Download 4-Energy Analysis of Closed Systems

Islamic Azad
University,
Karaj Branch
Energy Analysis of Closed
Systems
Instructor:
Dr.
M.
Khosravy
1
So far!
•  We ve considered various forms of energy
–  Q (heat)
–  W (work)
–  E (Energy)
•  We haven t tried to relate them during a
process
•  They are related through the
•  FIRST LAW OF THERMODYNAMICS
•  (CONSEVATION OF ENERGY PRINCIPLE)
2
Recall!
•  Energy can be neither created nor
destroyed; it can only change forms ( a
rock falling off the cliff)
•  Consider a system undergoing a series of
adiabatic processes from a specified state
1 to another specified state 2.
•  Obviously the processes can not involve
any heat transfer but they may involve
several kinds of work interactions
3
Recall!
•  Experimentally observed: all adiabatic
processes between two specified states of
a closed system, the net work done is the
same regardless of the nature of the
closed system and the details of the
process.
•  This principle is called the first law
•  A major consequences of the first law is
the existence and the definition of the
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property total energy E
Recall!
•  The net work is the same for all adiabatic processes
of a closed system between two specified states,
•  The net work must depend on the end states of the
system only.
•  Therefore it must correspond to a change in property
of the system (i.e. the total energy)
•  Simply the change in the total energy during an
adiabatic process must be equal to the net work
done.
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First low of thermodynamics
for Closed Systems
! 
Reminder of a Closed System.
! Closed
system = Control mass
! It is defined as a quantity of
matter chosen for study.
! No mass can cross its boundary
but energy can.
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Experience 1: W= 0
! 
! 
If we transfer 5 kJ of heat to a potato,
its total energy will increase by 5 kJ.
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Experience 2: W=0
! 
! 
! 
Heat a water in a pan by transferring 15 KJ from
the range
There is 3 KJ losses
This means E2-E1= 12 KJ
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Experience 3: Q=0
! 
! 
! 
Heat an insulated room with electric heater.,
W= 5KJ (electric)
means E2-E1= 5KJ.
Replace the electric
heater
with a paddle wheel.
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Experience 4: Q=0
! 
! 
! 
If we do a 10 kJ of boundary
work on a system,
the system s internal energy
will increase by 10 kJ.
This is because (in the absence
of any heat transfer (Q = 0), the
entire boundary work will be
stored in the air as part of its
total energy.
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Experience 5: Qin, Qout, Win
! 
! 
! 
! 
! 
If we do 6 kJ of shaft
work on system and
Transfer 15 kJ of heat
in.
Loose 3 kJ of heat
out.
Doing 4 kJ of Work
out.
Then the system
internal energy will
increase by 14 kJ.
Wb= 4 kJ
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Before proceeding further, let us fix the
direction of Energy Transfer
•  Heat transferred in to the system is positive
"  Qin is +ve
•  Heat transferred out of the system is negative
"  Qout is -ve
•  Work done on the system increases energy of the
system.
"  Win is +ve
•  Work done by the system decreases energy of the
system.
"  Wout is -ve
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•  Based on the previous experimental
observations, the conservation of energy
principle may be expressed as:
& Total Energy
# & Total Energy
# & Change in the total #
$$
!! ' $$
!! = $$
!!
% Entering the System " % Leaving the System " % system energy
"
Ein " Eout = !Esys
On a rate basis
E! in " E! out = !E! sys
Rate of net energy transfer
by heat, work and mass
Rate of change in total
energy of the system
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Let us discuss the right hand side of the
first low equation, i. e. !E
!E = !U + !KE + !PE
Usually the KE and PE are small
!E = !U = m!u
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Let us turn to the Left hand side of the
1st low equation
Ein " Eout = !E
Eout
Ein
0
0
Qin + Win + Emass ,in " ( Qout + Wout + Emass ,out ) = !E
Emass=0 for closed system
Qin + Win " ( Qout + Wout ) = !E
If we rearrange, we get
Qin " Qout + Win " Wout = !E
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The First low of thermodynamics
Qin " Qout + Win " Wout = !E
Assuming Qin>Qout and Wout> Win
Qnet ,in " Wnet ,out = !E
usually we drop the subscripts, hence
Q " W = !E
Q! " W! = !E!
General Form (KJ)
per unit time (or on a Rate basis) KJ/s= Watt
q " w = !e
per unit mass basis (KJ/kg)
"q ! "w = de
differential form
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KEY CONCEPT
Q " W = !E
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Energy Change for a system
undergoing a cycle
•  Q-W ="U
•  But "U = U2-U1=0
(initial state and final
states are the same)
•  Hence : Q=W
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Example (4-1):
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 800kj. During
the cooling process, the fluid loses 500kj of heat,
and the paddle wheel does 100 kj of work on the
fluid. Determine the final internal energy of the
fluid.
