IND Dang Minh Quan

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Czech Technical University in Prague

Faculty of Electrical Engineering

Department of Electroenergetics

Inuences of Distributed Generation on protectionin radial system

Author:Minh-Quan Dang

February 17, 2016Prague



Contents1 Theoretical background of Power system protection 1

1.1 Principle of protection of power system . . . . . . . . . . . . . . . . . . 11.2 System protection components . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 Radial system protection . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Coordinating overcurrent relays in a radial system 32.1 Given data and assumptions . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Short-circuit current calculations . . . . . . . . . . . . . . . . . . . . . 42.3 Setting for current tap of Overcurrent relays . . . . . . . . . . . . . . . 52.4 Setting of time-dial of relays . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Inuences of Distributed Generations on protection in a radial system 83.1 Given data and assumptions . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Short circuit analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.3 Setting of time-dial of relays . . . . . . . . . . . . . . . . . . . . . . . . 93.4 Simulation of inuences of DG . . . . . . . . . . . . . . . . . . . . . . . 10

4 Solutions and Conclusion 13

References 14



1 THEORETICAL BACKGROUND OF POWER SYSTEM PROTECTION

1 Theoretical background of Power system protec-tion

1.1 Principle of protection of power systemPower system protection design have these following criteria [ 1]:

1. Reliability: Whenever fault happen the protection system must operate (protec-tive relay system has reliability ≥99%)

2. Selectivity: Only the faulted section is isolated

3. Speed: Operate rapidly to minimize the affect of fault.

4. Economy: Maximum protection with minimum cost

5. Simplicity: Minimum protective equipment and circuitry to achieve the protec-tion objectives.

1.2 System protection componentsProtection systems consists of three basis elements [2], which are:

1. Instrument transformers: Current transformer (CT), voltage transformer (VT)

2. Relays: Overcurrent relays, Directional relays, Impedance relays, Differentialrelays...

3. Circuit breakers

Current/Voltage transformers reproduce a current/voltage in its secondarywinding. These secondary current/voltage has much smaller magnitude compare toprimary. For CT, I secondary ∈(0;5A) and for VT, V secondary ≈1V

The reasons for having small secondary output:

1. Safety: The personnel working with relays will work in a safer environment,which is isolated from the power system.

2. Economy: lower-level relay input allow to be used smaller, simpler and less ex-

pensive.3. Accuracy: accurately reproduce power current and voltages over wide operating

ranges.

Voltage Ratio1:1 2:1 5:1 4:1 5:1 20:1 40:160:1 100:1 200:1 300:1 400:1 600:1 800:1

1000:1 2000:1 3000:1 4500:1

Table 1: Standar VT ratio [2]

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1 THEORETICAL BACKGROUND OF POWER SYSTEM PROTECTION

Current Ratio50:5 100:5 150:5 200:5 250:5 300:5 400:5450:5 500:5 600:5 800:5 900:5 1000:5 1200:51500:5 1600:5 2000:5 2400:5 2500:5 3000:5 3200:54000:5 5000:5 6000:5

Table 2: Standard CT ratios

Overcurrent Relays detect short-circuit current in power system, when theI fault > I pickup the relay will signal circuit breaker to trip. There are two basic type of overcurrent relay: denite-time and inverse-time. The denite-time relay response tothe magnitude of input current. If the input current exceed the pickup current, then therelay contacts close immediately to energize the circuit breaker trip coil. The inverse-time relay has tripping time vary response to the magnitude of the input current. Thetripping time (TDS) of relay could be control by time-dial setting.

Figure 1: Denite and inverse-time characteristics

The overcurrent relay have two setting:

1. Current tap setting: The pickup current in amperes.

2. Time-dial setting: The adjustable amount of time delay.

1.3 Radial system protectionThe radial system is usually protected by inverse-time overcurrent relay. The op-

erating time of relay can be selected such that the breaker closest to the fault opens,while other upstream breakers with larger time delays remain closed. The coordinationtime interval is the interval between the primary and backup protective devices typicalfrom 0.2 to 0.5 seconds [2].

The process setting and coordinating inverse-time overcurrent relays in radial sys-tem will be present in the following chapter.

