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Question for the Week (Power PE)

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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:

Under the "Question For The Week" tab.

• 1

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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

 A. 457V B. 264V C. 269V D. 467V

Solution

= ( (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.

Edited by roy167

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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.

Edited by roy167

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18 hours ago, roy167 said:

View Question

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

 A. 457V B. 264V C. 269V D. 467V

Solution

= ( (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.

= ( (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.

Edited by Sdhabik

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17 hours ago, roy167 said:

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.

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=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.

Edited by roy167

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Week 2 Question For The Week can be found at:

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Week 3 Question For The Week﻿ posted at:

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Week 4 Question For The Week﻿ posted at:

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Week 5 Question For The Week﻿ posted﻿:

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Week 6 Question For The Week﻿ posted﻿:

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Two months until the exam.

Week 7 Question For The Week﻿ posted at:

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Week 8 Question For The Week﻿ posted at:

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Question below:

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Week 9 Question For The Week﻿ posted at:

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