THERMODYNAMICS

Prepared by: Engr. Allen Rey A. Nierva, MSME

 About your Faculty/ Facilitator:
Engr. Allen Rey A. Nierva, MSME
• Graduate of Saint Louis University (Baguio City) – BSME, 2014
• Mechanical Engineer Board Exam – 2015
• Graduate of Saint Louis University (Baguio City) – MSME, 2019
 THERMODYNAMICS

• Thermodynamics is that branch of physical science

that treats of various phenomena of energy and the
related properties of matter, especially of the laws of
transformation of heat into other forms of energy and
vice versa.
Example: converting of heat into electrical work (electrical
power generation)
• Converting electrical work into cooling (air conditioning and refrigeration)
 The Working Substance
• Working substance – a fluid in which energy can be stored or from
which energy can be removed.
• Fluid – a substance characterized by low resistance to flow and the
tendency to assume the shape of its container.
• Phase – refers to a quantity of matter that is homogeneous throughout in
both chemical composition and physical structure. (ex. solid phase, liquid
phase, vapor phase)
• Pure Substance – is one that is uniform and invariable in chemical
composition. It can exist in more than one phase but its chemical
composition must be the same in each phase. (ex. Water)
 The System (Thermodynamic System)
• A system is that portion of the universe, an atom, a galaxy, a certain
quantity of matter, or a certain volume in space, that one wishes to study.
It is a region enclosed by specified boundaries, which may be imaginary,
either fixed or moving.

• Surroundings or Environment – the region about the system,

- Anything external to the system
Boundary – it is a partition that separates the system from the surroundings.
 Types of System:
• Open system (control volume) - is one in which there is an exchange of
matter with the surroundings. Mass can cross its boundary and there is a
crossing of energy between system and surroundings.

• Closed system (control mass) – is one in which there is no exchange of

matter with the surroundings. Mass cannot cross its boundary however
energy can.
• Isolated system – is one that is completely impervious to its
surroundings- neither mass nor energy cross its boundary.
 Classification of Property
• Intensive Properties – those that are independent of the mass (ex.
Density, pressure, temperature)

• 2. Extensive Properties – those that are dependent of mass and are total
values (ex. Total volume, total internal energy)
 Systems of Units
Isaac Newton made the momentous statement that the acceleration of a particular body is directly proportional to the resultant force
acting on it and inversely proportional to its mass.
 Mass and Weight
• Mass (m) – is the absolute quantity of matter in a body. An unchanging
quantity when the speed of the mass is small compared to the speed of
light.
• Weight (W)– the force of gravity on the body.

Where g = gravitational acceleration = 9.81m/s2 = 981 cm/s2 = 32.2 ft/s2 at the surface of the earth

Note: At the surface of the earth (sea level), k and g are numerically equal, so are m
and Fg.
 UNITS OF MASS UNITS OF VOLUME
•1 short ton = 2,000 pounds (lbs) 1 quart = 2 pints
•1 long ton = 2,240 lbs 1 US gallon = 231 in3 = 3.7854 liters
•1 metric ton = 1,000 kg 1 US gallon = 4 quarts
•1 pound = 16 ounces 1 brit gallon = 277.42 in3
•1 kip = 1,000 lbs 1 m3 = 1,000 liters
•1 kg = 2.2 lbs 1 m3 = 35.31 ft3 = 264.2 US gallon
•1 kgf = 9.80665 newtons (N) = 9.81 N 1 ft3 = 7.48 US gallon = 28.32 liters
•1 newton = 0.1019 kgf 1 ft3 = 1,728 cubic inches
•1 pound = 0.453592 kg 1 ganta = 3 liters = 8 chupas
•1 poundal = 1 lbm – ft/s2 1 cavan = 25 gantas
•1 slug = 1 lbf – s2/ft
•1 slug = 32.174 lbm
 Sample Problem no. 1
• A system has a mass of 20 kg. Determine the external force necessary to
accelerate it 10 m/s2 .
• Horizontally along a frictionless plane, (N)
• Vertically in a region where g= 9.81 m/s2 if the external force is applied vertically
upward, (N)
• Vertically Downward (N)
• Solution:
• In this problem we are required to find the External forces given in the above
situation using Newton’s Law.
Density, Specific Volume, Specific Weight
Sample Problem # 7
Two liquids of different densities ( p1 = 1500 kgm /m3 ) , ( p2 = 500 kgm /m3 )
are poured together into 100 L tank, filling it. If the resulting density of the
mixture is 800 kgm /m3 , find the respective quantities of liquids used. Also,
find the weight of the mixture; where local g = 9.675 mps2

Solution: mass of mixture,

mm = pm vm
= (800 kgm/m3 )(0.100 m3) = 80 kgm
m 1 + m2 = m m
p1v1 + p2v2 = mm
1500 v1 + 500 v2 = 80 (eqn 1)
Fg = mg/k
v1 + v2 = 0.100 (eqn 2)
= (80 kgm )(9.675 m/s2)
v1 = 0.03 m 3
9.81 kgm – m
v2 = 0.07 m3 kgf – s2

m1 = p1v1 = (1500 kgm/m3) (0.03 m3) = 45 kgm = 78.8991 kgf

m2 = p2v2 = 500 kgm/m3)(0.07 m3) = 35 kgm

 Pressure (P)
• normal force applied to a unit area

• Atmospheric Pressure (Patm) – normal force exerted by the atmosphere on a unit

area.
• At sea level atmospheric pressure Patm = 101.325 kPa = 14.7 psi = 760 mm Hg =
29.92 in Hg
UNITS OF PRESSURE:

1 atmosphere = 760 mm Hg
= 14.7 psi
= 29.92 in Hg
= 1.033 Kg/cm2
1 bar = 100 kPa
1 Pa = 1 N/mm2
1 Kpa = 1 kN/m2
= 1000 Pa
1 Ksi = 1000 kips/in2
• Gage Pressure (Pg) – is the pressure difference between system’s absolute
pressure and surrounding /atmospheric pressure.

