THERMODYNAMICS Practice Workbook

Fundamental concepts of Thermodynamics|0

POLITECHNIC MALAYSIA

Fundamental concepts of Thermodynamics|1

Content

1. BASIC OF THERMODYNAMICS 2

2. PROPERTIES OF PURE SUBSTANCES 8

a. Steam

b. Ideal gas

3. FIRST LAW OF THERMODYNAMICS AND ITS PROCESSES 39

a. Non flow process

Isothermal, isometric, isobaric, adiabatic and polytrophic

b. Flow process
Boilers, condensers, Turbine, Nozzle, Throttle, Pump ….

4. SECOND LAW OF THERMODYNAMICS 76
a. Heat engine
b. Entropy and Isentropic
c. Carnot cycle
d. Rankine cycle

Fundamental concepts of Thermodynamics|2

Unit
Table 1.3 Multiplying factors

Prefix Coefficient factor multiples

Giga 1 000 000 000 109
Mega 1 000 000 106
kilo 1 000 103
centi 0.01 10-2
mili 0.001 10-3
micro 0.000 001 10-6
nano 10-12
0.000 000 000 001

Table 1.4 Unit Conversion

Physical quantities units
Mass 1kg = 1000g
Length 1m = 100cm = 1000mm
Time 1 hour = 60 minute = 3600 second
Volume 1m3 = 1000liter

Pressure 1bar = 10 5 Pa =100kPa
1Pa = 1N/m2
Temperature 0 oC = 273Kelvin
1Nm = 1 Joule
Energy 1J/s = 1Watt

Fundamental concepts of Thermodynamics|3
1.0 Basic terms of Thermodynamics

surrounding

system

boundry surrounding

Terms Figure 1.1 System, surroundings and boundary
System
Boundary Definitions
Surroundings
Quantity of matter or a region in space chosen for study.
Example:
The fluid contained by the cylinder head, cylinder walls and the
piston may be said to be the system.
Surface of separation between the system and its surroundings.

The mass or region outside the system.
The surroundings may be affected by changes within the
system.

Examples of Thermodynamics systems 2. Steam turbine
1. Compressor

Fundamental concepts of Thermodynamics|4

3. . Steam power plant 4. Refrigerator

5. internal combustion engine

Fundamental concepts of Thermodynamics|5

1.3 Open System and Close System

OPEN - Is a properly selected region in space.
- It usually encloses a device, which

involves mass flow
- such as a boiler, compressor, turbine or

nozzle.
- the device as the control volume.
- Both mass and energy can cross the

boundary of a control volume.

SYSTEM

CLOSE - Consists of a fixed amount of mass,
- and no mass can cross its boundary.
- But energy, in the form of heat or work

can cross the boundary,
- and the volume of a close system does

not have to be fixed.

1.4 Properties of Systems

Figure 1.2 Intensive and Extensive properties

Fundamental concepts of Thermodynamics|6

Properties

Macroscopic characteristics
of a system.
Example: Mass,volume,
energy, pressure, and
temperature

Intensive Extensive
independent of the size of depend on the size or
the system extent of the system
Example: Example:
Temperature,pressure and Mass, volume and total
density. energy

1.5 State and Equilibrium

Thermodynamics Condition

State Consider a system that is not undergoing any change. The
properties can be measured or calculated throughout the
entire system. This gives us a set of properties that
completely describe the condition or state of the system.
At a given state all of the properties are known; changing
one property changes the state.

State 1 State 2

Process Fundamental concepts of Thermodynamics|7

A process is a transformation from one state to another.
Processes may be reversible or irreversible.

Zeroth’s law of Zeroth Law of Thermodynamics: If two systems (A & B)
thermodynamics. are separately found to be in thermal equilibrium with a
third system (C), the first two systems are in thermal
equilibrium with each other.

Figure 1.3 State and Process

Fundamental concepts of Thermodynamics|8

Reversible and Irreversible Process
The reversible process is the ideal process which never occurs, while the irreversible process is the
natural process that is commonly found in nature. When we tear a page from our notebooks, we cannot
change this and ‘un-tear’. This is an irreversible process. Whereas when water evaporates, it can also

be condensed in the form of rains. This is a reversible process. Let us study more about them below.

Reversible Process
A thermodynamic process is reversible if the process can return back in such a that both the system and
the surroundings return to their original states, with no other change anywhere else in the universe. It
means both system and surroundings are returned to their initial states at the end of the reverse process.

In the figure above, the system has undergone a change from state 1 to state 2. The reversible process
can reverse completely and there is no trace left to show that the system had undergone
thermodynamic change. During the reversible process, all the changes in state that occur in the system
are in thermodynamic equilibrium with each other.

