Modul Thermodynamic

Question 1

Question 2
a) Superheated Steam Phase
b) Compressed Liquid Phase
c) Superheated Steam Phase
d) Compressed Liquid Phase

43

Question 3 uh x Fasa
Tp (m3/kg) (kJ/kg) (kJ/kg)
1.0228
(℃) (bar) × 10−3 292.95 292.98 0 Cecair Termampat
a) 70 2.0
0.4358 1500 1588.03 0.492 Campuran cecair &
b) 120.2 2.0 wap tepu
0.16125
c) 175.4 9.0 2120.83 2266.16 0.75 Campuran cecair &
0.16524 wap tepu
d) 179.9 10.0
e) 350 40.0 0.0665 2310.31 2475 0.85 Campuran cecair &
f) 80 5.0 1.0291 wap tepu
× 10−3
2826.7 3092.5 1 Wap Panas Lampau

334.86 334.91 0 Cecair Termampat

44

Question 4
Mark the dots in question 3 above in the T-υ and P-υ state diagrams.

45

Question 5

Given parameters;

 V = 500 liter = 500 × 10−3 = 0.5 3
 m = 2 kg
 = = 0.5 = 0.25 3⁄

2

 = 200 = 2

At P = 2 bar, = 0.8857 3⁄ ; >

∴ Phase of a mixture of liquid and saturated vapors, T = TS = 120.2℃

To determine the internal energy and mass of liquids and saturated vapors, the coefficient of dryness
must be determined. The coefficient of dryness is determined by;

0.25
= ; = = 0.8857 = 0.2822

Internal energy;

= 504.49 + (0.2822)(2529.5 − 504.49) = . /

Saturated liquids and vapors;

 =

+

 0.2822 = ; = 2 × 0.2822 = .

2

 = 2 − 0.5644 = .

46

CHAPTER 3
Energy, Heat And Work

OBJECTIVE
At the end of this chapter, student will be:

 Defines Energy, Heat And Work
 What is Energy, Potential energy,Kinetic

energy, Enthalpy and Specific heat ideal gas
 What is Heat, Heat Transfer and modes of

heat transfer
 What is work and function of work.

CHAPTER 3
Energy, Heat And Work

3.0 Introduction

The energy possessed by a system is defined as an extensive property that has the ability to
change the state of the system and its environment through the interaction that occurs through
the boundaries of the system.

= ( , , etc.)

Definition of energy
In physics, energy is the quantitative property that must be transferred to an object in order to
perform work on, or to heat, the object. Energy is a conserved quantity; the law of conservation
of energy states that energy can be converted in form, but not created or destroyed.

E U TK TU

48

3.1 Energy

3.1.1 Potential energy
 The energy possessed by the system is due to being in the field of ability or position of the body

( ). =

With m = mass, g = acceleration of gravity, h = position of the body. Refer figure 3.1
 Example; the change in energy by a body of mass m at a distance of h1 to h2 is given by;

Figure 3.1
∆
 Mass x Acceleration of grafity x Distance change = (ℎ2 − ℎ1)
 In thermodynamic analysis the potential energy is ignored because its value is small when
compared to other energies except the distance difference is very large. Example of pumping
water from the ground floor to the 20th floor using a pump.
Example 1
Determine the work done by a man who lifted 20kg of rice as high as 0.5m.
Solution
= (ℎ2 − ℎ1)

= 20 × 9.81 × 0.5
= .

49

3.1.2 Kinetic energy

 In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is

defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

Having gained this energy during its acceleration, the body maintains this kinetic energy unless

its speed changes.

 The energy possessed by the system is due to its movement relative to a datum such as the

earth which is considered stationary.

= ̅ ( )


 When a particle with mass m moves from the initial state with velocity ̅ 1to the final state with

̅ 2, the change of KE is given by.

∆ = ( ̅ − ̅ ) ( )


 This energy is not very important in the field of thermodynamics because of its small value. It

can be ignored unless the mass flow rate or large velocity changes such as jet or rocket engines.

Figure 3.2

Example 2

Determine the power in kW, which is required to accelerate a car of mass 1000kg from a break
to a speed of 100 km / h in 20s.

Solution

= ∆ = ( ̅ − ̅ )


= × × ([ × − )

]

= .
Power required

̇ = = 385.8 = .

20

50

3.1.3 Internal Energy

 The internal energy of a thermodynamic system is the energy contained within it. It is the energy
necessary to create or prepare the system in any given internal state.

 Quantity is influenced by the working fluid used (no general formula)
 Pure materials (liquids and vapors) - can refer to the table of properties of thermodynamic
materials
 Ideal gas - uses a specific formula

3.1.4 Internal energy, Enthalpy and Specific heat ideal gases

 The specific heat is the amount of heat per unit mass required to raise the temperature by one

degree Celsius. The relationship between heat and temperature change is usually expressed in

the form shown below where c is the specific heat. The relationship does not apply if a phase

change is encountered, because the heat added or removed during a phase change does not

change the temperature.

