basic concet of thermodynamics

Basics of Mechanical Engineering (B.M.E) Brown Hill College of Engg. & Tech.

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UNIT - 2

Basic Concept of Thermodynamics

Syllabus: Introduction, States, Work, Heat, Temperature, Zeroth, 1st 2nd and 3rd law of Thermodynamics,

Concept of internal energy, enthalpy and entropy. Problems Properties of Steam & Steam Generator: Formation

of steam at constant pressure, Thermodynamic properties of Steam, Use of steam tables, Measurement of dryness

fraction by throttling colorimeter

I n t r o d u c t i o n

Thermodynamics is the branch of science which deals with energy transfer & its effect on the state or

condition of a system. In this, we study:-

Interaction of system & surroundings

Energy & its transformation

Relationship b/w heat, work & physical properties such as pressure, volume & temperature of the

working substance employed to obtain energy conversion.

Feasibility of a process & the concept of equilibrium processes.

It is the science of 3 “Es”

a. Energy

b. Entropy

c. Equilibrium

The laws, principal & concept of thermodynamics are important & indispensable tools in the innovation, design,

development & improvement of engg... process, equipment & devices which deals with effective utilization of

energy. Application of engg... thermodynamics in the field of energy technology is:

Power producing devices eg internal combustion engine etc

Power consuming devices eg fans, blowers etc

Chemical producing plants & direct energy conversion devices.

a large number of processes in various fields such as agriculture, textile, dairy, drugs & pharmaceutical industry

are also governed by thermodynamic principles.

Thermodynamics is essentially based upon experimental result & observations of common experience;

there is no mathematical proof to the Zeroth, 1st & 2nd laws of thermodynamics.

L i m i t a t i o n s o f T h e r m o d y n a m i c s

1. The laws of thermodynamics are applicable only to matter in bulk, i.e. assemblage of large number of

molecules (macroscopic systems), and not to individual molecules of atoms. Thus thermodynamics is

independent of atomic and molecular structure of matter, i.e. it ignores the internal structure of atoms and

molecules.

2. Although thermodynamics predicts about the feasibility, direction and extent of a given process under

a given set of conditions, it does not tell anything about the rate at which a given process may proceed,

i.e. it provides no information regarding the time taken to reach equilibrium.

3. It concerns itself only with the initial and final states of a system and gives no information about the

path taken (mechanism) by a process.



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I m p o r t a n t T e r m s a n d D e f i n i t i o n s

1. System: A portion of universe which is under

investigation, e.g., portion of test tube where

reaction is taking place, is called system.

2. Surroundings: Everything else in the universe

except system is called surroundings, e.g., except

the portion of the test w041here reaction is taking

place is surrounding, i.e. above and around, the

test tube.

3. Boundary: the thermodynamic system &

surrounding separated by an envelope known as

boundary.

4. Open system: The system which can exchange matter and energy with the surrounding is called open

system, e.g., a cup of tea is an open system because it will become cold as well as its taste will also change,

i e., it is exchanging energy and matter with the surrounding.

5. Closed system: It can exchange energy but not matter with the surroundings, e.g., tea placed in a closed

kettle.

6. Isolated system: It can neither exchange heat nor matter with the surroundings, e.g., tea placed in thermos

flask.

7. Homogeneous system: A system is homogeneous when it is completely uniform throughout, i.e. consists

of one phase only, e.g., a pure solid, a pure liquid, a true solution, etc.

8. Heterogeneous system: A system is heterogeneous when it is not uniform throughout, i.e. it consists of

two or more phases, e.g., a mixture of two solids or two or more immiscible liquids, a solid in contact with

a liquid, a liquid in contact with its vapours.

9. Macroscopic system: When system consists of large number of atoms, molecules or ions, it is called

macroscopic system.

10. Microscopic properties: The properties associated with a macroscopic system, i.e. pressure, volume,

temperature, density, composition, viscosity, surface tension, refractive index, colour, etc.

11. State of the system: A system is said to be in definite state when its macroscopic properties have definite

values.

Or

It is the condition of the system at an instant of time as described or measured by its properties or each

unique condition of a system is called state.



