Td Notes by Chiranjeevarao

Thermodynamics Unit I Notes 1

GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela

Thermodynamics : “Thermi” literally means heat and “ Dynamis “ is nothing but Power, So thermodynamics is nothing but conversion of heat into power Applications of the thermodynamics :-

� I.C Engines � E.C Engines � Refrigeration system � Jet � Rockets. etc

Thermodynamic System

It is the region or space on which the attention is focused for study. Boundary



It is a real (or) imaginary surface. It separates the system from its surroundings. � If Nozzle is a system there will not be any real boundary at its ends. In such a case we

need to assume the boundary known as imaginary. � If any gas in a cylinder is bounded by a piston, whose upward movement is Restricted by

the stoppers is subjected to external heat source then the gas Will ties to expand. But the stoppers will not allow the expansion in such A case the boundary is a fixed boundary.

Surroundings

Everything beyond the boundary is known as surroundings. Surroundings will not have any boundary. Universe

A system, its boundary and the surrounding are combinedly known as universe.



Types of system A thermodynamic system can be classified into

i. Open system ii. Closed system iii. Isolated system

Open system

It is a thermodynamic system which allows both mass and energy to cross the boundary. It is also known as flow process (or) control volume.

Ex:- Turblnes, compressor, nozzles, I.C engines. Closed system

It is a thermodynamic system which does not allow mass to cross the boundary but the energy may cross the boundary. It is also known as control mass (or) Non flow process.

Ex:- Refrigeration cycle, closed cycle gas turbine power plant. etc



Isolated System It is a thermodynamic system which does not allow either mass (or ) energy to cross the boundary.

Ex’- Universe, Thermal flask, Adiabatic engine. Control Volume

It is a property selected region in space is nothing but control volume. Which allows the mass flow. An open system is nothing but a control volume.


GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela

macroscopic approach and microscopic view of approach

Concept of continuing

S.no Macro scopic approach Micro scopic approach

1. Studying a thermodynamic system

without bothering the events occurring

at molecule level.

Studying the system by considering the events

at molecular level.

2. The simple mathematic relations are

enough to analyze the system.

A simple mathematic relations are not enough

to analyze the systems

3. In case of macroscopic approach we can

measure a thermodynamic property in

laboratory.

It is very difficult to measure in laboratory.

4. Simple mathematical data is enough to

study the system.

Along with the mathematical data statistical

data is also essential.

5. This study is known as classical thermo

dynamics.

This study is known as statistical

thermodynamics.



Every substance is made from own molecules. In case of gaseous substance the molecules are

widely spreaded.

Consider a volume of δV around the point P which consists of a mass of δM. In the

region of continuum the ratio. δM is constant even though there is a change in volume. It will

be maintain up to the limit of volume δV. It indicates that the system is a continuum up to this

volume only. It the volume is still decrease beyond SV, some molecules may escape into the

surroundings and δM/δV may decrease. Otherwise with considerable decrease in volume the

SM/SV may increase that indicates that the substance is not continuum in the molecule region.



Thermodynamic Equilibrium

A system is said to be under thermodynamic equilibrium if it satisfies the following

condition of the equlibrium.

� Mechanical equilibrium

� Chemical equilibrium

� Thermal Equilibrium

Mechanical Equilibrium

A system is said to be under mechanical equilibrium if ther is no unbalanced force with in

the system (or) between system and surroundings.

Chemical Equilibrium

The system is said to be under chemical equilibrium if ther is no chemical reaction within

the system or systemate surroundings.

Thermal equilibrium

The system is said to be under thermal equilibrium if it is separated from the

surroundings by a diathermal wall adiabatic wall.

Note:-

Sometimes the system may not be under chemical equilibrium but it is under mechanical

andthermal equilibrium such a condition is knows as metastable equilibrium.

Property

It is the characteristic of the system . There are eight properties.



� Pressure ( p )

� Volume ( v )

� Temperature (T )

� Internal energy

� Enthalaphy (h )

� Entropy(S)

� Gibbs function(G)

� Helmotz’s function.( H )

Types of properties:-

1) Intensive property

2) Extensive property

� Intensive property:-

� These properties are independent of mass of the system.

Ex: Temperature:

� Extensive property:-

� These properties are dependent of mass of the system

Ex: Volume, energy

State

State is nothing but the physical condition of the system.

To define the state at least two properties are essential.



Process

A process is nothing but the change of state.

A substance which is being heated in a closed cylinder undergoes a non-flow process .

Closed systems undergo non-flow processes. A process may be a flow process in which mass is

entering and leaving through the boundary of an open system. In a steady flow process mass is

crossing the boundary from surroundings at entry, and an equal mass is crossing the boundary at

the exit so that the total mass of the system remains constant. In an open system it is necessary to

take account of the work delivered from the surroundings to the system at entry to cause the

mass to enter, and also of the work delivered from the system at surroundings to cause the mass

to leave, as well as any heat or work crossing the boundary of the system.



Cycle

Thermodynamic cycle (or) cycic process:-

A cycle is nothing but sequence of thermodynamic process

In which the end states are identical

The area under P-V diagram will give the work done.

Point function

The thermodynamic event which is the function of end states is called point function

Ex: temperature

All the thermodynamic properties are point functions.

Path function

A thermodynamic event which is a function of the path executed by the process.

Ex: heat, work

Work is a path function

Heat and work are in exact differentials.

∫i2dw =w2-w1



Consider three different processes a,b and c between the same end states. As you know the area

under P-V diagram will give the work done.

WA < WB < WC

Because the area under A < area under B < area under C from equation (1) it is evident that even

though the end states are same for all the three processes the work done is different from process

to process.i.e., the work done is independent of end state and it will dependent on the path of the

process.

Therefore work is a path function .

Heat is a path function

The area under TS diagram will give heat transfer.

The area under process A is less than the area under process B is less than the area under

process C, even though the end States are identical



QA < QB < QC

From the above equation we can say that heat transfer will depend on a process the path of a

process



Quasistatic process

Quasi literally means almost so the quasistatic process is the process in which the system

is said to be under equilibrium at each and every state in between the end states.

Consider certain amount of gas in a cylinder which is bounded by a piston when the load

W is acting on the piston the system is under equilibrium and is known as initial state of

equilibrium.

Let P1, V1, T1 are the properties at this condition. If the load W is removed suddenly the

pressurized gas will ties



To push out the piston the upward movement of piston is restricted by stoppers which are

provided at the other end of cylinder over a finite period of time again the system attains

equilibrium condition knows as second stage of equilibrium condition.

Even though the system is under equilibrium at stage (1) and (2), it may not be under

equilibrium in b/w these two states.

Consider the same type of system which is subjected to no. of small loads. At this stage

the system is under equilibrium. If one small load is removed there may not be any appreciable

change in properties. So, we can say that the system is under equilibrium i.e., there may not be

any change in equilibrium with every removal of small load. So, the system is said to be under

equilibrium at every states and the process will becomes a quasistatic process.

Note:-

In real life no process is a quasistatic process.

Reversible process

A reversible process is a process in which the System attains its initial state without any

effect on the rest of the environment.

` 1-2 → process

2-1 → Reversible process

A process can become reversible.



� If there is no involvement of friction ( can also be called as ideal process)

� If it is a quasistatic process

� If the process is very slow

Irreversible process

It is a process in which the initial state cannot be attained.

Note:-

In real life every process is irreversible.



Energy

It is the capacity to do work units- joules (or) kilojoules.

Types of Energy:-

1) Static form of energy ( Energy in state )

2) Dynamic form of energy ( Energy in Transission )

Static form of energy ( Energy in state )

It is the quantity of energy which is stored within the system.

Ex: potential energy kinetic energy, energy stored in full

Dynamic form of energy ( Energy in Transission )

It is the form of energy which cannot be stored within the system i.e., energy will cross

the boundery of system. Also called as energy Interaction (or) Energy transition.

Ex: Heat and work are energy in transition .

Heat It is the form of energy in transition which may cross the boundary by the virtue of

difference in temperature.

Denoted by Q.

Units – Joules (or) kilo joules

In adiabatic process there will not be any heat inflow or outflow.

Heat added to the system is always positive

And heat rejected by the system is always negative.

Work

If energy is crossing the boundary and if it is not heat then it is nothing but work. Work

input to the system is negative.

Work developed by system is positive.

Units-Joules (or) Kilojoules



Different types of work

1) Displacement work

Let P is pressure acting on the pision.

a = area of pision

dl = elemental displacement

… force acting on pision = P*Q

The elemental work dw = Pa.dt

= Pdv

W1-2 = ∫2pdv

Displacement work

2) Shaft work



W= F x S

= F x 2IIr (..for I revolution)

= F x 2IIr x N (..N revolutions)

= T/r x 2IIr x N

W= 2 IINT

Thermodynamics Unit II Notes 7


Thermal equilibrium:-

If two bodies of different temperatures are incontact heat will flow from the body at

higher temperature to the body at lower temperature both the bodies are said to be under thermal

equilibrium when they attains the same value of temperature,

Zeroth law of thermo dynamics:-

If two bodies are in thermal equilibrium with a third body separately then the all the three bodies

are in thermal equilibrium with one another.

Zeroth law is basis for temperature measurement.

Consider certain quantity of liquid in a vessel.

Let A is the liquid B is the Glass bulb and C is the mercury.

TA is the temperature of the liquid.

TB is the Bulb temperature.

TC is the mercury temperature.

The inner side of the moneter, over a finite period of time the glass bulb will attain the

thermal equilibrium condition with the liquid i.e., TA+TB



Aftert sometime the mercury in will attain the thermal equilibrium condition with respect

to the glass bulb

TB = TC

TA = TC

TA =TB =Tc

i.e., the mercury indicates the temperature of the liquid. At this incident.

TA=TB=TC

Which saticifies the zeroth law of thermo dynamics

Thermoemty:-

1) It is the art of measuring temperature.

2) Ice point

This is limit of temperature where the solid and liquid states are in the under equilibrium.

3) Steam point:

It is limit of temperature where a liquid and vapour under thermal equilibrium. Two

reference.po

Thermometric proerty:-

It is the properity which is influenced at by the temperature.

Ex: length, resistance. Pressure, volume

Two reference point temperature scale:-

Let T is the temperature, x is the thermometric properity

T œ x

T = ax →(1)

Here ‘a’ is constant

At the ice point the equation becomes T 1 = ax1→(2)

Similarly at the steam point T 2 = ax2→(3)



To find the temperature at the thermometric property x at

(3)/(1) → T 1 / T = x2 / x →(4)

(2)/(1) → T 2 / T =x1 / x →(5)

On subtracting (5) from (4) (4)-(5) ( T 2 - T 1 ) / T = x2-x1 / x

Some two point temperature scales:-

1) Celsius scale

2) Fahrenheit scale

3) Reamer Scale

4) Kelvin Scale

Celslus scale:-

Ice point-0 0C

Steam point-100 0 C

Scale is divided into 100 equal divisions.

Fanrenheat scale:-

Ice point-32F

Steam point – 212F

Scale is devided into 180 equalidivisions

Reamur scale:-

Ice point-80R

Steam point-80R

Scale is divided into 80 equal divisions

C / 100 = (F-32 ) / 180 = R / 80



Constant volume gas thermometer:-

It consists of two vertical grass tubes connected By a flexible tube. one end of the glass

tube is exposed to atmosphere and the other end is in Communication with the gas bulb to a

capillary tube,

The gas bulb is filled with any gas like oxyzen, nitrozed, hydrozen and the nanometer is

filled with mercury.

Let P0 is the atmospheric pressure

p the density of nanometric liquid H is the a differential head

Whenever the gas bulb is exposed to any temperature T (which is unknown) then the

pressure D is equals to atmospheric pressure + pressure due to differential head.

P = P0 + pgh



Now expose the gas bulb to the t ripple point temperature (273.16k) At this instaent the

pressure acting on the gas PTRP=P0 +Rgh

We know that T œ P

T = C P

T/P=constant

Ttrip / Ptrip = constant

T / P = Ttriple / Ptriple

T = P / Pptriple X Ttriple

If the a constant volume gas thermometer is operated with different gasses inside the gas bulb at

different instances the variation in temperature w.r to pressure is shown in fig

If these threads are extrapolated onto the negative X-axis all the lines are intersect with a

common point. Which indicates the temperature -273.100 C i.e.

it is the minimum temperature

which we can measure using a constant volume gas thermometer.



Joule’s experiment:-

Consider certain mass of water inside and adiabatic vessel. Which is

provided with a pedal wheel and a thermometer.

Let the initial temperature of wter is T1 by operating the pedal wheel we

can stir the water for a period of time we can find an observable rising temperature (with the

collision of water molecules).

Let dw is a elemental work supply to the system to stir the water.. The temperature of the

water will raise let this is process 1 to 2



If the system is exposed to the atmosphere by removing the cap, over a finite period of

time the system attains thermal equilibrium condition with the surroundings. Let this is process 2

to 1 and attains the initial state. During this process certain quantity of the heat is being rejected

to the atmosphere. The processs (1) to (2) and (2) to (1) combindly executes the cycle.

Joule connected thousands of experiments and the observed that dW œ dQ

dW = J dQ

Where J is the joule’s constant (or) mechanical equivalent. Of the work

First law of thermodynamics:-

Energy can be transformed from one formed to another,.

(or)

Energy neithery be created, nor be destroyed (or) algebric sum o work transfer equals

to the algebric sum of heat transfer.

First law applied to a System:-

Let us consider a system which is being supplied with different sources E1, E2 and E3

Let W1 , W2 and E4 are the energies Coming out the system.

Energy accumulated inside the system is



E1 +E2 +E3 = (W1 +W2 + E4 )

U = Q-W

Q= ∆ U+W→(Non flow energy equation)

The energy accumulated is nothing but ‘ internal energy.’

Thermodynamics Unit III Notes 12


LIMITATION OF I LAW OF THERMO DYNAMICS:-

1. It is not bothering about the physical change of a system.

2. Not considering about the direction of energy conversion

3. It may not be implied for all the forms of energy,

4. conversion of energy from one from to another is possible but omplete

conversion of energy is impossible.

THERMAL ENERGY RESERVE:-

It is a large body with infinite heat capacity Which can able to absorb or reject the heat

without any Appreacible change in thermodynamic co-ordinate. In General the thermal energy

reservoir from which the Energy is supplied to the engine is termed as source And the reservoir

to which the energy is rejected is Termed as sink.

Heat engine:-

Consider a certain amount of gas in a cylinder bounded by a poison byattatching a not body heat

will be transferred to the gas and it will expand on attatching the cold body the gas will reject

the heat to the coldbody and it is supposed to attain its initial state.

These two process will combindly executes a



Thermodynamic cycle and develops certain amount

Of useful work.

Such a devic is known as a heat engine.

Heat engine can be defined s a device which operates on thermodynamic cycle

anddevelops certain amount of useful work by transfer of heat from high temperature reservoir to

law temperature reservoir.

Efficiency of heat engine(n)

η =desired output/required input

=W/Qs

Large power planes=30.35%

SI η =25.30%

CI η =35.40%

an automobile engine develops 100 kw of power with an efficiency of 30% the amount

of energy released during the burning of fuel is 35000KJ1kg.determine the heat

rejected by the engine also find mass flow rate of the fuel.

Given date:

Effieciency =30%

Output power (W)=100KW

Amount of energy released (Qs)=35000KJ1kg

Heat rejected QR=?

=W/QS

W= *QS

=0.3*35000

=10500KJ1kg

Total output W=m*W

Mass flow rate(m)=100

0.0095Kg1sec

We have W=QS-QR



QS=QS-W

=35000-10500

Heat rejected (QR)=24500KJ1kg

Rerrigerator:-

It is a device which is used to transfer the heat from the body at lower temperature

reservoir to higher temperature reservoir.

Co-efficient of performance:-

COP=desired output/required in put

=Q2/W

= refrigeration effect/work done

= N/W

Cop is always greater than one (1)

Heat pump:-

Heat pump is a device which operates

On the same thermodynamic cycle as

Refrigerator and it is used to supply

(pump) the heat to the higher temperature

Reservoir from a lower temperature reservoir.

COPHP=QH/W

=QH/QH-QL



=MCTH/MCTH-MCTL

COP=TH/TH-TL

1) find the cop and or the heat rejected in the KJ1hr condenser of a refrigerator with the

heat removal rate is 1200 kJ1hr and work input is 0.75 KW.

