Thermodynamics Review for CFD: Compressible Flow Essentials

[Music] [Music] this is lecture number

six hello everyone welcome back today uh we are going to do some review of the ther Dynamics this is

essential because as we discussed in the earlier class that most of the compu flow dynamics that will be studying in

this course will be related to the compressive flows and uh for the compressive flow it's very important to

have a good understanding about the about the thermodynamics because you will have change in temperature and the

internal energy and there exchange of energy between uh the kinetic energy internal energy uh Etc so we'll review

some of the thermodynamics aspect which will be useful for us for the cfd course so before we go to the

thermodynamics okay I want to just uh recap the last class that we studied so uh we discussed in the last class that

uh uh the gases are more compressible so gases are more compressible than liquids

okay and we also found out that uh if you talk about this isothermal compressibility okay t t which would be

defined as - 1 by v d v by d p at constant temperature and the value of this was

for an ideal gas it will come to 1 by P and for a atmosphere temperature let's say pressure equal to 1 atmosphere which

is uh 1 1325 Pascal so the to Value will be around 10^ minus 5 m Square per Newton

right and we said that for the liquid this Stout is of the order of 10^ minus 10 around something like 5us

10 okay so there is at least four order of difference this is for water okay of difference between the

compressibility of water and the air and uh then we derived the expression that uh uh

our DV or D if you say it will become row to DP

right so because the top of air is higher for a given change in pressure you will have higher change in density

okay and we said that if the change in density uh is more than 5% that means your D by row is greater than 5% then it

will be a case of compressible flow we can no longer ignore

the uh the density here density change here right so uh finally uh we ended with the

thought that uh so this uh compressibility is a property of the fluid right so now we have found out

that uh the gases are more compressible than liquids at least four order more compressible than liquids but is there

any situation where we can treat uh the flow in a gas also as incompressible right that means the

density change is less than 5% right so uh I just want to give you uh one small example okay so there's a

difference between flow and fluid that we discussed the flow is something which is Flowing the flow when the fluid has

got motion then it's a flow right so if you have an arrow foil okay like

this okay and let's say you have some flow coming in okay and a low speed flow let's say I would say that this is

around 5 25 m/s okay and uh as it goes over there file it accelerates and let's say it goes to a value of 50 m/ second

Max of 50 m/ second right so this is my B1 and this is your V2 okay so what is the change in

pressure right so now there is a flow there's a flow which is coming in so the the medium is here let's say the medium

is here okay and the flow is coming in and then uh because of

that uh we want to find out what is the change in the pressure so that it can give us estimate of whether the flow can

be compressible or not right so typically you cannot use uh bar knowledge equation for the conversal

flow right but here since the speed is uh very low okay uh so we can uh try to use the B equation and just try to

estimate uh the change in pressure although it may not be that accurate because uh even if the the flow is

subsonic or the mark number is low or basically flow is at lowest speed then also you I mean the B equation ideally

is not valid it's valid only for purely incomp flow but the change will be not that high so let's say for this flow try

to use the B equation and you all know the B equation so if this is my point one and this is my point 2 right so this

is P1 + half row V1

squ is equal to P2 + half row V2 squ okay I'm assuming that uh the flow is nearly

incompressible right so this P1 minus P2 will become half row B2 s minus B1 squ and let's take uh for Simplicity

take row is equal to 1 okay this value is around 1.2 to5 or something like 1.3 kg per met CU but for Simplicity just

take r equal to 1 it will not change the conclusion that want to find out okay this will be 1X 2 into 1 into 50 s - 25

s okay so that is like uh 2500 - 625 so that will be around something around uh 1,00 by 2 so 950 okay so around it will

be some it will this value will be around 930 or 950 right so then if you do P1 minus P2 by P1 so P1 you can

assume that it's an atmospheric PR so which is by basically basically our DP by P so this will be 9:30 and P1 is

let's say one atmosphere so we can take it approximate to uh uh 10 5 Pascal right so you can see

uh the the value is 0.93 right so this is basically change is very very small the change in

pressure is around 93% okay not even 1% okay so that is why we are saying that if the speed is very very low your uh

change in pressure is very very small and uh despite uh the flow being compressible which you can see here uh