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1 2 . State variables: The macroscopic properties of a system are state variables since change in any of these

properties, causes the system to change into another state.

13. Properties of steam a) Extensive property of a system: It is the property which depends upon mass of the system, e.g.,

volume, pressure, internal energy, free energy, etc.

b ) Intensive property: The property which does not depend upon mass of the system, e.g.,

temperature, density, viscosity, refractive index, surface tension, specific heat, pressure etc.

14. Thermodynamic equilibrium: if the temperature & pressure at all points are same; there should be no

velocity gradient; the chemical equilibrium is also necessary. System under temperature & pressure

equilibrium but not under chemical equilibrium are sometimes said to be in metastable state equilibrium

conditions. It is only under thermodynamic equilibrium condition that the properties of a system can be

fixed.

Thus for attaining a state of thermodynamic equilibrium the following three types of equilibrium

states must be achieved:

a) Thermal equilibrium: A system is said to be in thermal equilibrium, when there is no

temperature difference between the parts of the system or between the system and the

surrounding.

b) Mechanical equilibrium: a system is said to be in mechanical equilibrium, when there is no

unbalance force acting on any part of the system as a whole.

c) Chemical equilibrium: a system is said to be in chemical equilibrium, when there no chemical

reaction within the system and also there is no movement of any chemical constituent from one

part of the system to the other.

15. Isothermal process: The process in which temperature of the system remains constant at each stage of the

process i.e., dT or ∆T = 0

16. Exothermic process: The process in which heat is given out to the surroundings is called exothermic

process. In this process, products are more stable than reactants because they have lower energy.

17. Endothermic process: The process in which heat is absorbed by the system from the surroundings. In this

process, products are less stable than reactants because they have higher energy.

18. Adiabatic process: The process in which no exchange of heat takes place between system and

surroundings, i.e. dq = 0 or q = 0

Note:

a. In adiabatic expansion of an ideal gas, the internal energy of the system decreases and the temperature

falls.

b. In adiabatic compression of an ideal gas, internal energy of the system increases and the temperature

rises.



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19. Isobaric process: The process which takes place at constant pressure is called an isobaric process, e.g.,

heating of water to its boiling point and its vaporization occurs at the same atmospheric pressure.

Expansion of a gas in an open system is also an example of isobaric process in which dP or ∆P = 0

20. Cyclic process: In cyclic process, the system in a given state goes

through a series of different processes, but in the end returns to its initial

state. dE or ∆E = 0, dH or ∆H = 0

21. Isochoric process: The process which is carried out at constant volume is called isochoric process, i.e., dV

or ∆V = O. e.g., ∆E, measurement at constant volume is isochoric process.

22. Reversible process: A reversible process is one which can be stopped at any stage and reserved so that the

system and surrounding are exactly restored to their initial states.Eg:

a) Frictionless relative motion.

b) Expansion and compression of spring.

c) Isothermal expansion or compression.

23. Irreversible process: An irreversible process is one in which heat is transferred through o finite

elements.eg:

a) Relative motion with friction.

b) Combustion.

c) Diffusion.

d) Heat transfer.

24. State function: The thermodynamic property which depends upon initial and final state of the system not

on the means how the state is reached are state functions.

Work It refers to the energy transferred because of a property difference, other than temp .., that exist b/w the system &

its surrounding.



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Piston cylinder arrangement

Consider a system formed by computer by a certain amount of a gas contained in a piston cylinder arrangement.

Let the piston moves outward through a small distance dx(dx 0) during a small interval dt. Since the piston

moves a small distance, the pressure p acting on the face of the face of the piston can be assumed to remain

constant. A single pressure gauge will then indicate the system pressure p, which also prevails over the entire

system boundary. The infinitesimal work done by the system is

δw = force * distance moved

δw = (pressure * area of piston) * distance moved

δw = p.A.dx

δw = p.dv 1

(dv = A.dx)

if the piston moves through a finite distance, the W.D by the piston can be evaluated by integrating 1

b/w the initial & final states.

The expansion of the system (gas) by outward movement of the piston can be represented on P-V

diagram. The shaded area represents the W.D due to small movement of the piston & the total W.D by the gas

during expansion process is

Work is a path function & not a point function.

