TAKORADI TECHNICAL UNIVERSITY
FACULTY OF ENGINEERING
MECHANICAL ENGINEERING DEPARTMENT
BTECH 1 (LECTURE NOTES)
BASIC THERMODYNAMICS (BME 312)
LECTURE 1
BASIC CONCEPTS OF THERMODYNAMIC
1.1
Introduction
Thermodynamics is concerned with relation between Heat, Work and Energy (also how to generate
work from heat).
Applications of thermodynamics

Stationary and Mobile power-producing Devices

Refrigeration and Air-Conditioning

Jet Engine and Rockets

Fluid Expanders, Compressors etc
Thermodynamics, basically entails four laws known as Zeroth, First, Second and Third law of
thermodynamics.
 The First law throws light on concept of internal energy.
 The Zeroth law deals with thermal equilibrium and establishes a concept of temperature.
 The Second law indicates the limit of converting heat into work and introduces the principle
of increase of entropy.
 The Third law defines the absolute zero of entropy.
These laws are based on experimental observations and have no mathematical proof. Like all physical
laws, these laws are based on logical reasoning.
1.2
Basic Definitions
• Thermodynamic system: A quantity of fixed mass under investigation or the object of
study under review
• Surroundings: Everything external to the system,
• System boundary: Interface separating system and surroundings.
• Universe: Combination of system and surroundings.
Using the figure below we can illustrate the system, surroundings, and system boundary for
a universe with the potato shaped system.
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Fig 1.1 System, Boundary and Surroundings
It can be deduced that two important interactions occur between the system and its
surroundings:
• heat can cross into the system (our potato can get hot), and
• work can cross out of the system (our potato can expand).
Similarly, the system boundaries can change, as illustrated above the potato might expand
on heating, but we can still distinguish the system and the surroundings.
Systems can be defined based on the interactions between the systems and its
surroundings. Systems can be:
•
Open System: System in which mass and energy can transfer between the system
and the surroundings. Example of open systems are boilers, turbines, condensers,
pumps, compressors, diffusers, nozzles, evaporators etc.
•
Closed System: System in which there is a transfer of energy between system and its
surrounding but not mass. Eg. Piston-cylinder arrangement without valves etc.
•
Isolated System: A system that is not influence by its surroundings, that is, neither
mass nor energy can transfer between the system and the surroundings. Eg. Thermos
flask, universe etc.
•
Adiabatic System: An adiabatic system is one which is thermally insulated from its
surroundings. It can, however, exchange work with its surroundings. If it does not, it becomes an
isolated system
.
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•
Phase. A phase is a quantity of matter which is homogeneous throughout in chemical
composition and physical structure.
•
Homogeneous System: A system which consists of a single phase is termed as homogeneous
system. Examples : Mixture of air and water vapour, water plus nitric acid and octane plus heptane
•
Heterogeneous System: A system which consists of two or more phases is called a
heterogeneous system. Examples : Water plus steam, ice plus water and water plus oil.
1.2.1
PURE SUBSTANCE
A pure substance is one that has a homogeneous and invariable chemical composition even though there
is a change of phase. In other words, it is a system which is (a) homogeneous in composition, (b)
homogeneous in chemical aggregation. Examples: Liquid, water, mixture of liquid water and steam,
mixture of ice and water. The mixture of liquid air and gaseous air is not a pure substance.
1.2.2
THERMODYNAMIC EQUILIBRIUM
A system is in thermodynamic equilibrium if the temperature and pressure at all points are same.
Systems under temperature and pressure equilibrium but not under chemical equilibrium are sometimes
said to be in metastable equilibrium conditions. It is only under thermodynamic equilibrium conditions
that the properties of a system can be fixed.
Thus for attaining a state of thermodynamic equilibrium the following three types of equilibrium states
must be achieved:
1.
Thermal equilibrium. The temperature of the system does not change with time and has same
value at all points of the system.
2.
Mechanical equilibrium. There are no unbalanced forces within the system or between the
surroundings. The pressure in the system is same at all points and does not change with respect to time.
3.
Chemical equilibrium. No chemical reaction takes place in the system and the chemical
composition which is same throughout the system does not vary with time.
1.2.3
PROPERTIES OF SYSTEMS
A property of a system is a characteristic of the system which depends upon its state, but not upon how
the state is reached. There are two sorts of property:
1.
Intensive properties. These properties do not depend on the mass of the system. Examples:
Temperature and pressure.
2.
Extensive properties. These properties depend on the mass of the system. Example: Volume.
Extensive properties are often divided by mass associated with them to obtain the intensive properties.
For example, if the volume of a system of mass m is V, then the specific volume of
𝑉
matter within the system is 𝑣 = 𝑚 which is an intensive property.
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STATE
State is the condition of the system at an instant of time as described or measured by its properties. Or
each unique condition of a system is called a state.
PROCESS
A process occurs when the system undergoes a change in a state or an energy transfer at a steady state.
A process may be non-flow in which a fixed mass within the defined boundary is undergoing a change
of state. Example : A substance which is being heated in a closed cylinder undergoes a non-flow
process . Closed systems undergo non-flow processes. A process may be a flow process in which mass
is entering and leaving through the boundary of an open system. In a steady flow process mass is
crossing the boundary from surroundings at entry, and an equal mass is crossing the boundary at the
exit so that the total mass of the system remains constant. In an open system it is necessary to take
account of the work delivered from the surroundings to the system at entry to cause the mass to enter,
and also of the work delivered from the system at surroundings to cause the mass to leave, as well as
any heat or work crossing the boundary of the system.