Fundamental concepts of Thermodynamics|9

PRACTICE 2

No Statement √/×

1 State means all properties of a system have fixed value

2 A thermodynamic process is reversible if the process can return back in such a that
both the system and the surroundings return to their original states

3 The word equiblirium implies a state of balance

4 Surrounding is not affected by system changes

5 A quantity of matter or region in space chosen for study, this statement refer to
system

6 Extensive properties are those that are independent of the mass of a system such as
density.

7 System undergoes from one equiblirium state to another is called process.

8 A boundary is real or imaginary surface that separete the system and their
surrounding.

9 A system is in thermal equiblirium if the temperature is the same throughout the
entire system.

10 The principles of conservation energy which says that energy can be neither created
nor be destroyed; however, it can change from one form to another.

11 System may be considered to be closed or open, depending on whether a fixed mass
or fixed volume in space chosen for study.

12 Zeroth law can be restated as two bodies are in thermal aquiblirium if both have the
same temperature reading even if they are not in contact.

13 The macroscopic form of energy are those a systems posseses as a whole with
respect to some outside reference frame, such as kinetic and potential energies.

Pure Substance (Steam) | 10

2.0 PROPERTIES OF PURE SUBSTANCES
2.1 Introduction to Steam
A substance that has a fixed chemical composition throughout is call pure substance, e.g water,
nitrogen, helium and carbon dioxide.

Phase-Change Process

Figure 2.1 Heating water and steam at constant pressure

T, oC

300 4

2 Saturated Superheated
100 mixture steam
3
Compressed liquid

20 1

v, m3/kg
Figure 2.2 P-v diagram for the heating process of water at constant pressure

Pure Substance (Steam) | 11

Summary of nomenclature:
Compressed liquid/ ( Between States 1 & 2 ).
The liquid state and the temperature of the fluid is below the saturation temperature..
Saturated liquid/ (State 2)
The condition is still liquid but starts to become steam
Saturated Liquid-Steam or Wet Steam Region / (Between States 2 & 3)
Fluid in a mixture of liquid and steam.
Dry Saturated steam /(State 3)
When all water has become a stim.
Superheated steam / (The right of State 3)
When the steam temperature is greater than the saturation temperature at a pressure.

Figure 2.3 The P-v graph shows the process of steam phase changes at different pressures, whereas
the higher the saturation temperature will increase

Pure Substance (Steam) | 12

When connected to all saturated temperature points, we will get a diagram that allows for easy
identification of different regions:

Figure 2.4 P-v diagram of a pure substance
2.2 Properties of a Wet steam region

total mass = 1 kg x kg of steam
(1 - x ) kg of liquid

Figure 2.5 Liquid-steam mixture

The dryness fraction is defined as follows; x  msteam
dryness fraction  mass of dry saturated steam mtotal

total mass
where mtotal = mliquid + msteam

Specific volume v = xvg

Specific enthalpy h = hf + xhfg

Specific Internal Energy u = uf + x(ug – uf )

Specific Entropy  s = sf + xsfg

Pure Substance (Steam) | 13

At point A, x = 0 Sat. steam Sat. steam
At point B, x = 1 Sat. liquid Sat. liquid
Between point A and B, 0  x  1.0
P x = 0.2 x = 0.8
Note that for a saturated liquid, x = 0;
and that for dry saturated steam, x = 1.

AB

ts v
vf vg

Figure 2.6 P-v diagram of a wet steam region of pure substance

2.3 The Use of Steam Tables

Below is a list of the properties normally tabulated, with the symbols used being those recommended
by British Standard Specifications.

Symbols Units Description

p bar Absolute pressure of the fluid
ts oC Saturation temperature corresponding to the pressure p bar
vf m3/kg Specific volume of saturated liquid
vg m3/kg Specific volume of saturated steam
uf kJ/kg Specific internal energy of saturated liquid
ug kJ/kg Specific internal energy of saturated steam
hf kJ/kg Specific enthalpy of saturated liquid
hg kJ/kg Specific enthalpy of saturated steam
hfg kJ/kg Change of specific enthalpy during evaporation
sf kJ/kgK Specific entropy of saturated liquid
sg kJ/kgK Specific entropy of saturated steam
sfg kJ/kgK Change of specific entropy during evaporation

Table 2.1 The property of steam tables

Steam table (steam tables = G.F.C Roger and Y.R. Mayhew)

Pure Substance (Steam) | 14

1. Answering questions based on temperature from 0.01oC to 100oC,

Example question
Determine the energy within the saturated steam at 19oC
[vg=61.34m3/kg]

Answer method
Take a reading from the steam table on saturated water and steam

Practice 2.1

a. Specific enthalpy of saturated steam at 17oC .......................
b. Specific entropy of saturated liquid at 70oC .......................
c. Specific enthalpy of saturated liquid at 60oC .......................
d. Specific volume of saturated steam at 85 oC .......................

