thermodynamics compiled vinodkallur

8/3/2019 Thermodynamics Compiled Vinodkallur

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e-Notes by Prof. Vinod Kalloru, RVCE, Bangalore

Thermodynamics

Basic ConceptsSession 1

Introduction:

Thermodynamics deals with heat inter-action and work inter-action with the

substances called systems. Work and heat are forms of energy. Transfer of heat or work

to a substance brings about certain changes in the substance and whatever change

happens is called a process. Thermo means heat. Since work is also a form of energy,

thermo is taken to mean heat and work. Dynamics refers to the changes that occur as a

result of heat or work transfer.

Biological systems are capable doing work. For example, micro-organism is

capable swimming in the body fluid of its host. It needs to do the work. Where does

energy for doing this work come from? It is the metabolic activity that converts some

form of energy (Nutrition that it takes form host is a form of chemical energy) into work.

It is important then to understand how this happens so that we can exploit this to our

engineering benefit.

In thermodynamics we have work transfer, heat transfer and then we have a

system for interaction which undergoes a process. Let us look at these basic terms.

System:

We need to fix our focus of attention in order to understand heat and work

interaction. The body or assemblage or the space on which our attention is focused iscalled system. The system may be having real or imaginary boundariesacross which the

interaction occurs. The boundary may be rigid and sometimes take different shapes at

different times. If the system has imaginary boundary then we must properly formulate

the idea of system in our mind.

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Surroundings:

Every thing else apart from system constitutes surroundings. The idea of

surroundings gets formulated the moment we define system. System and surroundings

together form what is known as universe.

Closed system:

If the system has a boundary through which mass or material cannot be

transferred, but only energy can be transferred is called closed system. In an actual

system, there may not be energy transfer. What is essential for the system to be closed is

the inability of the boundary to transfer mass only.

Open system:

If the system has a boundary through which both energy and mass can transfer,

then it is called open system.

Properties:

Variables such as pressure, temperature, volume and mass are properties. A

system will have a single set of all these values.

Intensive properties:

The properties that are independent of amount contained in the system are called

extensive properties. For example, take temperature. We can have a substance with

varying amount but still same temperature. Density is another example of intensive

property because density of water is same no matter how much is the water. Other

intensive properties are pressure, viscosity, surface tension.

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Extensive properties:

The properties that depend upon amount contained in the system are calledextensive properties. Mass depends upon how much substance a system has in it therefore

mass is an extensive property.

State:

It is defined as condition of a system in which there are one set of values for all its

properties. The properties that define the state of a system are called state variables.

There is certain minimum number of intensive properties that requires to be specified in

order to define the state of a system and this number is uniquely related to the kind of

system. This relation is phase rule which we shall discuss little later.

Process:

The changes that occur in the system in moving the system from one state to the other is

called a process. During a process the values of some or all state variables change. The

process may be accompanied by heat or work interaction with the system.

Heat:

It is a form of energy that exists only in transit. This transit occurs between two points

which differ in temperature. Since it exists only in transit, it should be accompanied by

changes that occur in the system. The moment this energy cease to move, it appears as

internal energy. We shall discuss internal energy when we deal with I law of

thermodynamics.

Work:

It is also a form energy that exists only in transit. The work cannot be stored. Work is

defined as the product of force and distance through the force moves. Mathematically,

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where f is the force and d the displacement. Work is a scalar quantity.

Let us look at the following piston and cylinder arrangement containing some gas. The

system considered is the gas in the cylinder. It is exerting some pressure on the piston in

the upward direction. This is balanced by the atmospheric pressure plus the weight and

therefore piston though it is free to move, does not move. If the weight is increased, then

the piston moves downwards. The pressure of the gas increases and the movement of the

piston stops when the pressures become equal. The additional weight is doing the work

on the gas. The additional weight has to push the force created by the gas inside to move

the piston. So there is force and distance. But as the piston moves, the pressure

continuously changes. Therefore consider small distance (dL) for which work done (dW)

is also small as shown in the figure which results in small volume change (dV). Then by

definition of work, we have, if A is the cross sectional area of the piston

dfW

= cosdf

=

dL

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Integrating between the two locations of the piston,

=

pdVW

This is the equation we use to determine the work. Please note that in general p cannot

be taken out of the integral sign as p may be continuously changing as above.

If we have a p vs. V curve, we can determine the integral as area under the curve.

The area under the curve a21b is the work done on the gas.

We can carry out a process differently with the same initial and final states given

by the points 1 and 2. Since it is a different process, the curve will be different. In that

case the area under the curve will also be different which means to say the work done is

different though the initial and final states are same. Thus the work depends upon theprocess or path or how the state is changing. Such quantities are calledpath functions.

Since heat is also energy in transit as the work is, heat must also be a path function. No

matter what process, as long we have same initial and final states, it brings about the

same pressure difference. Such properties which depend only on state are called state

properties.

1

2

a b

p

V

5

dLfdW =

=A

VdpA

pdV=


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Thermodynamics

Basic ConceptsSession 2

Equilibrium state:

A system is said to be in thermodynamic equilibrium if it satisfies the condition

for thermal equilibrium, mechanical equilibrium and also chemical equilibrium. If it is in

equilibrium, there are no changes occurring or there is no process taking place.

