NCEES Practice exam

[mvsapre]

I have a question from NCEES power practice exam. 

Q110. In a 3 phase 4 wire system loads are connected to phase B and C only. Phase-Phase voltage is 13.2kV, Load is 500kVA at 0.85 lagging PF. What is the magnitude of the line current in Phase C?

Solution is 500/13.2=37.9A

My question is why we are not dividing by PF here?

Please help.

Thanks

 

[Troll]

I have a question from NCEES power practice exam. 

Q110. In a 3 phase 4 wire system loads are connected to phase B and C only. Phase-Phase voltage is 13.2kV, Load is 500kVA at 0.85 lagging PF. What is the magnitude of the line current in Phase C?

Solution is 500/13.2=37.9A

My question is why we are not dividing by PF here?

Please help.

Thanks

because load is kVA and not kW

 

[Troll]

I have a question from NCEES power practice exam. 

Q110. In a 3 phase 4 wire system loads are connected to phase B and C only. Phase-Phase voltage is 13.2kV, Load is 500kVA at 0.85 lagging PF. What is the magnitude of the line current in Phase C?

Solution is 500/13.2=37.9A

My question is why we are not dividing by PF here?

Please help.

Thanks

how far into your studies are you? plowing through the practice exam prior to beginning your study?

shoot me a PM if you need study material

 

[TNSparky]

I have a question from NCEES power practice exam. 

Q110. In a 3 phase 4 wire system loads are connected to phase B and C only. Phase-Phase voltage is 13.2kV, Load is 500kVA at 0.85 lagging PF. What is the magnitude of the line current in Phase C?

Solution is 500/13.2=37.9A

My question is why we are not dividing by PF here?

Please help.

Thanks

A lot of these problems throw a power factor in there as a red herring. Be mindful of that.  

In this case, this problem goes back to the fundamentals of circuits:

1. What is the load connected to? Phase B and C. Phase A is another red herring and does not matter here.

2. What voltage does the load see? 13.2kV (Connected across B and C)

3. What is connected? A 500kVA apparent power load. The apparent power dissipated by any load is S = VI*. Power factor is not needed at all in this formula.

---> This is where the fx-115ES Plus comes in handy. Though you don't need it on this problem, if you were to have a voltage and current vector (magnitude and angle), you can simply flip the sign of the angle of the current quantity (positive becomes negative and vice versa) and multiply the 2 vectors directly in your calculator. This will give you the right magnitude and angle for apparent power.

4. Solve for current: S=500kVA, V=13.2kV, I=?. 500/13.2 = | I | = 37.9 A.

Hope this helped! Good luck!

 

[TNSparky]

A lot of these problems throw a power factor in there as a red herring. Be mindful of that.  

In this case, this problem goes back to the fundamentals of circuits:

1. What is the load connected to? Phase B and C. Phase A is another red herring and does not matter here.

2. What voltage does the load see? 13.2kV (Connected across B and C)

3. What is connected? A 500kVA apparent power load. The apparent power dissipated by any load is S = VI*. Power factor is not needed at all in this formula.

---> This is where the fx-115ES Plus comes in handy. Though you don't need it on this problem, if you were to have a voltage and current vector (magnitude and angle), you can simply flip the sign of the angle of the current quantity (positive becomes negative and vice versa) and multiply the 2 vectors directly in your calculator. This will give you the right magnitude and angle for apparent power.

4. Solve for current: S=500kVA, V=13.2kV, I=?. 500/13.2 = | I | = 37.9 A.

Hope this helped! Good luck!

Recall: In S = VI*. The * denotes a complex conjugate value. This, the need to flip the sign of the angle in the current vector prior to multiplying in a calculator.

 

[mvsapre]

Right. This was straightforward. Given power factor tricked me. Thank you very much for the explanation.  

 

[mvsapre]

Recall: In S = VI*. The * denotes a complex conjugate value. This, the need to flip the sign of the angle in the current vector prior to multiplying in a calculator.

All these days I didn't know that Casio 115ES can be used directly to multiply, add... two vectors. I just figured that out. Thanks a lot for mentioning it here.  

 

[rg1]

A lot of these problems throw a power factor in there as a red herring. Be mindful of that.  

In this case, this problem goes back to the fundamentals of circuits:

1. What is the load connected to? Phase B and C. Phase A is another red herring and does not matter here.

2. What voltage does the load see? 13.2kV (Connected across B and C)

3. What is connected? A 500kVA apparent power load. The apparent power dissipated by any load is S = VI*. Power factor is not needed at all in this formula.

---> This is where the fx-115ES Plus comes in handy. Though you don't need it on this problem, if you were to have a voltage and current vector (magnitude and angle), you can simply flip the sign of the angle of the current quantity (positive becomes negative and vice versa) and multiply the 2 vectors directly in your calculator. This will give you the right magnitude and angle for apparent power.

