LECTURE NOTE ON

THERMODYNAMICS (GEC 221)

Thermodynamics is the branch of science that treats the various phenomena of energy and

related properties of matter especially the relationship between heat, work and properties of

systems.

BASIC CONCEPTS AND DEFINITIONS

System: Any collection of matter contained within prescribed boundary, which is of interest for

a particular study or analysis, is referred to as a system. A boundary may be physical or

imaginary (either fixed or moving). The boundary separates the system from its surroundings or

environments. There are two kinds of system, closed and open system.

Closed System: A closed system is that which allows no exchange of matter with the

surroundings. A good example of a closed system is the gas contained within a cylinder that is

closed at one end and a movable piston as shown in fig.1.1. The inner surface of the piston and

cylinder form the boundary of this system and the boundary is a moving one.

Fig. 1.1 Closed System

Open system: An open system is that which allows exchange of matter with the surroundings.

i.e there is flow of mass in addition to work and heat across its boundary. With open systems the

boundary is normally specified as a control surface and the volume encompassed by the surface

as the control volume. The mass of matter within the control volume may be constant ( though

not the same molecular matter at any given instant). An example of an open system with constant

mass is a water nozzle and that with varying mass is air in a rubber tube undergoing inflation.

Fig. 1.2 Open System

State: At any instant or time a system is in a condition called state, which encompass all that can

be said about the result of any measurement and observation that can be performed on the system

at that time.

Process: That which changes the state of a system.

Path: path of a process is the series of state points passed through during process.

Cyclic process: is one in which the final state of the system is identical with its initial state. Such

a system is said to have undergone a cycle.

Property: any observable characteristics of a system is referred to as property of the system. The

thermodynamic properties are pressure, volume, temperature, internal energy, enthalpy and

entropy. Properties may be classified as intensive or extensive as they are independent or

dependent of mass of the system. For example, temperature, pressure, specific volume and

density are intensive properties, while total volume, total mass, total energy in a system, internal

energy, enthalpy and entropy are extensive properties. If a change in the value of a property

depends only on the initial and final states of the system, and is independent of the process

undergone by the system during a change of state, it is called a state/ point function. e.g

temperature, pressure and entropy. State functions are written as exact differentials e.g pressure

∫

= P2- P1 . Path functions have their values depend on the path followed during the process

of changing their states i.e their history is important in determining their values in the final

equilibrium state. Examples are heat and work written as inexact/partial differential as ∂Q and

∂w respectively. e.g heat ∫

≠ Q2 – Q1 but ∫

= Q1-2

Specific properties: are those for a unit mass and are extensive by definition e.g. specific

volume.

Energy: The general term is the capacity to produce an effect. It appears in many forms which

are related to each other by the fact that the conversion can be defined with the precision.

Internal Energy: the sum of various forms of energy that a system has is the internal energy of

the system. It is a property of the system. The absolute amount of internal energy that a system

has is never known, but internal energy changes can be measured from any convenient datum.

Isolated System : is one that is completely impervious to its surrounding neither mass nor

energy cross its boundary.

PROCESSES AND CYCLE

Thermodynamic Processes: is the path of successive states through which a system passes in

changing its states. When the processes are performed on a system is such a way that the final

state is identical with the initial state, the closed system is said to have undergone a

thermodynamic cycle or cyclic process.

Fig. 1.3 Thermodynamic process Fig. 1.4 Thermodynamic Cycle

REVERSIBILITY

Reversible Process: May be defined as a process between two states during which the system

passes through a series of equilibrium states. A reversible process between two states can be

drawn as a line on any diagram of properties. When a fluid undergoes a reversible process, both

the fluid and its surroundings can always be restored to their original states. In practice, the fluid

undergoing a process cannot be kept in its intermediate states and a continuous path cannot be

traced on diagram of properties. Such a real process is called irreversible process and is usually

represented by a dotted line joining the end states to indicate the intermediate states are

indeterminate.

The criteria for reversibility include:

Fig. 1.6 Irreversible process Fig. 1.5 Reversible process

(a) Absence of friction i.e no internal or mechanical friction

(b) No finite pressure difference between the fluid and its surroundings (i.e infinitely slow

process)

(c) No heat transfer across a finite temperature difference.

From the criteria stated above, it is apparent that no real process is reversible but a close

approximation to an internal reversibility may be obtained in many practical processes.

TEMPERATURE

Temperature is an intensive property, which determines the degree of hotness or the level of

heat intensity of a body. When a hot body and a cold body are brought in contact with each

other, heat energy is transferred from the hot body to the cold body until the two bodies are said

to have equal temperature if there is no change in any of their observable properties.

If a body A has the same temperature with body B when brought together and body B when in

contact with a third body C shows no change in any observable characteristics, then bodies A

and C when brought into contact will also show no change in their characteristics. That is, if two

bodies are each equal in temperature to a third body, they are equal in temperature to each other.

