Thermodynamics
Thermodynamics is the science that deals with energy, its forms and transformations, and the
interactions between energy and matter. Because of its generality, thermodynamics is the underlying
science that forms the framework for many engineering subjects: Heat Transfer, Gas Dynamics, Fluid
Mechanics, Combustion and Material Science. The objectives of this course are to study the
fundamentals of thermodynamics and to apply them to the solution of engineering problems. In this
class, we will develop a thorough understanding of the properties of thermodynamic systems and of the
laws of thermodynamics.
Thermodynamics= Therme
(Heat)
+
Dynamis
(Power)
Science that deals with heat and work and the changes they can produce. e.g. change of temperature
(T), pressure (P) etc.
Applications of Thermodynamics
1) All types of vehicles that we use, cars, motorcycles, ships, aeroplanes, and many other types work on
the basis of second law of thermodynamics and Carnot Cycle. They may be using petrol engine or
diesel engine, but the law remains the same.
2) All the refrigerators, deep freezers, industrial refrigeration systems, all types of air-conditioning
systems, heat pumps, etc. work on the basis of the second law of thermodynamics.
3) All types of air and gas compressors, blowers, fans, run on various thermodynamic cycles.
4) One of the important fields of thermodynamics is heat transfer, which relates to transfer of heat
between two media. The concept of heat transfer is used in wide range of devices like heat exchangers,
evaporators, condensers, radiators, coolers, heaters, etc.
5) Thermodynamics also involves study of various types of power plants like thermal power plants,
nuclear power plants, hydroelectric power plants, power plants based on renewable energy sources like
solar, wind, geothermal, water waves etc,
6) Renewable energy is an important subject area of thermodynamics that involves studying the
feasibility of using different types of renewable energy sources for domestic and commercial use.
Basis is experimental observations written down as laws. e.g.
1st law of thermodynamics:
Energy can change from one form to another but the total amount remains constant.
2nd law of thermodynamics:
Energy has quality (more or less useful) and quantity. Real changes occur only in a direction of
decreasing quality of energy.
Basic Principles
Thermodynamics involves the storage, transformation, and transfer of energy. Energy is stored as
internal energy (due to temperature), kinetic energy (due to motion), potential energy (due to
elevation), and chemical energy (due to chemical composition); it is transformed from one of these
forms to another; and it is transferred across a boundary as either heat or work. We will present
equations that relate the transformations and transfers of energy to properties such as temperature,
pressure, and density. The properties of materials thus become very important. Many equations will be
based on experimental observations that have been presented as mathematical statements or laws:
primarily the first and second laws of thermodynamics.
The system
To solve problems the parts of the problem must be enumerated. The first consideration is that there
must be something performing the energy transformation. This something is a substance. A substance
does not exist alone; it must be contained. This brings us to the concept of a system. In
thermodynamics a system is defined as any collection of matter or space of fixed identity, and the
concept is one of the most important in thermodynamics.
The system is whatever we want to study. It may be as simple as a free body or as complex as an entire
chemical refinery.
Thermodynamic System
A quantity of matter or a region in space chosen for study
Surroundings
Everything external to the system
Boundary
Surface that separates the system from the surrounding.
It may be fixed or movable
Universe:
System and its Surroundings
(Fig 1): Thermodynamic System
Types of systems
Open, closed, and isolated systems
Any system can be classified as one of three types: open, closed, or isolated. They are defined as
follows:
open system: Both energy and matter can be exchanged with the environment. Example: an open cup
of coffee.
closed system: energy, but not matter, can be exchanged with the environment. Examples: a tightly
capped cup of coffee.
isolated system: Neither energy nor matter can be exchanged with the environment {in fact, no
interactions with the environment are possible at all. Example (approximate): coffee in a closed, wellinsulated thermos bottle.
Closed System
Type of system
Mass flow
Isolated System
Work
Open System
Heat
Open
Closed
Isolated
(Figure 2) shows a gas in a piston–cylinder assembly. When the valves are closed, we can consider the
gas to be a closed system. The boundary lies just inside the piston and
cylinder walls, as shown by the dashed lines on the figure. The
portion of the boundary between the gas and the piston moves with
the piston. No mass would cross this or any other part of the
boundary.
State Functions
System properties, such as pressure (P), volume (V), and temperature
(T) are called state functions. The value of a state function
(Fig 2): Closed system
depends only on the state of the system and not on the way in which
the system came to be in that state.
A change in a state function describes a difference between the two states. It is independent of the
process or pathway by which the change occurs.
For example, if we heat a block of iron from room temperature to 100°C, it is not important exactly
how we did it. Just on the initial state and the final state conditions. For example, we could heat it to
150°C, and then cool it to 100°C. The path we take is unimportant, so long as the final temperature is
100°C.
Microscopic and Macroscopic Approach
To study the behavior (changes in T & P) of substances we have 2 approaches
Microscopic:
Study behavior of each atom & molecule (Quantum mechanics)
Macroscopic:
Study average behavior of many atom & molecule
a) Find the average behavior based on probability theory (Statistical thermodynamics)
b) Find the average behavior using instruments
Thermodynamic Properties of a system
A measurable quantity that defines the condition of a system e.g. temperature (T) pressure (P) mass
(m) volume (V) and density (𝝆)
Another useful property is Specific volume (v) defined as the volume of a unit mass.
Extensive and Intensive properties
Properties are one of 2 types
Intensive properties: Independent of mass. e.g. P, T, v, 𝝆
Extensive properties: Change with mass. e.g. m, V, Energy
(Fig 3): Extensive and Intensive properties
State and Equilibrium
A substance can be at various pressures & temperatures or in various states
State:
Condition of a system identified by properties (e.g. T, P, v). In a given state each property has 1 value.
(Fig 4): system at two different states
Equilibrium
No unbalance exists in the system, and values of properties (T, P etc.) remain the same when it is
isolated from the surroundings.
Thermal equilibrium: temperature of system does not change when it is isolated from surroundings
a
b
(Fig 5): Thermal equilibrium
Mechanical equilibrium: pressure of system does not change when it is isolated from surroundings.
(Fig 6): Mechanical equilibrium
Chemical equilibrium: chemical composition does not change when it is isolated from surroundings.
Processes and Cycles
A process is a change in the system state. Just as
b
there are an infinite number of ways to go between
two points (Figure 7), so too are there an infinite
number of ways for a system to change from state 1
to state 2. The path describes the infinite number of
system states that occur when a system undergoes a
particular process from state I to state 2. A
thermodynamic cycle is a collection of two or more
processes for which the initial and final states are
the same.
The cycle in Figure 7 illustrates a closed system that
(Fig 7): Graphical illustration of a path's
undergoes process A and process B. The system is
being a succession of system states
returned to state 1 from which it started.
The state of a system is defined when it is in equilibrium.
