thermodynamics chapter 1
TRANSCRIPT
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ThermodynamicsChapter 1
Diploma in EngineeringMechanical Engineering Division, Ngee Ann Polytechnic
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Chapter 1 Introduction to Thermodynamicso Contents
Section Sub-Section Description
1.1 Define Thermodynamics
1.1.1 Basic Thermodynamics and Applied Thermodynamics
1.2 Basic Concepts of Thermodynamics- System, Surroundings, Boundary, Properties of a
System, Thermal Equilibrium, Work Transfer, Heat Transfer
1.3 First Law of Thermodynamics
1.4 Derive the Non-flow Energy Equation
1.5 Derive the Steady Flow Energy Equation
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Chapter 1 Introduction to Thermodynamicso 1.1 What is ThermodynamicsSimple definitions:
A science which deals with the relationship between thermal energy (heat) and all other forms of energy (e.g. mechanical, electrical, etc.).
A study of thermal engineering.
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued)
Human activity requires energy Energy considered as something which “gives life” to
otherwise inanimate objects
Where does this energy come from?
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued)Energy source: The major source of energy for humanity is the
thermal energy received directly from the sun in the form of solar radiation
For engineering applications, the most common source of thermal energy is from combustion, that is, the burning of fuel such as wood, coal or oil.
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued) Primitive man discovered that fire provided him with
thermal energy to keep warm and to cook food
Modern man often makes direct use of thermal energy, it is a fact that energy in this form is less useful, and must first be converted into one of the other energy forms before it can be used to power many of our modern machines
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued) Similarly, it is often necessary to convert between all
the other forms of energy
Some of these energy conversion processes might require several steps to produce energy in the form it is needed.
Common energy conversion processes
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued)(a) Thermal energy to mechanical energy
An example of this energy conversion process is the motor car engine.
Fuel is burned to produce thermal energy which the engine then converts into the mechanical energy needed to move the car along the road
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued)(b) Mechanical energy to thermal energy
The reverse process occurs when a moving car is stopped by applying the brakes
The mechanical energy of the car is absorbed by the brakes and converted into thermal energy, evidenced by the fact that the brakes become hot after application
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued)(c) Thermal energy to electrical energy
In a power station fuel such as coal or oil is burned to produce thermal energy. This is used to produce steam which drives the machinery that produces electrical energy
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued)(d) Electrical energy to thermal energy
Electrical energy can be turned into thermal energy, as in the case of a domestic electric kettle
An interesting comparison is the use of electrical energy to remove thermal energy, as in the case of a refrigerator (or air-conditioning unit). A refrigerator removes the thermal energy from the food stored inside the insulated box
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued)(e) Electrical energy to mechanical energy
Electrical energy supplied to a motor causes the motor shaft to rotate. The resulting mechanical energy can, for example, be used to drive a machine tool or propel a vehicle
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued)(f) Mechanical energy to electrical energy
The rotating shaft of an internal combustion engine can be used to drive an electrical generator, thus providing a portable power source for use in remote areas
Regenerative braking
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Chapter 1 Introduction to Thermodynamicso 1.1 What is Thermodynamics (continued) Many more involving conversions between all the different
forms of energy Thermodynamics is the science that deals with these processes It is concerned with how they occur, with the laws which control
their occurrence, as well as with the engineering of the machinery needed to convert between the various energy forms
It is also concerned with what actually happens within this machinery, both microscopically and macroscopically
As such, Thermodynamics is a subject at the very core of engineering practice
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Chapter 1 Introduction to Thermodynamicso 1.1.1 Basic and Applied Thermodynamics
Thermodynamics is all about energy, and the various transformations between the different forms of energy
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Chapter 1 Introduction to Thermodynamicso 1.1.1 Basic and Applied Thermodynamics (continued) In Basic Thermodynamics, we are concerned not only with
the different forms of energy, but also with various process by which energy is changed from one form to another.
We are also concerned with the fundamental physical laws which must be obeyed during these transforming processes.
It is essential that we understand the important properties of the substances which are involved in the process, for it is by the changes in these properties that we measure the changes of energy.
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Chapter 1 Introduction to Thermodynamicso 1.1.1 Basic and Applied Thermodynamics (continued) In Applied Thermodynamics, we are concerned with
the application of basic thermodynamics principles to actual engineering problems.