This principle of thermal equilibrium is often called the Zeroth law of Thermodynamics.

Scale of Temperature: A temperature scale is an arbitrary thing. The Fahrenheit and Celsius

(Centigrade) scales are based on melting and boiling points for water at 1 atmosphere. The

respective temperatures are 32 and 212 degrees in the former and 0 and 100 degrees in the latter.

The thermodynamic temperature scale which is measured from a point of absolute zero will be

discussed later.

Thermometer: is a system with a readily observable characteristics termed a thermometric

property. The systems are equal in temperature when there is no change in any property when

they are brought into contact.

WORK AND HEAT

For a closed system to undergo change of state, this must be accompanied by the appearance of

work and heat at the boundary. In mechanics, work is defined as the product of force and

distance (d) moved in the direction of the force.

Mathematically, work done W= Fxd

Where Force = pressure x area.

The basic unit of work in SI system is in Nm or Joule. Work is a transient quantity which only

appears at the boundary while a change of state is taking place within a system. Work can

therefore be described in thermodynamics as a form of energy transfer which appears at the the

boundary when a system changes its state due to the movement of a part of the boundary under

the action of a force.

By convention, work done by the system on the surroundings is taken to be positive i.e. Wout =

+ve e.g when a fluid expands within a cylinder pushing a piston outwards. Conversely, work

done on the system by the surroundings is taken to be negative i.e. Win = -ve e.g when a piston

compresses a fluid within a cylinder.

Heat can also be described in an analogous manner to work as a form of energy transfer that

appears at the boundary when a system changes its state due to a difference in temperature

between the system and its surroundings. Heat is transitory energy like work and can never be

contained or possessed by a body

By sign convention, heat flowing into a system from the surroundings is taken to be positive i.e

Qin = +ve while heat flowing from the system to the surroundings is taken to be negative.ie Qout =

-ve. It is worth noting that both heat and work are not thermodynamic properties and therefore

they are path path functions whose values depend on the particular path followed during the

process. Hence heat and work are written as inexact differentials in the forms ∂Q and ∂W

respectively.

i.e ∫

= W1-2 and ∫

= Q1-2

REVERSIBLE WORK

Consider an ideal frictionless fluid contained in a cylinder behind a piston. Assume the pressure

and temperature of the fluid are uniform and that there is no friction between the piston and the

cylinder walls.

Let the cross-sectional area of the piston be A, and the fluid pressure be p, let the pressure of the

surroundings be (p-dp). The restraining force exerted by the surroundings on the piston is (p-

dp)A. Let the piston move under the action of the force exerted by the fluid a distance dl to the

right. Then, the work done by the fluid on the piston is given by the force times the distance

moved. ie. Work done = pA x dl = pdv

Where dv is a small increase in volume.

Fig. 1.7 Fluid in a cylinder undergoing expansion

Or for a mass m, dW = mpdv, where v is the specific volume. This is only true when criteria (a)

and (b) for reversibility hold.

This is only true when criteria (a) and (b) for reversibility hold.

When a fluid undergoes a reversible process, a sales state points can be joined up to form a line

on a diagram of properties. The workdone by the fluid during any reversible process is therefore

given by the area under the line of the process plotted on a p-v diagram as shown in Fig. 1.8.

Fig. 1.8 Reversible expansion process on a p-v diagram

Example 1.1

A fluid of unit mass at an initial pressure 6 bar and specific volume of 0.36m3/kg which is

enclosed in a cylinder behind a piston undergoes a reversible expansion process according to a

law pv2= constant. The final pressure is 1.2bar. Calculate the workdone during the expansion

process.

Solution

Work done = shaded area = 2

1pdvm where 2/vcp

2325

232252

11

2

21

2

2

1 2

//107776.0

/36.0/106

11112

1

kgmmNx

kgmxmNxvpc

pvcBut

vvmc

vmcd

vmcW

v

v

Also 2

2

2

22p

cvandcvp

23

2

23

25

25

2

2

/805.0

/102.1

107776.0

kgmv

kgmNmx

Nmx

p

cv

For unit mass m=1

kgKJ

kgNm

kgm

kgmNmx

vvCW

/404.119

/73.119403

/

/

805.0

1

36.0

1107776.0

113

2323

21

Example 1.2

A fluid of unit mass which is enclosed in a cylinder at an initial pressure of 30bar is allowed to

undergo a reversible expansion process behind a piston according to a law pv2= constant until the

volume is doubled. The fluid is then cooled reversibly at constant pressure until the piston

regains its original position. With the piston firmly locked in position, heat is then supplied

reversibly until the pressure rises to the original value of 30 bar. Calculate the net work done the

