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Thermodynami cs Chapter 1 Diploma in Engineering Mechanical Engineering Division, Ngee Ann Polytechnic

Thermodynamics chapter 1

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Page 1: Thermodynamics chapter 1

ThermodynamicsChapter 1

Diploma in EngineeringMechanical Engineering Division, Ngee Ann Polytechnic

Page 2: Thermodynamics chapter 1

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

Page 3: Thermodynamics chapter 1

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.

Page 4: Thermodynamics chapter 1

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?

Page 5: Thermodynamics chapter 1

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.

Page 6: Thermodynamics chapter 1

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

Page 7: Thermodynamics chapter 1

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

Page 8: Thermodynamics chapter 1

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

Page 9: Thermodynamics chapter 1

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

Page 10: Thermodynamics chapter 1

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

Page 11: Thermodynamics chapter 1

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

Page 12: Thermodynamics chapter 1

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

Page 13: Thermodynamics chapter 1

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

Page 14: Thermodynamics chapter 1

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

Page 15: Thermodynamics chapter 1

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.

Page 17: Thermodynamics chapter 1

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.

Page 18: Thermodynamics chapter 1

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

Page 20: Thermodynamics chapter 1

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.

Page 23: Thermodynamics chapter 1

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.

Page 24: Thermodynamics chapter 1

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)

Page 26: Thermodynamics chapter 1

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)

Page 27: Thermodynamics chapter 1

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

Chapter 1 Introduction to Thermodynamics

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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

Page 29: Thermodynamics chapter 1

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)

Chapter 1 Introduction to Thermodynamics

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Piston(mass=0)

p

water

Force = 500 N

A = 0.01 m2

Example:

Chapter 1 Introduction to Thermodynamics

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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

+-

Chapter 1 Introduction to Thermodynamics

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Absolute pressure, pabs

A pressure measured above the absolute zero pressure, which is a perfect vacuum.

Relationship:pabs = pgauge + patm

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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)

Chapter 1 Introduction to Thermodynamics

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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.

Chapter 1 Introduction to Thermodynamics

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Properties are broadly classified into two categories, namely extensive properties and intensive properties.

Chapter 1 Introduction to Thermodynamics

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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.

Chapter 1 Introduction to Thermodynamics

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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.

Chapter 1 Introduction to Thermodynamics

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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.

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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Density, ρ(kg/m3) Mass per unit volume

The density is also the reciprocal of specific volume

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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.

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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.

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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Chapter 1 Introduction to Thermodynamics

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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)

Chapter 1 Introduction to Thermodynamics

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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.

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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

Chapter 1 Introduction to Thermodynamics

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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

Page 64: Thermodynamics chapter 1

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

Page 67: Thermodynamics chapter 1

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

Chapter 1 Introduction to Thermodynamics

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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

Page 72: Thermodynamics chapter 1

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

Page 74: Thermodynamics chapter 1

» 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

Page 75: Thermodynamics chapter 1

Thank you!Q & A