unit-i basic concept of thermodynamics basic …
TRANSCRIPT
UNIT-I
BASIC CONCEPT OF THERMODYNAMICS
BASIC DEFINITION -Thermodynamics is the field of science which deals with the
relationship among heat, work and properties of system which are in equilibrium with one
another. It is a characteristic of the system. The system is identified by some quantise like
temperature, pressure, volume etc.
CLASSIFICATIONS OF THERMODYNAMICS
I. Classical Thermodynamics –Macroscopic Approach
II. Statistical Thermodynamics -Microscopic Approach
Macroscopic Approach-Macroscopic Approach is the study of thermodynamics which does not
require any knowledge of the behaviour of individual particles of the substance is called classical
thermodynamics
Microscopic Approach- Microscopic Approach is the study of thermodynamics based on the
behaviour of a large numbers of individual particles is called statistical thermodynamics.
INTENSIVE AND EXTENSIVE PROPERTIES
Intensive Properties -The properties are independent on mass of the system or independent on
size of the system.
Example: Pressure, Temperature, Specific volume etc.
Extensive properties- The properties are dependent of the mass of the system or extensive
properties are those which vary directly with the size or extent of the system.
Example: Total volume, total energy
SYSTEM SURROUNDINGS AND BOUNDARY
In thermodynamics we confine our attention to a particular part of the universe which we call our
system. A system is a region of space containing a quantity of matter whose behavior is being
investigated. This quantity of matter is separated from the surroundings by a boundary, which
may be a physical boundary like walls of a vessel, or some imaginary surface enveloping the
region. The term surroundings is restricted to those portions of the matter external to the system,
which are affected by changes occurring within the system. Before any thermodynamic analysis
is attempted, it is necessary to define the boundary of the system because it is across the
boundary that work, heat and mass are said to be transferred
Surrounding - The rest of the universe outside our system is call the surroundings.
Boundary-The system and the surroundings are separated by a boundary or a wall. They may,
in general, exchange energy and matter, depending on the nature of the wall.
TYPES OF THERMODYNAMICS SYSTEMS
Closed system – When the same matter remains within the region throughout the process under
investigation it is called closed system. In this case, only heat and work cross the boundary mass
does not transfer,
Open system- An open system is a region in space defined by a boundary across which the
matter may flow in addition to work and heat, it meant both mass and energy can transfer
Isolated system-The isolated system is one in which there is no interaction between the system
and surroundings. There is no mass or energy transfer across the system boundary.
Examples of open system- flow nozzles, steam turbine, boiler etc.,
Examples of closed system- mixer of water and steam in a closed vessel, a gas expanding in a
cylinder by displacing a piston. Hence, for a closed system, boundary need not be fixed; it may
contract or expand to accommodate any change in volume undergone by a fixed quantity of
fluid.
Examples of isolated system -entire universe of the system
EQUILIBRIUM OF THE SYSTEM
When a system is said to be in equilibrium, it involves no change with time. Also it can
be defined as a system is said to be in equilibrium, if it does not tend to undergo any change of
state on its own accord.
When a system is in Thermodynamic equilibrium, it should be satisfy the following three
conditions.
(a) Mechanical Equilibrium – Pressure remains constant. No pressure difference exists
between any two points within the system (Neglecting gravitational effects) and between the
system and surroundings, so that it is mechanically balanced.
(b)Thermal Equilibrium – Temperature remains constant. There should not be any temperature
difference within the system, so that the system is thermally balanced.
(c) Chemical Equilibrium – There is no chemical reaction of the system. No chemical reaction
is taking place, so that it is chemically balanced.
STATE, PROPERTY, PATH, PROCESS
Every system has certain characteristics by which its physical condition may be
described, e.g. volume temperature, pressure etc. Such characteristics are called properties of the
system. These are all macroscopic in nature. When all the properties of a system have definite
values, the system is said to exist at a definite state. Any operation in which one or more
properties of a system change is called a change of state. The succession of states passed through
during a change of state is called the path of the change of state. When the path is completely
specified, the change of state is called a process, e.g. a constant pressure process. The value of
property does not depend on process through which the fluid is passed. The change in the value
of property depends on the initial and final states of the system. Pressure, specific volume and
temperature are some examples of basic properties. Three more properties- internal energy,
enthalpy and entropy emerge as a consequence of First and Second Laws of Thermodynamics.
From these six properties, only two may be selected to determine the state of a closed system in
thermodynamic equilibrium and the remaining four values are then fixed. Care must be taken to
see that the two properties are independent of each other, i.e. it must be possible to vary one of
these properties without changing the other For example, when a liquid is in contact with its
vapour in a closed vessel it is found that the temperature at which the liquid and vapour in
equilibrium is always associated with a particular pressure and one cannot change one without
the other. Pressure and temperature cannot be used to determine the state of such systems.
However, pressure and specific volume may be used to define the state of such system. It follows
that the initial and final states of any closed system can be located as points on a diagram using
two properties as coordinates. Properties may be of two types. Intensive (Intrinsic) properties are
independent of mass of the system, e.g. pressure, temperature, etc. Extensive properties are
related to the mass of the system, e.g. volume, energy etc. A process that eventually returns to its
initial state is called a cyclic process.
ZEROTH LAW OF THERMODYNAMICS. Zeroth law of thermodynamics states that
when two systems are separately in thermal equilibrium with a third system , then they
themselves are in thermal equilibrium with each other.
QUASI - STATIC PROCESS
The process is said to be quasi-static, it should proceed infinite slow and follow
continuous serious of equilibrium states. Therefore, the quasi - static process may be a reversible
process.
A process is said to the reversible, it should trace the same path in the reverse direction
when the process is reversed, and it is possible only when the system passes through a
continuous series of equilibrium state if a system does not pass through continuous equilibrium
state, then the system is said to be irreversible.
POINT AND PATH FUNCTION OF THERMODYNAMICS SYSTEMS
Point function:
The quantity which is independent on the process or path followed by the system is
known as point function. Ex: Pressure, volume, temperature etc
Path function:
The quantity which is dependent on the process or path followed by the system is known
as path function. Ex: Heat transfer, Work transfer.
WORK DONE BY A GAS
Let us consider a closed system where a part of the boundary is allowed to move under
such conditions that the external restraining force is infinitesimally smaller than the force
produced by the pressure of the system. The area of the piston is A and the pressure of the fluid
at any instant is p. If p is assumed to be constant during an infinitesimal movement of the piston
over a distance dl, the work done by the fluid in moving the external force pA through this
distance is pA dl. But A.dl is dV, the infinitesimal change of volume, therefore
dW = pdV
Force acting on the piston = Pressure X Area
= pA
Therefore, Work done= Force X distance
= pA X dx
= pdV
where dV - change in volume.
if the expansion occurs from pressure p1 to a pressure p2 in such a way that the restraining force
is changed continuously, then the total work done can be found out by summing up all the
increments of work p dV, i.e. W = ∫ p dV
Work done by a rubber band
When we increase the length of a rubber band by a small length x by applying a pull F, then:
Work done on band = Fx
We could equally well write this as
Work done by band = – Fx
To see the point of this, suppose we allow the band to contract, exerting a pull F. In this case x is negative so the work done by the band is positive, which makes sense.
If we stretch the band a lot, then F will change significantly during the stretching. In this case we
add together all the bits of work, Fx, which means adding together the areas of all the narrow strips under the graph, from the initial extension x1 up to the final extension, x2. So:
Work done on (or by) band = area under Force-extension graph A positive amount of work is done on the band if x is increasing; a positive amount of work is done by the band if x is decreasing.
When a gas exerting a pressure p expands by a small volume V, then: Work done by gas = pV
[This is, in fact, just a more convenient way of writing Fx, in which F is the force on the piston
and x is the distance it moves.]
We could equally well write
Work done on gas = –pV
If we push the piston in a little way V is negative so a positive amount of work is done on the
gas – as expected, since we’ve had to do the pushing .If the gas expands a lot, then p will change
significantly. In this case we add together all the bits of work, pV, which means adding together
the areas of all the narrow strips under the graph (see diagram), from the initial volume V1 up to
the final volume, V2. So:
Work done by gas = area under pressure - volume graph
A positive amount of work is done by the gas if V is increasing; a positive amount of work
is done on the gas if V is decreasing.
Heat Energy
Heat is energy flowing from a region of higher temperature to a region of lower
temperature because of the temperature difference.
Compare with the flow of charge due to a potential difference…
The formula used to calculate heat transfer; Q = m Cv(T2-T1)
Where Q= heat transfer
m= mass of the body
T2-T1 = Temperature Difference
If there is no temperature difference, no heat will flow. [In the electrical case, if there is no
potential difference, no charge will flow.]
It takes time for a finite amount of heat to flow, though the rate of flow is greater the greater the
temperature difference (other things being the same).
Example 1: Rapid expansion of an ideal gas
Heat-If the expansion is really rapid the heat flow will be (almost) zero.
Work- As the gas expands work is done by the gas. The last term in the equation is negative.
Another way of seeing this is to write the equation as:
Whichever way we look at it, because the heat term is zero, the right hand side of the equation is
negative, so the gain in internal energy of the gas is negative, i.e. the internal energy decreases.
Example 2: Slow expansion of an ideal gas
Here it takes time to flow of heat. For each small increase in volume the gas temperature will
drop a little and heat will flow in from the surroundings – limiting further temperature drop. If
the expansion is really slow (and the cylinder walls conduct heat well) the temperature drop is
negligible, so the expansion is isothermal.
Gain in system’s
internal energy =
Net heat flowing into
system +
Net work done on
system
Gain in system’s
internal energy =
Net heat flowing into
system
Net work done by
system
The energy stored by the gas is internal energy. It isn’t heat and it isn’t work. Its decrease
could result in heat being given out or in work being done (or both).
Heat and work are both energy in transit.
U is a change in a property of the system, its internal energy, U. A positive value of U
means an increase in U; a negative value means a decrease in U.
Q is heat entering the system from (hotter) surroundings. A negative value of Q means
heat leaving the system (to cooler surroundings). We don’t have a ‘’ in front of Q, because heat
flow is not a change in heat. It’s energy in transit. It is not a function of the system’s state.
W is work done by the system. A negative value of W means a positive amount of work
done on the system. We don’t have a ‘’ in front of W for the same reason as for Q; namely, W is
not a change in work. It’s energy in transit. Work is not a function of the system’s state.
Specific heat capacity at constant volume and at constant pressure
Specific heat at constant volume of a substance is the amount of heat added to rise the temperature
of unit mass of the given substance by 1 degree at constant volume
From first law for a stationary closed system undergoing a process
dQ = pdV + dU or dq = pdv + du
For a constant volume process
dQ = dU or dq = du
or du=CvdT
Similarly specific heat at constant pressure is the quantity of heat added to rise the temperature of
unit mass of the given substance by 1 degree at constant pressure
where dQ = pdV + dU
dQ = pdV + d(H - PV)
dQ = pdV + dH- Vdp - pdV
dQ = dH - Vdp
For a constant pressure process dp = 0
Hence dQ = dH or dq= dh
or dh = CpdT
For solids and liquids, constant volume and constant pressure processes are identical and
hence, there will be only one specific heat.
The difference in specific heats Cp - Cv = R
The ratio of sp. heat Cp/Cv=
Since h and u are properties of a system, dh = CpdT and du=CvdT, for all processes.
FIRST LAW OF THERMODYNAMICS
First law of thermodynamics states that when system undergoes a cyclic process, net heat
transfer is equal to work transfer.
(ΣδQ) cycle ∝ (ΣδW)cycle
dQ = dW
In a cyclic process the system is taken through a series of processes and finally returned to its
original state. The end state of a cyclic process is identical with the state of the system at the
beginning of the cycle. This is possible if the energy level at the beginning and end of the cyclic
process are also the same. In other words, the net energy change in a cyclic process is zero.
Consider a system undergoing a cycle consisting of two processes A & B .Net energy change
EA
+EB 0
(QA W
A) + (Q
B W
B) 0
QA Q
B W
A W
B
(or) dWdQ
Hence for a cyclic process algebraic sum of heat transfers is equal to the
algebraic sum of work transfer.
First Law of Thermodynamics for a Change of State
The expression (ΣδQ) cycle = (ΣδW)cycle applies to a system undergoing a cycle. But if
the system undergoes a change of state during which both heat and work transfers are involved,
the net energy transfer will be stored or accumulated within the system. If Q is the amount of
heat transferred to the system and W is the amount of work transferred from the system during
the process, the net energy transfer (Q-W) will be stored in the system. This stored energy is
known as internal energy or simply energy of the system. Hence, the First Law for a change of
state is the net energy transfer during a process involving heat and work transfers is equal to the
change in increase in energy of the system.ie Q – W = ΔE
COROLLARIES FIRST LAW OF THERMODYNAMICS.
Corollaries I
There exists a property of a closed system such that a change in its value is equal to the
difference between the heat supplied and the work done during any change of state.
Corollaries II
The internal energy of a closed system remains unchanged system is isolated from its
surrounding.
Corollaries III
A perpetual motion machine of first kind is impossible.
In an isolated system, there is no change in internal energy.
For any isolated system, there is no heat, work and mass transfer.
Q = W = 0
According to the first law of thermodynamics,
Q = W + ΔU
ΔU = 0
CYCLIC PROCESSES
Suppose an ideal gas is taken through the cycle of changes ABCDA shown above on the left.
