0
\$\begingroup\$

I'm really confused in solving the following question.

An isolated 50 Hz synchronous generator is rated at 15 MW which is also the maximum continuous power limit of its prime mover. It is equipped with a speed governor with 5% droop. Initially, the generator is feeding three loads of 4 MW each at 50 Hz. One of these loads is programmed to trip permanently if the frequency falls below 48 Hz. If an additional load of 3.5 MW is connected then the frequency will settle down to ?

Here, is my understanding of the question. At 15 MW load the frequency should drop by 50*0.05 = 2.5 Hz so the load would be working at 47.5 Hz. Now the new load is 4*3 + 3.5 MW = 15.5 MW which is above the maximum capacity and also the frequency would have fallen below 47.5 Hz, so one of the load has to trip. So the new load = 4*2 + 3.5 MW = 11.5 MW.

If we consider linear droop from the generator the linear equation relating frequency to power would be

$$ f = -0.05P + Constant(K) $$ To find the constant we put f = 50 Hz at P = 12 MW (given in the question). So we get K = 50.6. So at 11.5 MW the frequency would be $$ f = -0.05 * 11.5 + 50.6 = 50.025 Hz $$. But the actual answer is 50.083 Hz. I know I differ slightly from the actual answer, I was wondering if my understanding of the question was right or wrong.

\$\endgroup\$

3 Answers 3

2
\$\begingroup\$

What we know:

  1. Droop is 5% at 100% load (15 MW)
  2. Frequency is set at 50 Hz with a 12 MW load (so the No-Load frequency is higher than 50 Hz)

Note: This is what would be done practically with a generator. They are never run at 100% load, so typically would be used at some under rating. For example if the generator is running at 80% load, the speed would be set to provide 50 Hz at this load point. I've assumed here that this is what is meant by 12 MW @ 50 Hz ...not that 50 Hz is the no load speed.

12/15 = 0.8 (12 MW is 80% of full load)

Therefore Droop at 12 MW is: 0.8 * 5 = 4%

If the speed at 12 MW (50 Hz) has drooped 4 %, then the speed is 100 - 4 = 96% of full speed.

No-Load frequency: (50/96) * 100 = 52.0833 Hz

Note: 50/96 = 0.520833r Hz ...this represents the frequency drop for a 1% change in load. Since it's 6 decimals I chose to use 50/96 elsewhere in the calculations.

We now need to decide if the under frequency trip will operate at 15.5 MW (The question was corrected for this to be 48 Hz)

15.5/15 = 1.03 (15.5 MW is 103% of full load)

Therefore Droop at 15.5 MW is: 1.03 * 5 = 5.17%

Max frequency drop: (50/96) * 5.17 = 2.693 Hz

Frequency at 15.5 MW: 52.0833 - 2.693 = 49.39 Hz .....(this seems to indicate the under frequency trip of 48 Hz would not trip)

**** Speculation now. Let's assume that you have the number wrong in the question and that the frequency trip does trip.

The load on the generator is now as you specified: 2 * 4 + 3.5 = 11.5 MW.

11.5/15 = 0.767 (11.5 MW is 76.7% of full load)

Therefore Droop at 11.5 MW is 0.767 * 5 = 3.833%

Frequency drop at 11.5 MW: (50/96) * 3.833 = 1.997 Hz (we can round this up to 2)

Frequency at 11.5 MW: 52.0833 - 2 = 50.0833 Hz

****This seems to give what you said was the correct answer.....however by my calculations the frequency trip would not actuate.

What is Droop?

enter image description here

Droop is a governor characteristic that reduces fuel/throttle as the load increases. It is fixed at only 1 point, it's defined point. It is a slope defined for the genset. For example a 1500 rpm genset may droop 5% at full load. So the speed will drop 75 rpm to 1425 at full load. Since the frequency of the generator is related to the number of poles and speed....the frequency drops by 5% at full load. At 50 percent load the frequency drop (droop) would be 2.5%.

With the values calculated we can plot the droop for the question:

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Yes you are right, the tripping frequency was wrong, I must have typed it wrong, it is 48 Hz. But why are you multiplying droop by (50/96) and why does droop keeps changing with load ? \$\endgroup\$
    – Vedanshu
    Jan 23, 2017 at 8:46
  • \$\begingroup\$ Droop is a constant at a defined level (typically 100%) of load. I've added to the answer. If the frequency dropout point is 48 Hz....then the load would not drop off, and I'd suggest that the answer is 49.39 Hz. I'd be arguing the question is wrong! \$\endgroup\$ Jan 23, 2017 at 17:23
  • \$\begingroup\$ I disagree with "Droop is a governor characteristic that reduces fuel/throttle as the load increases." It can not be right that the governor reduces the the fuel going in, as the load increases!! Perhaps something like "Droop is a governor characteristic that increases fuel/throttle as the load increases in a controlled way. Similar to a proportional control system. The net effect being a slowing of a generator as its load increases. This enables synchronised generators to share power without 'fighting' since they share the same characteristic." Overall a good explanation. \$\endgroup\$ May 22, 2022 at 13:19
  • \$\begingroup\$ @SteveRichards, most governors are simple mechanical proportional controllers, they are not PIDs. This means there is a constant error which implies Droop, This error between the amount of air/fuel set by the controller is the measured droop since it is always less than required to meet the setpoint. I can't think of another way to explain it ...the controller delivers less air/fuel than required. When you have an electronic controller (PID) you can adjust the droop towards zero ....but you NEED droop if you are going to parallel generators or you cannot load balance. \$\endgroup\$ May 24, 2022 at 15:31
0
\$\begingroup\$

solution to this GATE problemDroop characteristic shouldn't change. 4 MW will trip due to overload not due to frequency constrain. And you can solve it by drawing graph with 5% droop characteristic.

\$\endgroup\$
1
  • \$\begingroup\$ Welcome :-) Thanks for this answer. However, writing it on paper and photographing that paper, has several problems here e.g. (a) someone using a screen reader (e.g. blind) cannot "read" your answer; (b) Google (and search on this site) cannot index your answer; (c) users on a small screen (e.g. mobile) has more problems to read your answer, where text would be easier to read. These are some of the reasons why this style of answer is not encouraged. If you can replace the main image with a text version (and only add the required diagrams as images), that will be better for readers. Thanks :-) \$\endgroup\$
    – SamGibson
    Jan 23, 2019 at 15:37
0
\$\begingroup\$

It is mentioned in the question that frequency is 50 hertz at given load of 12 MW so frequency will change when you add additional load to this initial 12 MW

Solution:

For 15 mw droop is 5 percent so if we add additional load of 3.5 mw it will create an additional droop percentage of

(5/15) * 3.5

Therefore change in frequency due to increase in load will be 1.166/100(above result . I kept 100 in denominator as it is droop PERCENTAGE) * frequency = 0.583 New frequency = 50 - 0.583 = 49.417

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.