Quasi-static process. Quasi means ‘almost’. A quasi-static process is also called a reversible
process.
Other processes generally used
Isobaric - the pressure p of the system is constant
Monobaric – the external pressure po is constant
Isochoric – the volume V of the system is constant
Isothermal – the temperature T of the system is constant
Mono-thermal – the outside temperature To is constant
Adiabatic – the system evolves without heat exchange with its surroundings (no heat transfer between
the system and the surroundings
CYCLE
Any process or series of processes whose end states are identical is termed a cycle.
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p (Pressure)
2
1
3
V (Volume)
Fig 1.2 Cycle of Operation
1.3
TEMPERATURE
The temperature is a thermal state of a body which distinguishes a hot body from a cold body. The
temperature of a body is proportional to the stored molecular energy. Instruments for measuring
ordinary temperatures are known as thermometers and those for measuring high temperatures are
known as pyrometers.
It has been found that a gas will not occupy any volume at a certain temperature. This temperature is
known as absolute zero temperature. The temperatures measured with absolute zero as basis are called
absolute temperatures. Absolute temperature is stated in degrees centigrade. The point of absolute
temperature is found to occur at 273.15°C below the freezing point of water.
Then : Absolute temperature = Thermometer reading in °C + 273.15.
Absolute temperature is degree centigrade is known as degrees kelvin, denoted by K (SI unit).
1.3.1
ZEROTH LAW OF THERMODYNAMICS
‘Zeroth law of thermodynamics’ states that if two systems are each equal in temperature to a third,
they are equal in temperature to each other.
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Fig 1.3 Zeroth Law of thermodynamics
1.4
Energy
In thermodynamics, we deal with change of the total energy only. Thus, the total energy of a
system can be assigned a value of zero at some reference point. Total energy of a system has
two groups: macroscopic and microscopic
Macroscopic forms of energy: forms of energy that a system possesses as a whole with
respect to some outside reference frame, such as kinetic and potential energy. The macroscopic
energy of a system is related to motion and the influence of some external effects such as
gravity, magnetism, electricity, and surface tension
Kinetic energy: energy that a system possess as a result of its relative motion relative to
some reference frame, KE
𝑚𝑣 2
𝐾𝐸 = 2
(KJ)
1.1
Where V is the velocity of the system in (m/s)
Potential energy: is the energy that a system possess as a result of its elevation in a
gravitational field, PE
PE = mgz
(KJ)
1.2
Where g is the gravitational acceleration and z is the elevation of the center of gravity of the
system relative to some arbitrary reference plane.
Microscopic forms of energy: are those related to molecular structure of a system. They are
independent of outside reference frames. The sum of microscopic energy
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Is called the internal energy, U. The total energy of a system consists of the kinetic, potential,
and internal energies:
𝐸 = 𝑈 + 𝐾𝐸 + 𝑃𝐸 = 𝑈 + (
1.5
𝑚𝑉 2
2
) + mgz
(KJ)
1.3
Work and Heat
Work
Work is said to be done when a force moves through a distance. If a part of the boundary of a system
undergoes a displacement under the action of a pressure, the work done W is the product of the force
(pressure × area), and the distance it moves in the direction of the force. Fig. 2.31 (a) illustrates this
with the conventional
Sign Convention
If the work is done on the system by the surroundings, e.g., when a force is applied to a rotating handle,
or to a piston to compress a fluid, the work is said to be negative (-W)
Work input to the system = -W
If the work is done by the system on the surroundings, e.g., when a fluid expands pushing a piston
outwards, the work is said to be positive.
Work output of the system = + W
Heat
When a system changes its state due to the difference in temperature between the system and its
surroundings, the outcome is heat produced.
Sign convention
If the heat flows into a system from the surroundings, the quantity is said to be positive and, conversely,
if heat flows from the system to the surroundings it is said to be negative.
In other words :
Heat received by the system = + Q
Heat rejected or given up by the system = – Q.
1.5.1
Properties of Pure Substance
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A pure substance is a system which is (i) homogeneous in composition, (ii) homogeneous in chemical
aggregation, and (iii) invariable in chemical aggregation.
“Homogeneous in composition” means that the composition of each part of the system is the same as
the composition of every other part. “Composition means the relative proportions of the chemical
elements into which the sample can be analysed. It does not matter how these elements are combined.
“Homogeneous in chemical aggregation” means that the chemical elements must be combined
chemically in the same way in all parts of the system.
Invariable in chemical aggregation” means that the state of chemical combination of the system does
not change with time.
1.5.2
PHASE 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.
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. The piston starts moving upwards.
Fig. 1.4. Phase change of water at constant pressure from liquid vapour
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. 1.4 (ii). This state is
described by the line 2-3 in Fig. 1.5. 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 the volume increases rapidly and piston moves upwards
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Fig. 1.5
Fig. 1.6. Vapour pressure curve for water.
For a pure substance, definite relationship exists between the saturation pressure and saturation
temperature as shown in Fig., 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.
1.5) 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. 1.5 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. 1.4 (iii)].
The difference between the superheated temperature and the saturation temperature at the given
pressure is called the degree of superheat.

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.
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If the substance at the triple point is compressed until
Liquid
phase
there is no vapour left and the pressure on the
Critical
point
resulting mixture of liquid and solid is increased, the
temperature will have to be changed for equilibrium
toexist between the solid and the liquid.
Measurements of these pressures and temperatures
Solid
phase
Triple
point
Vapour
phase
give rise to a third curve on the p-T diagram, starting
atthe triple point and continuing indefinitely.
Temperature(T)
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 Fig. 3.5. p-T diagram for a substance such as water.
