NCEES Power Afternoon Problems 505, 508,511

Guys

I have a couple of questions that I hope some of you can shed some light to.

Problem 505: I have the NEC 2005 handbook and found the table they are referring to on the solution. HOwever, according to the NEC handbook, that table only pertains to Design B,C,D motors.

Problem 508 Solution: How does the absolute value of the Current for A & B = 29.3?

Problem 511 Solution: Why do we need to get the Phase voltage when the Distribution is a Balanced Delta? In a balanced Delta, the phase and line voltage are equal. What is with the " 12.5(30 degress)/sqrt(3)? This is for a wye system.

Please explain and thank so much

Pete

pete25 said:

Guys
I have a couple of questions that I hope some of you can shed some light to.

Problem 505: I have the NEC 2005 handbook and found the table they are referring to on the solution. HOwever, according to the NEC handbook, that table only pertains to Design B,C,D motors.

Problem 508 Solution: How does the absolute value of the Current for A & B = 29.3?

Problem 511 Solution: Why do we need to get the Phase voltage when the Distribution is a Balanced Delta? In a balanced Delta, the phase and line voltage are equal. What is with the " 12.5(30 degress)/sqrt(3)? This is for a wye system.

Please explain and thank so much

Pete

Click to expand...

Hang in there Pete. I have to have notes on those problems, but at home and I don't remember the problems at all now. I will post back to you about 5:30PM.

I'm actually in the same area as my reference materials. BringItOn (I'm still getting used to that) can add or correct as necessary.

Problem 505: The table used for this problem was moved and edited in the change to the 2005 code. If you have a copy of the 2002 NEC available, the question makes sense and is answerable as written.

Problem 508: I'm not sure if I understand your question. The absolute value symbols indicate that all you are really interested in is the magnitude of the current rather than both magnitude and phase angle. The only other thing to realize is that the neutral current is the sum of the two phase currents.

Problem 511: This problem can be worked without converting it to a wye configuration but I always manage to screw up the math any time I try it. The conversion to wye values sets the problem up such that it is a simple single phase loop analysis. For me, it makes the math easier and I'm much less likely to misapply the square root of 3 in the solution.

Hope this helps.

Jim

IFR_Pilot said:

I'm actually in the same area as my reference materials. BringItOn (I'm still getting used to that) can add or correct as necessary.
Problem 505: The table used for this problem was moved and edited in the change to the 2005 code. If you have a copy of the 2002 NEC available, the question makes sense and is answerable as written.

Problem 508: I'm not sure if I understand your question. The absolute value symbols indicate that all you are really interested in is the magnitude of the current rather than both magnitude and phase angle. The only other thing to realize is that the neutral current is the sum of the two phase currents.

Problem 511: This problem can be worked without converting it to a wye configuration but I always manage to screw up the math any time I try it. The conversion to wye values sets the problem up such that it is a simple single phase loop analysis. For me, it makes the math easier and I'm much less likely to misapply the square root of 3 in the solution.

Hope this helps.

Jim

Click to expand...

Jim

But the currents each have a magnitude of 29.3. The way the solution is written how do you get 29.3 from adding to magnitudes of 29.3 for current A and current B?

OK. Now that I think I understand the question:

IA=29.3 angle 26.6 deg. = 26.2 + j13.12

IB=29.3 angle -93.4 deg. = -1.75 - j29.25

IN = IA + IB = 29.3 A angle -33.4 deg.

Let me know if this does the trick.

Jim

IFR_Pilot said:

OK. Now that I think I understand the question:
IA=29.3 angle 26.6 deg. = 26.2 + j13.12

IB=29.3 angle -93.4 deg. = -1.75 - j29.25

IN = IA + IB = 29.3 A angle -33.4 deg.

Let me know if this does the trick.

Jim

Click to expand...

Jim

for the current in B, should the phase angle be <-30 because in a wye config, the phase voltage is V(line)<-30 is that not correct? IN the solution where does the 120 come from?

The reason they don't apply the -30 degrees when the find the phase voltage is that they are solving for the magnitude of the neutral current only. The reference point for the angle is arbitrary.

The 120 comes from the phase difference between Phase A-N and Phase B-N. They are 120 degrees apart. For the NCEES solution, they picked phase A-N voltage as the zero degree reference.

Jim

Pete,

For 505 I am pretty sure you already have the answer since Jim posted it.

