Answer :
## Question One
### a. Definitions
i. Protection Relay: A protection relay is an electromagnetic device designed to detect faults in an electrical circuit and to initiate the disconnection of the faulty section from the rest of the circuit, thereby protecting the electrical system from damage.
ii. Pick-Up Current of Relay: The pick-up current of a relay is the minimum current at which the relay begins to operate. It's a threshold level, above which the relay senses that there is a fault condition.
iii. PSM (Plug Setting Multiplier): The Plug Setting Multiplier (PSM) is a ratio used in protection relay settings. It is computed as the ratio of the fault current to the relay’s pick-up setting current after considering the current transformer (CT) ratio. It indicates how many times the actual fault current exceeds the pick-up current.
iv. Relay Operating Time: Relay operating time is the time interval from the moment the fault occurs to the moment the protection relay initiates the disconnection of the faulty circuit.
v. Secondary Protection: Secondary protection refers to backup protective measures that operate in case the primary protection fails. It ensures that faults are cleared even if the main protective device does not function as intended.
### b. Problem Solution
Given:
- Relay current rating, [tex]\( I_{\text{relay}} = 5 \, A \)[/tex]
- Relay set percentage, [tex]\( \text{Set} \% = 150 \% \)[/tex]
- Time Multiplier Setting, [tex]\( \text{TMS} = 0.4 \)[/tex]
- Current Transformer (CT) ratio, [tex]\( \frac{400}{5} \)[/tex]
- Fault current, [tex]\( I_{\text{fault}} = 6000 \, A \)[/tex]
Steps:
1. Calculate Pick-Up Current:
The pick-up current is defined as the rated current of the relay multiplied by the set percentage:
[tex]\[ I_{\text{pickup}} = I_{\text{relay}} \times \left( \frac{\text{Set} \%}{100} \right) \][/tex]
[tex]\[ I_{\text{pickup}} = 5 \, A \times \left( \frac{150}{100} \right) \][/tex]
[tex]\[ I_{\text{pickup}} = 7.5 \, A \][/tex]
2. Calculate Plug Setting Multiplier (PSM):
PSM is the ratio of the fault current adjusted for the CT ratio to the pick-up current:
[tex]\[ \text{PSM} = \frac{I_{\text{fault}} / \text{CT ratio}}{I_{\text{pickup}}} \][/tex]
[tex]\[ \text{CT ratio} = \frac{400}{5} = 80 \][/tex]
[tex]\[ \text{PSM} = \frac{6000 \, A / 80}{7.5 \, A} \][/tex]
[tex]\[ \text{PSM} = \frac{75}{7.5} = 10 \][/tex]
3. Determine Operating Time at TMS = 1 from the Table:
From the given table:
[tex]\[ \text{At PSM} = 10, \text{Operating Time at TMS} = 1 \text{ is } 2.8 \text{ seconds} \][/tex]
4. Calculate the Actual Operating Time:
The operating time with the given TMS can be obtained by multiplying the operating time at TMS = 1 with the TMS:
[tex]\[ \text{Operating Time} = \text{Operating Time at TMS} = 1 \times \text{TMS} \][/tex]
[tex]\[ \text{Operating Time} = 2.8 \, \text{seconds} \times 0.4 \][/tex]
[tex]\[ \text{Operating Time} = 1.12 \, \text{seconds} \][/tex]
So, the detailed solution leads us to the values:
- Pick-Up Current: 7.5 A
- PSM: 10
- Operating Time at TMS = 1: 2.8 seconds
- Actual Operating Time: 1.12 seconds
## Question Two
### a. Alternator Stator Faults
Three common alternator stator faults are:
1. Phase-to-Ground Faults: These occur when one of the stator windings makes contact with the ground.
2. Phase-to-Phase Faults: These occur between two different phase windings due to insulation failure.
3. Inter-turn Faults: These occur when there's a short circuit within the turns of a single stator winding.
### b. Neutral Earthed Through a Resistance
Given:
- Alternator voltage, [tex]\( V = 10,000 \, V \)[/tex]
- Resistance, [tex]\( R = 10 \, \Omega \)[/tex]
- Relay set to operate at out-of-balance current of [tex]\( I_{\text{relay}} = 1 \, A \)[/tex]
- CT ratio = 1000 / 5
To determine the percentage of the relay protected against earth faults:
i. Calculate the full load current on the primary side of the CT:
[tex]\[ I_{\text{primary}} = \frac{V}{R} \][/tex]
[tex]\[ I_{\text{primary}} = \frac{10,000 \, V}{10 \, \Omega} = 1000 \, A \][/tex]
ii. Calculate the secondary current of the CT corresponding to the primary full load current:
[tex]\[ I_{\text{secondary}} = \frac{I_{\text{primary}}}{\text{CT ratio}} \][/tex]
[tex]\[ I_{\text{secondary}} = \frac{1000 \, A}{1000/5} = 5 \, A \][/tex]
iii. Determine the relay setting percentage:
The relay is designed to operate at 1 A secondary current. Thus:
[tex]\[ \text{Relay Setting Percentage} = \left( \frac{I_{\text{relay setting}}}{I_{\text{secondary}}} \right) \times 100 \][/tex]
[tex]\[ \text{Relay Setting Percentage} = \left( \frac{1 \, A}{5 \, A} \right) \times 100 = 20\% \][/tex]
Thus, the percentage of the relay protected against earth faults is 20%.
