THE SECOND LAW OF THERMODYNAMICS For Mechanical and Industrial Engineerig
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Transcript of THE SECOND LAW OF THERMODYNAMICS For Mechanical and Industrial Engineerig
THE SECOND LAW OF
THERMODYNAMICS
1 By Meng Chamnan GIM
Lesson 5:
Direction of
Spontaneous
Process
By Meng Chamnan 2 GIM
Usefulness of the Second Law
1. Predicting the direction of processes.
2. Establishing conditions for equilibrium.
3. Determining the best theoretical performance of cycles, engines, and other devices.
4. Evaluating quantitatively the factors that preclude the attainment of the best theoretical performance level.
Additional uses of the second law include its roles in
5. Defining a temperature scale independent of the properties of any thermometric substance.
6. Developing means for evaluating properties such as u and h in terms of properties that are more readily obtained experimentally.
By Meng Chamnan 3 GIM
Clausius Statement of the 2nd Law
• It is impossible for any system to operate in such a
way that the sole result would be an energy transfer
by heat from a cooler to a hotter body.
By Meng Chamnan 4 GIM
By Meng Chamnan GIM 5
Example1:
A heat pump receives energy by heat transfer
from the outside air at 0oC and discharges
energy by heat transfer to a dwelling at 20oC.
Is this in violation of the Clausius statement of
the second law of thermodynamics? Explain.
Kelvin–Planck Statement of the 2nd Law
• It is impossible for any system
to operate in a thermodynamic
cycle and deliver a net amount
of energy by work to its
surroundings while receiving
energy by heat transfer from a
single thermal reservoir
By Meng Chamnan 6 GIM
Kelvin-Planck: Single Reservoir
0W
By Meng Chamnan 7 GIM
Kelvin-Planck: Double Reservoir
0W
By Meng Chamnan 8 GIM
By Meng Chamnan GIM 9
Example2:
Air as an ideal gas expands isothermally at 20oC
from a volume of 1 m3 to 2 m3. During this process
there is heat transfer to the air from the
surrounding atmosphere, modeled as a thermal
reservoir, and the air does work. Evaluate the
work and heat transfer for the process, in kJ/kg. Is
this process in violation of the second law of
thermodynamics? Explain.
Reversible and Irreversible
• A process is reversible if both the system
and surroundings can be returned to their
initial states
• A process is called irreversible if the system
and all parts of its surroundings cannot be
exactly restored to their respective initial
states after the process has occurred.
By Meng Chamnan 10 GIM
Irreversibilities
1. Heat transfer through a finite temperature difference
2. Unrestrained expansion of a gas or liquid to a lower pressure
3. Spontaneous chemical reaction
4. Spontaneous mixing of matter at different compositions or states
5. Friction—sliding friction as well as friction in the flow of fluids
6. Electric current flow through a resistance
7. Magnetization or polarization with hysteresis
8. Inelastic deformation
By Meng Chamnan 11 GIM
Internally Reversible Processes
• An internally reversible process is one in
which there are no irreversibilities within the
system.
• Internally reversible process consists of a
series of equilibrium states: It is a
quasiequilibrium process.
By Meng Chamnan 12 GIM
By Meng Chamnan GIM 13
Example3:
Methane gas within a piston–cylinder assembly is
compressed in a quasiequilibrium process. Is this
process internally reversible? Is this process
reversible?
Power Cycles Interacting with Two
Reservoirs
By Meng Chamnan 14 GIM
1 1cycle C
H H
W Q
Q Q
thermal efficiency of the cycle
The Second Law Corollary: Carnot Corollary
• The thermal efficiency of an irreversible power cycle is
always less than the thermal efficiency of a reversible power
cycle when each operates between the same two thermal
reservoirs.
• All reversible power cycles operating between the same two
thermal reservoirs have the same thermal efficiency.
By Meng Chamnan 15 GIM
By Meng Chamnan GIM 16
Example 4:
The data listed below are claimed for a power cycle
operating between reservoirs at 727 and 127oC.
For each case, determine if any principles of
thermodynamics would be violated.
