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Thermodynamics EGR 334 Lecture 01: Introduction to Thermodynamics.
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Transcript of Thermodynamics EGR 334 Lecture 01: Introduction to Thermodynamics.
ThermodynamicsEGR 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:
Homework Assignment:
• Read Chap 1. Sections 1 - 9
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.
System
Environment
Boundary
Interactions
System Definition Radiation
Radiation
System Definition Radiation
Radiation
System Definition
Q
Gas
System Definition
W
Gas Gas
Q
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
Pf, Vf, TfCombustion
Q
∆V →WCxHy+A O2 B CO2 + C H2O
Control Volume / Open System•Heat (Q), work (W), and matter (m) may
cross the boundary
Q
Gas
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:
S1 S4
S2
S1
S2
S3
P P
v v
Unit Review: Weight and MassWeight Mass
U.S. Customary pound , lbf pound of mass, lbm or
slug
Metric Newton, N kilogram, kg
Relationships: W =m g where g = 9.81 m/s2 g = 32.2 ft/s2
Conversions: 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
Unit review: Density and Specific Volume
Density ρ
Specific Volume v or v
U.S. Customary
mass basis [lbm / ft3 ] [ ft3 /lbm]
molar basis [ ft3 /mol]
Metricmass basis [kg/m3] [ m3/kg]
molar basis [m3/mol]
Relationships m = mass V = Volume
M = molecular weight
n = number of moles
m
V 1
M mn
M
Unit review: Pressure Pressure = Presure head
U.S. Customary [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 pressurepabs = patm - pvac
Unit review: TemperatureMetric 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
Fg= 15 lbf
mass= 35 lbm
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?
FAF= 15 lbf
Fg= ?
m= 35 lbm
32.21532.2
35
18.4
AF g
AF g AF AF
f m m
m f
F ma
F F ma
F F F mg Fa g
m m mlb lb ft s lb ft
alb lb s
fta
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.
312
12 805.25.1
20100
ft
psi
VV
PPm
Since the relationship is linear (y=mx+b)
psipsiftft
psiP
PVVmPVVmPP
52205.21.280 33
1111
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
g
PPLgLPP atm
atm
2.0L cm104.0 101.33 2.67
0.0267
guage atmP P P kPa
bar
b)
Solution: a)
b)
3 2 3 2 2
3 2 2
104.0 101.33 10 / 10 1 / (1 )
(13.59 / )(9.81 / ) 1 1 1 (100 )
kPa N m g kg m s mL
g cm m s kPa kg N cm
c)1
104.0 1.026101.33abs
atmP kPa atm
kPa
14.71.026 15.09
1
psiatm psi
atm