still the change in density is not enough to consider this flows compressible okay we can make uh uh this

relation more accurate but B equation that we have by considering the oiler equation which we'll discuss when we go

to the conservation of momentum right but this is one aspect that I wanted to tell you that when the speed is low your

change is uh the pressure with respect to the initial pressure is very very small let's say now you change it to uh

a higher value okay so let's say I want to say that uh it started with uh um yeah it started 400

m/s and this went to 500 m/s right so in this part particular case remember this is a compressible

case typically you might be knowing about this speed of sound and we'll cover that again today okay more

formally but that is around 3 30 m/s okay uh in the atmosphere for a temperature around

300 Kelvin and pressure one atmosphere right so that is speed of sound so these both are basically more than speed of

sound right so these These are the cases of your supersonic flow of course I'm considering uh uh the temperature and

pressure conditions to the SE level right so these are the cases where the velocities are more than speed of sound

so these are cases of High Mark number and you cannot use bur knowledge there you can use oer equation there to find

out what is the change in pressure due to the change in velocity right and you would find that this D row is very very

high in that particular case much higher than 5% and you cannot ignore the change inity if you are very simpleminded and

let's say applied the burol here for this case which is not applicable then also you'll see that this P1 minus P2 by

P1 will be much much higher than 5% right because simply is that the change in pressure Delta P varies as ch v²

right so B2 Square minus B1 square right so when you do like this you will find that the changes uh in the pressure is

very very high and that is why the density change is very very high okay so I hope I convince you that as

the speed increases as the I mean in other words the Run number increases or the mark number increases okay so uh you

would notice that uh the consideration of density becomes important okay the another thing uh that is important which

is more useful is what is the value to S discussed about two compressibility uh one is the isothermal compressibility

another is isotropic compressibility and by using p equal to row RT we basically found what is your isothermal

compressibility right uh but in many situation uh beer flow is not isothermal okay but it can it can be

isentropic especially let's say take the same example that we had okay we have a an arrow file okay and let's say this is

your bond layer okay this is very close this is your bond layer but if you go away from the bond layer okay let's say

in this region okay so flow is more likely to be uh isentropic okay so

isentropic you might have read in the literature it's a process which is both adiabetic and reversible right so why

the flow is going to as Tropic here because there's no heat exchange here and secondly the effect of viscosity

outside the bond is not enough okay so there is the flow can be we assumed to be reversible right so in this

particular case the flow is more likely to be entropic so what is important is what is the value of to S okay what is

the isentropic compressibility so before we go into what is the value of isentropic compossibility and how do we

calculate that let's look at the in tropy a little bit more formally okay and the entropic processes let's review

the laws of thermodynamics and then we'll come back and calculate this value of

TS okay so uh first very brief review for a perfect guess so for most of the

discussion we have we'll assume it to be calorically perfect gas okay so that means

your uh e will be equal to CVT and your H will be equal to CPT okay if it is thermally perfect we had

discussed that that will be D = to CV DT okay that means your CP and CV May VAR with the temperature but for normal

practical conditions of nearly atmospheric uh pressure and temperature less than 1,000 Kelvin we can use this

calorically perfect gas right so for that those cases we can have this uh uh perfect gas equation okay these are the

relations of internal energy and enthalpy okay with change in with the specific volume at constant volume

specific uh Heat at constant pressure okay and CP minus v as you know it's a it's a gas constant R okay and uh we

have discussed about the relation of R okay and the ratio of CP by CB is the ratio of a specific which is usually

denoted by gamma right and if you use this two relation so let's say CP minus CV = to R so divide the both sides by CV

so this is CP by CV minus 1 V will be R by CV so this is GMA - 1 = R by CV okay and that is basically CV equal

to R by gamma minus 1 okay and similarly because uh CP equal to CV + r so that will be gamma R by

gamma minus one remember this these relations because this we'll be using again and again when we go to the

compressible shaped equations uh let's say for example when we go to uh the energy conservation laws okay now

uh in the last class we discussed that uh the value of R okay it depends on the gas because this is basically Ru by

molecular mass okay so Ru is equal to 8.314 okay uh JW per mole Kelvin I think Jew or yeah I think this should be okay

