Question for the Week (Power PE)

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

spinup

Well-known member
Joined
Dec 22, 2011
Messages
145
Reaction score
16
The next exam is on April 5th, 2019. There are 15 weeks left. Each week until the exam a question will be posted on www.spinupexams.com/

Previous weeks will also be contained on the website.

Week 1 Question For The Week can be found at:

www.spinupexams.com/

Under the "Question For The Week" tab.

 
View Question


4004.png

The above 3 phase system has a transmission line with impedance and distance stated below (Z and D). The voltage source is 480V ∠0°, and the load is drawing 75A with a 0.85 power factor lagging. What is the voltage at the load end?
Z = 0.1 + j0.02 per 1,000 ft.
D = 1750 ft
 


c.png
A. 457V


B. 264V


C. 269V


x.png
D. 467V


 


Solution


Vload = vsource - (Iload)(Zload/line)
= ( (480∠-30° / √3) - (75∠-cos-10.85)(0.1 + j0.02)(1750 / 1000) )√3
=( (277∠-30°) - (75∠-31.8°)(0.1∠11.3°)(1750/1000) )√3
=( 277∠-30° - 13.4∠-20.5° )√3
=( 264∠-30.5° )√3
=457V

Can someone tell me why you would divide by √3 for Zload /line?  I thought impedance is usually given per phase and there is no need to multiply/divide 3 phase impedance  by √3 like we do to 3 phase delta/wye current and voltages. 

 
Last edited by a moderator:
Week_1_QFTW_Question_Solution.png


Why isn't VAN not equal 277 angle -30? The magnitude of answer would be same but angle would be different.  In case of Wye, Phase voltage lags line voltages by 30 degrees. 

The funny thing is, if you take 277<0  and multiply that with answer provided in B, ( S= VI *) then you do get 25 KVA,0.75  lagging. 24.9K<41.4 😡

Only inference I draw is, in practice this situation won't exist. Something has to change, either source has to be 277<-30,  or load information is not correct. 

 
Last edited by a moderator:
View Question


4004.png

The above 3 phase system has a transmission line with impedance and distance stated below (Z and D). The voltage source is 480V ∠0°, and the load is drawing 75A with a 0.85 power factor lagging. What is the voltage at the load end?
Z = 0.1 + j0.02 per 1,000 ft.
D = 1750 ft
 


c.png
A. 457V


B. 264V


C. 269V


x.png
D. 467V


 


Solution


Vload = vsource - (Iload)(Zload/line)
= ( (480∠-30° / √3) - (75∠-cos-10.85)(0.1 + j0.02)(1750 / 1000) )√3
=( (277∠-30°) - (75∠-31.8°)(0.1∠11.3°)(1750/1000) )√3
=( 277∠-30° - 13.4∠-20.5° )√3
=( 264∠-30.5° )√3
=457V

Can someone tell me why you would divide by √3 for Zload /line?  I thought impedance is usually given per phase and there is no need to multiply/divide 3 phase impedance  by √3 like we do to 3 phase delta/wye current and voltages. 
Vload = vsource - (Iload)(Zload/line)
= ( (480∠-30° / √3) - (75∠-cos-10.85)(0.1 + j0.02)(1750 / 1000) )) ---> (L-N)
= 263.8 <-30.48 ---> (L-N)
= √3 x 263.8 <(-30.48+30)---> (L-L)

=457<-0.48V---> (L-L)

 √3 there in the solution is just conversion from L-N to L-L.

 
Last edited by a moderator:
Week_1_QFTW_Question_Solution.png


Why isn't VAN not equal 277 angle -30? The magnitude of answer would be same but angle would be different.  In case of Wye, Phase voltage lags line voltages by 30 degrees. 

The funny thing is, if you take 277<0  and multiply that with answer provided in B, ( S= VI *) then you do get 25 KVA,0.75  lagging. 24.9K<41.4 😡

Only inference I draw is, in practice this situation won't exist. Something has to change, either source has to be 277<-30,  or load information is not correct. 
I agree with you. VAN shall be 277<-30. These kind of things we have to be careful during calculation. 

 
=457<-0.48V---> (L-L)

Yes that solution is correct. I was confused because they hadn't given whether source/load is wye/delta. Since you are first converting everything in per phase and then back to 3 phase. You still have to use -30, +30 phase shift. 

 
Last edited by a moderator:
Week 12 Question For The Week posted at:

www.spinupexams.com/

Under the "Question For The Week" tab.

Previous weeks are also contained on the website. 
Good little problem, I calculated it wrong (35.15% )by using angle. That goes to show there are so many ways one can be tipped even with simple problems! I have never come across a problem which gave an angle and asked to find voltage regulation.  

 
Wouldn't it be wonderful if all the questions were this easy. More protection questions is what I need.

 
Good little problem, I calculated it wrong (35.15% )by using angle. That goes to show there are so many ways one can be tipped even with simple problems! I have never come across a problem which gave an angle and asked to find voltage regulation.  
I saw similar question in one of the Complex Imag test. I did the same thing of including the angle & got the wrong answer.

 
Wouldn't it be wonderful if all the questions were this easy. More protection questions is what I need.
I agree. Not sure what % of questions on the actual exam are of this level.

 
I saw similar question in one of the Complex Imag test. I did the same thing of including the angle & got the wrong answer.


Good little problem, I calculated it wrong (35.15% )by using angle. That goes to show there are so many ways one can be tipped even with simple problems! I have never come across a problem which gave an angle and asked to find voltage regulation.  
For Voltage Regulation just remember it's the magnitude of the full load and no load voltage so you would not use the angle.

73fa1f88e0e93af18f22194273bf06da0484827b


 
Back
Top