Irrespective of exactly what is going on in the individual stages (AB, BC, CD, DA), or the exact
shapes of the curves, we can draw some general conclusions…
When the gas has undergone one cycle and is back at A, its internal energy is the same as it was
originally, that is U = 0. This is because internal energy is a function of state, and the gas is
back in the same state as originally.
Over the cycle as a whole, the gas has done a positive amount of work. This is because the area
under ABC represents the gas doing a positive amount of work. The area under CDA represents
work being done on the gas, but this area is smaller.
Thus applying the First Law, we see that for the cycle as a whole, there has been a net flow of
heat into the gas.
THE INTERNAL ENERGY OF A SYSTEM
The internal energy, U, of a system is the sum of the potential and kinetic energies of its
particles.
Internal energy of a gas is the energy stored in a gas due to its molecular interactions. It is
also defined as the energy possessed by a gas at a given temperature
There are two ways, then, that the total energy inside a system may change—heat and/or work.
We use the term internal energy for the total energy inside a system, and the symbol U. Q and
W will stand for heat and work, respectively. Energy conservation gives us the First “Law” of
Thermodynamics: WQU .
Now, we have to be careful with the algebraic signs. In this case, Q is positive as the heat
entering the system and W is positive as the work done on the system. So a positive Q and a
positive W both cause an increase of internal energy, U.
ENTHALPY OF THE SYSTEM
The total energy required to create a system of particles at sea level air pressure would include
the expansive work done in displacing the air. It is sum of internal energy and flow energy.
We define the enthalpy to be PVUH . The enthalpy is useful when a change takes place in
a system while pressure is constant.
other
other
WQH
VPWVPQVPWQVPUH
Now, if no other work is done, then QH exactly. In practice, tables of measured enthalpies
for various processes, usually chemical reactions or phase transitions are compiled. The text
mentions the enthalpy of formation for liquid water. Evidently, when oxygen and hydrogen
gases are combined to form a mole of liquid water, the change in enthalpy is -286 kJ. In other
words, burning hydrogen at constant pressure releases this much energy.
dW = − p dV
⇒ dQ = dU + pdV = d(U + pV) - Vdp
⇒ d(U + pV) = dQ + Vdp
We define H ≡ U + pV as the enthalpy of the system, and h = u + pv is the specific enthalpy. In
particular, for a constant pressure process,
ΔQ = ΔH
TERMODYNAMICS PROCESSESS
CONSTANT VOLUME PROCESS
For a constant volume process, work can only be done by some method of churning the
liquid. There cannot be any ∫pdV work as no external force has been moved through a distance.
Hence, W must be zero or negative. For a constant volume process, unless otherwise stated,
work done is taken as zero. Thus energy equation for a constant volume process is usually
written as
Q = u2 – u1
W= 0, Q= Δu
If, in addition to work being zero, it is stipulated that the heat is transferred by virtue of an
infinitesimally small temperature difference, then the process is reversible and the equation can
be written in differential form as dQ = du
CONSTANT PRESSURE PROCESS
A closed system undergoing at constant process is shown in Fig. The fluid is enclosed in a
cylinder by a piston on which rests a constant weight. If heat is supplied, the fluid expands and
work is being done by the system in overcoming the constant force; if the heat is extracted, the
fluid contracts and work is done on the system by the constant force. In the general case Q – W =
(u2 – u1)
If no paddle work is done on the system, and the process is reversible,
dQ – p dv = du
Since p is constant, this can be integrated to give
Q – p (v2 – v1) = (u2 – u1)
A further simplification is for constant pressure process if a new property is introduced
Since p is constant, p dv is identical to d(pv). Thus energy equation becomes
dQ – d(pv) = du (or)
dQ = d(u + pv) = dh
where
h = u + pv, known as enthalpy.
Since enthalpy is a combination of properties u, p and v, it itself is a property. Hence
h = u + pv
and for any mass of fluid m, in a state of equilibrium
H = U + pV
Using this derived property, the energy equation for a reversible constant pressure process
becomes
dQ = dh (or) in the integrated form
Q = (h2 – h1)
Thus heat added in a reversible constant pressure process is equal to increase of enthalpy,
whereas it is equal to increase of internal energy in the reversible constant
ISOTHERMAL & ADIABATIC PROCESSES
Imagine a system in thermal contact with its environment. The environment is much
larger than the system of interest, so that heat flows into or out of the system has no effect on the
temperature of the environment. We speak of the system being in contact with a heat bath so
that no matter what happens to the system, its temperature remains constant. If such a system is
compressed slowly enough, its temperature is unchanged during the compression. The system is
compressed isothermally. As an example, consider an ideal gas:
1W2 = 2
1
pdV
= 2
1
CdV where C=pV
= C 2
1
1dV
V
= C ln (V2/V1)
1= p1V1ln(V2/V1) (or) p2V2ln (V2/V1)
On the other hand, the compression may be so fast that no heat is exchanged with the
environment (or, the system is isolated from the environment) so that Q = 0. Such a process is
adiabatic. Naturally, the temperature of the system will increase.
An isotherm is a curve of constant temperature, T, on the PV diagram. An arrow
indicates the direction that the system is changing with time. For an ideal gas, an isotherm is
parabolic, sinceV
P1
; that’s a special case. A curve along which Q = 0 is called an adiabatic.
Work done for isothermal process; W= P1V1 ln(V2/V1)
Work done for adiabatic process= (P1V1-P2V2)/ (γ-1)
𝑊 = (𝑝1𝑣1 − 𝑝2𝑣2)/(𝛾 − 1)
POLYTROPIC PROCESS
Generally all processes are polytropic process, by changing the index of ‘n’ either
expansion or compression can get number of process. Then the process name is changed.
Any process can be represented by the general form pVn
of n, the process differs as given below; For other values of n, the process is known as
polytropic process. Figure 2.8 shows the polytropic process of various possible polytropic
index ‘n’ on p-V coordinates. Expression for displacements work for a polytropic process can
be obtained as follows:
1W2 = 2
1
pdV
= 2
1
dVV
Cn
where C = pVn
= C
2
1
dVV n
= C
2
1
1
1
n
V n
=
2
1
1
1
1
2
1
n
CVCVnn
=
1
1
111
1
222
n
VVpVVpnnnn
since C = p1V1n = p2Vn
2
1
=
1
1122
n
VpVp
APPLICATION OF FIRST LAW OF THERMODYNAMICS
The processes undergone in a closed system - Non-flow process
The processes undergone in an open system - Flow process.
FLOW ENERGY PROCESS
Flow energy is defined as the energy required to move a mass into the a control volume against a
pressure. Consider a mass of volume V entering into a control volume as given in the Figure.
The Flow energy = Work done in moving the mass
= Force X distance
= pA X dx
= p X (Adx)
= pV
Therefore, Enthalpy = Internal energy + Flow energy
First Law of Thermodynamics for a Control Volume
Mass simultaneously entering and leaving the system is a very common phenomenon in most of
the engineering applications. Control volume concept is applied to these devices by assuming
suitable control surfaces.
To analyze these control volume problems, conservation of mass and energy concepts are to be
simultaneously considered.
Energy may cross the control surface not only in the form of heat and work but also by total
energy associated with the mass crossing the boundaries. Hence apart from kinetic, potential and
internal energies, flow energy should also be taken into account.
Conservation of mass
volumecontrol
theofcontentmass
theinchangeNet
volumecontrol
theleaving
massTotal
volumecontrol
theentering
massTotal
Conservation of energy
volumecontrol
theofcontent
theenrgyin
changeNet
volumecontrolthe
leavingmass
thewithassociated
energyTotal
volumecontrolthe
enteringmass
thewithassociated
energyTotal
workand
heatofform
theinboundary
thecrossingenergyNet
As a rate equation, it becomes
CV
out
out
in
in EZgC
hmZgC
hmWQ
22
22
The Steady-state Flow Process
When a flow process is satisfying the following conditions, it is known as a steady flow process.
1. The mass and energy content of the control volume remains constant with time.
2. The state and energy of the fluid at inlet, at the exit and at every point within the control
volume are time independent.
3. The rate of energy transfer in the form of work and heat across the control surface is constant
with time.
Therefore for a steady flow process
also
Control Surface
.
Q
.
W
.
inm.
outm
Control
Volume
outin mm
For problem of single inlet stream and single outlet stream
gZZ
CChhmWQ 12
2
1
2
212
2)(
This equation is commonly known as steady flow energy equation (SFEE).
Application of SFEE
SFEE governs the working of a large number of components used in many engineering practices.
In this section a brief analysis of such components working under steady flow conditions are given
and the respective governing equations are obtained.
Turbines
Turbines are devices used in hydraulic, steam and gas turbine power plants. As the fluid
passes through the turbine, work is done on the blades of the turbine which are attached to a
shaft. Due to the work given to the blades, the turbine shaft rotates producing work.
General Assumptions
1. Changes in kinetic energy of the fluid are negligible
2. Changes in potential energy of the fluid are negligible.
)( 12 hhmWQ
Compressors
Compressors (fans and blowers) are work consuming devices, where a low-pressure fluid is
compressed by utilising mechanical work. Blades attached to the shaft of the turbine imparts
kinetic energy to the fluid which is later converted into pressure energy.
0 CVE
022
22
Zg
ChmZg
ChmWQ
out
out
in
in
Mass entering
Mass leaving
Shaft work
Control
Surface
General Assumptions 1. Changes in the kinetic energy of the fluid are negligible
2. Changes in the potential energy of the fluid are negligible
Governing Equation Applying the above equations SFEE becomes
)( 12 hhmWQ
Pumps
Similar to compressors pumps are also work consuming devices. But pumps handle
incompressible fluids, whereas compressors deal with compressible fluids.
General Assumptions
1. No heat energy is gained or lost by the fluids;
2. Changes in kinetic energy of the fluid are negligible.
Mass entering
Mass leaving
Shaft work
Control
Surface
Governing Equation
gZZhhmW 1212 )(
As the fluid passes through a pump, enthalpy of the fluid increases, (internal energy of the fluid
remains constant) due to the increase in pv (flow energy). Increase in potential energy of fluid is
the most important change found in almost all pump applications.
Nozzles
Nozzles are devices which increase the velocity of a fluid at the expense of pressure. A typical
nozzle used for fluid flow at subsonic* speeds is shown in Figure.
General Assumptions
1. In nozzles fluids flow at a speed which is high enough to neglect heat lost or gained
as it crosses the entire length of the nozzle. Therefore, flow through nozzles can be
regarded as adiabatic. That is = 0.
2. There is no shaft or any other form of work transfer to fluid or from the fluid; that is = 0.
3. Changes in the potential energy of the fluid are negligible.
Governing Equation
)(2
02
)(
21
2
1
2
2
2
1
2
212
hhCC
CChh
Diffusers
Diffusers are (reverse of nozzles) devices which increase the pressure of a fluid stream by
reducing its kinetic energy.
In Out
Control Surface
Hot fluid out
Cold fluid in
cold
fluid out
Hot fluid in
General Assumptions
Similar to nozzles, the following assumptions hold good for diffusers.
1. Heat lost or gained as it crosses the entire length of the nozzle. Therefore, flow through
nozzles can be regarded as adiabatic. That is 0Q
2. There is no shaft or any other form of work transfer to the fluid or from the fluid;
that is = 0.
3. Changes in the potential energy of the fluid are negligible
Governing Equation
Heat Exchangers
Devices in which heat is transferred from a hot fluid stream to a cold fluid stream are known as
heat exchangers.
General Assumptions
1. Heat lost by the hot fluid is equal to the heat gained by the cold fluid.
2. No work transfer across the control volume.
3. Changes in kinetic and potential energies of both the streams are negligible.
2)(
02
)(
2
2
2
112
2
1
2
212
CChh
CChh
Governing Equation
For both hot and cold streams
As per the assumption,
= coldhot QQ
The negative sign in the LHS is to represent that heat is going out of the system.
)()( 1221 hhmhhm ch
Throttling A throttling process occurs when a fluid flowing in a line suddenly encounters a restriction in the
flow passage. It may be Plate with a small hole as shown in (a), Valve partially closed as shown
in (b) Capillary tube which is normally found in a refrigerator as shown in (c)Porous plug as
shown in (d)
)( 12 hhmQ
QUESTIONS
PART-A
1.What is meant by thermodynamics system? How do you classify it?
2. What is meant by closed system? Give an example.
3. Define open system. Give an example
4. When a system is said to be in” Thermodynamic equilibrium “?
5. Define Zeroth law and first law of thermodynamics.
6. State corollaries first law of thermodynamics.
7. Prove that for an isolated system, there is no change in internal energy.
8. What is meant by open and closed cycle?
9. What is meant by reversible and irreversible process?
10. What is meant by point and path function?
.
PART -B
1. A piston and cylinder machine contains a fluid system which passes through a complete cycle
of four processes. During a cycle, the sum of all heat transfer is -170kJ. The system completes
100 cycles per min. Complete the following table showing the method for each item and
compute the net rate of work output in kW
2. A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and
0.105 MPa to a final state of 0.15m3 and 0.105 MPa, the pressure remaining constant during the
process. There is a transfer of 37.6 kJ of heat from the gas during the process. How much does
the internal energy of the gas change?
3. The internal energy of a certain substance is given by the following equation
U=3.56PV + 84
where U is given in kJ/kg, P is in kPa and V is in m3/kg.
A system composed of 3kg of this substance expands from an initial pressure of 500 kPa and a
volume of 0.22 m3 to a final pressure 100kPa in a process in which pressure and volume are
related by pv1.2=constant.