2. Answering questions based on saturated liquid phase
Example question
determine the internal energy of saturated liquid at a pressure of 40bar
so, the answer is u = uf =1082kJ/kg

dari soalan dinyatakan
fasa saturated liquid (cecair
tepu), jawapannya mestilah
merujuk kepada nilai
dalam steam table yang
mempunyai subscript f, i.e;
hf, vf, sf and uf

Pure Substance (Steam) | 15

Practice 2.2 ………………….
Determine
a. Specific Enthalpy of liquid saturated at pressure 5 bar

b. Specific Entropy of liquid saturated at 50bar pressure …………………..

c. Specific internal energy of liquid saturated at 150bar ....……...............

d. Specific enthalpy of liquid saturated at a pressure of 80bar. …………………

3. Answer questions in saturated steam phase.

Example question
determine the enthalphy of saturated steam at a pressure of 40bar
so, the answer is u = ug=2602kJ/kg

dari soalan dinyatakan fasa
saturated steam (steam tepu),
jawapannya mestilah merujuk
kepada nilai dalam steam table
yang mempunyai subscript g, i.e
hg, vg, sg dan ug

The diagram shows the point of saturated liquid and
saturated steam at 40bar pressure

Practice 2.3
a. The specific volume of saturated steam at P = 7bar. …………………
b. The specific enthalpy of saturated steam at pressure 70bar ………………..
c. Specific entropy of saturated steam at pressure 17bar .………………
d. Specific internal energy of saturated steam at P= 1.7bar ………………

Pure Substance (Steam) | 16

4. Answering questions based on x given

Example question
determine the internal energy of steam at a pressure of 20bar and x = 0.9

rujuk kepada soalan, jika x antara 0 dan 1, steam berada pada fasa stim basah (wet steam)
jawapan mesti menggunakan formula, samada

Specific volume v = xvg
Specific enthalpy h = hf + xhfg
Specific Internal Energy u = uf + x(ug – uf )
Specific Entropy s = sf + xsfg

Solution
An extract from the steam tables at 20bar and saturated steam

Question wish for internal energy value at x = 0.9 and P = 20bar, so use internal energy formula

Specific internal energy
u = uf + x( ug -uf )
= 907 + 0.9(2600 - 907)
= 2430.7 kJ/kg

Pure Substance (Steam) | 17

Practice 2.4

a. Determine specific entropy at 30bar and x = 0.75.
sf = 2.645kJ/kgK
sfg = 3.541 kJ/kgK
s = sf + xsfg

b. Determine specific entalphy at 3bar pressure and dryness fraction 0.75.

c. Determine specific internal energy at 13bar pressure and 90% steam quality.

d. Determine the specific volume at pressure of 40bar and x = 0.75.

Pure Substance (Steam) | 18

e. Determine the specific volume at pressure of 4bar and x = 0.75.

Answer :
a)5.301kJ/kgK, b)5.663 kJ/kgK, c)5.968 kJ/kgK, d)2373kJ/kg, e)2546kJ/kg

5. Find the value of dryness fraction based on the given property value

Setiap soalan untuk mencari nilai x ( dyness fraction), menunjukkan stim tersebut is in the Wet

steam phase then use one of the following formula to find the value of x

Specific volume v = xvg

Specific enthalpy h = hf + xhfg

Specific Internal Energy u = uf + x(ug – uf )

Specific Entropy  s = sf + xsfg

If in question based on pressure and specific enthalpy, use a specific enthalpy formula

Example question
Determine the dryness fraction of steam at 8 bar and specific internal energy 2450kJ/kg.

Solution
An extract from the steam tables,

At 8 bar, and = 2577 kJ / kg, since the actual specific internal energy is given as 2450kJ/ kg, the
steam must be in the wet steam state (u <u) must use formula specific internal energy.
u = uf + x(ug -uf)

Pure Substance (Steam) | 19

u = uf + x(ug -uf)
2450 = 720 + x(2577 - 720)
 x = 0.93

Practice 2.5

Determine dryness fraction of wet steam if the pressure is

a. 42 bar and u = 2000 kJ/kg [Ans: 0.6]

u = 2000 kJ/kg, uf=1097kJ/kg,

ug = 2601kJ/kg

u = uf + x ( ug - uf )

b. at 30bar pressure and specific entropy 5.301kJ / kgK [Ans: 0.75]

s = sfg=
sf=
s = sf + x ( sfg )

Pure Substance (Steam) | 20

c. at 3bar pressure and specific internal energy 5.663kJ / kgK [Ans:0.75]

d. at 13bar pressure and specific enthalpy 2787kJ /kg

Pure Substance (Steam) | 21

6. Answering steam property questions based on the value of the other properties
provided

example question
Determine the entropy of steam at 2.5bar and entalphy 2000kJ/kg

Suggestions
The question is to determine the value of entropy, based on the enthalpy at p = 2.5bar.

i. Firstly, determine the steam phase, if the phase is wet steam, obtain the the dryness
fraction using the enthalpy formula

ii. then use the entropy formula to find the required value,

At 2.5 bar, hg = 2717 kJ/kg, since the actual specific enthalpy is given as 2000kJ/kg, the steam
must be in the wet steam state caused by ( h <hg).

to find the value of x use the enthalpy then use the entropy formula to find the
formula entropy value
h = hf + xhfg , s = sf + xsfg
2000 = 535 +x(2182) s = sf + xsfg
x  2000  535  0.67
=1.607 +0.67(5.446)
2182 =5.226kJ/kgK

Practice 2.6

a. determine the specific entropy of steam at 42 bar and u = 2000kJ/kg. [Ans:4.760kJ/kgK]

Phase = ? Find entropy
find the value of x we use
u = uf + x(ug - uf )

Pure Substance (Steam) | 22

b. determine the specific enthalpy of steam at 30bar and specific entropy is 5.301kJ/kgK.
[Ans:2354kJ/kg]

c. determine the specific internal energy of steam at 5bar and specific enthalpy 2500kJ/kgK.
[Ans:6.236kJ/kg]

d. determine the specific volume at 13bar pressure and the specific entropy is 5.968 kJ / kgK.

