# Exercise_2 - ΑΣΚΗΣΗ 2 ∆ίνεται γραµµή...

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Unformatted text preview: ΑΣΚΗΣΗ 2 ∆ίνεται γραµµή µεταφοράς, 50 Hz, µε γενικευµένες σταθερές B = 50∠64o Ω A = D = 0,977∠0,6o C = 0,001∠90,2o Ω−1 Ζητούνται: α) Η ανύψωση τάσης εν κενώ, όταν η τάση αναχώρησης είναι 150 kV. (10%) β) Η τιµή της εγκάρσιας αντιστάθµισης ώστε η ανύψωση της τάσης εν κενώ να γίνει µηδέν. (60%) γ) Η τάση στην άφιξη της γραµµής και η αντίστοιχη διακύµανση τάσεως (µε και χωρίς αντιστάθµιση), όταν στο άκρο άφιξης συνδεθεί φορτίο Ζ=16×Β/Α. (30%) MVA := 1000kW MW := MVA f := 50 ⋅ Hz MVAr := MVA ω := 2πf ω = 314.159 Hz MVA := 1000kW MW := MVA f := 50 ⋅ Hz MVAr := MVA ω := 2πf ω = 314.159 Hz j⋅0.6deg j⋅64deg A := 0.977 ⋅ e j⋅90.2deg C := 0.001 ⋅ e Z := 16 ⋅ B A B := 50 ⋅ e ⋅S ⋅Ω D := A Z = 818.833 Ω arg ( Z) = 63.4 deg B = 0.061059151904 + 0.000639433318i Z B⎞ = 0.6 deg ⎝Z⎠ arg ⎛ ⎜ B = 0.061 Z A = 0.977 + 0.01i C = −3.491 × 10 −6 B = 21.919 + 44.94i Ω −4 + 10i × 10 S D = 0.977 + 0.01i A = 0.977 arg ( A) = 0.6 deg B = 50 Ω arg ( B) = 64 deg C = 1 × 10 −3 arg ( C) = 90.2 deg S D = 0.977 arg ( D) = 0.6 deg Ερώτηµα (α) j⋅0 ES := 150 ⋅ ( kV) ⋅ e ES = 86.603 kV 3 ES ER := (phase voltage) ER = 88.636 − 0.928i kV A arg ( ER) = −0.6 deg ER = 88.641 kV Ερώτηµα (β) V ohm IR := 0 ⋅ IR = 0 V ohm Ax := Re ( A) Ax = 0.977 Ay := Im ( A) Ay = 0.01 Bx := Re ( B) Bx = 21.919 Ω By := Im ( B) By = 44.94 Ω k1 := ( B )2 ⋅ ( ES )2 k1 = 1.875 × 10 4 8 12 13 6 kg m s A 13 k1 = 1.875 × 10 ( Ω ⋅ V) 2 ( PROBLEM) k2 := 2 ⋅ ( ES ) 2 ⋅ ( Bx ⋅ A y − By ⋅ A x ) 3 6 kg m 11 k2 = −6.552 × 10 9 s A 4 11 2 A )2 − ( k2 = −6.552 × 10 Ω ⋅ V k3 := ( )2 ⋅ ( ES 8 k3 = −3.41 × 10 2 ES )2 4 kg m 6 s A 8 k3 = −3.41 × 10 V 2 2 2 k2 + k2 − 4 ⋅ k1 ⋅ k3 XL1 := 2 ⋅ k3 XL1 = −28.204 Ω 2 k2 − k2 − 4 ⋅ k1 ⋅ k3 XL2 := 2 ⋅ k3 L1 := L2 := XL1 3 XL2 = 1.949 × 10 Ω L1 = −0.09 H ω XL2 L2 = 6.205 H ω 1 ⎞⎤ ES := ⎡A + ⎛ −B ⋅ j ⋅ ⋅ ES + B ⋅ IR ⎢ ⎜ XL1 ⎠⎥ ⎣ ⎝ ⎦ A + ⎛ −B ⋅ j ⋅ ⎜ ⎝ ES = 86.603 kV arg ( ES) = 128.058 deg 1 ⎞ = −0.616 + 0.787i XL1 ⎠ A + ⎛ −B ⋅ j ⋅ ⎜ ⎝ (per circuit & per phase) 1 ⎞ =1 XL1 ⎠ arg ⎡A + ⎛ −B ⋅ j ⋅ ⎢ ⎜ ⎣ ⎝ 1 ⎞⎤ ⎥ = 128.058 deg XL1 ⎠⎦ 1 ⎞⎤ ES := ⎡A + ⎛ −B ⋅ j ⋅ ⋅ ES + B ⋅ IR ⎢ ⎜ XL2 ⎠⎥ ⎣ ⎝ ⎦ A + ⎛ −B ⋅ j ⋅ ⎜ ⎝ arg ( ES) = −0.058 deg 1 ⎞ −3 = 1 − 1.013i × 10 XL2 ⎠ A + ⎛ −B ⋅ j ⋅ ⎜ ⎝ ES = 86.603 kV arg ⎡A + ⎛ −B ⋅ j ⋅ ⎢ ⎜ 1 ⎞ =1 XL2 ⎠ ⎣ ⎝ 1 ⎞⎤ = −0.058 deg XL2 ⎠⎥ ⎦ −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− k1 := ( A )2 − 1 k1 = −0.045471 k2 := 2 ⋅ ( Ax ⋅ By − Ay ⋅ Bx)k2 = 87.359 Ω k3 := ( B ) 2 3 k3 = 2.5 × 10 2 4 kg m 6 s A 4 2 −k2 + k2 − 4 ⋅ k1 ⋅ k3 XL1 := XL1 = −28.204 Ω 2 ⋅ k1 2 −k2 − k2 − 4 ⋅ k1 ⋅ k3 3 XL2 := XL2 = 1.949 × 10 Ω 2 ⋅ k1 L1 := L2 := XL1 ω XL2 ω 1 ⎞⎤ ES := ⎡A + ⎛ −B ⋅ j ⋅ ⋅ ES ⎢ ⎜ XL1 ⎠⎥ ⎣ ⎝ ⎦ 1 ⎞⎤ ES := ⎡A + ⎛ −B ⋅ j ⋅ ⋅ ES ⎢ ⎜ XL2 ⎠⎥ ⎣ ⎝ ⎦ L1 = −0.09 H L2 = 6.205 H (per circuit & per phase) ES = 86.603 kV arg ( ES) = 128.058 deg ES = 86.603 kV arg ( ES) = −0.058 deg Ερώτηµα (γ) j⋅0 ES := 150 ⋅ ( kV) ⋅ e 3 ER1 := ES = 86.603 kV ES ⎡A + ⎛ −B ⋅ j ⋅ 1 ⎞ + ⎢ ⎜ XL2 ⎠ ⎣ ⎝ ER1 = 81.619 kV arg ( ER1) = 0.02 deg ( ER − ER1 ER1 ER1 := ) ⋅ 100 = 8.604 ES ⎛A + B ⎞ ⎜ Z⎠ ⎝ ER1 = 83.427 kV arg ( ER1) = −0.6 deg ( ER − ER1 ER1 ) ⋅ 100 = 6.25 B⎤ Z⎥ ⎦ (phase voltage) ...
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## This note was uploaded on 10/02/2009 for the course G 001 taught by Professor Shmmygr during the Spring '07 term at National Technical University of Athens, Athens.

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