If we change the state very fast it is not in equilibrium during the process (non-equilibrium process)
If we change it slowly then the system is in equilibrium during the process (quasi-equilibrium process)
Quasi-equilibrium process (ideal process)
The system is very near to equilibrium in all successive states during the process.
Non-equilibrium process
The system is not in equilibrium during the process. States during the process are undefined.
We can only define the initial and final states.
Questions
Which of the following is not an extensive property?
(a) Kinetic Energy (b) Momentum
(c) Mass (d) Density
(e) None of these
Which of the following is not an intensive property?
(a) Velocity (b) Volume
(c) Pressure (d) Temperature
(e) None of these
The force
In physics, a force is any influence that causes an object to undergo a certain change, either
concerning its movement, direction, or geometrical construction. In other words, a force can cause an
object with mass to change its velocity (which includes to begin moving from a state of rest), i.e.,
to accelerate, or a flexible object to deform, or both. Force can also be described by intuitive concepts
such as a push or a pull. A force has both magnitude and direction, making it a vector quantity. It is
measured in the SI unit of newton’s and represented by the symbol F.
⃑𝑭 = 𝒎𝒂
⃑
𝑷=
𝑭
𝒅𝑭𝒏
𝒐𝒓 𝑷 =
𝑨
𝒅𝑨
Where:
P. is the pressure,
F. is the normal force,
A. is the area of the surface on contact.
Pressure
Pressure is defined as the normal force per unit area.
Pressure = F/A
The unit of pressure is Nm-2 and is called the Pascal (Pa).
If the force F is not normal to the surface on which it acts, the component of the force normal to the
surface is used in the calculation of pressure.
If the force F makes an angle 𝜽 with the surface is F sin 𝜽 .
𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 =
𝑭𝒔𝒊𝒏𝜽
𝑨
Thrust = P.A
Pressure in fluids
Every point inside the fluid is subjected to a pressure as a result of the weight of the fluid standing over
it.
Statements
I.
II.
III.
IV.
A fluid exerts pressure in all direction.
The pressure exerted by a fluid at a point is the same in all direction
Pressure exerted by a fluid at any point on the same horizontal surface is the same.
The pressure exerted by a fluid at a point is directly proportional to the depth of the point from
the surface of the fluid.
Pressure at point in a fluid
Consider a small area dA in the fluid at a depth h.
The force acting on dA is a measure of the weight of the fluid standing on dA.
The volume of fluid standing on dA
V = dA×h
The mass of fluid on dA
m= dA×h× 𝝆
𝝆 is the density of the fluid.
The weight of the fluid standing on dA
mg = dA × h × 𝝆 ×g
This is a measure of the thrust on the area dA
𝑷=
𝑻𝒉𝒓𝒖𝒔𝒕 𝒅𝑨 × 𝒉𝝆𝒈
=
= 𝒉𝝆𝒈
𝑨𝒓𝒆𝒂
𝒅𝑨
Measurement of pressure
We live in ocean of air. Air is a fluid and it exerts pressure in all directions. The pressure exerted by air
is called atmospheric pressure. The pressure exerted by atmosphere can be measured by a simple
mercury barometer. A long tube filled with mercury and the open end is immersed in a pool of mercury
in a dish holding the tube vertical. The mercury in the tube falls, but remains steady at around 760 mm
when it is set up at sea level. The space above the mercury in the tube is found to be perfect vacuum.
Why doesn’t the column of mercury in the tube fall further?
Consider two points A and B on the same horizontal level in the dish of mercury. The pressure at A =
the pressure at B, pressure at A is the atmospheric pressure. Pressure at B is the pressure exerted by
760 mm of mercury. Since these two pressures are balanced we conclude:
Atmospheric pressure = the pressure exerted by the column of
mercury in the tube.
Pressure exerted by 1mm of mercury = 1torr
Atmospheric pressure = 760 torr
1 torr = 13.3 millibars (mb)
1 atmosphere = 13.3 mb×760 mm=1011mb
Atmospheric pressure in absolute unit
We can now use the pressure equation to obtain the atmospheric
pressure at sea level in Pa
P=h𝝆g
𝝆𝒎𝒆𝒓𝒄𝒖𝒓𝒚 =13600Kgm-3
(Fig 8): Barometer
Pressure exerted by 760 mm of mercury is:
P =0.76 ×13600×9.8 = 1.01×105 Pa
When the atmospheric pressure falls, the height of mercury in
the tube will also fall. When atmospheric pressure rises, the
height of mercury will rise as well.
In most thermodynamic relations absolute pressure must be
used. Absolute pressure is gage pressure plus the local
atmospheric pressure:
(Fig 9): Manometer.
Pabs = Pgage + Pat
When the local atmospheric pressure is greater than the pressure of the system, the term vacuum
pressure is used.
P vacuum = patm- pabsolute
Engineers in the United States frequently use the
letters a and g to distinguish between absolute and
gage pressures. For example, the absolute and
gage pressures in pounds force per square inch are
written as Pisa and Psig, respectively. The
relationship among the various ways of expressing
pressure measurements is shown in Fig
Pressure Units
The SI unit of pressure and stress is the Pascal.
1pascal=1N/m2
(Fig 10): Relationships among the absolute, atmospheric,
gage, and vacuum pressures.
However, multiples of the Pascal: the kPa, the bar,
and the MPa are frequently used.
1kPa = 103 N/m2
1bar = 105 N/m2
……..
1MPa = 106N/m2
Since 1 bar (105 N/m2) closely equals one standard atmosphere, it is a convenient pressure unit despite
not being a standard SI unit. When working in SI, the bar, MPa, and kPa are all used in this text.
N/m2
Pascal
atm
psi
mm Hg
1 Pa = 1 N/m2
1 atm = 1 × 105 Pa
1 psi = 1 lb/inch2
1 atm = 760 mm Hg
Bourdon gauge
The Bourdon pressure gauge uses the principle that a
flattened tube tends to straighten or regain its circular form in
cross-section when pressurized. Although this change in crosssection may be hardly noticeable, and thus involving moderate
stresses within the elastic range of easily workable materials,
the strain of the material of the tube is magnified by forming
the tube into a C shape or even a helix, such that the entire
tube tends to straighten out or uncoil, elastically, as it is
(Fig 11): Bourdon gauge
pressurized. Eugene Bourdon patented his gauge in France in
1849, and it was widely adopted because of its superior sensitivity, linearity, and accuracy; Edward
Ashcroft purchased Bourdon's
American patent rights in 1852 and became a major manufacturer of gauges. Also in 1849, Bernard
Schaeffer in Magdeburg, Germany patented a successful diaphragm (see below) pressure gauge,
which, together with the Bourdon gauge, revolutionized pressure measurement in industry. But in 1875
after Bourdon's patents expired, his company Schaeffer and Budenberg also manufactured Bourdon
tube gauges.
