NCEES Sample Exam #111 - Wye Source or delta source

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wiliki

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I have the solution to this problem, but am currently confused. I understand the process it took to get the answer.

However, what is the "giveaway" in this problem that it is a wye-source serving the delta load? What wording gives the assumption that it's a wye-source versus a delta-source? 

NCEES 111.PNG

NCEES 111 Solution.PNG

 
Yes.  This one can definitely be confusing.  The problem does not tell you if the source is Delta or Wye.  Someone correct me if I'm wrong, but I don't think it matters if the source is Delta or Wye.  I think you would get the same answer.  The problem tells you the system is balanced and they give you the line impedence but not the load impedence.  However, they give you the load voltage, VAB.  

One way to solve this is to draw the delta system into a single phase equivalent circuit (see below).  You just have to remember to divide by sqrt of 3 and subtract by 30 degrees since the phase voltage lags the line voltage.   Then use KVL to solve for the phase voltage VAN.

Next, convert the phase voltage VAN to the line voltage (also equal to the phase voltage in a delta system) VAB.  Just remember to multiply by the sqrt of 3 when you do.

Someone please let me know if this is incorrect or if there is an even better approach to this problem.  Hope this helps!

image.png

 
Last edited by a moderator:
You can calculate VaA using ohms law, VbB has a 120 degree phase shift with it and then Vab=VaA+VAB+VBb

 
thanks MEtoEE! Your explanation makes a lot of sense. I'll have to make sure to tab this 1PH equivalent in my references.

 
leo2car said:
You can calculate VaA using ohms law, VbB has a 120 degree phase shift with it and then Vab=VaA+VAB+VBb
Click to expand...
When I attempt to solve it this way, I don't get the correct answer.

VAB=12500<0, VaA=70<-20 * (5+j10) = 782.6<43.4, VbB = 782.6<163.4, add them all together and I get Vab=12341.6<3.5

 
Last edited by a moderator:
MEtoEE said:
Yes.  This one can definitely be confusing.  The problem does not tell you if the source is Delta or Wye.  Someone correct me if I'm wrong, but I don't think it matters if the source is Delta or Wye.  I think you would get the same answer.  The problem tells you the system is balanced and they give you the line impedence but not the load impedence.  However, they give you the load voltage, VAB.  

One way to solve this is to draw the delta system into a single phase equivalent circuit (see below).  You just have to remember to divide by sqrt of 3 and subtract by 30 degrees since the phase voltage lags the line voltage.   Then use KVL to solve for the phase voltage VAN.

Next, convert the phase voltage VAN to the line voltage (also equal to the phase voltage in a delta system) VAB.  Just remember to multiply by the sqrt of 3 when you do.

Someone please let me know if this is incorrect or if there is an even better approach to this problem.  Hope this helps!

View attachment 13991
Click to expand...
Bingo. 

it doesn’t matter how the secondary of The upstream transformer feeding the load is connected. The power source can be delta or wye, it doesn’t make a difference. 
 

what does make a difference, is how the load is connected. 

 
Zach Stone said:
Bingo. 

it doesn’t matter how the secondary of The upstream transformer feeding the load is connected. The power source can be delta or wye, it doesn’t make a difference. 
 

what does make a difference, is how the load is connected. 
Click to expand...
This problem can be solved easily by applying KVL for loop (a-A-B-b) as it is without converting it into single phase values  where,

IaA = 65.78-j23.94, IbB = -53.62-j45 ( IbBI = 70, angle = -140),

Vab = (65.78-j23.94)*(5+j10) + 12500 -  (-53.62-j45)*(5+j10)

Vab = 12886.47 + j1299.3

IVabI = 12952 volt

 
Last edited by a moderator:
Mohammed Kamel said:
This problem can be solved easily by applying KVL for loop (a-A-B-b) as it is without converting it into single phase values  where,

IaA = 65.78-j23.94, IbB = 53.62-j45 ( IbBI = 70, angle = -140),

Vab = (65.78-j23.94)*(5+j10) + 12500 -  (53.62-j45)*(5+j10)

Vab = 12886.47 + j1299.3

IVabI = 12952 volt
Click to expand...
This is how I would do it as well.

 
Mohammed Kamel said:
IbB is equal to IaA but lagged by angle 120 degree, hence, IbB = 70 * (Cos(-140) + j Sin(-140)) = -53.623-j45.
Click to expand...
MO,

This is embarrassing to ask, the 120 IbB angle lagged is a principle of electrical engineering. Nothing in this problem tells us that, we just know 1bB is 120 degree lagged and 1cC is lagged 240 degrees? 

Thanks, 

 
SparkyBill said:
MO,

This is embarrassing to ask, the 120 IbB angle lagged is a principle of electrical engineering. Nothing in this problem tells us that, we just know 1bB is 120 degree lagged and 1cC is lagged 240 degrees? 

Thanks, 
Click to expand...
The problem states that it is a balanced 3 phase ABC system.

 

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