0.6 Ngắn mạch không đối xứng (Page 5/7)

Page 5 / 7

$[\begin{matrix} {\dot{U}}_{N0} \\ {\dot{U}}_{NA 1} \\ {\dot{U}}_{NA 2} \end{matrix}] = \frac{1}{3} [\begin{matrix} \overset{}{1} & 1 & 1 \\ \overset{}{1} & a & a^{2} \\ \overset{}{1} & a^{2} & a \end{matrix}] [\begin{matrix} {\dot{U}}_{NA} \\ {\dot{U}}_{NB} \\ {\dot{U}}_{NB} \end{matrix}] \Rightarrow {\dot{U}}_{NA 1} = {\dot{U}}_{NA 2} (7.7)$

$[\begin{matrix} {\dot{I}}_{N0} \\ {\dot{I}}_{NA 1} \\ {\dot{I}}_{NA 2} \end{matrix}] = \frac{1}{3} [\begin{matrix} \overset{}{1} & 1 & 1 \\ \overset{}{1} & a & a^{2} \\ \overset{}{1} & a^{2} & a \end{matrix}] [\begin{matrix} \overset{}{0} \\ {\dot{I}}_{NB} \\ - {\dot{I}}_{NB} \end{matrix}] \Rightarrow \begin{matrix} {\dot{I}}_{N0} = 0 (7.8) \\ {\dot{I}}_{NA 1} =- {\dot{I}}_{NA 2} (7.9) \end{matrix}$

Giải các phương trình từ (7.1) đến (7.9) ta có:

${\dot{E}}_{AS} - j {\dot{I}}_{NA 1} . X_{1S} = 0 - j {\dot{I}}_{NA 2} . X_{2S} = 0 + j {\dot{I}}_{NA 1} . X_{2S}$

Như vậy:

${\dot{I}}_{NA 1} = \frac{{\dot{E}}_{AS}}{j (X_{1S} + X_{2S})}$

***SORRY, THIS MEDIA TYPE IS NOT SUPPORTED.***

Hình 7.9

Ngắn mạch 1 pha:

Xét ngắn mạch 1 pha ở pha A (hình 7.10). Điều kiện ngắn mạch là: $\begin{matrix} {\dot{I}}_{NB} = 0 (7.10) \\ {\dot{I}}_{NC} = 0 (7.11) \\ {\dot{U}}_{NA} = 0 (7.12) \end{matrix}$ Thay vào phương trình thứ tự dòng:	*SORRY, THIS MEDIA TYPE IS NOT SUPPORTED.* Hình 7.10

$[\begin{matrix} {\dot{I}}_{N0} \\ {\dot{I}}_{NA 1} \\ {\dot{I}}_{NA 2} \end{matrix}] = \frac{1}{3} [\begin{matrix} \overset{}{1} & 1 & 1 \\ \overset{}{1} & a & a^{2} \\ \overset{}{1} & a^{2} & a \end{matrix}] [\begin{matrix} {\dot{I}}_{NA} \\ \overset{}{0} \\ \overset{}{0} \end{matrix}] \Rightarrow {\dot{I}}_{N0} = {\dot{I}}_{NA 1} = {\dot{I}}_{NA 2} = \frac{1}{3} {\dot{I}}_{NA} (7.13)$

Từ phương trình thứ tự áp ta có:

${\dot{U}}_{NA} = {\dot{U}}_{NA 1} + {\dot{U}}_{NA 2} + {\dot{U}}_{N0} = 0$

Và từ các phương trình cơ bản (7.1)  (7.3) ta có:

${\dot{E}}_{AS} - j {\dot{I}}_{NA 1} (X_{1S} + X_{2S} + X_{0S}) = 0$

Như vậy: ${\dot{I}}_{NA 1} = \frac{{\dot{E}}_{AS}}{j (X_{1S} + X_{2S} + X_{0S})}$

$\begin{matrix} {\dot{U}}_{N0} =- {jX}_{0S} {\dot{I}}_{N0} =- {jX}_{0S} {\dot{I}}_{NA 1}; {\dot{U}}_{NA 2} =- {jX}_{2S} {\dot{I}}_{NA 2} =-j X_{2S} {\dot{I}}_{NA 1} \\ {\dot{U}}_{NA 1} =- ({\dot{U}}_{N0} + {\dot{U}}_{NA 2}) = j {\dot{I}}_{NA 1} (X_{0S} + X_{2S}) \end{matrix}$

Dòng tại chỗ ngắn mạch, cũng là dòng đi qua đất IĐ:

${\dot{I}}_{NA} = {\dot{I}}_{å} = 3 {\dot{I}}_{NA 1}$

Áp tại chỗ ngắn mạch:

$\begin{matrix} {\dot{U}}_{NB} = {\dot{U}}_{N0} + a^{2} {\dot{U}}_{NA 1} + a {\dot{U}}_{NA 2} = j [(a^{2} - a) X_{2S} + (a^{2} - 1) X_{0S}] {\dot{I}}_{NA 1} \\ = \sqrt{3} {\dot{I}}_{NA 1} (X_{2S} - {aX}_{0S}) = \sqrt{3} {\dot{I}}_{NA 1} X_{2S} (1 - a \frac{X_{0S}}{X_{2S}}) \end{matrix}$

$\begin{matrix} {\dot{U}}_{NC} = {\dot{U}}_{N0} + a {\dot{U}}_{NA 1} + a^{2} {\dot{U}}_{NA 2} = j [(a - a^{2}) X_{2S} + (a - 1) X_{0S}] {\dot{I}}_{NA 1} \\ =- \sqrt{3} {\dot{I}}_{NA 1} (X_{2S} - a^{2} X_{0S}) =- \sqrt{3} {\dot{I}}_{NA 1} X_{2S} (1 - a^{2} \frac{X_{0S}}{X_{2S}}) \end{matrix}$

***SORRY, THIS MEDIA TYPE IS NOT SUPPORTED.***

Hình 7.11

Ngắn mạch 2 pha chạm đất:

Xét ngắn mạch 2 pha B, C chạm đất (hình 7.12). Điều kiện ngắn mạch là: $\begin{matrix} {\dot{I}}_{NA} = 0 (7.14) \\ {\dot{U}}_{NB} = 0 (7.15) \\ {\dot{U}}_{NC} = 0 (7.16) \end{matrix}$ Thay vào phương trình thứ tự áp:	*SORRY, THIS MEDIA TYPE IS NOT SUPPORTED.* Hình 7.12

$[\begin{matrix} {\dot{U}}_{N0} \\ {\dot{U}}_{NA 1} \\ {\dot{U}}_{NA 2} \end{matrix}] = \frac{1}{3} [\begin{matrix} \overset{}{1} & 1 & 1 \\ \overset{}{1} & a & a^{2} \\ \overset{}{1} & a^{2} & a \end{matrix}] [\begin{matrix} {\dot{U}}_{NA} \\ \overset{}{0} \\ \overset{}{0} \end{matrix}] \Rightarrow {\dot{U}}_{N0} = {\dot{U}}_{NA 1} = {\dot{U}}_{NA 2} = \frac{1}{3} {\dot{U}}_{NA} (7.17)$

Từ (7.14) ta có: ${\dot{I}}_{NA} = {\dot{I}}_{NA 1} + {\dot{I}}_{NA 2} + {\dot{I}}_{N0} = 0$

Và từ các phương trình cơ bản (7.1)  (7.3) ta có:

${jX}_{2S} {\dot{I}}_{NA 2} = {jX}_{0S} {\dot{I}}_{N0}$

Như vậy: ${\dot{I}}_{N0} =- {\dot{I}}_{NA 1} (\frac{X_{2S}}{X_{0S} + X_{2S}}); {\dot{I}}_{NA 2} =- {\dot{I}}_{NA 1} (\frac{X_{0S}}{X_{0S} + X_{2S}})$

Từ các phương trình cơ bản và (7.17) ta có:

${\dot{U}}_{NA 1} = {\dot{E}}_{AS} - j {\dot{I}}_{NA 1} X_{1S} = {\dot{U}}_{N0} =- j {\dot{I}}_{N0} X_{0S} = j {\dot{I}}_{NA 1} (\frac{X_{2S}}{X_{0S} + X_{2S}}) X_{0S}$

Do đó: ${\dot{I}}_{NA 1} = \frac{{\dot{E}}_{AS}}{j (X_{1S} + \frac{X_{2S} X_{0S}}{X_{2S} + X_{0S}})}$

Dòng tại chỗ ngắn mạch:

${\dot{I}}_{NB} = (a^{2} - \frac{X_{2S} + {aX}_{0S}}{X_{2S} + X_{0S}}) {\dot{I}}_{NA 1}; {\dot{I}}_{NC} = (a - \frac{X_{2S} + a^{2} X_{0S}}{X_{2S} + X_{0S}}) {\dot{I}}_{NA 1}$

Dòng đi qua đất IĐ là: ${\dot{I}}_{å} = 3 {\dot{I}}_{N0} =- 3 {\dot{I}}_{NA 1} \frac{X_{2S}}{X_{0S} + X_{2S}}$

Áp tại điểm ngắn mạch: ${\dot{U}}_{NA} = 3 {\dot{U}}_{NA 1} = 3j {\dot{I}}_{NA 1} \frac{X_{2S} X_{0S}}{X_{0S} + X_{2S}}$