Sign convention

+W = W.D by the system

-W = W.D on the system



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When a system does +ve work, its surrounding do equal amount of –ve work, the system does an equal

amount of –ve work. Thus in any process

Wsystem + Wsurrounding = 0 3

E.g.

A gas undergoes a reversible non-flow process according to the relation P = (-3V +15) where V is the volume in

m3 & P is the pressure in bar. Determine the W.D when the volume changes from 3 to 6 m

3 .

Solu :- given P = (-3V +15) bar = (-3V +15) * 105

N/ m2

= [(-40.5 + 45) *10

5

= 40.5*105

Nm (or J)

T e m p e r a t u r e

It is a thermal state of a body which distinguishes a hot body from a cold body. A particular molecules does not

have a temperature, it has energy. The gas as a system has temperature .

Instruments for measuring ordinary temperature are known as thermometer & those for measuring higher

temperature are known as pyrometers.

Heat

It is a form of energy that is transferred (without transfer of mass) occur the boundaries of a system because of

temperature difference b/w the system & surrounding, & it was done from higher to lower temperature body.

No heat interaction is possible if system & surrounding are in thermal equilibrium.

Sign convention

1) Heat transfer to the system = +ve

2) Heat transfer to the system = -ve

S p e c i f i c h e a t & h e a t c a p a c i t y

The specific heat may be defined as the quantity of heat that enters or leaves a unit mass of the substance

when it experience one degree change in temperature.

The quantity of heat rejects or received by a system during the process of heating or cooling is measured

in terms of the product of mass, specific heat & temperature change.

Q = m.c (t2-t1)

Where ,

Q = heat gained by system in KJ

M = mass of the substance in kg

(t2-t1) = temp.. rise in degree centigrade or Kelvin

C = specific heat in KJ/kg k

The product(m*c) is separately termed as heat capacity.

For water c = 4.186 KJ/kg-deg

For air cp = 1.005 KJ/kg-deg

Cv = 0.718 KJ/kg-deg

Adiabatic exponent (γ) =cp/cv for air γ = 1.4



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E.g.During a particular process, the specific heat of the working fluid comprising a system is given by the

relation C = (0.2 + 0.002T) KJ/kg K.

Where T is the absolute temperature in degree K. what amount of heat is required to be supplied to this

fluid system to raise its temperature from 300 to 400 K? Also determine the mean value of specific heat.

Solu: the total heat interaction for change in temperature from T1 to T2 :

Q = 180KJ

The specific heat is given by

Q = mcm(t2-t1)

180 = 2cm(400-300)

cm = 0.9 KJ/kg K

L a w s o f t h e r m o d y n a m i c s

There are four laws of thermodynamics which are:-

Zeroth Law of Thermodynamics

1st law of thermodynamics

Second law of thermodynamics

Third law of thermodynamics

Z e r o t h L a w o f T h e r m o d y n a m i c s Two bodies are said to be in thermal equilibrium if no transfer of heat takes place when they are placed in

contact. We can now state the Zeroth law of thermodynamics as follows:

If two bodies A and B are in thermal equilibrium and A and C are also in thermal equilibrium then B and

C are also in thermal equilibrium.

It is a matter of observation and experience that is described in the Zeroth law". It should not be taken as

obvious. For example, if two persons A and B know each other and A and C know each other, it is not necessary

that Band C know each other.

Thermal Equilibrium

Body A Body B

Thermal Equilibrium Thermal Equilibrium

Body C

The Zeroth law allows us to introduce the concept of temperature to measure the hotness or coldness of a body.

All bodies in thermal equilibrium are assigned 'equal temperature. A hotter body is assigned higher temperature

A

C

B



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than a colder body. Thus, the temperatures of two bodies decide the direction of heat flow when the two bodies

are put in contact. Bent flows from the body at higher temperature to the body at lower temperature.

F i r s t l a w o f t h e r m o d y n a m i c s

“It states that, energy can neither be created nor be destroyed, it changes only its form”.

Or

“When a system undergoes a thermodynamic cycle then the net heat supplied to the system from the

surrounding is equal to net work done by the system on its surrounding”.

