Practice English Speaking&Listening with: Module 3 lecture 2 Power System Operations and Control

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Welcome to lecture number 2 of module 3. In previous lecture, that is lecture number 1

we discussed the modeling of governing system. And now, in this lecture I will try to model

the turbine as well as the generators. And so that we can see how we can control the

frequency of the system and how the various loops are working in this ALFC or you can

say load frequency control.

So, to have this turbine model as I discussed we can have the steam turbines or we can have

the water turbines. Means the turbines which are used for the hydro power generators that

is called your hydro turbines here I have written this. And if we are using the steam

as an energy flow medium then it is called steam turbine. Steam turbine can have either

re heater or without heater they can work. If it is without re heater then we can run

the steam turbine in a very simple transfer function that is a G T G is the transfer function

and T denotes the turbine in the S domain. I can say here it is a KT that is a gain after

turbine over 1 plus stt and this TT is the time constant of the turbine. And normally

this tt is approximately that is a line between 0.1 to 0.5 second. For hydro turbines here

the transfer function is slightly different and here I can write the gain of the turbine

is KT that is a multiplied by 1 minus 2 STW divided by 1 plus STW.

And here TW is the time delay, why this time delay is involved? Because the water is coming

from very far end it may be through the and other things and it will be having the time

delay as well. But if your steam turbine is having the reheating facility then we have

to model that re heaters as well. So, the turbine with the re heater now, what we can

do if it is having the re heat facility means certainly your turbines will be having more

than one stage. That it will have HP turbine that is a high pressure turbine here I have

written this rating of high pressure turbine. Let us it is alpha I am talking in the per

unit system then if a total power rating of the turbine is unity then I can say the rating

of IP and LP stages is 1 minus alpha. So, the total is 1 per unit if we add alpha

and 1 minus alpha. You will get unity means if the total rating in the per unit is 1,

then IP sp stage is having alpha per unit then the remaining stages may have IP or and

LP it will be 1 minus alpha. So, the time constants for the re heater it is normally

4 to 40 second, because the steam from this outlet of HP will go to the boiler. And then

after re heated it will be coming back to the IP and or LP stages. So, it is a time

travel involves and that is normally 4 to 10 seconds. So, static gain again I have used

the KT as the static gain of the turbine and tt is the time constant of turbine that we

have used here you can see. So, using all these then we can have the time constant or

we can have the transfer function models.

Again this how the steam flow which is going even though I explained in the previous lecture

that here. You can say your HP turbine that is high pressure turbine here it is your intermediate

pressure turbine. And here it is your low pressure turbine and all these are connected

on the same shaft. And it is coupled with your generator G here that is called generator.

So, once you can see here the steam which is coming from the boiler it is going to the

SP and you can see here the expansion process is done. And due to the expansion process

this steam will do work and that work will be in form of the rotation of the shaft. And

after this is which is coming out that is steam after expansion means it will release

the energy which is in the steam. It will going back to the boiler again and here it

is basically reheated. And after once it is reheated it is called a re heating processor

process. And the finally, it is again here coming to

an IP state were again it is doing the work. Again here I want to mention that the output

of this SP turbine is reheated, because the energy consumes in SP is very very high and

then the steam which is coming out from the SP turbine it is having low pressure and thus

low temperature as well. So, here there will be if it is not reheated there will be possibility

that this steam will be converted to the water. And here and that will damage the blade and

the efficiency will be poor. So, if you are reheating then again we can utilize the remaining

energy in terms of here IP and SP turbines. This IP here again the steam is going out

and the finally, it is going to the LP here where expansion process taken place and then

after coming out from the LP it is this steam is condensed.

So, we try to reduce the temperature as you know this temperature difference normally

gets the work done. So, here if T 2 we can minimize that is the T 2 here it is a T 1

that relates the efficiency of this whole cycle that is a steam cycle here. So, we have

the cooling towers and we normally condense this steam we convert it to the water and

this water is still having the high temperature. Now, then we are having some pumps once it

is condensed then we are sending to hp heaters. Because some it again it is heated, because

the energy which is coming out the boiler it is heated here. And then this heater water

is passing through the boiler and there will be some leakage here this demineralization

water means normally it is called dematerialize there is no mineral is involved otherwise

that will damage the blade of the turbine. So, here thus the leakage etcetera will happen

some steam will be going out in the air. So, that will be compensated by the DM water and

this water is combine together and then we are using the BFP that is called boiler feed

pump. This boiler feed pump again it is passing through the IP heater. Because we there is

so many heaters stays, because the energy which is coming out from the boiler that is

utilized then we can increase the efficiency to HP heaters. And LP heaters are used and

then it is coming to your boiler here there is boiler it is heated and that is a steam.