fluid for an initial volume of 0.075m3

Solution

NmmmNvvp

Nm

mmmN

VVC

VcdV

vdV

v

cpdV

v

cpandpvcso

mmN

mmNxxVpC

mNx

xx

v

vpp

CVpVp

w

wv

v

56250/)15.0075.0(105.7)(

112500

1/

015.0

1

075.0

116875

11

11

,,

/16875

/075.01030

/105.7

4

1030

15.0

075.01030

325

23223

3

62

21

2

1

2

1 22

2

112

2

2

62

62252

11

25

52

5

2

2

112

2

22

2

11

2

1

= -56250Nm

W23 = 0 since the piston is locked in position dV= 0

Net work done = W12 + W23 +W31

= 112500-56250 + 0

= 56250Nm

Hence the net work done by the fluid is 56.25KJ

PRACTICE QUESTIONS

1. A fluid at 10bar is enclosed in a cylinder behind a piston, the initial volume being

0.05m .Calculate the work done by the fluid when it expands reversibly according to a law pv

3=

C to a final volume of 0.06m3 (7640Nm)

2. 1kg of a fluid is compressed reversibly according to a law pv = 0.25, where p is in bar and v is

in m3/kg. The final volume is one-fourth of the initial volume. Calculate the work done on the

fluid and sketch the process on a p-v diagram (3466Nm)

3. 0.05m3 of a gas at 6.9bar expands reversibly in a cylinder behind a piston according to the law

pv1.2

= Constant, until the volume is 0.08m3. Calculate the work done by the gas and sketch the

process on a p-v diagram (15480Nm)

4. A fluid is heated reversibly at a constant pressure of 1.05 bar until it has a specific volume of

0.1m3/kg. It is then compressed reversibly according to a law pv= Constant to pressure of 4.2bar,

then allowed to expand reversibly according to a law pv1.7

= Constant, and is finally heated at

constant volume back to the initial conditions. The work done in the constant pressure process is

1515Nm, and the mass of fluid present is 0.2kg. Calculate the network of the cycle and sketch

the cycle on a p-v diagram (781Nm)

2.0 THE FIRST LAW OF THERMODYNAMICS

The first law is concerned with the principle of conservation of energy as applied to

closed systems which undergo changes of state due to transfer of work and heat across the

boundary. It is in fact, a statement of the law of conservation of energy as applied to heat energy

and mechanical work that energy can neither be created nor destroyed though it can be

transformed from one form to another.

The law states that, when a closed system is taken through a thermodynamic cycle, the net heat

supplied to the system from its surroundings is equal to the net work done by the system on its

surroundings.

Mathematically, ∂Q = ∂W

Where represents summation for a complete cycle. This implies that network done can never

be greater than the heat supplied. `

EXAMPLE 2.1

In a certain steam plant the turbine develops 2000kW. The heat supplied to the steam in the

boiler is 3800kJ/kg, the heat rejected by the steam to the cooling water in the condenser is

3100kJ/kg and the feed-pump work required to pump the condensate back into the boiler is

10kW. Calculate the steam flow rate.

Solution

Steam Plant

∂Q = Qin- Qout =3800-3100=700KJ/kg

Let the steam flow rate be in kg/s

Then, ∂Q = m x 700 = 700mkw

∂W = Wout-Win = 2000-10 =1990KW

But ∂Q = ∂W

700 =1990

= 1990/700 = 2.84kg/s

i.e. mass flow rate of steam required = 2.84kg/s

COROLLARIES OF THE FIRST LAW

Corollaries are basic propositions or inferences about the behaviour of thermodynamic systems

deduced from the laws of thermodynamics. Corollaries of the first law of thermodynamics are

stated as follow:

COROLLARY 1

There exists a property of a closed system such that a change in its value is equal to the

difference between the heat supplied and the work done during any change of state.

Mathematically, (∂Q-∂W) = du

Where U denotes the property so discovered called internal energy of the system.

i.e. Q-W = (U2-U1) for a non-flow process.

This equation is called non-flow energy equation (NFEE)

Internal energy, which is the intrinsic energy of a body not in motion, being a property, can be

said to reside in the system and it can be increased or decreased by a change of state.

COROLLARY 2

The internal energy of a closed system remains unchanged if the system is isolated from its

surroundings.

COROLLARY 3

A perpetual motion medicine of the first kind is impossible. That is, a system cannot produce

work continuously without absorbing heat-energy from the surroundings.

NON FLOW ENERGY EQUATION

From the first corollary of the first law, the following can be rightly deduced.

Gain in internal energy = Net heat supplied- Network done

Mathematically, dU = 2

1

2

1WQ

For a single process between state 1 and state 2, we have, U2-U1 = Q-W for a non-flow process

or in differential form, dU= ∂Q - ∂W

Where U is the specific internal energy in kJ/kg

This equation is true for both reversible and irreversible processes.