 For pure materials, in the form of gas, the specific heat at constant volume, and the specific

heat at constant pressure, are given by;

 = and = ℎ (since u and h function temperature)


 Assume heat is of course constant and condensed

− = ( − ) and − = ( − )

Enthalpy is a thermodynamic property : H = U + pV and for ideal gas H = U + RT

 In the form of a unit of mass and can be differentiate by

dh = du+RdT or ℎ = +



= and = ℎ; so



= + R or − =

Specific ratio = so



= or =
− −

51

3.2 Heat

3.2.1 Heat Transfer

Heat transfer, also referred to simply as heat, is the movement of thermal energy from one thing
to another thing of different temperature. These objects could be two solids, a solid and a liquid
or gas, or even within a liquid or gas. The first 3 different ways the heat can transfer: conduction
(through direct contact), convection (through fluid movement), or radiation (through
electromagnetic waves). Heat transfer occurs when the temperatures of objects are not equal
to each other and refers to how this difference is changed to an equilibrium state.
Thermodynamics then deals with things that are in the equilibrium state.

=

̇ =
=

= = heat transfer per unit mass


3.2.2 Modes of Heat Transfer

Heat can be transferred in four ways; conduction, convection, radiation and advection.

 Conduction or diffusion The transfer of energy between objects that are in physical contact. Ex:
Solid bar

 Convection The transfer of energy between an object and its environment, due to fluid motion.
 Natural Convection-the fluid motion is caused by buoyancy forces which are induced by
density differences due to the variation of temperature in the fluid
 Force Convection-the fluid is forced to flow in a tube or over a surface by external means
such as fan, pump, or the wind

 Radiation The transfer of energy to or from a body by means of the emission or absorption of
electromagnetic radiation. electromagnetic waves (or photons) as a result of the changes in the
electronic configurations of the atoms or molecules

 Advection The transfer of energy from one location to another as a side effect of physically
moving an object containing that energy.

52

3.3 Work

In thermodynamics, work performed by a system is energy transferred by the system to its
surroundings, by a mechanism through which the system can spontaneously
exert macroscopic forces on its surroundings. In the surroundings, through suitable passive
linkages, the work can lift a weight, for example. Energy can also transfer from the surroundings
to the system; in a sign convention used in physics, such work has a negative magnitude. In
mechanical, work (W); force × distance.

 The application of the equation found that the work done on the mass, m is:

= ( 2 − 1)

3.3.1 Electrical work Figure 3.3

= ( ) Electrical power in terms of resistance R, current I, and
potential difference V.
 V = voltan
 N = cas elktrik 53

Electric power

̇ =
= 2

= 2 ( )



̇ = =


= ∆

3.3.2 Shaft work

 To rotate the shaft, a certain amount of torque, T must be applied


= ∴ =
ℎ = ;

= 2 ;

= ℎ

 shaft work is equal to :

ℎ = ( )
= × 2

 So the power of shaft is

̇ ℎ = ̇

̇ = ℎ

Figure 3.4
Example 3

Determine the power transferred by the car engine to the shaft when as much as 200 Nm of
torque is applied to the shaft. The shaft rotates at a speed of 3500 laps per minute.
Solution
̇ ℎ = 2πṅ T

3500
= 2 × ( 60 ) × 200 = 73304 = .

54

3.3.3 Spring work

 Spring - a flexible tool used to restore the position of a body to its original position when the
body is moving.

 An example of spring work is a car suspension system
 If pulled by force F and lengthened by a distance of dx then the work done is

 =

 To determine the work for the spring, the F & x relationship needs to be known

 = ; = ⁄

 = ( - )


3.3.4 Boundary Work

 Work is energy expended when a force acts through a displacement. Boundary work occurs
because the mass of the substance contained within the system boundary causes a force, the
pressure times the surface area, to act on the boundary surface and make it move. This is what
happens when steam, the "gas" in the figure below, contained in a piston-cylinder device
expands against the piston and forces the piston to move; thus, boundary work is done by the
steam on the piston. Boundary work is then calculated from

2

12 = ∫

1

 The result of the integral is equal to the area under the curve 1-2 in diagram p-V. refer figure
3.5

Figure 3.5
= ℎ ℎ

= ( )
 Positive work = performed by the system on its environment for the development process.
 Negative work = performed by the environment on the system for the compression process.

55

3.3.5 Boundary work for close system
Isometric Process

An Isometric process, also called a constant-volume process, an isovolumetric process, or
an Isochoric process, is a thermodynamic process during which the volume of the closed
system undergoing such a process remains constant. An isometric process is exemplified by
the heating or the cooling of the contents of a sealed, inelastic container: The thermodynamic
process is the addition or removal of heat; the isolation of the contents of the container
establishes the closed system; and the inability of the container to deform imposes the constant-
volume condition. The isometric process here should be a quasi-static process.

Volume change dV = 0

2

12 = ∫ ; = 0

1

∴ 12 = 0

Figure 3.6 (The area under the arch of the p-V diagram is zero)
Isobaric Process

In thermodynamics, an isobaric process is a type of thermodynamic process in which the
pressure of the system stays constant: ΔP = 0. The heat transferred to the system does work,
but also changes the internal energy (U) of the system. This article uses the physics sign
convention for work, where positive work is work done by the system. Using this convention, by
the first law of thermodynamics.

 The process occurs when the volume of the system changes. The volume increases
during the pressure expansion process and decreases during the pressure compression
process.