Thermal equilibrium:

There should not be any temperature difference between different regions or

locations within the system. If there are, then there is no way a process of heat transfer

does not take place. Uniformity of temperature throughout the system is the requirement

for a system to be in thermal equilibrium.

Surroundings and the system may be at different temperatures and still system

may be in thermal equilibrium.

Mechanical equilibrium:

There should not be any pressure difference between different regions or locations

within the system. If there are, then there is no way a process of work transfer does not

take place. Uniformity of pressure throughout the system is the requirement for a system

to be in mechanical equilibrium.

Surroundings and the system may be at pressures and still system may be in

mechanical equilibrium.

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Chemical equilibrium:

There should not be any chemical reaction taking place anywhere in the system,

then it is said to be in chemical equilibrium. Uniformity of chemical potential throughoutthe system is the requirement for a system to be in chemical equilibrium.

Surroundings and the system may have different chemical potential and still

system may be in chemical equilibrium.

Thermodynamic process:

A system in thermodynamic equilibrium is disturbed by imposing some driving

force; it undergoes changes to attain a state of new equilibrium. Whatever is happening to

the system between these two equilibrium state is called a process. It may be represented

by a path which is the locus all the states in between on a p-V diagram as shown in the

figure below.

For a system of gas in piston and cylinder arrangement which is in equilibrium, altering

pressure on the piston may be driving force which triggers a process shown above in

which the volume decreases and pressure increases. This happens until the increasing

p

V

1

2

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pressure of the gas equalizes that of the surroundings. If we locate the values of all

intermediate states, we get the path on a p-V diagram.

Phase rule:

The two gas samples below have the same density, viscosity and all other intensive

properties.

1 m3 of nitrogen at a pressure of 2 atmosphere and 300 K2 m3 of nitrogen at a pressure of 2 atmosphere and 300 K

If a gas behaves ideally, then

RT

p

V

n

nRTpV

=

=

If M is the molecular weight of the gas, then

RT

pM

V

nM

V

m ===

RT

pM=

Thus the density of an ideal gas depends upon only temperature and pressure as all other

quantities are constants. Sincep and Tare same for the above nitrogen samples, density is

same. If we take out 1 cc of sample from any of those two, by observing only this 1 cc

sample we can never from where we took the sample. This means to say the samples are

exactly same and indistinguishable. They are said to be in the same state.

In order to define this state of nitrogen, we need only T and p. Thus we need minimum

two variables to specify the state. This number is given by Gibbs phase rule which is

given by

NF +=

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where F : degrees of freedom

: Number of phases

N : Number of components

For the above example, number of phases is one; number of components is one therefore

the degrees of freedom are two.For a system containing liquid water and its vapor in

equilibrium, we get the degrees of freedom to be one.

Following is the phase diagram of water which describes in what phase or phases

it can exist for different temperature and temperature.

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T

p

Liquid

Solid

Vapor

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For the system to have two phases, if we mention the temperature alone given by

the vertical line, the pressure has to be the one given by horizontal line. Given the

pressure, its temperature gets fixed. Thus we have a freedom of fixing only one of them

which amounts to say degrees of freedom is one.

Heat reservoirs:

If the temperature of the system does not change if we remove a finite quantity of

heat from it, then the system is said to be heat reservoir. If the temperature of the system

does not change if we add a finite quantity of heat to it, then the system is said to be heat

sink.

Ocean can be made to act either as heat reservoir or heat sink.

Heat Engine:

Work and heat are forms of energy and are inter-convertible. Any device which

converts continuous supply of heat into useful work continuously is called heat engine.

For any system to undergo continuous process, it should be cyclic. The detailed

mathematical treatment of heat engines will follow in 5 th chapter on second law of

thermodynamics.

Irreversible process:

If a thermodynamic process takes place in such a way that we can retrace the path

exactly, the process is called reversible. Consider a piston and cylinder arrangement with

piston free to move at some pressure and temperature in equilibrium. The pressure of the

gas is balanced by a pressure equal to sum of atmospheric pressure and the weight on the

piston pan as shown. If the weight on the piston is reduced, then the piston moves up. Let

us just displace the part of the weight horizontally. This involves work. The piston moves

up as it is free to move until the pressures again become equal.

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Following is the p-V diagram for the process.

State -2

12

State -1


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In order for a process to be reversible, the driving force must be infinitesimally small. If

we remove the weight bit by bit and place it horizontally, every bit would be placed one

above the other and then piston moves upward. At any point, the piston and the bit

removed are at the same level. As a result, at any point, we can move the bit horizontally,

making the piston move a bit down where the next bit of weight is available at the same

level. Reversing is thus possible.

However there is one work that can never be reversed even in this process. That is the

work done on the bit to move it horizontally. But this is less irreversible compared to the

earlier process where large part of the weight was removed. Look at the state-2, in order

to reverse; the weight has to be lifted by some vertical distance which is missing in the bit

by bit process. We can only strive to make a process only closer to reversibility.