We are not only concerned with the physical details of the hardware, but also with how the thermodynamic processes actually occur inside the machinery, as well as with the factors which affect the performance and effectiveness of such machinery.
This is clearly a very practical area of study.
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Chapter 1 Introduction to Thermodynamicso 1.1.1 Basic and Applied Thermodynamics
(continued)
Some typical domestic examples to elaborate how Basic Thermodynamics applies to the machinery
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Chapter 1 Introduction to Thermodynamicso 1.1.1 Basic and Applied
Thermodynamics (continued)(a) Electric Fan Electrical energy is supplied to
a motor causing the shaft to rotate, and with it the fan blades, which in turn create a draft of air. Here electrical energy is converted into mechanical energy
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Chapter 1 Introduction to Thermodynamicso 1.1.1 Basic and Applied Thermodynamics
(continued)(b) Refrigerator Electrical energy is supplied to a
compressor which produces a flow of working fluid known as a refrigerant. This fluid is used to remove thermal energy from inside an insulated box to make a cold space where we can preserve our food. Here an input of electrical energy results in the removal of thermal energy. However, in making one place cold, we have to make another place hot, which means we can actually heat water using a refrigerator.
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Chapter 1 Introduction to Thermodynamicso 1.1.1 Basic and Applied
Thermodynamics (continued)(c) Air conditioner It works in much the same way as a
refrigerator, except that the refrigerant is used to remove the thermal energy from the air inside a room. This cools the air to a comfortable temperature for the occupants. Here again, an input of electrical energy results in the removal of thermal energy. When we cool a room with an air conditioner, we sometimes see that the windows in the room are covered with droplets of dew. Here we are actually extracting water from air.
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Chapter 1 Introduction to Thermodynamicso 1.1.1 Basic and Applied
Thermodynamics (continued)(d) Pressure cooker A little water is placed inside a sealed
container along with the food to be cooked. The cooker is placed over a gas burner where the chemical energy of the fuel is converted into thermal energy. This energy is supplied to the cooker to boil the water and produce pressurized steam inside the container, enabling us to cook the food quickly.
One result of using a pressure cooker is that we save fuel, and hence fuel cost at the same time.
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Chapter 1 Introduction to Thermodynamicso 1.1.1 Basic and Applied Thermodynamics (continued)
We can see from all these examples that a study of Basic Thermodynamics can help us to understand how many basic domestic appliances work.
The principles of thermodynamics are evident all around us and are not merely confined to the world of engineers.
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Chapter 1 Introduction to Thermodynamicso 1.2 Basic Concepts of ThermodynamicsBasic units of the S. I. System:
Quantity Unit Mass kg Length m Time s
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Chapter 1 Introduction to Thermodynamicso 1.2 Basic Concepts of Thermodynamics (continued)Derived units for thermodynamics in the S. I. System:
ForceA measure of the “push” or “pull” which is often exerted on a body.F=m a∙F: force, 1 N = 1 kg m/s∙ 2
m: mass (kg); a: acceleration (m/s2)
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Chapter 1 Introduction to Thermodynamicso 1.2 Basic Concepts of Thermodynamics (continued)Derived units for thermodynamics in the S. I. System:
WeightThe gravitational attractive force which the earth exerts on a mass. It depends on the acceleration due to gravity and varies with height and location on the earth.
w=m g∙
w: weight, 1 N = 1 kg m/s∙ 2
m: mass (kg); g: gravity (assumption: 9.81m/s2)
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o 1.2 Basic Concepts of Thermodynamics (continued) SystemA collection of matter within a prescribed region
System
Cylinder
Boundary
Piston
Surroundings
Surrounding: the region enclosing the systemBoundary: the surface that separates the system from the surroundings
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Chapter 1 Introduction to Thermodynamicso 1.2 Basic Concepts of Thermodynamics (continued)Properties of a system ( to define its condition)
Most common:PressureTemperatureVolume and specific volumeDensity and specific gravity
Additional:Internal energyEnthalpyEntropy
At least 2 independent
properties of a fluid known, then the state
of the system is known
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Pressure
p
p: pressure, the force exerted by a fluid on unit area (N/m2) 1 bar = 105 N/m2
F: force (N)A: area (m2)
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Piston(mass=0)
p
water
Force = 500 N
A = 0.01 m2
Example:
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Atmospheric pressure, patm
A pressure due to the atmosphere at the surface of the earth. It depends on the weight of air above the surface.