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.
Triple point
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, and on a U-V diagram it is a triangle.
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.
FORMATION OF STEAM
Consider a cylinder fitted with a piston which can move freely upwards and downwards in it. Let, for
the sake of simplicity, there be 1 kg of water at 0°C with volume vf m3 under the piston [Fig .1.7 (i)].
Further let the piston is loaded with load W to ensure heating at constant pressure. Now if the heat is
imparted to water, a rise in temperature will be noticed and this rise will continue till boiling point is
reached. The temperature at which water starts boiling depends upon the pressure and as such for each
pressure (under which water is heated) there is a different boiling point. This boiling temperature is
known as the temperature of formation of steam or saturation temperature.
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tsup = Temperature of superheated steam
v = Volume of dry and saturated steamg vsup=
Volume of superheated steam
Now, if supply of heat to water is continued it will be noticed that rise of temperature after the boiling
point is reached nil but piston starts moving upwards which indicates that there is increase is volume
which is only possible if steam formation occurs. The heat being supplied does not show any rise of
temperature but changes water into vapour state (steam) and is known as latent heat or hidden heat. So
long as the steam is in contact with water, it is called wet steam [ (iii)] and if heating of steam is further
progressed [as shown (iv)] such that all the water particles associated with steam are evaporated, the
steam so obtained is called dry and saturated steam. If vg m3 is the volume of 1 kg of dry and saturated
steam, then work done on the piston will be
p(vg – vf)
where p is the constant pressure (due to weight ‘W’ on the piston).
Again, if supply of heat to the dry and saturated steam is continued at constant pressure there will be
increase in temperature and volume of steam. The steam so obtained is called superheated steam and it
behaves like a perfect gas.
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Tsup
Evaporation
Ts
Latentheat
(hfg)
Heataddition
Fig. 1.8. Graphical representation of formation of steam.
TERMS RELATING TO STEAM FORMATION

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.

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
mg = Mass of dry steam contained in steam considered, and
mw = Weight of water particles in suspension in the steam considered
𝑚𝑔
Then 𝑥 = 𝑚 +𝑚
𝑠

𝑤
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
Any fluid which is used as working fluid, the six basic thermodynamic properties required are : p
(pressure), T (temperature), v (volume), u (internal energy), h (enthalpy) and s (entropy). These
properties must be known at different pressure for analysing the thermodynamic cycles used for work
producing devices. The values of these properties are determined theoretically or experimentally and
are tabulated in the form of tables which are known as ‘Steam Tables’
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The following should be noted
ℎ𝑓𝑔 = ℎ𝑔 − ℎ𝑓
change of enthalpy during evaporation
𝑆𝑓𝑔 = 𝑆𝑔 − 𝑆𝑓
change of entropy during evaporation
𝑉𝑓𝑔 = 𝑣𝑔 − 𝑣𝑓
change of volume during evaporation
Steam Table
Absolute
Temperature
Specific enthalpy
Specific entropy
Specific volume
pressure
°C
kJ/kg
kJ/kg K
m3/kg
bar, p
ts
hf
hfg
hg
sf
1.0
99.6
417.5
2257.9
2675.4 1.3027
50.0
263.9
1154.9
1639.7
2794.2 2.9206
100.0
311.1
1408.0
1319.7
2727.7 3.3605
sfg
sg
vf
vg
6.0571
7.3598
0.001043
1.6934
3.0529
5.9735
0.001286
0.00394
2.2593
5.6198
0.001452
0.01811
Tutorials
1. Determine enthalpy, specific volume, and entropy for a mixture with 15% quality at 2.9 bar,
Take 𝑣𝑓 = 0.001053 𝑚3/𝐾𝑔
2. Determine the enthalpy of steam at 3.0 bar using the steam tables for the following
conditions. a. Dryness fraction x = 0
b. Dryness fraction x = 0.5
c. Dryness fraction x = 1
1.6
Gas Laws
Boyle's Law
The Volume of a given mass of a perfect gas varies inversely as the absolute pressure when
the temperature is constant
Let P be the absolute pressure, V be the volume at pressure, T is the absolute temperature.
Therefore, V α 1/p when T is constant
PV = constant.
1.4
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Charles' Law
The Volume of a given mass of a gas varies directly as its absolute temperature if the pressure
remains unchanged.
Let P be the absolute pressure, V be the Volume of gas, T be the absolute temperature
Therefore, V α T when P is constant
V/T = constant
1.7
1.5
Equation of State
In Engineering practice, volume, pressure and temperature all vary simultaneously and
therefore, Boyle's or only Charles' law is not applicable. From the two Laws, we get a general
law:
V α 1/p when T is constant
1.6
V α T when P is constant
1.7
V α T/P when T and P both vary
1.8
PV = Ro T per unit mole where Ro is the universal gas constant, Ro = 8.314 KJ/Kg mol.K
For any particular gas
PV = mRT
1.9
Where R = specific or characteristic gas constant for the particular gas; R = Ro/M, M = molar
mass.