Or

“Cyclic integral of work transfer is equal to cyclic integral of heat transfer”

Experiment to verify firsrt law of thermodynamics

The work input to the paddle wheel as measured by the fall of weight; while corresponding temperature rise of

liquid in the insulated container is measured by the thermometer. it is already known to us from experiment on

heat transfer that temperature rise can also be produced by heat transfer. This experiment shows that:

a. A definite quantity of work is always required to accomplish the same temperature rise obtained with a

unit amount of heat.

b. Regardless of wheather temperature of liquid is raised by work transfer or heat transfer, the liquid can be

returned by heat transfer in opposite direction to the identical state from which it started.

E 2 = E 1 + q + w

E 2 – E 1 = q + w

∆ E = Q + W is mathematical statement of 1st law of thermodynamics

'q' is not a state function, w is not a state function but q + w = ∆E is a state function.

a). For isothermal process: ∆E = 0, Q = - W, i.e., heat absorbed by system is equal to work by the system.

b). For isochoric process: ∆V = 0, W = P ∆V = 0, ∆E = q", i.e., change in internal energy is e to heat

exchanged with the surroundings at constant volume.

c ) . For adiabatic process: Q = 0, ∆E = W, i.e., internal energy change is equal to work done on the system.



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L i m i t a t i o n : energy can flow from from higher temperature to low temperature body but not able to flow

from lower to higher without assistance it sets no limit to the amount of total energy of a system which can be

caused to flow out as work or does not specify the condition under which conversion of heat into work is

possible.

S e c o n d l a w o f t h e r m o d y n a m i c s

It accommodates two laws inside it which are clasius, Kelvin and Planck in slightly different words although

both statements are basically identical and based on irreversible process. The first consider transformation of heat

b/w two thermal reservoir while the second one transformation of heat into work.

Clasius statement : According to this theory “it is impossible for a self acting machine working in a cyclic

process, to transfer heat from a body at a lower temperature to a body at a higher temperature body

without the aid of external agency.”

Schematic representation of a heat pump/refrigerator



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Kelvin - Planck statement

According to this theory “it is impossible to construct

an engine working on a cyclic process, whose sole

purpose is to convert heat energy from a single

thermal reservoir into equivalent amount of work”

Schematic representation of a heat engine

PERPETUAL MOTION MACHINE OF THE SECOND KIND (PMM2)

A machine which violates the first law of thermodynamics is called perpetual motion machine of the first kind

(PMM1).such a machine creates its own energy from nothing and does not exist.

Q

W = Q

Without violating the first law, a machine can be imagined which would continuously absorb heat from a

single thermal reservoir and convert this heat completely into work. The efficiency of such a machine would be

100%.this machine is called PMM2.

A practical heat engine exchanges heat with two thermal reservoirs (source & sink) & its thermal

efficiency is given by:

T h i r d l a w o f t h e r m o d y n a m i c s

The degree of atomic or molecular activity of a substance depends upon its temperature. As absolute zero

temperature is approached, the randomness of molecules ceases & the entropy becomes equal to 0.

At absolute zero temperature, the entropy of all homogeneous crystalline (condensed) substance in a state

of equilibrium becomes 0.

Thermal

Reservoir

Perpetual motion

machine



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This implies that at absolute zero temperature, the entropy ceases to be a function of state & approaches a

constant value independent of the parameters of the state.

This law is helpful in

Measurement of chemical affinity, i.e., the action of chemical forces of the reacting substance.

Explaining the behaviour of solids at very low temperature.

Analyzing the chemical & phase equilibrium.

Sign conventions:

1 Q is +ve when heat is absorbed, Q is -ve if heat is evolved

2 W is +ve when work is done on the system, W is -ve when work is done by the system

C o n c e p t o f I n t e r n a l e n e r g y

It is sum of all the forms of energies that a system can possess. It made of 6 components:

a. Energy due to transitional motion of molecules

b. Energy due to rotational motion of molecules

c. Energy due to vibration motion of molecules

d. Bond energy

e. Energy due to intermolecular force of attraction

f. Energy due to electrostatic force of attraction between nucleus and electrons

Note:

a. Internal energy is an extensive property & state function..

b. It depends upon mass, temperature, pressure and physical state of the system.

c. The absolute value of internal energy of a substance cannot be determined.

d. ∆E can be experimentally determined at constant volume with the help of bomb calorimeter.

e. E = mol-I for monoatomic gas.