Sea level ≈ 1.01325 × 105 N/m2 ≈ 1.01325 bar
Gauge pressure, pgauge
A pressure measured with respect to the atmospheric pressure.
E.g Pressure= 2atm; Gauge Pressure= 1atmpatm
+-
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Absolute pressure, pabs
A pressure measured above the absolute zero pressure, which is a perfect vacuum.
Relationship:pabs = pgauge + patm
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Example:The gauge pressure of the air in a vessel is 10 kN/m2. Determine the absolute pressure of the air. Assume the atmospheric pressure is 100 kN/m2.
Solution:pabs = pgauge + patm
pabs = 10 + 100 = 110 kN/m2
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Temperature, T (K)For the degree of hotness or coldness of anything.
The same temperature interval Conversion between the two
temperature scales:
T(K) = t(˚C) + 273
Celsius Kelvin
boiling pt*
freezing pt*
*atmospheric pressure
100˚C
0˚C
-273˚C
373 K
273 K
0 K (absolute zero)
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A thermodynamics property may be defined as any quantity that describes the state of a system and, conversely, as any quantity, the value of which depends solely on the state of the system.
In other words, the thermodynamic property is independent of how the system reaches a given state.
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Properties are broadly classified into two categories, namely extensive properties and intensive properties.
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In a given state, the value of an extensive property depends on the amount of mass in the system. Volume is an example of this type of property. The greater the mass, the greater is the volume. Extensive properties are denoted by upper case letters. For example, volume is denoted by V.
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A property that is independent of the mass of the system is called an intensive property. Pressure and temperature come into this category.
Specific Property: At a given state, the volume (V) is directly proportional to the mass (m) of the system. The quantity V/m is a specific property known as specific volume and is denoted by v (lower case).
Similarly, all extensive properties expressed per unit mass become specific properties, and are denoted by lower case letters.
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Volume, V(m3) The amount of space which it occupies.
Specific volume, v(m3/kg) The volume occupied by unit mass of the
substance.
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Example:1.29 kg of the compressed air is contained in a rigid vessel of volume 1.0 m3. Determine the specific volume of the air.
Solution:
3
11.290.7752 /
air
VvmVvm
m kg
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Density, ρ(kg/m3) Mass per unit volume
The density is also the reciprocal of specific volume
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Solution:
Volume = Length × Width × height
Vair = 6.0×4.0×2.9 = 69.6 m3
Example:
4.0 m
6.0 m
2.9 m
A living room: density of air ρ=1.2 kg/m3
What is the mass of air?
air air air
mρ=V
m =ρ ×V=1.2×69.6=83.52kg
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Specific gravity, s.g.sub (-) A ratio of the density of a substance over the
density of water at 4˚C (1000 kg/m3)
Example:The density of sea water is 1025kg/m3. What is the specific gravity of sea water?Solution:
subtan
s.g. =4
1025s.g. =1000=1.025
sea water
density of subs cedensity of water at C
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System equilibrium
If a system is completely stable when one particular value of any property (e.g. pressure) is the same at all points throughout the system at that particular instant.
In practice, the properties of a system changes and it is usually assumed that the initial and final conditions are in states of equilibrium.
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Work transfer
Rectilinear motionWork transfer of a constant force F is the product of the force and the distance traveled by the force measured along the line of action of the force.
F F
S
Work transfer = F × S
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Work transfer
Angular motion» Work transfer of a constant torque is the product of
the torque and the angular displacement by the torque.
torque
Work transfer = r × θ
r: torque in N·mθ: angular displacement in rad
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Work transfer
Convention» Positive: when work energy is transferred from the
system to the surroundings» Negative: when work energy is transferred into the
system from the surrounding
system
surroundingswork output
(positive)
work input(negative)
boundary
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Heat transfer
A form of energy which crosses the boundary of a system during a change of state produced by a difference of temperature between the system and its surrounding.
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Heat transferConvention» Positive: when heat energy flows into the system
from the surroundings» Negative: when heat energy flows from the system
to the surroundings
system
surroundingsheat loss
(negative)
heat supplied(positive)
boundary
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The principle of the conservation of energy states:
Energy can neither be created nor destroyed
» The First Law of Thermodynamics refers to heat energy and work energy
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The First Law of Thermodynamics: When a system undergoes a thermodynamics cycle then
the net heat supplied to the system from its surroundings is equal to the net work done by the system on its surroundings.