Given date:-

Heat absorbed by the refrigerator=1200KJ1hr

Work input=0.075KW

=0.75*60*60KJ1h

=2700KJ1h

Determine

Cop and heat rejected in condenser

COP=Q/W

=1200/2700

=0.44

Heat rejected in condenser Q4=W+Q

=3900KJ1Hr

Prove that cop of heat pump is equal to it COPR:-

COPR=Q/W

=QL/QH-QL→(1)

COPHP=Q/W

=QH/QH-QL

=QH-QL+QL/QH-QL

=1+QL/QH-QL

COPHP=1+COPR



IInd law of themodynamics:-

Based on avarious limitations of first law thermodynamics. The second law of

thermodynamics is formulated with two different statements namely

1) clauslus statement

2) Kelvin-plank’s statement

Claslus statement:-

The heat cannot flow from the body at lower temperature reservoir. To da body at

higher temperature r eservoir without any external agenecy.

Heat cannot flow from lower temperature to higher temperature un aided.

Kelvin-plank’s statement:-

It is impossible to construct an engine to operate with a single reservoir which develops

net work done.(energy supplied)

It is impossible to construct an engine with 100% of efficiency.

Equallance of Kelvin-plank’s and clasius statements;-

Consider an engine E which violets. The second Law of thermodynamics Kelvin-plank’s

statement



i.e. the engine E,the work developed is equal to energy supplied consider a heat pump whose

sole purpose is to desiver the heat continuously to a higher temp reservoir from a lower

tempreservoir.

Let the work required to run the

Compressor is supplied by the engine

E. Also assuming that the heat pump

Violets a Kelvin plank’s statements.

The energy input should be equals to the energy leaving from the pump (energy leaving the

heat pump is equals to =W+QL

=QH+QL

NOTE:-

1) if a device violates both the statements (or) if a device obeys both the statements we can

conclude that the statements are identical.

Now consider the combination of a two devices in a single unit. Them the heat delivered to a

higher temperature reservoir will becomes QL

W+QL=QL

W=0



Corollaries to the 2nd law of thermodynamics: 1st Corollary: The Clausius statement of the second law is often treated as the 1st corollary. 2 nd Corollary: It is impossible to construct an engine operating between only two heat reservoirs that will have a higher efficiency than a reversible engine operating between the same two reservoirs. 3 rd Corollary: All reversible engines operating between the same two reservoirs have the same efficiency. 4 th Corollary: A scale of temperature can be defined which is independent of any particular thermometric substance, and which provides an absolute zero of temperature. 5 th Corollary: The efficiency of any reversible engine operating between more than two reservoirs must be less than two reservoirs must be less than that of a reversible engine operating between two reservoirs that have temperatures equal to the highest and lowest temperatures of the fluid in the original engine. 6 th Corollary: There exists a property of a closed system such that a change its value is equal to for any reversible process undergone by the system between state 1 and state 2. 7 th Corollary: The entropy of any closed system, which is thermally isolated from the surrounding either increase, or, if the process undergone by the system is reversible, remains constant. Perfectional motion machine(PMM-II):-



It is a device which fails to saticify the IInd law of thermodynamics. It is a device with

100% of efficiency.



The Carnot principles

• First Carnot principle: The efficiency of an irreversible heat engine is always less than the

efficiency of a reversible heat engine operating between the same two reservoirs.

(ηIrr<ηrev)

• Second Carnot principle: All reversible engines operating between the same two

reservoirs have the same efficiency.

η Rev1 =ηRev2

Proof of First Carnot principle

• Proof by contradiction: Assume ηIrr>ηRev

• Conclusion: Assumption ηIrr>ηRev is incorrect. Efficiency of a reversible engine is

higher than that of an irreversible engine.

,Work deliveredη=

Heat input from the hot reservoirnet out

H

W

Q=



Proof of Second Carnot principle

• Proof by contradiction: Assume ηRev1>ηRev2.

Conclusion so far: Assumption ηRev1>ηRev2 is incorrect

1 2 Rev Revη η≤∴

Proof by contradiction (continued): Assume ηRev1<ηRev2



• Final conclusion: ηRev1=ηRev2. All heat engines working between the

same reservoirs have the same efficiency.

CARNOT CYCLE:-

Assumptions:-

1) all the processes are reversible process. This condition can be attain considering the

friction b/w the cylinder and pision is negligible

2) the ways ofcylinder and the pision are perfect insulators

3) the cylinder head is a perfect conductor and it becomes perfect insulator onattatchin an

insulating cop

4) the mass inside the cylinder remainssome through out the cycle.

5) The specific heats wil remains same through out the cycle

Consider some amount of gas inside a cylinderwhich is under compression

Process a-b:-

Attach one hot body to the cylinder.

Heat is being transpired from the not

Body to the gas and it is supposed to

Expand to a volume of vb assuming

At constant temperature.



As it is isothermal expansion process a trend is represented by which a horizontal line on

T-S diagram. Let Vb/Va=re(expansion ratio)

Work done during the process

Wa-b=PAVAlog(VB/VA)

=MRT1 log re T1→higher temp

Heat supply QS=∆u+w

=mcv∆T+w

QS=W

Process b-c:-

Detach the hot body from the cylinder and attaching

One insulating cop even though there is no heat addition

The gas is supposed to expand for same further extend.

i) as the cylinder and pision are friction less

ii) with the internal existments.

It is called reversible adlabatic expansion process.

Heat transfer QS=0

Wb-c=PbVb-PcVc/γ-1

As it is isentropic processes trend is represented by a vertical in T-S diagram. After expansion

temp of gas is fall down to T2.

Process c-d:-

Remove the insulating cop

And attatch one cold body. Heat

Is being rejected by the gas considering at constant temperature

Let the ratio VC/Vd=rC(compression ratio)

Work done during the process

Wc-d=PCVC log (VC/VD)

W=P1V1 log rC

= MRT2 log rc



Heat rejected QR=Qc-d=∆u+w

=MCv∆T+W

=W

QR=MRT2 Log rc

Process d-a:-

Remove the cold body and attatch one insulating cop. As ther is no friction the pision will

not stop suddenly and it will move towards left for further extend hence we can treat this

process as isentropic compression process.

Work done

Wd-a=PdVd-PaVa/γ-1

Q=0

Heat transfer

Work done during the cycle

W=QS-QR

=MRT1log rc-MRT2log rc

=MR log r(T1-T2)

(…r=rc=rc)

Efficiency ŋ cannot = W/QS

=MR log r (T1-T2)/MR log r T1

Ŋ cannot = 1-T2/T1

16Thermodynamics Unit III Notes 16


Thermodynamic temparature scale

• A scale that is independent of the properties of any substance. • Lord Kelvin was aware of the Carnot principles and suggested in1848, that a

thermodynamic temperature scale could be based on the theoretical consideration that, during the operation of a reversible heat engine, the amounts of energy exchanged are related to the temperature of the reservoirs, but not to the properties of any substance.

Notation for the thermodynamic temperature scale to be developed : (T)

• To construct a temperature scale, it is convenient to know a quantity Y which varies linearly with temperature (q ), whether Y is

– a property of a particular substance (empirical scale t) – not a property of a particular substance (thermodynamic scale T)

• Value of Y at the triple point of water (qtp) is Ytp • Value of Y at the coldest temperature q0 is Y0

q can be calculated by: Example: Celsius scale q tp=0.01oC, q(0)= -273.150C If it is known with certaintly, that the quantity Y is such that: as q� q0 Y0 �0 Two fixed point scale(<1954)� Single fixed point scale(>1954)

00 0

0tp

tp

Y Y

Y Yθ θ θθ−+

− = −



CLAUSIUS INEQUALITY When a reversible engine uses more than two reservoirs the third or higher numbered reservoirs

will not be equal in temperature to the original two. Consideration of expression for efficiency of

the engine indicates that for maximum efficiency, all the heat transfer should take place at

maximum or minimum reservoir temperatures. Any intermediate reservoir used will, therefore,

lower the efficiency of the heat engine. Practical engine cycles often involve continuous changes

of temperature during heat transfer. A relationship among processes in which these sort of

changes occur is necessary. The ideal approach to a cycle in which temperature continually

changes is to consider the system to be in communication with a large number of reservoirs in

procession. Each reservoir is considered to have a temperature differing by a small amount from

the previous one. In such a model it is possible to imagine that each reservoir is replaced by a

reversible heat engine in communication with standard reservoirs at same temperature T0. Fig.

shows one example to this substitution.

For an irreversible process

2 1 2 1

2 1 2 1

' ' ' '0 or 0

Q Q đQ đQ

T T T T+ < + <

1 1 1 1 1

2 2 2 2 2

' '' 1 1

' '

Q Q Q Q T

Q Q Q Q Tη η= + < = + ⇒ < = −



đQdS

T⇒ ≥



Entropy Change (∆∆∆∆S)

Entropy is that property of a substance that determines the amount of randomness and

disorder of a substance. If during a process, an amount of heat is taken and is by divided by the

absolute temperature at which it is taken, the result is called the ENTROPY CHANGE.

dS = dQ/T → 31

and by integration

∆∆∆∆S = ∫dQ/T → 32

and from eq. 39

dQ = TdS → 33

Increase in Entropy Principle

• Consider a cycle consisting of an irreversible process followed by a reversible one:

The inequality can be turned into an equality by considering the “extra” contribution to the

entropy change as entropy generated by the irreversibilities of the process:

0

0 or

)inequality (Clausius 0

2

1irrev

12

21

2

1irrev

1

2revint

2

1irrev

∫

∫

∫∫

∫

>−∴

<−+

<

+

<

T

QSS

SST

Q

T

Q

T

QT

Q

δ

δ

δδ

δ

process impossible an is 0

process reversiblea for 0

process leirreversib anfor 0

where

gen

gen

gen

gen

2

112

<

=

>

+=− ∫

S

S

S

ST

QSS

δ



• The increase in entropy principle states that an isolated system (or an adiabatic closed

system) will always experience an increase in entropy since there can be no heat transfer,

i.e.,

• However, this principle does not preclude an entropy decrease, which may occur for a

system that loses heat (Q < 0)

( ) 0 genisolated12 ≥=− SSS



Availability or exergy

• is total energy useful?

• maximum work potential = availability or exergy

• work is generally a function of the:

Dead state and maximum work

• dead state = equilibrium with the surroundings • thermal • mechanical (potential and kinetic) • chemical, • magnetic......

“a system must go to the dead state at the end of the process in order to maximise the work

output”

Reversible work and availability

reversible work is the maximum amount of useful work that can be obtained as a system undergoes a process between the specified initial and final states when

• final state=dead state then

• availability=reversible work examples: furnace and windmill

Irreversibility Irreversibility = reversible work – useful work

I = W rev - W us

• for a work producing device irreversibility =lost opportunity to do work



• for a work consuming device irreversibility =wasted work



Thermodynamic potentials

Helmhotz and gibbs function:-

Q=∆U+W

W=Q-∆U

=T∆S-∆U

=T(S2-S1)-(U2-U1)

=(U1-TS1)=(U2-TS2)

W=A1-A2

A=U-TS is helmhotz function similary

Gibbs function G=W-TS

Maxwell relations:-

Maxwell belongs to u.k (from 1831-1872)

A seientist in physics and astrology has formulated different relations. Which as termed as max

well relations.

We have dq=du+dw

Du=dq-dw

=dq-pdv (…dw=pdv)

Du=tds-pdv →(1)

H=u+pd

Dh=du+d(pv)

=du+pdv+vdp

=Tds-pdv+pdv+vdp (ref(1))

Dh=Tds+vdp→(2)

G=HTS→GIBBS FUNCTION

G=h-Ts

Dg=dh-d(Ts)

Dg=dh-SdT-Tds



=Tds+Vdp-Tds-SdT (…ref (2))

Dg=Vdp-SdT→(3)

A=U-TS →HEIMHOTZ FUNCTION

A=U-TS

Da=DU-D(TS)

Da=DU-TDS-SDT

Da=(TDS-PDV)-TDS-SDT

Da=-SDT-PDV→(4)

(1),(2),(3) and (4) are called thermodynamic equations on right hand side of every equation the

properties are exact differentials and the equations are in the form Z=Mdx+Ndy to make to

left hand side is also an exact differential the equation has to saticify the Condition.

│∂m/∂y│x=│∂N/∂x│y

From (1)= ∂T/∂V│S=-∂p/∂S│U→(5)

Form (2)= ∂T/∂P│s =∂V/∂S│P→(6)

From (3)=- ∂S/∂P│T=∂v/∂T│P→(7)

From (4)= ∂s/∂p│T=∂p/∂T│V→(8)

The above 5,6,7 abd 8 equations are called Maxwell relations.

The Maxwell formulated the above relations to related the entropy with the other properties

like pressure and volume.

∂u/∂s│v=T

∂u/∂v│s=p ref (1)

∂h/∂p│s=u

∂h/∂s│p<T ref (2)

∂g/∂T│p=-S

∂g/∂p│T=v ref (3)

∂a/∂T│V=-S

∂a/∂v│T=-P ref (4)



T=∂u/∂s│V=∂h/∂s│p→(9)

-P=∂u/∂v│s=∂u/∂v│T→(10)

V=∂h/∂p│S=∂g/∂p│T→(11)

-S=∂g/∂T│P=∂u/∂T│V→(12)

The equation (9),(10),(11),(12), are called secondaru Maxwell relations.



The third law of thermodynamics

The third law of thermodynamics is sometimes stated as follows:

The entropy of a perfect crystal at absolute zero is exactly equal to zero.

At zero kelvin the system must be in a state with the minimum possible energy, and this

statement of the third law holds true if the perfect crystal has only one minimum energy state.

Entropy is related to the number of possible microstates, and with only one microstate available

at zero kelvin, the entropy is exactly zero.

A more general form of the third law applies to systems such as glasses that may have more than

one minimum energy state:

The entropy of a system approaches a constant value as the temperature approaches zero.

The constant value (not necessarily zero) is called the residual entropy of the system.

Physically, the law implies that it is impossible for any procedure to bring a system to the

absolute zero of temperature in a finite number of steps.

The third law was developed by the chemist Walther Nernst during the years 1906-1912, and is

therefore often referred to as Nernst's theorem or Nernst's postulate. The third law of

thermodynamics states that the entropy of a system at absolute zero is a well-defined constant.

This is because a system at zero temperature exists in its ground state, so that its entropy is

determined only by the degeneracy of the ground state.

In 1912 Nernst stated the law thus:

"It is impossible for any procedure to lead to the isotherm T = 0 in a finite number of

steps."

An alternative version of the third law of thermodynamics as stated by Gilbert N. Lewis and

Merle Randall in 1923:

If the entropy of each element in some (perfect) crystalline state be taken as zero at the

absolute zero of temperature, every substance has a finite positive entropy; but at the

absolute zero of temperature the entropy may become zero, and does so become in the

case of perfect crystalline substances.



This version states not only ∆S will reach zero at 0 K, but S itself will also reach zero as long as

the crystal has a ground state with only one configuration. Some crystals form defects which

causes a residual entropy. This residual entropy disappears when the kinetic barriers to

transitioning to one ground state are overcome.

With the development of statistical mechanics, the third law of thermodynamics (like the other

laws) changed from a fundamental law (justified by experiments) to a derived law (derived from

even more basic laws). The basic law from which it is primarily derived is the statistical-

mechanics definition of entropy for a large system:

where S is entropy, kB is the Boltzmann constant, and is the number of microstates consistent

with the macroscopic configuration.

Thermodynamics Unit IV Notes 21


Pure Substances It is a substance which is homogeneous in chemical composition and homogeneous in chemical aggregation

p-V-T- surfaces

• For a fixed number of molecules/moles The Ideal Gas Law forms the surface of a three-dimensional plot where the axis are Pressure, Volume, and Temperature.

• Lines of constant pressure, constant volume, and constant temperature form a coordinate system labeling the location of an ideal gas.

• Robert Boyle showed that the pressure of a low-density gas is inversely proportional to the volume of a gas when the temperature is held constant, P α 1/V for constant temperature. Blowup of PV diagram for isothermals.

• Jacques Charles and Gay-Lussac showed that the pressure of a low-density gas is proportional to the temperature of the gas when the volume is held constant, P α T for constant volume.



• The ideal gas law is only valid for low-density gas. Fortunately, most ordinary gases behave like an ideal gas because the sizes of the molecules are small compared to their separation. None the less, there is still a PVT surface for high-density gases only it is not the ideal equation of state PV = nRT. One such equation is that of the van der Waals equation.

T-S and h-s diagrams,



Mollier Charts Richard Mollier (1863-1935) spent most of his working life at the Technische Hochschule in

Dresden studying the properties of thermodynamic media and their effective representation in the

form of charts and diagrams. His major contribution was in popularizing the use of enthalpy. In

1904, Mollier devised the first enthalpy-entropy chart still most closely associated with his name.