Pure Substance (Steam) | 23

2.4 Superheated Steam Tables

Find the value of steam properties based on the pressure and superheated temperature in the

table.

Steam is called superheated when the temperature specified in the question is more than saturated

temperature tquestion > tsaturation

Example Question

Determine the degree of superheat, the specific volume and the specific enthalpy steam at
10bar pressure and 200C temperature

Suggestion

Determine the steam phase first, if superheated, read directly from the steam table.

Formula Degree of superheat = tsuperheat – tsaturation

Table 2.2 Superheated steam at a pressure of 10 bar

Steam and 10 bar and 200 oC are superheated due to tquestion > tsaturation

Degree of superheat DOS = tsuperheat – tsaturation
=200 oC – 179.9 oC
= 20.1 oC

Values for specific volume and specific enthalpy are available under column 200oC, ie v = 0.2061
m3/kg and h = 2829 kJ/kg.

Practice 2.7

Determine the: ……………………
a. specific enthalpy at 20bar pressure and 300oC.

b. specific entropy at 2bar pressure and 300oC .……………………

c. specific volume at 20bar pressure and 350oC. …………………….

d. the specific internal energy at 40bar and 400oC .……………………

Pure Substance (Steam) | 24

e. specific entropy at 40bar and specific enthalpy 3330kJ/kg. …………………
f. specific volume at 5bar pressure and 7,060kJ/kgK entropy. …………………
g. the specific enthalpy at 15bar and a specific volume of 0.2667kJ/kg. ……………………

7. Find the specific internal energy value for pressure over 70 bar,

Example question

Steam at 120 bar and 500 oC. Find the degree of superheat and specific internal energy.

Suggestion

Determine the steam phase first, if superheated,. use the formula u = h - pv and its value read

directly from the steam table

tquestion =500 oC tsaturation=324.6 oC, jadi ini fasa?

Degree of superheat DOS = tsuperheat – tsaturation

= 500 oC – 324.6 oC = 175.4 oC

So, at 120 bar and 500 oC, we have
v = 2.677 x 10-2 m3/kg
h = 3348 kJ/kg
u = h – Pv

= 3348 kJ/kg – (120 x 102 kN/m2)(2.677 x 10-2 m3/kg)
= 3026.76 kJ/kg

Practice 2.8 [Ans 2772kJ/kg]

Determine the :
a. specific internal energy of course at 100bar pressure and 375oC.

b. specific internal energy at 120bar and 425oC. Pure Substance (Steam) | 25

[Ans 3107kJ/kg]

c. specific internal energy at 200bar and 500oC.

d. specific internal energy at 140bar and 400oC

8. Finds the value of a property based on the value of one other property in the
superheated phase.
Example question
Steam at 100 bar has a specific volume of 2812 m3/kg. Find the temperature, degree of superheat,
and specific enthalpy
Solution
First, it is necessary to decide whether the steam is wet, dry saturated or superheated.
The steam is superheated kerana, vsoalan> vg iaitu pada 100 bar, vg = 0.01802 m3/kg. Specific
volume of 2.812 x 10-2 m3/kg.
Hence,. The state of the steam is at point A in the diagram below.

Pure Substance (Steam) | 26

The diagram shows the
position of point A at
100bar pressure at
Superheated phase with 425
oC superheat temperature

From the superheated table

at 100 bar, the specific volume is 0.02812m3 / kg below the 425 oC column. So

tsuperheat =425 oC dan ts=311.0 oC,
= tsuperheat – tsaturation
Degree of superheat, DOS
= 425 oC – 311 oC
= 114 oC

So, at 100 bar and 425 oC, we have
v = 2.812 x 10-2 m3/kg

h = 3172 kJ/kg

Practice 2.9

a. the specific internal energy of the pressure at 180bar and h = 2888kJ / kg.
[Ans 2674kJ/kg]

v = u = h – Pv
h= =

=

Pure Substance (Steam) | 27

b. the specific internal energy of the pressure at 160bar and entropy 6.919kJ/kgK.
[Ans 3404kJ/kg]

c. specific entropy at 40bar and enthalpy 2330kJ / kg.

d. specific volume at 5bar pressure and its entropy is 5.060kJ / kgK.

e. specific enthalpy at 15bar pressure and its specific volume is 0.1667kJ/kg

Pure Substance (Steam) | 28

2.5 Interpolation
Interpolation are used to search for unspecified property values in the steam table.

Example question
a. Determine the specific enthalpy at the pressure of 5bar and temperature of 275oC,
b. determine the specific entropy at a pressure of 85 bar and 350 oC temperature

