Prepared by: EFREN A. DELA CRUZ
E-mail Address: eadelacruz@clsu.edu.ph
Central Luzon State University
Science City of Muñoz 3120
Nueva Ecija, Philippines
Instructional Module for the Course
(MENGR 3100 – Basic Mechanical Engineering)
Module 1
Fundamental Concepts and
Definitions
Overview
Every science has its own foundation and unique
terminologies, as in the course of thermodynamics and heat
transfer. Accurate definition relating to basic concepts makes a
sound foundation for the development of a science and prevents
possible confusions. The goal of this topic is to introduce some
fundamental concepts and definitions used in the study of
engineering thermodynamics and heat transfer. In most instances
the discussions were brief, and further elaboration is provided in
the succeeding modules illustrations and sample problems were
also provided for clearer understanding of the course.
MENGR 3100 – Basic Mechanical Engineering
I.
Objectives
Upon successful completion of the module, students are expected to:
a. Explain some of the basic concepts of the course.
b. Define or recognize the terminologies used in thermodynamics and heat
transfer.
c. Solve property related problems with proper units and dimensions.
II.
Learning Activities
1 Basic Principle, Concepts and Definitions
The word thermodynamics was derived from Greek words “Therme” means heat
and “Dynamis” means power.
Thermodynamics is the science that deals with various phenomena of energy and
the related properties of matter, especially of the laws of transformation of heat into other
forms of energy.
Thermodynamics Substance - a fluid that receives, transports, and transforms
energy.
1.1 Thermodynamics System
An important step in any engineering analysis is to describe precisely what is being
studied. A system is defined as a quantity of matter or a region in space chosen for study.
It may be as simple as a closed, rigid-walled tank or as complex as an entire Industrial
refinery. The composition of the matter inside the system may be fixed or may be changing
through chemical or nuclear reactions. The shape or volume of the system being analyzed
is not necessarily constant, as when a gas in a cylinder is compressed by a piston or an
inflated balloon.
The mass or region outside the system is called the surroundings. The real or
imaginary surface that separates the system from its surroundings is called the boundary
(Fig. 1.1). The boundary of a system can be fixed or movable. Note that the boundary is the
contact surface shared by both the system and the surroundings. Mathematically speaking,
the boundary has zero thickness, and thus it can neither contain any mass nor occupy any
volume in space.
Fig. 1.1 System, Surroundings and Boundary
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MENGR 3100 – Basic Mechanical Engineering
Types of Systems
Two basic kinds of systems are studied in this material. Systems may be considered
to be closed or open, depending on whether a fixed mass or a fixed volume in space is
chosen for study. A closed system (also known as a control mass or just system when the
context makes it clear) consists of a fixed amount of mass, and no mass can cross its
boundary. That is, no mass can enter or leave a closed system, as shown in Fig. 1.2 (a). But
energy, in the form of heat or work, can cross the boundary; and the volume of a closed
system does not have to be fixed.
If, as a special case, even energy is not allowed to cross the boundary, that system
is called isolated system. Consider the piston-cylinder device shown in Fig. 1.2 (b). Let us
say that we would like to find out what happens to the enclosed gas when it is heated. Since
we are focusing our attention on the gas, it is our system. The inner surfaces of the piston
and the cylinder form the boundary, and since no mass is crossing this boundary, it is a
closed system. Notice that energy may cross the boundary, and part of the boundary (the
inner surface of the piston, in this case) may move. Everything outside the gas, including
the piston and the cylinder, is the surroundings.
(a)
(b)
Fig. 1.2 Closed System
An open system, or a control volume, as it is often called, is a properly selected
region in space. It usually encloses a device that involves mass flow such as a compressor,
turbine, or nozzle Fig. 1.3. Flow through these devices is best studied by selecting the
region within the device as the control volume. Both mass and energy can cross the
boundary of a control volume.
min
mout
Fig. 1.3 Open System
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MENGR 3100 – Basic Mechanical Engineering
Thermodynamics deals with the macroscopic as opposed to microscopic analysis
in describing the system behavior.
In microscopic analysis we must look and analyze the molecular action.
In macroscopic we can define the system behavior with the overall effect of the
molecular interaction.
1.2 Properties of Thermodynamics substance/ system.
Characteristic quality of the entire systems and depends not on how the system
change state but only on the final particular state of the substance/system.
Extensive Properties - those properties that vary directly with mass.
Ex. mass, volume of solids.
Intensive properties -Those properties that are independent of the mass.
Ex. Tem. , Pressure.
Extensive properties per unit mass (specific properties) are considered as
intensive properties
Ex. Specific volume.
1.3 Phase and state of Thermodynamic substance.
Phase- quantity of matter that is homogeneous throughout
(solid, liquid, gas)
Change of Phase
 Solid to Liquid
 Liquid to Solid
 Liquid to Gas
 Gas to liquid
 Solid to Gas
- Melting
- Solidification (Freezing)
- Evaporation
- Condensation
- Sublimation
State of Substance - condition of a substance as described by certain observable
macroscopic parameter called Properties.
1.4 Processes and Cycles
Process- Simply the change of state