Thermodynamics

Session 3

First law is a statement of law of conservation of energy. For a closed system,

since there is no mass exchange with the surroundings, the total energy of the system

must be always a constant. One form of energy may be converted into another. The

system may possess several forms of energy such as kinetic energy, potential energy,

internal energy.

Potential energy:

It is the energy possessed by a body by virtue of its position. An object kept at a

height z, possesses energy to do the work. A spring kept in a compressed state is also

capable of doing work. A body is said to be possessing energy only if it can lift weight by

some mechanical means. If we tie a rope to a body at higher level, pass the rope over a

pulley and tie the other end to another body whose mass is slightly lower, the second

body will be lifted. To one end of the compressed spring in the vertical position, tie a

weight and then release the spring. The weight goes up. Thus a body at higher level or

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spring in its compressed state possesses energy. The potential energy of a body of mass

m at height ofzfrom some arbitrary level (reference level) is given by

mgzEP =

Arbitrary level may by any level. As long as the same, the changes in energy will also be

same. The changes in potential energy as the level changes will be

zmgzzmgE ifP == )(

Kinetic energy:

It is defined as the energy possessed by a body by virtue of its velocity. For a

body of mass m which is moving with a velocity u, the kinetic energy is given by,

muEK =

Change in kinetic energy is

)(

umuumE ifK ==

Internal energy:

When some external work is transferred to a body, its temperature increases

indicating rise in internal energy. Internal energy is due to molecular energy of vibration

and rotation.

First law of thermodynamics for a closed system:

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IfQ is the amount of heat transfer to a system which does Wamount of work

(work transfer form the system), then the change in the total energy of the system will be

Q W. This change will have its implications on the kinetic, potential and internal

energy. Since law of conservation is to be valid, the change in internal energy Q W

must be equal to the changes occurring in all forms energy. Mathematically,

KP EEUWQ ++=

Generally, the changes in kinetic and potential energy will be negligible compared to heat

and work interaction,

UWQ =

This is first law of thermodynamics. Note that Q is the heat transfer to the system and W

is work done by the system.

Cyclic process:If the process occurs in such a way that periodically its state repeats then the

process is called cyclic process. The following p V diagram represents a cyclic process.

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The states 1, 2 and 3 keep repeating in the process in that order. Not only this, all the

states in between also will repeat. Every time process state changes from 1 to 2, 2 to 3

and 3 to 1 following the same paths. During such processes, there will be heat and work

interaction. All the properties will get the same values at state 1 regardless of number of

times cycle repeats. Thus

03333 ==== TpVU

First law applied to a cyclic process:

Since the change in internal energy is zero,

WQ

WQ

WQU

===0

This means that net heat transfer to the system is the net work done by the system.

p

V

12

3

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Cyclic process consists of several steps. In the cyclic process shown in the diagram, there

are three steps viz., 1-2, 2-3 and 3-1.

Let us have another cyclic process for which heat and work interaction are known as

below.

Step Heat interaction Work interaction

1-2 Q1-2 is added to the system No work interaction

2-3 Q2-3 is removed form the system W2-3 is the work done by the system

3-1 Q3-1 is added to the system W3-1 is the work done on the system

Net heat transfer to the system, += QQQQ

Net work done by the system, 33 = WWW and WQ =

Flow processes:

So far we considered closed systems only. Let us look at open systems. Mass

exchange can occur. If the process variables do not change with respect to time for the

system, the process is said to be under steady state conditions. Under such conditions the

mass of the system is constant. This is due to the equality of mass entering and leaving.

If the system as a whole is moving undergoing changes, it constitutes steady state flow

process only when the system at a fixed location will have fixed values.

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This diagram shows an open and steady state flow process. The equipment considered is

between 1 and 2. The gas enters at 1 and leaves at 2. Steady state flow process requires

mass entering at 1 is equal to mass leaving at 2. If certain mass of gas considered to be

the system, such as cylindrical shaped gas at 1, when it reaches a particular location such

as A, it will be at a state (having certain fixed values for all the properties). No matter

what time, the system at A will be in the same state. This is true for all locations.

However, state of the system differs from location to location. There is no variation of the

state with respect to time at a given location.

Heat in, Q

Work output, Ws

1

2

z1

z2A

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Thermodynamics

Session 4

First law applied to flow process:

Following is the equipment through which the steam is flowing.

Steam enters at 1 and leaves at 2. There is a heat exchanger which adds heat Q to

the steam making it superheated. The superheated steam runs the turbine and shaft work

is obtained. Thus the gas does the work Ws. The kinetic, potential and internal energy of

the gas changes as it goes from one location to the other. The entrance 1 is at height z1

and exit is at height z2 from some reference datum. The velocity of the gas varies as the

flow area varies along the path.