Figure a Figure b

The interpolation method is based on the concept of finding the coordinate on straight line using
the formula of the straight line equation
sketch the diagram as below with an unknown value into x axis and the value given to the y axis,
and enter in the formula
the value of x can be determined by:

x  x1  x2  x1 y

y  y1 y2  y1

x  y  y1 x2  x1   x1 y2 (x2 , y2)
y2  y1  y (x , y)
y1 (x1 , y1)

There are two methods of interpolation: x1 x x2 x
i. single interpolation
ii. double interpolation

Pure Substance (Steam) | 29

2.5.1 Single interpolation

Single interpolation is used to find the values in the table when one of the values is not tabulated.
For example, tentukan entropi tentu pada tekanan 85bar dan suhu 350oC, it is necessary to
interpolate between the values given in the table.
Example question

Determine the specific entropy at the pressure of 85bar and 350C (when the pressure is not in
the table)

Solution
The specific entropy value (sought value) at 85
bars (set value) is not available on Steam Tables.
So, the interpolation process needs to be made
between the nearest available value. 80 and 90bar

x  x1  x2  x1
y  y1 y2  y1

s  6.133  6.039  6.133
85  80 90  80

s  6.133   0.094
5 10

s  6.133  5  0.094
10

s  6.086 kJ kgK

Example question

Determine the specific volume of steam at 8 bar and 220oC. (when the superheat temperature is
not in the table)

Solution

The steam is at superheated condition as the temperature of the steam is 220oC > ts (170.4 oC)..
An extract from the Steam Tables,

Pure Substance (Steam) | 30

Nilai yang dicari berada pada paksi x, nilai ditetapkan berada pada paksi y, masukkan nilai
dalam formula

x  x1  x2  x1 v  0.2610 0.2933  0.2610
y  y1 y2  y1 220  200  250  200
v  0.2739 m3 kg
x  y  y1 x2  x1   x1
y2  y1 

Example question

Determine the specific enthalpy of dry saturated steam at 53bar. [2791kJ/kg]
(when the pressure value is not in the steam table)

2.5.2 Double Interpolation

Double interpolation is used when two properties (eg. temperature and pressure) not on Steam
Tables. For example to find enthalpy of superheated steam at 25 bar and 320oC, interpolation
between 20 bar and 30 bar and between 300oC and 350oC is required.

Example question
Determine the specific enthalpy of superheated steam at 25 bar and 320oC.

Solution
An extract from the Superheated Steam Tables:

t(oC) 300 320 Pure Substance (Steam) | 31
3025
p(bar) 2995 h1 350
20 h 3138
25 h2 3117
30

Firstly, find the specific enthalpy (h1) at 20 bar and 320
oC; At 20 bar,

h1  3025  3138  3025
320  300 350  300

h1  3070.2 kJ/kg

Secondly, find the specific enthalpy (h2) at 30 bar and
320 oC;At 30 bar,

h2  2995  3117  2995
320  300 350  300

h2  3043.8 kJ/kg

Now interpolate between h1 at 20 bar, 320oC, and h2 at h  h1  h2  h1
30 bar, 320oC in order to find h at 25 bar and 320oC. At 25  20 30  20
320oC

,

h  3070.2 3043.8  3070.2

25  20 30  20

h  3057kJ / kg

Pure Substance (Steam) | 32

Practice 2.10

1. Determine the specific volume, specific enthalpy and specific internal energy of wet

steam at 32 bar if the dryness fraction is 0.92.

[Ans:0.05746 m3/kg, 2661 kJ/kg, 2476 kJ/kg]

v = xvg h = hf + xhfg u = uf + x(ug – uf )

2. Find the dryness fraction, specific volume and specific internal energy of steam at 105 bar

and specific enthalpy 2100 kJ/kg

[Ans:0.52, 0.00882 m3/kg, 1998kJ/kg]

h = hf + xhfg v = xvg u = uf + x(ug – uf )

s = sf + xsfg

3. Steam at 120 bar is at 500 oC. Find the degree of superheat and specific internal energy.
[Ans:Dos = 175.4 oC,u = 3026.76 kJ/kg]

4. S.team at 160 bar has a specific enthalpy of 3139 kJ/kg. Find the temperature, degree
of superheat, and specific internal energy.
[Ans:Tos=450 oC, Dos = 102.7 oC, u = 2866.68kJ/kg]

Pure Substance (Steam) | 33

5. With reference to the Steam Tables determine
a. the temperature of superheat, specific volume and specific internal energy of steam
at 7 bar and enthalpy 3060 kJ/kg.
b. the specific volume of superheated steam at 12.5 MN/m2 and 650oC.
c. the degree of superheat and entropy of steam at 10 bar and 380oC.
d. the specific enthalpy of steam at 15 bar and 275oC.
(Ans: 300 oC, 0.3714x10-2 m3/kg, 2804kJ/kg,
v = 3.246 x 10-2 m3/kg Dos = 200.1oC, s=7.3988kJ/kg K, h=2982 kJ/kg)