56

2

12 = ∫

1

12 = ( 2 − 1)

Figure 3.7
Example 4

15 kg of water in the boiler cylinder system initially at a pressure of 10 bar and a temperature of
350 ℃ cooled at constant pressure until it becomes saturated vapor. Sketch the above process
on diagrams T-υ and p-υ and determine the work in progress.
Solution
Ts at 10 bar = 179.91℃ (superheated steam)
T> Ts
Table A3 at T=350 ℃ and P = 10 bar
1 = 0.2825 3/
Condition 2 : (saturated steam)
2 = 10 = 0.1944 3/
∴ 12 = ( 2- 1)
= 15 × (10 × 102) × (0.1944 − 0.2825)
= − .

57

Example 5

A corrugated cylinder system contains air at 200kPa and 30 ℃. In this case the piston is located
on a pair of retainers and the volume of the cylinder is 400 liters. To move the piston, a pressure
of 400kPa is required. The air is heated until its volume doubles. Using the model of ideal gas,
determine.

a) Temperature at final condition
b) The amount of work done

Given, = 0.718 / dan = 0.287 /

Solution

 The process that occurs in the cylinder consists of 2 parts, namely the process of volume during
the pressure increases by 400kPa & the process of pressure during the volume increases by 2
times.

Temperature at final condition

 1 1 = 2 2 = 3 3

1 2 3

3 = 3 3 1
1 1

= 400(2 1)(303)
200( 1)

=

Work for a whole process (processes 1-2) = 0. Hence the amount of work.
 = 23 = ( 3 − 2)

2 × 400 400
= [( 1000 ) − (1000)] 400
=

P-v diagram
58

Isothermal Process

In thermodynamics, an isothermal process is a type of thermodynamic process in which
the temperature of the system remains constant: ΔT = 0. This typically occurs when a system
is in contact with an outside thermal reservoir, and the change in the system will occur slowly
enough to allow the system to continue to adjust to the temperature of the reservoir
through heat exchange. In contrast, an adiabatic process is where a system exchanges
no heat with its surroundings (Q = 0).

Simply, we can say that in isothermal process

o pV = mRT

o mR = constant

o T = constant (temperature)

o pV = constant

 ∴ 1 1 = 2 2 =. . . . =

 12 = ∫ 1 2

 12 = 2 2 ( 2) ( )

1

 12 = ( 2) = ( 1) ( )

1 2

Example 6

A gas at 1.5 bar and 20 ℃ (R = 0.1890 kJ / kgK) in a closed system is compressed to 15 bar.
Assuming gas is the ideal gas, determine the work for this process.

Solution

 12 = 2 2 ( 2)

1

 1 =
1

0.5 × 0.1890 × (20 + 273)
= 1.5 × 102

= 0.1846 2

 2 =
1

0.5 × 0.1890 × (20 + 273)
= 15 × 102

= 0.01846 2

 12 = (15 × 102) × 0.01846 × ln (0.01846) = − .

0.1846

59

Polytropic Process

The term "polytropic" was originally coined to describe any reversible process on any open or
closed system of gas or vapor which involves both heat and work transfer, such that a specified
combination of properties were maintained constant throughout the process. In such a process,
the expression relating the properties of the system throughout the process is called
the polytropic path.

= ; = , =

12 = 1 1 − 2 2 ( )
− 1

Or for a unit of mass

12 = 1 1 − 2 2 ( / )
− 1

For the ideal gas is , pV=mRT

12 = ( 1 − 2) ( )
− 1

Or for a unit of mass

12 = ( 1 − 2) ( / )
− 1

The relationship of pressure, definite volume and temperature to the polytropic process of a gas
ideal can be derived by combining the equations of the ideal gas state with pVn=C

1. 1 = ( 2

2 1 )

2. 1 = ( 2) −1

2 1

−1

3. 1 = ( 1)

2 2

Example 7

1.5 kg of a pure material at 20 bar and 250 ℃ undergoes a polytropic expansion process to a
pressure of 5 bar. The polytropic index is 1.2. Determine the work if the pure material is

a) Water
b) Air considered ideal gas (R = 0.287 kJ / kgK)

60

Solution

For question a (water)

 = − = ( − )
− −

Determine the specific volume of air at the beginning and end

 1=0.1114 m3/kg
 2= 0.3540 m3/kg
 12= 345.0 kJ

For question b (Air considered ideal gases)

 Using the equation of ideal gas conditions

 1 = 1
1

= . ⁄

1

 2 = ( 1)1.2

2

=0.2384 m3/kg

 12 = ( 1 1− ⅔ ⅔)
−1

= .

