# Calculate apparent power when real power given in PU



## Sthabik PE (Feb 20, 2019)

Question states : "A 3-phase, 22 kV, 500 MVA star-connected synchronous generator has a reactance Xs of 1.5 pu and supplies 0.8 pu real power (P) at the rated voltage, with a lagging power factor of 0.6 to the load.Calculate the magnitude of the actual complex power consumed by the load in MVA."

My ans is 385MVA. 

I want to know if my ans is correct.


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## daydreambeliever (Feb 20, 2019)

Following......I got a different answer but I feel like I missed a step. I didn't use 0.8pu real power anywhere. I'll take another try at it on my lunch break but maybe someone else will be able to chime in.


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## roy167 (Feb 20, 2019)

Sdhabik said:


> Question states : "A 3-phase, 22 kV, 500 MVA star-connected synchronous generator has a reactance Xs of 1.5 pu and supplies 0.8 pu real power (P) at the rated voltage, with a lagging power factor of 0.6 to the load.Calculate the magnitude of the actual complex power consumed by the load in MVA."
> 
> My ans is 385MVA.
> 
> I want to know if my ans is correct.


They have given 0.8 PU real power(KW) and Power factor of 0.6. 

Therefore the apparent power(KVA) is 0.8/.6 = 1.333 PU X 500 MVA = 666.666 MVA actual. 

 Does the above makes sense or is this correct?


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## eatsleep (Feb 20, 2019)

roy167 said:


> They have given 0.8 PU real power and Power factor of 0.6.
> 
> Therefore the apparent power is 0.8/.6 = 1.333 PU X 500 MVA = 666.666 MVA actual.
> 
> Does the above makes sense or is this correct?


I just went through the question and got the same answer.


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## Sthabik PE (Feb 20, 2019)

roy167 said:


> They have given 0.8 PU real power(KW) and Power factor of 0.6.
> 
> Therefore the apparent power(KVA) is 0.8/.6 = 1.333 PU X 500 MVA = 666.666 MVA actual.
> 
> Does the above makes sense or is this correct?


@roy167 &amp; @eatsleep: I don't think it is correct. Here the power consumed by the inductive load is 667 MVA which is greater than the capacity (500 MVA) of Syn Gen.

Let me make few changes on your post.

They have given 0.8 PU real power and Power factor of 0.6 (Cos Q). 

Let Base MVA  =500 MVA

Base MW = 500 x Cos Q

Actual MW consumed by the load = 0.8 PU real power x Base MW = 0.8 x 500 x Cos Q

Actual Current (Iact) = Actual MW/(sqrt3 x V line x Cos Q) = (0.8 x 500 x Cos Q) / (sqrt3 x V line x Cos Q) = (0.8 x 500) / (sqrt3 x V line)

Actual MVA consumed by the load = sqrt3 x V line x I act = (sqrt3 x V line x 0.8 x 500) / (sqrt3 x V line) = 0.8 x 500 = 400 MVA

Does it make sense? Please correct me if i am wrong.

Here is my another approach with ans (385 MVA).




I hope we don't have to see both answers as our option in the exam.


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## Sthabik PE (Feb 20, 2019)

daydreambeliever said:


> Following......I got a different answer but I feel like I missed a step. I didn't use 0.8pu real power anywhere. I'll take another try at it on my lunch break but maybe someone else will be able to chime in.


Pleas let me have your views on this.


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## roy167 (Feb 20, 2019)

Okay so there are some keywords and on the 2nd thought here is how I would solve it.  Generator capacity does not equal load consumed. Load consumed depends upon the load. For e.g. You can have 500 MW generator but if load draws 200KW then that's what it is.  In this case, they are asking for apparent power consumed by load.

Power consumed by load  = Power generated - losses (losses are neglected here since no information was given)

Generator supplies real power of 0.8 PU = 500 MVA X 0.6 = 300 MW x 0.8 = 240 MW 

Pf  = 0.6

Therefore , S = 240/0.6 = 400 MVA

In other words, 500 MVA generator is loaded 80% (0.8 pu), therefore the real power and active power supplied to load must be 80% of respective power.


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## Sthabik PE (Feb 20, 2019)

roy167 said:


> therefore the real power and active power supplied to load must be 80% of respective power.


