EE_387_200809SP_homework_04_solution

# EE_387_200809SP_homework_04_solution - 1 a In Homework 3...

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1 Homework 04 Solution © Jeffrey Mayer 2005 – 2009. All rights reserved. EE 387 Spring 2009 In Homework 3 Problem 1, μ r = 1000; changing to r = 1100 from the linear approximation to the B-H saturation curve on Slide 59 yields: 0 7 1T 0.86 kA/m 41 0 H / m 925 r B H π = × = a. Changing to r = 925 yields: Data from the B-H saturation curve on Slide 59 b. 6 6 1000 (1.203 10 ) 1100 1.094 10 c ℜ= × 36 (10 10 )(1.094 10 ) 104.6 105 turns c NL =ℜ × = 6 6 1000 (1.203 10 ) 925 1.300 10 c × (10 10 )(1.300 10 ) 114.0 114 turns c × = B H = 0 r B H μμ = 1. Î

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2 Homework 04 Solution © Jeffrey Mayer 2005 – 2009. All rights reserved. EE 387 Spring 2009 2 230 0 V V =∠ ° ± 1 ~ V + + 0.3 j 0.65 0.003 j 0.0065 2400:240 ideal 2 I ± 1 I ± j 1 k 4.5 k 75 kW 0.85 lagging 22 2 tan( ) 75tan(31.7 ) 46.48 kvar QP ϕ = = * 2 2 2 * 75 46.48 kVA 230 326.1 202.1A 383.6 31.78 A S I V j j ⎛⎞ = ⎜⎟ ⎝⎠ + = =− −° ± ± 2 75 46.48 kVA SPj Q j =+ 1 1 cos ( ) cos (0.85) 31.7 pf = = 2. 2 75 kW P = 2 1 2 2 2400 (0.003) 240 0.3 N rr N ′ = = 2 1 2 2 2400 (0.0065) 240 0.65 ll N XX N ′ = = 2 1 240 383.6 31.78 A 2400 38.36 31.78 A N II N ′ = = ∠− ° ±± 10 230 0 2300 0 V Va V ′ = = ⋅∠ ° = ∠°
3 Homework 04 Solution © Jeffrey Mayer 2005 – 2009.

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EE_387_200809SP_homework_04_solution - 1 a In Homework 3...

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