The End

61

Exercise
Question 1
5 kg of water-saturated vapor is contained in a rigid tank closed at a pressure of 40 bar. The
water pressure drops to 20 bar when it is cooled. Determine the volume of the tank, in m3 and
the fraction of dryness at the final state. Sketch the process above the P-υ state diagram.
Question 2
4 kg of water vapor at 50 bar and 350 ℃ contained in the cylinder and piston systems. Water is
cooled at a constant volume until the temperature reaches 200 ℃. Then it is compressed at a
constant temperature to a pressure of 30 bar.
a) Sketch the processes that take place in the diagrams P-υ and T-υ and completely label
the temperature and pressure.
b) Determine the fraction of dryness and pressure at condition 2 and the volume of the
cylinder at condition 3.
Question 3
2 kg of air in the boiler cylinder is at a pressure of 1 bar and a temperature of 25 ℃. The air is
compressed to a pressure of 5 bar and a temperature of 50 ℃. Determine the volume of air at
the condition and end. Take the air specific gas constant as 0.287 kJ/kgK.
Question 4
The air in a vehicle tire is initially at a pressure of 380kPa and a temperature of 20 ℃ when the
tire volume is 0.120m3. When the tire is heated by sunlight, the pressure inside the tire increases
to 450kPa and the volume of the tire increases by 5 percent. Determine the temperature at the
final condition and the mass of air in the tire.
Question 5
A rigid tank contains 2kg of oxygen, initially at 40 bar, and 200K. This gas is cooled until the
pressure drops to 30 bar. Determine the volume of the tank, in m3 and the temperature of the
final condition in K. [R = 0.2598 kJ/kgK]

62

CHAPTER 4

The First Law of Thermodynamics

OBJECTIVE
At the end of this chapter, student will be:

 Defines the first law of thermodynamic
 What is principles of energy eternity and

closed system heat transfer
 How to calculated closed system heat

transfer.

CHAPTER 4
The First Law of Thermodynamics

4.0 Introduction

What is the first law of thermodynamics?
The First Law of Thermodynamics states that heat is a form of energy, and
thermodynamic processes are therefore subject to the principle of
conservation of energy. This means that heat energy cannot be created or
destroyed.
∑ =∑

Closed System
Does not involve the flow of mass in or out of the system through system
boundaries (constant mass throughout the process). Heat transfer and
work can occur depending on the process the system goes through.

64

4.1 Principles Of Energy Eternity
 The First Law of Thermodynamics is similar to the statement of the Principle of the Immortality

of Energy.
 Energy cannot be created or destroyed but it can change from one form to another.
 Energy changes occur, but the amount of change remains the same.
 Transformation of system energy is not accompanied by destruction or energy creation (total

system energy remains the same before, during or after the process)
 (System energy change) = (Incoming energy) - (Outgoing energy)
 ∆ = -
 This equation is widely used in energy analysis of thermodynamic systems (general - closed /

open systems)
 The only difference is the type of energy involved during the process.

Figure 4.1

65

4.2 First Law Of Thermodynamic

In general, the first law of thermodynamics for closed systems can be stated as follows.

[ ] – [ ] =The amount of energy coming in
the amount of energy coming out

[ ]The amount of energy change in the system.

Consider a closed system undergoing a process of heat transfer to the system & transfer of
work out of the system.

 The energy balance of a closed system can be specified :-
 − =
 Summarizing the equation, the change of energy from state 1 to state 2 is given by;
 − = −

Scan me for more about Figure 4.2
first law of thermodynamic

as we know = + + . Closed systems do not involve mass flow across borders, TK & TU can be
ignored.

• − = + +

So = − . If a process takes place from state 1 to state 2, then the perfection of the equation
will be;

 − = − Non-flow Energy
 ( − ) = − Equation
 For a unit of mass is − = −

66

 Non-flow Energy Equation
The rules governing non-flow systems are as follows.

 The volume of the boundary may change.
 No fluid crosses the boundary.
 Energy may be transferred across the boundary.
When the volume enlarges, work (-W) is transferred from the system to the surroundings. When
the volume shrinks, work (+W) is transferred from the surroundings into the system. Energy may also
be transferred into the system as heat (+Q) or out of the system (-Q). This is best shown with the
example of a piston sliding inside a cylinder filled with a fluid such as gas.

Figure 4.3
When a system undergoes a thermodynamic cycle consisting of several series processes, the
total change in thermodynamic properties is zero. (initial condition = final condition). Therefore
dU=0

 = −
 =
So for one thermodynamic cycle is ∮ =∮ (total heat transfer = system clean work)

67

Example 1

A closed system undergoes a thermodynamic cycle in the order of 1-2-1. Process 1 -2
according to path A and process 2 - 1 according to path B. Heat transfer and pre-determined
work are : 12 = 50 , 12 = −20 21 = −30 .
If the system energy is at a state 1 1 = 35 , calculate 21.

Solution − = −
For proses A: = − +
= (− ) − +
= −

For proses B: − = −
= − +
= − (− ) + (− )
= (heat supplied)

If one cycle is considered, it is known = so that

12 + 21 = 12 + 21.
21 = 12 + 21 − 12
21 = 50 + (−30) − (−20)
= 40

68

 Thermodynamic systems involve boundary work that is, work performed as a result of
boundary changes.

 Compression = reduced volume, work -ve (environment doing work on the system).
 Expansion = increased volume, work +ve (system does work on the environment).
 In certain cases it is possible that a process involves other types of work besides border work

such as electrical work, shaft work & so on.

Figure 4.4
Example 2

A perfectly insulated closed system contains air at a temperature of 30 ℃. The air is then heated
by a 1kW electric heater for 30 seconds which causes the boundary of the system to change
and the air temperature to 60 ℃. Determine the boundary work bg this process.

69

Solution
Wother involves electrical work supplied to the system. So his work is negative.