However, he published a number of other enthalpy-based charts and in recognition of his work,

the US Bureau of Standards recommended in 1923 that all such charts should be known as

Mollier diagrams.

Mollier's H-S diagram (Enthalpy v Entropy) was a logical extension of the T-S diagram

(Temperature v Entropy) first proposed by Gibbs, retaining the advantages of T-S diagrams but

introducing several new advantages. A typical H-S Mollier diagram for a thermodynamic fluid

such as steam is shown in Figure


GMRIT , Rajam Rajam Rajam Rajam Chiranhiranhiranhiranjeeva Rao jeeva Rao jeeva Rao jeeva Rao Seelaeelaeelaeela

Transformations – Triple point at critical state properties during change of phase,

Phase transformation:-

Consider a kg water to which heat is being added to convert the water into steam at the given

pressure.

With the addition of heat the volume of the water will increase and also the temperature.

Saturation temperature:-

It is the limit of temperature at which the water is supposed to convert to steam (or) vapour at

the given pressure. i.e., the water will turn into steam at a constant temperature the given

temperature.

Satured liquid:-

It is the liquid which is ready for phase transformation.

Saturated vapour:-

It is the condition of the vapour which is ready to convert into the liquid on condensation.



The point C indicates saturated vapour (or) dry steam. In between b and c the steam (or)

vapour is said to be wet steam.

Liquid enthalpy (or) sensible heat and enthalpy of water:-

It is the amount of heat added to the liquid to raise the temperature to its saturation value at a

given temperature.

Lated heat of vapourization (or) hidden heat (L or Hfg):-

It is the amount of heat added to convert the saturated liquid into the saturated vapour at the

given pressure (or) the amount of heat rejected to convert the saturated vapour into saturated

liquid at the given pressure (or) it is the heat added (or) rejected during the phase

transformation.

Latent heat of water L=Hfg=2256KJ1kg

Latent heat of Ice=-3331KJ1kg

Enthalpy of dry steam:-

Enthalpy of dry steam

H=hf+hfg

It is an amount of heatadded to convert water into dry steam at the given pressure.

Critical point It is the condition or state at which the volume of saturated liquied is equals to the volume of

saturated vapour.

Critical pressure for water=22.09MPQ

=225bar

Critical temperature for water=373.130c

Critical volume for water=0.00301m3



Dryness Fraction

It is the ratio of mass of dry steam in the given steam It is represerved by x or q X or q =mv/mv+mp mv→mass of dry steam mf→mass of liquid particles enthalpy of wet steam:- (h 1 = hf) h1 = hf+xhfg where x is dryness fractions



Clausius – Clapeyron Equation Property tables. consider the water at a pressure p1

Where the saturation temperature is T1

Latent heat of vapourisation is hfg1

Consider the water at another pressure P2 (saturation temperature is T2), so closer

To the previous pressure in such a way that enthalpy of Evapouration.

Hfg1=hfg2 i.e., hfg1=hfg2=hfg3

Maxwell relations ∂S/∂V/T=∂p/∂T│V

∂S/∂V=∂p/∂T

Sg-Sf/Vg-Vf=∂p/∂T

Where Sf is liquid entropy Sg is saturated vapours entropy Vf,Vg are specific volume of

saturated liquid and saturated vapour resp.

Sfg/Vg=∂p/∂T

Sfg=∂p/∂T xVg

Hfg/T=∂p/∂T.Vg

Hfg=∂p/∂T.T.vg (…Sfg=hfg/T)

(…pvg=Rt,Vg=RT/P)

=∂p/∂T.T.Rt/p

Hfg.1/Tv.∂T=R.∂p/p

hfg∫1/TV.∂t=R ∫21 ∂P/P

hgf(-1/T)2=R log (p2/p1)

hfg =-Rlog (P2/P1)/(1/T2-1/T1)

.



H-S Diagram (or) mollier chart:-

From the T-S diagram always it may not be possible to determine the properties.

From the relation Tds=dh-Vdp

Dh/ds∫p=T

Based on the above relation another chart considering enthalpy on Y-axis and entropy on X-axis

from which we can read the properties easily.



In the above chart the thick line represents the saturation condition. In the wet region the

constant pressue lines are the straign lines which are supposed to concide with constant

temperature lines. In the superheated region these pressure lines are diverged and takes the

from of an up ward curve.

The constant temperature lines which are deriated from pressure lines becomes horizonital in

super heated region.

The line which are approimartely parallel to saturation curve in wet region indicates the

quality of steam and it is represented with X or Q.

The horizontalal lines are constant enthalpy lines and vertical lines are constant entropy lines.

The constnt specific volume line are the curvd lines which are diverged up wards. In

General these lines on mollier diagram are chain lines



Thermodynamic processes

1. Constant pressure or Isobaric Process ( P = C): An Isobaric Process is an

internally reversible constant pressure process.

Closed System:(Nonflow)

Q = ∆U + W → 1 any substance

W = ∫PdV → 2 any substance

∆U = m(U2 - U1) → 3 any substance

W = P(V2 - V1) → 4 any substance

Q = ∆h = m(h2-h1) → 5 any substance

For Ideal Gas:

PV = mRT

W =mR(T2-T1) → 5

∆U = mCv(T2-T1) → 6

Q = ∆h = mCP (T2-T1) → 7



2. Constant pressure or Isochoric Process (V = C): An Isometric process is

internally reversible constant volume process.

Closed System: (Nonflow)


W = ∫PdV at V = C; dV = 0

W = 0

Q = ∆U = m(U2 - U1) → 2 any substance

∆h = m(h2-h1) → 3 any substance

For Ideal Gas:

Q = ∆U = mCv(T2-T1) → 4

∆h = mCP(T2-T1) → 5

For Ideal Gas:

-∫VdP = -V(P2-P1) = mR(T1-T2)

Q = ∆U = mCv(T2-T1) → 12

∆h = mCP(T2-T1) → 13

If ∆KE = 0 and ∆PE = 0

Q = ∆h + W → 14 any substance

W = - ∫VdP → 15

W = -∫VdP = -V(P2-P1) → 16 any substance



W = mR(T1-T2) → 16 ideal gas

∆h = mCP(T2-T1) → 17 ideal gas

3. Isothermal Process(T = C): An Isothermal process is reversible constant temperature process.

Closed System (Nonflow)




For Ideal Gas: dU = mCv dT; at T = C ; dT = 0

Q = W → 4

W = ∫PdV ; at PV = C ;

P1V1 = P2V2 = C; P = C/V

Substituting P = C/V to W = ∫PdV

W = P1V1 ln(V2/V1) → 5

Where (V2/V1) = P1/P2

W = P1V1 ln(P1/P2) → 6

P1V1 = mRT1



Isentropic Process (S = C): An Isentropic Process is an internally“Reversible Adiabatic”

process in which the entropy remains constant

where S = C and PVk = C for an ideal or perfect gas.





Q = 0 → 4

W = - ∆U = ∆U = -m(U2 - U1) → 5

1

2

1

1

1

2

1

2

2

22

1

−−

=

=

==

==

kk

k

k22

k11

11

k

V

V

P

P

T

T

VPVP and T

VP

T

VP

C PV and CT

PV Using



For Ideal Gas

∆U = mCV(T2-T1) → 6

From PVk = C, P =C/Vk, and substituting P =C/Vk

to W = ∫PdV, then by integration,

Q = 0

Polytropic Process ( PVn = C): A Polytropic Process is an internally reversible

process of an Ideal or Perfect Gas in which PVn = C, where n stands for any constant.

Closed System: (Nonflow

( )

−

−=∫

−

−==∫

−=∫=

−

−

11

11

1

1

1

211

1

1

21

kk

VP

kk

12

1122

P

P

kPdV

P

P

k

mRT

k-1

T-TmRPdV

k

VP-VPPdV W

1

2

1

1

1

2

1

2

2

22

1

−−

=

=

==

==

nn

n

n22

n11

11

n

V

V

P

P

T

T

VPVP and T

VP

T

VP

C PV and CT

PV Using



Q = ∆U + W → 1

W = ∫PdV → 2

∆U = m(U2 - U1) → 3

Q = mCn(T2-T1) → 4

∆U = m(U2 - U1) → 5

From PVn = C, P =C/Vn, and substituting

P =C/Vn to W = ∫PdV, then by integration,

( )

−

−=∫=

−

−==∫=

−=∫=

−

−

11

11

1

1

1

211

1

1

21

nn

VP

nn

12

1122

P

P

nPdVW

P

P

n

mRT

n-1

T-TmRPdVW

n

VP-VPPdV W


GMRIT, R R R Rajamajamajamajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela

Steam Calorimetry.

The presence of moisture in steam causes a loss, not only in the practical waste of the heat

utilized to raise this moisture from the temperature of the feed water to the temperature of the

steam, but also through the increased initial condensation in an engine cylinder and through

friction and other actions in a steam turbine. The presence of such moisture also interferes with

proper cylinder lubrication, causes a knocking in the engine and a water hammer in the steam

pipes. In steam turbines it will cause erosion of the blades.

The percentage by weight of steam in a mixture of steam and water is called the quality of the

steam.

The apparatus used to determine the moisture content of steam is called a calorimeter though

since it may not measure the heat in the steam, the name is not descriptive of the function of the

apparatus. The first form used was the “barrel calorimeter”, but the liability of error was so great

that its use was abandoned. Modern calorimeters are in general of either the throttling or

separator type.

THROTTLING CALORIMETER—Fig. below shows a typical form of throttling calorimeter. Steam

is drawn from a vertical main through the sampling nipple, passes around the first thermometer

cup, then through a one-eighth inch orifice in a disk between two flanges, and lastly around the

second thermometer cup and to the atmosphere. Thermometers are inserted in the wells, which

should be filled with mercury or heavy cylinder oil.

he instrument and all pipes and fittings

leading to it should be thoroughly insulated

to diminish radiation losses. Care must be

taken to prevent the orifice from becoming

choked with dirt and to see that no leaks

occur. The exhaust pipe should be short to

prevent back pressure below the disk.

When steam passes through an orifice from

a higher to a lower pressure, as is the case

with the throttling calorimeter, no external

work has to be done in overcoming a

resistance. Hence, if there is no loss from


GMRIT, R R R Rajamajamajamajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela

radiation, the quantity of heat in the steam will be exactly the same after passing the orifice as

before passing. If the higher steam pressure is 160 pounds gauge and the lower pressure that of

the atmosphere, the total heat in a pound of dry steam at the former pressure is 1195.9 B. t. u.

and at the latter pressure 1150.4 B. t. u., a difference of 45.4 B. t. u. As this heat will still exist in

the steam at the lower pressure, since there is no external work done, its effect must be to

superheat the steam. Assuming the specific heat of superheated steam to be 0.47, each pound

passing through will be superheated 45.4⁄0.47 = 96.6 degrees. If, however, the steam had contained

one per cent of moisture, it would have contained less heat units per pound than if it were dry.

Since the latent heat of steam at 160 , pounds gauge pressure is 852.8 B. t. u., it follows that the

one per cent of moisture would have required 8.5 B. t. u. to evaporate it, leaving only 45.4 - 8.5 =

36.9 B. t. u. available for superheating; hence, the superheat would be 36.9⁄0.47 = 78.5 degrees, as

against 96.6 degrees for dry steam. In a similar manner, the degree of superheat for other

percentages of moisture may be determined. The action of the throttling calorimeter is based

upon the foregoing facts, as shown below.

Let H = total heat of one pound of steam at boiler pressure,

L = latent heat of steam at boiler pressure,

h = total heat of steam at reduced pressure after passing orifice,

t1 = temperature of saturated steam at the reduced pressure, t2 = temperature of steam after expanding through the orifice in the disc,

0.47 = the specific heat of saturated steam at atmospheric pressure,

x = proportion by weight of moisture in steam.

H - h = xL + 0.47(t2 - t1)

Thermodynamics Unit V Notes 27


PERFECT GAS (OR) IDEL GAS:-

It is the gas. Which obeys all the lows of the gas at all pressure and temperatures.

In real practice no gas is a perfect gas some of the gases can be treated as perfect gs with

in the specified range of temperatures.

Laws of gas:-

1. boyle’s law

2. charle’s law

3. gaylussaes law

4. avagadro’s law

boyle’s law:-

at the given constant temp the volume of gas is inversely properational to the pressure.

Vœ1/P

Pv=constant (T-constant temp)

Charle’s Law

A. At Constant Pressure (P = C) If the pressure of a certain quantity of

gas is held constant, the volume V is directly proportional to the temperature T during a qua-

sistatic change of state

B. At Constant Volume (V = C) If the volume of a certain quantity of gas is held constant, the pressure P varies directly as the

absolute temperature T.

2

2

1

1

T

V

T

V

CT

VT;CV; T α V

=

==

2

2

1

1

T

P

T

P

CT

P; TCPT α P

=

== ;



Note:- the temp at which the volume becomes is zero is known as absolute zero temp.

gaylussaes law:-

At constant volume the pressure of gas is directly proportional to temp

PœT (volume is constant)

avagadro’s law:-

one mole of all gases will occupy same volume at the given pressure and temperature.

Avagadro’s hypothesis 1kg mole of gas will occupy 22.4136 literes at NTR and consists

of 6*1023 no.of molecules. The no 6*1023 is known as avagadro’s number.

NTR→normal temperature pressure

00c,760MM of hg



Equation of state (or) law of perfect gas:-

P œ 1/V →(1) (…boyle’s law)

P œ T →(2) (…gaylussa’s law)

From (1) & (2)

P œ T/V

PV = RuT (… Ru is the universal gas contant)

PV=RuT→(1)

Ru = 101325 X105 X224/273

Ru = 8313.84

Universal gas constant Ru=8314J1kg mole k for ‘n’ mole ≡> (1)

PV=nRuT

PV=MRT

R→ characteristic gas constant

R=Ru/M

M→molecular weight (mole)

MO2→32

MN2→28.06 MNH2→17

MH2→4 MH2→ 2.016

MCO2→44 Air → 29

Rair=Ru/m=8314/29=286.8=287J1kg md

1. a vessel of 10m3 volume can be filled with oxyzen and nitrogen and Co2 at a pressure of

15 bar and temperature of 400C These gases are filled individually. Determine the mass

of each gas.

Salutation:

Given data:

Consider N2

Volume V=10m3



Pressure P=15 bar

Temp T=400c=313K

Pv=MRT R=Ru/m

15X105X10=m(296.75)X313 =8313.4/28016

Mass of N2=161.48KJ =296.73J1kgk

For Co2:-

R=Ru/M=8313.4/44

=188.94J1kgk

Pv=MRT

15X105X10=M(188.94)(313)

M=253.6Kg

2. an aero stat ballon is filledwith W2 at a pressure of 100 kpa and 300K temp and it

occuples the volume of 1000m3 determine the pay load to lift the ballon.

Soluation:-

Give data:-

Pressure (P)=100KPa

Volume (V)=1000m3

Temperature (T)=300K

R=Ru/M=8313.4/2.016=4124J1kgk

Pv=MRT

(100X103)X1000=M(4124)(300)

M4=80.83Kg

Air pressure P=1.01325 bar

Volume of air displaced

V=volume of ballon =1000m3

T=300K



R=Ru/M

=8313.4/29

=286.6

287J1kg

Mass of air Ma=PV/RT

=1.01325X105X1000/287X300

=1176.82Kg

Down ward force applied by air Fa = mag

= 1176.82 X 9.81

Fa =11544.69N



Specific heats

Specific Heat or Heat Capacity is the amount of heat required to raise the temperature of

a 1 kg mass 1°C or 1°K

Specific heat at constant pressure

From: dh = dU + PdV + VdP

but dU + VdP = dQ ; therefore

dh = dQ + VdP → 1

but at P = C ; dP = O; therefore

dh = dQ → 2

and by integration

Q = ∆∆∆∆h → 3

considering m,

∆∆∆∆h = m(h2 - h1) → 4

Q = ∆∆∆∆h = m (h2 - h1) → 5

From the definition of specific heat, C = dQ/T

Cp = dQ /dt → 6

Cp = dh/dT, then

dQ = CpdT → 7

and by considering m,

dQ = mCpdT → 8

then by integration

Q = m Cp ∆∆∆∆T → 9

but ∆T = (T2 - T1)

Q = m Cp (T2 - T1) → 10

Specific heat at constant volume

At V = C, dV = O, and from dQ = dU + PdV

dV = 0, therefore

dQ = dU → 11



then by integration

Q = ∆∆∆∆U → 12

then the specific heat at constant volume Cv is;

Cv = dQ/dT = dU/dT → 13

dQ = CvdT → 14

and by considering m,

dQ = mCvdT → 15

and by integration

Q = m∆∆∆∆U → 16

Q = mCv∆∆∆∆T → 17

Q = m(U2 - U1) → 18

Q = m Cv(T2 - T1)→ 19

From:

h = U + Pυ and Pυ = RT

h = U + RT → 20

and by differentiation,

dh = dU + Rdt → 21

but dh =CpdT and dU = CvdT, therefore

CpdT = CvdT + RdT → 22

and by dividing both sides of the equation by dT,

Cp = Cv + R → 23



Relation between Specific heats

From:

h = U + Pυ and Pυ = RT

h = U + RT → 20

and by differentiation,

dh = dU + Rdt → 21

but dh =CpdT and dU = CvdT, therefore

CpdT = CvdT + RdT → 22

and by dividing both sides of the equation by dT,

Cp = Cv + R → 23

Ratio Of Specific Heats

k = Cp/Cv → 24

k = dh/du → 25

k = ∆∆∆∆h/∆∆∆∆U → 26

From eq. 32,

Cp = kCv → 27

substituting eq. 27 to eq. 24

Cv = R/k-1 → 28

From eq. 24,

Cv = Cp/k → 29

substituting eq. 29 to eq. 24

Cp = Rk/k-1 → 30



Non – flow processes

1. Isobaric Process ( P = C): An Isobaric Process is an internally

reversible constant pressure process.