Specific Heat Capacities
If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the temperature of
the gas by one degree whilst the volume remains constant, then the amount of heat energy
supplied is known as the specific heat capacity at constant volume Cv. If the pressure rather
remains constant, the heat supplied is the specific heat capacity at constant pressure Cp
𝑑𝑄 = 𝑚𝑐𝑑𝑇
Where m = mass
C = specific heat
dT = Temperature rise
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Therefore, two specific heats for gases
𝑑𝑄 = 𝑚𝑐𝑝 𝑑𝑇 For reversible non flow process at constant pressure = 𝑚𝑐𝑝 (𝑇2 − 𝑇1 )
𝑑𝑄 = 𝑚𝑐𝑣 𝑑𝑇 For reversible non flow process at constant volume = 𝑚𝑐𝑉 (𝑇2 − 𝑇1 )
ENTHALPY
One of the fundamental quantities which occur invariably in thermodynamic is the sum of internal
energy (u) and the pressure volume product (pv). The sum is called Enthalpy
h = u+pv
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LECTURE 2
FIRST LAW OF THERMODYNAMICS AND ITS APPLICATION
2.1
Introduction
In thermodynamics, there are two laws of very general validity, such as, they do not depend on the
details of the interactions or the systems being studied. Hence, they can be applied to systems about
which one knows nothing other than the balance of energy and matter transfer.
2.1.2
The First Law of Thermodynamics
The first law is an expression of the principle of the conservation of energy. With its aid, it is
possible to calculate the quantities of heat and work that cross the boundary. It is first stated
with reference to closed system with application to various processes undergone by such
system and the applied to open system. The first law states that:
When any closed system is taken through a cycle, the net work delivered to the
surrounding is equal to the net heat taken from the surrounding.
However, the converse is also true
When any open system is taken through a cycle, the net work done on the system is
equal to the net heat delivered to the surrounding
∑ δQ =∑δW
∑ (δQ – δW) = 0
2.1
COROLLARIES
1. There exists a property of a closed system such that a change in its value is equal to
the difference between the heat supplied and the work done during any change of state.
2. The internal energy of a closed system remains unchanged if the system is isolated
from its surrounding
3. A perpetual motion machine of the first kind is impossible. i.e. A machine producing
a continuous supply of work without absorbing energy from the surrounding
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From the Corollaries, the first law can be re –stated as:
The change of energy of a system during a change of state is numerically equal
to the net heat transfer during the process minus the net work transfer during
the process
If the property so discovered (corollary 1) is denoted by U, the corollary can be expressed
mathematically as
∑21 (𝛿𝑄 − 𝛿𝑊) = 𝑈2 − 𝑈1
2.2
Or writing Q and W for the net quantities of heat and work crossing the boundary during the
change of state
𝑄12 − 𝑊12 = 𝑈2 − 𝑈1
2.3
For a particular fluid, the change in internal energy is always given by
∆𝑈 = 𝑚𝑐𝑉 ( 𝑇2 − 𝑇1)
2.4
Where m = mass of fluid, Cv = specific heat capacity at constant volume.
a) Constant Volume Process
In this process, there is displacement of work therefore, W = 0 and with no paddle work.
𝑄 = 𝑈2 − 𝑈1
2.5
That is, the quantity of heat is equal to the change of internal energy. If in addition to the 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 energy equation can be written
in differential form as
dQ – du
2.6
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For a perfect gas as the working fluid
𝑄12 = 𝑚𝑐𝑣 (𝑇2 − 𝑇1)
2.7
For a perfect gas we also have the following relations between properties which are valid for
any two states of the same volume. From the equation of state:
Example 2.1
A fluid in a closed vessel of fixed volume 0.14m3 exerts a pressure of 10 bars at 250oC. If
the vessel is cooled so that the pressure falls to 3.5 bar, determine the final temperature and
the heat transferred using air as the working fluid.
b) Constant Pressure process
This is the process where the working fluid expands or contracts in the presence of a
constant pressure force under reversible condition,
𝑑𝑄 − 𝑝𝑑𝑣 = 𝑑𝑢
2.9
Since p is constant, this can be integrated to give
𝑄12 − 𝑝(𝑣2 − 𝑣1) = (𝑢2 − 𝑢1) … … … … … … . . 2.10
Also, since p is constant, pdv is identical with d(pv). Thus the energy equation can be
written as
𝑑𝑄 = 𝑑(𝑢 + 𝑝𝑣) = 𝑑ℎ … … … … … … … … 2.11
Or in the integrated form
𝑄12 = (ℎ 2 − ℎ1) … … … … … … … … … … … 2.12
Where h, a property, is called enthalpy
Example 2.2
A mass of 0.2 kg of fluid, initially at a temperature of 165 oC, expands reversibly at a constant
pressure of 7 bars until the volume is doubled, Find the final temperature, work, and heat
transfers. when the fluid is air. (Take R = 0.287 KJ/Kg.k)
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c)
Polytropic process
In this process both heat energy and work energy cross the boundary of the system. The
expansion and compression states are found to follow approximately a relation of the form pvn
= constant, where 'n' is the index of expansion or compression and p and v are the average
values of pressures and specific volume. For the reversible polytropic process, single values
of p and v can truly define the sate of a system, and dw = pdv. The work done per unit mass
during a change from state 1 to state 2 may then be found by integrating as follows. For initial,
final and any intermediate state,
Therefore
2.14
𝑊=
𝑃2 𝑉2 𝑉2 1−𝑛 −𝑃1 𝑉1 𝑉1 1−𝑛
1−𝑛
𝑃 𝑉 −𝑃1 𝑉1
= 22
1−𝑛
2.15
The integrated form of the energy equation for a reversible polytropic process may therefore
be written as
Example 2.3
A mass of 0.9 kg of fluid, initially at a pressure of 15 bar and a temperature of 250 oC, expands
reversibly and polytropically to 1.5 bar. Find the work if the index of expansion is 1.25 and V2
is 3.21 m3/Kg
d)
Adiabatic process
It is a process during which heat is prevented from crossing the boundary of the system. From
the first law
𝑄 − 𝑊 = 𝑈2 − 𝑈1 … … … … … . . (2.17)
𝑊 = 𝑈2 − 𝑈1 … … … … … … (2.18)
For a reversible adiabatic process (isentropic) involving a perfect gas, there is a definite
relation between p and v given by
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e)
Isothermal process
When the quantities of heat and work are so proportioned during an expansion or compression
that the temperature of the fluid remains constant, the process is said to be isothermal. Since
temperature gradients are excluded by definition, the reversibility of an isothermal process is
implied.