 2 − 1 = 12 − ( + )
 = ̇ × = 1 × 30 = −30
 = − ( 2 − 1) −

= −0.5(0.718)(60 − 30) − (−30)
= .
 This heating process causes the air to expand and the volume to increase. Border work is
positively valued.
Example 3
A rigid tank contains a fluid. An electric heater with 240V & 10A is used to heat the fluid for 15s.
At the same time a stirring wheel is used to stir so that the temperature of the fluid becomes
uniform. The work used by the stirring wheel to stir the fluid is 20kJ. During this process the
energy in the fluid increases by 30kJ. Determine heat transfer.
Solution
Since the tank is rigid, then we assume the boundaries do not change. Wboundary =0
The work involved is input work from electric heaters & stirrer wheels.
• ∆ = − ( + + ℎ )
• = ×
• = 240×10×15 = 36

1000

• = ∆ + + ℎ
• = 30 + (−36) − 20
• = −
 This indicates that heat transfer occurs from the fluid in the tank to the surrounding air. During

the heating process, the fluid temperature rises above the ambient temperature.

70

4.3 Closed System Heat Transfer

Using the statement of the First Principle of Thermodynamics equations that combine work &
heat can be published (relationship between changes in system properties & energy interactions
that occur at system boundaries). If the liquid must use a steam table while if the gas must use
the ideal gas condition equation. In this topic we only consider border work.

4.3.1 Isometric process

During a process, the boundary work value is zero.

 12 − 12 = 2 − 1

 12 = 0

 12 = 2 − 1 = ( 2 − 1)

 = ( − )

 Cv = specific heat at constant volume

 = −
−

 2 − 1 = ( 2 − 1)

Example 4

0.5 kg of water-saturated vapor in a closed system at a pressure of 10 bar is cooled at a constant
volume until the pressure drops to 5 bar. Determine the heat transfer and sketch the process
on the diagrams T-ν and p-ν.

Solution

− = ( − )

 Situation 1 (vapour phase)
 P1 = 10 bar
 ν1 = νg = 0.1944 m3⁄kg
 u1 = ug = 2583.6 kJ/kg

71

 Situation 1 (Mix phase)
 P2 = 5 bar
 2 = 1 = 0.1944 3⁄
 = 0.3749 3⁄
 = 0.1944 = 0.519

0.3749

 2 = + = 1636.9 /

 After being able to all the data continue to the equation
 12 − 12 = ( 2 − 1)
 12 = 0
 12 = ( 2 − 1)
= 0.5(1636.9 − 2583.6)

= − . /

` Diagrams P-ν Diagrams T-v

72

4.3.2 Isobaric process

Rearrange 12 − 12 = 2 − 1
 12 = 2 − 1 + 12 (Heat transfer of liquid)

It is known that the work for the isobar process is;

 = ( − )
 12 = 2 − 1 + ( 2 − 1)
 12 = ( 2 + 2) − ( 1 + 1) : H = U + pV
 12 = 2 − 1
 12 = (ℎ2 − ℎ1)

For ideal gas it is known that CP is equal to the specific heat at constant pressure :

 = ℎ2−ℎ1 jadi ℎ2 − ℎ1 = ( 2 − 1)
2− 1

 = ( − ) (Heat transfer for ideal gas)

Example 5

0.5 kg of water vapor in the condensed cylinder system initially at 10 bar and 400 ℃ is
compressed at constant pressure until it becomes saturated vapor. Determine the work and
heat transfer for this process and sketch the process on the diagrams T-ν and p- ν.

Given

 Mass = 0.5kg

 Situation 1 (superheated steam phase)

 P1 = 10 bar

 T1 = 400℃

 1 = 0.3066 3⁄
 1 = 2957.3 ⁄

 Situation 2 isobaric compression (saturated steam phase)
 P2 = 10 bar
 2 = = 0.1944 3⁄
 2 = = 2583.6 ⁄

73

Solution

12 = ( 2 − 1) = ( 2 − 1)
= 0.5 × 1000 × (0.1944 − 0.3066)

= − . Work is done on the system

12 = ( 2 − 1) + 12 work is done on the system
work is done on the system

= 0.5 × (2583.6 − 2957.3) + (−56.10)

= − . Heat is transferred to the environment

Diagrams P-ν Diagrams T-v

74

4.3.3 Isothermal process

 A certain amount of heat must be supplied to the system continuously for the isothermal

expansion process & removed from the system to the environment during the isothermal

compression process to ensure a constant temperature.

 The working formula for pure material in liquid & vapor form for isothermal process cannot be

published due to non-uniform isothermal line function.

 in contrast to the ideal gas when the pV is equal to zero. The change in internal energy, dU for

the ideal gas for isothermal processes is zero (∆ = ) When the constant temperature is

equals to the constant internal energy, dU = 0 (functioning temperature).

 12 − 12 = 2 − 1

 12 − 12 = 0
 = ( for ideal gas for temperature process )

12 = 1 1 ( 2) Working formula for ideal gas in
isothermal processes
1

*Specific formulas for heat transfer of pure liquid and vapor materials cannot be published.

Example 6

An embossed cylinder system contains 0.4 3 of air initially at 100kPa and 80 ℃. The air is
compressed at a constant temperature until the volume decreases to 0.1 3. Determine the heat
transfer for this process.