and molecular mass is is so if you want to find out R of oxygen or let's say R of air you want to find out that is Ru

by molecular mass of air which is around 29 okay and that will give you around 287 which is JW per kg Kelvin so far so

good uh so if you are simulating with a different gas you need to change the value of R right the gas constant in

your shd simulation what about Gamma okay so let's say I'm doing if I'm using nitrogen or oxygen okay your our value

of R will change okay but what will happen to our gamma will will the gamma change in this case or

not okay so as it so happens that the ratio of a specific heat basically gamma it depends on the degrees of freedom so

what is the degrees of freedom that what are the different kind of ways uh a a gas molecule can change the energy okay

so uh I without going into the discussion uh so we can have uh basically uh internal energy for per

degree of Freedom will be around RT by2 okay so if you have n degrees of freedom your internal energy will

be uh stored energy will be D equal to nrt by2 right so uh if you use the relation CB

equal to so CB equal to D by DT so that d by DT at constant volume okay so that will be

become uh NR by 2 right and if you use the CP = to CV + r that will become NR + 2 by 2

right and then if you take the ratio of CP by CV okay you'll get the expression which is equal

to n + 2 by n right where n is the degree of Freedom so for the Atomic gas okay uh for you will have three degrees

of freedom okay three ways it can exhibit the motion or it can uh uh basically exchange energy so in that

case your gamma will be 3 + 2 by uh by 3 that will be 5x 3 it will be 1.66 for the case of diatomic gas which

is hydrogen nitrogen and oxygen or any dimic gas okay so n will be five because you'll have two extra degrees of freedom

okay there is dimic molecule you'll have two extra degrees of freedom okay you can have vibrational energy okay so then

in this case you have Nal 5 and GMA = 7 5 and that will become GMA = to 1x 4 so this is the value that we use let's say

if we are doing a cfd simulation with air right so here we assume that the content of carbon dioxide or nitrogen

dioxide or n is very very small and it's largely basically uh 79% let's say N2 and 21% O2 is largely diatomic so uh and

uh the component of other gases is less than 1% so in that cases we can assume it to be diatomic and we can use this

gamma equal to 1x 4 however if let's say if we doing a simulation if you change the medium and do the simulation let's

say with ozone or carbon dioxide and so on and so forth in that case you will have the degrees of freedom even higher

that case n equal to 7 okay and then gamma will become 7 + 2 by 7 which is 9 by 7 and you have to use this gamma

value of 1.33 right so coming back to this question n22 if you change from N2 to O2

you have to change the r but the gamma will remain the same because your it is both are diatomic gases right however if

you go to like let's say your from this to let's say you go to helium okay in that case your both R will change as

well as gamma will change okay you have to use the gamma of 1.66 okay so these are some of the care that

we need to keep when you are trying to uh basically uh set up your CBD simulation okay so first of all you have

to understand whether you can assume it to be the calorically perfect gas okay if it is fine then what

kind of gas is it it's a diatomic triatomic monatomic and so on and so forth then you check up the gamma

according to that okay and then you can uh uh set up the value of r based on the molecular mass of the gas okay and then

you can uh assume the CP and CB to be constant if it is uh caloric per gas and set the value of CP and CB for that

particular uh medium that you're using okay now uh let's review the laws of thermodynamics very briefly because I

expect that you have gone through this course uh in undergrad uh first year or second year okay and uh you have basic

knowledge of this LW of thermodynamics okay so the if you go to the first law okay so the first law is simply it's a

energy conservation okay so basically if you add heat anywhere which is this Delta Q so it's a basically system and

the surrounding okay so if the heat is added to a system okay it is used to basically increase

the internal energy plus First Love of thermodynamics there are many different forms or many different ways of writing

it okay we'll just uh sometimes they take this Delta W to be negative okay that is

when the system is work done on the system okay so if there's work done on the system then uh the Delta W is

negative if the work done is by the system then Delta W is positive okay and for a reversible system okay if the

system is reversible that means there is no dissipation uh due to viscosity or other effects then you can uh take this