### a. Definitions
i. Protection Relay: A protection relay is an electromagnetic device designed to detect faults in an electrical circuit and to initiate the disconnection of the faulty section from the rest of the circuit, thereby protecting the electrical system from damage.
ii. Pick-Up Current of Relay: The pick-up current of a relay is the minimum current at which the relay begins to operate. It's a threshold level, above which the relay senses that there is a fault condition.
iii. PSM (Plug Setting Multiplier): The Plug Setting Multiplier (PSM) is a ratio used in protection relay settings. It is computed as the ratio of the fault current to the relay’s pick-up setting current after considering the current transformer (CT) ratio. It indicates how many times the actual fault current exceeds the pick-up current.
iv. Relay Operating Time: Relay operating time is the time interval from the moment the fault occurs to the moment the protection relay initiates the disconnection of the faulty circuit.
v. Secondary Protection: Secondary protection refers to backup protective measures that operate in case the primary protection fails. It ensures that faults are cleared even if the main protective device does not function as intended.
### b. Problem Solution
Given:
- Relay current rating, [tex]\( I_{\text{relay}} = 5 \, A \)[/tex]
- Relay set percentage, [tex]\( \text{Set} \% = 150 \% \)[/tex]
- Time Multiplier Setting, [tex]\( \text{TMS} = 0.4 \)[/tex]
- Current Transformer (CT) ratio, [tex]\( \frac{400}{5} \)[/tex]
- Fault current, [tex]\( I_{\text{fault}} = 6000 \, A \)[/tex]
Steps:
1. Calculate Pick-Up Current:
The pick-up current is defined as the rated current of the relay multiplied by the set percentage:
[tex]\[ I_{\text{pickup}} = I_{\text{relay}} \times \left( \frac{\text{Set} \%}{100} \right) \][/tex]
[tex]\[ I_{\text{pickup}} = 5 \, A \times \left( \frac{150}{100} \right) \][/tex]
[tex]\[ I_{\text{pickup}} = 7.5 \, A \][/tex]
2. Calculate Plug Setting Multiplier (PSM):
PSM is the ratio of the fault current adjusted for the CT ratio to the pick-up current:
[tex]\[ \text{PSM} = \frac{I_{\text{fault}} / \text{CT ratio}}{I_{\text{pickup}}} \][/tex]
[tex]\[ \text{CT ratio} = \frac{400}{5} = 80 \][/tex]
[tex]\[ \text{PSM} = \frac{6000 \, A / 80}{7.5 \, A} \][/tex]
[tex]\[ \text{PSM} = \frac{75}{7.5} = 10 \][/tex]
3. Determine Operating Time at TMS = 1 from the Table:
From the given table:
[tex]\[ \text{At PSM} = 10, \text{Operating Time at TMS} = 1 \text{ is } 2.8 \text{ seconds} \][/tex]
4. Calculate the Actual Operating Time:
The operating time with the given TMS can be obtained by multiplying the operating time at TMS = 1 with the TMS:
[tex]\[ \text{Operating Time} = \text{Operating Time at TMS} = 1 \times \text{TMS} \][/tex]
[tex]\[ \text{Operating Time} = 2.8 \, \text{seconds} \times 0.4 \][/tex]
[tex]\[ \text{Operating Time} = 1.12 \, \text{seconds} \][/tex]
So, the detailed solution leads us to the values:
- Pick-Up Current: 7.5 A
- PSM: 10
- Operating Time at TMS = 1: 2.8 seconds
- Actual Operating Time: 1.12 seconds
## Question Two
### a. Alternator Stator Faults
Three common alternator stator faults are:
1. Phase-to-Ground Faults: These occur when one of the stator windings makes contact with the ground.
2. Phase-to-Phase Faults: These occur between two different phase windings due to insulation failure.
3. Inter-turn Faults: These occur when there's a short circuit within the turns of a single stator winding.
### b. Neutral Earthed Through a Resistance
Given:
- Alternator voltage, [tex]\( V = 10,000 \, V \)[/tex]
- Resistance, [tex]\( R = 10 \, \Omega \)[/tex]
- Relay set to operate at out-of-balance current of [tex]\( I_{\text{relay}} = 1 \, A \)[/tex]
- CT ratio = 1000 / 5
To determine the percentage of the relay protected against earth faults:
i. Calculate the full load current on the primary side of the CT:
[tex]\[ I_{\text{primary}} = \frac{V}{R} \][/tex]
[tex]\[ I_{\text{primary}} = \frac{10,000 \, V}{10 \, \Omega} = 1000 \, A \][/tex]
ii. Calculate the secondary current of the CT corresponding to the primary full load current:
[tex]\[ I_{\text{secondary}} = \frac{I_{\text{primary}}}{\text{CT ratio}} \][/tex]
[tex]\[ I_{\text{secondary}} = \frac{1000 \, A}{1000/5} = 5 \, A \][/tex]
iii. Determine the relay setting percentage:
The relay is designed to operate at 1 A secondary current. Thus:
[tex]\[ \text{Relay Setting Percentage} = \left( \frac{I_{\text{relay setting}}}{I_{\text{secondary}}} \right) \times 100 \][/tex]
[tex]\[ \text{Relay Setting Percentage} = \left( \frac{1 \, A}{5 \, A} \right) \times 100 = 20\% \][/tex]
Thus, the percentage of the relay protected against earth faults is 20%.