Solution

 12 = 12 = 1 1 ( 2)

1

 1 = 100 ,

 1 = 0.4 3

 2 = 0.1 3

 12 = 1 1 ( 2)

1

0.1
= 100 × 0.4 × (0.4)

= − .

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Example 7

1.5 kg of air in a closed system, initially at 7 bar and a dry fraction of 0.7. During the process
1000kJ of heat is supplied to the system. Determine the temperature at the end of the process
and the work that produces the process in kJ. Sketch a diagram of the state of p-v and T-v

Solution

 isothermic process at the end of the process T2=T1=Ts
 12 − 12 = 2 − 1 (From Non-flow Energy Equation)
 Q has been given, W has to be searched, & the value of u can be obtained from the table.
 Condition 1 is a mixed phase, so :

u1=uf+xufg
= 696.44+0.7(2572.5-696.44)
= 2009.7 kJ/kg
 Determine the phase of condition 2. At P = 1.5bar Ts = 111.4 ℃ T2 > Ts then condition 2 is an
superheated steam.
 Energy in condition 2 is determined through an internal determinant at a pressure of 1.5bar.

 1.5kg of water
Condition 1

 P1 = 7bar
 X = 0.7
 mixed phase
 T1 = Ts = 165℃

Condition 2

 P2 = 1.5bar

 Phase = ?

 T2 = T1 = 165℃

 2 = [165−150] (2656.3 − 2579.9) + 2579.9 = 2602.8 /

200−150

 12 − 12 = 2 − 1

The End

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Exercises
Question 1
A car weighing 1300kg moves at a speed of 110 km / h. Determine its kinetic energy in kJ. If
the car is slowed to a speed of 60km / h, determine the change in kinetic energy in kJ.
Question 2
A rock with a mass of 50 kg is located on a hill with a height of 150m measured from the ground
surface. By taking the speed of gravity, g = 9.81 m / s2, determine the energy possessed by the
rock, kJ.
Question 3
0.5 kg of a gas in a closed system expands at constant pressure. The specific volumes at the
initial and final conditions are 0.2m3 / kg and 0.4m3 / kg respectively. The resulting work is 100
kJ. Calculate the pressure during the process, in the bar.
Question 4
3 kg of water-saturated vapor in a closed system at 200 kPa is heated at constant pressure until
the temperature reaches 400 ℃. Calculate the work done by the steam during the process.
Question 5
A gas expands from the initial state p1 = 500kPa and V1 = 0.1m3 to the final state where the
pressure, p2 = 100kPa. The relationship of pressure and volume during the process is pV =
constant. Sketch the process on p-V diagram and calculate the work in kJ.

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Answer

Question 1

 = 1300
 1 = 110 ⁄ = 30.56 ⁄
 2 = 60 ⁄ = 16.67 ⁄

1 = 1 12 = 1 × 1300 × 30.562 2⁄ 2 × 1 / = 607.04
2 2 (1000 2⁄ 2)

2 = 1 22 = 1 × 1300 × 16.672 2⁄ 2 × 1 / = 180.63
2 2 (1000 2⁄ 2)

∆ = 2 − 1 = 180.63 − 607.04 = − .
Question 2

 = 50
 ℎ = 150
 = 9.81 ⁄ 2

= ℎ = 50 × 9.81 ⁄ 2 × 150 = 73575 = .

Question 3

 = 0.5
 12 = 100
 1 = 0.2 3⁄
 2 = 0.4 3⁄

12 = ( 2 − 1)

100 = × 0.5 (0.4 − 0.2)

= 1000 =

78

Answer
Question 4
3 kg water saturated vapor
State 1
Phase = saturated vapor

 1 = 200
 1 = = 120.2℃ (temperature on saturated vapor)
 1 = = 0.8857 3⁄

State 2
Fasa = ?
 2 = 1 = 200
 2 = 400℃, = 120.2℃
 2 > ∴
 2 = = 1.5493 3⁄
Work for isobaric processes
12 = ( 2 − 1) = 200 × 3 × (1.5493 − 0.8857) 3⁄ = .

79

Answer
Question 5

 1 1 = 2 2

 2 = 1 1 = 500×0.1 = 0.5 3
2 100

 12 = 2 2 ( 2) = 100 × 0.5 × (0.5) = .

1 0.1

80

CHAPTER 5

The First Law of Thermodynamics
(Open System)

OBJECTIVE
At the end of this chapter, student will be:

 Defines principle of energy conservation
for open system

 What is steady flow equation and function
of the steady flow.

 How to calculated the steady flow equation
for open system

CHAPTER 5
The First Law of Thermodynamics

(Open System)

5.0 Introduction

What is Open System
 Involves the flow of mass in or out of the system through the system
boundary (volume of control)
 Heat transfer and work can occur depending on the process the system
goes through.
 Other energy is equal to the energy carried by the mass entering & exiting
 Kinetic energy and Potential energy can not be ignored
 Flow energy is the energy required by the mass for inward and outward
mass flow (pV)

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5.1 Principle of Energy Conservation for Open System

The first law of thermodynamics is a version of the law of conservation of energy, adapted
for thermodynamic processes, distinguishing two kinds of transfer of energy, as heat and
as thermodynamic work, and relating them to a function of a body's state, called Internal energy.