• Absolute Pressure (Pabs) – sum of atmospheric and gage pressures

Pabs = Patm + Pg
• Note: Pg is negative if it involves vacuum or negative reading relative to
atmospheric pressure.
• Fluid Pressure – pressure due to a column of a fluid or liquid.

P =yh ;
where h is the depth/Height of Fluid
Y = Specific Weight
P = Fluid Pressure
A 25 m vertical column of fluid whose specific volume is 5.32 x 10 -4 m3/kgm
is located where g = 9.65 mps2. Find the pressure at the base of the column.
P =1/v
P = 1/ 5.32 x 10 -4 m3/kgm = 1,879.6992 kgm /m3
Y =pg/k
1,879.6992 kgm/m3 (9.65 m/s2 )
(1 kgm – m / N – s2)
Y = 18,139.0973 N = 18.1391 KN/m3
Pressure = Yh = (18.1391 kN /m3)(25 m) = 453.4775 KN/m2
 Temperature
• Temperature – measure of the hotness or coldness of a body. It is also defined as the
measure of the internal energy of a body.

Conversion of temperature reading to another temperature scale:

t(oF) = 1.8 t(oC) + 32
t(oC) =
T(K) = t(oC) + 273
T(R) = t(oF) + 460
Conversion of change in temperature from one scale to another
temperature scale:
 Zeroth Law of Thermodynamics

• States that when two bodies are in thermal equilibrium

with a third body, the two are in thermal equilibrium
with each other.
 Sample Problem No. 2
• A 10 ft diameter x 15 ft height tank is receiving water at the rate of
350GPM and is discharging through a 6in. ID pipeline at a constant
velocity of 5 ft/s. At a given instant, the tank is half-full; find the water
level in the tank 15 min later. Required: the height of water inside the tank
after 15 mins.
Solution:
In this problem we are required to find the height of the water in the tank
whose half full
Note the tank has a suction and discharge, we can say that the water are
coming in and out of the tank.
(-) means that the discharge water is greater than the water flowing into the
tank, therefore we can say that the water after 15 minutes decreases
ft
𝐹𝑖𝑛𝑎𝑙h𝑖𝑒𝑔h𝑡=𝟓.𝟏𝟖𝟓𝟑 𝒇𝒕𝑎𝑛𝑠𝑤𝑒𝑟
The height of the tank after 15 minutes will be 5.1853 ft.
 Sample Problem no. 3
• A vertical composite fluid column whose upper end is open to the
atmosphere is composed of 50 cm of Hg (SG=13), 80 cm of oil (SG=0.8)
and 65 cm of water. Determine the pressure at the,
• Oil-water interface
• Water-mercury interface
• Base of the column
• Given the data we are required to find the pressure of the following
interface.
• Solution:

Solve for the specific weight of the following substances

For Oil – water interface
For water-mercury interface
At the base of the column
 Sample Problem:
• A pressure gage registers 50 psig in a region where the barometer is 14.25
psia. Find the absolute pressure in psia.
• Pabs = Patm + Pgage
• Pabs = 14.25psia + 50 psig
• Pabs = 64.25 psia
 Additional Problem Solving Exercise:
• How high does mercury barometer stands on a day when
atmospheric pressure is 98.6 kPa considering that the specific
gravity of mercury is 13.6.
• A manometer containing water and mercury (S.G= 13.5)
connects two pressure regions A and B as shown. If pressure at
B is 350 Kpa gage, find the pressure at A in KPa gage.
 Assignment: Short Bond Paper
1. A mass of 0.10 slug in space is subjected to an external vertical forces of 4 lb. If
the local g=30ft/s2 and if friction effects are neglected determine the acceleration of
the mass if the external vertical force is acting.
a. Upward
b. downward
2. A weatherman carried an aneroid barometer from the ground floor to his office
atop the Sear tower in Chicago, On the Ground level; the barometer reads 30.15 in
Hg absolute; topside it reads 28.60 in Hg absolute. Assume that average atmospheric
air density of 0.075 lb/ft3 ; estimate the height of the building, ft.
3. If a pump discharges 75 GPM of water whose specific weight is 61.5
lb/ft3 at local g= 31.9 ft/s2 , find a. Mass flow rate lb/min b. Time to fill a
drum 10ft diameter and 12 ft high, min.
4. A weatherman carried an aneroid barometer from the ground floor to his
office atop the Sear tower in Chicago, On the Ground level; the barometer
reads 30.15 in Hg absolute; topside it reads 28.60 in Hg absolute. Assume
that average atmospheric air density of 0.075 lb/ft3 ; estimate the height of
the building, ft.
 References:
• Thermodynamics by Hipolito Sta. Maria
• Thermodynamics by Cengel
• Thermodynamics by Moran