E n t h a l p y ( H )

It is defined as total heat content of the system i.e. sum of internal energy and product of pressure & Volume.

The total enthalpy of mass m, of a fluid can be

H = U + P V

H is state function. The enthalpy of a fluid is the property of the fluid , since it consist of the sum of a property &

the product of the product of the two properties. Since enthalpy is a property like internal energy, pressure,

specific volume & temperature, it can be introduced into any problem wheather the process is a floe or a non-

flow process.

The total enthalpy of mass, m, of a fluid can be

H = U + pV, where H = mh

Ratio of specific heats

It is the ratio of specific heat at constant pressure to the specific heat at constant volume & is represented by

γ ( g a m m a ) .



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i . e

Since cp = cv + R, it is clear that cp must be greater than cv for any perfect gas. It follows; therefore, that γ

is always greater than unity. Values of γ for

1 Monoatomic(argon, helium) γ = 1.6

2 diatomic(CO2, hydrogen, oxygen & nitrogen) γ = 1.4

3 Triatomic(CO2, SO2) γ = 1.3

E n t h a l p y c h a n g e ( ∆ H )

It is defined as heat exchanged with the surrounding at constant pressure i.e. when process is carried out

in open container.

∆H = H2 – H1

H1 = E1 + PV1 …(1)

and H2 = E2 + PV2 …(2)

substract (1) from (2), we get

H2 – H1 = E2 – E1 + P (V2 – V1)

∆H = ∆E + P∆V

(PV = n RT, P∆V - ∆n RT)

∆n is the difference of no. of moles of gaseous product & no. of moles of gaseous reactants)

If ∆n = 0, ∆H = ∆E

If ∆n > 0, ∆H > ∆E

If ∆n < 0, ∆H < ∆E

E n t r o p y

It is a function of a quantity of heat which shows the possibility of conversion of that heat into work.. If a

system has a temperature T (in absolute scale) and a small amount of heat ∆Q is given to it, we define the change

in the entropy of the system as

In general, the temperature of the system may change during a process is reversible, the change in entropy is

defined as

Where Sf = final entropy

Si = initial entropy

∆H = ∆E + ∆n RT

∆S =



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Entropy is a property of the system that remains constant during a reversible adiabatic process.

The reversible adiabatic process during which entropy remains constant is called isentropic process.

An interesting fact about entropy is that it is not a conserved quantity. More interesting is the fact that

entropy can be created but cannot be destroyed.

It is thermodynamic property which measures randomness or disorder of a system. The more disorder

or randomness, higher will be entropy, e.g., solid < liquid < gas. Entropy of a system is state function, i.e., it

depends upon initial and final states of the system, When the state of a system changes, the entropy also changes.

∆S = where „q‟ is heat supplied isothermally, T is absolute temperature.

For a irreversible spontaneous process ∆S = +ve

For a reversible change at equilibrium ∆S = 0

The entropy of a system increases with absorption of heat & decreases with heat rejection. Its unit is KJ/K

while specific entropy is KJ/KgK.

Thermodynamic Properties of Steam

There are six basic thermodynamic properties are required

1. P (pressure)

2. T (temperature)

3. V (volume)

4. U (internal energy)

5. H (enthalpy)

6. S(entropy)

These properties must be known at different pressure for analyzing the thermodynamic cycles used for work

producing devices.

The values of these properties are determined theoretically or experimentally and are tabulated in the form of

tables which are known as 'Steam Tables'.