The law of conservation of energy states that the total energy of an isolated system is constant;
energy can be transformed from one form to another, but can be neither created nor destroyed.

[ ℎ ] − [ ] = [ ℎ ℎ ]

− + (∑ − ∑ ) = ∆

= mass energy enters Flow energy (pv), internal energy, kinetic
= mass energy out energy, & potential energy

= internal energy, kinetic energy, & potential energy

 − + (∑ − ∑ ) = ∆

̇ − ̇ + ̇ 1 ( 1 + 1 1 + ̅ 12 + 1) − ̇ 2 ( 2 + 2 2 + ̅ 22 + 2) = ̇ ( + ̅ 2 + )
2 2 2

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83

̇ − ̇ + ̇ ( + ̅ + ) − ̇ ( + ̅ + ) = ̇ ( + ̅ + )


 ̇ = the rate of heat supplied ( )
 ̇ = power generated ( )
 ̇ = mass flow rate( / )
 ℎ = ℎ ( / )
 ̅ = ( / )

 =gravity acceleration = 9.81 ⁄ 2
 = distance from datum( )

5.2 The steady flow Equation.

A flow is said to be a steady flow if the following are met,
a. The properties (pressure, temperature, mass flow rate etc.) of the fluid during flow in or
out across the system boundary are constant at each point of the boundary. Refer
figure 5.1.

Figure 5.1
b. Incoming mass flow rate is equal to outgoing mass flow rate (total mass flow rate

covered by the boundary is constant). Refer figure 5.2

Figure 5.2
c. All reactions between the system and the environment occur at a steady rate. (heat

transfer & work occurs at a constant value that does not change over time).

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Refer to statement (b) ∑ ̇ = ∑ ̇ , Flow rate formula is

̇ = ̅ or ̇ = ̇ or ̇ = ̅ or = ̇



 ̇ = mass flow rate, kg / s
 A = cross-sectional area of pipe @ channel, m2
 ̅ = fluid velocity, m/s
 = specific volume of fluid,, m3/kg
 ̇ = volume flow rate m3/s

 = density, kg/m3

If fluid enters state 1 and exits state 2

 ̇ 1 = ̇ 2

 1 ̅ 1 = 2 ̅ 2 or Continuity equation

1 2

 ̇1 = 2̇ or

1 2

 1 1 ̅ 1 = 2 2 ̅ 2 or
 1 1̇ = 2 2̇

Steady flow, the mass flow rate is equal to the outflow mass rate.

̇ = ̇ 1 − ̇ 2 = 0

 ̇ − ̇ + ̇ ( + ̅ + ) − ̇ ( + ̅ + ) = ̇ ( + ̅ + )


 ( ̇ − ̇ ) = ̇ ((ℎ2 − ℎ1) + ( ̅ 22 − ̅ 12) + ( 2 − 1))

2 2

 ( − ) = ((ℎ2 − ℎ1) + ( ̅ 22 − ̅ 12) + ( 2 − 1))

2 2

 ( − ) = ((ℎ2 − ℎ1) + ( ̅ 22 − ̅ 12) + ( 2 − 1))

2 2

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5.3 Function The steady flow Equation.
\

5.3.1 Nozzle

A device used to increase the velocity or kinetic energy of a fluid. The velocity at the outlet of
the nozzle is higher than the velocity at the inlet of the nozzle. Example; nozzles mounted on
the ends of rubber tubes, & tubes used by firefighters.

Figure 5.3
5.3.2 Diffuser

Return from the nozzle; to reduce the velocity of a fluid. Example; openings for air out in an air-
conditioning system mounted on the ceiling.

Figure 5.4
Equation for Nozzle and Diffuser is,

 ( − ) = ((ℎ2 − ℎ1) + ( ̅ 22 − ̅ 12) + ( 2 − 1))

2 2

 0 = (ℎ2 − ℎ1) + ( ̅ 22 − ̅ 12)

2 2

 (ℎ2 − ℎ1) = ( ̅ 12 − ̅ 22)

2 2

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Example 1

Air enters a nozzle at a temperature of 200 ℃ and 100m/s and exits at a velocity of 200m/s.
Determine the air temperature at the exit. Take specific heat at constant pressure, cp for air as
1.005kJ/kgK.

Solution

 − = ( − )
( − ) = ( ̅ − ̅ )

 = ̅ − ̅ +


= 1002 − 2002 + (200 + 273)
2 × 1.005 × 103

 =

Example 2

Steam enters the nozzle at 40bar, 400 ℃ and 10m/s velocity and exits at 14bar and 665m/s
velocity. Heat transfer and energy changes can be ignored. The steam mass flow rate is 2kg/s.
Determine the cross-sectional area of the nozzle outlet, in m2.

̇ =2kg/s and find A2=? condition out to the
system
condition in to the
system P2 14bar

P1 40bar T2 ?
T1 400°C ̅ 2 665m/s
̅ 1 10m/s

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Solution

The cross-sectional area of the nozzle outlet can be determined using the equation;

= ̇ ;
̅

To obtain , the phase at condition 2 must be determined by obtaining another parameter

(known as P2 = 14bar). Using the nozzle equation, ℎ2 can be determined.

 ℎ2 = ℎ1 + ̅ 12− ̅ 22
2

 = 3213.6 + 102−6652

2000

= . /

Condition 2

 2 = 14 ,
 hg = 2790.0 kJ/kg
 h2 > hg phase = superheated steam

Internalization is performed to obtain the value of

h End of condition
2927.2 0.1635
2992.5 2 P 14bar
3040.4 0.1823 2

= . / hg 2790

ℎ2 2992.5

ℎ2>hg Phase SHS

 Luas, A = ̇
̅

 = 2(0.1743)

665

 = . × −

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5.3.3 Turbine

 A turbine is a device that harnesses the kinetic energy of some fluid - such as water, steam,
air, or combustion gases - and turns this into the rotational motion of the device itself. These
devices are generally used in electrical generation, engines, and propulsion systems and
are classified as a type of engine. They are classified as such because engines are simply
technologies that take an input and generate an output. A simple turbine is composed of a series
of blades - currently steel is one of the most common materials used - and allows the fluid to
enter the turbine, pushing the blades. These blades then spin and eject the fluid which now has
less energy it did than when it entered the turbine. Some of the energy is captured by the turbine
and used.

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Figure 5.5

For a simple analysis of turbines, here are the things that can be ignored.

 Heat transfer - (if small compared to the work produced)
 Potential Energy - (if the difference in inlet and outlet is too small)
 Kinetic energy - (if the change in velocity between the inlet & outlet is small)

( ̇ − ̇ ) = ̇ (( − ) + ( ̅ − ̅ ) + ( − ))

̇ = ̇ ( − )

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Example 3

Steam flows steadily on the turbine with a mass flow rate of 4600 kg/hour The turbine produced
produces 1000kW of energy. At the inlet, stimulate at 60 bar and 400 ℃ with a velocity of 10m/s.
At the outlet, the pressure is 0.1 bar, the dryness fraction is 0.9 and the velocity is 50m/s. Small
businesses can be ignored. Calculate the rate of heat transfer

400°C ̅ = 10 / ̇ = 1.278kg/s
60bar

= 1000kW

60bar
̅ = 50 /

Solution h1 = 3177.2 kJ/kg

Condition 1

 1 = 60
 1 = 400℃
 Phase = Superheated

Steam

Condition 2 h2 = 2345.4 kJ/kg

 = 0.9
 mixed phase

( ̇ − ̇ ) = ̇ ((ℎ2 − ℎ1) + ( ̅2 22 − ̅2 12) + ( 2 − 1))

̇ = ̇ [(ℎ2 − ℎ1) + ( ̅ 22 − ̅ 12)] + ̇
2

̇ = 4600 [(2345.4 − 3177.2) + 502 − 102 1 ⁄ + 1000
(3600) ( 2 ) (1000 2⁄ 2)]

̇ = − .

90

5.3.4 Compressors and Pumps
 Compressors
A compressor is a mechanical device that increases the pressure of a gas by reducing its
volume. An air compressor is a specific type of gas compressor. Examples: air conditioning, air
pump, refrigerator
Compressors are similar to pumps: both increase the pressure on a fluid and both can
transport the fluid through a pipe. As gases are compressible, the compressor also reduces the
volume of a gas. Liquids are relatively incompressible; while some can be compressed, the
main action of a pump is to pressurize and transport liquids.

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compressors, Please scan!!

 Pumps
A pump is a device that moves fluids (liquids or gases), or sometimes slurries, by mechanical
action, typically converted from electrical energy into Hydraulic energy. Pumps can be classified
into three major groups according to the method they use to move the fluid: direct lift,
displacement, and gravity pumps.
Pumps operate by some mechanism (typically reciprocating or rotary), and consume
energy to perform mechanical work moving the fluid. Pumps operate via many energy sources,
including manual operation, electricity, engines, or wind power, and come in many sizes, from
microscopic for use in medical applications, to large industrial pumps.

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91

Example 4

Air enters the compressor operating in a steady state at 1bar, 290K and velocity 6m/s through

a pipe with a cross-sectional area of 0.1m2. At the exit, the pressure is 7bar, the temperature

is 450K and the velocity is 2m/s. The heat transfer from the compressor to the environment is

180kJ/min. Using the ideal gas model, calculate the input power of the compressor in kW. [R =

0.287kJ/kgK and cp=1.005 kJ/kgK]  = 7
 = 450
 ̅ = 2 /

̇ = 180 /

 =
 = 290
 ̅ = 6 /
 = 0.1 2

Solution

 ( ̇ − ̇ ) = ̇ ((ℎ2 − ℎ1) + ( ̅ 22 − ̅ 12) + ( 2 − 1))

2 2

 ̇ − ̇ = ̇ ((ℎ2 − ℎ1) + ( ̅ 22 − ̅ 12)) because the workpiece is air

2 2

 ̇ = ̇ − ̇ ( ( 2 − 1) + ( ̅ 22 − ̅ 12))

2 2

 ̇ = 1 ̅ 1 = 1 ̅ 1 1 = 0.1×6×100 = 0.72 /

1 1 0.287×290

 ̇ = (− 180) − 0.72 (1.005(450 − 290) + 22−62)

60 2000

 = −3 − 0.72(160.784)
 = − .

92