ΣQ = ΣW
Where: ΣQ: the net heat supplied ΣW: the net work done/work output
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Example:A system undergoes a complete thermodynamic cycle. Determine the value of the work output, Wout.
system
surroundings
Qin=10 kJ boundary
Qout=3 kJ Wout
Win=2 kJ
Solution:
ΣQ = ΣWQin + Qout = Win + Wout
Wout = Qin + Qout - Win=10×103 + (-3 ×103) – (-2 ×103)=9×103 J (9 kJ)
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Non-flow Processes
» In a closed system energy may be transferred across the boundary in the form of work energy and heat energy but the working fluid itself never crosses the boundary
» Any process undergone by a closed system is referred to as a non-flow process.
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Non-flow ProcessesExample:
i. Suction stroke:» The working fluid
flows into the cylinder, which is then sealed by closing of the inlet valve
A cylinder of an internal combustion engine
T TFluid flows in
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Non-flow ProcessesExample:
ii. Compression stroke:» Whilst the cylinder is
sealed, the fluid is compressed by the piston moving into the cylinder
A cylinder of an internal combustion engine
T TNon-flow process
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Non-flow ProcessesExample:
iii. Working stroke:» Heat energy is
supplied so that the fluid possesses sufficient energy to force the piston downward and produce work output
A cylinder of an internal combustion engine
T TNon-flow process
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Non-flow ProcessesExample:
iv. Exhaust stroke:» The exhaust valve is
opened for the fluid to flow out of the cylinder
A cylinder of an internal combustion engineT TFluid flows out
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Non-flow Energy Equation» When the fluid in a closed system is undergoing a non-flow process
from State 1 to State 2, the internal energy of a fluid depends on pressure and temperature.
» The non-flow energy equation:
or U2 - U1 = Q12 – W12• U2 the internal energy of the fluid at State 2• U1 the internal energy of the fluid at State 1• Q12 the net heat energy transferred to the system
from the surrounding• W12 the net work energy transferred from the
system to the surrounding
Chapter 1 Introduction to Thermodynamics
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Non-flow Energy Equation» The fluid in a closed system produces work amounting to
600 kJ whilst heat energy amounting 1000 kJ is transferred into it. Determine the change of internal energy of the fluid and state whether it is an increase or decrease.
Solution:U2 - U1 = Q12 – W12
=1000 – 600=400 kJ
Since U2 > U1, the internal energy has increased.
Q=1000 kJ W=600 kJ
Chapter 1 Introduction to Thermodynamics
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Flow Processes» In an open system, in addition to energy transfers
taking place across the boundary, the fluid may also cross the boundary.
» Any process undergone by an open system is called a flow process.
» Steady flow processes, and unsteady flow processes
Chapter 1 Introduction to Thermodynamics
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Steady Flow ProcessesThe conditions:
» The mass of fluid flowing past any section in the system must be constant with respect to time
» The properties of the fluid at any particular section in the system must be constant with respect to time
» All transfers of work energy and heat energy which take place must do so at a uniform rate
Chapter 1 Introduction to Thermodynamics
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Steady Flow Processes:Example: A steam boiler under constant load
» To maintain the water level in the boiler:
» To maintain the production of team at this rate ( )at a steady pressure, the furnace will need to supply heat energy at a steady rate
» Under these conditions, the properties of the working fluid at any section within the system must be constant with respect to time
Boiler
Qin (from furnace)
Feed water in
Steam out
Boundary
wm
sm
Chapter 1 Introduction to Thermodynamics
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The Continuity of Mass Equation» In steady flow, the mass flow rate of fluid is the same
across any section in the system» Consider kg/s of fluid flowing through a system in
which all conditions are steady (i.e. under steady flow conditions)
m
Specific volume, v1
Velocity, c1
Specific volume, v2
Velocity, c2
Cross-sectional area, A1 Cross-sectional area, A2
𝑚·
=𝑐⋅ 𝐴𝑣Continuity of mass equation: in out
m m
Chapter 1 Introduction to Thermodynamics
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The Steady Flow Energy Equation» The working fluid flows along the inlet pipe at a constant rate and
enter the system at point 1. » Various energy transfers take place across the boundary of the
system.» The fluid flows out of the system at point 2 along the outlet pipe.