Examples
1. Calculate the force due to the pressure acting on the 1-m-diameter horizontal hatch of a submarine
submerged 600 m below the surface.
Sol
P = 𝝆gh = (1000 kg/m3) (9.81 m/s2) (600 m) = 5.89 MPa
The pressure is constant over the area; hence, the force due to the
pressure is given by
F = P A = (5.89 x 106 N /m2)⌊
𝝅(𝟏)𝟐
𝟒
𝒎𝟐 ⌋ = 𝟒. 𝟔𝟐 × 𝟏𝟎𝟔 𝑵
2. The air pressure in a tank is measured by an oil manometer. For a given oil-level difference
between the two columns, the absolute pressure in the tank is to be determined.
Properties: The density of oil is given to be ρ = 850 kg/m3.
Analysis The absolute pressure in the tank is determined from
P=Patm+ ρgh
= (𝟗𝟖𝒌𝑷𝒂) + (𝟖𝟓𝟎
𝒌𝒈
𝒎
𝟏𝒌𝑷𝒂
) (𝟗. 𝟖𝟏 𝟐 ) (𝟎. 𝟔𝟎𝒎) [
]
𝟑
𝟏𝟎𝟎𝟎𝑵
𝒎
𝒔
𝒎𝟐
=103kPa
3. Calculate the absolute pressure for a system, given a gauge pressure of 1.5 MPa and a barometric
pressure (atmospheric pressure) of 104 KPa.
Absolute pressure=atmospheric pressure + gauge pressure
Pabs =Patm +Pgauge
Given:
Patm=104 kPa (where kPa= kilo pascal)
And
Pgauge=1.5MPa (where MPa=mega Pascal)
=1.5×1000 kPa
=1500 kPa
=Pams=1500+104 =1604 KPa =1604/1000 MPa
Pams=1.604 MPa
4.
Differential pressure gauge mounted on a vessel shows 1.25 MPa and a local barometer gives
atmospheric pressure as 0.96 bar. Find the absolute pressure inside the vessel.
Solution:
Convert all pressures to units of kPa.
Pgauge = 1.25 MPa = 1250 kPa;
Patm = 0.96 bar = 96 kPa
P = Pgauge + Patm = 1250 + 96 = 1346 kPa
5. The gage pressure in a liquid at a certain depth is 28kpa. The gage pressure in the same liquid
at a different depth is to be determined. (h1 = 3m
h2 = 9m)
Assumptions The variation of the density of the liquid with depth is negligible.
Analysis The gage pressure at two different depths of a liquid can be expressed as
P1=ρgh1
and P2=ρgh2
Taking their ratio,
𝑷𝟐
𝝆𝒈𝒉𝟐
𝒉𝟐
=
=
𝑷𝟏
𝝆𝒈𝒉𝟏
𝒉𝟏
Solving for P2 and substituting gives
𝒑𝟐 =
𝒉𝟐
𝒉𝟏
𝒑𝟏 =
𝟗𝒎
𝟑𝒎
(28kpa) = 84 kPa
Discussion Note that the gage pressure in a given fluid is proportional to depth.
Temperature
Temperature is the MEASURE of the AVERAGE molecular motions in a system and simply has units
of (degrees F, degrees C, or K). Notice that one primary difference between heat and temperature is
that heat has units of Joules and temperature has units of (degrees F, degrees C, or K). Another
primary difference is that energy can be transported without the temperature of a substance changing
(e.g. latent heat, ice water remains at the freezing point even as energy is brought into the ice water to
melt more ice). But, as a general statement (ignoring latent heat), as heat energy increases, the
temperature will increase. If molecules increase in vibration, rotation or forward motion and pass that
energy to neighboring molecules, the measured temperature of the system will increase.
Temperature scale
Much of the world uses the Celsius scale (°C) for most temperature measurements. It has the same
incremental scaling as the Kelvin scale used by scientists, but fixes its null point, at 0°C = 273.15K,
approximately the freezing point of water (at one atmosphere of pressure). The United States uses
the Fahrenheit scale for common purposes, a scale on which water freezes at 32 °F and boils at 212 °F
(at one atmosphere of pressure).
For practical purposes of scientific temperature measurement, the International System of Units (SI)
defines a scale and unit for the thermodynamic temperature by using the easily reproducible
temperature of the triple point of water as a second reference point. The reason for this choice is that,
unlike the freezing and boiling point temperatures, the temperature at the triple point is independent of
pressure (since the triple point is a fixed point on a two-dimensional plot of pressure vs. temperature).
For historical reasons, the triple point temperature of water is fixed at 273.16 units of the measurement
increment, which has been named the kelvin in honor of the Scottish physicist who first defined the
scale. The unit symbol of the kelvin is K.
Absolute zero is defined as a temperature of
precisely 0 kelvins, which is equal to −273.15 °C
or −459.67 °F
Temperature Conversions
Modern chemistry works in the Celsius and
Kelvin scales. Conversion from Celsius to Kelvin
requires simple addition since the degrees are the
same size. The formula is:
K = °C + 273
(Fig 12) Comparison of the Fahrenheit,
Celsius, and Kelvin temperature scales.
Example:
The boiling point of ethylene glycol, a substance used as antifreeze in automobiles, is 198°C. Express
this temperature in K.
Solution:
This problem involves a conversion from °C to K. Use the equation
K = °C + 273
Substitute 198 for °C:
K = 198 + 273 = 471 K
Remember that there is no degree sign used with Kelvin temperatures. Generally, there is little need to
convert between Fahrenheit and Celsius; however, examples of these conversions are included for
information purposes only. Comparing the Celsius and the Fahrenheit temperature scales there is a
difference of 180 F° between the freezing point of water and its boiling point as compared with 100 C°.
The average size of a Fahrenheit degree is 100/180 or5/9 of a Celsius degree. To convert from
Fahrenheit to, we also have to take into account that there is a 32° difference in starting temperature.
Thus the formula for the temperature conversion from Fahrenheit to Celsius is:
0C
= 5/9 × (0F-32)
The formula for the temperature conversion from Celsius to Fahrenheit is:
0F
= (9/5× 0C) + 32
Conversion formulas between the three temperature scales are summarized in the table below.
Quiz Review
1. Is it possible to boil water at room temperature?
Answer: Yes. How?
2. Is it possible to freeze water at room temperature?
Answer: Maybe. How?
The zeroth law of thermodynamics.
When two systems (A, C) are each in thermal equilibrium with a third system (B), the first two systems
are in thermal equilibrium with each other. This property makes it meaningful to use thermometers as
the “third system” and to define a temperature scale.
Not
Exa
mpl
es:
1.
(Fig 13) Comparison of the Fahrenheit, Celsius, and Kelvin temperature scales.
degrees Kelvin?