Closed System:(Nonflow)




W = P(V2 - V1) → 4 any substance

Q = ∆h = m(h2-h1) → 5 any substance

For Ideal Gas:

PV = mRT

W =mR(T2-T1) → 5

∆U = mCv(T2-T1) → 6

Q = ∆h = mCP (T2-T1) → 7



2. Isometric Process (V = C): An Isometric process is internally reversible constant

volume process.

Closed System: (Nonflow)


W = ∫PdV at V = C; dV = 0

W = 0



For Ideal Gas:

Q = ∆U = mCv(T2-T1) → 4

∆h = mCP(T2-T1) → 5

For Ideal Gas:

-∫VdP = -V(P2-P1) = mR(T1-T2)

Q = ∆U = mCv(T2-T1) → 12

∆h = mCP(T2-T1) → 13



W = - ∫VdP → 15

W = -∫VdP = -V(P2-P1) → 16 any substance



W = mR(T1-T2) → 16 ideal gas

∆h = mCP(T2-T1) → 17 ideal gas




3. Isothermal Process(T = C): An Isothermal process is reversible constant temperature process.





For Ideal Gas: dU = mCv dT; at T = C ; dT = 0

Q = W → 4

W = ∫PdV ; at PV = C ;

P1V1 = P2V2 = C; P = C/V

Substituting P = C/V to W = ∫PdV

W = P1V1 ln(V2/V1) → 5

Where (V2/V1) = P1/P2

W = P1V1 ln(P1/P2) → 6

P1V1 = mRT1

4. Isentropic Process (S = C): An Isentropic Process is an internally“Reversible

Adiabatic” process in which the entropy remains constant








Q = 0 → 4

W = - ∆U = ∆U = -m(U2 - U1) → 5

For Ideal Gas

∆U = mCV(T2-T1) → 6

From PVk = C, P =C/Vk, and substituting P =C/Vk

1

2

1

1

1

2

1

2

2

22

1

−−

=

=

==

==

kk

k

k22

k11

11

k

V

V

P

P

T

T

VPVP and T

VP

T

VP

C PV and CT

PV Using



to W = ∫PdV, then by integration,

Q = 0

( )

−

−=∫

−

−==∫

−=∫=

−

−

11

11

1

1

1

211

1

1

21

kk

VP

kk

12

1122

P

P

kPdV

P

P

k

mRT

k-1

T-TmRPdV

k

VP-VPPdV W




Polytropic Process ( PVn = C): A Polytropic Process is an internally reversible

process of an Ideal or Perfect Gas in which PVn = C, where n stands for any constant.

Closed System: (Nonflow

Q = ∆U + W → 1

W = ∫PdV → 2

∆U = m(U2 - U1) → 3

Q = mCn(T2-T1) → 4

∆U = m(U2 - U1) → 5

1

2

1

1

1

2

1

2

2

22

1

−−

=

=

==

==

nn

n

n22

n11

11

n

V

V

P

P

T

T

VPVP and T

VP

T

VP

C PV and CT

PV Using



From PVn = C, P =C/Vn, and substituting

P =C/Vn to W = ∫PdV, then by integration,

( )

−

−=∫=

−

−==∫=

−=∫=

−

−

11

11

1

1

1

211

1

1

21

nn

VP

nn

12

1122

P

P

nPdVW

P

P

n

mRT

n-1

T-TmRPdVW

n

VP-VPPdV W


GMRIT, Rajam Rajam Rajam Rajam Chhhhiranjeeva Rao iranjeeva Rao iranjeeva Rao iranjeeva Rao Seelaeelaeelaeela

Throttling process:-

From SFEE

H1+V21+gZ1+Q=H21+V2

2+gZ2+W

P.E=0

Q=0

W=0

H1=H2

U1+P1V1=U2+P2V2

CV∆T+P1V1=CV∆T+P2V2

Consider an insulated pipe which is providedwith a disc having a small opening if the fluid under

high pressure is allowed toflow through the pipe then the pressures will fall down tremendously

at the exit of the disc.


GMRIT, Rajam Rajam Rajam Rajam Chhhhiranjeeva Rao iranjeeva Rao iranjeeva Rao iranjeeva Rao Seelaeelaeelaeela

We have SFEE

H1+v21+gz1+Q=h2+v2

2+gz2+W

Insulated Q=0

W=0

P.E=0

During the expansion of the fluid throught the disc the velocity remains contant as the whole acts

like as a small pipe.

….∆T=0

∆K.E=0

… the SFEE →h1=h2

During the throttling process the enthalpy remains constnt the pressure andtemperature will fall

down tremendously.

Free expansion process:-

Consider an insulated vessel which is devided into two compartments by a well. Let one

compartement is filled with a gas at a certain pressure and other compartment is complete

vocume. If the wall is punctured the system attains a equaliblium condtion over a period of

line as there is n6 restiction for expansion the expansion is known as free expansion process.



Flow processes

1. Isobaric Process ( P = C): An Isobaric Process is an internally reversible constant

pressure process.

Open System: Q = ∆h + ∆KE + ∆PE + W → 10 any substance W = - ∫VdP - ∆KE - ∆PE → 11 any substance - ∫VdP = 0 Q = ∆h → 12 W = - ∆KE - ∆PE → 13 If ∆KE = 0 and ∆PE = 0 W = 0 → 14 Q = mCP(T2-T1) → 15 Ideal Gas

2. Isometric Process (V = C): An Isometric process is internally reversible constant

volume process.



Open System: Q = ∆h + ∆KE + ∆PE + W → 7 any substance

W = - ∫VdP - ∆KE - ∆PE → 8 any substance

-∫VdP = -V(P2-P1) → 9 any substance



3. Isothermal Process(T = C): An Isothermal process is reversible onstant temperature process.



Open System (Steady Flow) Q = ∆h + ∆KE + ∆PE + W → 10 any substance


-∫VdP = -V(P2-P1) → 12 any substance


For Ideal Gas:

-∫VdP = -P1V1ln(P2/P1) → 14

-∫VdP = P1V1ln(P1/P2) → 15

P1/P2 = V2/V1 → 16

dh = CPdT; at T = C; dT = 0

∆h = 0 → 16



W = - ∫VdP = P1V1ln(P1/P2) → 18

For Ideal Gas

∆h = 0 → 19

Q = W = - ∫VdP = P1V1ln(P1/P2) → 20

4. Isentropic Process (S = C): An Isentropic Process is an internally “Reversible

Adiabatic” process in which the entropy remains constant




Open System (Steady Flow) Q = ∆h + ∆KE + ∆PE + W → 10 any substance



Q = 0

W = -∆h - ∆KE - ∆PE → 13

From PVk = C ,V =[C/P]1/k, substituting V to

-∫VdP, then by integration,

( )

( )

−

−=∫−

−

−==∫−

−=∫−

∫=∫−

−

−

11

11

1

1

1

211

1

1

21

kk

kk

12

1122

P

P

k

VkPVdP

P

P

k

kmRT

k-1

T-TkmRVdP

k

VP-VPkVdP

PdV kVdP




0 = ∆h + W → 17 any substance

W = - ∫VdP = - ∆h → 18 any substance


Q = 0

( )

( )

( )12P

kk

kk

12

1122

T-TmChW

P

P

k

VkPW

P

P

k

kmRT

k-1

T-TkmRW

k

VP-VPkPdV kVdPW

−=∆−=

−

−=

−

−==

−=∫=∫−=

−

−

11

11

1

1

1

211

1

1

21



5. Polytropic Process ( PVn = C): A Polytropic Process is an internally reversible process of an Ideal or Perfect Gas in which PVn = C, where n stands for any constant.

Open System (Steady Flow) Q = ∆h + ∆KE + ∆PE + W → 11

W = - ∫VdP - ∆KE - ∆PE → 12

∆h = m(h2-h1) → 13

Q = mCn(T2-T1) → 14

dQ = mCn dT

W = Q - ∆h - ∆KE - ∆PE → 15

From PVn = C ,V =[C/P]1/n, substituting V to

-∫VdP, then by integration,

1

2

1

1

1

2

1

2

2

22

1

−−

=

=

==

==

nn

n

n22

n11

11

n

V

V

P

P

T

T

VPVP and T

VP

T

VP

C PV and CT

PV Using



( )n

VP-VPnVdP

PdV nVdP

1122

−=∫−

∫=∫−

1

( )

−

−=∫−

−

−==∫−

−

−

11

11

1

1

211

1

1

21

nn

nn

12

P

P

n

VnPVdP

P

P

n

nmRT

n-1

T-TnmRVdP



Deviation from perfect gas model:

Real gases do not obey ideal gas equation under all conditions. They nearly obey ideal gas

equation at higher temperatures and very low pressures. However they show deviations from

ideality at low temperatures and high pressures.

The deviations from ideal gas behaviour can be illustrated as follows:

The isotherms obtained by plotting pressure, P against volume, V for real gases do not coincide

with that of ideal gas, as shown below.

It is clear from above graphs that the volume of real gas is more than or less than expected in

certain cases. The deviation from ideal gas behaviour can also be expressed by compressibility

factor, Z.

Compressibility factor (Z):

The ratio of PV to nRT is known as compressibility factor.

(or)

The ratio of volume of real gas, Vreal to the ideal volume of that gas, Vperfect calculated by

ideal gas equation is known as compressibility factor.

But from ideal gas equation:

PVperfect = nRT

or



Therefore

* For ideal or perfect gases, the compressibility factor, Z = 1.

* But for real gases, Z ≠1.

Case-I : If Z>1

* Vreal > Videal

* The repulsion forces become more significant than the attractive forces.

* The gas cannot be compressed easily.

* Usually the Z > 1 for so called permanent gases like He, H2.

Case-II: If Z < 1

* Vreal < Videal

* The attractive forces are more significant than the repulsive forces.

* The gas can be liquefied easily.

* Usually the Z < 1 for gases like NH3, CO2, SO2.

The isotherms for one mole of different gases, plotted against the Z value and pressure, P at

0oC are shown below:



* For gases like He, H2 the Z value increases with increase in pressure (positive deviation).

It is because, the repulsive forces become more significant and the attractive forces become

less dominant. Hence these gases are difficult to be condensed.

* For gases like CH4, CO2, NH3 etc., the Z value decreases initially (negative deviation) but

increases at higher pressures.

It is because: at low pressures, the attraction forces are more dominant over the repulsion

forces, whereas at higher pressures the repulsion forces become significant as the molecules

approach closer to each other.

* But for all the gases, the Z value approaches one at very low pressures, indicating the ideal

behaviour.

Also consider the following graphs of Z vs P for a particular gas, N2 at different

temperatures.

In above graphs, the curves are approaching the horizontal line with increase in the

temperature i.e., the gases approach ideal behaviour at higher temperatures.



Vanderwaal’s equation:-

Equaltion of state: pv=Rt → ( 1 ) Or perfect gas equation

In real pratice no gas is the perfect gas.

If V is the volume of the container the effective volume is V-b as the molecules themselves

occupies some volume.

P(V-b)=RT→(2) clausius equation of state

attractive forces vanderwall observed that on negeecting these forces the pressure will be raised

to P+a/V2

Then the equation number (2) becomes

P+a/Vr(V-b)=RT→ vander wall equation

Where a and b are constants

Compressible charts:-

Pv=MRT←equation of state (perfect gas equation)

In real practice no gas is the perfect gas

Pv=ZRT ←for real gas where Z ics compressible factor



Z=PV/RT

Let p,v and T are the properties of the gas at given condition Pc,Vc and Tc are the critical

properties of given gas.

The new properties Pr.Vr and Tr are defined such a way that

P/Pc=Pr≡> P=PcPr

C/Vc=Vr≡>V=VcVr

T/Tc=Tr ≡> T=TcTr

(1) ≡> Z=PV/RT

=PcPrVcVr/RTcTr

=PcVc/RTC PrVr/Tr

Z=Zc PrVr/Tr

Compressible chart is a chart prepared b/w PrVrZ considering the veduced temperature. These

charts will fesilitates to read the compressibility factor corresponding to the given pressure

and temperatue with minimum calculations.



Variation of Specific Heats Variation of Specific Heats with Temperature

Flow Rate Conversion Table

Nm³/hr scfh scfm slpm sccm 1,000 34,898 582 15,528 15,527,500 500 17,449 291 7,764 7,763,750 400 13,959 233 6,211 6,211,000 300 10,470 174 4,658 4,658,250 200 6,980 116 3,106 3,105,500 150 5,235 87 2,329 2,329,125 100 3,490 58 1,553 1,552,750



90 3,141 52 1,397 1,397,475 80 2,792 47 1,242 1,242,200 70 2,443 41 1,087 1,086,925 60 2,094 35 932 931,650 50 1,745 29 776 776,375 45 1,570 26 699 698,738 40 1,396 23 621 621,100 35 1,221 20 543 543,463 30 1,047 17 466 465,825 25 872 15 388 388,188 20 698 12 311 310,550 15 523 9 233 232,913 10 349 5.8 155 155,275 9.5 332 5.5 148 147,511 9.0 314 5.2 140 139,748 8.5 297 4.9 132 131,984 8.0 279 4.7 124 124,220 7.5 262 4.4 116 116,456 7.0 244 4.1 109 108,693 6.5 227 3.8 101 100,929 6.0 209 3.5 93 93,165 5.5 192 3.2 85 85,401 5.0 174 2.9 78 77,638 4.5 157 2.6 70 69,874 4.0 140 2.3 62 62,110 3.5 122 2.0 54 54,346 3.0 105 1.7 47 46,583 2.5 87 1.5 39 38,819 2.0 70 1.2 31 31,055 1.5 52 0.9 23 23,291 1.0 34.9 0.6 15.5 15,528 0.9 31.4 0.52 14.0 13,975 0.8 27.9 0.47 12.4 12,422 0.7 24.4 0.41 10.9 10,869 0.6 20.9 0.35 9.3 9,317 0.5 17.4 0.29 7.8 7,764 0.4 14.0 0.23 6.2 6,211 0.3 10.5 0.17 4.7 4,658 0.2 7.0 0.12 3.1 3,106 0.1 3.5 0.06 1.6 1,553 0.05 1.7 0.03 0.8 776

Moisture Conversion Table



Dew point °C °F

PPM on Volume Basis at 760 mm of Hg Pressure

Relative Humidity at 70°F%

PPM on Weight Basis in Air

-90 -130 0.0921 0.00037 0.057 -88 -126 0.123 0.00054 0.082 -86 -123 0.184 0.00075 0.11 -84 -119 0.263 0.00107 0.16 -82 -116 0.382 0.00155 0.24 -80 -112 0.562 0.00214 0.33 -78 -108 0.737 0.00300 0.46 -76 -105 1.01 0.00410 0.63 -74 -101 1.38 0.00559 0.86 -72 -98 1.88 0.00762 1.17 -70 -94 2.55 0.0104 1.58 -68 -90 3.43 0.0140 2.13 -66 -87 4.59 0.0187 2.84 -64 -83 6.11 0.0248 3.79 -62 -80 8.08 0.0328 5.01 -60 -76 10.6 0.0430 6.59 -58 -72 13.9 0.0565 8.63 -56 -69 18.2 0.0735 11.3 -54 -65 23.4 0.0948 14.5 -52 -62 30.3 0.123 18.8 -50 -58 38.8 0.157 24.1 -48 -54 49.7 0.202 30.9 -46 -51 63.3 0.257 39.3 -44 -47 80 0.325 49.7 -42 -44 101 0.410 62.7 -40 -40 127 0.516 78.9 -38 -36 159 0.644 98.6 -36 -33 198 0.804 122.9 -34 -29 246 1.00 152 -32 -26 305 1.24 189 -30 -22 376 1.52 234 -28 -18 462 1.88 287 -26 -15 566 2.3 351 -24 -11 692 2.81 430 -22 -8 842 3.41 523 -20 -4 1020 4.13 633 -18 0 1240 5.00 770 -16 3 1490 6.03 925 -14 7 1790 7.25 1110 -12 10 2150 8.69 1335 -10 14 2570 10.4 1596



-8 18 3060 12.4 1900 -6 21 3640 14.7 2260 -4 25 4320 17.5 2680 -2 28 5100 20.7 3170 0 32 6020 24.4 3640

Thermodynamics Unit VI Notes 39


Gaseous Mixtures

• Mole Fraction – the amount of molecules of a component (J) of a mixture expressed as a

fraction of the total amount of molecules (n) in the sample:

x j = nj / n

where n = n A + n B = ...

