NCEES Power Practice Exam #129

[ellen3720]

All,

A quick question about #129 in the NCEES practice exam.

This is a voltage drop problem with a main distribution panel (MDP) feeding another panel, Panel A. The problem gives the MDP voltage as 480V, a load current of 400A with power factor 0.80, and provides cable details for the 250' connection between MDP and Panel A.

The solution calculates the voltage at Panel A as follows: VpanelA = 277<0 - (400<-36.87)*(0.00725+j0.0120) = 271.94<-0.44 V

In the equation above, why is it 277<0, not 277<-30? Doesn't the conversion from phase to phase (480V) to phase to ground (277) always require a phase shift of -30 degrees? 

Thanks!

Ellen 

 

[Drewism]

I'm thinking that they just selected the voltage to neutral phase angle to be 0 at random for simplification. In fact, it could be anything. You're right that it would be -30 degrees with a line-to-line reference voltage at 0 degrees. In the end the voltage angle is irrelevant as shown in the answers. Notice how in the solution they neglect the voltage angle when getting the phase to phase voltage. You would just need the correct angles for current and impedance to get the right voltage value.

 

[ellen3720]

Thanks for the input. 

The difference in angle didn't turn out to be irrelevant for me - when I solved the panel A equation with 277<-30, the panel A voltage ends up about 2V higher, which was enough to get the question wrong. 

I just cant wrap my head around this one.

 

[Drewism]

R=0.049

X=j0.049

Z = (0.049+j0.048)(0.25 - distance factor) = 0.017<44.40

Vr = Vs-IZ -> (277> -30) - (400< -36.87)*(0.017< 44.40) = 271.63< -30.87 * (root 3)

= 470.49< -30.87V = 470V

Hope this helps.

 

[ellen3720]

@Drewism, where did R = 0.49 come from? It's R=0.29 in NEC table 9 and NCEES solutions.

 

[Drewism]

@Drewism, where did R = 0.49 come from? It's R=0.29 in NEC table 9 and NCEES solutions.

Whoops, sorry. Scratch that. I was in a bit of a rush.

R = 0.029

X = j0.048

Z = (0.029 + j0.048) * (250/1000) = 0.01403 < 58.56

Voltage due to impedance = (400 < -36.86) * (0.01403 < 58.86) = 5.612 < 22 V

Voltage drop = (277 < 0) - (5.612 < 22) = 271.80 < -0.443

Convert back to line voltage = 271.80 <- 0.443 * (root 3 < 30)

= 470.77 V < 29.557 = 471 < 29.557

The reference angle for the line to neutral voltage assumed to be 0 degrees, I guess to make things simple... When you include the 30 degree compensation back to line voltage is comes out to 470.77 V.

Look at question 30 in the morning sample exam of Graffeo's book for another example. It's done the same way. I guess it's just another thing to remember. Also, the answer just wants the voltage value. So if you if ignore the voltage angles it still comes out to 471 V.

Disclaimer: I am studying too and this is the way I construe it. I hope someone else will chime in as to why it's done this way.

 

[ellen3720]

I guess we can agree on the solution then: just remember 277<0 

 

[Drewism]

I guess we can agree on the solution then: just remember 277<0 

Yep. Just another one of those little things to remember. Do it per phase with a reference angle of 0 and then multiply by root 3 to get the line voltage.

Also, there is an additional formula for three-phase, short transmission lines to calculate the actual three phase voltage drop based on R and X values:

Voltage drop (Line-to-Line) = (R*cos(theta)) + (X*sin(theta)) * I * root (3)

R = Resistance of the line

X = Reactance of the line

 

[kris7o2]

In the question, it mentions the main distribution panel is 480 V, however, chapter 9, table 9 lists a voltage for 600 Volt cables. Why do we use the table for 600 V when the question mentions 480 V?