What
is a
tempe
ratur
e of –
50 C
in
Solution: The offset from Celsius to Kelvin is 273.15 K, so we get
TK = TC + 273.15 = -5 + 273.15
= 268.15 K
2. Room temperature is usually about 70°F. Express the temperature in °C.
Solution:
This problem involves a conversion from °F to °C. Use the equation:
0C
= 5/9 × (0F-32)
Substitute the Fahrenheit temperature for °F:
0C
= 5/9 × (70-32)
°C = 21.1°C
3. Pure aspirin melts at 135°C. Express this temperature in °F
Solution:
This problem involves a conversion from °C to °F. Use the equation:
0F
= (9/5× 0C) + 32
0F
= (9/5× 135) + 32 = 2750F
4. Liquid nitrogen boils at a temperature of 77 K. Express this temperature in Degrees Fahrenheit
and Celsius.
Solution:
We are given TK = 77 and must find Tf and TC using formulas in the above table. For Celsius this
is very easy since the Kelvin and Celsius scales have the same size of one "degree", but different
zeros (The lines in Figure 1 above are parallel). Thus T = TK - 273.12
C
= (77) - (273.16)
= -196.16°C.
For converting to Fahrenheit we have to account for different zeros and different size "degrees."
These are both accomplished by the formula T = (9/5) TK - 459.60 = (9/5) x (77) - (459.60)
C
= -321.0°F.
5. The temperature of a given gas is -100C. What are the equivalent Fahrenheit and absolute kelvin
scale readings?
Solution
(a) The Fahrenheit and Celsius scales are related by the following equation:
TF =32 + 9/5 TC
TF =32 + 9/5 (-10) = 32 + (-18) =14 0F
(b) 0 0C is equivalent to 273 K.
Therefore
TK = 273 + (-10) = 263 K
Energy
Energy is the ability to do work. One of the very important concepts in a study of thermodynamics is
the concept of energy. Energy is a fundamental concept, such as mass or force and, as is often the case
with such concepts, is very difficult to define. Energy is defined as the ability to produce an effect. It is
important to note that energy can be stored
within a system and can be transferred (as heat,
for example) from one system to another.
Thermal energy total potential and kinetic
energy in an object, It depends on mass,
temperature, and phase of an object.
As shown in (Fig 14) if both objects are in the
same phase & at the same temperature, which
one has MORE thermal energy?
Energy is measured in the same unit as work,
namely joule (J)
(Fig 14) Two objects with different mass
1000J = 1kJ
There are many types of energy forms, but we will discuss only three: gravitational potential, kinetic,
and internal energies -the ones that occur in the typical situations we will be analyzing.
The potential energy
Consider a book of mass m kg on the floor. Its weight mg acts vertically down. What is the force
required to lift it without causing an acceleration?
Since there is no acceleration causing by lifting, there is no net force on the
book.
Let F is the force required to lift it:
F – mg = 0
A force F = mg is required to lift the book without acceleration.
Let this force F lift the book by the force F + mg
W = F.h = mgh (J)
This is the same as saying that the energy used in doing this work is mgh. This energy in turn allows
the book to do work when you stop supporting it. The book placed h has a certain amount of energy
stored in it. This energy is called the Potential energy of the book.
The energy that a system possesses due to its elevation h above some arbitrarily selected datum is its
potential energy; it is determined from the equation PE = mgh and is measured in joules (J)
The change in potential energy…. ΔPE=mg (h2-h1)
Conservative and non – Conservative forces
Weight of an object is a measure of the gravitational force on it.
The gravitational force on the book is – mg. what is the work done by the
gravitational force when the book is raised through a height h?
The force –mg is displaced by h m (W = -mgh)
If the book is now allowed to fall, the gravitational force – mg is displaced by –h.
The work done by gravitational force now is: W = -mg(-h) = mgh.
The total work done by the force of gravity for the entire trip is:
W = -mgh + mgh = 0
A force is conservative if the total work it does on an object is zero when the object moves around in
any closed path returning to its initial position. Gravitational force is a conservative force. The
restoring force of spring is a conservative force.
Kinetic energy
Work done on an object can produce motion on the object.
A force F acting on an object of mass m displaces it through a distance x giving it a velocity vf = v
𝒗𝟐 = 𝒗𝟐𝒊 + 𝟐𝒂𝒙 Since vi = 0, v2 = 2ax
x = v2 / 2a
Work done by F is: W = F.x = ma.v2/2a = ½ mv2
F
vi =0
vf = v
x
This work done is stored in the object as energy of motion it is called kinetic energy of the object in
motion.
Kinetic energy is represented by KE and it is measured in joules (J)
KE = ½ mv2
Cheng in kinetic energy …………… ΔKE= 1/2m(vf2-vi3)
KE is the work done in giving an object a velocity v
Law of Conservation Energy
The law of conservation of energy states that the total energy in the universe is a constant. This means
that the net change in energy must be zero.
If ΔPE is the change in potential energy and ΔKE is the change in kinetic energy, than
ΔPE + ΔKE=0
The gain in one form of energy comes at the expense of the other so that
the net change is always zero.
Imagine 1-kg ball raised through as height of 5.1 m high.
The work done to raise it through this height is:
W = mgh = 1kg × 9.8 ms-2 × 5.1m = 50 J
This is a measure of the potential energy of the ball
PE = 50 J
Since the ball is at rest here, its kinetic energy KE = 0
Total energy at top of the stairs
E = PE + KE = 50 + 0 = 50 J
What happens if the ball is now allowed to fall to the ground?
All the potential energy it has at the top becomes kinetic energy at the bottom
The potential energy at the bottom is: PE = 0
In falling through a vertical height h, it acquires a velocity v such KE = 50J, PE = 0
v2 = 2gh = 2×9.8×5.1=100(ms-1)2
KE = 1/2mv2 = 1/2×1×100 = 50 J
Total energy at the bottom is:
Total energy at the bottom is:
E = PE + KE + = 0 + 50 = 50 J
Consider the ball at some intermediate position when it has fallen through a vertical height of 2m.
While falling through 2m it acquires a velocity v such that
V2 = 1/2gh = 2×9.8×2 = 39.2 (m/s) 2
Its kinetic energy at this position is:
KE = 1/2m v2 = 1/2×1×39.2 = 19.6 J
At this position, its height from the ground is 5.1-2=3.1m
The potential energy of the ball at this height is:
PE = mgh = 1 kg×9ms-2×3/1m = 30/4 J
The total energy at this position is:
E = PE + KE = 19.6 + 30.4 = 50 J
Irrespective of the position of the ball, the total energy is a constant.