• The sum of the mole fractions of the components of a mixture is unity, or:

1 = x A + x B = ...

• The partial pressure of any gas, not necessarily an ideal gas, is related to the mole

fraction by:

Pj = x j + P

Gravimetric and Volumetric Analysis of a Two Component System Gravimetric Analysis A two component mixture can be simultaneously analyzed for both compounds if both

compounds react completely to produce a common product. The data required are:

a. The total mass of the mixture to be analyzed

b. The two balanced chemical reactions for the conversion of each of the compounds to

the common product.

c. The total mass (or some other quantity that can be related to moles) of the common

product..

In this part of the experiment, a mixture of NaHCO3 and Na2CO3 is reacted with an excess

of hydrochloric acid, HCl, to form three common products: NaCl, CO2, and H2O.

NaHCO3 (s) + HCl (l) → NaCl (aq) + CO2 (g) + H2O (l)

Na2CO3 (s) + 2 HCl (aq) → 2 NaCl (aq) + CO2 (g) + H2O (l)

The CO2 escapes from the reaction mixture and the water and the excess HCl solution are

evaporated leaving solid NaCl. The stiochiometry of the two reactions allows you to relate the

masses of NaHCO3 and Na2CO3 in the original sample to the mass of the NaCl produced.



Grams of original mixture = Grams of NaHCO3 + Mass of Na2CO3

Total moles of NaCl = 1 (Moles of NaHCO3) + 2 (Moles of Na2CO3

Using the relationship of: MassFormula

GramsMoles

.=

and the stoichiometry of the reactions, the following relationships can be developed:

Moles NaCl = 1 (Moles NaHCO3) + 2 (Moles Na2CO3)

( )32

32

3

3

.

.2

.

.1

.

.

CONaFM

COgNaxmixtureg

NaHCOFM

NaHCOxg

NaClFM

NaClg −+=

The grams of each of the compounds and the percent composition of each in the mixture can be

calculated.

Volumetric Analysis: The stiochiometry of the two reactions also allows you to relate the masses of NaHCO3

and Na2CO3 in the original sample to the moles of HCl that react with the sample. While a direct

titration of the sample is possible. The sample is reacted with an excess of a standardized

solution of HCl and the excess HCl “back-titrated” with a standardized solution of NaOH. The

sample is warmed to increase the rate of reaction with the hydrochloric acid and to drive off the

carbon dioxide formed. The excess HCl left from the reaction with the NaHCO3 and Na2CO3 is

neutralized by NaOH in the following reaction:

HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)

The net ionic reaction is:

H+1 (aq) + OH-1 (aq) → H2O (l)

The solution must be heated to drive off the carbon dioxide because the CO2 reacts with water to

form carbonic acid which also reacts with the NaOH.

CO2 (g) + H2O (l) → H2CO3 (aq)



The moles of HCl used in the reaction with the unknown mixture equals the moles of HCl added

to the mixture minus the moles of excess HCl. The excess HCl is determined by a titration with

standardized NaOH. The moles of HCl used in the reaction with the unknown mixture are related

to the composition of the mixture by:

Moles HCl = 1 (Moles NaHCO3) + 2 (Moles Na2CO3)

( )

32

32

3

3

.

.2

.

.1.

CONaFM

COgNaxmixtureg

NaHCOFM

NaHCOxgHClMoles

−+=

The grams of each of the compounds and the percent composition of each in the mixture can

again be calculated.

in the mixture. Volumetric analysis gives the volumetric or molal fractions

i

i

i

i

i

ii

iii

Mx

Mx

y

My

Myx

Σ=

Σ=



DALTON’S LAW

The total pressure of a mixture P is equal to the sum of the partial pressure

that each gas would exert at mixture volume V and temperature T.

T1 = T2 = T3 = T

V1 = V2 = V3 = V For the components

For the mixture For the components

The total moles n

TR

PVn =

TR

VPn

TR

VPn

TR

VPn

33

22

11

=

=

=

321

321

321

321

PPPP

V

TR

TR

VP

TR

VP

TR

VP

TR

PV

TR

VP

TR

VP

TR

VP

TR

PV

nnnn

++=

++=

++=

++=



The mole fraction:

AMAGAT’S LAW Of ADDITIVE VOLUMES

The total volume of a mixture V is equal to the volume occupied by each component at the

mixture pressure P and temperature T.

P = P1 = P2 = P3

T = T1 = T2 = T3

For the components:

The total moles n:

P

Pyi

TR

PVTR

VP

y

n

ny

i

i

i

ii

=

=

=

TR

PVn ;

TR

PVn ;

TR

PVn 3

32

21

1 ===



The mole fraction

Equation of State

Mole Basis

Where:

R – Gas constant of a mixture

in KJ/kg-°K

- universal gas constant in

KJ/kgm- °K

321

321

321

321

VVVV

P

TR

TR

PV

TR

PV

TR

PV

TR

PV

TR

PV

TR

PV

TR

PV

TR

PV

nnnn

++=

++=

++=

++=

V

Vyi

TR

PVTR

PV

y

n

ny

i

i

i

ii

=

=

=

TRnPV =



Enthalpy of a Perfect Gas Mixture Therefore, the molar enthalpy Hig(T,P) of a pure ideal gas is independent of pressure:

Hig(T,pi) = Hig(T,P)

The total molar enthalpy of an ideal mixture of ideal gases is:

Hig = ∑ yi Hiig (T,pi)

or

Hig = ∑ yi Hiig (T,P)

Entropy of a Perfect Gas Mixture

We have seen that the entropy of a pure ideal gas is a function of T and P according to

Therefore, the entropy of a pure ideal gas at a partial pressure, pi relative to a total pressure P is

for a given temperature:

Or

The entropy of an ideal mixture of perfect gases is therefore,

The mixture specific internal energy

0dH

dPTV

TVdTCdH

T

ig

Pp

=

∂∂−+=

dPPR

dTT

CdS p −=

∫−=−ip

P

igii

igi dP

PR

)P,T(S)p,T(S

)P/pln(R)P,T(S)p,T(S iigii

igi −=

iiigii

iiigii

iigii

ig

ylnRy)P,T(Sy

)P/pln(Ry)P,T(Sy

)p,T(Sy)P,T(S

∑∑

∑∑

∑

−=

−=

=



The partial pressure of a component,

Pi, in the mixture (units: kPa) is:

∑∑ ===∑∑ =======

n

iii

n

i

iin

iii

n

i

ii hxm

hm

m

Hhux

m

um

m

Uu

1111

PyPP

P

RTPV

RTVP

n

ny ii

ii

ii ==== or



PSYCHOMETRY

It is the science which investigeates the properties measure and control the quantity of

moisture air and study the effect of moisture air on the comfort of metals and human.

Dry air:-

Air is the mixture of different gases. Dry air is nothing but the combination of O2,N2,H2,CO

and CO2 neglecting the traces of gases like argan.

Moist air:-

It is the mixture of dry air and water vapour.

Atmospheric pressure P=1.01325 bar= (pressure excerted by air+ pressre excerted by water

upour)

P=Pa+Pv (…P=Pb=Atm pressure (2) (barameter pressure)

Sensible heat:-

It is the amount of head added to raisethe temperature to its boiling point.

Latent heat:-

It is the amount of heat added (or) rejected during the phase transformation.



Properties of the air

The properties of the air are listed and explained below

Dry bulb temperature:- (DBT)

It is the temperature indicated by an ordinary thermometer

Wet bulb temperature:- (WBT)

It is a temperature indicated by the thermometer whose bulb is Covered with wet wick (or)

sock The moisture content in wet wick will try to Diffuse into dry air Always dry bulb

temperature is greater than That of wet bulb temperature.At the saturation condtion the dry

bu lemperature is equals to the wet bulb temperature.

Wet bulb depression:- (WBD)

It is the difference between dry bulb temperature and wet bulb temperature.

WBD=DBT-WBT

Dew point temperature

Let a represents the given condition Of the air and the temperature measured By the

thermometer is DBT On diffusing the moisture into th air, we can arrive saturation

condtion B. The corresponding temperature is WBT.

From A, is the temperature is supposed to decrease maintaining the pressure constant we can

arrive the saturation condtion C, the corresponding temperature is DPT.

Definition:-

It is the limit of temperature, where the moisture in the air is ready to condension.

Dew point depression:-

It is a difference b/w dry bulb temperature dew point temperature.

Humidity (or) specific numidity (or)humidity ratio:-

Mass of water vapour present in the given mass of moist air,

We have idel gas equn PV=MRT→(1)



On assuming that both dry air and moisture obeys the above law

Mass of water vapour MV=(PV/PT)V

Simiary mass of air MA=(PV/RT)a

From thedefinition specific humidity

W=MU/MA→(2)

For vapour Ru=8.3.14/18=0.461KJ1kgk

For air Ra=8.314/29=0.287KJ1kgk

On substitution W=(PV/RT)V/(PV/RT)A

W=(PV/0.46*T)v / (PV/.287*T) a

(..At the given volume and temperature)

W=0.622 Pv/Pa We have Patm=1.01325 bar,

Pa=P+Pv

Pa=P-Pv)

W=0.622Pv/P-Pv

Specific humidity units are kg of water vapour 1kg of air

Specific humidity/Absolute humidity:-

It is the mass of water vapour present in the given volume of moist air

Units kg1m3 (or) kg 1 lit



Relative humidity:-

It is ratio of mass of water vapour in the given air to the mass of water vapour when it is

saturated.

RH=MV/MVS

=(PV/RT)Water vapour/(PV/RT) Saturated with vapour

RH=PV/PVS (expressed in terms of %)

Note:-

1. the specific humidity (W)=0.622(PVS/P-PVS)

Vapour pressure

Vapor pressure or equilibrium vapor pressure is the pressure exerted by

a vapor in thermodynamic equilibrium with its condensedphases (solid or liquid) at a given

temperature in a closed system. The equilibrium vapor pressure is an indication of a liquid's

evaporation rate. It relates to the tendency of particles to escape from the liquid (or a solid). A

substance with a high vapor pressure at normal temperatures is often referred to as volatile.

The vapor pressure of any substance increases non-linearly with temperature according to

the Clausius–Clapeyron relation. Theatmospheric pressure boiling point of a liquid (also known

as the normal boiling point) is the temperature at which the vapor pressure equals the ambient

atmospheric pressure. With any incremental increase in that temperature, the vapor pressure

becomes sufficient to overcome atmospheric pressure and lift the liquid to form vapor bubbles

inside the bulk of the substance. Bubble formation deeper in the liquid requires a higher pressure,

and therefore higher temperature, because the fluid pressure increases above the atmospheric

pressure as the depth increases.

The vapor pressure that a single component in a mixture contributes to the total pressure

in the system is called partial pressure. For example, air at sea level, and saturated with water

vapor at 20 °C, has partial pressures of about 23 mbar of water, 780 mbar of nitrogen, 210 mbar

of oxygen and 9 mbar of argon.

degree of saturation

(U)=W/WS

it is the ratio of mass of water vapour in the given air of given volume to the mass of water



The Adiabatic Saturation Process

• Air having a relative humidity less than 100 percent flows over water contained in a well-

insulated duct. Since the air has < 100 percent, some of the water will evaporate and the

temperature of the air-vapor mixture will decrease.

Psychrometric chart

• the left, and the wet bulb temperature is read following a constant wet-bulb line from the

state-point to the saturation line.

• Dew point temperature tDP : This temperature is read by following a horizontal line from the

state-point to the saturation line.

• Specific volume v: It is shown from the constant-volume lines slanting upward to the left.

• Humidity ratio w: it is indicated along the right-hand axis of the chart.

• Enthalpy h: It is read from where the constant enthalpy line crosses the diagonal

scale above the saturation curve. The constant enthalpy lines, being slanted lines,

are almost coincidental as the constant wet-bulb temperature lines.



A psychrometric chart graphically represents the thermodynamic properties of moist air

It is very useful in presenting the air conditioning processes

� The psychrometric chart is bounded by two perpendicular axes and a curved line:

� 1) The horizontal ordinate axis represents the dry bulb temperature line t , in℃ ;

� 2) The vertical ordinate axis represents the humidity ratio line w , in kgw/kgdry.air

� 3) The curved line shows the saturated air, it is corresponding to the relative

humidity Ф=100% .

� The psychrometric chart incorporates seven parameters and properties.

� They are dry bulb temperature t , relative humidity Ф , wet bulb temperature tWB, dew

point temperature tDP , specific volume v, humidity ratio w and enthalpy h.

� Dry-bulb temperature t is shown along the bottom axis of the psychrometric chart.

• The vertical lines extending upward from this axis are constant-temperature lines.

• Relative humidity lines Ф are shown on the chart as curved lines that move

upward to the left in 10% increments.

• The line representing saturated air ( Ф= 100% ) is the uppermost curved line on the chart.

• And the line of Ф = 0% is a horizontal ordinate axis itself.

• Wet-bulb temperature tWB : On the chart, the constant wet-bulb lines slope a little upward to

• Only two properties are needed to characterize the moist air because the point of intersection

of any two properties lines defines the state-point of air on a psychrometric chart.

• Once this point is located on the chart, the other air properties can be read directly.

the psychrometric chart will medicate the properties of moist of air dry bulb temperature is

represented on X-axis, on Y-axis specific humidity is represented in separated scale.



Horizontal line representation for DPT inclined line representation for WBT curves are r

elative humidity curves.

1) air is at a dry blub temp of 350c and maintaining relative humidity of 80% determine

i) specific humidity

ii) DPT

iii) density of mixtrure

iv) degree of saturatin

v) enthalpy

soluation:-

give data:-

DBT=350C

RH=80%

Detremine W=?

DPT=?

PMIX=?

U=?

H=?