Let us consider unit mass of steam whose volume is the volume of cylinder shown

at location 1. This steam element enters the equipment, goes through it receiving heat Q

in the exchanger, doing work at the turbine and comes out at 2. There are changes

occurring in potential (as the height at which the steam element is present changes),

Heat in, Q

Work output, Ws

1

2

z1

z2

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kinetic energy (as the velocity changes) and internal energy. Let u1 and u2 be the

velocities at location 1 and 2. Change in kinetic energy, since m =1, is

)(

uuEK =

Change in potential energy of the steam from 1 to 2 is

)( zzmgEP =

Let the change in internal energy be .U We must account for all heat and work

interaction that occur during the process of steam element moving form 1 to 2. The heatQ is getting transferred in the heat exchanger to the steam element. Apart from work

output of the turbine, there are other fork forms. At location 1, the surrounding has to do

the work on the steam element to push it into the equipment. If the pressure of the steam

isp1, and the cross sectional area of the pipe is a1, then the distance moved by the steam

element is V1 / a1 where V1 is the specific volume of the steam at 1. The work done at the

entrance on the steam element at 1 is then,

)( Vp

a

VapW == (Work done on the system)

Similarly work needs to be done by the steam element at section 2 as it has to push the

surrounding which is given by

)( Vp

a

VapW == (Work done by the system)

The entrance and exit work are also called as flow work. The product of pressure and

volume is the flow work.

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If the steam element gets into the equipment, it receives work done by the previous steam

element and uses it to push the element next to it. Thus the net work is zero for all the

locations inside the equipment.

Thus the net work done by the element,

WWWW S +=

VPVPWW S +=

Substituting in first law of thermodynamics,

)()(

)(

VpUVpUEEWQ

VpVpUUEEWQ

EEVpVpWQ

UEEWQ

PKS

PKS

PKS

PK

++++=+++=++=+

++=

The group pVU+ occurs frequently in flow processes and is called enthalpy denoted

by the letter H.

pVUH +=

Since internal energy, pressure and volume are state functions, enthalpy is also state

function. The above equation becomes,

HEEWQ

HHEEWQ

PKS

PKS

++=++=

This equation is first law of thermodynamics applied to flow processes. If the changes in

potential and kinetic energy are very small then,

HWQ S =

Heat capacity:

Heat capacity of a substance is defined as the heat transfer necessary to bring

about a change in the temperature of unit amount of substance by one degree centigrade.

Since it is heat transfer which is a path function, it depends upon the way heating is done.

For example gases can be heated to increase the temperature by two different methods.

The unit quantity of gas taken in container with rigid wall, when heated its volume

remains constant. Another method is to have the wall which is flexible. If the piston is

movable in the piston and cylinder arrangement, gas when heated pushes piston and

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pressure will be constant. Even if we take unit amount of gas in both these heating

methods, it is observed the heat transfer is not the same. Thus we have heat capacity and

constant volume and heat capacity at constant pressure. Mathematically,

dT

dQC=

p

pdT

dQC =

V

VdT

dQC =

The heat capacity at constant volume will not be equal to that at constant pressure.

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Thermodynamics

Session 5

Prepared by Vinod Kallur, RVCE, Bangalore.

PVT Behavior of fluids

Pressure, temperature and volume of gases vary in a definite manner. Based upon the

conditions, matter can exist in solid, liquid and gas phase. This is represented in graphical

form as below. This diagram is for pure water. All pure substances will have more or less

similar diagrams.

Based upon the values of pressure and temperature, if the point lies in a region, the

substance will exist in the corresponding state represented. For example the point arepresents solid state and b represents vapor state. Any point lying on the curve

represents both the phases which are in equilibrium. For example, the point d represents

liquid water and its vapor in equilibrium. Thus there are numerous pair of pressure and

temperature values on the curve and all those points represent liquid and vapor in

equilibrium. Each curve will have always two phases in equilibrium.

T

p

24

Liquid

Solid

Vapor

Critical Point.

Fluid

Gas

a b

d

p Tdiagram for pure water


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The point where all the curves meet is called triple point and at this condition, all the

three phase exist in equilibrium. Critical point is the highest temperature and highest

pressure where the liquid and its vapor can exist in equilibrium.

For all pressures and temperatures higher than those at critical point, the liquid and vapor

properties become similar. This state is called fluid.

Phase is considered liquid if it can be vaporized by reduction in pressure keeping T

constant. This is represented by vertical downward arrow in the figure. The initial point is

in liquid phase region. Phase is considered gas if it can be condensed by reduction in Tkeepingp constant. This is shown by horizontal arrow pointing towards left. The point a

represents solid state which on increasing the temperature at constant pressure, only the

temperature of the solid increases until it reaches curve where first drop of liquid appears

and beyond this only vapor exists. This is shown as arrow a b.

The other way is representation using p V diagram for pure substance which is given in

the next figure.

S / V

L / V

LS

CpC

p

V

S / L

VC

TC

Gas

V

fluid

25

p V diagram for pure substance


26/84

There are regions where two phases co-exist in equilibrium. The dashed curve labeled Tc

represents an isotherm at critical temperature.

In thermodynamics, we deal largely with either vapor or gases. The significant portion of

the diagram is given below.

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The substance at point 1 is in liquid state. If we decrease pressure keeping temperature

constant, the volume decreases along the dashed curve. But this change is very small and

this is why liquids are considered incompressible. Volume change with pressure at

L / V

L

CpC

p

VVC

TC

Gas

V

fluid1

2 3

T1

4

27

p Vdiagram for pure substance


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constant temperature is negligible for liquids. Decreasing pressure from 1 at constant

temperature (T1), a point 2 will be reached where first bubble of vapor appears. Any

further attempt to decrease the pressure fails bringing pressure down, instead more and

more liquid becomes vapor and the pressure and temperature remain constant represented

by dashed horizontal line. Thus point such as 3 represent liquid and vapor in equilibrium.