I think you mean to say both active and reactive power.


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## roy167 (Feb 22, 2019)

Sdhabik said:


> I think you mean to say both active and reactive power.


Correct . Did you have the answer key to the question you posted?


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## Phenomenon083 (Feb 25, 2019)

Sdhabik said:


> reactance Xs of 1.5 pu and supplies 0.8 pu real power (P) ﻿at the rated voltage, with a lagging power factor of 0.6 to the load


I believe the answer should be 666.67 MVA. 

Problem says the generator supplies 0.8 pu real power with a lagging power factor of 0.6. So you calculate the apparent power based on this lagging power factor. which is 1.33 pu. And you find Actual complex power 666.67 MVA from this since we already know the base value is 500 MVA.

Reactance does not have anything to do for the solution of this problem, and we should not also confuse ourselves by multiplying base values with power factor.


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## roy167 (Feb 25, 2019)

Phenomenon083 said:


> I believe the answer should be 666.67 MVA.
> 
> Problem says the generator supplies 0.8 pu real power with a lagging power factor of 0.6. So you calculate the apparent power based on this lagging power factor. which is 1.33 pu. And you find Actual complex power 666.67 MVA from this since we already know the base value is 500 MVA.
> 
> Reactance does not have anything to do for the solution of this problem, and we should not also confuse ourselves by multiplying base values with power factor.


Generator parameters are 22KV, 500 MVA

500 MVA is the rated apparent power of Generator. So it can not supply more power than it's rating. I'm also confused by this simple problem. I have read somewhere MVA base and MW base are the same thing. So what that means is 0.8 PU real power would be .8*500 = 400 MW , with 0.6 power factor the apparent power would be 400/0.6 = 666.66 MVA.   

When you convert MVA base to MW base that is when I get 400 MVA answer.


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## Phenomenon083 (Feb 25, 2019)

roy167 said:


> Generator parameters are 22KV, 500 MVA
> 
> 500 MVA is the rated apparent power of Generator. So it can not supply more power than it's rating.


Seeing actual apparent power more than generator's rated apparent power confuses me as well. But I don't think if you can multiply base apparent power 500 MVA with PF to find actual real power. Correct me if I am wrong tho.


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## Sthabik PE (Feb 25, 2019)

roy167 said:


> Correct . Did you have the answer key to the question you posted?


Sorry to say i don't have the answer for this. I got this question from the internet "chegg.com".


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## Sthabik PE (Feb 25, 2019)

roy167 said:


> I have read somewhere MVA base and MW base are the same thing.


If you find the source for this statement please post it here for our reference.

My view: MVA &amp; MW are same only for unity pf.  So may be the case taken on that situation only. Need to see the whole statement for the conclusion.


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## Sthabik PE (Feb 25, 2019)

Phenomenon083 said:


> But I don't think if you can multiply base apparent power 500 MVA with PF to find actual real power.


At first, Base apparent power 500MVA multiply with pf will lead to base real power not actual real power. For actual real power, base real power shall be multiply with its pu value.

Regarding calculation, can we multiply or not?

I think we must. If we ignore it, our answer will be 667MVA which will not make sense since the generator parameter is 500MVA.


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## Phenomenon083 (Feb 26, 2019)

Sdhabik said:


> At first, Base apparent power 500MVA multiply with pf will lead to base real power not actual real powe﻿r. For actual real power, bas﻿e real power shall be multiply with its pu value.
> 
> Regarding calculation, can we multiply or not?
> 
> I think we must. If we ignore it, our answer will be 667MVA which will not make sense since the generator parameter is 500MVA.


Your solution makes sense in a way. But I am still not confident if we can multiply Rated apparent power with PF. Cause we get PF from the angle difference of V to I or the angle of power triangle is actually thetaZ. Also the problem says the generator supplies 0.8 pu real power (P) at the rated voltage, with a *lagging power factor of 0.6 to the load. *

I understand the answer 667 MVA looks dubious but overall this problem is confusing.


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## Zach Stone P.E. (Feb 26, 2019)

What book is this problem from?

*The answer is 667MVA. *Either the generator is operating in overload, or It appears that it was written without realizing that 0.8pu of real output power at 0.6 lagging power at an output rating of 500MVA results in an output that is greater than the rating of the generator.






