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    Handout no. 3

    Name Muhammad Waseem

    Reg. No 2010-EE-421

    Marks/Grade

    EXPERIMENT # 3

    Coordination of DTOC Relays in a Power System

    Objective:

    At the end of this lab session students will be able to Understand the concept of Relay Coordination and Selectivity. Calculate the time setting of the relays for relay coordination in a power system. Analyze the effects of absence of relay coordination. Understand the concept of back-up protection.

    Introduction:

    Definite Time Over-Current Relay: A definite time over-current relay operates like an

    instantaneous over-current relay coupled with a timer. Once current reaches the pickup value, it

    initiates the timing circuit. As long as current stays above this pickup value, the timer will continue to

    time. Once the definite time setting is reached, the relay gives trip signal to the circuit breaker.

    Application of DTOC: The DTOC relay is different from the instantaneous over-current

    relay in a way that it waits for a definite time once the fault is detected. The purpose of the time-

    delay setting is to enable relays to coordinate with each other. When a fault occurs in a power

    system, the relay closest to the fault should operate first so that minimum number of customers is

    affected.

    Where there are two or more series protective devices between the fault point and the power

    supply, these devices must be coordinated to insure that the device nearest the fault point will operate

    first. The other upstream devices must be designed to operate in sequence to provide backup

    protection, if any device fails to respond. This is called selective coordination or selectivity.

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    Consider a radial power line having two relays, one at the start of the line and one at the far

    end. The relay at the far end can be set to operate faster than the relay at the start using the delay time

    setting. If there is a fault at the far end of the line, a heavy current flows in the line which is picked

    up by both relays but we can ensure that the relay closest to the fault operates first and the relay at the

    start of the line only operates as a backup protection if the far end relay fails to operate.

    Time over-current protection is mainly applied to achieve discrimination, i.e. selective

    operations for faults on lines and in transformers in radial power systems. But where selectivity is not

    the main objective, a short delay does permit a lower pick-up setting (current-setting) and therefore

    higher sensitivity, if the application requires it.

    Laboratory Task:

    Simulate a small radial power system having a single generating source and multiple loads.

    Determine the normal operating current in various sections of the power system so that the Pickup

    setting for the protection relays can be determined. Model the DTOC relay as in lab session #2.

    Install the DTOC relay model for the protection of the system at two different places in the radial

    system. Determine the time setting of the relays to ensure relay coordination. Simulate this circuit for

    faults at different locations in the system and examine the operation of the relays.

    Procedure:

    Step 1:

    Draw a simple power system having a three phase source, three phase circuit breakers, three

    phase VI measurement units and three phase series RLC loads. Connect the blocks as shown below.

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    Set the properties of each block as mentioned below. Keep all other parameters as it is.

    System Frequency: 50 Hz

    Total Simulation Time: 0.5 s

    Solver: Ode23tb (stiff/TR-BDF2)

    Solver reset Method: Robust

    Three phase source:

    Voltage (Phase to Phase): 11e3 V

    Internal connection: Y grounded

    3 phase short circuit level: 600e6 VA

    Base voltage: 11e3 V

    Three-Phase Breakers:

    Initial status of breakers: closed

    Three-Phase Series RLC Load 1:

    Configuration: Y grounded

    Nominal voltage: 11e3 V

    Active Power (MW): 100e6 W

    Inductive reactive power: 100 VAR

    Capacitive reactive Power: 0 VAR

    Three-Phase Series RLC Load 2:

    Configuration: Y groundedNominal voltage: 11e3 V

    Active Power (MW): 150e6 W

    Inductive reactive power: 100 VAR

    Capacitive reactive Power: 0 VAR

    Constant (C.B Control):

    Value: 1

    Note: The constant block is connected to the C.B control to keep the circuit breaker closed initially. It

    will be replaced by the relay once we have implemented the relay model.

    Step 2:

    Measure the RMS value of current at Scope 1 and Scope 2. Calculate the Pick-up value of

    current for the relays to be installed.

    Current 1 RMS Value = I1 = 4599

    Current 2 RMS Value = I2 =11500

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    Here I1 is sum of currents drawn by the both loads and I2 is current drawn by load2. This

    thing can be observed in the above figure.

    Pick-up Value for Relay 1 = 1.4 x I1 = 6438.6

    Pick-up Value for Relay 2 = 1.4 x I2 = 16100

    Current-setting of relay is also calledPick-up value.

    Step 3:

    Make a Subsystem named DTOC Relay 1. Implement the model of DTOC relay in this

    subsystem as in lab Session 2. Connect this relay block to the circuit breaker. Create a copy of this

    subsystem to connect with the other breaker. Set the pick up and time delay setting in the relay.

    The details of this implementation are given below:-

    Three Phase Fault:Ground fault: Select this option

    External control of fault timing: Select this option

    Initial status of fault: [0 0 0]

    Simulate fault at 0.1s.

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    Current Setting (Pick-up value):

    Set the Current Pickup value of the relays as calculated in Step 2.

    Step 4:

    Set the Time-Delay (time setting) of relay 2 equal to 0.02 seconds. For Coordination the

    time setting of relay 1 is determined by:-

    Time Setting Relay 1 = Time Setting of Relay 2 + C.B operating time + a small constant

    Circuit Breaker interrupts current at zero crossing. So it will take some time to interrupt the

    current after operation. It will be the worst case when breaker operation starts just after the zero

    crossing of current. In this case breaker operating time will be approximately equal to the time taken

    by the half-cycle of the current i.e. half of the time period of current. A small constant delay is

    added to ensure no overlap. Put this constant equal 0.015s.

    Note:

    If the simulation is taking a long time to complete, change the powergui setting from

    continuous to Discrete and set sample time 50e-6. The powergui block is present on the top

    left corner of your main Simulink file. If powergui block is not present in the file, place it from

    Simpower Systems toolbox.

    Step 5:

    Simulate the following fault conditions and describe your results:

    1. A fault in section 2 of the power system with

    Time setting of relay 1 > Time Setting of relay 2

    Results:

    In this condition, as the fault is produced in the zone of Relay 2, so relay 2 will operate and

    isolate the faulty part of the power system.

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    2. A fault in section 2 of the power system with

    Time setting of relay 1 < Time Setting of relay 2

    Results

    If the time Setting of relay 1 is less the that of relay2, then if fault occured then the relay 2

    will not operate, relay 1 will operate and isolate the fault.

    3. A fault in section 1 of the power system with any time setting of relay 2.

    Results:

    In this case when fault occured no current flows through the section 2, so the relay 1 will

    operate according to the time setting or pick up time of the relay. In this case the relay2 have no

    importance

    Observations:

    1. Explain how Relay 1 works as a Back-up Protection for section 2 of the power system

    Their is two type of protection , primary and back up, in our work relay1 is the back up and

    relay2 is the primary relay. Because the time setting of relay1 is greater then that of relay 2, so when

    fault occured the relay 2 operate quicker then the relay 2. So the relay 2 is primary and relay 1 is

    primary.

    2. What is the necessary condition for Relay Coordination in this system?

    If the fault occured in a zone, then the protection in that zone should operate , if due to relay

    fault is not operate then the other zone relay should operate. It is the necessory condition for relay

    coordination

    3. What is the disadvantage of setting the DTOC time-delay to a very high value?

    DTOC relay some interval of time to operate. So if the fault current is very high magnitude it

    can destroy the System before the relay produce trip signal.

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    Waveforms

    Section 1

    Section 2

    DTOCR Circuit