# NCEES Power 71



## yaoyaodes (Dec 7, 2021)

Can anyone help me understand this. The question said that maximum power is 5000MW occurs at angle 90 degree. if the power angle is 90 degree, does it mean it's pure reactive power? if it's reactive power, why the maximum power is MW instead of MVAR? For solution, why we use 3* 8.165 square * 50? should we calculate with angles?


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## Zach Stone P.E. (Dec 7, 2021)

Hi @yaoyaodes,

I'll do my best to answer your questions:



> The question said that maximum power is 5000MW occurs at angle 90 degree.



The maximum power formula comes from the real power transformer formula:

Pmax = (|E1|•|E2| / X) sin(δ)​
Where δ is the angle between the two voltages E1 and E2​
The maximum value of sin(δ) is sin(90º) = 1.

An angle of anything other than δ = 90 degrees results in a value of sin(90º) < 1, which decreases the amount of power transformer.

Therefore, the maximum amount of power occurs when the electrical torque angle (how far the rotor leads the stator poles by) is equal to δ = 90º, and the resulting maximum power is calculated by:

Pmax = (|E1|•|E2| / X) sin(90º)​​Pmax = (|E1|•|E2| / X)•(1)​​Pmax = (|E1|•|E2| / X)​



> if the power angle is 90 degree, does it mean it's pure reactive power?
> 
> 
> > if it's reactive power, why the maximum power is MW instead of MVAR?



This formula calculates the amount of real power in watts transferred from the leading voltage source to the lagging voltage source, in this case from VA (the generator on the left) to VB (the generator on the right).




> For solution, why we use 3* 8.165 square * 50? should we calculate with angles?



I = 8,165A<45º is the amount of line current from generator A on the left to generator B on the right.

Using the power formula S1ø = |I|²Z, we can calculate the amount of *single-phase* power absorbed in the line. Since the line is purely reactive (Z = 0 + j50Ω), the power will be *reactive power* (S1ø = 0 +jQ1ø) in *VARs*:

S1ø = |I|²•Z​​Q1ø = |I|²•X​​Q1ø = |8,165A|²•(50Ω)​​Q1ø = 3,333.4 MVAR​
This is the amount of *single-phase* reactive power absorbed in just one out of the three (XL = 50Ω) line conductors since it is a three-phase system.

To calculate the total *three-phase *reactive power absorbed in the entire three-phase system, we have to account for not just one of the line conductors but by all three. We can do this by multiplying single-phase power by three (3):

Q3ø = 3•Q1ø​​Q3ø = 3(3,333.4 MVAR)​​Q3ø = 10,000 MVAR​​*The total amount of reactive power losses in the transmission line is approximately 10,000 MVAR.*


If you are looking for more help, these are the types of topics we explore in much more depth in our online class for the new CBT format of the PE exam. You can sign up today for the Free Trial by clicking on the button in my signature below, or by visiting www.electricalpereview.com.


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## HDPower (Dec 31, 2021)

Zach Stone P.E. said:


> Hi @yaoyaodes,
> 
> I'll do my best to answer your questions:
> 
> ...



Hi Zach,
I believe NCEES provided a wrong solution for this question, and you just simply follow their approach.

The active power and reactive power flow from Bus a to Bus b, while the reactance between the buses is X is as follow:

P = Va.Vb.sin(delta)/X
Q = Va.(Va-Vb.cos(delta))/X

If we consider that your approach is correct, the active power should be equal to zero: P=I^2*R, R=0 => P=0

Please see the answer in the attachment.


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## Zach Stone P.E. (Jan 4, 2022)

HDPower said:


> Hi Zach,
> I believe NCEES provided a wrong solution for this question, and you just simply follow their approach.
> 
> The active power and reactive power flow from Bus a to Bus b, while the reactance between the buses is X is as follow:
> ...



The question is asking for the amount of reactive power (Q), as opposed to active power (P) absorbed in the line during max power conditions. 

During max power conditions you'll have the max current flowing from bus A to bus B. 

This is the amount of current flowing through the line and is responsible for the reactive power absorbed in the line.


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