Examples
1. An object of known mass decelerates from a given initial velocity to a known final velocity. Determine
the change in kinetic energy of the object.
vi = 100m/s
ΔKE=1/2m(vi2 - vf2)
m=1000kg
vf = 20m/s
= ½(1000kg) (202 – 1002) = - 4800 kJ
2. Water drops from waterfall in height (100m) when the water crash with ground (80%) of energy steel in
it and the rest go to the river. Calculate the height of temperature?
The specific heat of water (4.2 kJ/kg. K)
0.8 PE = Q
𝜟𝑻 =
0.8×m g z = m c ΔT
𝟎. 𝟖 × 𝟗. 𝟖𝟏 × 𝟏𝟎𝟎
= 𝟎. 𝟏𝟖𝟕𝑲
𝟒. 𝟐 × 𝟏𝟎𝟑
3. A 2200-kg automobile traveling at (25 m/s) hits the rear of a stationary, 1000-kg automobile. After the
collision the large automobile slows to (13.89 m/s), and the smaller vehicle has a speed of (24.44 m/s). What
𝟏
has been the increase in internal energy, taking both vehicles as the system? 𝑲𝑬𝟏 = 𝟐 𝒎𝒂 𝑽𝟐𝒂𝟏 =
𝟏
(𝟐) (𝟐𝟐𝟎𝟎)(𝟐𝟓𝟐 ) = 𝟔𝟖𝟕𝟓𝟎𝟎𝑱
After the collision the kinetic energy is
𝑲𝑬𝟐 =
𝟏
𝟏
𝟏
𝟏
𝒎𝒂 𝑽𝟐𝒂𝟐 + 𝒎𝒃 𝑽𝟐𝒃𝟐 = ( ) (𝟐𝟐𝟎𝟎)(𝟏𝟑. 𝟖𝟗𝟐 ) + ( ) (𝟏𝟎𝟎𝟎)(𝟐𝟒. 𝟒𝟒𝟐 ) = 𝟓𝟏𝟎𝟗𝟎𝟎𝑱
𝟐
𝟐
𝟐
𝟐
The conservation of energy requires that
𝑬𝟏 = 𝑬𝟐
Thus,
𝑲𝑬𝟏 + 𝑼𝟏 = 𝑲𝑬𝟐 + 𝑼𝟐
𝑼𝟐 − 𝑼𝟏 = 𝑲𝑬𝟏 − 𝑲𝑬𝟐 = 𝟔𝟖𝟕𝟓𝟎𝟎 − 𝟓𝟏𝟎𝟗𝟎𝟎 = 𝟏𝟕𝟔𝟔𝟎𝟎𝑱 𝒐𝒓 𝟏𝟕𝟔. 𝟔 𝒌𝑱
Heat
Heat is defined as the form of energy that is transferred across the boundary of a system at a given
temperature to another system (or the surroundings) at a lower temperature by virtue of the
temperature difference between the two systems.
Heat, like work, is a form of energy transfer to or from a system. Therefore, the units for heat and for
any other form of energy as well, are the same as the units for work, or at least are directly
proportional to them. In the International System the unit for heat (energy) is the joule. In the English
System, the foot pound force is an appropriate unit for heat.
Heat transferred to a system is considered positive, and heat transferred from a system is considered
negative. Thus, positive heat represents energy transferred to a system, and negative heat represents
energy transferred from a system. The symbol Q represents heat. A process in which there is no heat
transfer (Q = 0) is called an adiabatic process.
From a mathematical perspective, heat, like work, is a path function and is recognized as an inexact
differential.
𝟐
∫𝟏 𝜹𝑸 = 1Q2
In words, 1Q2 is the heat transferred during the given process between states 1 and 2. The rate at which
heat is transferred to a system is designated by the symbol ˙Q :
̇ 𝜹𝑸
𝑸̇ =
𝒅𝒕
It is sometimes convenient to refer to the heat transfer per unit mass, or q, defi ned as
q =Q/m
Often we are interested in the rate of heat transfer, or in the heat flux, denoted by 𝑸̇ with units of J/s =
W.
There are three heat transfer modes: conduction, convection, and radiation. In general, each of these
three heat transfer modes refers to a separate rate of energy crossing a specific plane, or surface.
Often, engineering applications involving heat transfer must consider all three modes in an analysis.
In each case, we can write relationships between the rate of heat transfer and temperature. A course in
heat transfer is dedicated to calculating the heat transfer in various situations. In thermodynamics, it is
given in a problem or it is found from applying the energy equation.
Example:
How much heat is required to raise the temperature of 5 kg of water from 288K to
320K? (Assume the specific heat (cp) for water is constant at 1410 J/(kg.K.)
Solution:
Q = CpmdT
Q = 1410×5×(320-288) = 225600 J
Work
Work is usually defined as a force F acting through a
displacement x, where the displacement is in the direction of the
force. That is,
𝟐
𝑾 = ∫ 𝑭𝒅𝑳
𝟏
This is a very useful relationship because it enables us to find the
work required to raise a weight, to stretch a wire, or to move a
charged particle through a magnetic field.
(Fig 15): Piston Movement
We can consider this quasi-equilibrium process and calculate the
amount of work W done by the system during this process. The total force on the piston is PA, where P is
the pressure of the gas and A is the area of the piston. Therefore, the work δW is
δW = PA dL
But (A dL = dV), the change in volume of the gas. Therefore,
δW = P dV
*
The work done at the moving boundary during a given
quasi-equilibrium process can be found by integrating
Eq.* .However, this integration can be performed only if
we know the relationship between P and V during this
process. This relationship may be expressed as an
equation, or it may be shown as a graph.
On a pressure–volume diagram (Fig 16) usually referred
to as a P–V diagram. At the conclusion of the process the
piston is in position 2, and the corresponding state of the
gas is shown at point 2 on the P–V diagram. Let us
assume that this compression was a quasi-equilibrium
process and that during the process the system passed
(Fig 16): P-V Diagram
through the states shown by the line connecting states 1
and 2 on the P–V diagram. The assumption of a quasi- equilibrium process is essential here because
each point on line 1–2 represents a definite state, and these states correspond to the actual state of the
system only if the deviation from equilibrium is infinitesimal. The work done on the air during this
compression process can be found by integrating Eq.*
𝟐
𝑭𝒅𝑳
𝟏
1W2=∫
Further consideration of a P–V diagram, such as (Fig 16), leads to another important conclusion. It is
possible to go from state 1 to state 2 along many different quasi-equilibrium paths, such as A, B, or C.
Since the area under each curve represents the work for each process, the amount of work done during
each process not only is a function of the end states of the process but also depends on the path
followed in going from one state to another. For this reason, work is called a path function or, in
mathematical parlance, δW is an inexact differential.