So, this is basically a com complete cycle with the reheating facility.

Now, to, go for the derivation for the transfer function of the turbine with reheating facility.

Here I can say that this HP part the power output of this HP turbine that can be related

with the simple transfer function that is a KT 1 plus ST. And then it is multiplied

by here you can say alpha term and that is coming multiplied by the valve output. Means

here this is your turbine this is your changing the valve power. And here that is we are getting

change in the pth that is a high pressure turbine. So, this transfer function, because

here we are using the only the alpha times of that one so, here we are just multiplying

alpha. Now, for the LP and IP unitspower here now, 1 minus alpha is multiplied by this

transfer function. Because the remaining energy is going there plus here there is a some travel

time. This TR that is a reheating times constant tr that is added here and then it is your

P. So, the total power means here the total turbine

power is nothing but your change in the power here of your THP stage and plus here change

in your power at IP and LP stages. So, here you can see this is this expression. So, the

total output of the turbine including our various stages will be the summation of all

the power of the stages. So, here your HP stage here it is your LP and combination of

your IP stage. That is a intermediate pressure low pressure and high pressure turbines. So,

if will add and then you can simplify then you will find that KT 1 plus St is common

here we can take. And then we can go for here now, what we are getting? This stage here

alpha plus I can say 1 minus alpha here 1 plus STR and this is simplified and will get

this figure. So, you can see we are getting a complex transfer function of the complete

turbine which is having the reheating facility. To go for the simplicity let us take it is

not having the reheating facility. So, only we have to ignore this means we have

to put alpha is equal to unity means only HP part is there high pressure turbine is

there. So, this part here this part will vanish. So, will have a simple turbine transfer function

that is retain here and I have used this simple turbine without reheating facility. So, it

will come here this turbine model this is the governor model if you remember which I

derived in the lecture number 1. Now, this PV which is coming the here what is happening?

This your change in the PC is the change in the differentiating here change in the frequency.

That is coming through the drooping characteristic with the negative sign. It is now, coming

to your governing and governor is taking action whenever there is change in frequency or change

in the power reference this governor will act.

And it will try to change the input to the turbine that is a steam and that is you can

say valve power that is a change in PV. And this PV now it is steam power that is coming

to the turbine and finally, it is giving the turbine power or you can say rotational power

that is where your generator is connected. So, now, this is the model of your governing

system this model this is your turbine model. And now this is going to your generator where

generator is coupled and we will see the generator model later on. Normally we try to make in

such a fashion that the multiplication of Kg and KT here we normally keep it unity.

Means if you governor Kg gain and your here the turbine KT we normally rejoin in such

a fashion that we can make it unity. So, what will happen? Now, the multiplication of this

will be unity and we will see in the later part as well.

Now, let us see now keeping this KT and your Kg is unity we can write here this change

in the power. Means from here from this expression from this expression we can derive what will

be this change in the turbine power by in terms of you can say change in the reference

setting and in terms of frequency setting and we can write in the simple way. Because

this minus one upon R change in FS and that is multiplied by these 2 time constants or

you can say transfer functions. So, here you can see that is what we are getting this is

the transfer function of governor; this is the transfer function of turbine. That is

multiplied by here change in this reference setting minus change in the frequency divided

by R that is a regulation or cons slope or you can say drooping characteristic we will

see that later on again. So, we can write this is a complete this input

output relation. So, in this case input is your change in frequency and change in your

reference power. Now, to see the impact or you can say the change in the steady state

turbine output power that is a change here I normally say this change in PTSS the SS

denotes the steady state condition. You know this final value theorem. If there is a any

change in any of the input then as per final value theorem we can obtain the steady state

value by limiting S tends to 0 multiplying S with this your output here that is change

in turbine power in S domain. So, here once we are going for this and we are putting here

now, we can get the steady state value of the turbine that is a PTSS. First case I am

assuming there is no change in the frequency of the system. And we want to change the output

of the turbine by changing the step response. Means suddenly here you can see in this here

step change here time is change and the differentiating is changed up to this magnitude. So, this

is the step input response just I want to know here this change now, in the domain I

can say S domain the change in the PCs will be nothing but for this one. We know it is

the change in PC that magnitude here this is your magnitude divided by S is your step

response. So, after putting this value here and putting your change in FS is 0. So, and

putting S multiplied here we can get this steady state value in the turbine output.

That is we can see here what we are going to get here we are to change it here what

we will get here S and here this is this by s. So, S S will be cancelled and all this

S term will be 0s here 0. So, we are going to get here this expression.