We have specific humidity W=0.622(PV/P-PV)

Given RH=0.8=PV/PVS

Referring the steam table at 350 DBT

Corresponding pvs=0.05622 bar

0.8=PV/PVS

Pv=0.8*0.05622=0.044976 bar

W=0.622 PV/P-PV

W=0.028KJ1KGK

II)PV=0.0499 bar (from steam table)

Tsat=DPT=31.030C

III) PV=MRT

P=PaRT



Pa=P/RT=101325/287*(35+273)=1.146KG1M3

W=MV/MA=PV/PA

Pa=WPa=0.028*1.146=0.3KG1M3

PMIX=Pa+Pv=1.146+0.03=1.176kg1m3

IV) Degree of saturation U=W/WS→WS=0.622*PVS/P-PVS

=0.622*0.5622/(1.01325-0.05622)

=0.036Kg 1kg air

U=W/WS=.0.028/0.036=0.777

V) total enthalpy=enthalpy of air+enthalpy of vapour

=maha+mvhv

Specific entholply=h=ha+mv/ma,hv

=Ha+whv

=cpT+w((hfg+cpv(DBT-DPT))

Cpair=1.005KJ1kgk

Cpwater=1.68KJ1kgk

Latent heat of water=hfg=2558KJ1kg

... h=(1.005)(35+273)+(0.028)(25+1.88((35+273)-(31.03+2.13))

H=381.37KJ1kgair

TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444444444444

GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa

OOTTTTOO CCYYCCLLEE ::--

All the assumptions which are made during the cannot cycle are applied to the Otto cycle. Otto cycle consists of the following processes

� Process 1-2 (isentropic compression) � Process 2-3 (constant volume heat addition process) � Process 3-4 (isentropic expansion) � Process 4-1 (constant volume heat rejection process)

Process 1-2:- The air inside the cylinder gets compressed isentropically so the pressure, temperature are increased to from P1 , T1 TO P2, T 2, and The corresponding change in volume is V 1 to V2. The ratio v1 == rrcc iiss vv22 ccoommpprreessssiioonn rraattiioo.. Process 2-3:- Heat is being supplied to the air suddenly without any appreciable change in volume. So the pressure and temperature will attain their values of P3 and T3

Heart supplied Qs == CCVV ((TT33--TT22)) ffoorr kkgg ooff aaii rr TThhee rraattiioo pp33 == rrpp iiss tthhee pprreessssuurree rraattiioo .. pp22 Process 3-4:- TThhee aaii rr iiss ssuuppppoosseedd ttoo eexxppaanndd iisseennttrrooppiiccaall ll yy ttoo tthhee pprreessssuurree ooff PP44 aanndd ttoo tthhee iinnii ttiiaall vvoolluummee ((II ,,ee VV 44 == VV11 )).. TT VV44 == rree iiss tthhee eexxppaannssiioonn rraattiioo VV33 wwoorrkk ddoonnee ((WW)) == PP33 VV33 -- PP44 VV44

γγ--11



wwii tthh tthhee eexxppaannssiioonn pprroocceessss wwii ll ll tteemmppeerraattuurree ffaall ll ddoowwnn ttoo tthhee vvaalluuee ooff TT44.. ((iinn ggeenneerraall TT44 ggrreeaatteerr tthhaann tthhaatt ooff TT22 )) PPrroocceessss 44--11::-- TThhee aaii rr ww22ii ll ll rreejjeecctt tthhee hheeaatt ssuuddddeennll yy ,, wwii tthhoouutt aannyy aapppprreecciiaabbllee cchhaannggee ;;iinn vvoolluummee,, ssoo tthhee pprreessssuurree aanndd vvoolluummee aattttaaiinnss ii ttss iinnii ttiiaall ccoonnddii ttiioonn aanndd eexxeeccuutteess tthhee ccyyccllee.. HHeeaatt ssuuppppiieedd QQSS ==CCvv ((TT33 ––TT22)) HHeeaatt rreejjeecctteedd QQSS ==CCvv ((TT44 ––TT11)) EEffff iicciieennccyy nnoottttoo == ddeessii rreedd oollpp RReeqquuii rreedd iinnppuutt == WW QQSS

==QQSS –– QQRR

QQSS ==11 -- QQRR →→ ((11))

QQSS CCoonnssiiddeerriinngg 11--22 TT22//TT11 ==(( VV11//VV22))

γγ--11 == rr cc γγ--11 →→ ((22))

CCoonnssiiddeerriinngg 33--44 TT 33//TT44 ==((VV44 //VV33))

γγ--11 == rr cc γγ--11 →→ ((33))

FFrroomm eeqquuaattiioonnss ((22)) aanndd ((33)) ((22)) == ((33)) ::.. RRcc == RRcc ==RR

TT22 //TT11 == TT33 //TT44

→→ ((44)) ((11)) nnoottttoo == 11-- CCvv((TT44--TT11))



CCvv((TT44--TT11)) == 11-- TT11 ((TT44//TT11 -- 11)) TT22 ((TT33//TT22 -- 11)) ((::.. FFrroomm((44)))) == 11-- TT11 ((TT33//TT22 -- 11)) TT22 ((TT33//TT22 -- 11)) == TT11 -->> tteemmpp bbeeffoorree ccoommpprreessssiioonn == TT22 -->> tteemmpp ooff tteerr ccoommpprreessssiioonn

oottttoo ==== 11-- TT11לל TT22

oottttoo ==== 11-- 11לל

(( TT11//TT22))

::.. WWee hhaavvee -- TT22 // TT11 ==((VV 11// VV22)) γγ--11

== rr γγ--11 oottttoo ==== 11-- 11לל rr γγ--11

TThhee nneett wwoorrkk ddoonnee ((WW)) ==WW33--44++WW11--22

==(( PP33 VV33 -- PP44 VV44//γγ--11)) ++(( PP11 VV11 -- PP22 VV22//γγ--11)) == PP11 VV11 // γγ--11 [[ PP33 VV33 // PP11 VV11 -- PP44 VV44 // PP11 VV11 ++11-- ==PP22 VV22 // PP11 VV11 ]] ...... PP33 // PP11 == PP33 // PP22 ** PP22 PP11

==rrpp..rr γγ [[ ...... PP11 VV11

γγ == PP22 VV22 γγ

PP22 // PP11 == ((VV11 // VV22 ))

γγ == rr γγ ]] == PP11 VV11 // γγ--11 [[PP33 // PP11 .. VV22 // VV11 -- PP44 // PP11 .. VV11 // VV11 ++ 11 == PP22 // VV22 ,,

VV22 // VV11 ]] == PP11 VV11 // γγ--11 [[ rrpp rr γγ

11//rr --rrpp ++ 11-- rrpp rr γγ ..11//rr ]]

== PP11 VV11 // γγ--11 [[ rrpp .. rr γγ--11 --rrpp ++ 11-- rrpp rr γγ --11]]



WW == PP11 VV11 // γγ--11 [[ rr γγ--11 ((rrpp –– 11)) –– 11((rrpp --11))]] WW == PP11 VV11 // γγ--11 [[ ((rrpp--11)) ((rr

γγ --11 --11)) ]] OOnn ddiivviinniinngg nneett wwoorrkk ddoonnee wwii tthh ssttookkee vvoolluummeess WW // VVSS == PP11 VV11 // γγ--11 [[ ((rr γγ --11 --11)) ((rrpp--11)) ]] ((VV11--VV22)) == PP11 VV11 [[ ((rr

γγ --11 --11)) ((rrpp--11)) ]] γγ--11 VV11 ((11--VV22 // VV11)) == PP11 rr [[ ((rr

γγ --11 --11)) ((rrpp--11)) ]] ((γγ--11 )) ((rr--11))

MMeeaann eeff ffeeccttiivvee pprreessssuurree::-- DDeeff iinnii ttiioonn::-- TThhee rraattiioo ooff tthhee nneett wwoorrkk ddoonnee ttoo tthhee ssttrrookkee vvoolluummee iiss kknnoowwnn aass mmeeaann eeff ffeeccttiivvee pprreessssuurree..

PPmm == WW//VVSS

MMeeaann eeff ffeeccttiivvee pprreessssuurree ((PPmm)) :: AArreeaa ooff iinnddiiccaatteedd ddiiaaggrraamm ** ((iinnddiiccaatteedd ccoonnssttaanntt )) LLeennggtthh ooff iinnddiiccaatteedd ddiiaaggrraamm

NNoottee ::-- 11.. EEffff iicciieennccyy ooff oottttoo ccyyccllee == ללoottttoo ==== 11-- 11 rr γγ--11



IIss iinnddeeppeennddeenntt ooff pprreessssuurree ll iimmii tt 22.. EEffff iicciieennccyy wwii ll ll iinnccrreeaassee wwii tthh iinnccrreeaassiinngg ccoommpprreessssiioonn rraattiioo.. 33.. EEffff iicciieennccyy wwii ll ll iinnccrreeaassee iiss wwii tthh iinnccrreeaassee iinn AAddiiaabbaattiicc iinnddeexx.. PPrroovvee tthhaatt ffoorr tthhee mmaaxxiimmuumm wwoorrkk ddoonnee tthhee ccoommpprreessssiioonn rr== ((TT33//TT11)) 11 //22(( γγ--11)) WWhheenn tthhee ccyyccllee iiss ooppeerraattiinngg bbeettwweeeenn tthhee tteemmppeerraattuurree ll iimmii ttss ooff TT33 aanndd TT11 AAllssoo pprroovvee tthhee iinntteerrmmeeddiiaattee tteemmppeerraattuurree TT22== TT44 == √√ TT11TT33

WWee hhaavvee wwoorrkk ddoonnee WW==QQSS -- QQRR

==CCvv((TT33--TT22)) -- CCCC((TT44--TT11)) -- →→11 PPrroocceessss 11--22::-- TT22//TT11==(( VV22 // VV11 ))

γγ--11 == rr γγ--11 TT22 == TT11..rr

γγ--11

PPrroocceessss 33--22::-- TT33//TT44==(( VV44 // VV33 ))

γγ--11 == rr γγ--11

TT44 == VV33 // rr γγ--11

((11)) →→ WW ==CCvv((TT33--TT11 rr γγ--11 )) –– CCVV ((TT33 // rr

γγ--11 --TT11)) IInn tthhee aabboovvee eeqquuaattiioonn eexxcceepptt rr,, rreemmaaiinniinngg aall ll tthhee tteerrmmss aarree ccoonnssttaanntt vvaalluuee .. FFuunnccttiioonn iiss mmaaxxiimmuumm,, wwhheenn ii ttss ddeerriivvaattiivvee ww..rr..ttoo tthhee vvaarriiaabbllee iiss zzeerroo FFoorr mmaaxxiimmuumm wwoorrkk ddkkoonnee ddww//ddrr ==00 ddww//ddrr ==00 CC11 TT11 ((

γγ--11)) (( rr γγ--22)) -- CCVV TT33 ((11-- γγ )) (( rr --γγ)) ++ 00 00 == -- CC11 TT11 ((

γγ--11)) (( rr γγ--22)) -- CCVV TT33 ((-- γγ ++11)) rr --γγ

CCVV TT11 (( γγ--11)) rr γγ--22 == CCVV TT33 ((

γγ--11 )) rr ––γγ

TT33 //TT11 == rr γγ--22 // rr ––γγ

TT33 //TT11 == rr

22((γγ--11))



rr== (( TT33 //TT11 ))

11//22((γγ--11))

WWee hhaavvee TT22 == TT11 rr

γγ--11

== TT11[[ (( TT33 //TT11 )) 11//22((γγ--11))]] ((γγ--11))

== TT11 √√ TT33TT11

TT22== √√ TT11TT33

TT22 == TT33 // rr

γγ--11

== TT33 // [[ (( TT33 //TT11 ))

11//22((γγ--11)) ]] γγ--11 ==TT33 // (( TT33 //TT11 ))

½½

TT44 == TT33 (( TT33 //TT11 )) 11//22

TT44 == √√ TT11TT33



DDIIEESSEELL CCYYCCLLEE ::-- TThhee ddiieesseell ccyyccllee ((ccoonnssttaanntt pprreessssuurree ccyyccllee )) ccoonnssiissttss ooff tthhee ffooll lloowwiinngg pprroocceessss

PPrroocceessss 11--22 :: -- AAii rr ggeettss ccoommpprreesssseedd ffrroomm PP11,, VV11,, TT11 ttoo PP22,, VV22,, TT22 iisseennttrrooppiicc aall llyy ccoommpprreessssiioonn rraattiioo.. rr == VV11 //VV22 PPrroocceessss 22--33::--

HHeeaatt iiss aaddddeedd aatt ccoonnssttaanntt pprreessssuurree ,, ssoo tthhee tteemmppeerraattuurree wwii ll ll rraaiissee ttoo TT33..

TThhee rraattiioo VV33//VV22 ==℮℮iiss kknnoowwnn aass ccuutt ooff ff rraattiioo hheeaatt aaddddii ttiioonn QQss== mmccpp ((TT33//TT22))

PPrroocceessss 33--44;;--

TThhee aaii rr eexxppaannddss iisseennttrrooppiicc aall llyy ttoo tthhee PP44,, VV44,, TT44 .. VV44//VV33 ==rree iiss kknnoowwnn eexxppaannssiioonn rraattiioo PPrroocceessss 44--11::-- HHeeaatt iiss rreejjeecctteedd bbyy tthhee aaii rr aatt ccoonnssttaanntt vvoolluummee ssoo tthhee pprreessssuurree aanndd tteemmppeerraattuurree wwii ll ll rreeaacchh ii ttss iinnii ttiiaall vvaalluuee .. HHeeaatt rreejjeecctt QQRR ==mmccvvvv ((TT44//TT11)) NNoottee ::--

DDiieesseell ccyyccllee ccoonnssiissttss ooff ttwwoo iisseennttrrooppiicc pprroocceessss oonnee ccoonnssttaanntt vvoolluummee aanndd oonnee ccoonnssttaanntt pprreessssuurree pprroocceesssseess ddiieesseellלל == QQSS --QQSS // QQSS

== 11-- QQ RR // QQSS



== 11 –– CCVV ((TT44 -- TT11)) // CCpp ((TT33 –– TT22)) == 11 -- 11 // γγ ((TT44 -- TT11 // TT33 –– TT22))

PPrroocceessss 11--22 →→ TT22 == TT11 .. rr

γγ--11 WWee kknnooww VV11 //VV22 == rr VV33 //VV22 == pp VV44 == VV11 VV33 //VV44 == VV33 //VV11 == VV11 //VV22 ** VV22 //VV11 == PP.. 11//rr VV33 //VV44 == pp // rr 22--33 pprroocceessss :: VV22//TT33 == .. VV33//TT33

TT33 == TT22 VV33//VV22 == TT11 rr

γγ--11 .. ℓℓ

33--44 pprroocceessss :: TT33 VV33 γγ--11 == TT44 VV44

γγ--11

TT44 == TT33 ((VV33 VV44)) γγ--11

== TT33 ((pp // rr)) γγ--11

TT44 == VV11

rr γγ--11 .. ℓℓ γγ--11 // rr γγ--11

TT44 == TT11 pp γγ

γγ ((TT44 -- TT11)) // ((TT33 –– TT 22)) // 11 –– 11 == לל == 11 –– 11 // γγ ((TT11 pp

γγ -- TT11)) // ((TT11 rr γγ--11 ℓℓ

–– TT11 rr γγ--11 ))



ddiieesseellלל == 11 –– 11 // γγ TT11 // // rr

γγ--11 ** TT11 (( ℓℓ γγ--11 )) // ℓℓ

--11

ddiieesseellלל == 11 –– 11 // rr γγ--11 .. 11 // γγ (( ℓℓ γγ--11 // ℓℓ --11))

NNoottee :: 11)) AAss .. VV33>>VV22 ,,ℓℓ iiss aallwwaayyss ggrreeaatteerr tthhaann 11 aass aa rreeaassoonn

SSoo ffoorr tthhee ssaammee ccoommpprreessssiioonn rraattiioo rr,,

22)) wwii tthh iinnccrreeaassee iinn ccuutt ooff ff rraattiioo ((ℓℓ)) ,, tthhee eeff ff iicciieennccyy wwii ll ll ffeell ll ddoowwnn .. wwii tthh iinnccrreeaassee iinn ccoommpprreessssiioonn rraattiioo ((rr)),, tthhee eeff ff iicciieennccyy wwii ll ll iinnccrreeaasseess .. WW == PP22 ((VV33 –– VV22)) ++ PP33 VV33 -- PP44 VV44 // γγ –– 11 ++ PP11 VV11 -- PP22 VV22 -- γγ –– 11 == PP22 VV22 // γγ –– 11 [[ ((VV33 // VV22 -- 11)) ((γγ –– 11)) ++ PP33 // PP22 .. VV33 // VV22 -- PP44 // PP22 .. VV44 // VV22 ++ PP11 VV11 // PP22 VV22 -- 11 ]] ...... PP33 == PP22

VV44 == VV11

CCoonnssiiddeerr 33 -- 44 PP33 VV33

γγ == PP44 VV44 γγ

PP22 // PP11 == VV11

γγ // VV22γγ ==>> PP22 // PP11 == 11 // rrγγ

PP44 // PP33 == ((VV33 // VV44 ))

γγ ==>> PP44 // PP22 == ((PP // rr))γγ

CC == VV33 // VV44 == PP // rr



WW == PP22 VV22 // γγ –– 11 [[ ((ℓℓ -- 11)) ((γγ -- 11)) ++ ℓℓ -- (( pp // rr))γγ .. rr ++ ((11 // rrγγ .. rr)) -- 11]] WWee hhaavvee PP22 // PP11 == rrγγ PP22 == PP11 rr

γγ WWee hhaavvee VV11 // VV22 == rr VV22 == VV11 rr

--11 ==>> PP22 // VV11 == PP11// VV11 rr

γγ--11 WW == PP11 VV11 rr

γγ--11 [[ ((ℓℓ -- 11)) ((γγ -- 11)) ++ ℓℓ -- ℓℓγγ .. rr 11-- γγ ++ rr 11--γγ -- 11 ]] == PP11 VV11 rr

