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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
Thermodynamics Unit I Notes 1
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit I Notes 1
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit I Notes 1
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit I Notes 2
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.
Thermodynamics Unit I Notes 2
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit I Notes 3
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit I Notes 3
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
� 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.
Thermodynamics Unit I Notes 3
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit I Notes 4
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit I Notes 4
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit I Notes 4
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
QA < QB < QC
From the above equation we can say that heat transfer will depend on a process the path of a
process
Thermodynamics Unit I Notes 5
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit I Notes 5
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit I Notes 5
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
� 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.
Thermodynamics Unit I Notes 6
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit I Notes 6
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit I Notes 6
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit II Notes 7
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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)
Thermodynamics Unit II Notes 7
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit II Notes 8
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit II Notes 8
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit II Notes 9
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit II Notes 9
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit II Notes 9
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit III Notes 12
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit III Notes 12
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit III Notes 12
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
=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
Thermodynamics Unit III Notes 13
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit III Notes 13
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit III Notes 14
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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):-
Thermodynamics Unit III Notes 14
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
It is a device which fails to saticify the IInd law of thermodynamics. It is a device with
100% of efficiency.
Thermodynamics Unit III Notes 15
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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=
Thermodynamics Unit III Notes 15
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit III Notes 15
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• 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.
Thermodynamics Unit III Notes 15
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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
Thermodynamics Unit III Notes 15
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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
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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−+
− = −
16Thermodynamics Unit III Notes 16
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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= + < = + ⇒ < = −
16Thermodynamics Unit III Notes 16
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đQdS
T⇒ ≥
Thermodynamics Unit III Notes 17
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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
Thermodynamics Unit III Notes 17
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
• 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
Thermodynamics Unit III Notes 18
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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
Thermodynamics Unit III Notes 18
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• for a work consuming device irreversibility =wasted work
Thermodynamics Unit III Notes 19
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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
Thermodynamics Unit III Notes 19
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=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)
Thermodynamics Unit III Notes 19
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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.
Thermodynamics Unit III Notes 20
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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.
Thermodynamics Unit III Notes 20
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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
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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.
Thermodynamics Unit IV Notes 21
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• 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,
Thermodynamics Unit IV Notes 21
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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
Thermodynamics Unit IV Notes 22
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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.
Thermodynamics Unit IV Notes 22
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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
Thermodynamics Unit IV Notes 22
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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
Thermodynamics Unit IV Notes 23
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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)
.
Thermodynamics Unit IV Notes 23
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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.
Thermodynamics Unit IV Notes 23
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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
Thermodynamics Unit IV Notes 24
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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
Thermodynamics Unit IV Notes 24
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2. Constant pressure or Isochoric Process (V = C): An Isometric process is
internally reversible constant volume process.
Closed System: (Nonflow)
Q = ∆U + W → 1 any substance
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
Thermodynamics Unit IV Notes 24
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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)
Q = ∆U + W → 1 any substance
W = ∫PdV → 2 any substance
∆U = m(U2 - U1) → 3 any substance
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
Thermodynamics Unit IV Notes 25
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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.
Closed System (Nonflow)
Q = ∆U + W → 1 any substance
W = ∫PdV → 2 any substance
∆U = m(U2 - U1) → 3 any substance
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
Thermodynamics Unit IV Notes 25
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit IV Notes 25
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit IV Notes 26
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
Thermodynamics Unit IV Notes 26
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
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
=
== ;
Thermodynamics Unit V Notes 27
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 28
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 28
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 28
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 29
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 29
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 30
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 31
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
Non – flow processes
1. 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
Thermodynamics Unit V Notes 31
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
2. Isometric Process (V = C): An Isometric process is internally reversible constant
volume process.
Closed System: (Nonflow)
Q = ∆U + W → 1 any substance
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
Thermodynamics Unit V Notes 31
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
W = mR(T1-T2) → 16 ideal gas
∆h = mCP(T2-T1) → 17 ideal gas
Thermodynamics Unit V Notes 32
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
Non – flow processes
3. Isothermal Process(T = C): An Isothermal process is reversible constant temperature process.
Closed System (Nonflow)
Q = ∆U + W → 1 any substance
W = ∫PdV → 2 any substance
∆U = m(U2 - U1) → 3 any substance
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
where S = C and PVk = C for an ideal or perfect gas.
Thermodynamics Unit V Notes 32
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
Closed System (Nonflow)
Q = ∆U + W → 1 any substance
W = ∫PdV → 2 any substance
∆U = m(U2 - U1) → 3 any substance
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
Thermodynamics Unit V Notes 32
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 33
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
Non – flow processes
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
Thermodynamics Unit V Notes 33
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 34
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.