So, it shows that whenever you want to go for the steady state increase means there

is a step input it will be finally, it will be settled to its power turbine power will

be increased. So, when you are willing to increase the power this steady state value

will be increased of course, we have the time constant. So, it well may require some times

and it will settle down, but the final value will be change in PC. In another case let

us suppose I do not want to change the p reference that is a PC means I do not want to give raise

or lower command. So, and frequency is changed means suddenly if system frequency changes

and again I am talking about the step change in the frequency.

Here again I can say this is your time and change in frequency here that is your change

in f. So, this magnitude is changed. So, I want to see how much the turbine output power

is going to be changed during the steady state condition here suddenly change means there

will be some settling transient and finally, it will be settled. So, here what in this

case now this change in the PCs is 0 and we can put it 0. And we can get the expression

from here from this expression here we are getting change in the PTS will be 1 over this

governing transfer function here your turbine transfer function and multiplied by change

in f divided by SR. Because here that change in f is here change in f divided by S and

R is already regulation.

So, again using your final value theorem in S domain we can get this ptss here means we

can get again this change in ptss is nothing but limit S tends to 0 f into change in PTS.

So, using this we can get this expression final that is with the change in the frequency

in this case we did not consider the change in the reference setting. So, now, you can

see for change here the 0 means there is no change in the differentiating the increase

in turbine output power is directly proportional to the frequency drop what I want to say why

it is drop? Means here I am talking this increase here I am talking drop, because the terms

we are we are getting here is the negative. Means if your frequency falls down the turbine

output must increase to maintain the frequency as we know we require that and this is giving

that characteristic. So, whenever there is a increase in the frequency

then turbine power must fall here the change in here is the positive then change in the

turbine output is negative so, it is a vice versa. So, in other words I can say if you

are changing the frequency I can say your system frequency increases then your turbine

power output will reduced. Or this is reducing then turbine power must increase to compensate

the remaining energy. And therefore, we can maintain the system frequency. In this expression

here you can see the R is appearing and that is 1 over R and that here it is called the

droop of the governor. This characteristic is called the droop of governor and normally

the value of R is kept between 2 to 5 percent. I will come to that point what is the 2 to

5 percent? Basically the unit of R you can see here it is nothing but you can see the

unit of R here it is Hertz per mega watt why? Here R will go here means we can get here

this R is nothing but your change in f divided by change in power of course, it is negative

sign. So, unit is not related with the plus and

minus sign, but it related with the variables that is are coming in to the picture. So,

this is a frequency this is your power. So, it is hertz per mega watt is the unit in actual

value. But in the percentage it is we can write again I come later part you will see

how that we can change from per unit? To means suppose one we are taking per unit another

will taking actual value or both are we are taking per unit here both are taking actual

value and we will see that. So, this here gives a relation that your change in frequency

will be negative of the change in the turbine output. So, M power is increasing frequency

here it is increasing what will happen? The frequency of the system will be changed accordingly.

So, the increase in the turbine power system frequency falls means the power must increase

to compensate that. To see that let us see 1 characteristic here.

This characteristic is nothing but you are the droop characteristic and here normally

you can say this is the nominal frequency in our case it is 50 Hertz. And this is from

50 hertz to that is we have increased to 1.05 means here it is 1 per unit you can say. So,

this is the change in the frequency and here change in the turbine output power and this

slope which is showing one upon minus R. So, we have the family of curves this is 1 here

we have 2 here we have 3 curves and 4 curves what does it shows that? Here for 100 percent

power if this load is removed if this load suddenly comes to 0 your frequency will be

increased by 5 percent. So, your frequency will become 1.05 means change will be point

here now, the change will be 0.5 per unit. So, this gives that is if you are loading

100 percent your frequency if it is 100 per rated 250 hertz and if you are just changing

then again youre this will be there will be change here with the 5 percent. Means now,

the load is reduced to 0. Now, we are going to change in the frequency

here is 0.5 multiplied by 50 means there will be change in 255 hertz in the frequency. So,

we have the family of curves again they are having the slopes. So, now, suppose you want

to have this what will be? If you are your operating that you are maintaining controlling

your frequency for 75 percent here of the loading. If you are increasing here then how

much you are going to get how much the now, you can see this is the frequency which will

be if this value at this point if you are reducing then your how much frequency this

increase just you are going for have. Means for 70 percent increase here this will be

your frequency increase. If there for 50 percent here your frequency increase will be this

one and for 0 percent you will get this one. So, this relates the loading of the turbine

with here the frequency and again it depends upon this characteristic that is a slope here

R that is very important.

Now, as I discussed that is what this is? Let us suppose R 0.05 per unit as I said the

unit of R can be your hertz per mega watt or it can be your hertz per unit mega watt.