This concept leads to a brief consideration of point and path functions or, to use other terms, exact and
inexact differentials. Thermodynamic properties are point functions, a name that comes from the fact
that for a given point on a diagram (such as Fig 17) the state is fixed, and thus there is a definite value
for each property corresponding to this point. The differentials of point functions are exact
differentials, and the integration is simply
𝟐
∫ 𝒅𝑽 = 𝑽𝟐 − 𝑽𝟏
𝟏
Thus, we can speak of the volume in state 2 and the volume in
state 1, and the change in volume depends only on the initial and
final states.
Work, however, is a path function, for, as has been indicated,
the work done in a quasi-equilibrium process between two given
states depends on the path followed. The differentials of path
functions are inexact differentials, and the symbol δ will be used
in this book to designate inexact differentials (in contrast to d for
exact differentials). Thus, for work, we write
𝟐
Units for work
∫𝟏 𝜹𝑾 =1W2
≠ W2 –W1
(Fig 17): different quasi-equilibrium
paths
As already noted, work done by a system, such as that done by a gas expanding against a piston, is
positive, and work done on a system, such as that done by a piston compressing a gas, is negative.
Thus, positive work means that energy leaves the system, and negative work means that energy is added
to the system.
Our definition of work involves raising of a weight, that is, the product of unit force (one newton)
acting through a unit distance (one meter). This unit for work in SI units is called the joule (J).
1 J = 1Nm
Power is the time rate of doing work and is designated by the symbol 𝑾̇:
𝑾̇ =
𝜹𝑾
𝒅𝒕
The unit for power is a rate of work of one joule per second, which is a watt (W):
Sign conventions
Heat transfer to a system and work done by a system are positive; Heat transfer from a system and
work done on a system are negative
Examples
1. A hydraulic cylinder has a piston of cross sectional area 25cm 2 and a fluid pressure of 2 MPa. If the
piston is moved 0.25m, how much work is done?
S0l.
The work is a force with a displacement and force is constant: F=PA
𝑾 = ∫ 𝑭𝒅𝒙 = ∫ 𝑷𝑨𝒅𝒙 = 𝑷𝑨∆𝒙
2000kpa x 25 x 10-4m2 x 0.25m=1.23KJ
Work done to move the piston=1.23KJ
2. Calculate the necessary work in (KJ) units to fill balloon by air volume (0.568m3) at pressure (0.75
mmHg).
Internal Energy
One of the thermodynamic properties of a system is its internal energy, U, which is the sum of the
kinetic and potential energies of the particles that form the system. The internal energy of a system
can be understood by examining the simplest possible system: an ideal gas. Because the particles
in an ideal gas do not interact, this system has no potential energy. The internal energy of an ideal
gas is therefore the sum of the kinetic energies of the particles in the gas. It is the energy needed to
create the system. It may be divided into potential energy (Upot) and kinetic energy (Ukin)
components:
U = Upot + Ukin
Enthalpy
In the solution of problems involving systems, certain products or sums of properties occur with
regularity. One such combination of properties we define to be enthalpy H:
Enthalpy is a defined thermodynamic potential, that consists of the internal energy of the system (U)
plus the product of pressure (P) and volume (V) of the system:
H = U + PV
*
This property will come in handy, especially in a constant-pressure process, but also in other situations,
as we shall see in examples and applications in future chapters. The specific enthalpy h is found by
dividing by the mass: h = H/m. From Eq. (*), it is
h = u + Pv
Enthalpy is a property of a system and is also found in the steam tables. The energy equation can now
be written, for a constant-pressure equilibrium process, as
1Q2 =
H2 - H1
In a nonequilibrium constant-pressure process ΔH would not equal the heat transfer. It is only the
change in enthalpy or internal energy that is important; hence, we can arbitrarily choose the datum
from which to measure h and u. We choose saturated liquid at 0°C to be the datum point for water;
there h = 0 and u = 0.
The First Law of Thermodynamics
The total amount of energy (and mass) in the universe is constant.
(Energy can neither be created nor destroyed; it can only transfer from one form to another.)
During an interaction between a system and its surroundings, the amount of energy gained by the
system must be exactly equal to the amount of energy lost by the surroundings. A rock falling off a
cliff, for example, picks up speed as a result of its potential energy being converted to kinetic energy.
Suppose that a closed system of unit mass takes in a certain quantity of thermal energy Q, which it can
receive by thermal conduction and/or radiation. As a result the system may do a certain amount of
external work W. The excess of the energy supplied to the body over and above the external work done
by the body is Q − W. It follows from the principle of conservation of energy that the internal energy of
the system must increase by Q − W.
That is,
∆U = Q − W
Where ∆U is the change in internal energy of the system.
In differential form this becomes
dU = dQ – dW
Where dQ is the differential increment of heat added to the system, dW the differential element of work
done by the system, and dU the differential increase in internal energy of the system.
Joules experiment
James P. Joule carried out his famous
experiment; he placed known amounts of water,
oil, and mercury in an insulated container and
agitated the fluid with a rotating stirrer. The
amounts of work done on the fluid by the stirrer
were accurately measured, and the temperature
changes of the fluid were carefully noted. He
found for each fluid that a fixed amount of
work was required per unit mass for every
degree of temperature rise caused by the
(Fig 18): Joules experiment
stirring, and that the original temperature of the
fluid could be restored by the transfer of heat through simple contact with a cooler object.
In this experiment you can conclude there is a relationship between heat and work or in other word
heat is a form of energy.
The cycle consists of two processes, one an adiabatic work transfer followed by heat transfer
(∑ 𝒘)𝒄𝒚𝒄𝒍𝒆 = 𝑱(∑ 𝑸)𝒄𝒚𝒄𝒍𝒆 𝒐𝒓 ∮ 𝒅𝑾 = ∮ 𝒅𝑸
Historically, work was measured in mechanical units of force times distance, such as foot pounds force
or joules, and heat was measured in thermal units, such as the British thermal unit or the calorie.
Measurements of work and heat were made during a cycle for a wide variety of systems and for various
amounts of work and heat. When the amounts of work and heat were compared, it was found that they
were always proportional. Such observations led to the formulation of the first law of thermodynamics,
which in equation form is written
∮ 𝜹𝑾 = ∮ 𝜹𝑸
The symbol ∮ 𝜹𝑸, which is called the cyclic integral of the heat transfer, represents the net heat
transfer during the cycle, and ∮ 𝜹𝑾, the cyclic integral of the work, represents the network during the
cycle(∮ 𝜹𝑾 = ∮ 𝜹𝑸) is the state’s equation of the first law of thermodynamics for a control mass
during a cycle. Many times, however, we are concerned with a process rather than a cycle. We now
consider the first law of thermodynamics for a control mass that undergoes a change of state. We begin
by introducing a new property, energy, which is given the symbol E.
Consider a system that undergoes a cycle in
which it changes from state 1 to state 2 by
process A and returns from state 2 to state 1 by
process B. This cycle is shown in (Fig 19) on a
pressure (or other intensive property)–volume
(or other extensive property) diagram.