Isothermal
Isobaric
Isometric
Adiabatic
Isenthalpic
Polytropic
Types of processes
- constant temperature.
- constant pressure
- constant volume
- constant entropy (PVk = C)
- constant enthalpy
– PVn = C
Cycle - a series of processes one after the other such that the initial and final states are the
same.
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MENGR 3100 – Basic Mechanical Engineering
1.5 Dimensions and System of Units
When engineering calculations are performed, it is necessary to be concerned with
the units of the physical quantities involved. A unit is any specified amount of a quantity
by comparison with which any other quantity of the same kind is measured. For example,
meters, kilometers, feet, and inches are all units of length. Seconds, minutes, and hours are
alternative time units.
SI Units
SI is the abbreviation for Système International d’Unités (International System of
Units), which is the legally accepted system in most countries. The SI base units for mass,
length, and time are listed in Table 1.1. They are, respectively, the kilogram (kg), meter
(m), and second (s). The SI unit of force, called the newton, is defined in terms of the base
units for mass, length, and time. Newton’s second law of motion states that the net force
acting on a body is proportional to the product of the mass and the acceleration, written F
α ma. The newton is defined so that the proportionality constant in the expression is equal
to unity.
F = kma or
k = ma/F
Where k is proportionality constant
Table 1.1 SI Units for Mass, Length, Time and Force
Quantity
mass
length
time
force
Unit
kilogram
meter
second
newton
(= 1 kg m/s2)
Symbol
kg
m
s
N
Other Units
Although SI units are the worldwide standard, at the present time many segments
of the engineering community regularly use some other units. A large portion of market
stock of tools and industrial machines and much valuable engineering data utilize units
other than SI units. For many years to come, engineers will have to be conversant with a
variety of units. Accordingly, in this section we consider the alternative units for mass,
length, time, and force listed in Table 1.2.
Table 1.2 OtherUnits for Mass, Length, Time and Force
Quantity
Unit
Symbol
mass
pound mass
lb
slug
slug
length
foot
ft
time
second
s
force
pound force
lbf
(= 32.174lb ft/s2
= 1 slug ft/s2)
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MENGR 3100 – Basic Mechanical Engineering
SI is based on a decimal relationship between units. The prefixes used to express
the multiples of the various units are listed in Table 1.3. They are standard for all units, and
the student is encouraged to memorize them because of their widespread use.
Table 1.3 Standard Prefixes in SI Units
Factor
Prefix
Symbol
24
10
yotta
Y
1021
zetta
Z
18
10
exa
E
1015
peta
P
1012
tera
T
9
10
giga
G
106
mega
M
3
10
kilo
k
102
hecto
h
1
10
deka
da
10-1
deci
d
-2
10
centi
c
-3
10
milli
m
10-6
micro
μ
-9
10
nano
n
10-12
pico
p
-15
10
femto
f
10-18
atto
a
-21
10
zepto
z
10-24
yocto
y
System of units where k is unity but not dimensionless.
CGS System:
1-dyne force accelerates 1-gram mass at 1 cm./s2
& SI:
1 Newton force accelerates 1 kg. mass at 1 m/s2
English:
1 lb force accelerate 1 slug mass at 1 ft/s2



For CGS
o k = 1gm.cm/dyne. s2
For SI Or MKS:
o k = 1 kgm.m/Newton.s2
For English:
o k = 1slug.ft/lbf. s2
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MENGR 3100 – Basic Mechanical Engineering
System of units where k is neither unity nor dimensionless.
If the same word were used for both mass and force in a given system
CGS System:
1-gram force accelerates 1-gram mass at 980.665 cm./s2
MKS:
1-kg force accelerates 1 kg- mass at 9.80665 m/s2
English:
1-lb force accelerate 1 lb-mass at 32.174 ft/s2



