. Using advanced wireless communication and sensing devices could improve the control and monitoring of the entire system. Intelligent Electronic Devices (IED) for monitoring and control to improve the technology of smart grids, and. IEDs can be installed and distributed within the system to be used for protection and monitoring [24–28]. These devices are

Network monitoring and control by supervisory control and data acquisition systems (SCADA).

$$\mathbf{I\_B} = \left| \mathbf{I\_{mk1}} + \mathbf{I\_{nk}} - \mathbf{I\_{f1}ch}(\mathbf{y\_1}\mathbf{l\_{fk}}) \right| \tag{27}$$

The effects of the capacitive current can be cancelling with any of the use of the following techniques. Shunt Reactor, Phasor Compensation algorithm, Capacitor Current Compensation, and Bergeron line Model.

In reality, the proposal is a reliable method for the protection system. Since each current transformer and circuit breaker can only declare a line, it is usual to protect by the use of two side current transformers and circuit breakers. The currents on both sides were compared. Under normal conditions or for defects outside the protected area, the current Bas-Bar A is equal to the current Bas-Bar B. Thus, the currents in the secondary current transformer were equal; no current flowed through the relay current. If a fault occurs in the protected zone, the secondary transformer currents of Bas-Bar A and Bas-Bar B will not be the same, and there will be a current flowing through the relay current, as shown Figure 6.

The differential protection ratio, by using a multi-slope feature of the relay was inserted into the new relay for its excellent compromise between reliability and sensitivity. The components of the relay compared the different current Idiff (also called operating current), and the restraining current Ibias are expressed in Eqs. (28) and (29) [20–24].

$$\mathbf{I}\_{\rm diff} = \left| \mathbf{I}'\_{\rm As} + \mathbf{I}'\_{\rm Bs} \right| \tag{28}$$

$$\mathbf{I}\_{\rm bias} = \left( \left| \mathbf{I}\_{\rm As}' \right| + \left| \mathbf{I}\_{\rm Bs}' \right| \right) / 2 \tag{29}$$

where I<sup>0</sup> As and I<sup>0</sup> Bs are secondary RC phasor currents.

Figure 6. Differential relay currents at the external fault.

Innovative Differential Protection Scheme for Microgrids Based on RC Current Sensor DOI: http://dx.doi.org/10.5772/intechopen.85473
