Your operation in division was wrong you cant divide polar form to rectanguLarform. Convert tha polar form to rectangular form then divide bythedinominTor by. j 0.25 simply 1(cos0+j sin0) multiply both numerator and dinominator by -j0.25
hye sir. i want to clarify.For the second example If = net current = 0.3 + j3.48 pu after considering the load current, Io?. without conaidering the load, If -j4 pu?
A 250 MVA, 13.8 kV star connected with the neutral connected to the earth through a reactance, is connected to a 250 MVA, 13.8 kV/220 kV transformer. Primary and secondary windings of the transformer are star connected with neutral point is solidly grounded. Generator reactances are : X” =0.2 pu, X’=0.3 pu, X2=0.2 Xo=0.1 p.u., Xn=0.05 p.u. Positive, negative and zero sequence reactance of the transformer is 11%.A single line to ground fault is developed at the high voltage side of the transformer . Calculate: a) Fault current during sub-transient period b) Voltage at the generator terminals during sub transient period c) Fault current during transient period d) Voltage at the generator terminals during transiwnt period How solve this question
sir,in the last question,we have used Xd" for both prefault and post fault condition (but you said earlier that Xd" is used post fault only).Can you clarify?
I have a long-pending query. 1. For a balanced 3 Phase, positive phase sequence components current in Fig. (or diagram) show the direction of rotation in ANTI-CLOCKWISE direction and mention RYB. But, in moving ANTI-CLOCKWISE direction, sequence to face first IR1 further IB1 and at last IY1 2. Similarly for balanced negative sequence components current in Fig. (or diagram) show the direction of arrow ANTI-CLOCKWISE direction and mention RBY. But, in moving ANTI-CLOCKWISE, sequence to face IR2 further IY2 and at last IB2 It is not understandable. Till now, in every power system book, this is mentioned. For your ready reference, Page 424 from Power System VK Mehta.
at 30:33, you calculate pu of real power (12 Mw) and at 31.19 you are considering it as apparent power.....i think you need to correct your question ...12 Mw should be 12 MVA ..if i m not wrong.
hye sir.at 33:47, you use the formula I= Sconj/ Vconj.Just wondering since we are doing 3 phase calculation should there be a square root of 3 somewhere? Same for LF I saw S= VIconj. No square root of 3.
I used the formula I_pu = conj(S_pu)/conj(V_pu) that means i am calculating everything in per unit. So sqrt(3) is not required. Root 3 will come when you want to convert the pu value to actual value.
thanks sir.at first I thought it was due to the fact balance 3 phase can be analysed using 1 phase circuit. As long as we use pu, no square root of 3 involves?
In Q1 we are getting fault current with negative sign.Does that mean current is flowing from ground to the generator.And if yes how this is possible?? if time permits please reply sir
Thanks for this good lecture first ! I have a quetion at 33:15 when you caclulate the line current I, for this 3-phase system it is supposed that V is the line-to-line voltage, therefore S=sqrt(3)*V*I_conj. So when we calculate the complex current, I should = S_conj/(sqrt(3)*V_conj).....not S_conj/V_conj
sir at 43.41 u said that we have taken bus volatge at node B u have said to consider the potential at 1.0 pu and the overall potential differece will come out to be zero volts..i do agree with that sir...but why havent we considered the potential at bus b when we were calculating for yV2 ..please explain sir..?
Superposition theory states that any voltage source must be considered individually (which he has done) and whilst doing so, all voltage sources should be replaced by a short circuit, considering explicitly only one source at a time and adding up the individual contributions. Therefore it is in fact xV2 that in my opinion has not been calculated correctly. Pre fault Voltage at bus b should be short circuit for purposes of superposition analysis, exactly as was done when he calculated YV2.
I didn't understand the way you transformed load power into PU! I think you assumed it as MVA not MW! Please, I'll appreciate your clarifying Sir! Thanks for your magnificent examples and explanations!
thank you so much may allah protect you please make exemples for unbalanced fault analysis please i need it in my project please respect from your algeria student
Simple, clear and amazing.! Keep making them good videos. They truly help a lot of ppl
is it an sign error when you add the 0.3+i0.52-i10.23 = 0.3+j9.71 or I am missing some concept ?
sir in problem 1 the impedance of the generator need to be neglected right , why cannot we calculate the line reactance taking into account
Your operation in division was wrong you cant divide polar form to rectanguLarform. Convert tha polar form to rectangular form then divide bythedinominTor by. j 0.25 simply 1(cos0+j sin0) multiply both numerator and dinominator by -j0.25
Thank you Sir. You lectures have rescued me and my career. You're much more than just being Awesome :)
hye sir. i want to clarify.For the second example If = net current = 0.3 + j3.48 pu after considering the load current, Io?.
without conaidering the load, If -j4 pu?
Sir, why you kept the bus voltage on the generator side 1
A 250 MVA, 13.8 kV star connected with the neutral connected to the earth through a reactance, is
connected to a 250 MVA, 13.8 kV/220 kV transformer. Primary and secondary windings of the
transformer are star connected with neutral point is solidly grounded. Generator reactances are : X”
=0.2 pu, X’=0.3 pu, X2=0.2 Xo=0.1 p.u., Xn=0.05 p.u. Positive, negative and zero sequence reactance of
the transformer is 11%.A single line to ground fault is developed at the high voltage side of the
transformer . Calculate:
a) Fault current during sub-transient period
b) Voltage at the generator terminals during sub transient period
c) Fault current during transient period
d) Voltage at the generator terminals during transiwnt period
How solve this question
sir,in the last question,we have used Xd" for both prefault and post fault condition (but you said earlier that Xd" is used post fault only).Can you clarify?
