Mathcad - 02-18

Mathcad - 02-18 - kVAR kW kVA kW define P2-18 A...

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|S| = 261 kVA P = 250 kW Q = 75 kVAR I ph 181.255A = I ph I L 3 = the phase current is the line current divided by the square root of three I L 313.943A = I L S 3ph 3 V L = b. and c.: It is actually easier to find the line current first F p 0.958 = F p P 3ph S 3ph = so the power factor is: P 3ph 250kW = P 3ph Re S 3ph ( 29 = the real power is: S 3ph 261.008 kVA = the apparent power is: V L 480 V = S 3ph 250 j 75 + ( ) kVA = given a. The power factor can be found since the load is specified in kW and kVARs. Several possible methods can be used Solution
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Unformatted text preview: kVAR kW ≡ kVA kW ≡ define P2-18. A three-phase load consists of three, identical impedances connected in delta. The total three phase load is (250 +j 75)KVA. The line voltage is 480 volts. a. Find the power factor of the circuit. b. Find the magnitude of the phase current. c. Find the magnitude of the line current. d. Draw the power triangle for this load and label the values for Real, Reactive, and Apparent power....
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