For 508 the 0 and 120 are the phase angles of VA and VB. The 29.3 comes from the sum of the currents magnitudes and angles. Do as Jim says and you will be OK.

If you still need 511 let us know. I have some notes but right now I gotta run to pick up my son at the TriRail station.

Good Luck!!!!!!!!!

OOOOOOOOOOOOOOOOOOOOPPPPS...Jim is here...You are in good hands.

BringItOn said:

Pete,
For 505 I am pretty sure you already have the answer since Jim posted it.

For 508 the 0 and 120 are the phase angles of VA and VB. The 29.3 comes from the sum of the currents magnitudes and angles. Do as Jim says and you will be OK.

If you still need 511 let us know. I have some notes but right now I gotta run to pick up my son at the TriRail station.

Good Luck!!!!!!!!!

OOOOOOOOOOOOOOOOOOOOPPPPS...Jim is here...You are in good hands.

Click to expand...

Bring it on

Can you please email me some notes for problem 511. email is "[email protected]"

Thanks

Pete

pete25 said:

Bring it on
Can you please email me some notes for problem 511. email is "[email protected]"

Thanks

Pete

Click to expand...

I have to admit I must be missing something. Pete did you ever get this? Jim if I convert the Delta to a Wye (which makes sense and I agree) the Yz's are 1/3 the Delta Zs. If I do that and set it up as a Voltage drop calc V=IZ or (70/_-20) (3.726/_63.4) I get 260 Volts then what...wait...if I take 260 Volt add it to 7217=7477 * Sqrt 3=12.95....is that it right?

jdd18vm said:

I have to admit I must be missing something. Pete did you ever get this? Jim if I convert the Delta to a Wye (which makes sense and I agree) the Yz's are 1/3 the Delta Zs. If I do that and set it up as a Voltage drop calc V=IZ or (70/_-20) (3.726/_63.4) I get 260 Volts then what...wait...if I take 260 Volt add it to 7217=7477 * Sqrt 3=12.95....is that it right?

Click to expand...

John,

You don't have to convert the Load impedances to wye values. The known values are the line voltage (VLL=VAB) and the line current.

The line current (IaA) will remain unchanged regardless of the configuration of the load impedance.

If you try to solve this without converting to phase voltage, KVL around the loop should get you the answer also. The problem is that you have to be **** sure you have the angle correct for current IbB. It turns into a much bigger PITA at that point than it needs to be. There will be a sqrt 3 that will mathematically appear through the current angles.

This is pretty much just an expansion on how the NCEES solution is put together:

Known Values:

VLL=VAB=12.5kV at 0 deg.

IL=IaA=70A at -20 deg.

ZLine=5+j10 ohms

Step one, convert line voltage into phase voltage: VAN=VAB/sqrt 3 at -30 deg. = 7217 V at -30 deg

IL will remain at 70A at -20 deg. regardless of if you are using line or phase voltage at the load.

The next thing you need to know is VaA. Since we are now working on a phase voltage basis we don't need to concern ourselves with VbB.

VaA=IL*ZLine=783 V at 43.4 deg.

Now we can determine VaN.

VaN = VaA + VAN = (7217 ang. -30) + (783 ang. 43.4 deg.) = 7.48kv ang. -24.3 deg.

Converting back to line voltage values:

Vab = sqrt 3 * VaN = 12.95 kV

Jim

IFR_Pilot said:

John,
You don't have to convert the Load impedances to wye values. The known values are the line voltage (VLL=VAB) and the line current.

The line current (IaA) will remain unchanged regardless of the configuration of the load impedance.

If you try to solve this without converting to phase voltage, KVL around the loop should get you the answer also. The problem is that you have to be **** sure you have the angle correct for current IbB. It turns into a much bigger PITA at that point than it needs to be. There will be a sqrt 3 that will mathematically appear through the current angles.

This is pretty much just an expansion on how the NCEES solution is put together:

Known Values:

VLL=VAB=12.5kV at 0 deg.

IL=IaA=70A at -20 deg.

ZLine=5+j10 ohms

Step one, convert line voltage into phase voltage: VAN=VAB/sqrt 3 at -30 deg. = 7217 V at -30 deg

IL will remain at 70A at -20 deg. regardless of if you are using line or phase voltage at the load.

The next thing you need to know is VaA. Since we are now working on a phase voltage basis we don't need to concern ourselves with VbB.