6. With reference to the Steam Tables determine
a. the temperature of superheat, specific volume and specific internal energy of
steam at 7 bar and enthalpy 3060 kJ/kg.
b. the specific volume of superheated steam at 12.5 MN/m2 and 650oC.
c. the degree of superheat and entropy of steam at 10 bar and 380oC.
d. the specific enthalpy of steam at 15 bar and 275oC.
(Ans : 300 oC, 0.3714x10-2 m3/kg, 2804kJ/kg,
v = 3.246 x 10-2 m3/kg Dos = 200.1oC, s=7.3988kJ/kg K, h=2982 kJ/kg)

Pure Substance (Steam) | 34

7. Referring to Steam Tables, determine the phase of each point and determine the value

Point Phase Value
A u = 1149kJ/kg
B v
C h
D T=
E s
F p

8. With reference to the Steam Tables
a. Determine the saturation temperature at 87 bar,
b. Determine the specific enthalpy of saturated steam at 103bar.
c. find the specific entropy of steam at 30 bar and 380oC
d. find the specific entropy of steam at 2.5 bar and 300oC. .

[Ans: 301 oC;2719kJ/kg, 6.850kJ/kgK;7.797kJ/kgK]

Pure Substance (Steam) | 35

9. With reference to the Steam Tables, Sketch the P-v diagram and determine
a. Specific internal energy of steam at 5bar and 350oC.
b. Specific enthalpy of steam at 60 bar and x = 0.8.
c. Specific internal energy of steam at 80 bar and 350oC.
d. Specific entropy of dry saturated steam at 100bar pressure.
(Ans:2883kJ/kg, 2468kJ/kg, 2866.68kJ/kg, 5.615kJ/kgK)

P u r e S u b s t a n c e ( I d e a l g a s ) | 36

2.6 Ideal gas

Actually there is no perfect gas, but under certain temperature and pressure limits, for example when

conditions exceed critical temperature and at very low pressure, hydrogen, oxygen, nitrogen, air, gas

and so on it can be a perfect gas. Each gas can behave as a perfect gas when it complies with the

perfect gas characteristic equation i.e pv  R or written as pv  RT , or written, where R is a
T

gas constant.

Gas constant
a. Each gas has a different constant value.
b. For a gas it has the same value for each condition (state).

c. Where R  R and for gas constants are J / kgK or J / kgK.
M

d. It depends on the molecular weight (M) of each gas and
e. universal gas constant, Ro = 8.314 kJ / kgKf.
f. For air R = 287J / kgK or 0.287kJ / kgK

substance formula Molar Mass Gas Constant
M R
1 Carbon dioxide CO2
2 Nitrogen N2 Kg/kmol kJ/kgK
3 Oxygen O2 44.01 0.1889
4 Water H2O 0.2968
5 Hydrogen H2 28.013 0.2598
6 Argon Ar 0.4615
31.999
0.2081
18.015

2.016

A.Cengel

For example, what is the value of the gas constant for hydrogen (H2).

Found Mhydrogen=2.016 kJ/kgK

R Hidrogen R  8.314  4.214 kJ kgK
M 2.016

For example, what is the molar mass of Argon if Rargon = 0.2081kJ / kgK

P u r e S u b s t a n c e ( I d e a l g a s ) | 37

2.6.1 Universal gas

Referring to the process diagram above, each situation, all the gas

follows the perfect gas characteristic equation pv  RT , atau 1

pV=mRT with mkg gas 2

So on condition 1 p1V1  mRT1 Figure: prosess

And on condition 2 p2V2  mRT2
when m1 = m2 (because the closed system) and R must be the same
as the properties of the gas do not change

so p1v1  p2v2 , therefore, at
T1 T2

constant pressure process, p1  p2  v1  v2
T1 T2

constant volume process, v1  v2  p1  p2
T1 T2

constant temperature process, T1  T2  p1v1  p2v2

Specific Heat is the amount of heat energy needed to raise the temperature of a unit of mass to
increase 1Kelvin - unit (kJ / kgK)

If the process occurs at constant pressure the heat is written Cp, with the sum of heat as

Q  mCp (T2  T1)

If the process occurs at constant volume of heat it is written Cv, with the sum of heat as

Q  mCv (T2  T1)

Table 2.3 : Examples of some specific heat capacity at 300K

gas Cp ( kJ/kgK ) Cv ( kJ/kgK )

Air 1.005 0.718

Carbon dioxide 0.846 0.657

hidrogen 14.307 10.183

Oksigen 0.918 0.658

A.Cengel pg-905

Cp  Cv  R Cp  where γ is the adiabatic index
Cv
and

P u r e S u b s t a n c e ( I d e a l g a s ) | 38

Practice 2.11

1. The gas masses 0.18kg at 15oC and the 103kN / m2 pressure occupies 0.15m3 volume

space, calculate the gas constant.

V=0.15m3 Dari pV=mRT

T=15+273=288K R  pV  103103  0.15  0.296kJ / kgK
mT 0.18 288
m=0.18kg
p= 103kN/m2

R=?

2. A perfect gas has a mass of 0.04kg and a volume of 0.0072m3 at 6.76bar pressure and
127oC temperature. Calculate the molecular weight of the gas.

v=0.0072m3 Dari pV=mRT R  pV  ----
T=127+273=400C mT
m=0.04kg
p= 6.76bar = 6.76x102kN/m2 M  Ro
M=? R

M  Ro
R

R=?

3. Determine the specific volume of nitrogen gas (M = 28kg / kmol) at 20bar and
temperature 300K.

T= R  Ro , = --------------- =
M= M
p=
v=? v  RT
p
v  RT
p

R=?

R  Ro
M

P u r e S u b s t a n c e ( I d e a l g a s ) | 39

4. A gas has a pressure of 3.5 bar when a volume of 0.03m3 and a temperature of 35oC. The
gas constant is 0.29 kJ / kgK. Calculate the mass of the gas. [ Ans :0.118kg]

v=
T=
m=
p=
R=?

5. 0.16kg gas has a volume of 0.4m3 at a pressure of 4 bar. The weight of the gas molecule is

32. Calculate the temperature of the gas. [ Ans :385K]

6. If the air pressure, p = 1.5x106 N / m2 and T = 300 oC. Determine the specific volume.

P u r e S u b s t a n c e ( I d e a l g a s ) | 40

7 If the gas (M=24kg/kmol) has a value of Cv = 0.820 kJ/kg K, calculate the:
[ Ans :R = 0.346kJ/kgK, ,Cp=1.166kJ/kgK, γ=1.42]

a. gas constant
b. specific heat at constant pressure,Cp
c. specific heat ratio, γ

8. A mass of 0.18 kg gas, 0.17m3 at temperature of 15 oC and pressure 130 kN/m2. If the gas
has a value of Cv = 720 J/kg K, calculate the:
a. gas constant, R
b. molecular weight, M
c. specific heat at constant pressure, Cp

d. specific heat ratio,

[ Ans :R = 0.426kJ/kgK, M = 19.52kg/kmol,Cp=1.146kJ/kgK, γ = 1.59]

First Law – Non Flow Process /41

3.0 FIRST LAW OF THERMODYNAMICS AND ITS PROCESSES

3.1 Basic Thermodynamic

Thermodynamics – science of energy, heat to power

Source of energy - coal, animal, human, solar, water, gravity, wind, gas, fuel, tidal, wave, etc.
Form of energy - kinetic energy, potential, heat, sound, light, elastic, electromagnetic etc.

Energy conversion. – Energy can neither be created nor destroyed; it can only change forms.

Ein = Eout or Einitial = Efinal.

The First Law of Thermodynamics can be stated as follows:

In symbols,  dQ =  dW

where  represents the sum of a complete cycle.

Difference of Open system and open system

Open system Close system

Fluid Heat system
Inlet

Work

system Work

Boundary Heat

Boundary

Fluid
Outlet

Heat and work can go out through the borders Heat and work can go out through the
borders
Mass can pass through the boundary
The mass can not go through the borders

Control volume Control mass

 

min  mout m0

Open system for a cycle, Close system for a cycle,
ΣQ = ΣW + ∆ke +∆pe +∆h ΣQ = ΣW
under going ΣQ = ΣW + ∆Esystem Undergoing ΣQ = ΣW + ∆U

Figure 3.1 : Difference of Open system and open system

First Law – Non Flow Process /42

3.2 Work and Heat Transfer

Work transfer is defined as a product of the force and the distance moved in the direction of the
force. ( w = F x d )

Heat energy transfer between body two systems at different temperatures. This energy is called
heat, or thermal energy, and the term "heat flow" refers to an energy transfer as a consequence of
a temperature difference.

Sign Convention for Work Transfer

out in
-ve
Work (W) +ve +ve

Heat (Q) -ve

Figure 3.2 Sign convention for work and heat transfer

. QA = Qout = (-50kJ) WB = Win = -70kJ

. QD = Qin = +40kJ WC = Wout = 90kJ

Example question
The figure above shows a certain process, which undergoes a complete cycle of operations.
Determine the value of the work output for a complete cycle, Wout.

Qin = 10 kJ Wout = ?

SYSTEM

Qout = 3 kJ

Win= 2 kJ

Figure 3.3 Work and heat transfer

Solution to Example 1 Hence Q - W = 0
Q = Qin + Qout = (10) + (-3) = 7 kJ W = Q
W = Win + Wout = (-2) + (Wout)
(-2) + (Wout) = 7
Wout = 9 kJ

First Law – Non Flow Process /43

Example Question

The figure above shows a certain process, which undergoes a complete cycle of operations.
Determine the value of the work output for a complete cycle, Wout.

Qin = 15 kJ Wout = x?