Means here the mega watt is in per unit. So, just I am writing, because the per unit is

very confusion, because per unit of hertz will be different than per unit of mega watt.

So, it is it is better write the per unit mega watt means power we are using the per

unit quantity. That we can also write here pu hertz over this pu mega watt and this is

called completely pu means your change in the frequency in the per unit here and the

per unit mega watt it gives you per unit value. Now, for this you are let us supposing your

base power is 100 mega watt and your frequency is 50 hertz then R is equal to 0.05 per unit

is nothing but we can write here 0.5 per unit hertz divided by per unit mega watt.

Now, we want to calculate in the actual value. So, here 0.05 now, from per unit to actual

value we have to multiply by it is the base value and that is 50 Hertz. Now, we are getting

here hertz divided by here hertz that is a power 100 and now, we are getting mega watt.

So, we are getting here 2.5 divided by 100 here that are hertz per mega watt. So, now,

this is the way of converting and finally, you can see this value just we are going to

get 0.25 hertz per mega watt. So, this is very important you can say we started in the

per unit 0.5, but this value has actual value is this one means if you are changing 1 mega

watt power here there is a change of 0.25 hertz in the system frequency.

So, if you are increasing then it will be reducing as I said we had the relation. That

we are going to the change in here is minus or change in turbine power. So, it is with

the negative sign. So, when you want to increase the turbine power that is increased system

frequency. Here if you are increasing system frequency will increase, but we want that

reverse action. So, if the frequency falls down this turbine will automatically try to

increase the this output because here the R sign is there. So, this is basically the

concept for this R that is a regulation. So, we saw the combine effect of governor and

turbine modeling. Now, let us model another very important component in this load frequency

control that is generator.

So, the generator modeling that consists here we can now understand that this is your mechanical

power or mechanical energy input through the turbine which is rotating here this is rotating.

So, the torque is developed and that is called the T mechanical torque. Now, on the shaft

your electrical that is a generator is connected which is giving output in terms of electrical

energy. And you know the turbine generator system is nothing but we are from mechanical

energy we are converting to the electrical energy. Although in the whole process it is

not only the mechanical energy is converted to electrical energy. But he fuel energy that

is the coal energy or you can say water energy that is in hydro power stations that is coming

to the turbine. So, that energy is converted to the mechanical energy and from mechanical

energy we normally get to this electrical energy.

So, you can see this turbine torque T here that is coming this T mechanical that is going

in 1 direction here this. Let us suppose this rotating trying to rotating, but the electrical

that is output it will try to oppose and that is a basically the equilibrium condition.

If this torque is in the same direction what will happen? This will keep on accelerating

and system will be out of step. So, always had the steady state condition when the T

mechanical energy is equal to your electrical energy in that case here this T mechanical

will be equal to your the T electrical and the speed of this rotating mask will be the

constant. If there is a balance imbalance here means if 1 quantity will increase let

us suppose the T mechanical is increasing more what does that mean? That your mechanical

energy is more than your electrical energy this machine will try to accelerate. Means

if your T mechanical is greater than T electrical the speed of rotation body will increase by.

Because this energy going output here output energy input energy some energy is here kept

as you know the energy cannot be wasted. So, as per energy conservation law that energy

is going from one form to another form means that mechanical energy since it is more we

are getting less in energy. So, that energy is mechanical energy is going to convert it

into the rotational energy and that is nothing but your kinetic energy and that kinetic energy

you know that speed will increase and then speed is increased and that is we have the

here. Reverse is also true if we are feeding less mechanical power, less mechanical energy

and your electrical output is more then it is a reverse situation. Means you are drawing

more power from this whole system this is your complete system where rotating bodies

is there. So, if you are getting more then what will happen? Whatever the energy which

is stored in the system here that will be coming out. Means the stored kinetic energy

is released in the electrical energy and this if it is ki k is reduced means the change

in frequency will be reduced and then frequency will fall.

So, in I want to say that T mechanical that is torque acts to increase the rotational

speed it will try. Because input you are giving it will try to increase whereas, this T electrical

will acts to slow down the rotational speed and thus it will do some work and then electrical

energy will be generated and that is in three phase power that is you are going to get.

To model this we have to have this rotating body concept, because there is a some energy

that is a input and output. It has some inertia that is a inertia constant I can say I which

I have used I is a inertia constant and here it is rotating that is we have the angular

speed omega will also use this angular acceleration alpha, because if there is a any mismatch

this there will be changed in the omega and this alpha is your angular acceleration. And

now, another term that is called m. Means angular momentum which is defined as the inertia

constant multiplied by your speed that is omega. So, the basic relationship just I want

to derive the Ttnet is nothing but Tnet I can say it is your T mechanical minus T electrical.

