Question 4: Protection Systems and Power Economic 4.1 By considering a radial power system shown below, overcurrent IDMT protection relays are to be installed as shown in the Figure. Assume that the fault current of 18.69 kA is taking place and the relay time-setting multiplier ranges from 0.1 to 1 in steps of 0.02 and the relay plug setting ranges from 25% to 250% in steps of 15%. Plug setting multipliers for Relay 1 and Relay 2 are 7 and 15, respectively. For a given fault current, Relay 2 must trip within 0.3 seconds and a grading margin of 0.6 seconds should be allowed between the relays. 1100/5A 1100/5A Relay 1 Relay 2 IFault 41. Determine the plug setting current for relay 2 (A). 42. Determine the plug setting current for relay 1 (A). 43. Determine the plug setting percentage for relay 1 (%). 44. Determine the plug setting percentage for relay 2 (%).

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Question 4: Protection Systems and Power Economics
4.1 By considering a radial power system shown below, overcurrent IDMT protection
relays are to be installed as shown in the Figure. Assume that the fault current of 18.69
kA is taking place and the relay time-setting multiplier ranges from 0.1 to 1 in steps of
0.02 and the relay plug setting ranges from 25% to 250% in steps of 15%. Plug setting
multipliers for Relay 1 and Relay 2 are 7 and 15, respectively. For a given fault current,
Relay 2 must trip within 0.3 seconds and a grading margin of 0.6 seconds should be
allowed between the relays.
1100/5A
1100/5A
Relay 1
Relay 2
IFault
41. Determine the plug setting current for relay 2 (A).
42. Determine the plug setting current for relay 1 (A).
43. Determine the plug setting percentage for relay 1 (%).
44. Determine the plug setting percentage for relay 2 (%).
45. Determine the time-setting multiplier for relay
Datormina the tima aattina muItinlior for rolou A
Transcribed Image Text:Question 4: Protection Systems and Power Economics 4.1 By considering a radial power system shown below, overcurrent IDMT protection relays are to be installed as shown in the Figure. Assume that the fault current of 18.69 kA is taking place and the relay time-setting multiplier ranges from 0.1 to 1 in steps of 0.02 and the relay plug setting ranges from 25% to 250% in steps of 15%. Plug setting multipliers for Relay 1 and Relay 2 are 7 and 15, respectively. For a given fault current, Relay 2 must trip within 0.3 seconds and a grading margin of 0.6 seconds should be allowed between the relays. 1100/5A 1100/5A Relay 1 Relay 2 IFault 41. Determine the plug setting current for relay 2 (A). 42. Determine the plug setting current for relay 1 (A). 43. Determine the plug setting percentage for relay 1 (%). 44. Determine the plug setting percentage for relay 2 (%). 45. Determine the time-setting multiplier for relay Datormina the tima aattina muItinlior for rolou A
4.2 Eskom uses a 1MVA transformer to supply electricity to Willows community. The
electricity cost price is R2.50 per kWh. The load supplied by the transformer varies as
shown below. The purchasing price of the transformer is R350 000. The transformer
has a full load copper loss of 3.83 kW and the iron loss of 6.13 kW.
A certain load varies as follows for 269 days per annum:
550 KVA at a power factor of 0.819 lagging for eleven hours per day.
120 KVA at a power factor of 0.836 lagging for six hours per day.
450 kVA at unity power factor for remaining hours of a day.
For the remaining time of the year, the load varies as follows:
O 290 kVA at a power factor of 0.822 lagging for eight hours per day.
O 25 KVA at a power factor of 0.866 lagging for eight hours per day. O No
load for eight hours per day.
47. Determine the daily copper losses at 450 kVA (kWh)
48. Determine the daily copper losses at 290 kVA (kWh)
49. Determine the annual cost of iron losses (Rands)
50. Determine the average value of the load per annum (kWh).
Transcribed Image Text:4.2 Eskom uses a 1MVA transformer to supply electricity to Willows community. The electricity cost price is R2.50 per kWh. The load supplied by the transformer varies as shown below. The purchasing price of the transformer is R350 000. The transformer has a full load copper loss of 3.83 kW and the iron loss of 6.13 kW. A certain load varies as follows for 269 days per annum: 550 KVA at a power factor of 0.819 lagging for eleven hours per day. 120 KVA at a power factor of 0.836 lagging for six hours per day. 450 kVA at unity power factor for remaining hours of a day. For the remaining time of the year, the load varies as follows: O 290 kVA at a power factor of 0.822 lagging for eight hours per day. O 25 KVA at a power factor of 0.866 lagging for eight hours per day. O No load for eight hours per day. 47. Determine the daily copper losses at 450 kVA (kWh) 48. Determine the daily copper losses at 290 kVA (kWh) 49. Determine the annual cost of iron losses (Rands) 50. Determine the average value of the load per annum (kWh).
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