Following are the thermodynamic properties of steam which are tabulated in the form of table:

p = Absolute pressure (bar or KPa)

ts = Saturation temperature (0C)

hf = Enthalpy of saturated liquid (kJ/kg)

hfg = Enthalpy or latent heat of vaporization (kJ/kg)

hg = Enthalpy of saturated vapour (steam) (kJ/kg)

Sf = Entropy of saturated liquid (kJ/kg K)

Sfg = Entropy of vaporization (kJ/kg K)

Sg = Entropy of saturated vapour (steam) (kJ/kg K)

Vf = Specific volume of saturated liquid (m3/kg)

Vg = Specific volume of saturated vapour (steam)

(m3/kg)

Also, hfg = hg - hf ……Change of enthalpy during evaporation

Sfg = Sg - Sf ……Change of entropy during evaporation

Vfg = Vg - Vf ……Change of volume during evaporation

Table as per pressure

Absolut

e

pressur

e (p) in

Tempe

rat-ure

(t) in oc

Specific volume

in m3/kg

Specific enthalpy in KJ/kg Specific entropy in KJ/kg

k

Water

(vf)

Steam

(vg)

Water

(hf)

Evaporation

(hfg)

Steam

(hg)

Water

(sf)

Evapor

a-tion

Steam

(sg)



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bar (sfg)

0.010 6.983 0.001000 129.21 29.3 2485.1 2514.4 0.106 8.871 8.977

0.015 13.04 0.001001 87.982 54.7 2470.8 2525.5 0.196 8.634 8.830

0.20 60.09 0.001017 7.649 251.5 2358.4 2609.9 0.832 7.077 7.909

Table as per temperature

Temper

at-ure

(t) in oc

Absolu

te

pressur

e (p) in

bar

Specific volume

in m3/kg

Specific enthalpy in KJ/kg Specific entropy in KJ/kg

k

Water

(vf)

Steam

(vg)

Water

(hf)

Evaporation

(hfg)

Steam

(hg)

Water

(sf)

Evapor

a-tion

(sfg)

Steam

(sg)

0 0.00611 0.001000 206.31 0.0 2501.6 2501.6 0.000 9.158 9.158

5 0.00872 0.001000 147.16 21.0 2489.7 2489.7 0.076 8.951 9.027

10 0.01227 0.001000 106.43 42.0 2477.9 2477.9 0.151 8.751 8.902

F o r m a t i o n o f S t e a m

Consider a cylinder fitted with a piston which can move freely upwards and downwards in it. Let, 1 kg of water

at O°C with volume Vf m3 under the piston. Further let the piston is loaded with load W to ensure heating at

constant pressure. Now if the heat is imparted to water, a rise in temperature will be noticed and this rise will

continue till boiling point is reached. The temperature at which water starts boiling depends upon the pressure

and as such for each pressure (under which water is heated) there is a different boiling point. This boiling

temperature is known as the temperature of formation of steam or saturation temperature. During heating up

to boiling point that there will be slight increase in volume of water due to which piston moves up and hence

work is obtained but this work is so small that it is neglected.

Now, if supply of heat to water is continued it will be noticed that rise of temperature after the boiling point is

reached nil but piston starts moving upwards which indicates that there is increase is volume which is only

possible if steam formation occurs. The heat being supplied does not show any rise of temperature but changes

water into vapour state (steam) & is known as latent heat. So long as the steam is in contact with water, it is

called wet steam (fig iii) & if heating is further progressed (fig iv) such that all the water particles associated

with steam are evaporated , the steam so obtained is called dry & saturated steam.

If vg m3

is the volume of 1kg of dry and saturated steam then work done on the piston will be

Where p is the constant pressure (due to piston weight „W‟)

WD = P(vg-vf)



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P h a s e o f F o r m a t i o n s t e a m

Again, if supply of heat to the dry & saturated steam is continued at constant pressure there will be

increase in temperature & volume of steam.the steam so obtained is called superheated steam & behaves like a

perfect gas.

Graphical Representation of Formation of Steam

1 . S e n s i b l e H e a t o f W a t e r (hf): It is defined as the quantity of heat absorbed by 1kg of water when it is

heated from 00C (freeing point) to boiling point.also called total heat (or enthalpy) of water or liqid or lliquid

heat invariably.

h s u p = h f +h f g +c p s ( t s u p - t s )



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2 . L a t e n t H e a t of vaporization (hfg): It is the amount of heat required to convert water at a given

temperature and pressure into steam at the same temperature and pressure.

