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protection - KRISHNA CHAITANYA NUTUKURTHI MVA Load 1 Load 2...

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KRISHNA CHAITANYA NUTUKURTHI MVA KV Load Current(A) Load 1 2.5 13.2 109.3466419 Load 2 3 13.2 131.2159703 Load 3 2.5 13.2 109.3466419 Load 4 2.5 13.2 109.3466419 Load 5 1.5 13.2 65.60798514
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KRISHNA CHAITANYA NUTUKURTHI 1) a) Calculation of load currents. Given the MVA rating of each load is listed in the table below and the voltage of the feeder is given as 13.2KV. The load current is calculated by using the following formula and the results are tabulated =( * ) * I MVA 1000 3 KVA Table 1: Load Currents at each load b) Calculation of line section currents The load current at load 5 will be the line section current flowing in the section 4-5.i.e 65.60798514 A. Now the line current in the section 3-4 will be equal to the sum of the currents flowing in load 4 and load 5. Similarly the line section currents for each section is calculated and tabulated as follows: KV Line Currents(A) Section (S-1) 13.2 524.8638811 Section (1-2) 13.2 415.5172392 Section (2-3) 13.2 284.3012689 Section (3-4) 13.2 174.954627 Section (4-5) 13.2 65.60798514 Table 2: Line section currents 2. First we calculate the voltage drop at each section. This is tabulated below. R Ω/mile X Ω/mile Z Ω/mile No of miles Total Impe dence Section Currents(A) Voltage drop(V)
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KRISHNA CHAITANYA NUTUKURTHI (Ω) Section (S-1) 0.271 0.445 0.52102399 2 0.5 0.260 51199 6 524.863881 136.7333372 Section (1-2) 0.271 0.445 0.52102399 2 0.5 0.260 51199 6 415.517239 108.2472253 Section (2-3) 0.271 0.445 0.52102399 2 1.25 0.651 27999 284.301269 185.1597275 Section (3-4) 0.271 0.445 0.52102399 2 1 0.521 02399 2 174.954627 91.15555814 Section (4-5) 0.271 0.445 0.52102399 2 1.5 0.781 53598 8 65.60798514 51.27500146 Table 3: Voltage drops in each section Now the total Voltage drop is given by the sum of all the voltage drops at each section. Voltage drop = 136.7333372 + 108.2472253 + 185.1597275 + 91.15555814 + 51.27500146 = 572.5708496 V = 0.572 KV Now the voltage at the head of the feeder is given by 13.2 + 0.572 = 13.772KV………………(1) The 5% voltage = 13.2 + 5% of 13.2 = 13.2 + (0.05*13.2) = 13.86KV ……………………………………………………………………. ..(2) Hence from 1 and 2 we can see that the voltage at the head of the feeder is within the 5% of the voltage. 3) Calculation of 3-phase fault currents
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KRISHNA CHAITANYA NUTUKURTHI The 3-phase fault current at each bus is calculated by using MVA method. In this method we calculate the Short Circuit (SC) MVA of every component that is required till the fault bus. The components include generator, transformer, feeder line etc. SC MVA of Source SC MVA of Transformer SC MVA of both Zline SC MVA of line Net MVA of each bus Fault current(kA) Bus S 150 300 100 0 0 100 4.373865676 Bus 1 150 300 100 0.2605 668.8675624 86.9938589 3.804994535 Bus 2 150 300 100 0.2605 668.8675624 76.98153221 3.367068814 Bus 3 150 300 100 0.65 268.0615385 59.80641175 2.615852116 Bus 4 150 300 100 0.521 334.4337812 50.73375262 2.219026192 Bus 5 150 300 100
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protection - KRISHNA CHAITANYA NUTUKURTHI MVA Load 1 Load 2...

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