Transcript Reflection of Buddhism in Contemporary Cinema

ENGR 2213 Thermodynamics
F. C. Lai
School of Aerospace and Mechanical
Engineering
University of Oklahoma
Lesson from First Law
Second Law of Thermodynamics
The first law of thermodynamics requires that energy
be conserved during a process, but place no
restriction on the direction of a process.
Q
Q
Soda
4 ºC
Satisfying the first law does not guarantee that a
process will actually occur.
Second Law of Thermodynamics
The inadequacy of the first law to identify whether
a process can take place is remedied by introducing
the second law of thermodynamics.
A process will not occur unless it satisfies both the
first and second laws of thermodynamics.
The second law asserts that
1. Processes occur in a certain direction.
2. Energy has quality as well as quantity.
Second Law of Thermodynamics
Thermal Energy Reservoir
A hypothetical body with a relatively large thermal
energy capacity that can supply or absorb finite
amount of energy as heat without undergoing any
change in temperature.
Source
A reservoir that supplies energy in the form of heat
Sink
A reservoir that absorbs energy in the form of heat
Second Law of Thermodynamics
Heat Engines
Devices that are used to convert heat to work.
Characteristics of Heat Engines
1. They receive heat from a high-temperature source.
2. They convert part of this heat to work.
3. They reject the remaining waste heat to a lowtemperature sink.
4. They operate on a cycle.
Second Law of Thermodynamics
High-temperature Reservoir at TH
QH
W
HE
W = QH – QL
QL
Low-temperature Reservoir at TL
Second Law of Thermodynamics
Performance
Desired Output
Performance 
Re quired Input
Thermal Efficiency
W
QH  QL
QL


 1
< 100 %
QH
QH
QH
Automobile Engine 20%
Diesel Engine
30%
Gas Turbine
30%
Steam Power Plant 40%
Example 1
Heat is transferred to a heat engine from a furnace
at a rate of 80 MW. If the rate of waste heat
rejection to a nearby river is 50 MW, determine the
net power output and the thermal efficiency.
QH  80 MW, QL  50 MW
W  QH  QL  80  50  30 MW
W 30


 0.375
QH 80
Second Law of Thermodynamics
Kelvin-Planck Statement
It is impossible for any device that operates on a
cycle to receive heat from a single reservoir and
produce an equivalent amount of work.
No heat engine can have a thermal efficiency of
100%
The impossibility of having 100% efficiency heat
engine is not due to friction or other dissipative
effects.
Second Law of Thermodynamics
Refrigerators/Heat Pumps
Devices that are used to transfer heat from lowtemperature medium to high-temperature one.
Like heat engines, they are cyclic devices.
Refrigerators and heat pumps operate on the same
cycle but differ in their objectives.
Refrigerators – maintain the refrigerated space at a
low temperature.
Heat pumps – maintain the heated space at a high
temperature.
Second Law of Thermodynamics
High-temperature Reservoir at TH
QH
W
QL = QH - W
Ref
QL
Objective
Low-temperature Reservoir at TL
Second Law of Thermodynamics
Objective
High-temperature Reservoir at TH
QH
W
QH = W + QL
HP
QL
Low-temperature Reservoir at TL
Second Law of Thermodynamics
Performance
Desired Output
Performance 
Re quired Input
Coefficient of Performance (COP)
Refrigerators
Heat Pumps
QL
QL


W QH  QL
1
QH
QH

>1


W QH  QL 1  QL / QH
Second Law of Thermodynamics
Energy Efficient Rating (EER)
The amount of heat removed from the cooled space
in BTU’s for 1 Watt-hour of electricity consumed.
QL
Air-conditioners  
W
1 Wh = 3.412 BTU
EER = 3.412 COPR
Most air conditioners have an EER between 8 and 12.
Example 2
The food compartment of a refrigerator is
maintained at 4 ºC by removing heat from it at a
rate of 360 kJ/min. If the required power input is
2 kW, determine the COP and the rate of heat
discharged.
QL  360 kJ/ min  6 kW, W  2 kW
QL 6
COP 
 3
W 2
QH  W  QL  2  6  8 kW
Second Law of Thermodynamics
Clausius Statement
It is impossible to construct a device that operates
on a cycle and produce no effect other than the
transfer of heat from a low-temperature body to a
high-temperature body.
Equivalence of the two statements
A violation of one statement leads to the violation
of the other statement.
Equivalence of the Two Statements
High-temperature Reservoir at TH
QH
Net QOUT = QL
QH + QL
HE
Ref
HE + Ref
W = QH
QL
Low-temperature Reservoir at TL
Net QIN = QL
Equivalence of the Two Statements
High-temperature Reservoir at TH
QL
Net QIN = QH - QL
QH
Ref
HE + Ref
HE
W = QH – QL
QL
QL
Low-temperature Reservoir at TL
Net W = QH - QL