Homework 05

# Homework 05 - EE 3410 Homework#5 Solution Calculate...

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Unformatted text preview: EE 3410, Homework #5. Solution Calculate parameters of a 20 kVA, 480/120 Vtransformer equivalent model recalculated to the primary side, if the short circuit voltage is U I“ = 7% of U 1,, short-circuit power P,C = 2,5 % of Sn open circuit current, I ,0 = 4 % of I], and the open circuit power P0 = 1.5% of S,. Calculate transformer supply voltage U, needed to provide voltage U; = 120 V to the load if the transformer is loaded with the power a)S;=P;=20kW b) s, = 20 e’“ kVA c) s) = 20 e1“ [WA For each case draw the diagram of complex RMS values of voltages and currents AAAAAAMAAMAAAAAMAAAAMAAAAAAAAM Rated primary current RMS value: = i _ 20- 103 U1, ‘ 480 Parameters of the equivalent circuit recalculated to primary side: Open circuit admittance = 1,0 = o.o4><11r _o.04x41.7_ . -3 1'10 U—lr ——U1r _———7480 _3.510 s I1r =41.7A Magnetizing circuit conductance: 3 GFC=Lg=EOJ¥L=w= 1.3-10‘3 S Ulr U1: 430 Magnetiz'mg circuit susceptancc: 3m: — Y1?)- 0%6 =—«/3.51—1.32 x10‘3=—3.2-10'3s _U15c=0.07xU1t_0.07x430_ 21“ lr 1” ‘ 41.7—“181‘2 Short circuit impedance: Series resistance: 3 R1+R2_Pszc =A0.0252xS, = 0.025ng 10 :02” 11r 11, 41.7 Series reactance: X11+X'2,=‘lZI2N-(R1+R;)2 = 0.312 —o.292 =0176Q Hence R,=R'2 =0.14Q, X1,=X;, =0.3SQ, Lu= 1:21=%=%7i78=1.0-10’3=1.0mI-I RFe=GFLc=769Q Lm=- Eligwam The supply voltage U1 calculation, assuming that (1'2 = 480 e190“ v At the load power S; EL = 20 kW Since s2 = U'2(1'2)‘, 2 1'2 = = L2. = = 41.7e19°°A (02) (02) '1 48° Complex RMS value E1 of the transformer intemal voltage e: E] = U; + (123+ J'X;,)I'2 = j480 +(0.14+j0.38)xj4l.7 = 486.1 e9‘-9° v Complex RMS value 110 of the transformer open circuit (idle) current: 110 = (GFe+ij) E1 = (1.3 -j3.2)-10‘3x 486.129”, =1.7e24-°"A Complex RMS value of the primary current, i] 11: 110 + 1'2 =1.7e24:°°+ 41.7e19°"= 42.4e1'37-9° A Complex RMS value U1 of the transformer primary voltage u]: U1: 151+ (121+ jX11)Il= 486.1 e91-9°+(0. 14 + j0.38) x42.4e1'37-9° = 493.4 em" v At the load power sZ = 20 a“ kVA Complex RMS value E1 of t he transformer internal voltage 2: E1: U; + (125+ ng,)1’2= j480 +(0.14 + j0.38) x41.7ef45° = 495.4 e903” V Complex RMS value I m of the transformer open circuit (idle) current: 110 = (GFe+ij) E1 = (1.3 -j3.2)-10‘3 x 495.4 e903" = 1.7.9229" A Complex RMS value of the primary current, i1 11: 110 + 15:1.7e22~9°+ 41.7e1'45° = 43.3ef44-1” A Complex RMS value U1 of the transformer primary voltage 141'. U1: El + (121+ jX1,)Il= 495.4 e9°~3"+ (0.14 + j0.38) x43.3ef44-1° = 511.6 e91-6" v At the load ﬂower s_2 = 20 e-I'“ kVA . . 3145“ . 3145“ . a 12: 32 =w__=w_=41,7e1135A I t (02) (U‘J —J480 Complex RMS value E1 of the transformer internal voltage 2: E1: U'2 + (162+ ng,)I'2 = j480+(0.14 + j0.38)x41.7ef”5" = 473.2 e9119°v Complex RMS value 110 of the transformer open circuit (idle) current: 110 = (GFc+ij) E1=(1.3 .1'3.2)-10'3 x 473.2 e91-9" =1.6e24-°°A Complex RMS value of the primary current, i1 11: I10 + ['2 =1.6224'0u+ 4171:1135" = 41.1ej132'9uA Complex RMS value U1 of the transformer primary voltage ulr U1: El + (121+ jX1,)Il= 473.2 e91-9"+ (0.14 + j0.38) x41. 1eJ‘132-9" = 467.6 e933" v R Load Symbols 11, — Rated primary current U1, — Rated primary voltage S, — Rated power Ylo — Idle admittance 110 — Idle primary current Po ~ Idle active power GR * Equivalent conductance of the core Brn — Magnetizing susceptance U m — Short circuit primary voltage lec — Short circuit primary impedance R '2 — Recalculated secondary winding resistance R1 — Primary winding resistance X ‘2, — Recalculated secondary winding reactance X1; — Primary winding reactance ’2, — Recalculated secondary winding inductance L1, - Primary winding inductance R1:c — Equivalent resistance of the core Lm — Magnetizing inductance S2 — Complex apparent power of the load 0‘; - Recalculated secondary voltage CRMS 1’2 — Recalculated secondary current CRMS value U1 — Primary voltage CRMS value E, — Internel voltage CRMS value of the transformer [10 — Idle current CRMS value [1 — Primary current CRMS value ...
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Homework 05 - EE 3410 Homework#5 Solution Calculate...

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