Delta W as equal to pdb and there are several derivations available in the third course please go through that and

for that you can write Delta Q equal to D plus pdv okay so this is your first law of

thermodynamics okay for a reversible system now why we are I mean talking about the reversible system as we have

discussed that uh if we are away uh from the places where the viscous effects are important or D is important uh you

likely find the process to be reversible okay so that's why many of the theories that are developed in the literat is for

the reversible in fact reversible adiabetic flow which is the isentropic flow and uh the definition of entropy

comes from the second L of thermodynamics okay where it says that the change in entropy which is uh for

any system okay which is given by your Delta Q basically you're supplying some heat divided by the temperature plus uh

there are some losses due to irreversibility okay which is given by DS

irreversible okay so uh it should be greater than equal to zero that means that uh I mean

the DS reversible irreversible okay it will be basically uh greater than equal to zero

okay so the the overall even if uh there is no hit transfer okay because of this uh uh dissipation your entropy may

increase right so this is the the statement of the second law and it basically dictates which direction the

heat will flow okay it will flow in the direction where the entropy is always going to increase because this DS

irreversible is always either either at the best case it will be equal to zero if there's no dissipation there's no

loss otherwise it will be greater than zero right so if uh we are talking about a adab reversible system if the system

is adiabetic then Delta Q equal to0 if the system is reversible then D is irreversible equal to Z okay and then in

that case your D equal to Z that means you can uh say that my entropy remains constant S2 equal to S1 for any

reversible tic process from 1 to two Okay there won't be any change in entropy okay so that is what uh it says

about the second says about the uh the isentropic system it also gives you the definition of entropy it

gives you a handle which is called entropy and which is a state variable okay so you can calculate the entropy at

two entropy at 1 and if you subtract the value of S2 - S1 you can find out what is the Ching entropy and it it has to be

a positive number okay S2 - S1 has to be a positive because either positive or zero because the that's a law that it

always has to increase the entropy has to increase okay in this case I forgot to mention

one thing so here the D is basically a state variable so you can calculate a variable at the end point and the first

point uh but Delta q and Delta W they are not State variables they are path variables it depends on the path okay so

now we will basically combine Delta q and uh first law and second law and derive some very useful

relations okay so we'll combine both the first laws and Second Law first and second laws okay and we assume the

system to be reversible okay right now it's not adiabetic reversible only reversible okay so in that case you can

write Delta W as pdv okay uh and Delta Q can be written as TDS because we are assuming that

DS irreversible equal to Z because a reversible system so when you use

that so you have TDS Plus is equal to cvdt plus p p

DV okay so then it can be written as D is equal to CV

DT by t + P by T D B now we can use uh uh this uh ideal gas equation which is pbal to

PB = to RT right so then it will give you P by T is equal to R by V right so if you use this two relation DS will

become CV DT by T plus r DV by V right and if you integrate it okay

let's say from point 1 to point 2 and assume it to be theorically perfect gas okay 1 2

2 1 2 2 1 2 2 you'll get this will become S2 - S1 will be become CV Ln T2 by T1 and plus r

Ln B2 by B1 okay so one interesting thing you can see here okay is that even for the clor

equally perfect gas the f s depends on both T and V okay it's a function of both T and V

unlike uh our energy and enthalpy which is only function of either pressure or volume right here function of both TR b

or TR P whatever you want to say right so it depends on T and depends on V okay and if we use the the relation which is

basically in terms of T and P okay and then you can write your H equal to okay let mention here H equal

to e + p v right and this will become DH equal to D+ pdv plus

VDP right so TDS can be written as

D+ pdv okay and and that will be DH minus VDP right and that can be written as TDs

is equal to cpdt D equal to cpdt minus VDP okay and this can be written as DS equal to CP DT by

T and plus minus so now V by T you can use the same uh law PB = to RT so that will be R by P minus

R DP by P okay and if you integrate this you'll get this relation okay S2 - S1 = to Ln T2 - T1 and R Ln P2 P1 so uh your

entropy uh depends on both temperature as well as pressure even for the caloric gas okay unlike

the other case where uh unlike the energy internal and enthalpy which depends on only the pressure or the

volume okay so with this uh we will stop for uh the class today in the next class we'll discuss about the isentropic

system we'll use this relation and see some interesting relations thank you [Music]

[Music] [Music]