Electrical means what is coming in and what is coming out that difference is called Tnet.

So, and that can be related with this here this inertia constant I multiplied by here

the acceleration that is a power. So, this torque here divide multiplied by the acceleration

and in inertia constant that is related by the Tnet angular momentum already I derived.

Here this M will be equal to your omega dot i. Now, this Tnet as you know this torque

if it is multiplied by speed then it will give the power so, the torque here and the

power relationship with this expression. Now, if we will put the Tnet value from here this

first equation we can put here. And now, we can use this omega into I here this relation.

So, we can derive M into angular acceleration that is alpha will be nothing but your Pnet.

Here we are talking all the values in actual quantity you must remember this because in

the per unit this Pnet will be equal to Tnet. Here we are talking this p is in megawatt

this is in Newton meter and this is in radiant per second. So, here everything here I am

deriving in actual quantities later on will convert to the per unit if we want to calculate

in the per unit otherwise you can also us in actual quantities.

Now, So, as I said the Tnet is nothing but it is the T mechanical minus T electrical

and this is also called net accelerating torque and sometimes people called the Ta. So, Ta

Tnet in different books you will find the different names. So, this is nothing but the

difference in the input torque to the output torque. So, that is the net which is inside

the rotating mask and that is called net accelerating torque. M is angular momentum of the machine;

I as I defined as the moment of inertia of machine; alpha that is a rotational acceleration

in radiant per second square and omega is your rotational speed in radiant per second.

The difference in electrical and mechanical torque will cost to accelerate or decelerate

as I said if Ta is positive then machine will accelerate as already I discuss here. That

here this concept is the difference here if it is more, then is to be accelerate and if

it is a difference is negative then it will try to de accelerate.

So, here the difference in the electrical and mechanical torque will cost to machine

accelerate or decelerate. The speed of the machine under acceleration will be that is

a omega naught plus what is your acceleration? Acceleration here the rotation acceleration

multiplied by time. So, this will be your the speed of machine under the acceleration.

If there is no acceleration then the speed will be your omega naught omega naught is

the base or you can say constant speed where both are balanced the T mechanical is equal

to your T electrical. So, this relation is very widely used and will see in the later

part of this lecture. We also know that this omega is nothing but this here the phase angle

of rotating machine is delta then we can relate this omega with your change rate of change

of angular phase angle that is of rotation machine or in other words I can say this delta

is the integration of omega with respect to time that is deceleration.

Now, let us see here. How we can so, I want to calculate what is the change in the angular

or you can say phase angle of the rotating mask. Here what I want to derive let us suppose

this is your reference value. Means you can talk with the some any reference value and

here this is your rotating mask which is rotating. Normally in this rotating system here if it

is a reference from here this is rotating both are rotating with the same speed then

this angle is fixed and it is you can say locking torque angle. But if this router will

try to accelerate more than this delta will try to increase and this is also rotating

this is rotating as omega naught or you can say at the synchronous speed. Both are normally

in the steady state they are operating at the synchronous speed. So, that this torque

will be this delta angle is fixed. But if it is accelerating what will happen? This

angle will increase if it is decelerating then this angle will decrease and vice versa.

So, change in this angle I can say this acceleration minus with their initial that is a phase angel

of the reference angle. So, change in the delta that is here machine absolute phase

angle here that is we are going for this minus with the reference how much it has reached?

And from this we can subtract then we can get the change. So, change in the phase angle

is retained as the integration of omega naught plus omega T and this here we are into dt.

And here basically I am sorry this is alpha not omega as I said we wrote this expression

you remember this is your omega is nothing but omega naught plus alpha T which is I am

using here minus here that is a phase angle of the difference angle here omega naught

or you can say synchronous speed. So, if you are integrating here you will get omega naught

t, because this is a constant. So, this T term will be appearing here alpha

is constant. So, only in the case it is a constant if it is a change in alpha then it

is a different. So, we have to alpha is also changing we are assuming this alpha is constant.

So, this alpha here now T could T square upon 2 will be there minus omega naught T here

that we can get it. So, finally, what is happening? We are getting the change in the phase angle

is nothing but 1 over 2 alpha T square. Now, the deviation from the nominal speed what

we are trying to do? This change is nothing but it is we can see here 1 over delta we

have to derivate this means change in the angle divided by dt and if you are derivating

this you are getting this. So, we are getting the change in speed is nothing but your alpha

T or we can also if you remember I said here omega is nothing but your omega naught plus

alpha t. You will take this side so, omega minus omega naught that will be equal to your

alpha T and this is nothing but your change in omega.