For reversible non-flow process dw = pdv = mpdv

But ∂Q = dU+∂W = du + pdv for unit mass

Q = (u2-u1) + 2

1pdv

The above equation is true for ideal reversible non-flow processes.

EXAMPLE 2.2

In the compression stroke of an internal combustion engine, the heat rejected to the cooling water

is 145kJ/kg and the work input is 190kJ/kg. Calculate the change in specific internal energy of

the working fluid stating whether it is a gain or loss.

SOLUTION

Qout = 145KJ/kJ, Win = 190KJ/kg

kgkJenergyinernalingain

kgKJ

WQUU

/45

/45

190145

190145

12

Example 2.3

In the cylinder of an air motor, the compressed air has a specific internal energy of 4200KJ/kg at

the beginning of the expansion and a specific internal energy of 2050KJ/kg after expansion.

Calculate the heat flow to or from the cylinder when the work done by the air during the

expansion is 1050KJ/kg.

SOLUTION

U1 = 4200kJ/kg, U2=2050kJ/kg, Wout= +1050kJ/kg

10502150

105042002050

12

Q

Q

WQUU

Q = - 2150 + 1050

Q = -1100kJ/kg

i.e. heat rejected by the air = 1100kJ/kg

PRACTICE QUESTIONS

1. An air compressor which compresses air at constant internal energy rejects 80kJ of heat to

the cycling water for every kilogram of air. Calculate the workdone during compression

stroke per kilogram of air (80kJ/kg)

2. In the compression stroke of a gas engine the work done on the gas by the piston is

90KJ/kg and the heat rejected to the cooling water is 52kJ/kg. Calculate the change in

specific internal energy (38kJ/kg)

3. The gases in the cylinder of an internal-combustion engine has a specific internal energy

of 950kJ/kg and a specific volume of 0.08m3/kg at the beginning of expansion. The

expansion of the gases may be assumed to take place according to a reversible law pv1.4

=

constant, from 50bar to 1.6bar. The specific internal energy after expansion is 180 KJ/kg.

Calculate the heat rejected to the cylinder cooling water per kilogram of gases during the

expansion stroke (-144 KJ/kg).

THE STEADY FLOW ENERGY EQUATION

The governing equation for an open system with mass transfer across its boundary is referred to

as steady flow energy equation (SFEE). This is analogous to non-flow energy equation for a

closed system.

For a flow to be regarded as steady, the mass flow must be constant and the same at inlet and

outlet and the fluid properties at any point in the open system must not vary with time.

Consider a unit mass of a fluid with specific internal energy u flowing steadily and moving with

velocity C through a seam power plant shown in Fig. 1.9 below

Fig. 1.9 Section of a steam power plant

The system above constitutes an open system with its boundary shown cutting the inlet pipe at

section 1 and outlet pipe at section 2.

This boundary is often referred to as a control surface and the system as a control volume.

Assuming a steady flow of heat Q per kg of fluid is supplied at the boiler and each kg of fluid

does work W at the turbine.

Also considering an element of fluid at the entry of the boiler shown below.

Fig. 2.0 section at inlet of boiler

Energy required to push fluid element across the boundary

= p1A1 x l = p1A1l = p1 x volume of fluid element

Energy required to move fluid element across the boundary at the eating = p1v1 v1=specific

volume at inlet.

Similarly,

Energy required to move fluid element across the boundary at the exit-p2v2; v2 = specific volume

at exit.

Total energy entering the system consists of the energy of the flowing fluid at inlet, energy term

p1v1 and the heat supplied Q i.e.

QvpgzcuEin 111

2

11 21

Total energy leaving the system consists of energy of the flowing fluid at the exit, energy term

p2v2 and the output work W

QvpgzcuEout 222

2

22 21

Since the flow of fluid into and out of the system with heat and work crossing the boundary are

steady, then Ein = Eout. Thus

WvpgzcuQvpgzcu 222

2

21111

2

11 21

21

The sum of specific internet energy and pv term is given the symbol h and is called specific

enthalpy i.e. h=u+pv

By writing h for (u+pv), the equation reduces to

12

2

1

2

212

222

2

221

2

12

21

21

21

zzgcchhWQ

WvpgzchQgzch

Where Q and W are the heat and work transfers per unit mass flow through the system.

The above equation is called the steady flow energy equation (SFEE) and provides the basic

means for studying most open systems of importance in engineering.

In nearly all problems in practice, the potential energy term g(z2-z1) is either zero or negligible

compared with other terms. It has been included in the equation for the sake of completeness and

hence can be omitted in analysis.

For steady flow, the mass flow rate over a cross section of the flow at inlet is constant and equal

to the mass flow rate at exist.

i.e. in = out =

and mass flow rate is the ratio of volume flow rate (AC) and specific volume ® hence, mass flow

rate = ACv

CA

v

CA

2

22

1

11

This equation which expresses the principle of conservation of mass in steady flow is known as

continuity equation.