(a) If the expansion is quasi static, find Q, and W for the process.
(b) In another process the same system expands according to the same pressure volume
relationship as in part (a), and from the same initial state to the same final state as in part (a), but
the heat transfer in this case is 30kJ. Find the work transfer for this process.
(C) Explain the difference in work transferin part (a) and (b)
4. Air flows steadily at the rate of 0.5kg/s through an air compressor, entering 7m/s velocity,
100kPa pressure and 0.95 m3/kg volume, and leaving at 5m/s, 700k Pa and 0.19 m3/kg. The
internal energy of the air leaving is 90kJ/kg greater than that of the air entering. Cooling water in
the compressor jacket absorbs heat from the air at the rate of 58kW. (a) Compute the rate of shaft
work input to the air in kW. (b) Find the ratio of the inlet pipe diameter to outlet diameter
5. Air at a temperature of 150C passes through a heat exchanger at a velocity of 30m/s where its
temperature is raised to 8000C. It then enters a turbine with the same velocity of 30m/s and
expands untill the temperature falls to 6500C. On leaving the turbine, the air is taken at a velocity
of 60m/s to a nozzle where it expands untill the temperature has fallen to 5000C. If the air flow
rate is 200kg/s, calculate
(a) the rate of heat transferto the air in the heat exchanger,
(b) the power output from the turbine assuming no heat loss and
(c) the velocity at exit from the nozzle, assuming no heat loss.
Take the enthalpy of the air as h=CpT,
where Cp is the specific heat equal to 1.005 kJ/kgK and T the temperature.
Solutions
1. A piston and cylinder machine contains a fluid system which passes through a complete cycle
of four processes. During a cycle, the sum of all heat transfer is -170kJ. The system completes 100
cycles per min. Complete the following table showing the method for each item and compute the
net rate of work output in kW
2. A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 0.105
MPa to a final state of 0.15m3 and 0.105 MPa, the pressure remaining constant during the process.
There is a transfer of 37.6 kJ of heat from the gas during the process. How much does the internal
energy of the gas change?
Solution: First law for a stationary system in a process gives
3. The internal energy of a certain substance is given by the following equation
U=3.56PV + 84
where U is given in kJ/kg, P is in kPa and V is in m3/kg.
A system composed of 3kg of this substance expands from an initial pressure of 500 kPa and a
volume of 0.22 m3 to a final pressure 100kPa in a process in which pressure and volume are
related by pv1.2=constant.
(a) If the expansion is quasi static, find Q, and W for the process.
(b) In another process the same system expands according to the same pressure volume
relationship as in part (a), and from the same initial state to the same final state as in part (a), but
the heat transfer in this case is 30kJ. Find the work transfer for this process.
(C) Explain the difference in work transferin part (a) and (b)
Solution:
4. Air flows steadily at the rate of 0.5kg/s through an air compressor, entering 7m/s velocity,
100kPa pressure and 0.95 m3/kg volume, and leaving at 5m/s, 700k Pa and 0.19 m3/kg. The
internal energy of the air leaving is 90kJ/kg greater than that of the air entering. Cooling water in
the compressor jacket absorbs heat from the air at the rate of 58kW. (a) Compute the rate of shaft
work input to the air in kW. (b) Find the ratio of the inlet pipe diameter to outlet diameter
Solution: Figure shows the detail of the problem
5. Air at a temperature of 150C passes through a heat exchanger at a velocity of 30m/s where its
temperature is raised to 8000C. It then enters a turbine with the same velocity of 30m/s and
expands untill the temperature falls to 6500C. On leaving the turbine, the air is taken at a velocity
of 60m/s to a nozzle where it expands untill the temperature has fallen to 5000C. If the air flow
rate is 200kg/s, calculate
(a) the rate of heat transferto the air in the heat exchanger,
(b) the power output from the turbine assuming no heat loss and
(c) the velocity at exit from the nozzle, assuming no heat loss.
Take the enthalpy of the air as h=CpT,
where Cp is the specific heat equal to 1.005 kJ/kgK and T the temperature.
Solution:
A shown in figure write the S.F.E.E for the heat exchanger and eliminating the terms not relevant
UNIT-II
SECOND LAW OF THERMODYNAMICS
INTRODUCTION In this chapter the idea of cycle efficiency in introduced and the Second Law is then
stated and distinguished from the First Law. Formal definition of a reversible process is made and its implications both for non-flow and steady-flow processes are discussed
A process is said to the reversible, it should trace the same path in the reverse direction when the process is reversed, and it is possible only when the system passes through a continuous series of equilibrium state if a system does not pass through continuous equilibrium state, then the system is said to be irreversible.
The direction of spontaneous change for a ball bouncing on a floor. On each bounce some of its potential energy is degraded into the thermal motion of the atoms of the floor, and that energy disperses into the atoms of the floor. The reverse has never been observed to take place. The reverse, if it occurs, does not violate the1st Law as long as the energy is conserved Recall also that only a small amount of thermal energy is required to make the ball jump very high. Hence, the first Law only states that the net work cannot be produced during a cycle without some supply of heat. However, First Law never says that some proportions of heat supplied to an engine must be rejected. Hence, as per the First Law, cycle efficiency can be unity, which is impossible in practice. All that First Law states that net work cannot be produced during a cycle without some supply of heat, i.e. that a perpetual motion machine of the first kind is impossible So, the 1st Law is not enough. Something is missing! What is missing? A law that can tell us about the direction of spontaneous change. The Second Law of Thermodynamics tells us about the directionality of the process. We need it to ensure that systems we design will work. As we will see later, ENTROPY is a property that we have invented (like internal energy) that will allow us to apply the 2nd Law quantitatively.
REVERSIBLE AND IRREVERSIBLE PROCESSES
A process is said to the reversible, it should trace the same path in the reverse direction when the process is reversed, and it is possible only when the system passes through a continuous series of equilibrium state if a system does not pass through continuous equilibrium state, then the system is said to be irreversible
In the course of this development, the idea of a completely reversible process is central, and we can recall the definition, ``a process is called completely reversible if, after the process has occurred, both the system and its surroundings can be wholly restored by any means to their respective initial states’’. Especially, it is to be noted that the definition does not, in this form, specify that the reverse path must be identical with the forward path. If the initial states can be restored by any means whatever, the process is by definition completely reversible. If the paths are identical, then one usually calls the process (of the system) reversible, or one may say that the state of the system follows a reversible path. In this path (between two equilibrium states 1 and 2), (i) the system passes through the path followed by the equilibrium states only, and (ii) the system will take the reversed path 2 to 1 by a simple reversal of the work done and heat added.
Reversible processes are idealizations not actually encountered. However, they are clearly useful idealizations. For a process to be completely reversible, it is necessary that it be quasi-static and that there be no dissipative influences such as friction and diffusion. The precise (necessary and sufficient) condition to be satisfied if a process is to be reversible is the second part of the Second Law.
The criterion as to whether a process is completely reversible must be based on the initial and final states. In the form presented above, the Second Law furnishes a relation between the properties defining the two states, and thereby shows whether a natural process connecting the states is possible.
SECOND LAW OF THERMODYNAMICS Kelvin –Planck statement: It is impossible to construct an engine working on a cyclic process which converts all the heat energy supplied to it into equivalent amount of useful work. Clausius statement: Heat cannot flow from cold reservoir to hot reservoir without any external aid. But heat can flow from hot reservoir to cold reservoir without any external aid. HEAT ENGINE, HEAT PUMP, REFRIGERATOR ENERGY RESERVOIRS : Thermal energy reservoirs (TER) is defined as a large body of infinite heat capacity, which is capable of absorbing or rejecting an unlimited quantity of heat without suffering appreciable changes in its thermodynamic coordinates. SOURCE: TER from which heat is transferred to the system operating in a heat engine cycle. SINK: TER in which heat is rejected from the cycle during a cycle. HEAT ENGINE A heat engine is a device which is used to convert the thermal energy into mechanical energy. Heat supplied is input work done is the output. Hence efficiency of the engine
Work output Efficiency of the engine = --------------------------------- Heat supplied
€=(T1-T2)/T1
REFRIGERATOR : A device which operating in a cycle maintains a body at a temperature lower than the temperature of the surroundings
HEAT PUMP: Heat pump is a device which operating in a cycle process maintains the temperature of a hot body at a temperature of a hot body at a temperature higher that the temperature of surrounding. Different between heat pump and refrigeration states that heat pump is a device which operating in a cycle process maintains the temperature of a hot body at a temperature of a hot body at a temperature higher that the temperature of surrounding.
Coefficient of performance is defined as the ratio of heat of heat extracted of rejected to work input. Heat extracted or rejected COP = --------------------------------- Work input. Expression for COP of heat pump and a refrigerator. COP for heat pump: Heat rejected T2 COPHP = ----------------------- = ---------- Work Input T2-T1 COP for refrigerator: Heat Extracted T2 COP Ref = ----------------------- = ---------- Work Input T2-T1
The Second Law states that some heat must be rejected during the cycle and hence, the cycle efficiency is always less than unity Thus the First Law states that net work cannot be greater than heat supplied, while the Second Law goes further and states that it must be less than heat supplied. If energy is to be supplied to a system in the form of heat, the system must be in contact with a reservoir whose temperature is higher than that of the fluid at some point in the cycle. Similarly, if heat is to be rejected, the system must be at some time
be in contact with a reservoir of lower temperature than the fluid. Thus Second Law implies that if a system is to undergo a cycle and produce work, it must operate between two reservoirs of different temperatures. A machine which will work continuously, while exchanging heat with only single reservoirs, is known as a perpetual motion machine of the second kind (PMM II); such a machine contradicts Second Law. It is now possible to see why a ship could not be driven by an engine using the ocean as a source of heat, or why a power station could not be run using the atmosphere as a source of heat. They are impossible because there is no natural sink of heat at a lower temperature than the atmosphere or ocean, and they would therefore be PMM II.
It should be noted that Second Law does not restrict that work cannot be continuously and completely converted to heat. In fact, a fluid in a closed vessel may have work done on it and the heat thus generated is allowed to cross the boundary. The rates of work and heat may be made equal and the internal energy of the system remaining constant. An important consequence of Second law is work is a more valuable form of energy transfer than heat as heat can never be transformed continuously and completely to work, whereas work can always be transformed continuously and completely to heat. The following statements summarise the obvious consequences of the Second Law: a) If a system is taken through a cycle and produces work, it must be exchanging heat with at least two reservoirs at different temperatures, b) If a system is taken through a cycle while exchanging heat with one reservoir, the work done must be zero or negative, c) Since heat can never be continuously and completely converted into work whereas work can always be continuously and completely converted into heat, work is more valuable form of energy transfer than heat. THE KEIVIN-PLANK’S STATEMENT OF THE SECOND LAW It is impossible to construct a system, It is impossible to construct a system, which will operate in a cycle, extract heat from a reservoir and do an equivalent
THE CLAUSIUS STATEMENT OF THE SECOND LAW It is impossible to construct a system, which will operate in a cycle and transfer heat from a cooler to a hotter body without work being work done on the system by the surrounding
EQUIVALENCE KELVIN PLANK’S AND CLAUSIUS STATEMENTS. Proof: Suppose the converse of the Clausius’ proposition is true. The system can be
represented by a heat pump for which W = 0. If it takes Q units of heat from the cold reservoir, it must deliver Q units to the hot reservoir to satisfy the First Law. A heat engine could also be operated between the two reservoirs; let it be of such a size that it delivers Q units of heat to the cold reservoir while performing W units of work. Then the First Law states that the engine must be supplied with (W + Q) units of heat from the hot reservoir. In the combined plant, the cold reservoir becomes superfluous because the heat engine could reject its heat directly to the heat pump. The combined plant represents a heat engine extracting (W + Q) – Q = W units of heat from a reservoir and delivering an equivalent amount of work. This is impossible according to Kelvin-Plank’s statement of Second Law. Hence converse of Clausius’ statement is not true and the original proposition must be true.
CARNOT’S THEOREM. No heat engine operating in a cycle process between two fixed temperatures can be more efficient that a reversible engine operating between the same temperature limits. COROLLARIES OF CORNOT THEOREM i. All the reversible engines operating between the two given thermal reservoir with fixed temperature have the same efficient. ii.The efficient of any reversible heat engine operating between two reservoir is independent of the nature of the working fluid and depends only on the temperature of the reservoirs. CLAUSIUS INEQUALITY
Consider two heat engines operating between two reservoirs kept at temperature TH
and TL as shown in the Figure. Of the two heat engines, one is reversible and the other is
irreversible.
For the reversible heat engine it has already been proved that
0
0
rev
L
L
H
H
L
H
L
H
T
dQ
T
Q
T
Q
T
T
Q
Q
As discussed earlier, the work output from the irreversible engine should be less than that of the reversible engine for the same heat input QH. Therefore QL,Irrev will be greater than
QL,Rev . Let us define
QL,Irrev =QL,Rev + dQ
then
0
0
,
,
L
LL
revL
H
H
L
IrevL
H
H
Irrev
T
dQ
T
dQ
T
Q
T
Q
T
Q
T
Q
T
dQ
By combining this result with that of a reversible engine we get
0
IrrevT
dQ
This is known as Clausius inequality.
The following conditions of clausius inequality ƒdQ/T ≤ 0 is known as inequality of clausius. If 1. ƒ dQ/T = 0,the cycle is reversible. 2. ƒ dQ/T < 0,the cycle is irreversible and possible. 3. ƒ dQ/T > 0,the cycle is impossible.