I have a long-pending query.
1. For a balanced 3 Phase, positive phase sequence components current in Fig. (or diagram) show the direction of rotation in ANTI-CLOCKWISE direction and mention RYB. But, in moving ANTI-CLOCKWISE direction, sequence to face first IR1 further IB1 and at last IY1
2. Similarly for balanced negative sequence components current in Fig. (or diagram) show the direction of arrow ANTI-CLOCKWISE direction and mention RBY. But, in moving ANTI-CLOCKWISE, sequence to face IR2 further IY2 and at last IB2
It is not understandable.
Till now, in every power system book, this is mentioned. For your ready reference, Page 424 from Power System VK Mehta.
Thank you Sir, greetings from Ankara,Turkey
at 30:33, you calculate pu of real power (12 Mw) and at 31.19 you are considering it as apparent power.....i think you need to correct your question ...12 Mw should be 12 MVA ..if i m not wrong.
I thought the same thing.
Diwash Bhandari exactly it must be ratio of 24 MVA/20MVA = 1.2 pu for load then further we will solve
yes it has to MVA/MVA
yes me too find the same mistake
I think ur right.... What he got was p.u of apparent power I think
Could you explain why symmetrical fault
L-L-L-G and L-L-L is same?
hye sir.at 33:47, you use the formula I= Sconj/ Vconj.Just wondering since we are doing 3 phase calculation should there be a square root of 3 somewhere? Same for LF I saw S= VIconj. No square root of 3.
I used the formula I_pu = conj(S_pu)/conj(V_pu) that means i am calculating everything in per unit. So sqrt(3) is not required. Root 3 will come when you want to convert the pu value to actual value.
thanks sir.at first I thought it was due to the fact balance 3 phase can be analysed using 1 phase circuit. As long as we use pu, no square root of 3 involves?
see this th-cam.com/video/BGor7OKJSyw/w-d-xo.html
tqsm for the video
at 30:00, isnt P= 12 MW? why have we taken it to be S?
In Q1 we are getting fault current with negative sign.Does that mean current is flowing from ground to the generator.And if yes how this is possible??
if time permits please reply sir
Thanks for this good lecture first ! I have a quetion at 33:15 when you caclulate the line current I, for this 3-phase system it is supposed that V is the line-to-line voltage, therefore S=sqrt(3)*V*I_conj. So when we calculate the complex current, I should = S_conj/(sqrt(3)*V_conj).....not S_conj/V_conj
Hye Sir.
To clarify; if load is considered we use superposition method and if load is not considered we use normal thevenin theorem?
Sir just want to clarify,the load power is in MW and if we divide with Sbase we will obtain the real power right???
Yes
sir at 22.30 why j0.1 is included after bus 1 , but actually it has to be before bus1 am i right sir ??
again I made a mistake, bus 1 should be after j0.1.
sir but that will hardly effect the calculation as the fault to be found out is at bus2 or at the load end..(load here is 0) ?
sir ,why we neglect the effect of saliency during fault analysis??
At 47:02 why you neglect the left side ?
Very nice explanation. Highly appreciated
at 48.26 at net current angle 2.79 is not considered why sir?
You are write... I made a mistake in the video. that 2.79 degrees should be considered. In fact I made 2 mistakes at end.
at 36 min of lecture .can u please explain the flow of current and reactive power
sir at 43.41 u said that we have taken bus volatge at node B u have said to consider the potential at 1.0 pu and the overall potential differece will come out to be zero volts..i do agree with that sir...but why havent we considered the potential at bus b when we were calculating for yV2 ..please explain sir..?
Superposition theory states that any voltage source must be considered individually (which he has done) and whilst doing so, all voltage sources should be replaced by a short circuit, considering explicitly only one source at a time and adding up the individual contributions. Therefore it is in fact xV2 that in my opinion has not been calculated correctly. Pre fault Voltage at bus b should be short circuit for purposes of superposition analysis, exactly as was done when he calculated YV2.
*all other voltage sources
Sir can you make a video for unbalanced fault
I mean for Unsymmetrical fault analysis??
I didn't understand the way you transformed load power into PU! I think you assumed it as MVA not MW! Please, I'll appreciate your clarifying Sir! Thanks for your magnificent examples and explanations!
do you have an idea for the reason why he did that?
Hi Pradeep, I like your lecture
When calculating voltage drop why did you do this... I*j0.15 ...i got lost there
thank you sir i understand
my report next week is about fault analysis
thank you so much may allah protect you please make exemples for unbalanced fault analysis please i need it in my project please respect from your algeria student
Thanks for your help
Thank and it is so usefull for me
i saw some of ur tutorial but u doing wrong calculation again and again... first see some tutorial how to do calculation
this video us nothing but helpfull
new vids...nice
@47:43 error
thannks
Thank u sir
thanks