3-amaliy mashg’ulot Kuchlanishi 6-10 kV li kabel liniyalarni ximoyasini xisoblash

3-amaliy mashg’ulot
Kuchlanishi 6-10 kV li kabel liniyalarni ximoyasini xisoblash

Elektr prinsipial sxemaga mos ekvivalent almashtirish sxemasini tuzib, taqsimlash qurilmalari shinalarida qabul qilingan va xisoblanishi shart bulgan K1 – K9 nuqtalarda qisqa tutashish toklarini belgilaymiz. (2-rasm)

Elektr ta’minoti sistemasining maksimal va minimal ish rejimda qisqa tutashish toklarining qiymatini aniqlaymiz.
T1 va T2 transformatorlar parallel ishlaganda maksimal qisqa tutashish toklari xosil buladigan rejim xisoblanadi. (2-rasm) K1 nuqta uchun S = 500 MVA qabul qilingan.
Minimal ish rejimidagi qisqa tutashish toki bitta transformator ishlaganda xosil buladigan ta’minlovchi podstansiya yuqori kuchlanish tomonidagi minimal Tula quvvat S _{k. t. k.} = 400 MVA.

a) sistemaning maksimal ish rejimida (T1 va T2 parallel ishlaydi, S _{k. t. maks.} = 500 MVA)

Sistema qarshiligini quyidagicha xisoblaymiz:

X_sist = = = 0,8 va Z_c = j 0.8

Manba berilgan joyda qisqa tutashish nuqtasigacha bulgan qarshiliklar yigindisi:

Z _e = Z_c = j 0.8

Qisqa tutashish nuqtasida bazis kuchlanishdagi tok quyidagi formula yordamida aniqlanadi:


I _k1 = = = 2510 A
K-2 nuqta ( 2 ta transformator ishlaganda)

Z_E2 = Z_E1 + Z_T1/2 = j 0.8 + 0.12 + j 2.096 = 0.12 + j 2.896 = /2.898/

I _k2 = = = 2154 A


K-3 nuqta

Z_E3 = Z_E2 + Z_L1/2 = 0.12 + j 2.898 + 0,863 + j 0.65 = 0.983 +j 3.546 = /3.68/

I _k3 = = = 1696 A
K – 4 nuqta

Z_E4 = Z_E3 + Z_L3 = 0.983 + j 3.546 + 1.81 + j 1.012 = 2.793 + j 4.566 = /5.352/

I _k4 = = = 1166 A


K – 5 nuqta

Z_E5 = Z_E3 + Z_L4 = 2.793 + j 4.566 + 2.813 + j 16 = 5.606 + j 20. 566 = /21.316/

I _k5 = = = 27085 A


K – 6 nuqta

Z_E6 = Z_E3 + Z_L3 = 0.983 + j 3.546 + 0.84 + j 7.45 = 1.823 + j 10.996 = /11.146/

I _k6 = = = 1973 A


K – 7(8) nuqta

Z _{E7 (8)} = Z_E6 + Z _{L4 (5)} = 1.823 + j 10.996 + 7.1 + j 0.758 = 8.923 + j 11.754 = /14.757/

I _{k7 (8)} = = = 1490 A


K – 9 nuqta

Z _E9= Z_E8 + Z _L6= 8.923 + j 11.754 + 5.849 + j 0.467 = 14.772 + j 12.22 = /19.172/

I _k9 = = = 1147 A

b) sistema ishlarini minimal rejim ( bir transformator ishlaganda S _{k. t. k.} = 400 MVA)

Sistema qarshiligi
X_s = = = 1 Z_c = j 1
K – 1 nuqta

Z_E =Z_c = j 1

I _k1 = = = 2008 A


K – 2 nuqta
(bitta transformator ishlaydi)

Z _E2= Z_E1 + Z _T1= j 1 + 0.24 + j 4.192 = 0.24 + j 4.192 = /5.198/

I _k2 = = = 1201 A


K – 3 nuqta

Z _E3= Z_E2 + Z _K1/2 = 0.24 + j 5.192 + 0.863 + j 0.65 = 1.103 + j 5.842 = /5.945/

I _k3 = = = 1050 A


K – 4 nuqta

Z _E4= Z_E3 + Z _L1= 1.0103 + j 5.842 + 1.81 + j 1.012 = 2.913 + j 6.854 = /7.447/

I _k4 = = = 838 A


K – 5 nuqta

Z _E5= Z_E4 + Z _T4= 2.913 + j 6.854 + 2.813 + j 16 = 5.726 + j 22.854 = /23.56/

I _k5 = = = 24505 A


K – 6 nuqta

Z _E6= Z_E3 + Z _T3= 1.103 + j 5.842 + 0.84 + j 7.45 = 1.943 + j 13.292 = /13.433/

I _k6 = = = 1637 A


K – 7 (8) nuqta

Z _{E7 (8)} = Z_E6 + Z _L4(5)= 1.943 + j 13.292 + j 0.758 = 9.043 + j 14.05 = /16.709/

I _{k7 (8)} = = = 1316 A


K – 9 nuqta

Z _E9 = Z_E8 + Z _L6= 9.43 + j 14.05 + 5.849 + j 0.467 = 14.892 + j 14.517 = /20.797/

I _k9 = = = 1058 A

Ikki fazali qisqa tutashish tokini quyidagi formula buyicha aniqlaymiz:

I²_kt = _*I ⁽³⁾ _k
Xisoblangan qisqa tutashish toklar natijasini A jadvalga yozamiz.