EXAMPLE 2.4

In a steam turbine unit, the mass flow rate of stream through the turbine is 17kg/s and the power

developed by the turbine is 14000kw. The specific enthalpies of the steam at inlet and exit of the

turbine are 1200KJ/kg and 360KJ/kg while the velocities are 60m/s and 150m/s respectively.

Calculate the rate at which heat is rejected from the turbine. Also find the area of the inlet pipe

given that the specific volume of the steam at inlet is 0.5m3/kg.

SOLUTION

Given

In = 17kg/s, W = 14000kw, h1= 1200KJ/kg, h2 = 360KJ/kg

C1 = 60m/s, C2 = 150m/s, v1=0.5m3/kg

To determine Q

From SFEE, neglecting the potential energy term

Q-W = in (h2-h1) + ½ in ( )2

1

2

2 cc

kWQ

kW

Q

35.119140035.14119

35.1411965.16014280

60150172/1120036017000,14 22

i.e. heat rejected = 119.35 kW

To determine inlet area A1

From the continuity equation,

2

1

111

1

111

1

111

142.0

60

5.017

m

x

C

inA

C

AinAand

CAin

i.e. inlet area 2

1 142.0 mA

EXAMPLE 2.5

At the inlet to a nozzle employed for increasing the velocity of a steadily flowing fluid, the

specific enthalpy and the velocity of the fluid are 3025KJ/kg and 60mls respectively. The

specific enthalpy of the fluid at the exit is 760KJ/kg. The nozzle is horizontal and there is a

negligible heat loss from it. Calculate the fluid velocity at the exit, and also find the mass flow

rate of the fluid when the inlet area is 0.1m2 and inlet specific is 0.19m

3/kg. If the specific

volume at the nozzle exit is 0.5m3. Determine the exit area of the nozzle.

SOLUTION

Solution

Given

2

3

2

3

1

2

1

211

mindet

/5.0,/19.0,1.0

/2790,/60,/3025

Ceer

kgmvkgmvmA

kgkJhsmCkgkJh

From SFEE, neglecting the potential energy term

Once the nozzle is horizontal i.e. z1=z2

2

1

2

212 2/1 CChhWQ

But no heat and work cross the boundary

i.e. Q = 0 and W = 0

`/19.688

104736

106.473

108.2362

2/1102790108.3026

2/1102790602/1103025

2/12/1

2

3

2

3

32

2

2

2

33

2

2

323

2

22

2

11

smC

xC

x

xC

Cxx

Cxx

ChCh

i.e. fluid velocity at exit = 688.19m/s

From continuity equation,

19.0

601.0

1

11 x

r

CA

Inlet mass flow rate= 31.59kg/s

Also, = 2

22

1

11

v

CA

v

CA

2

2

22

02295.0

19.688

5.59.31

m

x

C

mvA

i.e. the exit area A2 = 0.02295m2

PRACTICE QUESTIONS

1. A steady flow of steam enters a condenser with a specific enthalpy of 2300KJ/kg and a

velocity of 350m/s. The condensate leaves the condenser with a specific enthalpy of

160KJ/kg and a velocity of 70m/s. calculate the heat transfer to the cooling fluid per

kilogram of steam condensed (02199KJ/kg)

2. A steam turbine receives a steam flow of 4.35kg/s and the power output is 500kw. The

heat loss from the casing is negligible. Calculate the change in specific enthalpy across

the turbine when the velocity at entrance is 60m/s, the velocity at exist s 360NS, and the

inlet pipe is 3m above the exhaust pipe. (433KJ/kj)

3. A turbine operating under steady-flow conditions receives steam at the following state:

pressure 13.8bar, specific volume 0.143m3/kg, specific internal energy 2590KJ/kg,

velocity 30m/s. The state of the steam leaving the turbine is as follows: pressure 0.35bar,

specific volume 4.37m3/kg, specific internal energy 2360KJ/kg, velocity 90m/s Heat is

rejected to the surroundings at the rate of 0.25kw and the rate of steam flow through the

turbine is 0.38kg/s.

Calculate the power developed by the turbine (102.7kw)

REVERSIBLE NON-FLOW PROCESS

CONSTANT VOLUME PROCESS (ISOCHORIC PROCESS)

In a constant volume process, he working substance is contained in a rigid vessel, hence the

boundaries of the system are immovable, so work cannot be done on or by the system other than

parallel which work input which is negative and this can be carried out in practice by some

method of during the fluid. It will be assumed that constant volume indies zero work unless

otherwise stated.

From the non-flow energy equation for unit mass, since no work is done, i.e. W = 0, we therefore

have Q = u2-u1

Or for mass m, of the working substance

Q= U2 – U1

That is, all the heat supplied in a constant volume process goes to increasing the internal energy.