***SORRY, THIS MEDIA TYPE IS NOT SUPPORTED.***

Hình 7.13

Bảng 7.3: TÓM TẮT BIỂU THỨC ĐỐI VỚI CÁC DẠNG NGẮN MẠCH

Dạng NM	Dòng	Áp
N(2)	$\begin{matrix} {\dot{I}}_{NA 1} = \frac{{\dot{E}}_{AS}}{j (X_{1S} + X_{2S})} \\ {\dot{I}}_{NA 2} =- {\dot{I}}_{NA 1}; {\dot{I}}_{N0} = 0 \end{matrix}$	$\begin{matrix} {\dot{U}}_{NA 1} = {jX}_{2S} {\dot{I}}_{NA 1} \\ {\dot{U}}_{NA 2} = {\dot{U}}_{NA 1}; {\dot{U}}_{N0} = 0 \end{matrix}$
N(1)	$\begin{matrix} {\dot{I}}_{NA 1} = \frac{{\dot{E}}_{AS}}{j (X_{1S} + X_{2S} + X_{0S})} \\ {\dot{I}}_{NA 2} = {\dot{I}}_{N0} = {\dot{I}}_{NA 1} \end{matrix}$	$\begin{matrix} {\dot{U}}_{NA 1} = j (X_{2S} + X_{0S}) {\dot{I}}_{NA 1} \\ {\dot{U}}_{NA 1} + {\dot{U}}_{NA 2} + {\dot{U}}_{N0} = 0 \end{matrix}$
N(1,1)	$\begin{matrix} {\dot{I}}_{NA 1} = \frac{{\dot{E}}_{AS}}{j (X_{1S} + \frac{X_{2S} X_{0S}}{X_{2S} + X_{0S}})} \\ {\dot{I}}_{NA 1} + {\dot{I}}_{NA 2} + {\dot{I}}_{N0} = 0 \end{matrix}$	$\begin{matrix} {\dot{U}}_{NA 1} = j {\dot{I}}_{NA 1} (\frac{X_{2S} X_{0S}}{X_{0S} + X_{2S}}) \\ {\dot{U}}_{NA 2} = {\dot{U}}_{N0} = {\dot{U}}_{NA 1} \end{matrix}$

Qui tắc đẳ̉ng trị thứ tự thuận:

Qua bảng 7.3 thấy rằng các thành phần đối xứng của dòng và áp tỷ lệ với dòng thứ tự thuận ở chỗ ngắn mạch, do vậy nhiệm vụ tính toán một dạng ngắn mạch không đối xứng bất kỳ trước hết là tìm dòng thứ tự thuận ở chỗ ngắn mạch. Để tính toán người ta đưa ra qui tắc đẳng trị thứ tự thuận như sau:

“ Dòng thứ tự thuận của một dạng ngắn mạch không đối xứng bất kỳ được tính như là dòng ngắn mạch 3 pha ở một điểm xa hơn điểm ngắn mạch thực sự một điện kháng phụ X(n). Trị số của X(n) không phụ thuộc vào tham số của sơ đồ thứ tự thuận mà chỉ phụ thuộc vào X2 và Xo.”

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?

Aislinn Reply

tijani

what is titration

John Reply

what is physics

Siyaka Reply

A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance

Jude Reply

Can you compute that for me. Ty

Jude

what is the dimension formula of energy?

David Reply

what is viscosity?

David

what is inorganic

emma Reply

what is chemistry

Youesf Reply

what is inorganic

emma

Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes

Adjei

please, I'm a physics student and I need help in physics

Adjanou

chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.

Pedro

A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity

Krampah Reply

2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.

Sahid Reply

you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s

Samuel Reply

can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?

Joseph Reply

Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time

Joseph

Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing

Joseph

"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true

Ryan

what's motion

Maurice Reply

what are the types of wave

Maurice

answer

Magreth

progressive wave

Magreth

hello friend how are you

Muhammad Reply

fine, how about you?

Mohammed

Mujahid

A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?

yasuo Reply

Who can show me the full solution in this problem?

Reofrir Reply

Got questions? Join the online conversation and get instant answers!

Jobilize.com Reply

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Giáo trình ngắt mạch trong hệ thống điện. OpenStax CNX. Jul 30, 2009 Download for free at http://cnx.org/content/col10820/1.1

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Giáo trình ngắt mạch trong hệ thống điện' conversation and receive update notifications?

Ask

	34 Biology 34 Animal Nutrition Digestive System By OpenStax Start Quiz
©flickr: Eduardo	Nursing Education NU589 Program Outcomes By Karen Gowdey Start Quiz
	1 Psychology Concept Test By John Gabrieli Start Test
	Gastrointestinal Pathophysiology 2006 By Tamsin Knox Start Exam
	7 Arts Society: Theater 7 By Jonathan Long Start Quiz
©flickr:	Biology 1 By Jill Zerressen Start Quiz
	17 Sociology 17 Government and Politics MCQ By OpenStax Start Quiz
	1 Microbiology Final 1 By Madison Christian Start Quiz
	21 AP 21 Lymphatic Immune System Essay By OpenStax Start Flashcards
	3 Arts Society: Theater 3 By Jonathan Long Start Quiz