γγ--11 // γγ -- 11 [[ ((ℓℓ -- 11)) ((γγ -- 11)) ++ rr 11-- γγ ((--ℓℓ γγ ++ 11)) ++ ((ℓℓ -- 11))]] == PP11 VV11 rr

γγ--11 // γγ -- 11 [[ ((ℓℓ -- 11)) ((γγ -- 11)) ++ rr 11-- γγ ((--ℓℓ γγ ++ 11)) ++ ((ℓℓ -- 11))]] == PP11 VV11 rr

γγ--11 // γγ -- 11 [[ ((ℓℓ -- 11)) ((γγ –– 11 ++ 11)) ++ rr 11-- γγ (( 11 -- ℓℓ γγ )) ]] WW == PP11 VV11 rr

γγ--11 // γγ -- 11 [[ γγ ((ℓℓ -- 11)) ++ rr 11-- γγ (( 11 -- ℓℓ γγ )) ]] ]] MMEEAANN EEFFFFEECCTTIIVVEE PPRREESSSSUURREE::-- PPmm == WW // VVSS == [[ PP11 VV11 rr

γγ--11 // γγ -- 11 [[ γγ ((ℓℓ -- 11)) ++ rr 11-- γγ (( 11 -- ℓℓ γγ )) ]] ]] (( VV11 -- VV22 )) == [[ PP11 VV11 rr

γγ--11 // γγ -- 11 [[ γγ ((ℓℓ -- 11)) ++ rr 11-- γγ (( 11 -- ℓℓ γγ )) ]] ]] VV11 (( 11 -- VV22 // VV11 )) == [[ PP11 VV11 rr

γγ--11 // γγ -- 11 [[ γγ ((ℓℓ -- 11)) ++ rr 11-- γγ (( 11 -- ℓℓ γγ )) ]] ]] (( 11 -- 11 // rr )) ...... [[ VV11 // VV22 == rr ]]



PPmm == PP11 rr

γγ // [[ γγ ((ℓℓ -- 11)) ++ rr 11-- γγ (( 11 -- ℓℓ γγ )) ]] ((γγ -- 11)) ((rr -- 11))

11.. IInn aann aaii rr ssttaannddaarrdd ddiieesseell ccyyccllee

AAtt tthhee bbeeggiinnnniinngg ooff ccoommpprreessssiioonn tthhee pprreessssuurree iiss 00..11 mmppaa ;; tteemmpp 440000cc .. HHeeaatt aaddddeedd

iiss 11..667755mmjj ddeetteerrmmiinnee tthhee mmaaxx.. tteemmppeerraattuurree,, ccuutt ooff ff rraattiioo,, iiss118800-- tthheerrmmaall eeff ff iicciieennccyy ,, tteemmpp aatt tthhee eenndd ooff 1188 eennttrrooppiicc eexxppaannssiioonn aanndd mmeeaann eeff ffeeccttiivvee pprreessssuurree CCoommpprreessssiioonn rraattiioo rr == 1155 IInnii ttiiaall pprreessssuurree PP11 == 00..11mmppaa == 00..11** 110066 NNllmm22 IInnii ttiiaall tteemmppeerraattuurree TT11 == 440000

CC == 33113311 kk QQ33 == 11..667755 MMJJ == 11..667755 ** 1100 JJ TToo DDeetteerrmmiinnee::-- MMaaxx .. tteemmppeerraattuurree == ?? CCuutt--ooff ff rraattiioo ==?? IIssootthheerrmmaall eeff ff iicciieennccyy == ?? TTeemmpp.. aatt eenndd ==?? PPmm == ?? WWoorrkk ddoonnee // ccyyccllee == ?? AAssssuummee CCPP == 11..000055 KKJJ // kkgg kk



RR == 00..228877 KKJJ // kkgg kk

PPrroocceessss 11--22::-- TT22 == TT11 rr γγ--11

== 331133 ((1155))11..44--11

== 992244..6655 kk HHeeaatt ssuuppppll iieedd QQSS == CCPP ((TT33 -- TT22)) 11..667755 ** 110066 == 11..000055 ** 110033 ((TT33 –– 992244..6655))

TT33 ==22559911..33 kk

PPrroocceessss 22--33::-- VV22 // TT22 == VV33 // TT33

VV33 // VV22 == TT33 // TT22 == ℓℓ TT33 // TT22 == ℓℓ == 22..8800 PPrroocceessss 33--44::-- TT33 // TT44 == ((VV44 // VV33 ))

γγ--11-- VV44 // VV33 == VV11 // VV33 [[

...... VV44 // VV11]]

== VV11 // VV22 ** VV22 // VV33

== ((rr // ℓℓ)) TT33 // TT44 == ((rr // ℓℓ))γγ –– 11

22559911..33 // TT44 == ((1155 // 22..88)) 11..44--11



TT44 ==11332244..11 kk

HHeeaatt rreejjeecctteedd ((QQRR)) == CCVV((TT44 –– TT11)) == 00..771188 ** 110033 ((11332244..11 ..331133)) == 772266..0022 kk WWoorrkk ddoonnee ((ww)) ==QQSS -- QQRR

== 11..667755 ** 110066 –– 772266..0022 ** 110033 == 994488..997788 ** 110033

EEffff iicciieennccyy (( ללddiieesseell )) == 11-- 11// rr γγ--11 .. 11//γγ ..(( ℓℓγγ –– 11 )) // ((ℓℓ -- 11)) == 5566..6655 %% [[ ..

.... cchheeeekk לל == WW // QQSS

WW == לל ** QQSS

== 00..55665566** 11..667755** 110066 == 994488..997788** 110033 ]] PP11 VV11 == RRTT11

VV11 == 228877 ** 331133 // 00..11** 110066 == 00..8899 MM33

VV11 // VV22 == rr == 1155 ==>>VV22 == 00..8899 // 1155 == 00..005599 MM33

SSttookkee vvoolluummee VV11 –– VV22 == VVSS == 00..8833 MM33

MMeeaann eeff ffeeccttiivvee pprreessssuurree == WW // VVSS == 994488..997788 ** 110033 // 00..8833



PPMM == 1111..4422 bbaarr

[[ ....

.. cchheecckk :: PPMM == PP11rrγγ [[ γγ ((ℓℓ -- 11)) –– rr11--γγ ((ℓℓ γγ -- 11)) ]] // ((rr -- 11)) ((γγ -- 11))

== 00..11** 110066 ** 115511..44 [[ ((11..44)) ((22..88 -- 11)) –– 115511--00..44 ((22..8811..44 --11)) // ((1155 -- 11)) ((11..44--11)) PPmm == 1111..2299 bbaarr MMaaxx.. tteemmpp TT22 == 992244..6655 kk ℓℓ == 22..8800



DDUUEELL CCOOMMBBUUSSTTIIOONN ((LLIIMMIITTEEDD PPRREESSSSUURREE CCYYCCLLEE))::--

TThhiiss dduuaall ccyyccllee ccoonnssiissttss ooff tthhee ffooll lloowwiinngg pprroocceessss PPrroocceessss 11--22::-- TThhee aaii rr ggeettss ccoommpprreesssseedd iisseennttrrooppiicc aall ll yy ffrroomm PP11,, VV11,, TT11 ttoo PP22,, VV22,, TT22

PPrroocceessss 22--33::-- TThhee ppaarrttiiaall qquuaannttii ttyy ooff tthhee hheeaatt iiss aaddddeedd aatt ccoonnssttaanntt vvoolluummee .. ssoo tthhee pprreessssuurree aanndd tteemmppeerraattuurree wwii ll ll rriissee ttoo PP33,,TT22,, VV11,,== VV22

.. PP33//PP22 ==pprreessssuurree rraattiioo // eexxpplloossiioonn rraattiioo ==rrpp

HHeeaatt ssuuppppll iieedd QQ11 == CCvv ((TT33//TT22 )) PPrroocceessss 33--44::-- AAtt PP33 rreemmaaiinniinngg qquuaannttii ttyy ooff tthhee hheeaatt iiss aaddddeedd ((aatt ccoonnssttaanntt pprreessssuurree)) .. tthhee vvoolluummee ii tt mmaaxxiimmuumm vvaalluuee .. VV11==VV33

.. PP22//PP22 == pprreessssuurree rraattiioo eexxppaannssiioonn rraattiioo ==rrpp



HHeeaatt ssuuppppll iieedd QQ11 ==CCvv ((TT33//TT22 )) ((VV11//VV22 )) ==((VV44//VV33 )) ccuutt ooff ff rraattiioo == ℓℓ HHeeaatt ssuuppppll iieedd QQ22 ==CCpp ((TT44//TT33 ))

PPrroocceessss 44--55::-- AAii rr wwii ll ll eexxppaanndd iisseennttrrooppiicc aall llyy ttoo tthhee pprreessssuurree ooff PP55 AAnndd tthhee ccoorrrreessppoonnddiinngg pprrooppeerrttiieess aarree PP55 aanndd TT55 PPrroocceessss 55--11::-- HHeeaatt iiss rreejjeecctteedd wwii tthhoouutt ff iinnddiinngg aannyy aapppprreecciiaabbllee cchhaannggee iinn vvoolluummee HHeeaatt rreejjeecctteedd QQRR == CCvv ((PP55 --TT11)) ddiiaall == 11 –– QQRR // QQSS לל

== 11 –– QQRR // QQ11 ++ QQ22

== 11-- CCvv((TT55 –– TT11)) // CCvv((TT33 –– TT22)) ++ CCpp((TT44 –– TT33)) ddiiaall == 11 –– ((TT55 –– TT11)) //((TT33 –– TT22)) ++ γγ ((TT44 –– TT33)) לל ((CCpp // CCvv == γγ )) CCoonnssiiddeerr 11--22::-- VV11 // VV22 == rr →→ ccoommpprreessssiioonn rraattiioo TT22 // TT11 == ((VV11 // VV22 ))

γγ--11

TT22 == TT11 .. rr γγ--11

CCoonnssiiddeerr 22--33::-- ((ccoonnssttaanntt vvoolluummee pprroocceessss))



PP22 // TT22 == PP33 // TT33

TT33 == TT22 .. PP33 // PP22

==>> TT33 == TT11 .. rr

γγ--11 .. rrpp ...... PP33 // PP22 == rrpp

CCoonnssiiddeerr 33--44 pprroocceessss ::-- VV33 // TT33 == VV44 // TT44

TT 44 == TT22 .. VV44 // VV33

TT44 == TT11 .. rr

γγ--11 == .. rrpp .. ℓℓ VV44 // VV33 ==ℓℓ CCoonnssiiddeerr 44--55 pprroocceessss::--

TT55 // TT44 ((VV55 // VV44)) γγ--11

VV55 == VV11

VV55 // VV44 == VV11 // VV44

VV55 // VV44 == VV11 // VV22 ** VV33 // VV44 == rr ** 11// ℓℓ TT44 // TT55 == ((rr // pp )) γγ--11 TT55 == TT44 ℓℓ

γγ--11 // rr γγ--11

== ((TT11 .. rr

γγ--11 .. rrpp .. pp)) ℓℓγγ--11 // rr γγ--11

TT55 == TT11 rrpp ℓℓ

γγ

γγ ((TT44 // TT33)) ++ ((TT33 –– TT22)) // ((TT55 –– TT11)) –– 11 == לל



== 11 –– ((TT11 rrpp ℓℓ

γγ -- TT11)) // ((TT11rr γγ--11 ..rrpp -- TT11 .. rr

γγ--11)) ++γγ((TT11rr γγ--11 .. rrpp))

== 11 –– TT11 (( rrpp pp

γγ -- 11)) // TT11rr γγ--11 [[ (( rrpp -- 11 ))

++γγ ++γγ(( rrpp ℓℓ -- rrpp)) ]] dduuaall == 11-- 11// rr γγ--11 ((rrpp)) ((pp לל

γγ--11 -- 11)) // [[ ((rrpp -- 11)) ++ γγ.. RRpp ((ℓℓ -- 11)) == 11-- 11// rr γγ--11 .. 11//γγ .. ℓℓγγ –– 11 // ((ℓℓ -- 11)) ℓℓ == 11 rrpp == 11

NNoottee ::--

11.. II ff tthhee ccuutt--ooff ff rraattiioo pp==11 tthheenn eeff ff iicciieennccyy ooff dduuaall ccyyccllee ==eeffff iicciieennccyy ==eeffff iicciieennccyy ooff oottttoo ccyyccllee

== 11-- 11// rr γγ--11

22.. II ff tthhee pprreessssuurree rraattiioo rrpp ==11 tthheenn eeff ff iicciieennccyy ooff dduuaall ccyyccllee== eeff ff iicciieennccyy ooff ddiieesseell ccyyccllee

== 11-- 11// rr γγ--11 .. 11//γγ .. ℓℓγγ –– 11 // ((ℓℓ -- 11)) NNeett wwoorrkk ddoonnee WW == WW33--44 ++ WW44--55 ++ WW11--22

== PP33 ((VV44 -- VV 33)) ++ PP44VV44 –– PP55 VV55 // γγ –– 11 ++ PP 11 VV22 –– PP22 VV22// γγ –– 11

MMeeaann eeff ffeeccttiivvee pprreessssuurree PPmm == WW//VVss == WW //VV11 -- VV22

dduuaall == 11-- 11// rr γγ--11 .. 11//γγ .. ℓℓγγ –– 11 // ((ℓℓ -- 11)) לל ...... NNeett wwoorrkk ddoonnee WW== PP33 ((VV44 -- VV 33)) ++ PP44VV44 –– PP55 VV55 // γγ –– 11 ++ PP 11 VV22 –– PP22 VV22// γγ –– 11 PPmm == WW //VV11 -- VV22



EExxtt::-- AAnn eennggiinnee iiss ooppeerraattiinngg wwii tthh aa ccyyll iinnddeerr ooff ddiiaammeetteerr 220000mmmm aanndd

SSttoocckk lleennggtthh 330000mmmm aanndd ii tt iiss ooppeerraattiinngg oonn ddiieesseell ccyyccllee wwii tthh aa ccoommpprreessssiioonn rraattiioo ooff 1155.. tthhee pprreessssuurree aanndd tteemmppeerraattuurreess aatt tthhee bbeeiinngg ooff ccoommpprreessssiioonn iiss 11 bbaarr aanndd 227700

cc rreessppeeccttiivveellyy.. TThhee ccuutt ooff ff rraattiioo iiss 88%% ,, tthhiiss ssttoocckk vvoolluummee .. ddeetteerrmmiinnee tthhee pprreessssuurree vvoolluummee,, tteemmppeerraattuurree aatt ssaall iieenntt ppooiinnttss..

ii .. TThheerrmmaall eeff ff iicciieennccyy

ii ii .. MMeeaann eeff ffeeccttiivvee pprreessssuurree

ii ii ii .. IIddeeaall ppoowweerr ddeevveellooppeedd nnoo.. ooff ccyycclleess aarree 330000 ppeerr mmiinnii ttee SSooll tt::-- GGiivveenn ddaattaa::-- ddiiaammeetteerr dd == 220000mmmm == 00..22mm ssttrrookkee lleennggtthh ((ll )) == 330000mmmm == 00..33mm ccoommpprreessssiioonn rraattiioo == 1155 == rr iinnii ttiiaall pprreessssuurree PP11 == 11 bbaarr ==110055 nnllmm22 iinnii ttiiaall tteemmppeerraattuurree == TT11 == 2277oo

CC == 330000 kk CCuutt ––ooffff rraattiioo ℓℓ == 88%% ooff ssttookkee vvoolluummee tthheerrmmaall eeff ff iicciieennccyy == ?? PPrreessssuurree ,, vvoolluummee ,, tteemmppeerraattuurreess,, iiddeeaall ppoowweerr ddeevveellooppeedd ssttrrookkee vvoolluummee VVSS == ∏∏ // 44 ** DD22

** LL == ∏∏ // 44 ((00..22))22 ((00..33)) VVSS == 99..44 ** 1100--33 MM33

VV11 –– VV22 == VVSS == 99..44 ** 1100--33 MM33

CCuutt -- ooff ff == 88%% ssttookkee vvoolluummee .. ((VV33 -- VV33)) == 88 // 110000 VVSS == 77..553366 ** 1100--44 mm