ENTROPY
The measure of irreversibility when the energy transfer takes place within the system or between system and surrounding is called as change of entropy. It is simply known as unaccounted heat loss.The change entropy of the system with respect to ambient conditions or any other standard reference conditions is known as absolute entropy.
Clausius inequality forms the basis for the definition of a new property known as entropy.
Consider a system taken from state 1 to state 2 along a reversible path A as shown in Figure. Let the system be brought back to the initial state 1 from state 2 along a reversible path B. Now the system has completed one cycle. Applying Clausius inequality we get
1
2
2
1
0
0
BA T
dQ
T
dQ
T
dQ
Instead of taking the system from state2 to state1 along B, consider another reversible path C. Then for this cycle 1-A-2-C-1, applying Clausius inequality:
1
2
2
1
0
0
CA T
dQ
T
dQ
T
dQ
Comparing, Hence, it can be concluded that the quantity is a point function, independent of the path followed. Therefore it is a property of the system
Principle of increasing entropy
Applying Clausius inequality, For any infinitesimal process undergone by a system, change in entropy dS ≥ dQ/T For reversible, dQ = 0 hence dS = 0 For irreversible, dS> 0
Consider a system interacting with its surroundings. Let the system and its surroundings are included in a boundary forming an isolated system. Since all the reactions are taking place within the combined system, whenever a process occurs entropy of the universe (System plus surroundings) will increase if it is irreversible and remain constant if it is reversible. Since all the processes in practice are irreversible, entropy of universe always increases
ie., (∆s)universe>0
QUESTION PART -A
1.State the Kelvin –Planck statement of second law of Thermodynamics. 2. State the Clausius statement of second law of Thermodynamics 3. Write the two statement of second law of Thermodynamics 4. State Cornot’s Theorem. 5. What are the corollaries of cornot theorem? 6. Define – PMM of second kind? 7. What is different between heat pump and refrigeration? 8. What is meant by heat engine? 9. Define the term COP? 10. Define change of entropy .How is entropy compared with heat transfer and absolute temperature? 11. Define the term source, sink and heat reservoir. 12. Why the performance of refrigerator and heat pump are given in terms of C.O.P and not in terms of efficiency? 13. What is meant by principle of increase of entropy? 14. What do you mean by “clausius inequality”?
PART-B 1.A cyclic heat engine operates between a source temperature of 8000C and a sink temperature of 300C. What is the least rate of heat rejection per kW net output of the engine? 2. A reversible heat engine operates between two reservoirs at temperatures of 6000C and 400C. The engine drives a reversible refrigerator which operates between reservoirs at temperatures of 400C and -200C. The heat transfer to the heat engine is 2000kJ and the net work output of the combined engine transfer plant is 360kJ. (a) Evaluate the heat transfer to the refrigerant and the net heat transfer to the reservoir at 400C. (b)Reconsider (a) given that the efficiency of the heat engine and the COP of the refrigerator are each 40% of their maximum possible values. 3. It is proposed that solar energy be used to warm a large collector plate. This energy would, in turn, be transferred as heat to a fluid within a heat engine, and engine would reject energy as heat to the atmosphere. Experiments indicate that about 1880 kJ/m2h of energy can be collected when the plate is operating at 900C. Estimate the minimum collector area that would be required for a plant producing 1kW of useful shaft power. The atmospheric temperature may be assumed to be 200C. 4. One kg of ice at -500C is exposed to the atmosphere which is at 200C. The ice melts and comes into thermal equilibrium with the atmosphere. (a) Determine the entropy increases of the universe . (b) what is the minimum amount of work necessary to convert the water back into ice ar -50C? Cp of ice is 2.093 kJ/kgK and the latent heat of fusion of ice is 333.3 kJ/kg. 5. A fluid undergoes a reversible adiabatic compression from 0.5 MPa, 0.2 m3 to 0.05m3 according to the law, pv1.3=constant. Determine the change in enthalpy, internal energy and entropy, and the heat transfer and work transfer during the process. 6. Calculate the available energy in 40kg of water at 750C with respect to the surrounding at 500C, the pressure of water being 1atm.
7. Air enters a compressor at 1 bar, 300C, which is also the state of environment. It leaves at 3.5 bar, 1410 and 90m/s. Neglecting inlet velocity and P.E. effect, determine (a) whether the compression is adiabatic or polytropic, (b) If not adiabatic, the polytropic index, (c) the isothermal efficiency, (d) the minimum work input and irreversibiity and (e) second law efficiency. Take Cp of air =1.0035kJ/kgK
Solutions 1.A cyclic heat engine operates between a source temperature of 8000C and a sink temperature of 300C. What is the least rate of heat rejection per kW net output of the engine? Solution: For a reversible engine, the rate of heat rejection will be minimum
2. A reversible heat engine operates between two reservoirs at temperatures of 6000C and 400C. The engine drives a reversible refrigerator which operates between reservoirs at temperatures of 400C and -200C. The heat transfer to the heat engine is 2000kJ and the net work output of the combined engine transfer plant is 360kJ. (a) Evaluate the heat transfer to the refrigerant and the net heat transfer to the reservoir at 400C.
(b)Reconsider (a) given that the efficiency of the heat engine and the COP of the refrigerator are each 40% of their maximum possible values. Solution: (a) Maximum efficiency of the heat engine cycle is given by
3. It is proposed that solar energy be used to warm a large collector plate. This energy would, in turn, be transferred as heat to a fluid within a heat engine, and engine would reject energy as heat to the atmosphere. Experiments indicate that about 1880 kJ/m2h of energy can be collected when the plate is operating at 900C. Estimate the minimum collector area that would be required for a plant producing 1kW of useful shaft power. The atmospheric temperature may be assumed to be 200C. Solution: The maximum efficiency for the heat engine operating between the collector plate temperature and the atmospheric temperature is
4. One kg of ice at -500C is exposed to the atmosphere which is at 200C. The ice melts and comes into thermal equilibrium with the atmosphere. (a) Determine the entropy increases of the universe . (b) what is the minimum amount of work necessary to convert the water back into ice ar -50C? Cp of ice is 2.093 kJ/kgK and the latent heat of fusion of ice is 333.3 kJ/kg.
The entropy-temperature diagram for the system at -50C converts to water at 200C is shown in fig.
Fig.7.3.2
Fig.7.3.3
5. A fluid undergoes a reversible adiabatic compression from 0.5 MPa, 0.2 m3 to 0.05m3 according to the law, pv1.3=constant. Determine the change in enthalpy, internal energy and entropy, and the heat transfer and work transfer during the process. Solution: TdS = dH - VdP
6. Calculate the available energy in 40kg of water at 750C with respect to the surrounding at 500C, the pressure of water being 1atm. Solution: If the water is cooled at a constant pressure of 1 atm from 750C to 50C (as shown in fig) the heat given up may be used as a source for a series of carnot engines each using the surrounding as a sink. It is assumed that the amount of energy received by any engine is small relative to that in the source and the temperature of the source doesnot change while heat is being exchanged with the engine. Let us consider that the source has fallen to temperature T, at which level there operates a carnot engine which takes in heat at this temperature and rejects heat at T0=278K. If del S is change in entropy water, the work obtainable is
With a very great number of engines in the series, the total work (maximum) obtainable when the water is cooled from 348K to 278K would be
7. Air enters a compressor at 1 bar, 300C, which is also the state of environment. It leaves at 3.5 bar, 1410 and 90m/s. Neglecting inlet velocity and P.E. effect, determine (a) whether the compression is adiabatic or polytropic, (b) If not adiabatic, the polytropic index, (c) the isothermal efficiency, (d) the minimum work input and irreversibiity and (e) second law efficiency. Take Cp of air =1.0035kJ/kgK Solution: (a) After isentropic compression
Since this temperature is higher than the given temperature of 1410C, there is heat loss to the surroundings. The compression cannot be adiabatic. It must be polytropic.
.
Unit 3: Properties of Pure Substances
3.1 PURE SUBSTANCE
A pure substance is a system which is having homogeneous composition throughout its mass. For
example a system comprising steam and water, is homogeneous in composition, since chemical analysis
would reveal that hydrogen and oxygen atoms are presents in the ratio 2 : 1 whether the sample be
taken from the steam or from the water.
3.2PHASE CHANGE OF A PURE SUBSTANCE
Let us consider 1 kg of liquid water at a temperature of 20°C in a cylinder fitted with a piston, which
exerts on the water a constant pressure of one atmosphere (1.0132 bar) as shown in Fig. 3.2
Fig 3.2 Phase change of water at constant pressure from liquid to vapour phase
As the water is heated slowly its temperature rises until the temperature of the liquid water becomes
100°C. During the process of heating, the volume slightly increases as indicated by the line 1-2 on the
temperature-specific volume diagram (Fig. 3.3). The piston starts moving upwards. Fig. 3.2.
If the heating of the liquid, after it attains a temperature of 100°C, is continued it undergoes a change in
phase. A portion of the liquid water changes into vapour as shown in Fig. 3.2 (ii). This state is described
by the line 2-3 in Fig. 3.3. The amount of heat required to convert the liquid water completely into
vapour under this condition is called the heat of vapourisation. The temperature at which vapourisation
takes place at a given pressure is called the saturation temperature and the given pressure is called the
saturation pressure.
During the process represented by the line 2-3 (Fig. 3.3) the volume increases rapidly and piston moves
upwards Fig. 3.2 (iii). For a pure substance, definite relationship exists between the saturation pressure
and saturation temperature as shown in Fig. 3.4, the curve so obtained is called vapour pressure curve.
— It may be noted that if the temperature of the liquid water on cooling becomes lower than the
saturation temperature for the given pressure, the liquid water is called a subcooled liquid. The point ‘1’
(in Fig. 3.3) illustrates this situation, when the liquid water is cooled under atmospheric pressure to a
temperature of 20°C, which is below the saturation temperature (100°C).
— Further, at point ‘1’ the temperature of liquid is 20°C and corresponding to this temperature, the
saturation pressure is 0.0234 bar, which is lower than the pressure on the liquid water, which is 1
atmosphere. Thus the pressure on the liquid water is greater than the saturation pressure at a given
temperature. In this condition, the liquid water is known as the compressed liquid. The term
compressed liquid or sub-cooled liquid is used to distinguish it from saturated liquid. All points in the
liquid region indicate the states of the compressed liquid.
— When all the liquid has been evaporated completely and heat is further added, the temperature of
the vapour increases. The curve 3-4 in Fig. 3.3 describes the process. When the temperature increases
above the saturation temperature (in this case 100°C), the vapour is known as the superheated vapour
and the temperature at this state is called the superheated temperature. There is rapid increase in
volume and the piston moves upwards [Fig. 3.2 (iii)]. The difference between the superheated
temperature and the saturation temperature at the given pressure is called the degree of superheat.
— If the above mentioned heating process is repeated at different pressures a number of curve similar
to 1-2-3-4 are obtained. Thus, if the heating of the liquid water in the piston cylinder arrangement takes
place under a constant pressure of 12 bar with an initial temperature of 20°C until the liquid water is
converted into superheated steam, then curve 5-6-7-8 will represent the process.
— In the above heating process, it may be noted that, as the pressure increases the length of constant
temperature vapourisation gets reduced.
From the heating process at a constant pressure of 225 bar represented by the curve 9-10-11 in Fig. 3.3,
it can be seen that there is no constant temperature vapourisation line. The specific volume of the
saturated liquid and of the saturated vapour is the same, i.e., vf = vg. Such a state of the substance is
called the critical state. The parameters like temperature, pressure, volume, etc. at such a state are
called critical parameters.
— The curve 12-13 (Fig. 3.3) represents a constant pressure heating process, when the pressure is
greater than the critical pressure. At this state, the liquid water is directly converted into superheated
steam. As there is no definite point at which the liquid water changes into superheated steam, it is
generally called liquid water when the temperature is less than the critical temperature and
superheated steam when the temperature is above the critical temperature.
3.3. p-T (Pressure-Temperature) DIAGRAM FOR A PURE SUBSTANCE
If the vapour pressure of a solid is measured at various temperatures until the triple point is reached
and then that of the liquid is measured until the critical point is reached, the result when plotted on a p-
T diagram appears as in Fig. 3.5.
If the substance at the triple point is compressed until there is no vapour left and the pressure on the
resulting mixture of liquid and solid is increased, the temperature will have to be changed for
equilibrium to exist between the solid and the liquid. Measurements of these pressures and
temperatures give rise to a third curve on the p-T diagram, starting at the triple point and continuing
indefinitely.
The points representing the coexistence of (i) solid and vapour lie on the ‘sublimation curve’, (ii) liquid
and vapour lie on the ‘vapourisation curve’, (iii) liquid and solid lie on the ‘fusion curve’. In the particular
case of water, the sublimation curve is called the frost line, the vapourisation curve is called the steam
line, and the fusion curve is called the ice line. The slopes of sublimation and the vapourisation curves
for all substances are positive. The slope of the fusion curve, however may be positive or negative. The
fusion curve of most substances have a positive slope. Water is one of the important exceptions.
The triple point is merely the point of intersection of sublimation and vapourisation curves. It must be
understood that only on p-T diagram is the triple point represented by a point. On p-V diagram it is a
line.The pressure and temperature at which all three phases of a pure substance coexist may be
measured with the apparatus that is used to measure vapour pressure.
For Water triple point is 273.16K 4.58mm of Hg.