(a) QH = 600 kJ, Wcycle = 200 kJ, QC = 400 kJ.
(b) QH = 400 kJ, Wcycle = 240 kJ, QC = 160 kJ.
(c) QH = 400 kJ, Wcycle = 210 kJ, QC = 180 kJ.
By Meng Chamnan GIM 17
Example 5:
A power cycle operating between two reservoirs
receives energy QH by heat transfer from a hot
reservoir at TH = 2000 K and rejects energy QC by
heat transfer to a cold reservoir at TC = 400 K. For
each of the following cases determine whether the
cycle operates reversibly, irreversibly, or is
impossible:
(a) QH = 1200 kJ, Wcycle = 1020 kJ.
(b) QH = 1200 kJ, QC = 240 kJ.
(c) Wcycle = 1400 kJ, QC = 600 kJ.
(d) η = 40%.
By Meng Chamnan GIM 18
Example 6:
An inventor claims to have developed a power cycle
capable of delivering a net work output of 410 kJ for
an energy input by heat transfer of 1000 kJ. The
system undergoing the cycle receives the heat
transfer from hot gases at a temperature of 500 K and
discharges energy by heat transfer to the atmosphere
at 300 K. Evaluate this claim.
Refrigeration and Heat Pump Cycles
Interacting with Two Reservoirs
By Meng Chamnan 19 GIM
coefficient of performance(COP)
for a refrigeration cycle
coefficient of performance for
a heat pump cycle
C C
cycle H C
Q Q
W Q Q
H H
cycle H C
Q Q
W Q Q
Corollaries for Refrigeration and Heat Pump
Cycles
• The coefficient of performance of an irreversible
refrigeration cycle is always less than the coefficient
of performance of a reversible refrigeration cycle
when each operates between the same two thermal
reservoirs.
• All reversible refrigeration cycles operating between
the same two thermal reservoirs have the same
coefficient of performance
By Meng Chamnan 20 GIM
Kelvin Temperature Scale
By Meng Chamnan 21 GIM
C C
revH Hcycle
Q T
Q T
where “rev cycle” emphasizes that the expression
applies only to systems undergoing reversible cycles
while operating between thermal reservoirs at TC and TH
Performance of Thermodynamic Cycle
By Meng Chamnan 22 GIM
max 1 C
H
T
T Reversible thermal efficiency for
power cycle “Carnot Efficiency”
Reversible COP for refrigeration cycle
maxC
H C
T
T T
Reversible COP for heat pump cycle
maxH
H C
T
T T
By Meng Chamnan GIM 23
Example 7: By steadily circulating a refrigerant at low
temperature through passages in the walls of the freezer
compartment, a refrigerator maintains the freezer
compartment at -5oC when the air surrounding the refrigerator
is at 22oC.
The rate of heat transfer from the
freezer compartment to the refrigerant
is 8000 kJ/h and the power input
required to operate the refrigerator is
3200 kJ/h. Determine the coefficient of
performance of the refrigerator and
compare with the coefficient of
performance of a reversible
refrigeration cycle operating between
reservoirs at the same two
temperatures.
By Meng Chamnan GIM 24
Example 8: The refrigerator shown in
the below picture operates at steady
state with a coefficient of performance
of 4.5 and a power input of 0.8 kW.
Energy is rejected from the refrigerator
to the surroundings at 20oC by heat
transfer from metal coils attached to
the back. Determine
(a) the rate energy is rejected, in kW.
(b) the lowest theoretical temperature
inside the refrigerator, K.
By Meng Chamnan GIM 25
Example 9:
A dwelling requires 6 ×105 Btu per day to maintain its
temperature at 70oF when the outside temperature is 32oF.
(a) If an electric heat pump is used to supply this energy,
determine the minimum theoretical work input for one day of
operation, in Btu/day.
(b) Evaluating electricity at 8 cents per kW·h, determine the
minimum theoretical cost to operate the heat pump, in $/day.
Carnot Cycle
By Meng Chamnan 26 GIM
Carnot cycle introduced in this section provides a specific
example of a reversible power cycle operating between
two thermal reservoirs