In the study of thermodynamics, it is convenient
to consider the bulk kinetic and potential energy
separately and then to consider all the other
energy of the control mass in a single property
that we call the internal energy and to which we
give the symbol U. Thus, we would write
(Fig 19): Demonstration of the existence of
thermodynamic property E.
E = Internal energy + kinetic energy + potential energy
E = U + KE + PE
The First Law for a Closed System
Generic Statement of the First Law for a Closed System
𝒓𝒂𝒕𝒆 𝒆𝒏𝒆𝒓𝒈𝒚
𝒓𝒂𝒕𝒆 𝒆𝒏𝒆𝒓𝒈𝒚
𝑻𝒊𝒎𝒆 𝒓𝒂𝒕𝒆 𝒐𝒇 𝒄𝒉𝒂𝒏𝒈𝒆
⟦
⟧ = ⟦𝒆𝒏𝒕𝒆𝒓𝒔 𝒔𝒚𝒔𝒕𝒆𝒎⟧ − ⟦
⟧
𝒍𝒆𝒂𝒗𝒆𝒔
𝒔𝒚𝒔𝒕𝒆𝒎
𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 𝒘𝒊𝒕𝒉𝒊𝒏 𝒔𝒚𝒔𝒕𝒆𝒎
ΔEsystem = Ein - Eout
Note: ΔEsystem - ΔE surroundings = 0
Specific Statement of First Law for a Closed System
1. Energy is conserved.
2. Energy can cross the boundary of a closed system by only two mechanisms: heat transfer and work
transfer.
3. The change in energy of a closed system is equal to the net heat transferred to the system minus the
net work performed by the system
The first law of thermodynamics is simply an expression of the conservation of energy principle, and it
asserts that energy is a thermodynamic property.
Energy can cross the boundary of a closed system in two distinct forms: heat and work. It is important
to distinguish between these two forms of energy. Therefore, they will be discussed first, to form a
sound basis for the development of the first law of thermodynamics.
We can use the principle of conservation of energy to define a function U called the internal energy.
When a closed system undergoes a process by which it passes from state A to state B, if the only
interaction with its surroundings is in the form of transfer of heat Q to the system or performance of
work W on the system, the change in U will be
ΔU = Qin.net - Wout.net
Or
Qin.net = ΔU + Wout.net
For a cyclic process:
∮ 𝜹𝑼 = 𝟎
During a non flow process the change in internal energy is calculated assuming the closed’s system
volume remains constant, the following equation is used
ΔU = m.Cv. ΔT
Where Cv is the specific heat capacity of the fluid, and ΔT is the temperature difference during the
process
The PV diagram
 Used when the pressure and volume are known at
each step of the process
 The work done on a gas that takes it from some
initial state to some final state is the
negative of the
area under the curve on the PV diagram.
 This is true whether or not the pressure stays
constant
 The curve on the diagram is called the path taken
between the initial and final states
(Fig 20): The PV diagram
Examples
1. In the cylinder of an air motor the compressed air has an internal energy of 420kJ/kg at the beginning
of the expansion and an internal energy of 200kJ/kg after expansion. Calculate the air heat reject
during the work expansion 100kJ/kg.
q = (u2 – u1) + w
q = (200 – 420) + 100 = - 220 + 100 = - 120kJ/kg
2. In the compression stroke of an internal - combustion engine the heat rejected to the cooling water is
45kJ/kg and the work input is 90kJ/kg. Calculate the change in internal energy of the working fluid
whether it is a gain or a loss.
Solution
q= - 45kJ/kg
- heat is rejected
w= - 90kJ/kg
- a work input to the system
q = (u2-u1) + w
- 45 = 𝜟u - 90
𝜟U = 45kJ/kg
3. Calculate the work necessary to compress air in an insulated cylinder from a volume of 2 m 3 to a
volume of 0.2 m3.The initial temperatures and pressure are20°C and 200 kPa, respectively.
Solution
We will assume that the compression process is approximated by a quasiequilibrium process, which is
acceptable for most compression processes, and that the process is adiabatic due to the presence of the
insulation (we usually assume an adiabatic process anyhow since heat transfer is assumed to be negligible).
The first law is then written as
−𝑾 = 𝒎(𝒖𝟐 − 𝒖𝟏 ) = 𝒎𝑪𝒗 (𝑻𝟐 − 𝑻𝟏 )
The mass is found from the ideal-gas equation to
𝑷𝑽
𝟐𝟎𝟎 × 𝟐
𝒎=
=
= 𝟒. 𝟕𝟓𝟕𝒌𝒈
𝑹𝑻 𝟎. 𝟐𝟖𝟕(𝟐𝟎 + 𝟐𝟕𝟑)
The final temperature T2 is found for the adiabatic quasiequilibrium process
𝑽𝟏 𝜸−𝟏
𝟐 𝟏.𝟒−𝟏
𝑻𝟐 = 𝑻𝟏 ( )
= 𝟐𝟗𝟑 (
)
= 𝟕𝟑𝟔 𝑲
𝑽𝟐
𝟎. 𝟐
Finally
6. An insulated, constant-volume system containing 1.36 kg of air receives 53 kJ of paddle work. The initial
temperature is 27oC. Determine
a) The change of internal energy.
b) The final temperature.
Assume a mean value Cv = 0.718 kJ/kgK.
Solution:
a) Q -W = ΔU
Q = 0 (insulated system)
W = -53 kJ (externally inputted work)
The change in internal energy ΔU is
ΔU = -W = +53 kJ
a)
since Q = 0
ΔU = m Cv ΔT
∴ 53 = 1.36 x 0.718 (T2 - 27)
T2=27+53/1,36×0.718=81.30C
7. An ideal gas occupies a volume of 0.5 m 3 at a temperature of 340 K and a given pressure. The gas
undergoes a constant pressure process until the temperature decreases to 290 K.
Determine
a) The final volume,
b) The work if the pressure is 120 kPa
Solution:
a) Since P = constant
𝑽𝟏
𝑻𝟏
𝑽
= 𝑻𝟐
V2=V1×T2/T1=0,5×290/340=0.426m3
𝟐
b) W = ∫P.dV
For a constant pressure process,
W = p (V2 - V1) = 120 (0.426 - 0.5) = - 8.88 kJ
The First Law of Thermodynamics: Control Volumes
Here we will extend the conservation of energy to systems that involve mass flow across their
boundaries, control volumes. Any arbitrary region in space can be selected as control volume. There
are no concrete rules for the selection of control volumes. The boundary of control volume is called a
control surface.
Conservation of Mass
Like energy, mass is a conserved property, and it cannot be created or destroyed. Mass and energy can
be converted to each other according to Einstein’s formula: E = mc2, where c is the speed of light.
