Short Transmission Line Problem

kduff70 · Jul 17, 2017

Hi, can anybody help me figure out this short transmission line problem. I've attach the problem but I don't understand how the equation was able to solve for Vr can some please held me understand how Vr was calculated? thank you for any help on this problem

Kduff70

View attachment short transmission.pdf

rg1 · Jul 17, 2017

kduff70 said:
Hi, can anybody help me figure out this short transmission line problem. I've attach the problem but I don't understand how the equation was able to solve for Vr can some please held me understand how Vr was calculated? thank you for any help on this problem

Kduff70

View attachment 9868

I am giving my prima facie input may be it gives you a clue. Solve the problem by KVL.

You try solving it by assuming the receiving end voltage as Vr with angle zero i.e. Vr<0 and then you get Ir in terms of Vr , from the load given. Calculate Voltage drop in the line= Ir*Z ( in short line there is no capacitance so Ir=Is). Equate IVsI=IVr+Voltage dropI. Equate the magnitudes of the RHS and LHS. You have a quadratic equation in one variable that is Vr. Solve for it. you get the answer. I will try to do it during the day.

kduff70 · Jul 17, 2017

Thank you Rg1

I'm still confused perhaps if I can see the steps I would probably get it

rg1 · Jul 17, 2017

kduff70 said:
Thank you Rg1

I'm still confused perhaps if I can see the steps I would probably get it

Given Vs (line)=3.3kV (magnitude only), Line Z = 0.3+j0.4, Pr=300kw at .8pf per phase; Vr ,Ir?

Let Vr (phase)=Vr<0 so Ir=300/(0.8*Vr)<-36.86, {Reason- V*I*cosphi=P}

Voltage drop= Ir*Z line= {300/(0.8Vr)<-36.86}*{0.3+0.4j}= 300/(.8Vr)*0.5<(-36.86+53.13)=187.5/Vr<16.27=180/Vr+j52.5/Vr

IVsI=IVr+Voltage dropI= 3.3**2/3=(180/Vr)**2+(52.5/Vr)**2

Can you do the maths further? also check for corrections in the my maths? I am directly doing it on my computer, may make mistakes. Steps are correct, I am sure you must get the answer.

kduff70 · Jul 17, 2017

rg1

I appreciate your help but I stuck at right after you do the voltage drop I don't see where you have 300/(.8Vr)*.5<(-36.86+53.13)

I don't see how you come up with 187.5/Vr<16.27= 180/Vr+j52,5/Vr

rg1 · Jul 17, 2017

kduff70 said:
rg1

I appreciate your help but I stuck at right after you do the voltage drop I don't see where you have 300/(.8Vr)*.5<(-36.86+53.13)

I don't see how you come up with 187.5/Vr<16.27= 180/Vr+j52,5/Vr

Vpoltage drop =I*Z; I P/Vcos phi with angle -inverse cos 0.8 & Z <theta is given . This is only multiplication of phasors

kduff70 · Jul 17, 2017

rg1

I follow you up to the point now where you have 187.5/Vr<16.27=180/Vr+j52.5/Vr but I calculated it to be 227/Vr<16.24 then 218/Vr<16.24=j63.5/Vr the last steps Im not sure after that

rg1 · Jul 17, 2017

rg1 said:
kduff70 said:

rg1

I follow you up to the point now where you have 187.5/Vr<16.27=180/Vr+j52.5/Vr but I calculated it to be 227/Vr<16.24 then 218/Vr<16.24=j63.5/Vr the last steps Im not sure after that

Click to expand...

May be my maths wrong. I am not doing it on paper. Then you add Vr to this Voltage drop and equate the magnitude of both sides. You must get the answer. I will take up further in the evening.

IVsI=IVr+Voltage dropI= 3.3**2/3=Real part square+j part square

rg1 · Jul 17, 2017

Here is the complete solution.

Given Vs (line)=3.3kV (magnitude only), Line Z = 0.3+j0.4, Pr=300kw at .8pf per phase; Vr ,Ir?

Let Vr (phase)=Vr<0 in kV so Ir=300/(0.8*Vr)<-36.86, {Reason- V*I*cosphi=P}

Voltage drop in V= Ir*Z line= {300/(0.8Vr)<-36.86}*{0.3+0.4j}= 300/(.8Vr)*0.5<(-36.86+53.13)=187.5/Vr<16.27=180/Vr+j52.5/Vr ( This is in Volts)

IVsI=IVr+Voltage dropI

Vs=3300/sqrt3=1905V

I1905<aI=1000Vr<0+ 180/Vr+j 52.5/Vr ( Vr is in kV so multiply by 1000 to make it V)

= (1000Vr+180/Vr)+j52.5/Vr

Now equate the magnitudes on both sides of the eq.