So, we also get the same thing from that expression and also we can derive from here this expression.

So, this is equally valid and this is the equation. Means change in angular speed is

nothing but your time multiplied by your acceleration angular acceleration this I am talking. We

also had the 1 relation that Tnet is nothing but your I that is angular here I is your

moment of inertia and alpha is your angular acceleration that we can multiply together

so, we are getting this. Now, this alpha this alpha is nothing here you can say what is

alpha? Alpha is omega upon here t. So, I can say here alpha is nothing but your change

in omega over change in time. So, we can write here I into the change in speed over change

in the time and we can get this. And it will put here this value omega in this difference

here we can get I multiplied by the double differentiation of change in phase angle delta.

So, we are getting 1 relation here. This means in the simple way I can say here I into d

square over dt square change in phase angle will be equal to your tnet. Now, here just

we have derived in a change in angle is very good. We want to here see the change that

is a Tnet from your the differentiating means here this we have Pnet. Pnet is nothing but

your I can say the p mechanical minus your p electrical. Or in similar fashion I can

say this Tnet is nothing but your T mechanical of course, minus T electrical means here this

the Tnet is the difference between mechanical and electrical and a power net is your mechanical

p mechanical minus p electrical. We know that this during this initial case this I can say

it is nothing but your Pnet at the where it is the difference actually we are talking

plus some change in Pnet or we can say how much thus we are changing from one value to

another value? So, this is your Pnet that is we are change in the Pnet we are getting,

but the p mechanical is equal to your p mechanical not which was earlier plus your p mechanical.

Or in other words I can say here we can derive this expression I can use some space here

what is happening? As I said your the speed net it is as I said your p mechanical minus

p electrical and we know that this p mechanical p mechanical. I can write is equal to your

p mechanical that is during the steady state minus sorry plus change in p mechanical. Similarly,

we can write this p electrical is equal to your p electrical not plus change in p electrical.

So, what happens if you want to calculate this change in Pnet if you put these values

here and as here? So, you will get the change in the Pnet that is the real power will be

change in the mechanical power from the not value, from the steady state value, from the

during the steady state value just I am talking. So, this minus here you have to change in

the p electrical value. So, also this Pnet since we have a change in the Tnet means the

accelerating power has changed means it is not 0.

So, the speed will change and the speed will change from your omega naught. So, omega naught

plus change in speed that will be multiplied by your Tnet that is a which is a difference

between the T mechanical minus T electrical and thus here we can define it. So, the Pnet

will be the multiplication of this factor multiplied by the tnet. And this Tnet is nothing

but your Tnet not that is you are the in steady state Tnet and in a steady state it was 0.

Because the mechanical was completely equal to your electrical if we ignore the losses.

So, that plus change in the Tnet how much that is the changing and that Tnet change

is nothing but your g Tnet change in the mechanical minus change in T electrical. So, if we will

simplify this means Pnet can be also retain in the Pnet not plus change in Pnet that will

be equal to you can multiply here. So, you will get the omega naught this factor

is multiplied by this factor. So, this is this factor this factor multiplied by this

factor will get this factor. This omega will be even change in omega is multiplied by the

here this factor we are getting this another one. And omega naught if multiplied by the

change in the Tnet we are getting this factor. So, will expand this will get the 4 terms

now, what is Tnet naught? Tnet naught or Pnet not they are 0 why I have written here? Because

in the steady state, what was this p mechanical was equal to your p electrical? So, this net

is 0. So, here the change in the Pnet at the not in steady state it was 0. Similarly, this

Tnet will be also 0, because if p is 0 the T will be 0, because the p is related with

omega. So, if will put these values so, what will happen? This will be your cancelled out

this will be your cancelled out and this will be your cancelled out.

So, what you are getting? You are getting change in Pnet that will be equal to your

this term omega naught into the Tnet plus here you are getting omega into change in

Tnet so, you are getting 2 terms. Now, change in the Pnet normally you can say this term

is very small and this is also small. So, we are just changing the speed change multiplied

by change in the net. So, this value is very small. So, normally we ignore this one also,

because this is having very less impact, because this value is very fraction. So, we can ignore

and we can put it 0 here already I have put 0 here 0 here it is 0 So, we are getting the

relation the change in Pnet here that will be equal to the omega naught that is the best

speed multiplied by the change in Tnet. So, from this equation here already I have use

this one. You can see his Tnet change in the change in mechanical power minus change in

electrical power here we have retain. That will be equal to your omega naught and this

is nothing but Tnet is your change in mechanical torque minus change in electrical torque and

we are having this expression. Now, here what we are now, we are going to

relate these term with our previous one here our intention is to relate in the Tnet form

here. So, to see that now, Tnet what will be the Tnet will be? Tnet is the T mechanical

minus T electrical torques that will be here, this mechanical plus change in the mechanical.

