# parallel generators problem



## PE blues (Mar 29, 2013)

Can someone please show the solution?

G1 has no load frequency of 61.5hz and a slope of 1MW/Hz. G2 has no load frequency of 61hz and a slope of 1MW/Hz

The two generators are supplying a real load totalling 2.5 MW at 0.8pf lagging.

a) at what frequency is system operating and how much power is supplied by each of the two generators

here's how i tried to solve it. If speed of one increases, speed of the other machine should decrease

gen 1 droop =(61.5-n)/61.5

gen 2droop =(61-n)/60.

the generator ratings are not given so how would you calculate the load shared by each generator

b)load is increased by 1 MW, what is new system frequency and how much power is supplied by each generator

c) with system configuration in part b, what will be the system frequency and generator powers be if governor set points of G2 are increased by 0.5 hz

I would appreciate any response.


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## EItoPE (Mar 29, 2013)

I found an iteresting article on this subject. Check Fig 5 and 7 of the following:

http://www.wecc.biz/library/WECC%20Documents/Documents%20for%20Generators/Governor%20Tutorial.pdf

Now, to answer your question,

a) G1 output is 0 MW at 61.5 hz. It follows that G1 output is 1.5 MW at 60 hz

G2 output is 0 MW at 61 hz. It's output is 1 MW at 60 hz.

So at 60 hz, total output = 2.5 MW

b) In a similar manner, let's just check what's happening at 59.5 hz.

G1 ouput is 2 MW at 59.5 hz

G2 output is 1.5 MW at 59.5 hz

So at 59.5 hz, total output = 3.5 MW

c) Total output = 3.5 MW

G2 share of the output is 1.5 MW from (b). Now it's governor increases speed to match 60 hz at the same output of 1.5 MW.

In doing so, the droop curves for both the generators are now same.

So now at 60 hz, total output = 1.5 + 1.5 = 3 MW. But the actual total load is 3.5 MW!

Since both generators have the same characteristics, they will share the extra load.

Result: at 59.75 hz, each generator carries 1.75 MW.


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## PE blues (Mar 30, 2013)

how did you calculate the 59.5Hz frequency?


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## nasir (Mar 30, 2013)

Mr. PE Blues!.

Here is easy way of calculating. Below is the link.

P = Slope(Fnl-Fsystem) = amount of power supplied by the generator.

See this link for clarity.

http://www.ece.ualberta.ca/~knight/electrical_machines/synchronous/parallel/fpvq.html

Thanks For asking this question. I did not know until i read your question. Probably it is one of those easy questions that we don't want to miss

in the exam.

Good Luck.

Nasir.


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## Wael (Mar 31, 2013)

EItoPE, I didn't get the solution. lets start by a) how did you conclude that G1 is 1.5MW at 60HZ. what formula did you use?


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## EItoPE (Mar 31, 2013)

I used a simple graphical method with Frequency (Hz) as x axis and Power output (MW) as y axis. Please see attached. Curve [1] is for the 1st generator and [2] for 2nd one. Point [a] is for case a and * for case b.**Parallel generators Sheet1.pdf*


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## EItoPE (Mar 31, 2013)

EItoPE said:


> I used a simple graphical method with Frequency (Hz) as x axis and Power output (MW) as y axis. Please see attached. Curve [1] is for the 1st generator and [2] for 2nd one. Point [a] is for case a and * for case b.*
> 
> 
> 
> ...


*
*

*Sorry the attachment didn't show up right. Here is the attachment again.**Parallel generators.pdf*


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## PE blues (Mar 31, 2013)

here's how i calculated MW sharing, P=slope (fnl-fsys)


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