system
1
2
pressure, p1
inlet velocity, c1
specific internal energy, u1
pressure, p2
outlet velocity, c2
specific internal energy, u2
inQ
outQ
outW
inWZ1
Z2
Chapter 1 Introduction to Thermodynamics
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Energy balance:
The total amount of energy entering the
system
The total amount of energy leaving the system=
in outE E
or
The total energy entering the system :• Rate of heat energy entering the system per second,• Rate of work energy entering the system per second,• The rate of energy of the fluid entering the system
Internal energy, Potential energy, Kinetic energy,
• The rate of work energy required to push the fluid across the boundary and enter the system,
inE
inQ
inW
1 1m u
1 1m gZ
21 1
12m c
1 1 1m pv
Chapter 1 Introduction to Thermodynamics
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Therefore:
Similarly:
Since:
And
Hence:Steady flow energy equation
21
1 1 1 1 1( )2
in inincE Q W m u pv gZ
22
2 2 2 2 2( )2
out outoutcE Q W m u p v gZ
in outE E
h u pv
2 21 2
1 21 1 2 2( ) ( )2 2
in outin outc cQ W m h gZ Q W m h gZ
Specific enthalpy, h, is a measure of the total energy of a thermodynamic system. It includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure.
Chapter 1 Introduction to Thermodynamics
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Example» In an open system, the fluid flow through the system at 17 kg/s
and the power developed by the system is 14 000 kW. The specific enthalpy of the fluid at inlet and outlet are 1 200 kJ/kg and 360 kJ/kg respectively, and the velocities of the fluid at inlet and outlet are 60 m/s and 150 m/s respectively. Assuming the change in potential energy is negligible.
Determine:
a. the rate at which heat is rejected from the system b. the area of inlet pipe, given that the specific volume of the fluid at inlet is 0.5 m3/kg
Chapter 1 Introduction to Thermodynamics
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Solution:a. Applying the continuity of mass equation
Applying steady flow energy equation
outQ
outW
outlet
inlet
1 21 2
0
0
14000
in
in
out
Q
W
W kW
m gZ m gZ
Hence:
Chapter 1 Introduction to Thermodynamics
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Applying
Since
Hence the cross-sectional area of inlet pipe
V A cA cvm
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Example » Fluid flows steadily at the rate of 0.4 kg/s through an open
system, entering at 6 m/s with a pressure of 1 bar and a specific volume of 0.85 m3/kg and leaving at 4.5 m/s with a pressure of 6.9 bar and a specific volume of 0.16 m3/kg. The specific internal energy of fluid leaving is 88 kJ/kg, greater than that of the fluid entering. 59 kJ/s of heat is rejected from the system to its surrounding and assuming the change in potential energy is negligible.
Determine:
a. the power required to drive the system b. the inlet and outlet pipe cross-sectional area
Chapter 1 Introduction to Thermodynamics
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Solution:a. Applying the continuity of mass equation
Applying steady flow energy equation
outQ
outlet
inlet
Hence:
1 2 0.4 /m m m kg s
inW
1 21 2
0
0
in
out
Q
W
m gZ m gZ
2 21 2
1 21 2( ) ( )2 2
in outc cW m h Q m h
Chapter 1 Introduction to Thermodynamics
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Solution:Hence:
Since
Hence:
2 1 2 2 2 1 1 1
2 1 2 2 1 1
3 5 5
3
( )( ) ( )
88 10 (6.9 10 0.16 1 10 0.85)
113.4 10 /
h u pvh h u p v u p vu u p v p v
J kg
2
2 21 2
1 21 2
2 22 1
2 1
23 3
3
( ) ( )2 2
( ) ( )2 2
4.5 659 10 0.4 113.4 10 0.4 ( )2 2
104.36 10 /
in out
out
c cW m h Q m h
c cQ m h h m
J s
Chapter 1 Introduction to Thermodynamics
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» b) applyingAnd
Hence, the cross-sectional area of inlet pipe
The cross-sectional area of inlet pipe
211
1
0.85 0.4 0.05676
v mA mc
222
2
0.16 0.4 0.01424.5
v mA mc
Chapter 1 Introduction to Thermodynamics
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Thank you!Q & A