And here minus torque electrical naught at this base case plus change in the T electrical

what is happening? This and this will be cancelled out because they are in a steady state case

both are equal. So, we are getting T mechanical here is it this change in T electrical. So,

here this is what we are getting this is nothing but your tnet. So, this we can write the Tnet

now from above the previous equation here this Tnet that is equal to I can write here

this expression or we can we can say here this change in this or you can say Tnet will

be equal to I into d upon dt change in omega.

So, again we can simplify here from equation star I can say this change in the mechanical

power now I am writing in terms of power, because all the quantity we represented in

the power represent torque. In the power system it is a torque has no meaning, because it

is very difficult to calculate the things in the torque. It is better to go in the mega

watts and the frequency terms so that we can easily solve the problem. So, here the change

in the mechanical and change in electrical powers that will be equal to be here I have

multiplied this and this is your nothing but the tnet. So, here omega naught is multiplied

and then omega naught into I is nothing but it is your M that is your angular momentum.

So, now, I can see here we are having expression that M d over dt change in omega that will

be equal to your change in p mechanical minus change in p electrical so, this p mechanical

what is this? This p mechanical is nothing but your turbine output.

So, this is your turbine output that is coming and if you are taking here the laplase transform

of this expression. We can have in the S domain. Change in the mechanical power in the S domain

minus change in electrical power in S domain that will be equal to M that is angular momentum

multiplied here that differentiation is there so, S term will be appearing in change in

omega s. So, what we are getting? We are getting here the expression for this we can write

in the transfer function. Here you can see the change in the mechanical and that change

in mechanical power is nothing but the output of the turbine. That is a change in PT which

we used in the modeling of the turbine minus here the change in electrical. What is the

change in electrical power? It is nothing but that is a load on the alternator. So,

that is also is related with the change in pl will see later on this pl is not directly

this electrical there is some damping also so, approximately we can say it is a equal

to the pl. And that is coming to here is one over S if it is multiplied then we will get

the change in here the speed. We also get 1 relation the very carefully

what we had? We had this change in omega is nothing but it is your d over dt change in

delta. So, if we want to write here omega S that is nothing but your S into change in

delta s. or here what we can do if you want to write in terms of delta. So, this is multiplied

by a I can say delta S here I can say it is 1 upon s. So, here we want to go for even

through delta terms also we can get it, but our concern here is the frequency rather than

delta. So, this is nothing but your change in omega is nothing but your changes in frequency

both are same thing there is no difference. So, here I can write here this omega S change

always here it will be your change in the frequency if we are using in the per unit

system. If you are not using the per unit system then this omega is radiant per second

it is in hertz both are different will see some problem how we can going to relate it?

Now, we have to now, combine all these models. Models means model of governor turbine and

this generator let us see what we are going to have.

Now, we had here if remember here it was your the PC that is coming your change in the pcs

here it was positive then we had some here the term with the negative. And we are had

some drooping characteristic that is 1 upon R with the negative sign and here it was coming

your change in fs. Now, this was coming and it was coming to the governor. So, now, the

governor time constant and I have to now, write the governor and the turbine together,

because I said that kg multiplied by KT is kept unity. So, both now, coming together

so, I can write 1 here Kg into KT. I can write 1 plus Tg that is a governor time constant

S here 1 plus your TT S here of your turbine. And here we are getting your change in PT

and that change in PT is nothing but is the change in mechanical power and here again

we are going to add this. And here we are getting change in this is your plus this is

minus and this is nothing but your change in the load. Means change in electrical power

and then here we are getting a turbo 1 over ms and here we are getting change in omega

s. So, this is the model that we are going to

get this is also related with the change in the frequency if we are changing the unit.

So, this frequency here is nothing but is coming here. So, the value here the change

in the frequency if we are using in the per unit this change in here will be there. Otherwise

you have to change with some constant here k and then it is fed to this means omega is

related to your frequency and then it is a loop. Now, you can see this is a loop and

we have to go for the solving and you have to see the system. And this is normally called

this primary control and will see other secondary and tertiary controls later on But here this

load is not constant it is also affected by the change in frequency in the power session

there are various loads they are also change if the frequency is changed how? For example,

let us suppose your induction machine is there. And once the frequency is changed its torque

is reduced or even the torque is changed and once the torque is changed means the output

of that machine is changed. So, we have to model the system we have some

damping coefficient that d that is related with your frequency and then you are this

load. So, we have to relate the frequency and this electrical load with that and then

we will see how much we are going? So, that is why I said here it is nothing but it is

approximity. So, we will see this we will try to model that this change in that is the

change in p electrical here it is your change in the load plus some other factor and that

basically the modeling of load will be the system here. Now, let us see in the power

system the there is not only one generator there are so, many generators. So, and they

are having the different R characteristic that is regulation constants R will be the

different for the different one. How they will share the load if they are operating

in the parallel? To see this as I said the characteristic that is we can draw here.