3.4. p-V-T (Pressure-Volume-Temperature) SURFACE
Fig 3.6 p-V-T (Pressure-Volume-Temperature) SURFACE
A detailed study of the heating process reveals that the temperature of the solid rises and then during
the change of phase from solid to liquid (or solid to vapour) the temperature remains constant. This
phenomenon is common to all phase changes. Since the temperature is constant, pressure and
temperature are not independent properties and cannot be used to specify state during a change of
phase.
The combined picture of change of pressure, specific volume and temperature may be shown on a three
dimensional state model. Fig. 3.6 illustrates the equilibrium states for a pure substance which expands
on fusion. Water is an example of a substance that exhibits this phenomenon.
All the equilibrium states lie on the surface of the model. States represented by the space above or
below the surface are not possible. It may be seen that the triple point appears as a line in this
representation. The point C.P. is called the critical point and no liquid phase exists at temperatures
above the isotherms through this point. The term evaporation is meaningless in this situation.
At the critical point the temperature and pressure are called the critical temperature and the critical
pressure respectively and when the temperature of a substance is above the critical value, it is called a
gas. It is not possible to cause a phase change in a gas unless the temperature is lowered to a value less
than the critical temperature. Oxygen and nitrogen are examples of gases that have critical
temperatures below normal atmospheric temperature.
3.5. PHASE CHANGE TERMINOLOGY AND DEFINITIONS
Suffices : Solid - i Liquid - f Vapour – g
No Phase change Name Process Process suffix
1 Solid-liquid Fusion Freezing, melting if
2 Solid-vapour Sublimation Frosting, defrosting ig
3 Liquid-vapour Evaporation Evaporating, Condensing
fg
Triple point—The only state at which the solid, liquid and vapour phases coexist in equilibrium.
Critical point (C.P.). The limit of distinction between a liquid and vapour.
Critical pressure. The pressure at the critical point.
Critical temperature. The temperature at the critical point.
Gas—A vapour whose temperature is greater than the critical temperature.
Saturation temperature. The phase change temperature corresponding to the saturation pressure.
Sometimes called the boiling temperature.
Saturation pressure. The phase change pressure.
Compressed liquid. Liquid whose temperature is lower than the saturation temperature. Sometimes
called a sub-cooled liquid.
Saturated liquid. Liquid at the saturation temperature corresponding to the saturation pressure. That is
liquid about to commence evaporating, represented by the point f on a diagram.
Saturated vapour. A term including wet and dry vapour.
Dry (saturated) vapour. Vapour which has just completed evaporation. The pressure and temperature
of the vapour are the saturation values. Dry vapour is represented by a point g on a diagram.
Wet vapour. The mixture of saturated liquid and dry vapour during the phase change.
Superheated vapour. Vapour whose temperature is greater than the saturation temperature
corresponding to the pressure of the vapour.
Degree of superheat. The term used for the numerical amount by which the temperature of a
superheated vapour exceeds the saturation temperature.
Fig.3.7 Phase change terminology
3.6. PROPERTY DIAGRAMS IN COMMON USE
Besides p-V diagram which is useful because pressure and volume are easily visualised and the T-s chart
which is used in general thermodynamic work, there are other charts which are of practical use for
particular applications. The specific enthalpy-specific entropy chart is used for steam plant work and the
pressure-specific enthalpy chart is used in refrigeration work. Sketches of these charts are shown in Fig.
3.8. These charts are drawn for H2O (water and steam) and represent the correct shape of the curves
for this substance.
Fig 3.8
3.7. FORMATION OF STEAM:
The process of formation of steam is shown in fig.3.9 and 3.10
Fig 3.9 Formation of steam
Fig 3.10
3.8. IMPORTANT TERMS RELATING STEAM FORMATION
1. Sensible heat of water (hf ). It is defined as the quantity of heat absorbed by 1 kg of water when it is
heated from 0°C (freezing point) to boiling point. It is also called total heat (or enthalpy) of water or
liquid heat invariably. It is reckoned from 0°C where sensible heat is taken as zero. If 1 kg of water is
heated from 0°C to 100°C the sensible heat added to it will be 4.18 × 100 = 418 kJ but if water is at say
20°C initially then sensible heat added will be 4.18 × (100 – 20) = 334.4 kJ. This type of heat is denoted
by letter hf and its value can be directly read from the steam tables. Note. The value of specific heat of
water may be taken as 4.18 kJ/kg K at low pressures but at high pressures it is different from this value.
2. Latent heat or hidden heat (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. It is expressed by the
symbol hfg and its value is available from steam tables. The value of latent heat is not constant and
varies according to pressure variation.
3. Dryness fraction (x). The term dryness fraction is related with wet steam. It is defined as the ratio of
the mass of actual dry steam to the mass of steam containing it. It is usually expressed by the symbol ‘x’
or ‘q’. If ms = Mass of dry steam contained in steam considered, and mw = Weight of water particles in
suspension in the steam considered,
(3.2)
Thus if in 1 kg of wet steam 0.9 kg is the dry steam and 0.1 kg water particles then x = 0.9.
4. Total heat or enthalpy of wet steam (h). It is defined as the quantity of heat required to convert 1 kg
of water at 0°C into wet steam at constant pressure. It is the sum of total heat of water and the latent
heat and this sum is also called enthalpy.
In other words, h = hf + xhfg ...(3.3)
If steam is dry and saturated, then x = 1 and hg = hf + hfg.
5. Superheated steam. When steam is heated after it has become dry and saturated, it is called
superheated steam and the process of heating is called superheating. Superheating is always carried out
at constant pressure. The additional amount of heat supplied to the steam during superheating is called
as ‘Heat of superheat’ and can be calculated by using the specific heat of superheated steam at constant
pressure (cps), the value of which varies from 2.0 to 2.1 kJ/ kg K depending upon pressure and
temperature.
If Tsup., Ts are the temperatures of superheated steam in K and wet or dry steam, then (Tsup – Ts) is
called ‘degree of superheat’.
The total heat of superheated steam is given by hsup = hf + hfg + cps (Tsup – Ts ) ...(3.4)
Superheated steam behaves like a gas and therefore it follows the gas laws. The value of n for this type
of steam is 1.3 and the law for the adiabatic expansion is pv1.3 = constant.
The advantages obtained by using ‘superheated’ steam are as follows :
(i) By superheating steam, its heat content and hence its capacity to do work is increased without
having to increase its pressure.
(ii) Superheating is done in a superheater which obtains its heat from waste furnace gases which would
have otherwise passed uselessly up the chimney.
(iii) High temperature of superheated steam results in an increase in thermal efficiency.
(iv) Since the superheated steam is at a temperature above that corresponding to its pressure, it can be
considerably cooled during expansion in an engine before its temperature falls below that at which it
will condense and thereby become wet. Hence, heat losses due to condensation of steam on cylinder
walls etc. are avoided to a great extent.
6. Volume of wet and dry steam. If the steam has dryness fraction of x, then 1 kg of this steam will
contain x kg of dry steam and (1 – x) kg of water. If vf is the volume of 1 kg of water and vg is the volume
of 1 kg of perfect dry steam (also known as specific volume), then volume of 1 kg of wet steam = volume
of dry steam + volume of water.
= xvg + (1 – x)vf ...(3.5)
Note. The volume of vf at low pressures is very small and is generally neglected.
Thus is general, the volume of 1 kg of wet steam is given by, xvg and density 1 /xvg kg/m3 .
= xvg + vf – xvf = vf + x(vg – vf )
= vf + xvfg ...[3.5 (a)]
= vf + xvfg + vfg – vfg = (vf + vfg) – (1 – x) vfg = vg – (1 – x)vfg ...[3.5 (b)]
7. Volume of superheated steam. As superheated steam behaves like a perfect gas its volume can be
found out in the same way as the gases.
If, vg = Specific volume of dry steam at pressure
Ts = Saturation temperature in K,
Tsup = Temperature of superheated steam in K, and
vsup = Volume of 1 kg of superheated steam at pressure p,
----- 3.6
8. ENTROPY OF WET STEAM.
The total entropy of wet steam is the sum of entropy of water (sf ) and entropy of evaporation (sfg).
9. ENTROPY OF SUPERHEATED STEAM
Total entropy of superheated steam above the freezing point of water.
Ssup = Entropy of dry saturated steam + change of entropy during superheating
3.9. THERMODYNAMIC PROPERTIES OF STEAM AND STEAM TABLES
Following are the thermodynamic properties of steam which are tabulated in the form of table:
p = Absolute pressure (bar or kPa) ;
t s = Saturation temperature (°C) ; hf = Enthalpy of saturated liquid (kJ/kg) ;
hfg = Enthalpy or latent heat of vapourisation (kJ/kg) ;
hg = Enthalpy of saturated vapour (steam) (kJ/kg) ;
sf = Entropy of saturated liquid (kJ/kg K) ; sfg = Entropy of vapourisation (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.
The above mentioned properties at different pressures are tabulated in the form of tables as under: The
internal energy of steam (u = h – pv) is also tabulated in some steam tables.
3.10. ENTHALPY-ENTROPY (h-s) CHART OR MOLLIER DIAGRAM
Fig 3.11 .ENTHALPY-ENTROPY (h-s) CHART OR MOLLIER DIAGRAM
3.11 RANKINE CYCLE :
Rankine cycle is the theoretical cycle on which the steam turbine (or engine) works.
Fig 3.12 Rankine cycle
The Rankine cycle is shown in Fig. 3.12. It comprises of the following processes :
Process 1-2 : Reversible adiabatic expansion in the turbine (or steam engine).
Process 2-3 : Constant-pressure transfer of heat in the condenser.
Process 3-4 : Reversible adiabatic pumping process in the feed pump.
Process 4-1 : Constant-pressure transfer of heat in the boiler.
Fig. 3.12 shows the Rankine cycle on p-v, T-s and h-s diagrams (when the saturated steam enters the
turbine, the steam can be wet or superheated also).
Considering 1 kg of fluid : Applying steady flow energy equation (S.F.E.E.) to boiler, turbine, condenser
and pump :
(i) For boiler (as control volume), we get
hf4 + Q1 = h1
∴ Q1 = h1 – hf4
(ii) For turbine (as control volume), we get h1 = WT + h2, where WT = turbine work
∴ WT = h1 – h2
(iii) For condenser, We get
h2 = Q2 + hf3
∴ Q2 = h2 – hf3
(iv) For the feed pump, we get
hf3 + WP = hf4 , where, WP = Pump work
∴ WP = hf4 – hf3
Now, efficiency of Rankine cycle is given by
The feed pump handles liquid water which is incompressible which means with the increase in pressure
its density or specific volume undergoes a little change. Using general property relation for reversible
adiabatic compression, we get
Tds = dh – vdp
ds = 0 ∴ dh = vdp or ∆h = v ∆p ...... (since change in specific volume is negligible)
or hf4 – hf3 = v3 (p1 – p2)
When p is in bar and v is in m3/kg, we have
hf4 – hf3 = v3 (p1 – p2) × 105 J/kg
The feed pump term (hf4 – hf3 ) being a small quantity in comparison with turbine work, WT, is usually
neglected, especially when the boiler pressures are low. Then,
3.12 REGENERATIVE CYCLE
In the Rankine cycle it is observed that the condensate which is fairly at low temperature has an
irreversible mixing with hot boiler water and this results in decrease of cycle efficiency. Methods are,
therefore, adopted to heat the feed water from the hot well of condenser irreversibly by interchange of
heat within the system and thus improving the cycle efficiency. This heating method is called
regenerative feed heat and the cycle is called regenerative cycle.
The principle of regeneration can be practically utilised by extracting steam from the turbine at several
locations and supplying it to the regenerative heaters. The resulting cycle is known as regenerative or
bleeding cycle. The heating arrangement comprises of : (i) For medium capacity turbines—not more
than 3 heaters ; (ii) For high pressure high capacity turbines—not more than 5 to 7 heaters ; and (iii) For
turbines of super critical parameters 8 to 9 heaters. The most advantageous condensate heating
temperature is selected depending on the turbine throttle conditions and this determines the number of
heaters to be used. The final condensate heating temperature is kept 50 to 60°C below the boiler
saturated steam temperature so as to prevent evaporation of water in the feed mains following a drop
in the boiler drum pressure. The conditions of steam bled for each heater are so selected that the
temperature of saturated steam will be 4 to 10°C higher than the final condensate temperature.
Fig. 3.13 (a) shows a diagrammatic layout of a condensing steam power plant in which a surface
condenser is used to condense all the steam that is not extracted for feed water heating. The turbine is
double extracting and the boiler is equipped with a superheater. The cycle diagram (T-s) would appear
as shown in Fig. 3.13 (b). This arrangement constitutes a regenerative cycle.
Fig 3.13 REGENERATIVE CYCLE
Advantages of Regenerative cycle over Simple Rankine cycle :
1. The heating process in the boiler tends to become reversible.
2. The thermal stresses set up in the boiler are minimised. This is due to the fact that temperature
ranges in the boiler are reduced.
3. The thermal efficiency is improved because the average temperature of heat addition to the cycle is
increased.
4. Heat rate is reduced.
5. The blade height is less due to the reduced amount of steam passed through the low pressure stages.
6. Due to many extractions there is an improvement in the turbine drainage and it reduces erosion due
to moisture.
7. A small size condenser is required.
Disadvantages:
1. The plant becomes more complicated.
2. Because of addition of heaters greater maintenance is required.
3. For given power a large capacity boiler is required.
4. The heaters are costly and the gain in thermal efficiency is not much in comparison to the heavier
costs.
