Power Systems | Tutorial - 7 | Solved Problems on Short Transmission Lines (Part-2)

[@ragaen3698]

Indian teachers are awesome. Thanks for all Sir.

[@EngineeringDevotion]

All the very best!!

[@thechannel7072]

thank you so much practicing these numericals have helped me so much to learn very appreciating work god bless you and your family

[@EngineeringDevotion]

Wonderful!

[@msh-1877]

Sir i have question at 3:47 why we dont dived by 3 to get the line current as we do on example three on the previous video , thanks

[@EngineeringDevotion]

Your question, however, is unclear to me. I just want to point out that the division or multiplication of 3 and the square root of 3 are affected by a variety of factors such as whether the system is single or three phase, whether the connection is star or delta, and so on. Please be more specific with your query.

[@revanthsibiram3342]

Sir at 3:27 By solving the equation we get two values of Vr why are we taking 18435

[@EngineeringDevotion]

The other root is 616 V. Since Vs is 19052 V, Vr should be less and close to it. Taking 616 V means there is a huge loss in the TL, which will not the case. In TL, resistance is very less, the governing parameter is the inductance. So, 18435 V is chosen.

[@nkumar2065]

Receiving end power/ sending end power krke efficiency nhi nikl skta hai kya sir 🙏🏾🙏🏾🙏🏼

[@EngineeringDevotion]

Yes, you can but need to know the losses. Otherwise, Pinput = Poutput + losses.

[@dogaceyhan4471]

Sir, I couldn't understand the solution of problem 7. There is load current part of the solution which is P=3*Vp*Ip*cos(theta) like you mentioned. But I couldn't see the cos(theta) in the equation of load current. Can you explain? Thank you.

[@EngineeringDevotion]

I think you are confused with real power (P) and apparent power (S). Note: P=3*Vp*Ip*cos(theta) and S = 3*Vp*Ip. Use accordingly. Hope it helps !!

[@levikekman5979]

Sir i have a doubt , In the equation we took Vr as reference . But why did we equate the magnitude of Vs to Vr + Iz and not the vector ? because if Vr is the reference then Vs should be a vector right ?

[@EngineeringDevotion]

Please mention problem number to identify your doubt.

[@cadrawdaily]

@@EngineeringDevotion I have the same doubt it's on solution 6 equation for find I at 6:12 please reply ASAP

[@EngineeringDevotion]

@@cadrawdaily Please see the Lecture-17 of the power system lecture playlist to understand the complete equation. As in solution it is mentioned approximate expression which means second term is not taken into consideration as its value will not impact the solution and can be easily solved.

[@katemuthoniwairimu1901]

Isn't that the phase current? Because Power/phase=Vphase*Iphase*PF Please clarify

[@EngineeringDevotion]

Yes, you are correct, it is phase current to be specific. Generally, any balanced 3-phase system is solved per-phase basis so for 1-phase we can say line current = phase current. After solving per phase basis we can back transformed the voltage, current and losses as per 3-phase system equations. In the solution, I should have written phase current it is a typo. Thanks for pointing out.

[@dogaceyhan4471]

Sir, I calculated Vs in the solution of problem 7 from the equation which is Vs= Vr+ I*Z and I found 7320.673 . Is it correct?

[@EngineeringDevotion]

Hope you have taken Z = R+jX with appropriate magnitude and angle. The solution mentioned is approximate Vs.

[@parmarriddhi4272]

Ans of current is single phase current or not?😢