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2 COORDINATING OVERCURRENT RELAYS IN A RADIAL SYSTEM

2 Coordinating overcurrent relays in a radial sys-tem

2.1 Given data and assumptionsThe 50-Hz radial system of Figure 2, data of this system are given in Table 2.1

and Table 4. Selection of current tap settings (TSs) and time-dial setting (TDSs) isrequired to protect the system from faults. Along with the given data some assumptionare made to study the process of setting and coordinating overcurrent relays in a radialsystem.

The rst assumption is all OC relays for each circuit breaker (CB) is normal inverse-time relay [3]. Each breaker has three relays for each phase, with a 0.2-second coordi-nation time interval. The relays for each breaker are connected in the way, so that allthree phases of the breaker open when a fault is detected on any one phase.

Second assumption is a line-to-line 10-kV at all buses during normal operation. Thegrid is considered as innite bus. Future load growth is included in the Table 2.1, suchthat maximum loads over the operating life of the radial system.

Figure 2: Single-line diagram of a 10-kV radial system

Bus SMVA Lagging p.f

Grid 200 -1 11 0.952 4 0.953 6 0.95

Table 3: Maximum loads

Type R [Ω/km ] X[Ω/km ] I nom [A]OH-line: DINGO 19/.132 0.218 0.311 525

Table 4: Over head line parameters of the radial system

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Breaker Breaker Operating Time CT Ratio RelayB1 5 cycle 1500:5 NIB2 5 cycle 600:5 NIB3 5 cycle 600:5 NI

Table 5: Breaker, CT and relay data

2.2 Short-circuit current calculationsThe maximum faulted current is assumed as bolted three-phase (3 φ) fault, which

occurs on buses [4]. The equivalent diagram of the network when fault occurs at bus3 as following gure.

Figure 3: Equivalent circuit of network

The grid impedance is calculated as:

Z grid =V 2grid

S grid= j 0.5 [Ω]

The impedance of over-head transmission line is calculated as:

Z OHL = ( r OHL + jx OHL ).length

Z 1 = 0.218 + j 0.311 [Ω]

Z 2 = 0.872 + j 1.244 [Ω]

Z 3 = 0.654 + j 0.933 [Ω]

If the fault occurs at bus 3 as the Figure 3, the total 3 φ short-circuit current is:

I k 3 φ = V grid

√ 3.(Z grid + Z 1 + Z 2 + Z 3)

I k 3 φ = 841.208 − j 1441.24 = 1668.78 −59.73 [A]

With the same process, the I k 3 φ at the bus 1 and bus 2 will be found and results ispresented in the Table

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BusMaximum Fault Current

(Bolted Three-Phase)A

1 6874.95 −74.95432 2481.96

−62.0579

3 1668.78 −59.7293

Table 6: Symmetrical fault currents

2.3 Setting for current tap of Overcurrent relaysThe current tap should be set, such the way the relays do not operate for maximum

load currents. Starting at bus 3, the primary and secondary CT currents for maximumload L3 are:

I L 3 = S L 3

√ 3V 3=

6×106

√ 3.(10 ×103

)= 346.41 [A]

I L 3 = I L 3

CT ratio =

346.41(600/ 5)

= 2.89 [A]

TS at B3 is selected as 3-A, which is the lowest TS above 2.89 A.The primary and secondary CT currents for maximum load at B2 are:

I L 2 = S L 2 + S L 3

√ 3V 2=

(4 + 6) ×106

√ 3.(10 ×103 )= 577.35 [A]

I L

2 = 577.35

(600/ 5) = 4.81 [A]

TS at B2 is selected as 5-A, which is the lowest TS above 4.81 A.The primary and secondary CT currents for maximum load at B1 are:

I L 1 = S L 1 + S L 2 + S L 3

√ 3V 1=

(11 + 4 + 6) ×106

√ 3.(10 ×103 )= 1212.44 [A]

I L 1 = 1212.44(1500/ 5)

= 4.04 [A]

TS at B1 is selected as 5-A, which is the lowest TS above 4.04 A.