This is a load that is a p that is they are of generator 1 that is you can say pg 1 and

another generator here it is I can say pg 2. Now, 1 generator is having this slope this

is very deep and another is having very less slope so, what happens? Let us suppose they

are operating here. At the some this is the frequency operation frequency here and since

they are connected parallely so, the operation of the frequency will be same. And let us

take this is your f naught that is the nominal frequency means they are operating at the

rated frequency. So, the loading of this generator will be nothing but this will be sharing here

p 1 and this will be sharing your load p 2. Now, you can also see here those are having

here this less slope those are loaded much. Means they are taking more load here from

you can say from 0. Now, if there is a suddenly release in the load or you can say let us

we put more load on the system is very important. I just I change a change in the loading that

is electrical power just we have increased this value just it is a positive what will

happen? Means now, that load must be shared by these 2 generators.

So, and they must share in a such a fashion that steady state frequency is again the constant.

And that is will be fixed and it will be of course, the lower than that, because the power

is increased. So, what will happen? Here you can see now let us suppose here this frequency

has said and that will be parallel here it is some by loop f prime. Now, you can see

this will be changed from here p 1 to p 1 prime here this will be changed from p 2 to

p 2 prime. So, the change in the loading means that is the difference here in the p 1 prime

minus p 1 plus that is your p 2 prime minus p 2 that is nothing but your change in the

load which has been taken place. So, this change in the here output of 1 change in the

output it is nothing but here I can say change in pg 1 plus change in pg 2. So, you can see

here this generator is sharing more compare to this one. So, those are having here less

slope it will be loaded much compare to this. So, this is the parallel combination of this

machine. So, you can see here this is similarly we can go for if load is released. Mean some

loads are shed of then frequency will rise from this nominal. Here this is your nominal

frequency thus I wrote here we are starting from 0 here.

So, the sharing also here the system frequency will be constant and then we will have that

system frequency. Based on that we can calculate what will be the loading of the system and

generators that and they will take care of that. So, you can see the change in here from

p 1 to p 1 prime and here it is change from p 2 to p 2 prime. So, this is the load that

is a if it is increased so, this value that they have to go for. If reduced then we can

also similarly what will happen? If this is a negative means we have reduced the load

what will happen? System frequency will go up and you can see here f double prime what

will happen the now, net release you can say here now, we are getting here p 1 double prime

and here we are having p 2 double prime. So, we have again you can say now we have shifted

means we have reduced the loading here and from here to here and you can see that the

reduction of load in this case is in 2 case it is much higher than your this case.

So, this is your frequency raise and again we can write your this change in now pl that

is negative. It is nothing but how much we are going for this is here p 1 minus here

p 1. Again here this will be the negative I can say this double prime minus p 1 plus

here p 2 double prime minus p 2 what will happen? Now, this quantity will become negative

this will be negative and the finally, the change in load is negative means we are shedding

of the load. So, this is the parallel combination of that. So, in this lecture now, I can in

this lecture we saw the modeling of turbine and; however, we used for this analysis purpose

or you can say simplicity. I have used only the turbine without the reheating, but if

you are using if you are using some computers and other things then you can use the detailed

modeling. Later on also I will use the turbine modeling without reheating. So, that we can

see the performance of the system or you can say the various loops that are a load frequency

control loops. I also model the generators along with the rotating mask of the turbine.

So, it is your generator plus turbine rotating mask and then we related that turbine output

to the electrical output and that is electrical output related with the frequency of the system.

And then I combine together all these 2 and then we saw that the complete block diagram

of governing system, which I discuss here the governing system in the lecture 1 governing

then turbine here the load and then we saw this performance. Another term is another

model is left out that is your load model which I will discuss in the next lecture.

And then your complete 1 loop that is a primary load frequency control loop is completed.

And then will add the load model as well and then we will have a complete this ALFC primary

loop. And we will see the performance that is steady state performance as well as the

dynamic performance. And then we will analyze the loops and we will see what else are required

for maintaining the frequency to its rated value. So, we will discuss in your next lecture.

The Description of Module 3 lecture 2 Power System Operations and Control