1905**2=(1000Vr+180/Vr)**2+52.5/Vr**2=1000000Vr**2+32400/Vr**2+360000+2756.25/Vr**2

3629025=1000000Vr**2+35156/Vr**2+360000

1000000Vr**4-3269025Vr**2+35156.25=0=Vr**4-3.27Vr**2+.035

If Vr**2=X

X**2-3.27X+.035=0

This gives X=3.25 which Gives Vr=1.805 kV or 1805V.

And Current will be 300/1.8/.8=207.85A

Hope it is okay.

Actually this you can understand by drawing phasors and equating them. I am not able to post the diagram I have drawn. I could do earlier but recently I am not able to post any file here. So sorry.

rg1 · Jul 17, 2017

This is good question but as it takes a lot of time to solve, I do not know if it can be asked in PE, experts can let us know. However same question can be made simpler in calculations by changing the figures. Then It becomes a very good question. The concept here is very simple. Apply KVL and equate the magnitudes when stuck to solve the equation involving phasors.

TNPE · Jul 17, 2017

No, you shouldn't see a question like this on the PE, not framed this way, anyway. It definitely takes too much time for the scope of this test. As rg1 alluded to, it is a concept that you most certainly will see in some aspect. That said, best way to get an idea of what topics will be on the exam, look at the outline provided by NCEES prior to the session you plan to sit for.

kduff70 · Jul 18, 2017

Thank you for all the help with this problem. I understand it now and your right this is definitely too long to be a PE problem this takes more then 6 minutes . I'm just trying to find and do all the problem I can, so I can be prepare for what ever type of transmission problem they my give me. The NCEES provide a few problem but I trying to find different problems that are manipulated to find different information they may ask base on the outline and the practice exam problems they provide.

Thank you very much

TNPE · Jul 18, 2017

Now that you've solved and understand this problem, drill down into concepts surrounding these types of problems. This test is heavily weighted on concepts rather than finding an equation to plug and chug. It is a certainty that you will see problems that will require no mathematical solutions at all. This is a good "conceptual" topic, and so are motors and generators. The concepts surrounding each of these topics are almost boundless.

cos90 · Jul 19, 2017

TNPE said:
Now that you've solved and understand this problem, drill down into concepts surrounding these types of problems. This test is heavily weighted on concepts rather than finding an equation to plug and chug. It is a certainty that you will see problems that will require no mathematical solutions at all. This is a good "conceptual" topic, and so are motors and generators. The concepts surrounding each of these topics are almost boundless.

All the sample tests seem to disagree with you. This is a voltage drop problem. That is an autotransformer problem. Is this a bad way to prepare for the exam? The only way to grasp the concepts is to continue drilling, in my opinion.

TNPE · Jul 19, 2017

cos90 said:
All the sample tests seem to disagree with you. This is a voltage drop problem. That is an autotransformer problem. Is this a bad way to prepare for the exam? The only way to grasp the concepts is to continue drilling, in my opinion.

I wasn't implying that you should study concepts only. If I came off as such, my apologies. I'm merely throwing ideas out at what topics are heavy on concepts. We all know circuits, but do we understand what affects voltage, current and power delivered with complex systems?

For instance, synchronous machines and how they operate. Stay with me. A synchronous machine's speed is dictated by the supply frequency, therefore, it is the most viable machine when constant speed is necessitated. Suppose a synchronous generator is supplying a large transmission system, where voltage fluctuations occur perpetually. How is this regulated so as to maintain as constant a terminal voltage as possible? Also, speed/torque characteristics are highly conceptual.

This is only a single scenario, but yes, they are boundless with these machine types, induction motors and transmission lines. I can't say too much, for obvious reasons, but think about what happens on a transmission line when you change the power angle. What about sending end? Receiving end? Both? What happens with MVAR and MW when the angle is changed on either end?

Hopefully these suggestions steer your studying habits, but by no means, do not disregard mathematical/numerical applications.

TNPE · Jul 19, 2017

Also, under normal operation for each respective machine, what are the differences between an induction and synchronous machine? Why are there any differences? Why would we use one machine for this application, but use that machine for another one?

I'm posing these scenarios to expand your understanding and to help groom your studying in prepping for this exam. It is not a game of "Where's Waldo", lets plug these numbers in and determine if his sweater is actually red/white striped....

kduff70 · Jul 20, 2017

Good Advice

I'm in the line of thinking I have to have the basic concept down for each topic then used that as guideline to approach problems so that I don't veer of what a problem might be asking me to solve.

thank you all

speedysc2 · Jul 22, 2017

kduff70 said:
Hi, can anybody help me figure out this short transmission line problem. I've attach the problem but I don't understand how the equation was able to solve for Vr can some please held me understand how Vr was calculated? thank you for any help on this problem

Kduff70

View attachment 9868

@kduff70 Can you post Figure 4-11 & 4-12 from this book or let us know what book this problem is in? Thanks.

kduff70 · Jul 27, 2017

it was from a Schaum's outline book Electric Powers Systems

Short Transmission Line Problem

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