Bus CT Ratio I primary

AI secondary

ATSA

1 1500/5 1212.44 4.04 52 600/5 577.35 4.81 53 600/5 346.41 2.89 3

Table 7: Current tap setting of CT

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2.4 Setting of time-dial of relaysCharacteristic of Normal-inverse relays are described by formula [ 3]:

t tripping = T DS

×

0.14

(I/Ip )0 .02

−1Starting at B3, the largest fault current through B3 is 2481.96 A, which occurs forthe 3φ fault at bus 2 (right side of B3). Neglecting CT saturation, the fault-to-pickupcurrent ratio at B3 for this fault is:

I 3 Fault

T S 3 =

2481.96/ (600/ 5)3

= 6.89

Since this bus is the most remote bus on the radial system, the fault should be clearas rapid as possible. For that reason, the relay operating time is chosen T3 = 50 ms.Then the relay TDS is:

T DS 3 = 0.05 ×( 0.146.890 .02 −1

)− 1 = 0 .0140604

Adding the breaker operating time (5 cycle = 0.1 s), primary protection clears thisfault in T 3 + T breaker = 0.05 + 0.1 = 0.15s

For the same fault, the fault-to-pickup current ratio at B2 is

I 2 Fault

T S 2 =

2481.96/ (600/ 5)5

= 4.14

Adding the B3 relay operating time ( T 3 = 0.05s), breaker operating time (0 .1s),and 0.2s coordination time interval, we want a B2 relay operating time:

T 2 = T 3 + T breaker + T coordination = 0.05 + 0.1 + 0 .2 = 0.35s

ThenT DS 2 = 0.35 ×(

0.144.140 .02 −1

)− 1 = 0 .0720536

Next select the TDS at B1. The largest fault current though B2 is 6874.95 A, forthe 3φ fault at bus 2 (right side of B2). Neglecting CT saturation, the fault-to-pickupcurrent ratio at B3 for this fault is:

I 2 Fault

T S 2 =

6874.95/ (600/ 5)

5 = 11.46

Then for this fault the operating time of B2 relay:

T 2 = 0.072 × 0.14

11.460 .02 −1 ≈0.2 [s]

Fault-to-pickup ratio of B1 relay

I 1 Fault

T S 1 =

6874.95/ (1500/ 5)5

= 4.58

T 1 = T 2 + T breaker + T coordination = 0.2 + 0 .1 + 0 .2 = 0.5 [s]

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ThenT DS 1 = 0.5 ×(

0.144.580 .02 −1

)− 1 = 0 .110364

Bus TDS1 0.112 0.0723 0.014

Table 8: Time-dial settings of relays

Figure 4: Setting of protection selectivity in classical system

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3 INFLUENCES OF DISTRIBUTED GENERATIONS ON PROTECTION IN ARADIAL SYSTEM

3 Inuences of Distributed Generations on protec-tion in a radial system

3.1 Given data and assumptionsThe distributed generator (DG) is connected to the radial system from the Chapter

2. The DG is connected at the distance d from Bus 1 and at the Bus 2 a three fault isoccurred [5]. The relative value are used for convenience:

l = d4

Figure 5: Radial system with Distributed generation connected

3.2 Short circuit analysisThe short-circuit current when fault occurs at bus 3 is smaller than when it occurs

at bus 2, then the maximum fault current ow through CB3 is short-circuit currentwhen fault at bus 2. For that reason only fault at bus 2 is considered. The short circuitcurrent when fault occurs at bus 1 has the same value as the previous analysis (Table6).

When the fault occurs in at the bus 2 then the total fault current I ktot is thecombination of fault current from the grid I kgrid and fault current from DG I kDG :

I ktot = I kgrid + I kDG [A]

The system could be demonstrated by equivalent circuit as follow

Figure 6: Equivalent impedance diagram when fault at bus 2

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In this case, the V DG is assumed equal to V grid then the Thevenin equivalent of thesystem is

Figure 7: Thevenin equivalent of Figure 6

The Thevenin impedance is:

Z th = (Z s + Z 1 + l.Z 2 ).Z g(Z s + Z 1 + l.Z 2 ) + Z g

+ (1 −l).Z 2 [Ω]

Where: Z s Z 1 and Z 2 are calculated by given data (see Sec .2.2).Z g is DG’s impedance and could be found as:

Z g = j.x g .V 2grid

S DG[Ω]

From the equation above, the Z th is function of location l of the DG as well aspower rating of DG (S DG ).

