Final_Solution_AU04

Final_Solution_AU04 - 1 EE740 Final, Au04 Problem 1 (25...

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Unformatted text preview: 1 EE740 Final, Au04 Problem 1 (25 points) Consider a 3- distribution feeder as shown below: Load #1 is rated 133.3 kVar with power factor of 0.0 leading. Load #2 is rated 100 kW with power factor of 0.6 lagging. 1) Calculate the power factor at the load bus. (15 points) 2) Describe the real power flow direction in the system. (5 points) 3) Describe the reactive power flow direction in the system. (5 points) Solution: 1) Load #1: P 1 = 0, Q 1 = -133.33 MVar. Load #2: P 2 = 100 kW, cos 2 = 0.6, 3 . 133 ) 6 . sin(cos 6 . 100 1 2 = = Q kVar tan tan 1 2 1 2 1 1 = = + + = P P Q Q pf = cos0 = 1 2) Real power flows from source to Load 2. 3) Reactive power flows from Load 1 to Bus and then to Load 2. Problem 2 (45 points) Consider the power system given below: Given: Z G(1) = Z G(2) = j 0.2 p.u., Z G(0) = j 0.1 p.u. Z AB(1) = Z AB(2) = j 0.3 p.u., Z AB(0) = j 0.8 p.u. Z BC(1) = Z BC(2) = j 0.3 p.u., Z BC(0) = j 0.8 p.u....
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This note was uploaded on 10/28/2009 for the course ECE 740 taught by Professor Keyhani during the Spring '07 term at Ohio State.

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Final_Solution_AU04 - 1 EE740 Final, Au04 Problem 1 (25...

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