However, except for nuclear reactions, the conservation of mass principle holds for all processes. For a
control volume undergoing a process, the conservation of mass can be stated as:
∑ 𝒎𝒊 − ∑ 𝒎𝒆 = ∆𝒎𝑪𝑽
(Fig 21): Conservation of mass principle for a CV.
The conservation of mass can also be expressed in the rate form:
∑ 𝒎̇ 𝒊 − ∑ 𝒎𝒆 = ̇𝒅𝒎𝑪𝑽 /𝒅𝒕
The amount of mass flowing through a cross section per unit time is called the mass flow rate and is
denoted by m°. The mass flow rate through a differential area dA is:
𝒅𝒎̇ − 𝝆𝑽𝒏 𝒅𝑨
The conservation of mass can also be expressed in the rate form:
∑ 𝒎̇𝒊 − ∑ 𝒎̇𝒆 = 𝒅𝒎𝑪𝑽 /𝒅𝒕
The amount of mass flowing through a cross section per unit time is called the mass flow rate and is
denoted by 𝒎̇. The mass flow rate through a differential area dA is:
d𝒎̇=ρVn dA
Where Vn is the velocity component normal to dA. Thus, the mass flow rate for the entire cross‐section
is obtained by:
𝒎̇ = ∫𝑨 𝝆𝑽𝒏 𝒅𝑨 (Kg/s)
Assuming one‐dimensional flow, a uniform (averaged or bulk) velocity can be defined:
𝒎̇= ρ V A (kg/s)
Where V (m/s) is the fluid velocity normal to the cross sectional area. The volume of the fluid flowing
through a cross‐section per unit time is called the volumetric flow,𝑽̇:
𝒎𝟑
( )
𝑺
𝑨
The mass and volume flow rate are related by:𝒎̇ = 𝝆𝑽̇ = 𝑽̇/𝒗.
𝑽̇ = ∫ 𝑽𝒏 𝒅𝑨 = 𝑽𝑨
Conservation of Energy
For control volumes, an additional mechanism can change the energy of a system: mass flow in and
out of the control volume. Therefore, the conservation of energy for a control volume undergoing a
process can be expressed as
Total energy crossing
Boundary as heat and +
work
total energy of _= _total energy
+
mass entering CV of mass leaving CV
=
net change in
energy of CV
𝑸 − 𝑾 = ∑ 𝑬𝒊𝒏 𝒎𝒂𝒔𝒔 − ∑ 𝑬𝒐𝒖𝒕 𝒎𝒂𝒔𝒔 = ∆𝑬𝑪𝑽
This equation is applicable to any control volume undergoing any process. This equation can also be
expressed in rate form:
𝑸̇ − 𝑾̇ = ∑ 𝒅𝑬𝒊𝒏 𝒎𝒂𝒔𝒔 /𝒅𝒕 − ∑ 𝒅𝑬𝒐𝒖𝒕 𝒎𝒂𝒔𝒔 /𝒅𝒕 = 𝒅𝑬𝑪𝑽 /𝒅𝒕
(Fig 22): Energy content of CV can be changed by
mass flow in/out and heat and work interactions.
Steady‐State Flow Process
A process during which a fluid flows through a control volume steadily is called steady-state process. A
large number of devices such as turbines, compressors, and nozzles operates under the same conditions
for a long time and can be modeled (classified) as steady‐flow devices.
The term steady implies no change with time. The term uniform implies no change with location over a
specified region.
A steady flow is characterized by the following:
1. No properties within the CV change with time. Thus, volume, mass, and energy of CV remain
constant. As a result, the boundary work is zero. Also, total mass entering the CV must be equal to total
mass leaving CV.
2. No properties change at the boundary of the CV with time. It means that the mass flow rate and
the properties of the fluid at an opening must remain constant during a steady flow.
3. The heat and mass interactions between the CV and its surroundings do not change with time.
Using the above observation, the conservation of energy principle for a general steady-flow system with
multiple inlets and exits can be written as:
𝑸̇ − 𝑾̇ = ∑ 𝒎̇ [𝒉𝒆 +
𝑽𝟐𝒆
𝑽𝟐𝒊
+ 𝒈𝒛𝒆 ] − ∑ 𝒎̇ 𝒊 [𝒉𝒊 +
+ 𝒈𝒛𝒊 ]
𝟐
𝟐
𝑸̇ − 𝑾̇ = ∑ 𝒎̇ 𝒆 𝜽𝒆 − ∑ 𝒎̇ 𝒊 𝜽𝒊
Examples
1. Steam at 2000 kPa and 600°C flows through a 60-mm diameter pipe into a device and exits
through a 120-mm-diameter pipe at 600 kPa and 200°C. If the steam in the 60-mm section has
a velocity of 20 m/s, determine the velocity in the 120-mm section. Also calculate the mass flow
rate.
Solution
From the superheat Table C.3 we find
𝝆𝟏 =
𝟏
𝟏
=
= 𝟓. 𝟎𝟏𝒌𝒈/𝒎𝟑
𝒗𝟏 𝟎. 𝟏𝟗𝟗𝟔
𝝆𝟐 =
𝟏
𝟏
=
= 𝟐. 𝟖𝟒𝒌𝒈/𝒎𝟑
𝒗𝟐 𝒐. 𝟑𝟓𝟐𝟎
The continuity equation of state is used to write
𝝆𝟏 𝑨𝟏 𝑽𝟏 = 𝝆𝟐 𝑨𝟐 𝑽𝟐
5.01× π× 0.032 × 20 = 2.84 ×π×0.062×V2
∴ 𝑽𝟐 = 𝟖. 𝟖𝟐𝒎/𝒔
The mass flow rate is
𝒎̇ =𝝆𝟏 𝑨𝟏 𝑽𝟏 = 𝟓. 𝟎𝟏 × 𝝅 × 𝟎. 𝟎𝟑𝟐 × 𝟐𝟎 = 𝟎. 𝟐𝟖𝟑𝒌𝒈/𝒔
2. Steam enters a converging‐diverging nozzle operating at steady state with P1 = 0.05 MPa, T1 =
400 °C and a velocity of 10 m/s. The steam flows through the nozzle with negligible heat transfer
and no significant change in potential energy. At the exit, P2 = 0.01 MPa, and the velocity is 665
m/s. The mass flow rate is 2 kg/s. Determine the exit area of the nozzle, in m2.
Steady state operation of the nozzle
Solution:


Work and heat transfer are negligible, 𝑸̇ = 𝑾̇ = 0.
Change in potential energy from inlet to exit is negligible, ΔPE = 0.
The exit area can be calculated from the mass flow rate m°:
A2 =𝒎̇ v2 / V2
We need the specific volume at state 2. So, state 2 must be found. The pressure at the exit is given; to
fix the state 2 another property is required. From the energy equation enthalpy can be found: