Transcript Chapter 2: First Law of Thermodynamics, Energy

PTT 201/4 THERMODYNAMICS
SEM 1 (2012/2013)
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Energy can exist in
lightforms:
numerous
Macroscopic forms of energy
light
Thermal
Mechanical
Kinetic
Potential
Electric
Magnetic
Chemical
Nuclear
Those related
light to the
molecular structure of a
system and the degree of the
molecular activity.
Their sum
constitutes
the total
energy, E
of a system
The total energy of a system
on a unit mass:
Elight
e
(kJ/kg)
m
Microscopic forms of energy
Related to the motion & the
influence of some external effects
such as :
gravity, magnetion, electricity &
surface tension
Internal energy, U
light
The sum of all
the microscopic
forms of energy.
Usually ignored.
Kinetic energy, KE
Potential energy, PE
The energy that a system
possesses light
as a result of its
motion relative to some
reference frame.
The energy that a system
possesses light
as a result of its
elevation in a gravitational
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field.
Kinetic energy, KE
Potential energy, PE
Velocity
Gravitational acceleration
PE  mgz
mV 2
KE 
(kJ)
light
2
(kJ)
Elevation of the center of
light
gravity of a system relative to
some arbitrarily selected
reference level
Kinetic energy per unit mass:
V2
ke 
(kJ/kg)
2
Kinetic energy per unit mass:
pe  gz (kJ/kg)
Total energy of a system
mV 2
E  U  KE  PE  U 
 mgz
2
light
(kJ)
Total energy of a system per unit mass:
V2
e  u  ke  pe  u 
 gz
2
(kJ/kg)
3
Flow of steam in a pipe
Energy flow rate
Mass flow rate
light

  ρV  Ac Vavg
m
(kg/s)
e
Em
light
(kJ/s @ kW)
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The various forms of
microscopic energies
light
that make
up sensible
energy.
The internal energy of a
system is the sum of all
light
forms of the
microscopic
energies
Sensible
& latent
energy
Thermal
Energy
Chemical
energy
Nuclear
energy
INTERNAL ENERGY
Sensible energy
The portion of the internal energy
light with the
of a system associated
kinetic energies of the molecules.
Latent energy
The internal light
energy associated
with the phase of a system.
Chemical energy
The internal light
energy associated
with the atomic bonds in a
molecule.
Nuclear energy
The tremendous amount of
lightwith the strong
energy associated
bonds within the nucleus of the
atom itself.
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Wind Energy
A site evaluated for a wind farm is observed to have steady winds at a speed of 8.5 m/s.
Determine the wind energy (a) per unit mass, (b) for a mass of 10 kg, and (c) for a flow
rate of 1154 kg/s for air.
a)
V 2 8.5 m/s 2  1 J/kg 
e  ke 