This equation can be written in differential form as ∂Q = dU

For a perfect gas, Q = mcv (T2-T1)

The figures below show the p-v diagrams of constant volume process for a vapour and a perfect

gas

CONSTANT PRESSURE PROCESS (ISOBARIC PROCESS)

A closed system undergoing a process at constant pressure is illustrated in the figure below

The fluid is enclosed in a cylinder by a piston on which rests a constant, weight. If heat is

supplied, the fluid expands and work is done by the system in overcoming the constant force.

Heat is extracted, the fluid contracts and work is done on the system by the constant force .

From the non-flow energy equation,

Q = (u2-u1) + W

Work might also be done in the system simultaneously by churning the fluid with a parallel, and

this negative quantity of work must be included in the term W if no paddle work is done on the

system and the process is reversible i.e. W = ∫pdv = pdv, we have

∂Q = du + pdv

Or ∂Q – pdv = du

Since p is constant, this can be integrated to give

1212 uuvvpQ

A further implication of the energy equation is possible if a new property is introduced. Since p

is constant, pdv is identical with d(pv)

Thus the energy equation can be written as

pvudQ

dupvdQ

But specific enthalpy h = u+pv

i.e. ∂Q = dh

or in the integrated form

Q = h2-h1

or for mass m, of a fluid

Q = H2-H1 (reversible process only)

That is the heat added in a reversible constant pressure process is equal to the increase in

enthalpy. The above equation does not apply if the process is irreversible.

For a perfect gas, Q = mcp (T2-T1)

The figures above show the p-v diagrams of constant pressure process for a vapour and a perfect

gas

NB: The areas shaded represent the work done by the fluid i.e. (v2-v1)

CONSTANT TEMPERATURE PROCESS (ISOTHERMAL PROCESS)

When the quantities of heat and work are so proportioned during an expansion or compression

that the temperature of the fluid remains constant the process is said to be isothermal when a

fluid in a cylinder behind a piston expands from a high pressure to a low pressure, there is

tenancy for the temperature to fall. In isothermal expansion heat must be added continuously in

order to keep the temperature at the initial value.

Similarly in an isothermal compression heat must be removed from the fluid continuously during

the process. It is possible to show that for a reversible isothermal process, a certain definite

relationship must exist between p and r, and consequently, the work done has a definite value

that can be produced.

For any reversible process the energy equation is ∂Q – pdv = du

Or Q - 2

112 uupdv

The figures below show the p-v diagrams of isothermal process for a vapour and a perfect gas

For isothermal process of a perfect gas which is assumed to have an equation of state.

pv=RT=constant, since T=constant

i.e pv = constant

or p1v1 = p2v2

The work done by a perfect gas which undergoes a reversible isothermal expansion from state 1

to state 2 as shown in the figure above is given by the shaded area and can be evaluated as:

That is the internal energy remains constant in an isothermal process for a perfect gas

From non-flow energy equation, we have

W = 2

1pdv

That is, the quantity of heat supplied is equal to the workdone in an isothermal process for a

perfect gas only.

ADIABATIC PROCESS

An adiabatic process is the one in which heat is prevented from crossing the boundary of the

system. i.e. no heat is transferred to or from the fluid during the process. That is, an adiabatic

process is one undergone by a system which is thermally insulated from its surroundings.

From the non-flow energy equation

Q = u2-u1 + W

For an adiabatic process Q = 0, we have

-W = (u2-u1)

Or W= (u1-u2)

This equation is true for an adiabatic non-flow process whether or not the process is reversible.

In an adiabatic compression process all the work done on the fluid goes to increasing the internal

energy of the fluid. Similarly in an adiabatic expansion process, the work done by the fluid is at

the expense of a reduction in the internal energy of the fluid.

A reversible adiabatic process is called isentropic (constant entropy) process.

For a vapour undergoing a reversible adiabatic process the work done can be found from

equation given above by evaluating u1 and u2 from tables. For a perfect gas, a low relating p and

v has been detained for a reversible adiabatic process as pv = constant

Each perfect gas undergoing a reversible adiabatic process have its own value of

POLYTROPIC PROCESS

The constant volume and constant pressure processes can be regarded as limiting cases of a more

general type of process in which both the volume and pressure change, but in a certain specified

manner.

In many real processes it is found that the states during an expansion or compression can be

described approximately by a relation of the form pvn = constant where n is a constant called the

index of expansion or compression, and p and r are average values of pressure and specific

volume for the system.