Further attempt in decreasing the pressure brings the state to the point 4 where all the

liquid becomes vapor. Further decrease in pressure at constant will follow the dashed

curve in the gas region. Thus the dashed line represents an isotherm (T1). Given in the

figure there is one more isotherm at a temperature less than T1. The dumb bell shaped

curve represents the locus of all points on all isotherms where the points similar to 2 and

3 lie. The isotherms of higher temperature lie on higher and saturated liquid and vapor

volume become closer and closer and finally merge at the critical point. This is point

where liquid and vapor cannot be distinguished. Isotherm TChas a point of inflexion at

the critical point.

A point on the curve is called point of inflexion if there is a tangent at that point which

lies on either side of the curve as shown below.

How do we represent this complicated behavior mathematically? This is a challenge

which led to lot of research and we have several mathematical equations. All these

equations do have limitations. We will use simple mathematical equation. This is not a

limitation as no matter what equation all it does is relate pressure, temperature and

volume of gases.

Ideal gas equation: For one mol of any gas,

Internal energy is a function of temperature only.

28

RTpV =

=

T

TV

dTCU

=

T

TP

dTCH


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No matter what process, the changes in internal energy and enthalpy are given by the

above expression not only for ideal gases even for real gases.

This law is valid for low pressures, large volumes and low temperatures.

At constant pressure

At constant volume

Isochoric Process:

Any process in which volume of the system does not change is called isochoric process.

Work done is zero.

Isobaric process:

pdVdQdU =

29

,dTCdQ

dT

dQC

P

p

P

=

=

,dTCdQ

dT

dQC

V

V

V

=

=

WQU =

QU =

)(

TTCdTCU V

T

T

V ==

QTTCdTCH P

T

T

P == )(

dTCdQ P=


30/84

)()( 1212 VVpTTC

pdVdTCdU

P

P

==

)( 12 TTCQH P ==

)( 12 VVpW =

Relation between heat capacities:

Isothermal Process:

Since the temperature is constant there will not be any change in internal energy.

RdTdUdH

RTUH

PVUH

+=

+=

+=

RdTdTCdTC VP +=

RCC VP +=

RCC VP =

VP CC >

30

dWdQ

dWdQdU

=

=


31/84

2

1

1

2 lnlnppRT

VVRTWQ ===

0== UH

Adiabatic Process:

There will not any heat exchange, Q = 0.

dWdU =

=2

1

V

V

pdV

=2

1

V

V

dVV

RT

2

1lnp

pRT=

2

1lnp

pRT=

==2

1

V

V

pdVWQ

dVV

RTdTCV =

dVV

RTdTCV =

V

dV

C

R

T

dT

V

=

31

T

Vp

T

Vp =

VV

P

C

R

C

C=1

VV

P

C

R

C

C =

=VC

R


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The temperature and volume change according to the above equation.

The pressure, temperature and volume all vary in the adiabatic process.

Let us get a relation between the other variables.

Since the gas is ideal,

V

dV

T

dT)1( =

V

dV

T

dT)1( =

=2

1

2

1

)1(

V

V

T

TVdV

TdT

1

2

1

2 ln)1(lnV

V

T

T=

1

2

1

1

2

=

V

V

T

T

1

2

1

1

2 lnln

=

V

V

T

T

32

T

Vp

T

Vp =

Vp

Vp

T

T =

=

V

V

V

V

p

p

2211VpVp =

pTT

VTV

p

H

==

=

T

T

VdTCWU

dWdU


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)(

1

12 TTR

=

1

2211

=

VpVpW

=

=

2

1

2

1

dTCU

dTCH

V

P

33

)( TTCW V =

1

12

=

RTRT

=

VpVp

VpVp=

=

VV

pp

=

p

p

V

V

=

p

p

p

pVpW

=

p

p

p

pVp

=

p

pVpW

( )

;

/

=

p

pVpW

( )

=

/

p

pVpU


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Polytropic Process:

It is a representation of all the processes in a mathematical form as below.

Sl. No n Process

1. 0 Isobaric

2. 1 Isothermal

3. Isochoric

Generally n lies between 1 and .

34

constpV n =


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Thermodynamics

Session 6

Prepared by Vinod Kallur, RVCE, Bangalore.

Guidelines to represent processes on p V diagram:

In the figure above there are two isotherms shown. For these isotherms, .12 TT > There

is one isotherm passing through every point on p V diagram. Higher the temperaturehigher will be location of the isotherm. As they go downwards they converge as shown.

No two isotherms will ever intersect.

p

V

T1

T2

35


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Similarly every point has an adiabatic curve passing through it. The adiabatic curve

follows the relation constpV = . Thus every point is an intersection of an isotherm and

an adiabatic curve as shown below. The adiabatic curve is steeper.

No two adiabatic curves will ever intersect.