4. Solve for current: S=500kVA, V=13.2kV, I=?. 500/13.2 = | I | = 37.9 A.

Hope this helped! Good luck!

Very nice explanation.

 

[jslogsdon]

A lot of these problems throw a power factor in there as a red herring. Be mindful of that.  

In this case, this problem goes back to the fundamentals of circuits:

1. What is the load connected to? Phase B and C. Phase A is another red herring and does not matter here.

2. What voltage does the load see? 13.2kV (Connected across B and C)

3. What is connected? A 500kVA apparent power load. The apparent power dissipated by any load is S = VI*. Power factor is not needed at all in this formula.

---> This is where the fx-115ES Plus comes in handy. Though you don't need it on this problem, if you were to have a voltage and current vector (magnitude and angle), you can simply flip the sign of the angle of the current quantity (positive becomes negative and vice versa) and multiply the 2 vectors directly in your calculator. This will give you the right magnitude and angle for apparent power.

4. Solve for current: S=500kVA, V=13.2kV, I=?. 500/13.2 = | I | = 37.9 A.

Hope this helped! Good luck!

Thank you for this...I wish the solutions manuals were written like this.

 

[TNPE]

I have a question from NCEES power practice exam. 

Q110. In a 3 phase 4 wire system loads are connected to phase B and C only. Phase-Phase voltage is 13.2kV, Load is 500kVA at 0.85 lagging PF. What is the magnitude of the line current in Phase C?

Solution is 500/13.2=37.9A

My question is why we are not dividing by PF here?

Please help.

Thanks

This problem can be solved using PF, but there is certainly no need to complicate it.  However, I will illustrate this below so that you can use this approach if you get hung up:

From our power equations, we know

P=VIcos(theta)

Conversely,

Q=VIsin(theta) and S=VI or VI*

We are interested in the magnitude here, therefore, PF is not necessary.  However, if PF were used, recall that P=Spf, thereby, P=(500KVA)(0.85).  Once you have determined P, then you can use the above solution with PF, namely:

425kW/(13.2kVx0.85)=37.9A  (or I = P/(VLLxPF)

Easy enough, but entirely unnecessary to solve this problem.  As others have mentioned, know what you're solving for and analyze what data is given (write everything down and put a question mark beside the unknown...sounds remedial, but hey, if it helps to pass, so what).  For instance, to carry this another step further, rarely ever will you have to use HP to calculate electrical power, amps, etc. (generally speaking, the HP that is usually given is in terms of mechanical/shaft HP, not electrical HP).  It can be done with generalized formulas and some elbow grease, but this test is about how well do you understand concepts, not how well can you regurgitate numbers and manipulate a problem algebraically to arrive at a solution. 

 

[JCM]

There is another question here, the question is asking about the line current in phase C and not phase current, so
Line current = phase current X square root of 3
Or it means phase current because he says in phase C???????

 

[JCM]

Or it is because it is the only load?

 

[Dude847]

There is another question here, the question is asking about the line current in phase C and not phase current, so
Line current = phase current X square root of 3
Or it means phase current because he says in phase C???????

Because the problem statement states that it is a 3 phase, 4 wire system you assume that it is a wye system (got to love when NEC does that). In a wye system, Line Current = Phase Current.

Also, the problem is essentially reduced to a more basic single phase circuit because the only load is between Phase B & C.

I am by no means an expert, just my understanding. Hope it helps.

 

[JCM]

Because the problem statement states that it is a 3 phase, 4 wire system you assume that it is a wye system (got to love when NEC does that). In a wye system, Line Current = Phase Current.

Also, the problem is essentially reduced to a more basic single phase circuit because the only load is between Phase B & C.

I am by no means an expert, just my understanding. Hope it helps.

First thank you for your reply. It problem is still confusing, since the question states that the load is connected between phase B and C then the load is connected in a delta configuration (the system or source is Y connection). That’s why the line voltage (phase to phase) of the system was considered as the phase voltage for the load between phases B&C. The answer of the book was based on the phase voltage for the load is equal to the line system voltage (which proof that he considered the load connected as delta not Y) the problem states that the phase to phase voltage ( line voltage) is 13.2KV and the answer
KVA(phase load) = V(phase)x I(phase)
500= 13.2x I(phase)
I(load phase)= 500/13.2= 37.88A
I think because this is the only load connected and it is considered as a single phase problem. No other loads or current in the other phase so the phase current is the same current in the line.

 

[Alex Kabeer]

I have a question from NCEES power practice exam.

Q110. In a 3 phase 4 wire system loads are connected to phase B and C only. Phase-Phase voltage is 13.2kV, Load is 500kVA at 0.85 lagging PF. What is the magnitude of the line current in Phase C?

Solution is 500/13.2=37.9A

My question is why we are not dividing by PF here?

Please help.

Thanks

Because line current I flowing through phase C is, I = S/V = 500/13.2 = 37.9A, where S is apparent power, and V is line voltage