VaA=IL*ZLine=783 V at 43.4 deg.

Now we can determine VaN.

VaN = VaA + VAN = (7217 ang. -30) + (783 ang. 43.4 deg.) = 7.48kv ang. -24.3 deg.

Converting back to line voltage values:

Vab = sqrt 3 * VaN = 12.95 kV

Jim

Click to expand...

thanks again Jim, I got it now.

pete25 said:

Jim
for the current in B, should the phase angle be <-30 because in a wye config, the phase voltage is V(line)<-30 is that not correct? IN the solution where does the 120 come from?

Click to expand...

Jim

I still dont understand. You state in your response that we are looking for the neutral current. In a wye config, the neutral and phase current are the same and they differ from the line voltages by 30 degrees. Why do they choose 120 degress this difference is only for Line values i.e. Line A voltage differs from Line B voltage by 120 degrees. We are dealing with Phase values in this problem not Line values. Why the 120 and not the 30 degrees

pete25 said:

In a wye config, the neutral and phase current are the same and they differ from the line voltages by 30 degrees.

Click to expand...

The neutral current is the equal to the SUM of the phase currents. For a balanced three-phase system, the neutral current will be zero. Since only two phases are loaded for this problem, there will be an imbalance resulting in a neutral current.

pete25 said:

Why do they choose 120 degress this difference is only for Line values i.e. Line A voltage differs from Line B voltage by 120 degrees. We are dealing with Phase values in this problem not Line values. Why the 120 and not the 30 degrees

Click to expand...

There is a 30 degree phase difference between LINE voltage VAB and PHASE voltage VAN.

There is a 120 degree phase difference between PHASE voltage VAN and PHASE voltage VBN.

The 30 degree shift only applies when converting between phase and line voltages.

Jim

IFR_Pilot said:

The neutral current is the equal to the SUM of the phase currents. For a balanced three-phase system, the neutral current will be zero. Since only two phases are loaded for this problem, there will be an imbalance resulting in a neutral current.There is a 30 degree phase difference between LINE voltage VAB and PHASE voltage VAN.

There is a 120 degree phase difference between PHASE voltage VAN and PHASE voltage VBN.

The 30 degree shift only applies when converting between phase and line voltages.

Jim

Click to expand...

Jim

You are converting a line voltage to phase voltage arent we?

Thanks

pete25 said:

Jim
You are converting a line voltage to phase voltage arent we?

Thanks

Click to expand...

If I'm not making sense here, somebody let me know.

At the point in the problem where the conversion from line voltage to phase voltage takes place, the reference angle is arbitrary. It simplifies the math if you use VAN as the reference voltage at 0 deg. This voltage will actually be 30 degrees behind the line voltage (VAB). However, the reference voltage is a choice based on whatever works best for you.

If you start the problem with VAN at 7.62kV and -30 degrees, the phase shift between VAN and VBN is still 120 degrees. The relative angles will remain the same and you will still get the same answer for the magnitude of the neutral current.

Jim

Can someone explain where the -30degrees comes from when converting line to phase voltage on the delta side.

It's simply a vector sum. Technically, there is no "phase" (Line-Neutral) voltage on a delta connected system. However, mathematically, it is often convenient to reduce 3-phase circuits to single-phase equivalents. The 30 degree phase shift is simply a vector summation of two voltages.

Flyer_PE said:

It's simply a vector sum. Technically, there is no "phase" (Line-Neutral) voltage on a delta connected system. However, mathematically, it is often convenient to reduce 3-phase circuits to single-phase equivalents. The 30 degree phase shift is simply a vector summation of two voltages.

Click to expand...

Is it always going to be -30? Is this in the EERM anywhere so I can read up on it? I don't remember this part from school. The negative 30 is an important part of the problem/solution and in my mind I need to understand that part . Will it always be negative 30 degrees in regards to all phases, such as BN & CN for wye systems. When I was looking in the EERM I didn't see any phase shifts for wye transforms.

bambooi said:

Is it always going to be -30? Is this in the EERM anywhere so I can read up on it?

Click to expand...

EERM 34-2, section 4: "with delta connected balanced loads, the phase and line currents differ in phase by 30 deg." Simliarly, EERM 34-2, section 2 explains, in wye connected loads, the phase and line voltages differ by 30 deg. (see equations 34.7-9.)