3.13 REHEAT CYCLE
For attaining greater thermal efficiencies when the initial pressure of steam was raised beyond 42 bar it
was found that resulting condition of steam after, expansion was increasingly wetter and exceeded in
the safe limit of 12 per cent condensation. It, therefore, became necessary to reheat the steam after
part of expansion was over so that the resulting condition after complete expansion fell within the
region of permissible wetness.
The reheating or resuperheating of steam is now universally used when high pressure and temperature
steam conditions such as 100 to 250 bar and 500°C to 600°C are employed for throttle. For plants of still
higher pressures and temperatures, a double reheating may be used.
In actual practice reheat improves the cycle efficiency by about 5% for a 85/15 bar cycle. A second
reheat will give a much less gain while the initial cost involved would be so high as to prohibit use of two
stage reheat except in case of very high initial throttle conditions. The cost of reheat equipment
consisting of boiler, piping and controls may be 5% to 10% more than that of the conventional boilers
and this additional expenditure is justified only if gain in thermal efficiency is sufficient to promise a
return of this investment. Usually a plant with a base load capacity of 50000 kW and initial steam
pressure of 42 bar would economically justify the extra cost of reheating.
The improvement in thermal efficiency due to reheat is greatly dependent upon the reheat pressure
with respect to the original pressure of steam.
Fig. 3.14 shows a schematic diagram of a theoretical single-stage reheat cycle. The corresponding
representation of ideal reheating process on T-s and h-s chart is shown in Figs. 3.14 (a and b).
a b
Figs 3.14 REHEAT CYCLE
Advantages of ‘Reheating’ :
1. There is an increased output of the turbine.
2. Erosion and corrosion problems in the steam turbine are eliminated/avoided.
3. There is an improvement in the thermal efficiency of the turbines.
4. Final dryness fraction of steam is improved.
5. There is an increase in the nozzle and blade efficiencies.
Disadvantages :
1. Reheating requires more maintenance.
2. The increase in thermal efficiency is not appreciable in comparison to the expenditure incurred in
reheating.
3.14 BINARY VAPOUR CYCLE
If we use steam as the working medium the temperature rise is accompanied by rise in pressure and at critical temperature of 374.15°C the pressure is as high as 225 bar which will create many difficulties in design, operation and control. It would be desirable to use some fluid other than steam which has more desirable thermodynamic properties than water. An ideal fluid for this purpose should have a very high critical temperature combined with low pressure. Mercury, diphenyl oxide and similar compounds, aluminium bromide and zinc ammonium chloride are fluids which possess the required properties in varying degrees. Mercury is the only working fluid which has been successfully used in practice. It has high critical temperature (588.4°C) and correspondingly low critical pressure (21 bar abs.). The mercury alone cannot be used as its saturation temperature at atmospheric pressure is high (357°C). Hence binary vapour cycle is generally used to increase the overall efficiency of the plant. Two fluids (mercury and water) are used in cascade in the binary cycle for production of power. Fig. 3.15 shows the schematic line diagram of binary vapour cycle using mercury and water as working fluids. The processes are represented on T-s diagram as shown in 3.16
Fig. 3.15 BINARY VAPOUR CYCLE
Fig. 3.16 BINARY VAPOUR CYCLE ON T-S DIAGRAM
PROBLEMS 1. A vessel having a capacity of 0.05 m3 contains a mixture of saturated water and saturated steam at a temperature of 245°C. The mass of the liquid present is 10 kg. Find the following : (i) The pressure, (ii) The mass, (iii) The specific volume, (iv) The specific enthalpy, (v) The specific entropy, and (vi) The specific internal energy.
2. Determine the amount of heat, which should be supplied to 2 kg of water at 25°C to convert it into steam at 5 bar and 0.9 dry.
3. What amount of heat would be required to produce 4.4 kg of steam at a pressure of 6 bar and temperature of 250°C from water at 30°C ? Take specific heat for superheated steam as 2.2 kJ/kg K.
3.1000 kg of steam at a pressure of 16 bar and 0.9 dry is generated by a boiler per hour. The steam passes through a superheater via boiler stop valve where its temperature israised to 380°C. If the temperature of feed water is 30°C, determine : (i) The total heat supplied to feed water per hour to produce wet steam. (ii) The total heat absorbed per hour in the superheater. Take specific heat for superheated steam as 2.2 kJ/kg K.
4. A pressure cooker contains 1.5 kg of saturated steam at 5 bar. Find the quantity of heat which must be rejected so as to reduce the quality to 60% dry. Determine the pressure and temperature of the steam at the new state.
5.Find the internal energy of 1 kg of steam at 20 bar when (i) it is superheated, its temperature being 400°C ; (ii) it is wet, its dryness being 0.9. Assume superheated steam to behave as a perfect gas from the commencement of superheating and thus obeys Charle’s law. Specific heat for steam = 2.3 kJ/kg K.
Solution. Mass of steam = 1 kg Pressure of steam, p = 20 bar Temperature of superheated steam = 400°C (Tsup = 400 + 273 = 673 K) Dryness fraction, x = 0.9 Specific heat of superheated steam, cps = 2.3 kJ/kg K
6. 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 passingthrough the calorimeter is at 1 bar pressure and 115°C. Calculate the dryness fraction of the steam in the main. Take cps = 2.1 kJ/kg K.
6.In a steam power cycle, the steam supply is at 15 bar and dry and saturated. The condenser pressure is 0.4 bar. Calculate the Carnot and Rankine efficiencies of the cycle. Neglect pump work.
7. In a steam turbine steam at 20 bar, 360°C is expanded to 0.08 bar. It then enters a condenser, where it is condensed to saturated liquid water. The pump feeds back the water into the boiler. Assume ideal processes, find per kg of steam the net work and the cycle efficiency.
8. A simple Rankine cycle works between pressures 28 bar and 0.06 bar, the initial condition of steam being dry saturated. Calculate the cycle efficiency, work ratio and specific steam consumption.
9. In a single-heater regenerative cycle the steam enters the turbine at 30 bar, 400°C and the exhaust pressure is 0.10 bar. The feed water heater is a direct contact type which operates at 5 bar. Find : (i) The efficiency and the steam rate of the cycle. (ii) The increase in mean temperature of heat addition, efficiency and steam rate as compared to the Rankine cycle (without regeneration). Pump work may be neglected.
10. Steam at a pressure of 15 bar and 250°C is expanded through a turbine at first to a pressure of 4 bar. It is then reheated at constant pressure to the initial temperature of 250°C and is finally expanded to 0.1 bar. Using Mollier chart, estimate the work done per kg of steam flowing through the turbine and amount of heat supplied during the process of reheat. Compare the work output when the expansion is direct from 15 bar to 0.1 bar without any reheat. Assume all expansion processes to be isentropic.
UNSOLVED EXAMPLES 1. Find the specific volume, enthalpy and internal energy of wet steam at 18 bar, dryness fraction 0.9. [Ans. 0.0994 m3/kg ; 2605.8 kJ/kg ; 2426.5 kJ/kg] 2. Find the dryness fraction, specific volume and internal energy of steam at 7 bar and enthalpy 2600 kJ/kg. [Ans. 0.921 ; 0.2515 m3/kg, 2420 kJ/kg] 3. Steam at 110 bar has a specific volume of 0.0196 m3/kg, find the temperature, the enthalpy and the internal energy. [Ans. 350°C ; 2889 kJ/kg ; 2673.4 kJ/kg]
4. Steam at 150 bar has an enthalpy of 3309 kJ/kg, find the temperature, the specific volume and the internal energy. [Ans. 500°C ; 0.02078 m3/kg ; 2997.3 kJ/kg] 5. Steam at 19 bar is throttled to 1 bar and the temperature after throttling is found to be 150°C. Calculate the initial dryness fraction of the steam. [Ans. 0.989] 6. Find the internal energy of one kg of steam at 14 bar under the following conditions : (i) When the steam is 0.85 dry ; (ii) When steam is dry and saturated ; and (iii) When the temperature of steam is 300°C. Take cps = 2.25 kJ/kg K. [Ans. (i) 2327.5 kJ/kg ; (ii) 2592.5 kJ/kg ; (iii) 2784 kJ/kg] 7. Calculate the internal energy of 0.3 m3 of steam at 4 bar and 0.95 dryness. If this steam is superheated at constant pressure through 30°C, determine the heat added and change in internal energy. [Ans. 2451 kJ/kg ; 119 kJ ; 107.5 kJ/kg] 8. Water is supplied to the boiler at 15 bar and 80°C and steam is generated at the same pressure at 0.9 dryness. Determine the heat supplied to the steam in passing through the boiler and change in entropy. [Ans. 2260.5 kJ/kg ; 4.92 kJ/kg K] 9. A cylindrical vessel of 5 m3 capacity contains wet steam at 1 bar. The volume of vapour and liquid in the vessel are 4.95 m3 and 0.05 m3 respectively. Heat is transferred to the vessel until the vessel is filled with saturated vapour. Determine the heat transfer during the process. [Ans. 104.93 MJ] 10. A pressure cooker contains 1.5 kg of steam at 5 bar and 0.9 dryness when the gas was switched off. Determine the quantity of heat rejected by the pressure cooker when the pressure in the cooker falls to 1bar. [Ans. – 2355 kJ] 11. A simple Rankine cycle works between pressure of 30 bar and 0.04 bar, the initial condition of steam being dry saturated, calculate the cycle efficiency, work ratio and specific steam consumption. [Ans. 35%, 0.997, 3.84 kg/kWh] 12. A steam power plant works between 40 bar and 0.05 bar. If the steam supplied is dry saturated and the cycle of operation is Rankine, find : (i) Cycle efficiency (ii) Specific steam consumption. [Ans. (i) 35.5%, (ii) 3.8 kg/kWh] 13. Compare the Rankine efficiency of a high pressure plant operating from 80 bar and 400°C and a low pressure plant operating from 40 bar 400°C, if the condenser pressure in both cases is 0.07 bar. [Ans. 0.391 and 0.357] 14. A steam power plant working on Rankine cycle has the range of operation from 40 bar dry saturated to 0.05 bar. Determine : (i) The cycle efficiency (ii) Work ratio (iii) Specific fuel consumption. [Ans. (i) 34.64%, (ii) 0.9957, (iii) 3.8 kg/kWh] 15. In a Rankine cycle, the steam at inlet to turbine is saturated at a pressure of 30 bar and the exhaust pressure is 0.25 bar. Determine : (i) The pump work (ii) Turbine work (iii) Rankine efficiency (iv) Condenser heat flow (v) Dryness at the end of expansion. Assume flow rate of 10 kg/s. [Ans. (i) 30 kW, (ii) 7410 kW, (iii) 29.2%, (iv) 17900 kW, (v) 0.763] 16. In a regenerative cycle the inlet conditions are 40 bar and 400°C. Steam is bled at 10 bar in regenerative heating. The exit pressure is 0.8 bar. Neglecting pump work determine the efficiency of the cycle. [Ans. 0.296]
1
UNIT 4 IDEAL AND REAL GASES, GAS MIXTURES AND THERMODYNAMIC RELATIONS
Gas mixtures - Properties of ideal and real gases - Equations of state - Avagadro’s law- Vanderwaal’s
equation of state - compressibility factor - compressibility chart - Dalton’s law of partial pressure - Exact
differentials - T-ds relations - Maxwell’s relations - Clausius Clapeyron equations - Joule - Thomson
coefficient.
4.1THE CHARACTERISTIC EQUATION OF STATE
At temperatures that are considerably in excess of critical temperature of a fluid, and also at very low
pressure, the vapour of fluid tends to obey the equation
Pv/ T = constant = R
In practice, no gas obeys this law rigidly, but many gases tend towards it. An imaginary ideal gas which
obeys this law is called a perfect gas, and the equation pv /T = R, is called the characteristic equation of a
state of a perfect gas. The constant R is called the gas constant. Each perfect gas has a different gas
constant.
Units of R are Nm/kg K or kJ/kg K.
Usually, the characteristic equation is written as pv = RT or for m kg, occupying V m3 pV = mRT ...
The characteristic equation in another form, can be derived by using kilogram-mole as a unit.
The kilogram-mole is defined as a quantity of a gas equivalent to M kg of the gas, where M is the
molecular weight of the gas (e.g., since the molecular weight of oxygen is 32, then 1 kg mole of oxygen is
equivalent to 32 kg of oxygen).
As per definition of the kilogram-mole, for m kg of a gas, we have m = nM
where n = number of moles. Substituting for m from eqn. gives
pV = nMRT or MR = pV nT
According to Avogadro’s hypothesis the volume of 1 mole of any gas is the same as the volume of 1
mole of any other gas, when the gases are at the same temperature and pressure.
Therefore, V/n is the same for all gases at the same value of p and T. That is the quantity pV/nT is a
constant for all gases. This constant is called universal gas constant, and is given the symbol, R0.
i.e., MR = R0 = pV/nT or pV = nR0T .
Since MR = R0, then R = Ro /M
It has been found experimentally that the volume of 1 mole of any perfect gas at 1 bar and 0°C is
approximately 22.71 m3. Therefore from eqn.
2
R0 = pV/ nT = 1 x105x 22. 71 /1 x273.15 = 8314.3 Nm/mole K
Using eqn.the gas constant for any gas can be found when the molecular weight is known.