Thermodynamics Unit V Notes 34
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.
Thermodynamics Unit V Notes 35
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit V Notes 35
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
Open System: Q = ∆h + ∆KE + ∆PE + W → 7 any substance
W = - ∫VdP - ∆KE - ∆PE → 8 any substance
-∫VdP = -V(P2-P1) → 9 any substance
Q = ∆U = m(U2 - U1) → 10 any substance
∆h = m(h2-h1) → 11 any substance
3. Isothermal Process(T = C): An Isothermal process is reversible onstant temperature process.
Thermodynamics Unit V Notes 35
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
Open System (Steady Flow) Q = ∆h + ∆KE + ∆PE + W → 10 any substance
W = - ∫VdP - ∆KE - ∆PE → 11 any substance
-∫VdP = -V(P2-P1) → 12 any substance
∆h = m(h2-h1) → 13 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
If ∆KE = 0 and ∆PE = 0
Q = ∆h + W → 17 any substance
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
where S = C and PVk = C for an ideal or perfect gas.
Thermodynamics Unit V Notes 35
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
Open System (Steady Flow) Q = ∆h + ∆KE + ∆PE + W → 10 any substance
W = - ∫VdP - ∆KE - ∆PE → 11 any substance
∆h = m(h2-h1) → 12 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
Thermodynamics Unit V Notes 35
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
If ∆KE = 0 and ∆PE = 0
0 = ∆h + W → 17 any substance
W = - ∫VdP = - ∆h → 18 any substance
∆h = m(h2-h1) → 19 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
Thermodynamics Unit V Notes 35
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 35
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
( )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
Thermodynamics Unit V Notes 36
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 36
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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:
Thermodynamics Unit V Notes 36
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
* 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.
Thermodynamics Unit V Notes 37
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 37
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit V Notes 38
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 38
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 38
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit V Notes 38
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
-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
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit VI Notes 39
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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)
Thermodynamics Unit VI Notes 39
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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=
Thermodynamics Unit VI Notes 40
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
++=
++=
++=
++=
Thermodynamics Unit VI Notes 40
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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 ===
Thermodynamics Unit VI Notes 40
GMRIT , Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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 =
Thermodynamics Unit VI Notes 41
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
∑∑
∑∑
∑
−=
−=
=
Thermodynamics Unit VI Notes 41
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit VI Notes 42
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit VI Notes 42
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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)
Thermodynamics Unit VI Notes 42
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit VI Notes 43
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit VI Notes 43
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit VI Notes 43
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit VI Notes 43
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit VI Notes 43
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
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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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444444444444
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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))
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444444444444
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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]]
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444444444444
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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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444444444444
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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))
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444444444444
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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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444455555555
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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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444455555555
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== 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 ))
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444455555555
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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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444455555555
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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 ]]
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444455555555
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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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444455555555
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444455555555
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444455555555
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444466666666
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444466666666
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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))
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444466666666
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444466666666
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
== 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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444466666666
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444466666666
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444466666666
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444466666666
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444477777777
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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::--
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444477777777
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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]]
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444488888888
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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
TTTTTTTThhhhhhhheeeeeeeerrrrrrrrmmmmmmmmooooooooddddddddyyyyyyyynnnnnnnnaaaaaaaammmmmmmmiiiiiiiiccccccccssssssss UUnniitt VVIIII NNNNNNNNooooooootttttttteeeeeeeessssssss 4444444488888888
GGMMRRIITT,, RRaajjaamm CChhiirraannjjeeeevvaa RRaaoo SSeeeellaa
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::
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Thermodynamics Unit VIII Notes 49
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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).
Thermodynamics Unit VIII Notes 49
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit VIII Notes 50
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit VIII Notes 50
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
region after the expansion in the turbine, [1] which reduces the energy removed by the
condensers.
Thermodynamics Unit VIII Notes 51
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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)
Thermodynamics Unit VIII Notes 51
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit VIII Notes 52
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit VIII Notes 52
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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 )
Thermodynamics Unit VIII Notes 52
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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.
Thermodynamics Unit VIII Notes 53
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
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
Thermodynamics Unit VIII Notes 53
GMRIT, Rajam Rajam Rajam Rajam Chiranjeeva Rao hiranjeeva Rao hiranjeeva Rao hiranjeeva Rao Seelaeelaeelaeela
• 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
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