The mathematical relation between pressure, volume and temperature applicable all

ranges of values has not been possible. However, there are equations which are valid over

certain ranges.

Conditions for any equation to be equation of state:

1. The equation should reduce to

RTpV =

As p 0.

2. p-Vcurve should have a point of inflexion for isothermal at critical point C,

0=

CV

pand 02

2

=

CV

p

Van der Waals equation of state:

( ) RTbVVa

p =

+ 2

Here a and b are constants that depend upon the gas. If we apply the partial derivative

conditions, we get,

C

C

C

C

p

RTb

p

TRa

8;

64

2722

==

36

p

V

Isotherm

Adiabatic curve


37/84

The Van der Waals equation is cubic in V. Therefore it should have three roots for a

given set of values of pressure and temperature and for a given substance. For any

temperature higher than critical temperature, there is one positive root. For critical

temperature and critical pressure, all the three roots will be equal which gives critical

volume. For temperature and pressure less than critical values, there will be three roots,

least one gives the molar volume of saturated liquid and highest gives that of saturated

vapor which are volumes given by the two ends of horizontal section of the isotherm.

Redlich Kwong Equation:

TbVV

abV

RTp)( +

=

Applying the partial derivative condition, we get

C

C

C

C

p

RTb

p

TRa

0867.0;

4278.02/52

==

Note that the above equations reduce to the equation ,RTpV = asp 0.

Other equations of state are Peng Robinson and Virial equations.

Numerical problem:

Air is compressed from an initial state of 1 bar and 298 K to a final state of 5 bar and 298

K through two mechanically reversible processes in a closed system. First heating at

constant volume followed by cooling at constant pressure. Calculate W, Q, change in

enthalpy and internal energy. Use 2

5

,2

7 R

C

R

C VP == .

37

a2

V

p

Isotherm at 298 K


38/84

The path is shown in the figure. 1 - a represents heating at constant volume and a - 2

represents cooling at constant pressure. Note that initial and final states lie on a single

isotherm.

Let us do the calculations for one mol of gas.

For the path 1 a, since volume is constant, work done is zero.

dQdU =

dTCdU V=

= aT

T

Va dTCU

1

1

To use this we need to know temperature at a. Since it is closed system and volume is

same at 1 and a,

11TpTp aa =

Since pressure at a is same as that at 2,

Kp

Tp

p

TpTa

a 14901

)298(5

2

1111 ====

= aT

T

Va dTCU

1

1

kJ

J

R

dTR

775.24

7.24775

)2981490(2

)314.8(5

)2981490(2

5

2

51490

298

==

=

=

=

Thus Q1-a = U1-a = 24.775 kJ.

38

1


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kJJdTCH Pa 686.3434686)2981490(2

)314.8(71490

298

1 ====

You can use the formula

)()( 111111 VpVpUpVUH aaaaaa +=+=

to calculate change in enthalpy.

For the step a 2,

dWdQdU =

= 2

2

T

T

Va

a

dTCU

kJ

J

R

dTR

775.24

7.24775

)1490298(2

5

2

5298

1490

==

=

=

kJJdTCH Pa 686.3434686)1490298(2

)314.8(7298

1490

2 ====

pdVdW =

=2

2

a

a pdVW

Along the path a 2, since pressure is constant and equal to 5 bar,

=2

2

a

a pdVW

)(105 25

aVVx =

We should know the volume at a.

From gas law, at the initial condition,3

5102477.0

101

)198(314.8m

xV ==

Note that we have taken one mol.

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1

11

2

22

T

Vp

T

Vp =

The temperatures are equal,

2

11

2 p

Vp

V =

3004954.0

5

)02477.0(1

m=

=

Similarly,

3

5 1245.0101

)1498(314.8

mxp

RT

Va

a

a ===

kJJxVVpW aa 773.5959773)1245.0004954.0(105)(5

22 ====

Since the volume is decreasing, work is done on the system.

kJQ

Q

WQU

a

a

aaa

548.84

)773.59(775.24

2

2

222

==

=

For the entire process,

kJQQQ aa 773.59)548.84(775.242121 =+=+=

0686.34686.34

0775.24775.24

2121

2121

==+===+=

aa

aa

HHH

UUU

kJWWW aa 773.59)773.59(02121 =+=+=

Since there are no changes in internal energy and enthalpy, work done on the system is

given out as heat.

40


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Thermodynamics

Session 7

Determination of changes in internal energy and enthalpy:

No matter what is the process the internal energy and enthalpy are determined using the

following formulae.

dTCdH

dTCdU

P

V==

If the process is isochoric, then .dQdTCdU V =If the process is isobaric, then .dQdTCdH P =Let us recall that

1. For ideal gases internal energy is a function of temperature only.2. The change in internal energy is a state function.

Let us take an adiabatic process and determine change in internal energy.

41

p

V

1

2


42/84

Consider two states (1) and (2) on an adiabatic path. There is one isotherm for each state

as shown. Adiabatic curve at (1) is steeper than the isotherm. Whether the process

follows adiabatic path or another curve shown in the figure or any arbitrary path, as long

the initial and final state are same, we get the same change in internal energy.