4.2VAN DER WAALS’ EQUATION
Van der Waals’ equation (for a real gas) may be written as:
The constants a and b are specific constants and depend upon the type of the fluid considered,‘v’ represents the volume per unit mass and R is the gas constant. If the volume of one mole is considered then the above equation can be written as
The units of p, v , T, R, a and b are as follows :p (N/m2), v (m3/kg-mol), T (K) and R = 8314 Nm/kg mol K, a [Nm4/(kg-mol)2], b (m3/kgmol).
Constants of Van der Waals’ Equation
4.3REDUCED PROPERTIES
The ratios of pressure, temperature and specific volume of a real gas to the corresponding critical values are called the reduced properties.
4.4 LAW OF CORRESPONDING STATES
If any two gases have equal values of reduced pressure and reduced temperature, then they have same values of reduced volume ; i.e., vR = f(Tr , pr) for all gases and the function is the same. This law is most accurate in the vicinity of the critical point.
3
4.5 COMPRESSIBILITY CHART
The compressibility factor (Z) of any gas is a function of only two properties, usually temperature and pressure, so that Z = f(Tr, pr) except near the critical point. The value of Z for any real gas may be less or more than unity, depending on pressure and temperature conditions of the gas. The general compressibility chart is plotted with Z versus pr for various values of Tr. This is constructed by plotting the known data of one or more gases and can be used for any gas. Such a chart is shown in Fig. 8.10. This chart gives best results for the regions well removed from the critical state for all gases.
Fig 4.1 Generalized Compressibility Chart
4.6 DALTON’S LAW
The pressure of a mixture of gases is equal to the sum of the partial pressures of the constituents. The
partial pressure of each constituent is that pressure which the gas would exert if it occupied alone that
volume occupied by the mixtures at the same temperature.
By the consideration of mass, m = mA + mB
By Dalton’s law, p = pA + pB
Dalton’s law is based on experiment and is found to be obeyed more accurately by gas mixtures at low
pressures. The law can be extended to any number of gases,
i.e., m = mA + mB + mC + ...... or m = Σ mi
where mi = Mass of a constituent.
4
Similarly p = pA + pB + pC + ...... or p = Σpi where pi = The partial pressure of a constituent
4.7 GIBBS-DALTON LAW
The internal energy, enthalpy, and entropy of a gaseous mixture are respectively equal to the sums of
the internal energies, enthalpies, and entropies, of the constituents.
Each constituent has that internal energy, enthalpy and entropy, which it could have if it occupied alone
that volume occupied by the mixture at the temperature of the mixture.
This statement leads to the following equations:
mu = mAuA + mAuB + ..... or mu = Σ mi ui ... and
mh = mAhA + mBhB + ...... or mh = Σ mi hi ... and
ms = mAsA + mBsB + ...... or ms = Σ mi si .....
4.8 VOLUMETRIC ANALYSIS OF A GAS MIXTURE
Let us consider a volume V of a gaseous mixture at a temperature T, consisting of three constituents A, B and C. Let us further assume that each of the constituents is compressed to a pressure p equal to the total pressure of the mixture, and let the temperature remain constant. The partial volumes then occupied by the constituents will be VA, VB and VC.
5
We know
Thus, the volume of a mixture of gases is equal to the sum of the volumes of the individual constituents when each exists alone at the pressure and temperature of the mixture. This is the statement of another empirical law, the law of partial volumes, sometimes called Amagat’s law or Leduc’s law. The analysis of mixtures, often, is simplified if it is carried out in moles. The mole is given by the equation
where, n = Number of moles, m = Mass of gas, and M = Molecular weight. According to Avogadro’s law, the number of moles of any gas is proportional to the volume of the gas at a given pressure and temperature. Referring to Fig. ,the volume V contains n moles of the mixture at p and T. In Fig., the gas A occupies a volume VA at p and T, and this volume contains nA moles. Similarly there are nB moles of gas B in volume VB and nC moles of gas C in volume VC.
The total number of moles in the vessel must equal the sum of the moles of the individual constituents,
6
4.9 THE APPARENT MOLECULAR WEIGHT AND GAS CONSTANT
An apparent molecular weight is defined by the equation
M = m/n where, m = Mass of the mixture, and n = Number of moles of mixture. The apparent gas constant (similarly as above) is defined by the equation
It can be assumed that a mixture of perfect gases obeys all the perfect gas laws. In order to determine the gas constant for the mixture in terms of the gas constants of the constituents let us consider the equation pV = mRT both for the mixture and for a constituent as follows
We know pi V = ni R0T, and vi = pi/p x v Combining two equations and applied to the mixture (i.e., pV = nR0T), we have
This means that the molar analysis is identical with the volumetric analysis, and both are equal to the ratio of the partial pressure to the total pressure. The apparent molecular weight can be also be determined by the following method. Let us apply characteristic equation to each constituent and to mixture, we have
7
Also
4.10 SPECIFIC HEATS OF A GAS MIXTURE
PROBLEMS
1. The volume of a high altitude chamber is 40 m3. It is put into operation by reducing pressure from 1 bar to 0.4 bar and temperature from 25°C to 5°C. How many kg of air must be removed from the chamber during the process ? Express this mass as a volume measured at 1 bar and 25°C.
8
2. A vessel of capacity 3 m3 contains 1 kg mole of N2 at 90°C. (i) Calculate pressure and the specific volume of the gas. (ii) If the ratio of specific heats is 1.4, evaluate the values of cp and cv. (iii) Subsequently, the gas cools to the atmospheric temperature of 20°C ; evaluate the final pressure of gas. (iv)Evaluate the increase in specific internal energy, the increase in specific enthalpy, increase in specific entropy and magnitude and sign of heat transfer.
9
3. A container of 3 m3 capacity contains 10 kg of CO2 at 27°C. Estimate the pressure exerted by CO2 by using : (i) Perfect gas equation (ii) Van der Waals’ equation.
10
4. Calculate the density of N2 at 260 bar and 15°C by using the compressibility chart.
5. A vessel of 0.35 m3 capacity contains 0.4 kg of carbon monoxide (molecular weight = 28) and 1
kg of air at 20°C. Calculate : (i) The partial pressure of each constituent, (ii) The total pressure in the vessel, and The gravimetric analysis of air is to be taken as 23.3% oxygen (molecular weight = 32) and 76.7% nitrogen (molecular weight = 28).
11
6. A mixture of ideal gases consists of 4 kg of nitrogen and 6 kg of carbon dioxide at a pressure of 4 bar and a temperature of 20°C. Find : (i) The mole fraction of each constituent, (ii) The equivalent molecular weight of the mixture, (iii) The equivalent gas constant of the mixture, (iv) The partial pressures and partial volumes, (v) The volume and density of the mixture, and (vi) The cp and cv of the mixture. If the mixture is heated at constant volume to 50°C, find the changes in internal energy, enthalpy and entropy of the mixture. Find the changes in internal energy, enthalpy and entropy of the mixture if the heating is done at constant pressure. Take γ : for CO2 = 1.286 and for N2 = 1.4.
12
13
7. Following is the gravimetric analysis of air : Constituent Percentage Oxygen 23.14 Nitrogen 75.53 Argon 1.28 Carbon dioxide 0.05 Calculate the analysis by volume and the partial pressure of each constituent when the total pressure is 1 bar
14
4.11 FUNDAMENTALS OF PARTIAL DIFFERENTIATION
15
4.12 SOME GENERAL THERMODYNAMIC RELATIONS
16
The equations are known as Maxwell relations
17
4.13 ENTROPY EQUATIONS (Tds Equations)
18
4.14 EQUATIONS FOR INTERNAL ENERGY AND ENTHALPY
19
4.15 CLAUSIUS-CLAPERYON EQUATION Clausius-Claperyon equation is a relationship between the saturation pressure, temperature, the enthalpy of evaporation, and the specific volume of the two phases involved. This equation provides a basis for calculations of properties in a two-phase region. It gives the slope of a curve separating the two phases in the p-T diagram.
Fig.4.2 P-T Diagram
20
4.16 THROTTLING PROCESS AND JOULE-THOMPSON POROUS PLUG EXPERIMENT
Throttling process involves the passage of a higher pressure fluid through a narrow constriction. The effect is the reduction in pressure and increase in volume. This process is adiabatic as no heat flows from and to the system, but it is not reversible. It is not an isentropic process. The entropy of the fluid actually increases. Such a process occurs in a flow through a porous plug, a partially closed valve and a very narrow orifice. The porous plug is shown in Fig.
Fig. 4.3 The Joule-Thomson porous plug experiment
21
22
4.17JOULE-THOMPSON AND JOULE CO-EFFICIENTS When a real gas undergoes a throttling process a change in temperature takes place. Let us perform a series of the experiments on the same gas, keeping p1 and T1 constant, by varying the pressure downstream of the plug to various values p2, p3, p4 etc. After throttling let T1, T2, T3, T4 etc. be the corresponding temperatures. Now if a graph is plotted between p and T (Fig. 4.42), a smooth curve drawn through these points will be a curve of constant enthalpy because h1 = h2 = h3 = h4 etc. It may be noted that this curve does not represent the process executed by the gas in passing through the plug, since the process is irreversible and the gas does not pass through a sequence of equilibrium states. The slope of a constant enthalpy line or a p-T diagram at a particular state may be p
The slope of a constant enthalpy line or a p-T diagram at a particular state may be positive, zero or negative value. The slope is called Joule-Thompson co-efficient, μ and is given by
If we carry out other series of experiments similar to described above starting from different initial states, we can obtain a family of constant enthalpy curves as shown in Fig. 4.43. The states where μ = 0 are called ‘inversion states’ and locus of these states is called the inversion curve.
Fig. 4.4. Inversion curve.
23
The region inside the inversion curve is the cooling region since μ is positive, and temperature falls with fall in pressure. The region outside the inversion curve is the heating region since μ is negative and temperature rises with fall in pressure. Cooling can take place only if the initial temperature before throttling is below the maximum inversion temperature. This temperature is about 5Tc. The maximum inversion temperatures of some gases are given below
For free expansion of gases the experimental data obtained is limited. From the data available it appears that η is positive (i.e., cooling accompanies a fall in pressure or increase in specific volume).
UNIT 5 : FUELS AND COMBUSTION
Types of fuel, Combustion equations, Stoichiometric air, Excess air, Air-fuel ratio by volume & weight.
Calorific value - HCV & LCV. Bomb and Boy’s gas calorimeter, Proximate and ultimate analysis of fuel-
mass basis and volume basis, analysis of exhaust and flue gases.
5.1 INTRODUCTION
Fuel may be chemical or nuclear. Here we shall consider briefly chemical fuels only. A chemical fuel is a substance which releases heat energy on combustion. The principal combustible elements of each fuel are carbon and hydrogen. Though sulphur is a combustible element too but its presence in the fuel is considered to be undesirable. In a chemical reaction the terms, reactants and the products are frequently used. ‘Reactants’ comprise of initial constituents which start the reaction while ‘products’ comprise of final constituents which are formed by the chemical reaction. 5.2. CLASSIFICATION OF FUELS Fuels can be classified according to whether: 1. They occur in nature called primary fuels or are prepared called secondary fuels ; 2. They are in solid, liquid or gaseous state. The detailed classification of fuels can be given in a summary form as follows:
TABLE 5.1 TYPES OF FUELS
TYPE OF FUEL NATURAL (PRIMARY) PREPARED (SECONDARY)
Solid Wood ,Peat ,Lignite coal Coke Charcoal Briquettes
Liquid Petroleum Gasoline Kerosene Fuel oil Alcohol Benzol Shale oil
Gas Natural gas Petroleum gas Producer gas Coal gas Coke-oven gas Blast furnace gas Carburetted gas Sewer gas
5.3 BASIC CHEMISTRY Atoms. It is not possible to divide the chemical elements indefinitely, and the smallest particle which can take part in a chemical change is called an ‘atom’. If an atom is split as in nuclear reaction, the divided atom does not retain the original chemical properties. Molecules. It is rare to find elements to exist naturally as single atom. Some elements have atoms which exist in pairs, each pair forming a molecule (e.g. oxygen), and the atoms of each molecule are held together by stronger inter-atomic forces. The isolation of a molecule of oxygen would be tedious, but possible; the isolation of an atom of oxygen would be a different prospect. The molecules of some substances are formed by the mating up of atoms of different elements. For example, water has a molecule which consists of two atoms of hydrogen and one atom of oxygen. The atoms of different elements have different masses and these values are important when a quantitative analysis is required. The actual masses are infinitesimally small, and the ratios of the masses of atoms are used. These ratios are indicated by atomic weight quoted on a scale which defines the atomic weight of oxygen as 16. The symbols and molecular weights of some important elements, compounds and gases are given in Table 5.2
TABLE 5.2 SYMBOLS AND MOLECULAR WEIGHTS
5.4. COMBUSTION EQUATIONS
Equation (1)
Since the oxygen is accompanied by nitrogen if air is supplied for the combustion, then this nitrogen should be included in the equation. As nitrogen is inert as far as chemical reaction is concerned, it will appear on both sides of the equation. With one mole of oxygen there are 79/21 moles of nitrogen, hence equation (1) becomes,
-------- (2) 2. Combustion of carbon
-------- (3)
---------- (4)
------ (4a)
----- (4b)
------ (5)
------ (6)
----- (6a)
----- (7)
----- (7a) 5.5. THEORETICAL AIR AND EXCESS AIR The minimum amount of air that supplies sufficient oxygen for the complete combustion of all the carbon, hydrogen, and any other elements in the fuel that may oxidise is called the “theoretical air”. When complete combustion is achieved with theoretical air, the products contain no oxygen. In practice, it is found that complete combustion is not likely to be achieved unless the amount of air supplied is somewhat greater than the theoretical amount. Thus 150 per cent theoretical air means that air actually supplied is 1.5 times the theoretical air. The complete combustion of methane with minimum amount of theoretical air and 150 percent theoretical air respectively is written as:
--------- (8)
----- (9) The amount of air actually supplied may also be expressed in terms of per cent excess air. The excess air is the amount of air supplied over and above the theoretical air. Thus 150 per cent theoretical air is equivalent to 50 per cent excess air. The amount of excess air supplied varies with the type of fuel and the firing conditions. It may approach a value of 100% but modern practice is to use 25% to 50% excess air. 5.6. STOICHIOMETRIC AIR FUEL (A/F) RATIO Stoichiometric (or chemically correct) mixture of air and fuel is one that contains just sufficient oxygen for complete combustion of the fuel. A weak mixture is one which has an excess of air. A rich mixture is one which has a deficiency of air.
