Transcript MEGR 324 – Heat Transfer

ChemE 260
Foundation of Thermodynamics
Dr. William Baratuci
Senior Lecturer
Chemical Engineering Department
University of Washington
TCD 1: All
CB 1: ALL
March 29, 2005
Classical Thermodynamics
• Large Groups of Molecules – Continuum Scale
• The Laws of Thermodynamics
– 1st Law: Energy can neither be created nor destroyed. It can only change
form.
– 2nd Law: Energy in the form of heat only flows spontaneously from
regions of higher temperature to regions of lower temperature.
• Forms of Energy
EP  m
g
z
gC
1 v2
EK  m
2 gC
ˆ  g z
E
P
gC
2
ˆE  v
K
2 gC
– Internal:
U
Uˆ
– Heat
Q
ˆ
Q
– Work
W
ˆ
W
– Gravitational Potential :
– Kinetic :
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ChemE 260
March 29, 2005
Dimensions & Units
• Dimensions: Mass, Length, Time
• Units: m, ft, kg, lbm, J, Btu
• Force
– IS a fundamental unit in the AE System
– Is NOT a fundamental unit in the SI System
gC F  ma
– Newton’s 2nd Law of Motion:
– SI:
– AE:
lb m  ft
lb f  s 2
kg  m
gC  1
N  s2
g C  32.174
• Conversion Factors
– Online: “The Foot Rule” website
• http://www.FootRule.com
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ChemE 260
March 29, 2005
Terminology or Nomenclature
• System: The material or volume that we are studying
– Systems have boundaries
– Closed Systems: Mass does not cross the boundary
– Open Systems: Mass does cross the boundary
• Properties
– Intensive vs. Extensive Properties
• Extensive properties depend on the size of the system, intensive
properties do not.
V
– Molar Properties: per mole.
Molar volume:
– Specific Properties: per kg or per lbm. Specific volume: Vˆ
• States
– The condition of a piece of matter or system as determined by its
intensive properties.
– If ANY intensive property is different, then the system is in a
different state.
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ChemE 260
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More Nomenclature
• Process
– A change in the state of a system
• Process Path
– The series of states that a system moves through on its
way from the initial state to the final state.
• Special Types of Processes
– Isobaric – constant pressure
– Isothermal - constant temperature
– Isochoric - constant volume
• Cycles
– Special process paths in which the initial state is the
same as the final state
– Thermodynamic cycles are a key topic in this course
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ChemE 260
March 29, 2005
Equilibrium
• A system is in equilibrium when no unbalanced
potentials or driving forces exist within the system
boundary.
–
–
–
–
Thermal: no temperature driving forces
Chemical: no chemical driving forces
Phase: no mass transfer driving forces
Mechanical: no unbalanced mechanical forces
• Quasi-Equilibrium Processes
– A process during which the system only deviates from
equilibrium by an infinitessimal amount.
– Every state along the process path is very nearly an
equilibrium state.
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ChemE 260
March 29, 2005
Pressure, Volume & Temperature
• Volume:
– SI: L, m3, mL=cm3
– AE: ft3
V
V
n
ˆ V
V
m

• Pressure: acts in all directions  to all surfaces
– SI: Pa, kPa, MPa, bar, atm
– AE: psia
– Absolute, Gage and Vacuum Pressures
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ChemE 260
March 29, 2005
m 1

ˆ
V V
Manometers
• Barometer Eqn:
g
P2  P1  f
h
gc
• Manometer Eqn:
Pin  Pout  f
• Differential
Manometer Eqn:
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ChemE 260
March 29, 2005
g
h
gc
Pup  Pdown   m  f 
g
h
gc
Temperature
• Thermometers and Thermocouples
• Temperature conversions are
straightforward
• T (oC) = T(K) and T (oF) = T(oR)
• Ideal Gas Thermometry
– Must be calibrated
– Tedious, but extremely accurate
– IG T-scale is identical to the Kelvin Scale !
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ChemE 260
March 29, 2005
Next Class
• Problem Solving Session
– Ch 1
– Homework #1, due Friday 4/1 !
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ChemE 260
March 29, 2005
Example #1
• Force Required to Accelerate a Rocket
– Calculate the force necesssary to accelerate a
20,000 lbm rocket vertically upward at a rate of
100 ft/s2.
– Ans.: Ftotal = 82,200 lbf
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ChemE 260
March 29, 2005
Example #2
• Relationships between Different Types of
Pressures
– Complete the following table if Patm = 100 kPa
and Hg = 13.6 H2O .
Pgage(kPa)
a.)
b.)
c.)
d.)
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ChemE 260
March 29, 2005
Pgage(kPa)
Pgage(kPa)
Pgage(kPa)
5
150
30
30
Example #2 - Answers
• Relationships between Different Types of
Pressures
– Complete the following table if Patm = 100 kPa
and Hg = 13.6 H2O .
Baratuci
ChemE 260
March 29, 2005
Pgage(kPa)
Pgage(kPa)
Pgage(kPa)
Pgage(kPa)
a.)
5
105
787
0.510
b.)
50.0
150
1120
5.10
c.)
-96.0
4.00
30
-9.79
d.)
294
394
2960
30
Example #3
• A horizontal 2 m diameter man-hole is located in
the bottom of a water tank as shown here.
Determine the extra upward force, F, that a man or
machine must exert on the man-hole cover to just
barely lift it.
– Ans.: F = 154 kN
5m
Baratuci
ChemE 260
March 29, 2005
F
Example #4
• Gravity is given as a function of altitude by
g = 9.81 - 3.32 x 10-6 h (m/s2), where h is
the altitude above sea level. An airplane is
traveling at 900 km/h at an elevation of 10
km. If its weight at sea level is 40 kN,
determine:
– Its kinetic energy
– Its potential energy relative to sea level
• Ans.: Ekin = 127 MJ
• Ans.: Ekin = 399 MJ
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ChemE 260
March 29, 2005