For CGS
o k = 980.665 gm.cm/gf. s2
For SI Or MKS:
o k = 9.80665 kgm.m/kgf.s2
For English:
o k = 32.174 lbm.ft/lbf. s2
1.6 Properties and Property Relations
Mass And Weight
The mass of a body is the absolute quantity of matter in it.
Mass equivalent based on gram:
 Gram- 1 g
 lb- 454 g
 kg- 1000 g
 slug- 14600 g
The weight of the body is the force exerted by gravity on the given mass.
Relation of Mass and Weight
Fg = mg/k
from
f = ma/k
Or
W= mg/k
Where:
W- weight
m- mass
g- Acceleration due to gravity If observe gravity (go) is not given, then, used
the
standard gravity (gs)
32.174 ft/s2
32.2-ft/s2
9.80665 m/s2
9.81 m/s2
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MENGR 3100 – Basic Mechanical Engineering
Other units related to mass and weight
For English & MKS
W= m go /gs
Only for kgf & lbf
kgm & lbm
Where;
go - observe gravity
gs - standard gravity
If go = gs (m = w) for:
1lbm= 1 lbf & 1kgm= 1kgf
Mass Fundamentals
Newton’s Physics- mass is constant anywhere in the universe.
Law of conservation of mass states that mass is indestructible, provided that there
is no nuclear process involved.
min= m change + mout
And
if Δm= O
Then
min= mout
Density, Specific Volume, and Specific Weight
Density (ρ)of any substance is its mass per unit volume.
ρ= m/V
Where:
V- Volume occupied by matter
Unit of Density
English
lbm/ft3
lbm/Gal
MKS
kgm/m3
kgm/l
SI
kgm/m3
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MENGR 3100 – Basic Mechanical Engineering
Density of Water at Standard Condition
English
62.4 lbm/ft3
8.33 lbm/Gal
MKS
1000 kgm/m3
1 kgm/l
SI
1000 kgm/m3
Specific Volume is the volume of substance per unit mass
v = V/m = 1/(m/V) = 1/ρ
For water non at standard condition look @ steam table
Specific weight (γ) of any substance is the force of gravity (weight) per unit volume.
γ = W/V
Units of γ
English
lbf/ft3
lbf/Gal
MKS
kgf/m3
kgf/l
SI
N/m3
For water at standard condition
English
γ = 62.4 lbf/ft3
= 8.33 lbf/Gal
MKS
γ = 1000 kgf/m3
= 1 kgf/l
SI
γ = 9806.65 N/m3
= 9.80665 KN/m3
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MENGR 3100 – Basic Mechanical Engineering
Ex.1
Two liquids of different densities (ρ1 = 1500 kgf/m3 and ρ2 = 500 kgf/m3) are poured
together into 100 L tank, filling it. If the resulting density of the mixture is 800 kgf/m3 find
the respective quantities of liquids used. Also, find the weight of the mixture.
Solution.
For the mass of mixture
mm = ρmVm = (800 kg/m3)(0.1m3) = 80 kg
mm = m1 + m2 = ρ1V1 + ρ2V2
80kg = 1500 V1 + 500 V2
Fr:
(1)
V1 + V2 = 0.1
V1 = 0.1 - V2
(2)
Substituting (2) to (1)
80 kg = 1500(0.1 – V2) + 500 V2
Or
(1500 – 500)V2 = 150 – 80
V2 = 70 kg / (1000 kg/m3)
V2 = 0.07 m3
And V1 = 0.03 m3
Also
m1 = ρ1V1 = 1500 kg/m3 (0.03 m3) = 45 kg
m2 = ρ2V2 = 500 kg/m3 (0.07 m3) = 35 kg
and
Wm = mm gs = 80 kg (9.81 m/s2) = 784.80 N
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MENGR 3100 – Basic Mechanical Engineering
For fluid passing through a given section (Applying the law of conservation of mass)
Q = AV
ṁ = Q/v = AV/v = AVρ
Where:
Q - volume flow rate
A - cross sectional area of section
V - average speed (velocity)
ṁ – mass flow rate
ṁin = ṁout
AVρin = AVρout
Problem to solve
Two gaseous streams enter a combining tube and leave as a single mixture. These
data apply at the entrance section:
A1 = 75 in2, V1 = 500 fps, v1 = 10 ft3/lb, A2 = 50 in2, ṁ 2 = 16.67 lb/s, ρ2 = 0.12 lb/ft3
At exit, V3 = 350 fps, and v3 = 7 ft3/lb. Find the speed at section 2, the flow rate and area
at section 3.
Sol’n.
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MENGR 3100 – Basic Mechanical Engineering
Specific Gravity Or Relative Density
For solids and liquids
Ratio of weight to the weight of water of equal volume
SG = W/Wwater = γ V/ γwater Vwater = γ / γwater
= ρ.g/ ρwaterg = ρ / ρwater
= vwater/ v
For Gasses
SG = MW/MWa
Where
MWa = 28.97 = 29
Temperature
Measures of the degree of hotness or coldness of a body.
Always use absolute temp. in the analysis of thermodynamics.
For absolute temp.
T = 460 + oF in oR
= 273 + oC in oK
Conversion of oF & oC
o
F = 1.8 oC + 32
o
C = (oF- 32)/1.8
Problem to solve
Determine the temperature for which a thermometer with degrees Fahrenheit is
numerically twice the reading of the temperature in degrees Celsius.
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MENGR 3100 – Basic Mechanical Engineering
Pressure
For solids
P = F/A (Stress)
For liquids and gasses
Measuring pressure
1) Using manometers open to atmosphere
a) Absolute press. is greater than atmospheric press.
P= Patm + γ h
b) Absolute press. is less than atmospheric press.
P = Patm - γ h
Where
P- Absolute press.
Patm- atmospheric press.
γ - specific weight of fluid in the manometer
h- height of fluid
2) Using pressure gages
a) absolute press. is greater than atmospheric
P = Patm + Pg
b) Absolute press. is less than atmospheric
P= Patm - Pg
Where
Pg- gage press.
Patm or 1atm =101.325 kPa
=14.7 Psi
= 760 mm Hg
= 29.92 in Hg
Problem to solve
A vertical column of water will be supported to what height by standard
atmospheric pressure?
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MENGR 3100 – Basic Mechanical Engineering
1.7 Laws of Thermodynamics