Compressions and expansion of the form pvn=constant are called polytropic processes. When

n=0, the relation reduces to p= constant (isobaric process) and when n-0, it can be seen to reduce

to r=constant by writing it in the form pnr= constant

For any reversible process, W = pdv

For a process in which pvn = constant, we have nv

cp

Where c is a constant

Thus,

2

1

2

1

2

1

1

1v

v

v

v

v

v

n

nn n

vc

v

dvdv

v

cW

11

1

1

1

2

1

1

1

2

n

vvc

n

vvc

nnnn

1

11

1

2

1

2

1

1

1

2

1

1

1

2

1

2

n

vvvpvp

n

vvc

n

vvc

nnn

nnnn

Since constant c = p1 v1n = p 2 v2

n

1

2211

n

vpvpW

This equation is true for any working substance undergoing a reversible polytropic process. In

follows also that for any polytropic process, we can write

n

v

v

p

p

1

2

2

1

The integrated form of the energy equation for a reversible polytropic process can therefore be

written as

121211

1uu

n

vpvpQ

In a polytropic process the index ‘n’ depends on in the heat and work quantities during the

process. The various processes considered previously are special cases of the polytropic process.

When n=0, pvo=c or p= constant (isobaric process)

When n = , pv = C or pv1/ = C i.e. v = constant (isochoric)

When n=1, pv = C i.e. T= constant (isothermal process

Since pv/T=constant for a perfect gas

When n= , = constant i.e. reversible adiabatic process

This is illustrated on a p-v diagram shown below

State 1 to A is constant pressure cooling (n=0)

State 1 to B is isothermal compression (n=1)

State 1 to C is reversible adiabatic compression (n=r)

State 1 to D is constant volume heating (n= )

Similarly,

State 1 to A

Is isobaric heating

State 1 to B’ is isothermal expansion

State 1 to C is reversible adiabatic expansion

State 1 to D is isochoric cooling

3.0 THE SECOND LAW OF THERMODYNAMICS

THE HEAT ENGINE

A heat engine is a thermodynamic system operating in a cycle and across the boundaries

of which flows only heat and work.

Fig. 3.1 Symbolic representation or Heat Engine

Example of Heat Engine

The Steam Turbine Plant: Represented diagrammatically in fig. 3.2.

Applying the first law

21 QQWW

dQdW

px

Fig. 3.2 Steam Turbine Plant

Note: - Open system is not a heat Engine. b/cos of heat & mass flow across the boundary.

3.1.2 Closed system gas Turbine plant: Represented diagrammatically by fig. 3.3.

Fig 3.3 Close System Gas Turbine Plant

21 QQWW cx

The open system gas turbine plant and the reciprocating types of internal combustion engine

have fuel and air mixture crossing their boundaries and so are not heat engine.

HEAT ENGINE PERFORMANCE

The greater the proportion of heat supplied that is converted to work the better is the

engine.

The ratio pliedheat

outputworknet

supis termed the cycle efficiency denoted by

1Q

WW px where

Q1 is heat supplied per mass of the system

Wx is work done per mass by the system

Wp is work done per mass on the system.

If Q2 is heat rejected per mass by the system, by the first law

Q1 - Q2 = Wx - Wp

1

2

1

21 1Q

Q

Q

QQ

This shows that the cycle efficiency will be unity (i.e. 100%) if Q2 = 0. That is, Q2 = 0 there will

be 100% conversion of heat into work.

Heat Reservoir: Heat Source and Heat Sink

The term heat reservoir (or reservoir) is used in thermodynamics to mean a heat source or heat

sink of uniform temperature and of infinite capacity such that unlimited quality of heat can flow

across its boundary without changing its temperature.

When heat flow is at a high temperature, it is referred to a hot reservoir and when heat flow is at

a low temperature, it is referred to as a cold reservoir.

STATEMENT OF THE SECOND LAW OF THERMODYNAMTICS

It is impossible to construct an engine which will work in a complete cycle and produce no

effect, except to raise a weight and exchange heat with a single reservoir.

The first law of thermodynamics shows that net work cannot be produced during a cycle without

some supply of heat. It went supplied. The second law emphasizes the fact that some heat must

always be rejected during the cycle, i.e. the net work must be less than the heat supplied.

PERPETUAL MOTION MACHINES

Any machine which creates its own energy is called a perpetual motion machine of the

first kind (PMM 1). Such a machine contradicts the first law since the first law rejects the

possibility of creating or destroying energy. Similarly, any machine which produces work

continuously while exchanging heat with only a single reservoir is know as perpetual motion

machine of the second kind (PMM 2). This machine contradicts the second law. From the

foregoing both propositions are impossible.

If the second law were not true it would have been possible

a. To drive a ship across the ocean by taking heat from the ocean.

b. To run a power station by extracting heat from the surrounding air.

There is nothing in the first law to say that the above projects are not possible (interchange

between heat and work). The projects are, however, impossible and so the second law must be

true. There is not natural sink of heat at a lower temperature than the atmosphere or ocean.

DIRECT AND REVERSED HEAT ENGINES:

Direct heat engine Reversed heat engine

Produces net out-put of Directions of work and

Work and heat from net heat flow is reversed

Input of heat. (Heat pump refrigerator)

Fig. 3.6 (A) Fig 3.6 (B)

Refrigerator withdraws heat at a lower temperature since its principal purpose is the

extraction of heat from a cold space.

