Lecture 01: Introduction to Thermodynamics and Units Review
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Transcript Lecture 01: Introduction to Thermodynamics and Units Review
Thermodynamics
EGR 334
Lecture 01:
Introduction to Thermodynamics
Today’s Objectives:
• Distribute and understand syllabus
• Take quick tour of thermodynamics topics covered in this
course
• Understand difference between system and control volume.
• Review units of thermodynamics related quantities
Reading Assignment:
• Read Chap 1. Sections 1 - 9
Homework Assignment:
From Chap 1: Problems 19, 37, 51, 59
Thermodynamics
Definition
-- Thermodynamics is the science of energy and entropy
-- Thermodynamics: the branch of physics dealing with
the relationships between heat, work, and forms of
energy.
From the Greek:
- θερμη, therme, meaning “heat“ → Energy
- δυναμις, dynamis, meaning “power“→ Transport
To the Engineer…Thermodynamics means:
- Understanding the 4 Laws of Thermodynamics
- Learn to work in with four different temperature
scales.
- Learn to balance energy, heat, and work with
respect to open and closed systems.
- Learn about common thermodynamcis devices
and applications and how the principles can be
used to predict system performance and
efficiencies.
Thermo….a Quick Survey
-- Properties of systems:
Temperature, Pressure, Specific Volume,
Phase, Quality, Density, Enthalpy, Entropy
-- Processes and Cycles:
State to state transitions.
Carnot cycle, Rankine cycle, Otto cycle,
Diesel cycle, Brayton cycle, Refrigeration cycle
-- Work-Energy-Heat Balance:
Applying the 1st Law of thermodynamics
Thermo…A Quick Survey…continued
-- Entropy Production and Accounting:
Applying the 2nd Law of thermodynamics
-- Thermodynamics devices:
Turbines, Heat Exchangers, Condensers, Engines
Pumps, Cooling Towers, Compressors, Diffusers
-- Psychrometrics:
HVAC, heating, ventilation, air conditioning,
and humidity
-- Combustion and Power Production:
Chemical energy production and balances
A basic concept: System
• A System is whatever it is that we want to study.
• In thermodynamics, the first step in defining any
problem is to define exactly what is to be
monitored, examined, measured, etc.
Environment
Interactions
System
Boundary
System Definition
Radiation
Radiation
System Definition
Radiation
Radiation
System Definition
Gas
Q
System Definition
Gas
Gas
Q
W
Types of Systems
• Isolated Systems – matter and energy may not
cross the boundary.
• Adiabatic Systems – heat may not cross the
boundary.
• Diathermic Systems - heat may cross boundary.
• Closed Systems – matter may not cross the
boundary.
• Open Systems – heat, work, and matter may cross
the boundary
(more often called a Control Volume (CV)).
Types of Systems
• Isolated Systems – matter and energy may not
cross the boundary.
• Adiabatic Systems – heat must not cross the
boundary.
• Diathermic Systems - heat may cross boundary.
• Closed Systems – matter may not cross the
boundary.
• Open Systems – heat, work, and matter may cross
the boundary
(more often called a Control Volume (CV)).
Closed System
• Matter (m) may not cross the boundary.
• Heat (Q) and Work (W) may cross the boundary
which change the energy (ΔE) of the system.
Po, Vo, To
Combustion
CxHy+A O2 B CO2 + C H2O
Pf, Vf, Tf
Q
∆V →W
Control Volume / Open System
• Heat (Q), work (W), and matter (m) may cross
the boundary
Ga
s
Q
System Properties
• The State of a system is defined by its properties (T, V,
P, E, ρ, v, u, h, s)
• Extensive properties – Depends on mass
• Intensive properties – Does not depend on mass.
•
Give some examples of
Extensive Properties
Intensive properties
Process vs. Cycle.
• A Process is when properties of the system
undergo a change. A process moves from one
“state” to another “state”.
• If State i = State f then system is said to be
Steady State.
• If a Process undergo changes that eventually
bring it back to the original state, these
transitions together are called a Cycle.
Process:
Cycle:
P
P
S1
S2
S2
S1
S4
S3
v
v
Unit Review: Weight and Mass
Weight
Mass
U.S. Customary
pound , lbf
pound of mass, lbm
or
slug
Metric
Newton, N
kilogram, kg
W =m g
where
g = 9.81 m/s2
g = 32.2 ft/s2
1 N = 0.2248 lbf
1 lbf = 4.4482 N
1 kg = 2.205 lbm
1 lbm = 0.4536 kg
1 slug = 32.2 lbm = 71.0 kg
Relationships:
Conversions:
Unit review: Density and Specific Volume
Density
ρ
mass basis
U.S.