WWee hhaavvee ccoommpprreessssiioonn rraattiioo RR == VV11 // VV22

VV11 == 1155 VV22

WWee hhaavvee (( VV11 // VV22 )) == 99..4422 ** 1100--33MM33

1155 VV11 // VV22 == 99..4422 ** 1100--33MM33

VV33 ––VV22 == 77..553366 ** 1100--44MM33

VV33 == 11..4455 ** 1100--33MM33

PPrroocceessss 11--22::-- TT22 == TT11 .. rr

γγ--11 == 330000 ** 115511..44--11

TT22 == 888866 .. 2255 kk PP22 // PP11 == ((VV11 // VV22))

γγ ==>> PP22 ==PP11 ((rr)) γγ

== 110055 ((1155..44))11..44

PP22 == 4444..33 bbaarr

PPrroocceessss 22--33::-- ccuutt –– ooff ff rraattiioo ℓℓ == VV33 //VV22 == 11..4455 ** 1100--33 // 66 ..7722 ** 1100--44 ==>> ℓℓ == 22..1111 VV22 // TT22 == VV33 // TT33 ‘‘ ..’’ CCoonnssttaanntt pprreessssuurree pprroocceessss

VV22 == 66..7722** 1100--44MM33



TT33 == TT22 ..VV33 //VV22 ==>> TT22.. ℓℓ == 888866..2255 ** 22..1111

TT33 == 11887788..8855 kk

PP33 == PP22 == 4444 .. 33 bbaarr

PPrroocceessss 33--44::-- TT33 // TT44 == ((VV44 //VV33))

γγ--11

TT44 == TT33 ((VV33//VV11)) γγ--11

TT44 == 885599..3322 kk

PP44 // TT44 == PP11 // TT11 PP44 == PP11 .. TT44 // TT11 == 110055 ** 885599..3322 // 330000

PP44 == 228866 bbaarr

ddiieesseell == 11-- 11// rr γγ--11 .. 11//γγ ..(( ℓℓγγ –– 11 )) // ((ℓℓ -- 11)) לל

== 11-- 11 //115511..44--11 .. 11 // 11..44 .. ((22..111111..44 -- 11)) // ((22..1111 --11)) == 00..55997777 == 5599..7777%% MMeeaann eeff ffeeccttiivvee pprreessssuurree PPmm == PP11rr

γγ [[γγ((ℓℓ--11)) ++rr11--γγ((11--ℓℓγγ))]] ((rr--11)) ((ℓℓ -- 11)) == 77..3355 bbaarr MMeeaann eeff ffeeccttiivvee pprreessssuurree == WW //VVSS



WWoorrkk ddoonnee ((ww)) == PPmm ** VVSS == 77..3355 ** 110055 ** 99..4433 ** 1100--33

WW == 66993311 .. 0055 jj // ccyyccllee

IIddeeaall ppoowweerr ddeevveellooppeedd== ww** nnoo..ooff ccyycclleess 11ppeerr sseecc ==ww** 330000//6600

PP == 3344665555..2255 WW



CCOOMMPPAARRIISSIIOONN OOFF CCYYCCLLEESS ::--

11)) FFoorr ssaammee ccoommpprree33ssssiioonn rraattiioo aanndd hheeaatt iinn ppuutt

22)) FFoorr ssaammee mmaaxx pprreessssuurree aanndd hheeaatt aaddddii ttiioonn::--

ii .. AAss tthhee hheeaatt rreejjeeccttiioonn iiss mmoorree iinn ddiieesseell aanndd dduuaall ccyycclleess ccoommppaarreedd ttoo oottttoo..

ii ii .. EEvveenn tthhoouugghh tthhee qquuaannttii ttyy ooff hheeaatt aaddddeedd iiss ssaammee ffoorr aall ll tthhee ccyycclleess ,, tthhee wwoorrkk ddoonnee iiss ddii ff ffeerreenntt ((AArreeaa ooff pp--vv ddiiaaggrraamm)) aass tthhee hheeaatt iiss bbeeiinngg aaddddeedd aatt tthhee ddii ff ffeerreenntt ccoonnddii ttiioonnss..

SStteerrll iinngg ccyyccllee::--



AAss ii tt iiss iissootthheerrmmaall pprroocceessss QQSS == PP11 VV11 lloogg VV22 // VV11 lloogg rr == RR TT44 lloogg rr QQRR == PP33 VV33 lloogg VV33//VV44 == RRTT33 lloogg rr == RR TT22 lloogg rr QQRR // QQSS--11 == לל

== 11-- RR TTcc lloogg rr // RRTTHH lloogg rr

TTLL // TTHH --11 == לל

AAttcckkiinnssoonn ccyyccllee ::--

QQSS == CCPP ((TT33 -- TT22 ))

QQRR == CCPP ((TT44 –– TT11))

QQRR // QQSS --11 == לל

== 11 -- CCPP ((TT44 –– TT11)) // CCPP ((TT33–– TT22)) ]] γγ --11 == לל <<== rree-- rrcc // rree

γγ --rrccγγ]]



EErriicc ssoonn ccyyccllee ::--

EErriiccssoonn eennggiinnee EErriiccssoonn CCyyccllee HHeeaatt ssuuppppll iieedd QQSS == CCPP ((TT33 -- TT22 )) HHeeaatt RReejjeecctteedd QQRR == CCPP ((TT44 –– TT11))

QQRR // QQSS --11 == לל

== 11 -- CCPP ((TT44 –– TT11)) // CCPP ((TT33–– TT22)) LLeennooiirr CCyyccllee



CCoonnssttaanntt vvoolluummee hheeaatt aaddddiittiioonn ((11--22))

IInn tthhee iiddeeaall ggaass vveerrssiioonn ooff tthhee ttrraaddii ttiioonnaall LLeennooii rr ccyyccllee,, tthhee ff ii rrsstt ssttaaggee ((11--22)) iinnvvoollvveess tthhee aaddddii ttiioonn

ooff hheeaatt iinn aa ccoonnssttaanntt vvoolluummee mmaannnneerr.. TThhiiss rreessuull ttss iinn tthhee ffooll lloowwiinngg ffoorr tthhee ff ii rrsstt llaaww ooff

tthheerrmmooddyynnaammiiccss::

IIsseennttrrooppiicc eexxppaannssiioonn ((22--33))

TThhee sseeccoonndd ssttaaggee ((22--33)) iinnvvoollvveess aa rreevveerrssiibbllee aaddiiaabbaattiicc eexxppaannssiioonn ooff tthhee ff lluuiidd bbaacckk ttoo ii ttss oorriiggiinnaall

pprreessssuurree.. IItt ccaann bbee ddeetteerrmmiinneedd ffoorr aann iisseennttrrooppiicc pprroocceessss tthhaatt tthhee sseeccoonndd llaaww ooff tthheerrmmooddyynnaammiiccss

rreessuull ttss iinn tthhee ffooll lloowwiinngg::

CCoonnssttaanntt pprreessssuurree hheeaatt rreejjeeccttiioonn ((33--11))

TThhee ff iinnaall ssttaaggee ((33--11)) iinnvvoollvveess aa ccoonnssttaanntt pprreessssuurree hheeaatt rreejjeeccttiioonn bbaacckk ttoo tthhee oorriiggiinnaall ssttaattee.. FFrroomm

tthhee ff ii rrsstt llaaww ooff tthheerrmmooddyynnaammiiccss wwee ff iinndd::

TThhee oovveerraall ll eeff ff iicciieennccyy ooff tthhee ccyyccllee iiss ddeetteerrmmiinneedd bbyy tthhee ttoottaall wwoorrkk oovveerr tthhee hheeaatt iinnppuutt,, wwhhiicchh ffoorr

aa LLeennooii rr ccyyccllee eeqquuaallss

..

NNoottee tthhaatt wwee ggaaiinn wwoorrkk dduurriinngg tthhee eexxppaannssiioonn pprroocceessss bbuutt lloossee ssoommee dduurriinngg tthhee hheeaatt rreejjeeccttiioonn

pprroocceessss..

Thermodynamics Unit VIII Notes 49


Brayton cycle The Brayton cycle is a thermodynamic cycle that describes the workings of the gas

turbine engine, basis of the airbreathing jet engine and others.

It is named after George Brayton (1830–1892), the American engineer who developed it,

although it was originally proposed and patented by Englishman John Barber in 1791.[1] It is also

sometimes known as the Joule cycle. The Ericsson cycle is similar but uses external heat and

incorporates the use of a regenerator. There are two types of Brayton cycles, open to the

atmosphere and using internal combustion chamber or closed and using a heat exchanger

The term Brayton cycle has more recently been given to the gas turbine engine. This also has three components:

1. a gas compressor 2. a burner (or combustion chamber) 3. an expansion turbine

Ideal Brayton cycle:

• isentropic process - ambient air is drawn into the compressor, where it is pressurized. • isobaric process - the compressed air then runs through a combustion chamber, where

fuel is burned, heating that air—a constant-pressure process, since the chamber is open to flow in and out.

• isentropic process - the heated, pressurized air then gives up its energy, expanding through a turbine (or series of turbines). Some of the work extracted by the turbine is used to drive the compressor.

• isobaric process - heat rejection (in the atmosphere).



The efficiency of the ideal Brayton cycle is

Closed Brayton cycle

A closed Brayton cycle recirculates the working fluid, the air expelled from the turbine is

reintroduced into the compressor, this cycle use a heat exchanger to heat the working fluid

instead of an internal combustion chamber. The closed Brayton cycle is used for example

in closed-cycle gas turbine and space power generation.



Rankine Cycle

There are four processes in the Rankine cycle. These states are identified by numbers (in brown)

in the above Ts diagram.

• Process 1-2: The working fluid is pumped from low to high pressure. As the fluid is a

liquid at this stage the pump requires little input energy.

• Process 2-3: The high pressure liquid enters a boiler where it is heated at constant

pressure by an external heat source to become a dry saturated vapor. The input energy

required can be easily calculated using mollier diagram or h-s chart or enthalpy-entropy

chart also known as steam tables.

• Process 3-4: The dry saturated vapor expands through a turbine, generating power. This

decreases the temperature and pressure of the vapor, and some condensation may occur. The

output in this process can be easily calculated using the Enthalpy-entropy chart or the steam

tables.

• Process 4-1: The wet vapor then enters a condenser where it is condensed at a constant

temperature to become a saturated liquid.

In an ideal Rankine cycle the pump and turbine would be isentropic, i.e., the pump and turbine

would generate no entropy and hence maximize the net work output. Processes 1-2 and 3-4

would be represented by vertical lines on the T-S diagram and more closely resemble that of the

Carnot cycle. The Rankine cycle shown here prevents the vapor ending up in the superheat



region after the expansion in the turbine, [1] which reduces the energy removed by the

condensers.



combined cycles

In electric power generation a combined cycle is an assembly of heat engines that work

in tandem off the same source of heat, converting it into mechanical energy, which in turn

usually drives electrical generators. The principle is that the exhaust of one heat engine is used as

the heat source for another, thus extracting more useful energy from the heat, increasing the

system's overall efficiency. This works because heat engines are only able to use a portion of the

energy their fuel generates (usually less than 50%). In an ordinary (non combined cycle) heat

engine the remaining heat (e.g., hot exhaust fumes) from combustion is generally wasted.

Combining two or more thermodynamic cycles results in improved overall efficiency,

reducing fuel costs. In stationary power plants, a widely used combination is a gas

turbine (operating by the Brayton cycle) burning natural gas or synthesis gas from coal, whose

hot exhaust powers a steam power plant (operating by the Rankine cycle). This is called a

Combined Cycle Gas Turbine (CCGT) plant, and can achieve a thermal efficiency of around

60%, in contrast to a single cycle steam power plant which is limited to efficiencies of around

35-42%. Many new gas power plants in North America and Europe are of this type. Such an

arrangement is also used for marine propulsion, and is called a combined gas and

steam (COGAS)plant. Multiple stage turbine or steam cycles are also common.

1-Electric generators, 2-Steam turbine, 3-Condenser, 4-Pump, 5-Boiler/heat exchanger, 6-Gas

turbine)



A single shaft combined cycle plant comprises a gas turbine and a steam turbine driving a

common generator. In a multi-shaft combined cycle plant, each gas turbine and each steam turbine has

its own generator. The single shaft design provides slightly less initial cost and slightly better efficiency

than if the gas and steam turbines had their own generators. The multi-shaft design enables two or more

gas turbines to operate in conjunction with a single steam turbine, which can be more economical than a

number of single shaft units.

The primary disadvantage of multiple stage combined cycle power plants is that the number of

steam turbines, condensers and condensate systems - and perhaps the number of cooling towers and

circulating water systems - increases to match the number of gas turbines. For a multi-shaft combined

cycle power plant there is only one steam turbine, condenser and the rest of the heat sink for up to three

gas turbines; only their size increases. Having only one large steam turbine and heat sink results in low

cost because of economies of scale. A larger steam turbine also allows the use of higher pressures and

results in a more efficient steam cycle. Thus the overall plant size and the associated number of gas

turbines required have a major impact on whether a single shaft combined cycle power plant or a multiple

shaft combined cycle power plant is more economical.



Bell coleman Cycle

The components of the air refrigeration system are shown in Fig.. In this system, air is taken into

the compressor from atmosphere and compressed. The hotcompressed air is cooled in heat

exchanger upto the atmospheric temperature (in idealconditions). The cooled air is then



expanded in an expander. The temperature of the aircoming out from the expander is below the

atmospheric temperature due to isentropicexpansion. The low temperature air coming out from

the expander enters into the evaporator and absorbs the heat. The cycle is repeated again. The

working of airrefrigeration

cycle is represented on p-v and T-s diagrams

Process 1-2 represents the suction of air into the compressor.

Process 2-3 represents the isentropic compression of air by the compressor.

Process 3-5 represents the discharge of high pressure air from the compressor into the heat

exchanger. The reduction in volume of air from v3 to v5 is due to the cooling of air in the heat

exchanger.

Process 5-6 represents the isentropic expansion of air in the expander.

Process 6-2 represents the absorption of heat from the evaporator at constant pressure.

Assumptions: 1) The compression and expansion processes are reversible adiabatic

processes.

2) There is a perfect inter-cooling in the heat exchanger.

3) There are no pressure losses in the system.

COP = Net refrigeration effect / Net work sup plied

Work done per kg of air for the isentropic compression process 2-3 is given by,

WC = Cp (T3 - T2 )

Work developed per kg of air for the isentropic expansion process 5-6 is given by,

WE = Cp (T5 - T6 )

Net work required = Wnet = (WC - WE ) = Cp (T3 - T2 ) - Cp (T5 - T6 )

Net refrigerating effect per kg of air is given by,

Rnet = Cp (T2 - T6 )



Advantages:

a) Air is a cheaper refrigerant and available easily compared to other refrigerants.

b) There is no danger of fire or toxic effects due to leakage.

c) The total weight of the system per ton of refrigerating capacity is less.

Disadvantages:

(a) The quantity of air required per ton refrigerating capacity is far greater than other systems.

(b) The COP is low and hence maintenance cost is high.

(c) The danger of frosting at the expander valves is more as the air taken into the system always

contains moisture.



Vapor compression Refrigeration Cycles

It is composed of the following 4 processes:

1. reversible heat addition at pe = const. in evaporation to saturated vapor (from point 4 to

point 1)

2. isentropic compression from saturated vapor to the condensing pressure pc (from point1

to point 2);

3. reversible heat rejection at pc=const. desuperheating and condensation to saturated liquid,

(from point 2 to point 3);

4. throttling (irreversible process) from high pressure pc to lower pressure pe, (from point 3

to point 4).

• The process 1-2 is a reversible, adiabatic (isentropic) compression of the refrigerant. So

the specific work input by the compressor is:

Wm = h 2 – h 1

• The process 2-3 is an internally reversible constant pressure heat rejection process, in

which the refrigerant is desuperheated and then condensed to a saturated liquid at point 3.

During this process, the refrigerant rejects heat to condensing media. So the specific

condensation heat load per unit mass flow rate of refrigerant is:

Qc = h 2 – h 3



• The process 3-4 is an irreversible throttling process, in which the temperature and

pressure of refrigerant both decrease at constant enthalpy:

h 3 = h 4

• The process 4-1 is an internally reversible constant pressure heat admission process, in

which the refrigerant is evaporated to a saturated vapor at state point 1.

• The heat necessary for evaporation of the refrigerant is supplied by the substance to be

cooled.

• The rate of heat transferred to the refrigerant in the evaporator is called the refrigeration

capacity.

• The specific refrigerating effect, i.e. the refrigerating capacity per unit mass flow rate of

refrigerant can be obtained as follow:

Qe = h 1- h 4

COP = Net refrigeration effect / Net work sup plied

Documents

Td Notes by Chiranjeevarao