Or It is the quantity of heat required to convert 1 kg of water at saturation temp for a given pressure into dry &

saturated steam at that temperature & pressure. The value of latent heat is not constant, it decreases as the

pressure increases & becomes zero when the critical pressure is reached.

Depending upon the condition of steam, the enthalpy or heat values are defined as :

(i) Total heat of wet steam : It represents the quantity of heat required to covert 1 kg of water at 0oC into

wet steam at constant pressure

hwet = hf + x hfg where „x‟ is the dryness fraction of wet steam

The sensible heat increases with pressure.

(ii) Total heat of dry saturated steam : It represents the quantity of heat required to covert 1 kg of water at

0oC into dry saturated steam at constant pressure

hs = hf + hfg = hg The enthalpy of dry saturated steam increases with pressure.

(iii) Total heat of superheated steam : It represents the quantity of heat required to covert 1 kg of water at

0oC into superheated steam at constant pressure

hsup = hf + hfg +cps(tsup + ts) the value of specific heat of steam at constant pressure cps depends upon the degree of superheat &

the pressure of steam generation. Its average value is taken from 2.0 to 2.1 kJ/kg.

3 . D r y n e s s F r a c t i o n (x): It is defined as the ratio of the mass of dry saturated steam to the Total mass of

mixture.

Then,

where, ms = Mass of dry saturated steam, and

mw = Mass of water vapour is suspension with saturated dry steam,

ms + mw = Total mass of mixture

5 . S a t u r a t e d S t e a m : When the molecules escaping from the liquid becomes equal to the molecules

returning to it.

6 . S a t u r a t e d L i q u i d : The liquid at its boiling point at a specified pressure is called a saturated liquid.

7 . S a t u r a t e d V a p o u r : Saturated steam may be dry or wet, when the saturated vapour contains particles of

liquid evenly distributed over the entire mass of vapour, it is called wet saturated steam.

8.superheated steam:-when steam is heated after it has become dry & saturated , it is called superheat steam &

the process of heating is called superheating. It is carried out at constant pressure. & the additional heat supplied

during this is „heat of superheat‟. It can be calculated as

h s u p = h f +h f g +c p s ( t s u p - t s )



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T h r o t t l i n g C a l o r i m e t e r

The dryness fraction of wet steam can be determined by using a throttling calorimeter.

Working: The steam to be sampled is taken from the pipe by means of suitable positioned and dimensioned

sampling tube. It passes into an insulated container and is throttled through an orifice to atmospheric pressure.

Here the temperature is taken & the steam ideally should have about 5.5K of superheat.

Throttling calorimeter Throttling process

Throttling process is shown on enthalpy & entropy i.e. h-s diagram by the line 1-2. If steam initially wet is

throttled through a sufficiently large pressure drop, then the steam at state 2 will become superheated. State 2 can

then be defined by the measured pressure and temperature. The enthalpy, h2 can then be found and hence

h2 = h1 = (hf1 + Xl hfg1) at P1

[Where h2 = hf2 + hfg2 + cps (Tsup2 – Ts2)]

X1 = (h2 - hf1)

hfg1

Hence the dryness fraction is determined and state 1 is defined.

E.g. A throttling calorimeter is used to measure the dryness fraction of the steam in the steam main which has

steam flowing at a pressure of 8 bar. The steam after passing through the calorimeter is at a pressure & 115oC.



18

Calculate the dryness fraction of the steam in the main. Take cps = 2.1kJ/kg K.

Solu : Condition of steam before throttling :

P1 = 8 bar, x1 =?

Condition of steam after throttling:

P2 = 1 bar, t2 = tsup = 115 oC

As throttling is a constant enthalpy process

h1 = h2

i.e., hf1 + xl hfg1 = hf2 + hfg2 + cps (Tsup2 – Ts2)

[Tsup = 115+273 = 388 K, Ts2 = 99.6 + 273 = 372.6(at 1 bar)]

720.9 + xl * 2046.5 = 417.5 + 2257.9 + 2.1(388 – 372.6)

720.9 + xl * 2046.5 = 2707.7

Hence, dryness fraction of steam in the main = 0.97.

basic concet of thermodynamics

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