For a perfect gas, the expansion and compression states follow the relation pv = constant and
so
2.1.3
Steady- Flow Energy Equation
Let us consider the steady-flow of a fluid through the device illustrated below. Energy in the
form of heat is supplied steadily to this device and mechanical work is done by it. Rotating
shaft. This external mechanical work is usually 'shaft work'. The conditions which must be
satisfied by these processes are:
Assumptions
1. The flow is steady and continuous, that is the rate at which mass enters the region
considered equals the rate at which mass leaves the region
2. Conditions at any point between the inlet and outlet sections do not vary with time
3. Heat and shaft works are transferred to or from the fluid at a constant rate
4. Quantities are uniform over the inlet and outlet cross sections
5. Energy due to electricity, magnetism, surface tension or nuclear reaction is absent.
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Boundary
C=
l
2 A 11
A2
l2
System
C1
2
p2
l
22
p1
A1
Z2
l1
Z1
Datum plane
Fig 2.1 Flow Device
To analyze a steady flow process the fact is used that energy cannot be created or destroyed but only
converted from one form to another. The resulting energy balance for a steady flow process is called
the Steady Flow Energy Equation.
For a system undergoing a steady flow process, the working fluid will flow along the inlet pipe at a
constant rate and enter the system, where various energy transfers take place, depending on the fraction
of the system (e.g. for a boiler, heat energy would cross the boundary and enter the system; for an
engine, work energy would cross the boundary and leave the system). When these energy transfers have
been completed, the fluid will flow out of the system along the outlet of the pipe.
To obtain an energy balance for the system, consider all the amounts of energies crossing the boundary
per unit time.
Let suffix (1) denote conditions at inlet to the system
Let suffix (2) denote conditions at outlet to the system
The conservation of energy equation for an open system undergoing a process can be expressed as:
(Total energy crossing boundary as heat and work) + (Total Energy of mass entering system) – (Total
Energy of mass leaving system) = (Net change in energy of system)
𝑄 − 𝑊 + ∑ 𝐸𝑖𝑛 − ∑ 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡𝑒𝑚
2.17
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During a steady- flow process the total energy content of an open system remains constant i.e ∆𝐸𝑠𝑦𝑠𝑡𝑒𝑚,
therefore for a steady- flow process
The total amount of energy entering the system per second must be equal to the total amount of
energy leaving the system per second
The energy entering the system will be made up of

Any heat energy transfers crossing the boundary into the system –Q

The energy of the fluid flowing into the system per second. This will include internal energy
𝑈1 , potential energy 𝑔𝑧1 , and kinetic energy (𝐶1 )2 / 2

In order for the fluid to enter the system, it must displace the fluid immediately preceding it
which is just inside the system i.e. the fluid entering the system must transfer work energy into
the system. Since the pressure remains constant at P1 N/m2 , the amount of work energy
transferred to the system per second per unit mass flow rate will be equal to 𝑝1 𝑣1
Therefore the total energy entering the system
2
𝑐
= 𝑄 + 𝑚 ̇ (𝑢1 + 𝑔𝑧1 + 21 ) + 𝑚𝑝1 𝑣1…………………………………. (2.18)
Where m is the mass flow rate in kg/s
The energy leaving the system will be made up of

Any work energy transfers crossing the boundary out of the system

The energy of the fluid flowing out of the system per second. This will include internal energy
𝑈1 , potential energy, 𝑔𝑧2 , and kinetic energy,( 𝐶2 )2 / 2

In order for the fluid to leave the system, it must displace the fluid immediately preceding it
which is just outside the system, i.e. the fluid leaving the system must transfer work energy to
the surroundings. Since the pressure remains constant at P2 N/m2, the amount of work energy
transferred to the surroundings per second per unit mass flow rate will be equal to 𝑝2 𝑣2
The total energy leaving the system per second
= 𝑊𝑠 + 𝑚̇ (𝑈2 + 𝑔𝑧2 +
𝐶22
) + 𝑚𝑝2 𝑣2
2
2.19
Since energy entering system per sec = energy leaving system per sec.
2
2
2
2
𝑐
𝐶
𝑄 + 𝑚 ̇ (𝑢1 + 𝑔𝑧1 + 1 ) + 𝑚𝑝1 𝑣1 = 𝑊𝑠 + 𝑚̇ (𝑈2 + 𝑔𝑧2 + 2 ) + 𝑚𝑝2 𝑣2
2.20
Therefore:
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𝑄 − 𝑊𝑠 = 𝑚̇ ((𝑈2 + 𝑃2 𝑉2 ) − (𝑈1 + 𝑃1 𝑉1 ) +
𝐶22 − 𝐶11
+ (𝑔𝑧2 − 𝑔𝑧1 ))
2
2.21
Since the potential energy term is generally small compared with the other terms, it is usually neglected
and the equation is normally written as:
𝑄 − 𝑊𝑠 = 𝑚̇(ℎ2 − ℎ1 ) +
𝐶22 − 𝐶11
… … … … … … … … … . .2.22
2
Applications of Steady Flow Energy Equation
The steady flow energy equation may be applied to any apparatus through which fluid is flowing,
provided the conditions stated before are applicable. Some of the most common cases found in
engineering practices are
(i)
Boiler
In a boiler operating under steady conditions, water will be pumped into the boiler along the feed line
at the same rate at which steam leaves the boiler along the steam main, and heat energy will be supplied
at the steady rate.
