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EE 255 - Chapter 2a

# EE 255 - Chapter 2a - 23 Chapter 2 Transformers 2-1 The...

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Unformatted text preview: 23 Chapter 2 : Transformers 2-1. The secondary winding of a transformer has a terminal voltage of ( ) 282.8 sin 377 V s v t t = . The turns ratio of the transformer is 100:200 ( a = 0.50). If the secondary current of the transformer is ( ) ( ) 7.07 sin 377 36.87 A s i t t =- ° , what is the primary current of this transformer? What are its voltage regulation and efficiency? The impedances of this transformer referred to the primary side are eq 0.20 R = Ω 300 C R = Ω eq 0.750 X = Ω 80 M X = Ω S OLUTION The equivalent circuit of this transformer is shown below. (Since no particular equivalent circuit was specified, we are using the approximate equivalent circuit referred to the primary side.) The secondary voltage and current are 282.8 0 V 200 0 V 2 S = ∠ ° = ∠ ° V 7.07 36.87 A 5 -36.87 A 2 S = ∠ - ° = ∠ ° I The secondary voltage referred to the primary side is 100 0 V S S a ′ = = ∠ ° V V The secondary current referred to the primary side is 10 36.87 A S S a ′ = = ∠ - ° I I The primary circuit voltage is given by ( ) eq eq P S S R jX ′ ′ = + + V V I ( )( ) 100 0 V 10 36.87 A 0.20 0.750 106.2 2.6 V P j = ∠ ° + ∠ - ° Ω + Ω = ∠ ° V The excitation current of this transformer is EX 106.2 2.6 V 106.2 2.6 V 0.354 2.6 1.328 87.4 300 80 C M j ∠ ° ∠ ° = + = + = ∠ ° + ∠ - ° Ω Ω I I I EX 1.37 72.5 A = ∠ - ° I 24 Therefore, the total primary current of this transformer is EX 10 36.87 1.37 72.5 11.1 41.0 A P S ′ = + = ∠ - ° + ∠ - ° = ∠ - ° I I I The voltage regulation of the transformer at this load is 106.2 100 VR 100% 100% 6.2% 100 P S S V aV aV-- = × = × = The input power to this transformer is ( )( ) ( ) IN cos 106.2 V 11.1 A cos 2.6 41.0 P P P V I θ = = ( )( ) IN 106.2 V 11.1 A cos 43.6 854 W P = ° = The output power from this transformer is ( )( ) ( ) OUT cos 200 V 5 A cos 36.87 800 W S S P V I θ = = ° = Therefore, the transformer’s efficiency is OUT IN 800 W 100% 100% 93.7% 854 W P P η = × = × = 2-2. A 20-kVA 8000/480-V distribution transformer has the following resistances and reactances: Ω = 32 P R Ω = 05 . S R Ω = 45 P X 0.06 S X = Ω Ω = k 250 C R Ω = k 30 M X The excitation branch impedances are given referred to the high-voltage side of the transformer. (a) Find the equivalent circuit of this transformer referred to the high-voltage side. (b) Find the per-unit equivalent circuit of this transformer. (c) Assume that this transformer is supplying rated load at 480 V and 0.8 PF lagging. What is this transformer’s input voltage? What is its voltage regulation? (d) What is the transformer’s efficiency under the conditions of part (c) ? S OLUTION (a) The turns ratio of this transformer is a = 8000/480 = 16.67. Therefore, the secondary impedances referred to the primary side are ( ) ( ) 2 2 16.67 0.05 13.9 S S R a R ′ = = Ω = Ω ( ) ( ) 2 2 16.67 0.06 16.7 S S X a X ′ = = Ω = Ω 25 The resulting equivalent circuit is 32 Ω 250 k Ω j 45 Ω j 30 k Ω j 16.7 Ω 13.9 13....
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EE 255 - Chapter 2a - 23 Chapter 2 Transformers 2-1 The...

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