 36.1 J/kg

2 2 
2
2
1
m
/s


b)
E  me  (10 kg)(36.1 J/kg)  361 J
c)
1 kW 
 e  (1154 kg/s)(36.1 J/kg) 
E  m
  41.7 kW
 1000 J/s 
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The total energy of a system,
can be contained or stored in
a system, and thus can be
viewed as the static forms of
energy.
The forms of energy not stored in a system can
be viewed as:
the dynamic forms of energy @ energy
interactions.
are recognized at the system
boundary aslight
they cross it
Represent the energy gained or lost
by a systemlight
during a process.
The macroscopic kinetic energy is an organized form
of energy and is much more useful than the
disorganized microscopic kinetic energies of the
molecules.
The only two forms of energy interactions associated
with a closed system are :
1)
2)
heat transfer
work.
The difference is
An energy interaction is heat transfer if its
driving force is a temperature difference.
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Otherwise it is work.
Heat
The form of energy that is
transferred between
light two systems
(or a system and its
surroundings) by virtue of a
temperature difference.
Energy can cross the boundaries of
a closed system in the form of heat
and work.
Temperature difference is the driving
force for heat transfer. The larger the
light
temperature difference, the higher is
the rate of heat transfer.
Energy is recognized as heat
transfer onlylight
as it crosses the
system boundary.
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Heat transfer per
unit mass
lightprocess
Adiabatic
light
Q
q
(kJ/kg)
m
A process during which
light
there is no heat transfer
Amount of heat transfer
when heat transfer rate
changes with time
light
 dt
Q  Q
t2
(kJ)
The system is well insulated
so that only
a negligible
light
amount of heat can pass
through the boundary
t1
Both the system and
surroundings are at the same
temperature
and therefore
light
there is no driving force (temp
diff) for heat transfer
Amount of heat transfer when
heat transfer rate
is constant
light
 Δt
QQ
(kJ)
During an adiabatic process, a
system exchanges no heat
with its surroundings.
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•
Kinetic theory: Treats molecules as tiny balls
that are in motion and thus possess kinetic
energy.
•
Heat: The energy associated with the random
motion of atoms and molecules.
Heatlight
Transfer
Mechanisms
Conduction:
The transfer of energy from the
more energetic particles of a
light
substance to the adjacent less
energetic ones as a result of
interaction between particles.
Convection:
The transfer of energy
between a solid surface and
the adjacent
fluid that is in
light
motion, and it involves the
combined effects of
conduction and fluid motion.
Radiation:
The transfer of energy
due to the
emission of
light
electromagnetic waves
(or photons).
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Work: The energy transfer associated with a force acting through a distance.
– A rising piston, a rotating shaft, and an electric wire crossing the
system boundaries are all associated with work interactions
Work done per
unit mass
Power is the
work done per
unit time (kW)
Specifying the directions of
heat and work.
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•
The first law of thermodynamics (the conservation of energy principle)
provides a sound basis for studying the relationships among the various forms
of energy and energy interactions.
•
The first law states that energy can be neither created nor destroyed during
a process; it can only change forms.
Energy cannot
be created or
destroyed; it
can only
change forms.
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The net change (increase or decrease) in the total energy of the system during a
process is equal to the difference between the total energy entering and the total energy
leaving the system during that process.
(Qin – Qout) + Wsh = ∆Esystem
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Esystem
(Qin – Qout ) + Wsh = ∆U + ∆KE + ∆PE (Energy Balance for Closed Systems)
Internal, kinetic, and
potential energy changes
light
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Energy Balance for Closed Systems
Water flows over a waterfall 100 m in height. Take 1 kg of the water as the system, and
assume that it does not change energy with its surroundings.
(a) What is the potential energy of the water at the top of the falls with respect to the
base of the falls?
(b) What is the kinetic energy of the water just before it strikes bottom?
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The 1 kg of water exchanges no energy with the surroundings. Thus,
for each part of the process
∆Esystem = ∆U + ∆KE + ∆PE = 0
a) Determine the potential energy, ∆PE
∆PE = mg (z2 – z1)
=
1 kg
9.81 m
(100-0) m
s2
= 981 kg m2/s2 = 981 N.m = 981 J
b) Determine the kinetic energy, ∆KE
∆U + ∆KE + ∆PE = 0
During the free fall of the water no mechanism exists for conversion
of potential or kinetic energy into internal energy. Thus, ∆U must be
zero.
= 0 + ∆KE + ∆PE = 0
= 0 + (KE2 – KE1) + (PE2 – PE1) = 0
Let KE1 and PE2 = 0
Thus, we get KE2 = 981 J
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Cooling of a Hot Fluid in a Tank
A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel.
Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid
loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine
the final internal energy of the fluid. Neglect the energy stored in the paddle wheel.
Ans:
U2=400 kJ
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Ein and Eout
•
•
•
Heat transfer
Work transfer
Mass flow
Energy Balanced for Open Systems:
SI unit: (Qin – Qout) + Wsh = ∆H + ∆u2+ g ∆z (kJ/kg)
2
English system unit: (Qin – Qout) + Wsh = ∆H + ∆u2 + g ∆z (Btu/Ibm)
2gc gc
The energy
content of a
control volume can
be changed by
mass flow as well
as heat and work
interactions.
Energy balance for cycle
When ∆E =0, Qnet=Wnet
Wnet,out = Qnet,in
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Energy Balance for Open Systems
Air at 1 bar and 25˚C enters a compressor at low velocity, discharges at 3 bar, and
enters a nozzle in which it expands to a final velocity of 600 m/s at the initial
conditions of pressure and temperature. If the work of compression is 240 kJ per
kilogram of air, how much heat must be removed during compression?
Write energy balance for open system:
Q + Wsh = ∆H + ∆u2+ g ∆z (kJ/kg)
2
Assumptions:
1. No enthalpy change, ∆H, of air due to the air returns to its initial temperature and
pressure.
2. The potential energy, ∆PE, is presumed negligible due to no change in elevation
involved
3. Neglecting the initial kinetic energy; u12
2
Ans:
Q= -60 kJ/kg
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EXAMPLE 2-11
EXAMPLE 2-12
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THANK YOU..
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