The steady –flow energy equation is.
𝑄 − 𝑊𝑠 = 𝑚̇(ℎ2 − ℎ1 ) +
𝐶22 − 𝐶11
2
2.23
(i)
𝐶22 −𝐶11
will be small compared with the other term and may usually be neglected.
2
(ii)
𝑊𝑠 Will be zero since a boiler has no moving parts capable of affecting a work transfer.
The steady-flow energy equation for this system reduces to
𝑄 = 𝑚̇(ℎ2 − ℎ1 )
ℎ2 > ℎ1
2.24
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Fig: 2.2
Condenser
A condenser is a boiler in reverse. Where as in a boiler, heat is supplied to convert the liquid into vapour,
in a condenser heat energy is removed in order to condense the vapor into a liquid. In a steady state, the
amount of liquid, usually called condensate. Leaving the condenser may be equal to the amount
entering.
The steady flow energy equation gives
𝑄 = 𝑚̇(ℎ2 − ℎ1 )
ℎ2 < ℎ1
2.25
Water out
t w2
Water in
t w1
Condensate out
iii)
Turbine (Steam and Gas)
A turbine is a device which uses a pressure drop to produce work energy which is used to drive an
external load. The steady flow energy equation gives
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𝐶 2 −𝐶 1
𝑄 − 𝑊𝑠 = 𝑚̇ ((ℎ2 − ℎ1 ) + 2 2 1 )
2.26
Note
(a) The average velocity of flow of fluid through a turbine is normally high and the fluid passes
quickly through the turbine. It may be assumed that, because of this, heat energy does not have
time to flow into or out of the fluid during it passage through the turbine and hence Q = 0
(b) Although velocities are high, the difference between them is not large, and the term representing
the change in kinetic energy may be neglected. The flow equation becomes
−𝑊𝑠 = 𝑚̇((ℎ2 − ℎ1 ))
ℎ1 > ℎ2
(2.27)
Compressor
The action of a compressor is the reverse of that of a turbine i.e. it uses external work energy to produce
a pressure rise. Rotary compressor can be regarded as reversed turbine. Work is done on the fluid by
bladed rotor driven from an external source.
This increases the velocity of the fluid. The velocity is then reduced in a set of fixed diffusers to some
value approximately to that at inlet to the compressor and the pressure is increased. In applying the
steady flow energy equation to the rotary compressor, exactly the same arguments are used as for a
turbine and the equation becomes
−𝑊𝑠 = 𝑚̇((ℎ2 − ℎ1 ))
ℎ1 < ℎ2
2.28
Since heat transfer does occur in a reciprocating compressor because the flow occurs at relatively low
speed, and the surface area of the cylinder is appreciable. The difference between the kinetic energies
at inlet and outlet, however, can be neglected, hence
𝑄 − 𝑊𝑠 = 𝑚̇(ℎ2 − ℎ1 )
2.29
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2
1
Receiver
Air
in
Compressor
Q
W
Nozzle
A nozzle is a duct of varying cross sectional area so designed that a drop in pressure from inlet to outlet
accelerates the flow. It is a device that increases the velocity of fluid at the expense of pressure and is
commonly utilized in jet engines, rockets and spacecraft. The cross-sectional area of a nozzle decreases
in the flow direction for subsonic flows and increases for supersonic flows
The flow equation gives
𝐶 2 −𝐶11
)
2
𝑄 − 𝑊𝑠 = 𝑚̇ ((ℎ2 − ℎ1 ) + 2
2.30
Note
(i)
The average velocity of flow through a nozzle is high; hence the fluid spends time in the
nozzle. For this reason, it may be assumed that there is insufficient time for heat energy to
flow into or out of the fluid during its passage through the nozzle, i.e Q = 0.
(ii)
The nozzle has no moving parts, hence Ws = 0, often C1 is negligible compared to C2. In
this case the equation becomes
𝐶2
0 = 𝑚̇ ((ℎ2 − ℎ1 ) + 22 )
2.31
𝐶2 = √2 (ℎ1 − ℎ2 )
2.32
Diffuser
A diffuser is the reverse of a nozzle; it uses a kinetic energy drop to increase the pressure. i.e it
increases the pressure of a fluid by slowing it down. The cross sectional area of a diffuser increases in
the flow direction for subsonic flows and decreases for supersonic flows. They are commonly utilized
in jet engines, rockets and spacecraft. It has the same analysis as the nozzle.
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A throttling process is one in which the fluid is made to flow through restriction. E.g. a partially open
valve or orifice, causing a considerable drop in the pressure of the fluid.
The steady flow energy equation gives:
𝑄 − 𝑊𝑠 = 𝑚̇ ((ℎ2 − ℎ1 ) +
𝐶22 − 𝐶11
)
2
Note
1. Throttling takes place over a very small distance and the available area through which heat
energy can flow is very small. It is therefore normally assumed that no energy can be transfer.
i.e. Q = 0
2. There is no moving parts i.e. W = 0
3. The difference between C1 and C2 is n ot great and consequently the change in K.E is normally
neglected. The steady flow energy equation becomes
0 = 𝑚̇ (ℎ2 − ℎ1 ) 𝑜𝑟 ℎ2 = ℎ1
2.33
Tutorials
1. Air passes through a gas turbine system at the rate of 4.5Kg/s. It enters the turbine system with
Velocity of 90 m/s and a specific volume of 0.85 m3/Kg. It leaves the turbine system with a
specific volume of 1.45 m3/Kg and a Velocity of 110 m/s. The exit area of the turbine is 0.38m2.