Heat pump. The principal purpose of the heat pump is to supply heat to the hot space.

The Coefficient o performance of Refrigerators and Heat Pumps:

The coefficient of performance of a refrigerator (known as the performance energy ratio ref is

defined as

plyWork

eTemperaturLoweratTransferHeat

sup

W

Qref

2 Q2 = heat extracted

The coefficient of performance of a heat pump (performance energy ratio)

hp is defined as

SupplyWork

eTemperaturHigheratTransferHeat

W

Qhp

1

The relation between these two coefficients of performance can be established by applying the

first law.

21

21

QWQ

WQQ

ref

221 11

W

Q

W

QW

W

Qhp

ref 1 hp

PRACTICE QUESTIONS

1. The work done on a reversed heat engine is 100kg whilst the heat transfer to the engine

from the low temperature is 300kJ. Determine the heat transfer to the high temperature reservoir

and the COP as a refrigerator and as a heat pump.(400kJ, 3, 4)

2. A heat pump picks up 1000kJ of heat from well water at 100C and discharges 3000kJ of

heat to a building to maintain it at 200C. What is the COP of the heat pump? What is the

minimum required cycle net work? (1.5, 2000kJ)

4.0 ENTROPY

From thermo temp. Scale 2

2

1

1

T

q

T

q numerically

2

2

1

1

T

q

T

q Algebraically

2

2

1

1

T

q

T

q = 0

i.e. summation of ratio of heat transfer to absolute temp. in reversible cycle = 0

two property diagram for any Reversible cycle may be replaced by elemental Carnot cycle

Thus, for any reversible cycle;

0 T

qr

But

isT

qR 0 a property = entropy = S

Note: property

propertyNon = property

For a reversible process,

ST

qr

In any adiabatic process, 0ad

q

In any reversible adiabatic process

0)( adR

T

q

cIsentroppi

QR =Tds = area under T-S curve

For an irreversible cycle, R 1

1

2

1

2 11T

T

q

q

Hence ynumericallT

qand

T

T

q

q

1

2

1

2

1

2

Algebraically,

02

2

1

1 T

q

T

q

0 T

q

[= 0 reversible ] Clausius

[< 0 irreversible] Inequality.

Path 132 0

2

1

1 T

q

Path 241 21

2

1

SST

qR

Path 13241 1

1

T

q

12

21

1

2

2

1

1

00

SS

SS

T

q

T

q R

Note: If part of a cycle is irreversible, the whole cycle must be irreversible.

Irreversible adiabatic process incurs rise in entropy.

All Reversible adiabatic process incurs no rise in Entropy. Rise in S. if non – adiabatic, heat flow

across boundary, and rise in S for system + environment e.g. heat flow q from system at T1 to

environment at lower To.

System: entropy loss, 1T

q

Environment entropy gain 0T

q

Since T1 > To, gain > loss.

s positive

A decrease in entropy can only be temporary and local with a greater increase elsewhere.

There is no process that cannot be reversed if we accept a greater irreversibility elsewhere.

Since entropy is a property (extensive) s . Same irrespective of path (rev. and irrev.) but

casereversibleinonlyT

q

T

qs

actual

R

One spontaneous tendency in nature is towards minimum energy.

Another spontaneous tendency in nature is towards maximum disorder (of motion or position) as

in changes from solid to liquid, to gas, to gas at lower pressure to a larger number of molecules,

mixing of different fluids, dissipation of energy by friction or viscosity

Entropy is a measure of disorder

In an adiabatic device, reversible (Reversible adiabatic) process is most efficient. Hence,

isentropic efficiency s can be devised.

(N.B. a process efficiency not cycle effy.)

For work absorber (compressor or pump)

Isentropic Efficiency ( s ) = W

W

workactual

workisentropic s

For work producer (turbine)

s = SW

W

For a flow device (nozzle)

s = k

k

Eoutletisentropic

Eoutletactual

EnergyKineticE 2

c.f. in a slow moving device (large piston compressor) time for heat transfer

W

W

idealisothermal

TT

4.1 Third law states that “the entropy of a pure substance in its most stable/state approaches

zero as the temperature approaches zero i.e. lim S = 0

Practice Questions

1. A system consists of a pure dissipative element and a pure thermal element with a heat

capacity of 50kJ/K. It experiences an irreversible work transfer interaction which takes the

system from state 1 at a temperature of 300K to state 2 at a temperature of 310K. If no heat is

transferred during this irreversible process, calculate the change in entropy of the system for this

process (1.6395kJ/kgK).

2. A carnot heat engine rejects 230kJ of heat at 250C. The net cycle work is 375kJ. Determine

the thermal efficiency and the cycle high temperature.