Another Complex Imaginary Question (Variation 3 #7)

[vdubEE]

I have been working back through the CI exams I have to make sure I follow along with all the details and also double check their answers since they seem to have quite a few errors.

On problem #7, I am not sure I agree with their method on obtaining the solution. See the image below for the problem, their solution method, and then mine. While the solutions are pretty close to each other, it could very well make the difference on the NCEES exam if the two solution methods produce two different answer selections.

 

[Insaf]

I have been working back through the CI exams I have to make sure I follow along with all the details and also double check their answers since they seem to have quite a few errors.

On problem #7, I am not sure I agree with their method on obtaining the solution. See the image below for the problem, their solution method, and then mine. While the solutions are pretty close to each other, it could very well make the difference on the NCEES exam if the two solution methods produce two different answer selections.

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Looks like CI considering 1-phase system and you are considering 3-phase system. Considering 3-ph sys, let try with a slightly different way:

Z (effective)= R* cos(pf)+X*sin(arc(cos(pf))= 0.41 * 0.96+0.047 * sin(16.3)=0.41

Z(eff-70 ft)=0.41*70/1000=0.0284

Now Voltage drop, Vd=sqrt(3)*70*0.0284=3.45 V ..... Voltage at relay= 208-3.45=204.55 volt.

If 1-ph sys: Vd= 70*0.0284=2.0 volt -------- Voltage at relay= 208-2.0=206.00 volt.

My understating the CI failed to account sqrt(3) in 3-phase system.

Thanks,

 

[vdubEE]

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Looks like CI considering 1-phase system and you are considering 3-phase system. Considering 3-ph sys, let try with a slightly different way:

Z (effective)= R* cos(pf)+X*sin(arc(cos(pf))= 0.41 * 0.96+0.047 * sin(16.3)=0.41

Z(eff-70 ft)=0.41*70/1000=0.0284

Now Voltage drop, Vd=sqrt(3)*70*0.0284=3.45 V ..... Voltage at relay= 208-3.45=204.55 volt.

If 1-ph sys: Vd= 70*0.0284=2.0 volt -------- Voltage at relay= 208-2.0=206.00 volt.

My understating the CI failed to account sqrt(3) in 3-phase system.

Actually, I forgot to put the 3-phase symbol under the AC generator symbol in my notes. I did the same Z effective calculation as you and got the same answer. I was using an example out of the NCEES practice exam that is very similar as a basis for my solution. I will update my original post to show it is a 3-phase generator.

I think this is another example of where CI failed to account for the sqrt(3) in the 3-phase calculations.

EDIT: Picture is updated and shows that the generator is actually 3-phase.

 

[DK PE]

Now Voltage drop, Vd=sqrt(3)*70*0.0284=3.45 V ..... Voltage at relay= 208-3.45=204.55 volt.

If 1-ph sys: Vd= 70*0.0284=2.0 volt -------- Voltage at relay= 208-2.0=206.00 volt.

My understating the CI failed to account sqrt(3) in 3-phase system.

I think if this is a single phase system you are missing a "2" factor in your voltage drop.... although at 208V system that should be a given. However you if given a 240V system you should remember the factor of 2.

 

[Insaf]

Now Voltage drop, Vd=sqrt(3)*70*0.0284=3.45 V ..... Voltage at relay= 208-3.45=204.55 volt.

If 1-ph sys: Vd= 70*0.0284=2.0 volt -------- Voltage at relay= 208-2.0=206.00 volt.

My understating the CI failed to account sqrt(3) in 3-phase system.

I think if this is a single phase system you are missing a "2" factor in your voltage drop.... although at 208V system that should be a given. However you if given a 240V system you should remember the factor of 2.

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2 factor because 240 volt system have two phase conductors. But in case of line - neutral voltage, no multiplication factor of 2.

Thanks,