•  Assumptions:
Tank is stationary and thus KE=PE=0.
•  Analysis:
Q-W=U2-U1
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Example (4-2):
Electric Heating of a Gas at Constant Pressure
•  A piston-cylinder device contains 25 g of
saturated water vapor that is maintained at
a constant pressure of 300 kPa. A
resistance heater within the cylinder is
turned on and passes a current of 0.2 A for
5 min from a 120-V source. At the same
time, a heat loss of 3.7 kJ occurs.
•  Show that for a closed system the
boundary work Wb and the change in
internal energy "U in the first-law
relation can be combined into one
term, "H, for a constant pressure
process.
•  Determine the final temperature of the
steam.
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Example (4-3):
Unrestrained Expansion
of water into an
Evacuated Tank
A rigid tank is divided into two equal parts
by a partition. Initially, one side of the tank
contains 5 kg of water at 200 kPa and 25oC,
and the other side is evacuated. The
partition is then removed, and the water
expands into the entire tank. The water is
allowed to exchange heat with its
surroundings until the temperature in the
tank returns to the initial value of 25oC.
Determine a) the volume of the tank, b) the
final pressure, and c) the heat transfer for
this process.
Expansion a against a vacuum
involves no work and thus no
energy transfer
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Example (4-6):
Heating of a Gas at
Constant Pressure
A piston-cylinder device initially contains air at 150 kPa
and 27oC. At this state, the piston is resting on a pair of
stops, as shown in the figure above, and the enclosed
volume is 400 L. The mass of the piston is such that a 350kPa pressure is required to move it. The air now heated
until its volume has doubled. Determine (a) the final
temperature, (b) the work done by the air, and c) the total
heat transferred to the air.
Answers: a) 1400 K, b) 140 kJ, c) 766.9 kJ
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1st Law of Thermodynamics
•  The first law of thermodynamics
is essentially an expression of the
conservation of energy principle.
Energy cannot be created or
destroyed. Energy can only
transform from one form to
another.
•  In closed system, energy can
cross the boundaries the form of
heat or work.
•  If the energy transfer across the
boundaries of a closed system is
due to a temperature difference, it
is heat; otherwise, it is work.
•  Work is energy expanded to lift a
weight.
Example 4-1
Complete the table given below for a closed system under going
a cycle.
Process
1-2
2-3
3-1
Cycle
Qnet kJ
+5
+20
-5
Wnet kJ
-5
+10
U2 – U1 kJ
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Heat Transfer
•  Heat is the form of energy that is transferred between two systems
or a system and its surrounding by virtue of a temperature
difference.
•  Heat is transferred in three ways:
•  Conduction is the transfer of energy from the more energetic
particles of a substance to the adjacent less energetic ones as a
result of interactions between the particles.
•  Convection is the mode of energy transfer between a solid surface
and the adjacent liquid or gas that is in motion, and it involves the
combined effects of conduction and fluid motion.
•  Radiation is the energy emitted by matter in the form of
electromagnetic waves (or photons) as a result of the changes in
the electronic configurations of the atoms or molecules.
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Heat Transfer
Heat transfer is
recognized at system
boundary
No heat transfer or adiabatic
process
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Modes of Heat Transfer
•  The three modes of heat transfer are
expressed as:
•  Fourier s Law, Newton s Law of Cooling and
Stefan-Boltzmann Law
•  Net heat transfer to a system is :
Q! net = ! Q! in " ! Q! out
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Work
•  If the energy crossing a system boundary is not heat, it
must be work.
•  Work is energy transfer associated with a force acting
through a distance
! !
! W = F • ds = Fds cos"
•  Energy in transition across system boundary and is done
by a system if the sole effect external to the boundary
could have been the raising of weight
•  Work is path dependent function
2
2
W = ! ! W = ! F ds " W # W
12
2
1
1
1
•  Therefore work is not a property of a system
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Work is Path Function
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Various Forms of Work
•  Electrical work:
(kJ)
•  Boundary work:
(kJ)
•  Gravitational work (=DPE):
(kJ)
•  Acceleration work (=DKE):
(kJ)
•  Shaft work:
(kJ)
•  Spring work:
(kJ)
Moving Boundary Work
Gas enclosed in piston-cylinder
device
!Wb = Fds = PAds = PdV
Expansion ~ dV => + dW
work done by the system on
surrounding
Compression ~ dV => - dW
work done on the system by
surrounding
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Moving Boundary Work
2
Wb = ! PdV
1
Work is dependent on process =>
To get W, relationship between P
and V required for the process
2
2
1
1
Area Under Curve = A = ! dA = ! PdV
The area under process curve
on P-V diagram is equal, to the
work done
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Cyclic Process Producing Net Work
Since boundary work is
path dependent, it is
possible for cyclic
process to produce net
work
Shaded region is the net
work produced in the
cyclic process
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Constant volume
•  If the volume is held constant, dV = 0, and the boundary
work equation becomes
P
1
2
V
P-V diagram for V = constant
•  If the working fluid is an ideal gas, what will happen to
the temperature of the gas during this constant volume
process?