Let us replace adiabatic path by two steps

1. 1 a constant volume process followed by

2. a 2 isothermal processto accomplish the same change in the state.

2121 += aa UUU

a

42

p

V

1

2


43/84

Since step 1 a is a constant volume process, dTCdU V= will have to be used. The

step a 2 is isothermal and therefore dU = 0.

0

1

21+=

aT

T

VdTCU

Since ,TTa 2=

=2

1

21

T

T

VdTCU

Any process between any two states can be replaced by either an isothermal process

followed by constant volume process or constant volume process followed by an

isothermal process. The total change in internal energy is the change in internal energy

for constant volume step only. Thus no matter what process the change in internal energy

is given by dTCdU V= .

Same idea may be extended to change in enthalpy.

Deviations of real gases form ideal behavior:

The deviation from ideal behavior is measured using what is known as

compressibility factor defined as the ratio of volume determined using ideal gas law

to the actual volume at any given temperature and pressure.

For ideal gases, Z = 1. Higher the value Z away from one higher will be the non-ideality.

Following figure gives the value of Z as a function pressure for different gases. All gases

approach ideal gas behavior at low pressures. This figure is called compressibility chart.

There will be as many curves as there are number of gases. The chart that is given here is

for one temperature. Therefore this figure becomes highly cumbersome if applied for all

gases and all temperatures.

43

RT

pV

V

VZ

ideal

actual ==


44/84

Theory of corresponding states:

It states that all fluids, when compared at the same reduced temperature and reduced

pressure, will have same compressibility factor and all deviate from ideal behavior to the

same extent.

N2

Z

1.0

p

44


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The reduced pressure and temperature are defined by

45

C

r

C

r

T

TT

p

pp

=

=


46/84

Here all gases with the same reduced temperature and pressure lie on the same curve.

This chart is called generalized compressibility chart. Different gases are indicated by

different points on the curve. There will be as many curves as there are temperatures.

This is less cumbersome compared to compressibility chart.

Z

pr

T

52.Tr =

46


47/84

The above table gives two gases with the same Z for different T and p. The reason is that

the critical values are different.

T p Tc pc Tr pr

N2

189.

3

83.7

5

126.

2

33.

5

1.

5

2.

5CH

4

286.0

5

114.

5

190.

7

45.

8

1.

5

2.

5

47


48/84

Thermodynamics

Session 8

Standard Heat of reaction:

It is defined as the enthalpy change accompanying a reaction when both reactants and

products are at their standard states at 298 K and is denoted by 2980H .

)g(CO)g(O)s(C 222

+ kJ.H 222129 80

=

If the stoichiometry of the above reaction is represented as

Then,

kJ.kJ.

H 61102

2221298

0 ==

Standard Heat of Combustion:

The standard heat of combustion of a substance is defined as the enthalpy change

accompanying the reaction when one mol of the substance undergoes combustion when

both reactants and products are at 298K.

The heat of combustion is 1168.12 kJ.

The standard heat of formation:

Standard heat of formation of a substance is defined as the change in enthalpy

accompanying the formation of 1 mol of the substance from the constituent elements

when the reactants and products are at 298K.

)g(CO)g(O)s(C 222

+ kJ.H 22212980

=

Standard Heat of Formation of Carbon monoxide is kJ.H f 61100

= .

48

)g(CO)g(O)s(C +2

2

2

;OH)g(CO)g(O)l(CHOCH2222

2222 ++ . k222222H =2222


49/84

Hesss Law of constant heat summation:

The net enthalpy involved in a reaction is the same whether the reaction takes place in a

single step or in a series of steps.

CBA +

Let A and B react to form C with heat of reaction H. Let there be one more method of

carrying out this reaction in two steps as given below.

ABBA + 1H

CAB 2H

Treating arrow as an equal sign, adding these two reactions gives,

CBA + , 21 HH +

It has been found that 21 HH + is same as .H

This idea is used to determine heat of reaction from heat of formation or heat of

combustion data.

For example if we want to determine heat of formation of sulfuric acid, we must

determine this from the reaction,

No matter what we do, the above reaction cannot be carried out. Let us carry out other

possible reactions such that the over-all reaction is the formation reaction as below.

49

H

22222 SOHOSH ++


50/84

22SOOS + 1H ----------(1)

32222 SOOSO + 2H ----------(2)

4223SOHOHSO + 3H -----------(3)

OHOH 2222

1 + 4H ----------(4)

2(1), 2(3) and 2(4) gives,

22222 SOOS + 12 H ----------(1)

32222 SOOSO + 2H ----------(2)

4223222 SOHOHSO + 32 H -----------(3)

OHOH 222 22 + 42 H ----------(4)

Adding these equations along with heat of reactions,

and heat of this reaction will be 24312 H)HHH( +++

Therefore heat of formation of sulfuric acid is2

2

431

HHHH +++

This is how Hesss law can be used to determine the heat of reaction of a reaction from

heat of reaction of other reactions.

50

22222222 SOHOSH ++


51/84

Thermodynamics

Session 9

First deals with law of conservation of energy and defines internal energy.