Customary
[lbm / ft3 ]
molar basis
mass basis
Specific Volume
v or v
[ ft3 /lbm]
[ ft3 /mol]
[kg/m3]
[ m3/kg]
Metric
molar basis
[m3/mol]
Relationships
m = mass
V = Volume
m
V
M = molecular weight
n = number of moles
n
m
M
1
M
Unit review: Pressure
Pressure =
U.S. Customary
Presure head
[psi ] or [lbf /in2 ]
[psf] or [lbf /ft2 ]
[in of Hg ]
[ in of H20 ]
[ft of H20]
Metric
[N/m2]
[ Pa]
[bar]
[mm of Hg ]
[ mm of H20 ]
[cm of H20]
Other
atm
Relationships:
1 atm = 14.69 psi = 1.01325 bar
= 100 324 Pa = 760 mm of Hg
= 29.92 in of Hg = 33.96 ft of H2O
Gage vs. Absolute
pressure
pabs = patm + pgage
Vacuum vs. absolute
pressure
pabs = patm - pvac
Unit review: Temperature
Metric
U.S.
Relative
Temperature
Celcius
[ oC ]
Fahrenheit
[ o F]
Absolute
Temperature
Kelvin
[K]
Rankine
[ oR ]
Relationships
K = oC + 273.15
oR = oF + 459.67
oC =( oF - 32)/1.8
oF= 1.8 oC + 32
oR = 1.8 K
Example 1: An object whose mass is 35 lb is subjected to an
applied upward force of 15 lbf. The only other force acting
on the object is the force of gravity. Determine the net
acceleration of the object in ft/s2 assuming the acceleration
of gravity is constant (g = 32.2 ft/s2). Is the net acceleration
up or down?
FAF= 15 lbf
mass= 35 lbm
Fg= 15 lbf
Example 2: A system consists of N2 in a piston-cylinder
assembly, initially at P1 = 20 psi, and occupying a volume of
2.5 ft3. The N2 is compressed to P2 = 100 psi and a final
volume of 1.5 ft3. During the process, the relationship
between P and V is linear. Determine the P in psi at an
intermediate state where the volume is 2.1 ft3 and sketch
the process on a graph of P vs V.
P1= 20 psi
V1 =2.5 ft3
P2= 100 psi
V2 =1.5 ft3
Example 3: A monometer is attached to a tank of gas in which the pressure is
104.0 kPa. The manometer liquid is mercury, with a density of 13.59 g/cm3.
If g = 9.81 m/s2 and the atmospheric pressure is 101.33 kPa, calculate
(a) The difference in mercury levels in the manometer, in cm.
(b) The gage pressure of the gas in kPa and bar
(c) The absolute pressure of the gas kPa, atm, and psi
Example 1: An object whose mass is 35 lbm is subjected to
an applied upward force of 15 lbf. The only other force
acting on the object is the force of gravity. Determine the
net acceleration of the object in ft/s2 assuming the
acceleration of gravity is constant (g = 32.2 ft/s2). Is the net
acceleration up or down?
F ma
FAF= 15 lbf
m= 35 lbm
Fg= ?
FAF Fg ma
FAF Fg
FAF mg
a
m
m
15 lb f 32.2 lbm ft
a
lb f
35 lbm
a 18.4
ft
s
FAF
g
m
s
lbm ft
32.2
s
Example 2: A system consists of N2 in a piston-cylinder
assembly, initially at P1 = 20 psi, and occupying a volume of
2.5 ft3. The N2 is compressed to P2 = 100 psi and a final
volume of 1.5 ft3. During the process, the relationship
between P and V is linear. Determine the P in psi at an
intermediate state where the volume is 2.1 ft3 and sketch
the process on a graph of P vs V.
Since the relationship is linear (y=mx+b)
P2 P1 100 20
psi
m
80
3
V
V
1
.
5
2
.
5
ft
2 1
P P1 mV V1 P mV V1 P1
psi
P 80 3 2.1 2.5 ft 3 20 psi 52 psi
ft
Example 3: A monometer is attached to a tank of gas in which the pressure is
104.0 kPa. The manometer liquid is mercury, with a density of 13.59 g/cm3.
If g = 9.81 m/s2 and the atmospheric pressure is 101.33 kPa, calculate
(a) The difference in mercury levels in the manometer, in cm.
(b) The gage pressure of the gas in kPa and bar
(c) The absolute pressure of the gas atm, and psi
Solution: a)
P Patm gL L
104.0 101.33 kPa
P Patm
g
103 N / m 2 103 g 1 kg m / s 2
(1m) 2
L
(13.59b)
g / cm3 )(9.81m / s 2 ) 1 kPa
1 kg
1N
(100 cm) 2
L 2.0cm
b)
c)
Pguage P Patm 104.0 101.33 2.67 kPa
0.0267 bar
1 atm
Pabs 104.0 kPa
1.026 atm
101.33 kPa
14.7 psi
1.026 atm
15.09 psi
1 atm