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Isobaric (constant pressure)
Process
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Constant temperature, ideal gas
•  If the temperature of an ideal gas system is held
constant, then the equation of state provides the
pressure-volume relation.
P=
mRT
V
•  Then, the boundary work is
Note: The above equation is the result of applying the ideal
gas assumption for the equation of state. For real gases
undergoing an isothermal (constant temperature) process,
the integral in the boundary work equation would be done
numerically.
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Isothermal (constant temperature)
Process
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General Polytropic Process
•  For the polytropic process (Pvn = constant)
of real gases, the boundary work can be
expressed as:
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Some Common Process
Process
Constant pressure
Constant volume
Isothermal & ideal gas
Adiabatic & ideal gas
Exponent n
0
!
1
k = CP/CV
Here, k is the ratio of the specific heat at constant pressure CP to
specific heat at constant volume CV. The specific heats will be
discussed later.
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Polytropic Process
•  The boundary work done during the polytropic
process is found by substituting the pressurevolume relation into the boundary work equation.
The result is
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Polytropic Process
•  For an ideal gas under going a polytropic
process, the boundary work is
Notice that the results we obtained for an ideal gas undergoing a polytropic
process when n = 1 are identical to those for an ideal gas undergoing the
isothermal process.
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First Law of Thermodynamics
•  For all adiabatic process between 2 specified
states of a closed system, the net work done is
the same regardless of the nature of the closed
system and the details of the process.
•  This gives rise to another thermodynamics
property => Total Energy, E
•  1st Law ~ Change in Total Energy in an
adiabatic process must be equal to the net work
done
•  Basically statement of conservation of energy
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Specific Heats (C)
• 
Specific heat (C): Energy required to raise the temperature of 1kg of a substance by one degree It
is defined when there is no phase change i.e. only solid, only liquid or only vapor 2 types
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Formal Definitions of Cv and Cp
•  The amount of energy
needed to raise the
temperature of a unit of
mass of a substance by
one degree is called the
specific heat at constant
volume Cv for a constantvolume process and the
specific heat at constant
pressure Cp for a constant
pressure process. They are
defined as:
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Simple Substance
The thermodynamic state of a simple, homogeneous substance
is specified by giving any two independent, intensive properties.
Let's consider the internal energy to be a function of T and v
and the enthalpy to be a function of T and P as follows:
u = u (T , v)
and
The total differential of u is
h = h(T , P )
The total differential of h is
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Ideal Gas Cp and Cv
For an ideal gas: u=f(T)
u does not depend on 2 properties it depends only on T
u=f(T) only for ideal gas
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Properties for Ideal Gases
Pv = nRT
Pv = RT
PV = mRT
Requirements:
R
R=
M
Z !1
P ! Pc
T " Tc
The Ideal Gas Model:
u = u(T )
h = h(T ) = u (T ) + RT
du
dT
dh
c p (T ) =
dT
cv (T ) =
When specific
heats are
u2 !assumed
u1 = cv (T2 ! T1 )
constant
Table
h2 !A-20(E):
h1 = c p (T2 ! T1 )
Internal energy, enthalpy, and specific
heats of Ideal gases for ideal gas
H = U + P !V
h = u + P!v
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Specific Heats for Some Gases
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Internal energy, enthalpy, and
specific heats of Ideal gases
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Approximations for liquids and solids
•  Using Saturated
Liquid Data
( Compressed Liquid
Rule )
v ! vf
•  Using
Incompressible
Substance Model
c p = cv = c
h ! hf
u2 ! u1 = c(T2 ! T1 )
u ! uf
h2 ! h1 = c(T2 ! T1 ) + v ( P2 ! P1 )
s ! sf
h2 ! h1 " c(T2 ! T1 )
General System Energy Balance
•  The energy balance for any system undergoing any
process can be expressed as:
•  In rate form :
•  Substituting for heat transfer and work :
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Closed System 1st Law (Summary)
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Thermodynamics Solution Procedures
Identify the
system and
draw a sketch
of it.
List the given
information on
the sketch.
Check for
special
processes.
State any
assumptions.
Apply the
conservation
equations.
Draw a
process
diagram.
Determine the
required
properties
and
unknowns.
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