First Law – also known as the Conservation of Energy principle, states that
energy can neither be created not destroyed, it can only change forms.

Second Law - deals with the quality of energy (energy degradation). There
are two classical statements of this law:
 Kelvin-Planck statement: It is impossible to construct a device that will
operate in a cycle and produce no effect other than the raising of a weight and
the exchange of heat with a single reservoir.
 Clausius statement: It is impossible to construct a device that operates in a
cycle and produces no effect other than the transfer of heat from a coolerbody to a hotter body.

Third law – states that the entropy of a perfect crystal is zero at the absolute
zero of temperature.

Zeroth law - when two bodies have equality of temperature with a third
body, they in turn have equality of temperature with each other.
III. Assessment
I.
Identification: Read the following statements carefully and try to identify the answer for
each item. Write your answer on the space provided before each number.
_________________________ 1. Fluid that receives, transports, and transforms energy.
_________________________ 2. Quantity of matter with fixed mass and identity in
which attention is focused for study.
_________________________ 3. Also known as control volume.
_________________________ 4. Change on a system state, if a system undergoes from
one equilibrium state to another.
_________________________ 5. Characteristic quality of the entire systems/substance.
Page 14 of 16
MENGR 3100 – Basic Mechanical Engineering
II.
Problem Solving – Clearly and neatly solve each problem in a separate sheet/s of short
bond paper. No solutions and/or units will be considered incorrect.
1. A pump discharges into a 3-m-per-side cubical tank. The flow rate is 300 liters/min,
and the fluid has a density 1.2 times that of water. Determine (a) the flow rate in
kg/s; (b) the time it takes to fill the tank.
2.
A new temperature scale is desired with freezing of water at 100 X and boiling
occurring at 1000 X in atmospheric pressure. Derive a conversion between degrees Celsius
and degrees X. What is absolute zero in degrees X?
0
0
3.
A 3 ft high 4.653 ft diameter cylindrical tank contains 3 lbs. of gas at 80 F and gage
pressure of 47 in Hg. (a) What is the temperature in C, R, and K? (b) What is the gas
specific volume and density? (c) What is the absolute pressure in Psi and in kPa.?
o
o
o
o
4.
A pressure cooker operates by cooking food at a higher pressure and temperature
than is possible at atmospheric conditions. Steam is contained in the sealed pot, with a
small vent hole in the middle of the cover; allowing steam to escape. The pressure is
regulated by covering the vent hole with a small weight, which is displaced slightly by the
escaping steam. Atmospheric pressure is 100 kPa, the vent hole area is 7 mm , and the
Pressure inside should be 250 kPa. What is the mass of the weight?
2
5.
A cylindrical tank is 50 in. long, has a diameter of 16 in., and contains 1.65 lb of
water. Calculate the specific volume and density of the water.
m
III.
References
Moran, M. J. and Shapiro, H. N. 2006. Fundamentals of Engineering Thermodynamics 5th
edition. SI version. John Willey & Sons. England.
Cengel, Yunus A. 2008. Introduction to Thermodynamics and Heat Transfer. McGrawHill Inc. New York.
Moran et al. 2003. Introduction to Thermal Systems Engineering: Thermodynamics, Fluid
Mechanics, and Heat Transfer. John Wiley & Sons, Inc. New York
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MENGR 3100 – Basic Mechanical Engineering
Burghardt, M. D. and Harvbach, J. A. 1993. Engineering Thermodynamics 4rth edition.
Harper Collins. New York.
Thermodynamics Online references and lectures (Yale Open courseware etc..)
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