In its passage through the turbine, the specific enthalpy of air is reduced by 200 KJ/Kg and
there is a heat loss of 40 KJ/Kg. Determine
i.
Power developed by the turbine system in KW
2. A steam turbine operating under steady flow conditions receives 3600 Kg of steam per hour.
The steam enters the turbine at a velocity of 80 m/s, an elevation of 10 m and specific enthalpy
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of 3276 KJ/Kg. It leaves the turbine at a velocity of 150 m/s, an elevation of 3 m and a specific
enthalpy of 2465 KJ/Kg. Heat losses from the turbine to the surrounding amount to 36 MJ/hr.
Estimate the power output of the turbine.
3. In an air motor cylinder, the compressed air has an internal energy of 450 kJ/kg at the
beginning of the expansion and an internal energy of 220 kJ/kg after expansion. If the work
done by the air during the expansion is 120 kJ/kg, calculate the heat flow to and from the
cylinder.
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LECTURE 3
SECOND LAW OF THERMODYNAMICS
3.1
Introduction
With the first law we dealt with the conservation of mass and energy, which are fine
principles that enable us to quantify and predict many phenomena that are seen in nature.
However, there are limitations to this principle and hence the need to provide a single allencompassing statement that expands thermodynamic theories so as to be capture all
phenomena in nature. However, the second law has been described by many in different
terms which can be stated as below:
ENTROPY-BASED STATEMENT: The second law of thermodynamics states that
the entropy of an isolated system can never decrease with time.
CLAUSIUS STATEMENT: Heat cannot of itself pass from a colder to a hotter body.
KELVIN-PLANCK STATEMENT: Its impossible for any system to operate in a
thermodynamic cycle and deliver a net amount of work to its surroundings while
receiving an energy transfer by heat from a single thermal reservoir. This in simple
terms is that “you can turn all the work into heat, but you cannot turn all the heat into
work”.
3.2
REVERSIBLE AND IRREVERSIBLE PROCESSES
REVERSIBLE PROCESS: A process in which it is impossible to return both the
system and surroundings to their original states.
IRREVERSIBLE PROCESS: A process in which it is impossible to return both the
system and surroundings to their original states. It is important to note that it may be
possible to restore the system to its original state but not the surroundings.
3.2.1 Heat engine
Heat engine is a device used for converting heat into work as it has been seen from nature that
conversion from work to heat may take place easily but the vice-versa is not simple to be realized. Heat
and work have been categorized as two forms of energy of low grade and high grade type. Conversion
of high grade of energy to low grade of energy may be complete (100%), and can occur directly whereas
complete conversion of low grade of energy into high grade of energy is not possible. For converting
low grade of energy (heat) into high grade of energy (work) some device called heat engine is required.
Thus, heat engine may be precisely defined as “a device operating in cycle between high temperature
source and low temperature sink and producing work”. Heat engine receives heat from source,
transforms some portion of heat into work and rejects balance heat to sink. All the processes occurring
in heat engine constitute cycle.
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3.2.2
Thermal Efficiency
An index of performance of a work-producing machine or a heat is expressed by its [thermal efficiency]
defined as the ratio of the net work output to the heat input. Only a fraction of the heat input is converted
into work and the rest is rejected.
Let Q1 = heat transferred to system, Q2 = heat transferred from system, Wnet = net work done by system.
𝑊
ɳth = 𝑄𝑛𝑒𝑡
3.1
1
T1 Source
Q1
HE
W = Q1 –Q2
Q2
T2 Sink
Heat Engine
But from first law
𝑄1 − 𝑄2 = 𝑊𝑛𝑒𝑡
Therefore:
ɳth =
𝑊𝑛𝑒𝑡 𝑄1 −𝑄2
𝑄
=
= [1 − 2 ]
𝑄1
𝑄1
𝑄1
3.2
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3.2.3
Heat pumps and Refrigerator
Heat pump refers to a device used for extracting heat from a low temperature surroundings and sending
it to high temperature body, while operating in a cycle. In other words heat pump maintains a body or
system at temperature higher than temperature of surroundings, while operating in cycle. Block diagram
representation for a heat pump is given below:
Body, T1
Q1
T1 ˃T 2
W
HP
Q2
LowTemp T2
Heat pump
As heat pump transfers heat from low temperature to high temperature, which is non-spontaneous
process, so external work is required for realizing such heat transfer. Heat pump shown picks up heat
Q2 at temperature T2 and rejects heat Q1for maintaining high temperature body at temperature T1. For
causing this heat transfer heat pump is supplied with work W as shown. As heat pump is not a work
producing machine and also its objective is to maintain a body at higher temperature, so its performance
can’t be defined using efficiency as in case of heat engine. Performance of heat pump is quantified
through a parameter called coefficient of performance (C.O.P). Coefficient of performance is defined
by the ratio of desired effect and net work done for getting the desired effect.
For Heat Pump: Net Work =W
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Desired effect = heat transferred Q1 to a high temperature body, at temperature T1
(COP)HP =
Q1
W
𝑄1 − 𝑄2 = 𝑊𝑛𝑒𝑡
(COP)HP =
Q1
Q1 −Q2
3.3
Refrigerators and heat pumps are essentially the same devices. They differ in their objectives. The
objective of the refrigerator is to maintain the refrigerated space at a low temperature by removing heat
from it. The objective of the heat pump is to maintain heated space at a high temperature. Therefore,
the COP for refrigerator is
(COP)ref =
Q2
Q2
=
W
Q1 −Q2
3.4
TUTORIALS
1. Determine the heat to be supplied to a Carnot engine operating between 400ºC and 15ºC and
producing 200 kJ of work.