The first step is to calculate the real power output from the given apparent power rating of the machine, using it as base, and the real power output in PU:












Pout = Pout_pu•Sb











Pout = 0.8•500MVA











Pout = 400MW







It's a little confusing because the base is in units of MVA and real power output is in the units of MW, the key here is remembering that base values are magnitudes only and can be multiplied by the real or imaginary component of a complex number like this:


if Sout = Pout +jQout,
and: Sout_pu = Pout_pu + jQout_pu,
 
then: Sout = Sout_pu•|Sb|,



         Sout = (Pout_pu + jQout_pu)•|Sb|
         Sout = Pout_pu•|Sb| + jQout_pu•|Sb|



          Pout = Pout_pu•|Sb|   and Qout = Qout_pu•|Sb|
 


In this problem, we are given Pout_pu, |Sb|, and Power factor. We are solving for |Sout|.

Now that we know Pout = 400MW, we can find |Sout| using the given power factor. Remember apparent power is always the largest power quantity when power factor is not unity, so to make Pout larger to solve for |Sout|, we need to divide Pout by PF:

|Sout| = Pout/PF
|Sout| = 400MW/0.6
|Sout| = 667MVA


*The answer is 667MVA, assuming that the apparent power output of the machine is consumed by the load neglecting any increase in VA due to the reactive power absorbed in the synchronous reactance of the machine itself. *


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## Phenomenon083 (Feb 26, 2019)

Zach Stone said:


> What book is this problem from?
> 
> *The answer is 667MVA. *Either the generator is operating in overload, or It appears that it was written without realizing that 0.8pu of real output power at 0.6 lagging power at an output rating of 500MVA results in an output that is greater than the rating of the generator.
> 
> ...


 It's a little confusing because the base is in units of MVA and real power output is in the units of MW, the key here is remembering that base values are magnitudes only and can be multiplied by the real or imaginary component of a complex number like this:


if Sout = Pout +jQout,
and: Sout_pu = Pout_pu + jQout_pu,
 
then: Sout = Sout_pu•|Sb|,



         Sout = (Pout_pu + jQout_pu)•|Sb|
         Sout = Pout_pu•|Sb| + jQout_pu•|Sb|



          Pout = Pout_pu•|Sb|   and Qout = Qout_pu•|Sb|
 


In this problem, we are given Pout_pu, |Sb|, and Power factor. We are solving for |Sout|.

Now that we know Pout = 400MW, we can find |Sout| using the given power factor. Remember apparent power is always the largest power quantity when power factor is not unity, so to make Pout larger to solve for |Sout|, we need to divide Pout by PF:

|Sout| = Pout/PF
|Sout| = 400MW/0.6
|Sout| = 667MVA


*The answer is 667MVA, assuming that the apparent power output of the machine is consumed by the load neglecting any increase in VA due to the reactive power absorbed in the synchronous reactance of the machine itself. *



Thanks Zach for your insight. Someone found this problem on chegg.com.


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## Sthabik PE (Feb 27, 2019)

Zach Stone said:


> Sout_pu = Pout_pu + jQout_pu,
> 
> then: Sout = Sout_pu•|Sb|,


 

         Sout = (Pout_pu + jQout_pu)•|Sb|
         Sout = Pout_pu•|Sb| + jQout_pu•|Sb|



          Pout = Pout_pu•|Sb|   and Qout = Qout_pu•|Sb|
 




I really appreciate your explanation. Thank you @Zach Stone, P.E.  for enlighten us.

The above statement elucidate my confusion.

Sorry guys for making it more complicated from my side.

However, learned a good lesson from the discussion.

Thank you all.


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## Zach Stone P.E. (Mar 1, 2019)

Sdhabik said:


> I really appreciate your explanation. Thank you @Zach Stone, P.E.  for enlighten us.
> 
> The above statement elucidate my confusion.
> 
> ...


My pleasure, I'm glad it was helpful.


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## daydreambeliever (Mar 2, 2019)

Sdhabik said:


> Pleas let me have﻿ your views﻿ on th﻿is﻿.


Sorry for the late response. We had strep throat and the flu passed around our house. I had hard enough time keeping up with work, today is my first day back to studying. 

Zack's explanation makes sense. I approached the problem completely wrong.


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