The percentage of excess air is given as under:
------- (10) The ratios are expressed as follows: For gaseous fuels By volume For solid and liquid fuels By mass For boiler plant the mixture is usually greater than 20% weak ; for gas turbines it can be as much as 300% weak. Petrol engines have to meet various conditions of load and speed, and operate over a wide range of mixture strength. The following definition is used:
--------- (11) The working value range between 80% (weak) and 120% (rich). Note. The reciprocal of the air fuel ratio is called the fuel-air (F/A) ratio. 5.7. AIR-FUEL RATIO FROM ANALYSIS OF PRODUCTS When analysis of combustion products is known air-fuel ratio can be calculated by the following methods: 1. Fuel composition known (i) Carbon balance method (ii) Hydrogen balance method (iii) Carbon-hydrogen balance method. 2. Fuel composition unknown (i) Carbon-hydrogen balance method. 1. Fuel composition known (i) Carbon balance method. When the fuel composition is known, the carbon balance method is quite accurate if combustion takes place with excess air and when free (solid) carbon is not present in the products. It may be noted that the Orsat analysis will not determine the quantity of solid carbon in the products. (ii) Hydrogen balance method. This method is used when solid carbon is suspected to be present. (iii) Carbon-hydrogen balance method. This method may be employed when there is some uncertainty about the nitrogen percentage reported by the Orsat analysis. 2. Fuel composition unknown When the fuel composition is not known the carbon-hydrogen balance method has to be employed. 5.8 CONVERTION OF VOLUMETRIC ANALYSIS TO WEIGHT ANALYSIS The conversion of volumetric analysis to weight analysis involves the following steps: 1. Multiply the volume of each constituent by its molecular weight. 2. Add all these weights and then divide each weight by the total of all and express it as percentage. 5.9 CONVERTION WEIGHT ANALYSIS TO VOLUMETRIC ANALYSIS 1. Divide the weight of each constituent by its molecular weight. 2. Add up these volumes and divide each volume by the total of all and express it as a percentage.
5.10 WEIGHT OF CARBON IN FLUE GASES The weight of carbon contained in one kg of flue or exhaust gas can be calculated from the amounts of CO2 and CO contained in it. In eqn. [4 (b)], it was shown that 1 kg of carbon produces 11/3 kg of CO2 when completely burnt. Hence 1 kg of CO2 will contain 3/11 kg of carbon. In eqn. [6 (a)], it can be seen that 1 kg of carbon produces 7/3 kg of CO, hence 1 kg CO contains 3/7 kg of carbon. Therefore, weight of carbon per kg of fuel
Where CO2 and CO are the quantities of carbon dioxide and carbon monoxide present in 1 kg of flue or exhaust gas. 5.11 WEIGHT OF FLUE GASES PER KG OF FUEL BURNT
Due to supply of air, the weight of flue gas or exhaust gas is always more than that of fuel burnt. The actual weight of dry flue gases can be obtained by comparing the weight of carbon present in the flue gases with the weight of carbon in the fuel, since there is no loss of carbon during the combustion process. As the analysis of the exhaust gases is volumetric, so this must first be reduced to weight analysis. Also, total weight of carbon in one kg of flue gas is
Therefore the weight of flue gas/kg of fuel burnt
5.12 ANALYSIS OF EXHAUST AND FLUE GAS The most common means of analysis of the combustion products is the Orsat apparatus which is described below: Construction : An Orsat’s apparatus consists of the following : (i) A burette (ii) A gas cleaner (iii) Four absorption pipettes 1, 2, 3, 4.
Fig. 5.1 Orsat apparatus
Pipette 1 : Contains ‘KOH’ (caustic soda) to absorb CO2 (carbon dioxide) Pipette 2 : Contains an alkaline solution of ‘pyrogallic acid’ to absorb O2 (oxygen) Pipette 3, 4 : Contain an acid solution of ‘cuprous chloride’ to absorb CO (carbonmonoxide) Furthermore the apparatus has a levelling bottle and a three way cock to connect the apparatus either to gases or to atmosphere. Procedure. 100 cm3 of gas whose analysis is to be made is drawn into the bottle by lowering the levelling bottle. The stop cock S4 is then opened and the whole flue gas is forced to pipette 1. The gas remains in this pipette for some time and most of the carbon dioxide is absorbed. The levelling bottle is then lowered to allow the chemical to come to its original level. The volume of gas thus absorbed is read on the scale of the measuring bottle. The flue gas is then forced through the pipette 1 for a number of times to ensure that the whole of the CO2 is absorbed. Further, the remaining flue gas is then forced to the pipette 2 which contains pyrogallic acid to absorb whole of O2. The reading on the measuring burette will be the sum of volume of CO2 and O2.The oxygen content can then be found out by subtraction. Finally, as before, the sample of gas is forced through the pipettes 3 and 4 to absorb carbon monoxide completely. The amount of nitrogen in the sample can be determined by subtracting from total volume of gas the sum of CO2, CO and O2 contents. Orsat apparatus gives an analysis of the dry products of combustion. Steps may have been taken to remove the steam from the sample by condensing, but as the sample is collected over water it becomes saturated with water. The resulting analysis is nevertheless a true analysis of the dry products. 5.13 CALORIFIC OR HEATING VALUES OF FUELS The “calorific value or heating value” of the fuel is defined as the energy liberated by the complete oxidation of a unit mass or volume of a fuel. It is expressed in kJ/kg for solid and liquid fuels and kJ/m3 for gases. If a fuel contains hydrogen water will be formed as one of the products of combustion. If this water is condensed, a large amount of heat will be released than if the water exists in the vapour phase. For this reason two heating values are defined the higher or gross heating value and the lower or net heating value. The higher heating value, HHV, is obtained when the water formed by combustion is completely condensed. The lower heating value, LHV, is obtained when the water formed by combustion exists completely in the vapour phase. Thus : (HHV)p = (LHV)p + m hfg ...(12) (HHV)v = (LHV)v + m(ug – uf) ...(13) where m = Mass of water formed by combustion, hfg = Enthalpy of vaporisation of water, kJ/kg, ug = Specific internal energy of vapour, kJ/kg, and uf = Specific internal energy of liquid, kJ/kg. In almost all practical cases, the water vapour in the products is vapour, the lower value is the one which usually applies.
5.14 DETERMINATION OF CALORIFIC OR HEATING VALUES The calorific value of fuels can be determined either from chemical analysis or in the laboratory. Solid and Liquid Fuels Dulong’s formula. Dulong suggested a formula for the calculation of the calorific value of the solid or liquid fuels from their chemical composition which is as given below. Gross calorific value or H.H.V. =
----- (14)
where C, H, O and S are carbon, hydrogen, oxygen and sulphur in percentages respectively in 100 kg of fuel. In the above formula the oxygen is assumed to be in combination with hydrogen and only extra surplus hydrogen supplies the necessary heat. Laboratory method (Bomb calorimeter)
Fig. 5.2 Bomb calorimeter
The calorimeter is made of austenitic steel which provides considerable resistance to corrosion and enables it to withstand high pressure. In the calorimeter is a strong cylindrical bomb in which combustion occurs. The bomb has two valves at the top. One supplies oxygen to the bomb and other releases the exhaust gases. A crucible in which a weighted quantity of fuel sample is burnt is arranged between the two electrodes as shown in Fig.5.2. The calorimeter is fitted with water jacket which surrounds the bomb. To reduce the losses due to radiation, calorimeter is further provided with a jacket of water and air. A stirrer for keeping the temperature of water uniform and a thermometer to measure the temperature up to accuracy of 0.001°C is fitted through the lid of the calorimeter. Procedure. To start with, about 1 gm of fuel sample is accurately weighed into the crucible and a fuse wire (whose weight is known) is stretched between the electrodes. It should be ensured that wire is in close contact with the fuel. To absorb the combustion products of sulphur and nitrogen 2 ml of water is poured in the bomb. Bomb is then supplied with pure oxygen through the valve to an amount of 25 atmosphere. The bomb is then placed in the weighed quantity of water, in the calorimeter. The stirring is started after making necessary electrical connections, and when the thermometer indicates a steady temperature fuel is fired and temperature readings are recorded after 1/2 minute intervals until maximum temperature is attained. The bomb is then removed; the pressure slowly released through the exhaust valve and the contents of the bomb are carefully weighed for further analysis. The heat released by the fuel on combustion is absorbed by the surrounding water and the calorimeter. From the above data the calorific value of the fuel can be found in the following way: Let wf = Weight of fuel sample (kg), w = Weight of water (kg), C = Calorific value (higher) of the fuel (kJ/kg), we = Water equivalent of calorimeter (kg), t1 = Initial temperature of water and calorimeter, t2 = Final temperature of water and calorimeter, tc = Radiation corrections, and c = Specific heat of water. Heat released by the fuel sample = wf × C Heat received by water and calorimeter = (ww + we) × c × [(t2 – t1) + tc]. Heat lost = Heat gained wf × C = (w + we) × c × [(t2 – t1) + tc]
[Value of c is 4.18 in SI units and unity in MKS units.]
Bomb calorimeter measures the higher or gross calorific value because the fuel sample is burnt at a constant volume in the bomb. JUNKER’S GAS CALORIMETER The calorific value of gaseous fuels can be determined by Junker’s gas calorimeter. Fig. 11.6 illustrates Junker’s gas calorimeter. Its principle is somewhat similar to Bomb calorimeter ; in respect that heat
evolved by burning the gas is taken away by the water. In its simplest construction it consists of a combustion chamber in which the gas is burnt (in a gas burner). A water jacket through which a set of tubes called flues pass surrounds this chamber. Thermometers are incorporated at different places (as shown in Fig. 11.6) to measure the temperatures. Procedure: A metered quantity of gas whose calorific value is to be determined is supplied to the gas burner via a gas meter which records its volume and a gas pressure regulator which measures the pressure of the gas by means of a manometer. When the gas burns the hot products of combustion travel upwards in the chamber and then downwards through the flues and finally escape to the atmosphere through the outlet. The temperature of the escaping gas is recorded by the thermometer fitted at the exit and this temperature should be as close to room temperature as possible so that entire heat of combustion is absorbed by water. The cold water enters the calorimeter near the bottom and leaves near the top. Water which is formed by condensation of steam is collected in a pot. The quantity of gas used during the experiment is accurately measured by the meter and temperature of ingoing and outgoing water is indicated by the thermometers. From the above data the calorific value of the gas can be calculated.
Fig. 5.3 JUNKER’S GAS CALORIMETER
PROBLEMS 1. A coal sample gave the following analysis by weight, Carbon 85 per cent, Hydrogen 6 per cent,
Oxygen 6 per cent, the remainder being incombustible. Determine minimum weight of air required per kg of coal for chemically correct composition.
2. The percentage composition of sample of liquid fuel by weight is, C = 84.8 per cent, and H2 = 15.2 per cent. Calculate (i) the weight of air needed for the combustion of 1 kg of fuel ; (ii) the volumetric composition of the products of combustion if 15 per cent excess air is supplied.
3. Percentage volumetric analysis of a sample of flue gases of a coal fired boiler gave CO2 = 10.4 ; CO = 0.2 ; O2 = 7.8 and N2 = 81.6 (by difference). Gravemetric percentage analysis of coal was C = 78, H2 = 6, O2 = 3 and incombustible = 13. Estimate : i) Weight of dry flue gases per kg of fuel ii) Weight of excess air per kg of fuel.
4. A single cylinder was supplied with a gas having the following percentage volumetric analysis ;
CO = 5, CO2 = 10, H2 = 50, CH4 = 25, N2 = 10. The percentage volumetric analysis of dry gases was CO2 = 8, O2 = 6 and N2 = 86. Determine the air-fuel ratio by volume.
5. The following is the ultimate analysis of a sample of petrol by weight: Carbon = 85 percent ; Hydrogen = 15 percent. Calculate the ratio of air to petrol consumption by weight if the volumetric analysis of the dry exhaust gas is: CO2 = 11.5 percent; CO = 1.2 percent ; O2 = 0.9 per cent ; N2 = 86 per cent. Also find percentage excess air.
6. A sample of fuel has the following percentage composition : Carbon = 86 per cent ; Hydrogen = 8 per cent ; Sulphur = 3 per cent ; Oxygen = 2 per cent ; Ash = 1 per cent. For an air-fuel ratio of 12 : 1, calculate : (i) Mixture strength as a percentage rich or weak. (ii) Volumetric analysis of the dry products of combustion.