SYSTEM

Win= 2 kJ Qout = 3 kJ

Solution to Example 2 Hence Q = W
Q = Qin + Qout = ( ) + ( ) = ___ kJ
W = Win + Wout = ( ) + (x)

3.3 Internal Energy

Internal energy is the sum of all the energies a fluid possesses and stores within itself.
U2 – U1 = +ve ( increased )
U2 – U1 = -ve ( decreased)

Figure 3.4 Added work and heat raise the internal energy of a close system

First Law – Non Flow Process /44

Practice 3.1 Q  W 
1. Determine the value of x for the system below Q  W

2. Determine the value of x for the system below Q  W 
Q  W

3. Determine the internal energy change for the system Q  W 
below U2 - U1  Q  W

4. Determine the internal energy change for the system below

Q = 27kJ W = 60kJ

system

W=25kJ

Q = 80kJ

First Law – Non Flow Process /45

5. The table shows processes that occur in closed systems. Fill in the space with the correct
answer.Q – W = U2 – U1

Proses Q kJ W kJ U2 kJ U1 kJ
a +50 -30 -70
b -50 -35 +100
c +130 -70 100
d +20 -120 -50
e -70 120 +30

6. In a diesel engine, the inner energy at the beginning is U1  300kJ/kg
300kJ / kg and at the end of the stroke, the energy in the U 2  200kJ/kg
value of 200 kJ / kg. the resulting work is 76 kJ.kg. W  76kJ/kg
Determine the value of heat transferred. Q?

7. A gas engine, heat transferred to coolant water is worth Q  150kJ / kg
150kJ / kg and the work done by piston on gas is 300 kJ
/ kg. Calculate the change in internal energy. W  300kJ/kg

U2 U1  ?

8. In the engine's internal combustion compression stroke,
heat thrown to cooling water is 45kJ / kg and input
work is 90kJ / kg. Calculate the internal energy change
of work fluid.

First Law – Non Flow Process /46

9. The cylinder has excellent heat insulation having a Q  0kJ / kg

mass of gas with energy in 950 kJ. Gas expands behind U1  950kJ / kg
the piston so internal energy is 825kJ. Determine the U 2  825kJ / kg
work done.

W  .............kJ / kg

10. A thermodynamic system undergoes a process in which its internal energy decreases by 300

kJ. If at the same time, 120 kJ of work is done on the system, find the heat transferred to or

from the system. [Ans : - 420 kJ]

11. The internal energy of a system increases by 70 kJ when 180 kJ of heat is transferred to the

system. How much work is done by the gas? [Ans : 110kJ]

12. During a certain process, 1000 kJ of heat is added to the working fluid while 750 kJ is

extracted as work. Determine the change in internal energy and state whether it is increased

of decreased. [Ans : 250 kJ]

First Law – Non Flow Process /47

3.4 Non-flow process
3.4.1 Constant Pressure Process

If the change in pressure during a process is very small then that process may be approximated as a
constant pressure process. For example, heating or cooling a liquid at atmospheric pressure may be
analyzed by assuming that the pressure remains constant.

In constant pressure process, the
temperature is proportional to the
volume

Consider the fluid in the piston cylinder as
shown in Figure. If the piston free moving
from state 1 to state 2 the pressure will
remain constant.

The general property relation between the

initial and final states of a perfect gas is applied as:

p1V1  p2V2
T1 T2

If the pressure remain constant during the process, p2 = p1 and then the above relation becomes

V1  V2 or T2  V2
T1 T2 T1 V1

Formulae: Heat transfer:
Work transfer: Q – W = U2 – U1
Q = h2 – h1
W  P(v2  v1)  mR(T2  T1)
Q  mCp (T2  T1)
U 2  U1  mCv(T2  T1)

First Law – Non Flow Process /48

Example question
Example 1
A quantity of gas at volume 0.54 m3 and temperature 345oC undergoes a constant pressure process
that causes the volume of the gas to decreases to 0.32m3. Calculate the temperature of the gas at
the end of the process.

V1 = 0.54m3,
T1=345oC
p1=p2
V2 = 0.32 m3
T2=?

Example 2
1.0kg of air and 0.02m3 volume expanded reversibly at constant pressure of 4.0bar to 270oC and
0.07m3 volume. Sketch the pv diagram and calculate the amount of heat transfer.

First Law – Non Flow Process /49

3.4.2 Constant Volume Process, Isometric(V2=V1)

If the change in volume during a process is very

small then that process may be approximated as a

constant volume process.

If the volume remain constant during the process,

V2 = V1

p1  p2 or T2  p2
T1 T2 T1 p1

From this equation it can be seen that an increase
in pressure results from an increase in
temperature. In other words, in constant volume
process, the temperature is proportional to the
pressure.

Formulae: Heat transfer:
Q = U2 – U1
.Work transfer:
W=0 Q  mCv(T2  T1 )
U 2  U1  mCv(T2  T1)

This result,the net amount of heat is equal to the change in the internal energy of the fluid

Example Questions

Example 3
Air mass of 1kg in rigid container is at 4.8 bar and 150oC. The container is heated to a
temperature of 200°C. Calculate the final pressure and heat supplied during the process. [Ans
:5.37bar, 35.9 kJ/kg ]