CI Exam 2, Problem 10: VD % Equation

I have a fundamental question on the voltage drop % equation, not the actual understanding of voltage drop. 

CI Problem 10 Exam 2 solves for the voltage magnitude drop, then divides that by the nominal voltage. My initial solving used the magnitude difference of the load voltage and feeder voltage and divided that by the nominal feeder voltage. Which is correct if the PE exam would ask for VD %? The feeder voltage drop, or (|No Load|-|Load|)/|No Load|?

Thanks!

Not everyone has this book and not everyone willing to help (who does) has this book handy.

I suggest you take a screenshot and post the problem and solution

HopefulPE said:

I have a fundamental question on the voltage drop % equation, not the actual understanding of voltage drop. 

CI Problem 10 Exam 2 solves for the voltage magnitude drop, then divides that by the nominal voltage. My initial solving used the magnitude difference of the load voltage and feeder voltage and divided that by the nominal feeder voltage. Which is correct if the PE exam would ask for VD %? The feeder voltage drop, or (|No Load|-|Load|)/|No Load|?

Thanks!

Click to expand...

 % voltage regulation of a line or transformer is given in (|NL|-|FL|)/|FL| . The difference is very less in the final figures for practical problems and hence making an intelligent choice out of options also helps sometimes. 

Will do. The question reads "What is the voltage drop for a 120V, single-phase copper feeder that is 220 ft long with a maximum load current of 40<-10A? The impedance per 1000ft of feeder is 0.78+j*0.052 Ohms"

Ans. 11%

Book Soln: Rline=0.34 Ohms, Xline=0.023 Ohms. VD= 40 [ 0.34cos(-10)+0.023sin(-10)] = 13.2  . VD (%) = 13.2/120*100 = 11%.

In my way I do 120- (40<-10)*(0.34+j*0.023) = 106.4568<0.78.   VD(%) = (120-106.4568)/120 *100= 13.54/120 *100 = 11.28%. 

In this example the difference is negligible, but it may not be on the exam, or the answers may be close. Which way would you guys calculate the voltage drop? Also, does anyone know why there is a discrepancy between the two? I think the book soln formula comes from the IEEE red book, but that's just what I saw online. 

Thanks!

HopefulPE said:

Will do. The question reads "What is the voltage drop for a 120V, single-phase copper feeder that is 220 ft long with a maximum load current of 40<-10A? The impedance per 1000ft of feeder is 0.78+j*0.052 Ohms"

Ans. 11%

Book Soln: Rline=0.34 Ohms, Xline=0.023 Ohms. VD= 40 [ 0.34cos(-10)+0.023sin(-10)] = 13.2  . VD (%) = 13.2/120*100 = 11%.

In my way I do 120- (40<-10)*(0.34+j*0.023) = 106.4568<0.78.   VD(%) = (120-106.4568)/120 *100= 13.54/120 *100 = 11.28%. 

In this example the difference is negligible, but it may not be on the exam, or the answers may be close. Which way would you guys calculate the voltage drop? Also, does anyone know why there is a discrepancy between the two? I think the book soln formula comes from the IEEE red book, but that's just what I saw online. 

Thanks!

Click to expand...

Both are right. The examiner should not give option answers very close to each other because there will always be small differences in answers. Here comes luck.

But in this question there is one more catch. The Impedance is given for feeder and not for one conductor still it is being taken twice for two wires. Feeder actually means two wires in single phase so according to me the answer should have been 5.7%.

rg1 said:

Both are right. The examiner should not give option answers very close to each other because there will always be small differences in answers. Here comes luck.

But in this question there is one more catch. The Impedance is given for feeder and not for one conductor still it is being taken twice for two wires. Feeder actually means two wires in single phase so according to me the answer should have been 5.7%.

Click to expand...

You are right that a single phase feeder is two wires (hot and neutral [plus ground, the non-current carrying conductor]), which is why the impedance is doubled. Current goes to load on hot and returns to source on neutral; therefore, the voltage drop across both line and neutral have to be considered. So since they say the "impedance per 1000 ft of feeder," I can see your point; however, this problem seems to be designed to test your knowledge of the difference in 1PH feeder impedance calculations vs. 3PH feeder impedance calculations.

The feeder impedance is calculated as [(2*220)/1000]*(0.78+j0.052).

The book solution the OP provides doesn't seem correct. It appears they are using the "approximated" impedance method from NEC Table 9, fpn 2, which is cool (both methods will yield the same result), but there appears to be no account for the fact that the line length is less than 1000-ft, so there seems to be a problem here with the solution provided.

Oh, I see. Looks like the calculation of impedance for 440-ft was left out of the post for brevity. 

Thanks guys. So if the question reads feeder on the exam tomorrow for single phase, then take that as the entire loop impedance with no need to double then, correct? 

Appreciate the helpful hint- and I'll just calculate it the way I've been doing in terms of VD.