A stone falls decreasing its potential energy and converting it into kinetic energy. Just

before it strikes the ground, it has maximum kinetic energy. From there why does it not

go up decreasing its kinetic energy and converting it into potential energy?

If we keep two bodies at different temperature in contact, why there is no heat transfer

from a body of low temperature to a body at higher temperature?

When we remove the wall which separates two gases, the two gases mix together without

any external work done on them. Why then the two gases in a mixed state separate

themselves without any external work?

For all the processes mentioned above there is absolutely no violation of I law. In spite of

that, we do not observe these processes. It appears that there is some sense of direction

for spontaneous processes. There is nothing in the first law to indicate any such direction

to any process.

Is there any difference between heat and work though they are both forms of energy? If

they inter-convertible, then any cyclic process should be able to convert all the heat

supplied into work completely as change in internal energy is zero. Does it happen? It has

been found that we cannot have a device whose sole effect is conversion of heat supplied

into work completely. Let us look at why this is so with an example of a process.

In the following figure, there is a heat exchanger where heat is supplied ( Q1) to the

system which is water in this case and generates steam at high pressure. The steam flows

into a turbine and rotates it producing useful workWS. From the turbine, we get steam at

51


52/84

relatively lower pressure than with which it entered. The objective here is to convert heat

Q1 into work. Since this heat is available, can we send the steam from turbine directly

into the heat exchanger in order to make it work continuously so that we get continuous

supply of work? This is shown by a dotted arrow.

The moment we establish this connection, whole process comes to an abrupt stop! We

have lower pressure steam line going into the heat exchanger in which there is higher

pressure steam. If there is connection like this, why does steam preferentially go into the

turbine line? It will enter into both the lines and stop the turbine. Therefore, in order to

have this going on, we must condense the steam bring about a change in its state so that it

can be fed to the heat exchanger. This state must be same as that of feed water to the heat

exchanger.

Thus there is no way Q2 = 0. We cannot device any process which converts Q1 completely

into work. The work obtained Ws is less than Q1by an amount equal to Q2. By first law,since for a cycle change in internal energy is zero,

21S QQW Practically Q2 cannot be zero.

If we analyze any cyclic process, there will be several heat and work interactions, such

that net work done is equal to net heat supplied. There will be always one step in which

Q1

Q2

WSHeat Exchanger

Condenser

Turbine

52


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there is heat removed which brings down the amount of heat that is getting converted into

work.

This fact is the basis of statement of II law of thermodynamics. There are several

statements for the II law which are equivalent. In fact we can deduce one from any other.

All the statements are given below.

Second law of thermodynamics:

1. Heat cannot by itself pass from a cold to a hot body.

2. All spontaneous processes are to some extent irreversible and are

accompanied by degradation of energy.

3. It is impossible to construct a heat engine that operates continuously in a cycle

to produce no effect other than conversion of heat supplied completely into

work. This is called Kelvin Planck statement.

4. It is impossible to construct a heat pump (reverse heat engine) that operates

continuously to produce no effect other than transfer of heat from low

temperature body to a high temperature body.

Carnot cycle and Carnot engine:

The Carnot cycle consists of four steps and all the four steps are reversible. The cycle is

shown in the diagram below.

53


54/84

There are two isotherms and two adiabatic processes as shown. The isotherms are at TH

and TC.

AB Reversible isothermal expansion

BC Reversible adiabatic expansion

CD Reversible isothermal compression

DA Reversible adiabatic compression

According to first law, it can be shown that

B

AHABAB

p

plnRTWQ =

)TT(CW CHVBC D

CCCDCD

p

plnRTWQ =

)TT(CW HCVDA The net work done is given by

p

V

A

B

CD

TH

TC

54


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DACDBCABnet WWWWW + )TT(C

p

plnRT)TT(C

p

plnRT HCV

D

CCCHV

B

AH

For adiabatic process BC,

/)1(

C

B

C

H

p

p

T

T

=And for the adiabatic process DA,

/)1(

A

D

H

C

p

p

T

T

=From these two,

B

A

C

D

p

p

p

p =Net work done becomes,

B

ACHnet

p

pln)TT(RW

Heat supplied with an objective of converting into work is

B

AHABAB

p

plnRTWQ =

The efficiency is defined as

H

CH

B

AH

B

ACH

AB

net

T

TT

p

plnRT

p

pln)TT(R

Q

W

==

It is clear that efficiency of Carnot engine depends upon only the temperatures between

which it is operating and not on the working substance undergoing the cycle. Since

Carnot engine is reversible, it must have maximum efficiency.

From first law,

55


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57/84

Thermodynamics

Session 10

Efficiency of engines:

The efficiency of reversible Carnot Engine is given by

H

CH

H

CH

T

TT

Q

QQ

==The maximum efficiency of any engine operated between two temperatures is always

given by

H

CH

T

TT

=as the efficiency of a reversible engine is maximum.

Efficiency of real engine is always less than that of reversible engine. If the engine is

irreversible then efficiency is calculated from

H

CH

Q

QQ

=and not from temperatures.

For any irreversible engine,

H

CH

H

CH

T

TT

dQ

dQdQ

thermodynamics compiled vinodkallur

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