The total 3 φ short-circuit current is calculated by:

I ktot = V th√ 3.Z th

[A]

The grid contributes to short circuit current:

I kgrid = Z g

(Z s + Z 1 + l.Z 2 ) + Z g.I ktot [A]

3.3 Setting of time-dial of relaysThe Tap Setting ratio of CT at every relay unchanged (see Sec. 2.3 and Table 7).The normal-inverse relay has characteristic, which described at Sec. 2.4.With the same process for setting TDS as in the case without DG, the TDS for

relay of each CB are coordinated.Starting with CB3 relay, the maximum fault current ow though this relay is I ktot .

From this data the fault-to-pickup current is found.

I ktot

T S 3

The desired tripping time for CB3 relay is T 3 = 0.05s, then by applying the relay’scharacteristic formula (see Sec. 2.4), the TDS3 is found.

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We want a CB2 relay operating time:

T 2 = T 3 + T breaker + T coordination = 0.05 + 0.1 + 0 .2 = 0.35s

However, the fault current ows though CB2 relay is only I kgrid . The fault-to-pickupcurrent ratio of CB2 relay:I kgrid

T S 2From that ratio and relay operating time, the TDS2 could be found.Next select the TDS for CB1 relay. The largest fault current though CB2 is

6874.95A as calculated before. Then the operating time for CB2 relay T2 with thisfault current could be found from the relay’s characteristic. Operating time of CB1relay is T 1 = T 2 + T breaker + T coordination . From that the TDS1 is determined.

3.4 Simulation of inuences of DGTo demonstrate the inuences of DG on the protection system two parameters of

DG are modied, which is generator size and its location in the system. The locationl will varies from 0 to 1, which represents for DG’s location is connected to bus 1to bus 2. The generator size is increased from 1 MW to 10 MW. The program onsoftware Wolfram Mathematica c 10 - Student Version [ 6] is used to execute iterativecalculations.

Figure 8: Result of TDS setting

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Figure 9: Changing of TDS1 with different location and size of DG



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From Figure .8, the DG has the same effect on CB1 and CB2 relays. Maximuminuence of DG on the system when location of DG l = 0.24, and larger the DG sizethe inuence is stronger. DG will cause the TDS value smaller than normal case upuntil location l larger than 0.8. Then TDS value become larger than normal case. Inthe case of CB3 relay the TDS value increase in all circumstances as Figure. 11.

Figure 12: Coordinated time characteristic of relays in the case l = 0.24 and S DS =0 M W

Figure 13: Coordinated time characteristic of relays in the case l = 0.24 and S DS =10 M W

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4 SOLUTIONS AND CONCLUSION

4 Solutions and ConclusionThe paper deals with coordinating inverse-time overcurrent relay in radial system.

The process coordination is demonstrated in Chapter .2.With the same process, Chapter.3 demonstrate the inuence of Distributed Gener-

ator when it is connected to the system.From the simulation’s results, existence of DG will require changing in relay setting.

The problem could appear when the DG is connected in the system, but the relaysetting is unchanged.

There are two solutions for this problem. First, the DG must be disconnected fromthe system as soon as possible when fault occurs. Second, the relay setting shouldadapt to the status the DG (location and size) on the system.

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REFERENCES

References[1] J Lewis Blackburn and Thomas J Domin. Protective relaying: principles and ap-

plications . CRC press, 2006.

[2] J Duncan Glover, Mulukutla Sarma, and Thomas Overbye. Power System Analysis & Design, SI Version . Cengage Learning, 2011.

[3] AG SIEMENS. Applications for siprotec protective relays–, 2005.

[4] Paul M Anderson. Faulted power systems, 1995.

[5] Edward Jeroen Coster. Distribution grid operation including distributed genera-tion. Eindhoven University of Technology, The Netherlands , 2010.

[6] Wolfram language & system: Documentation center, 2016.

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