2. A refrigerator operates on reversed Carnot cycle. Determine the power required to drive the
refrigerator between temperatures of 42ºC and 4ºC if heat at the rate of 2 kJ/s is extracted from
the low temperature region.
The CARNOT Heat Engine
The hypothetical heat engine that operates on the reversible Carnot cycle is called the Carnot heat
engine. The thermal efficiency of any heat engine, reversible or irreversible
𝑄
𝜂𝑡ℎ = 𝑄 𝐿
𝐻
ENTROPY
Entropy is a thermodynamic property that provides a quantitative measure of the disorder of
a given thermodynamic state. Entropy is simply a mathematical convenience and a
theoretical construct.
Entropy helps engineers to summarize important design concepts for thermal systems such as
steam power plants, automobile engines, refrigerators, heat pumps etc.
3.5
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Considering the figure above which will be used to develop the thermodynamic property of
entropy, we will restrict the analysis to reversible processes. Now starting from 1, proceeding
on path A to 2 and returning to 1 via path B, the cyclic integral of
decomposes to;
3.6
Performing the same starting from 1 to 2 on path A and returning on path C, WE HAVE;
3.7
Subtracting equation 3.7 from 3.6 , we get;
3.8
3.9
From equation 3.9 we can define an extensive thermodynamic property S as the entropy of
the system. Defined mathematically as:
3.10
Note the units of S must be kJ/K in the SI system. We also can scale by the constant mass
m to get the corresponding intensive property s = S/m
3.11
Therefore the units for s are KJ/KG/K; note they are the same as 𝑐𝑝,𝑐𝑣 𝑎𝑛𝑑 𝑅. In differential
form;
3.12
This leads to
𝑇𝑑𝑠
𝛿𝑞𝛿𝑞
==
𝑇𝑑𝑠
Integrating both sides
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2
2
∫ 𝛿𝑞 = ∫ 𝑇𝑑𝑠
1
1
2
𝑞1−2 = ∫ 𝑇𝑑𝑠
3.13
1
This is the heat transfer component as can be recalled from the first law. Therefore, if the
process lies on this path then an isentropic process is applicable. Isentropic process is a process
in which the entropy, s, is constant.
For problems in which no chemical reactions are present, we will find ourselves interested only
in entropy differences. For problems with chemical reactions, the absolute values of entropy
will be important. Such values can be obtained by consideration of the third law as depicted by
Lewis and Randall
• Third law of thermodynamics: “every substance has a finite positive entropy, but at the
absolute zero of temperature the entropy may become zero, and does so become in the case of
perfect crystalline substances,”.
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LECTURE 4
HEAT TRANSFER
4.1
Introduction
Heat is a form of energy transferred across the boundary of a system at a given temperature to
another system (or the surroundings) at a different temperature by virtue of the temperature
difference between the two.
We also note that when two bodies are at the same temperature, there can be no heat transferred
between the two bodies. The subject of heat transfer considers the details of the heat transfer
process. There are three fundamental classes of heat transfer:
•
Conduction
Conduction is defined as the transfer of heat occurring through intervening matter without bulk
motion of the matter. This type of heat conduction can occur through a turbine blade in an
engine. This is the transfer of energy in solids.
Conduction is due to local effects and its characterized by the Fourier law;
𝑞 = −𝑘∇𝑇
4.1
Where, q is the heat flux vector with units J/s/𝑚2=W/𝑚2, k is the thermal conductivity with
units J/s/m/K-W/m/K and ∇𝑇 is the vector representing the gradient of temperature. The
negative sign indicates that the thermal energy flows in the direction of most rapid temperature
decrease.
By the Fourier law,
4.2
Multiplying by the local cross sectional area, 𝑄̇ = 𝑞𝐴 and
4.3
•
Convection
This is heat transfer due to a flowing fluid. The fluid can be liquid or gas. For some systems,
convection effects are modelled by the Newton’s law of cooling:
𝑞 = ℎ(𝑇ℎ𝑜𝑡 − 𝑇𝑐𝑜𝑙𝑑)
4.4
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𝑄̇ = 𝑞𝐴 = ℎ𝐴(𝑇ℎ𝑜𝑡 − 𝑇𝑐𝑜𝑙𝑑)
4.5
Where h is a constant with units W/𝑚2/𝐾
•
Radiation
It is the transmission of energy through space without the necessary presence of matter. The
earth is heated by the sun via radiation effects, not conductive energy diffusion.
𝑞 = 𝜎(𝑇4ℎ𝑜𝑡 − 𝑇4𝑐𝑜𝑙𝑑)
4.6
𝑄̇ = 𝑞𝐴 = 𝜎𝐴(𝑇4ℎ𝑜𝑡 − 𝑇4𝑐𝑜𝑙𝑑)
4.7
Where 𝜎 is the Stefan-Boltzmann Constant, given as 𝜎 = 5.67𝑥10−8𝑊/𝑚2/𝐾4
Tutorials
1. During a steady state operation, a gearbox receives 60KW through the input shaft and delivers
power through the output shaft. In the gearbox as a system, the rate of energy transfer is by
convection. For the gearbox, evaluate the heat transfer rate and the power delivered through
the output shaft in KW. Take;
h = 0.171kW/m2, heat transfer coefficient
A = 1.0m2 is the outer surface area of the gearbox
T = 270C, temperature of the outer surface
T = 200C, temperature of the surrounding air away from the gearbox
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