First Law of Thermodynamics
The first law of thermodynamics is often called the
Law of Conservation of Energy
. This law
suggests that
energy
can be transferred from one
system
to another in many forms. Also, it
cannot
be
created
or
destroyed
. Thus, the total amount of energy available in the Universe is
constant. Einstein's famous equation
(written
below
) describes the relationship between energy
and
matter
:
E = mc
2
In the equation above,
energy
(
E
) is equal to
matter
(
m
) times the square of a constant (
c
).
Einstein suggested that ener
gy and matter are interchangeable. His equation also suggests that
the quantity of energy and matter in the Universe is fixed
Corollaries of First Law
Corollary 1: System Executing a Process
For a system executing a process, change in stored energy of the
system is given as
ΔE = δQ
–
δW
Where E = energy stored in the system
Corollary 2: Isolated system
For an isolated system,δQ = 0 and δW = 0
Hence ΔE = 0 or E = constant
Thus, the energy of an isolated system remains unchanged.
Therefore, the first law of
thermodynamics may also be stated as follows, “Heat and work are
mutually convertible but since energy can neither be created nor destroyed, the total energy
associated with an energy conversion remains constant”.
Corollary 3: Perpetual Motion Machine of F
irst Kind (PMM

1)
Definition:

It is an imaginary device which produces a continuous supply of work without
absorbing
any energy from the surrounding or from the system. Such a machine in effect creates energy
from
nothing and violates the first law of th
ermodynamics.
It is impossible to construct a perpetual motion machine of first kind. The PMM

1 violates the
first law.
As per the law of conservation of energy, no engine can produce mechanical work continuously
without
some
other form of energy disappearing simultaneously
First Law of Thermodynamics for a Control Mass Undergoing a Cycle
Quick
The first law of thermodynamics states that during any
cycle
a system (control mass)
undergoes, the cyclic integral of the
heat
is
proportional
to the cyclic
integral of the
work
.
Equation
(Eq2)
∮
δQ
=
=
=
∮
δW
=
basic
statement of the first law of thermodynamics
Nomenclature
∮
δ
n
=
the=cyclic=integ牡l=of=the=heat
=
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=
∮
δW
=
the=cyclic=integ牡l=of=the=wo牫,=牥p牥sents=the=net=wo牫=du物ng=the=cycle
=
g
=
p牯po牴ionality
=
facto爠that=
depends=on=the=units=used=fo爠wo牫=and=heat
=
=
=
aetails
=
=
=
Conside爠as=a=cont牯l=mass=the=gas=in=the=containe爠as=shownW
=
=
=
=
=
=
iet=this=system=go=th牯ugh=a
=
cycle
=
that=is=made=up=of=two
=
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esK=fn=the=fi牳t=p牯cess=wo牫=is=
done=on=the=system=by=the=paddle=that=tu牮s=as=the=weight=is=lowe牥dK=iet=the=system=then=
牥tu牮=to=its=initial=state=by=t牡nsfe牲rng=heat=fr
om=the=system=until=the=cycle=has=been=completedK
=
=
=
l物ginally,
=
wo牫
=
was=measu牥d=in=mechanical=units=of
=
fo牣
e
=
times=distance,=such=as=foot=pounds=
fo牣e=o爠joules,=and=heat=was=measu牥d=in=the牭al=units,=such=as=the=䉲Btish=the牭al=unit=⡂Eu⤠o爠
the=calo物eK=Measu牥ments=of=wo牫=and=heat=we牥=made=du物ng=a=cycle=fo爠a=wide=va物ety=of=
systems=and=fo爠va物ous=amoun
ts=of=wo牫=and=heatK=then=the=amounts=of=wo牫=and=heat=we牥=
compa牥d,=it=was=found=that=they=we牥=always=p牯po牴ionalK=卵ch=obse牶ations=led=to=the=
fo牭ulation=of=the=fi牳t=law=of=the牭odynamics,=which=in=equation=fo牭=is=w物ttenW
=
=
⡅E1F
††=
=
g
=
∮
δQ
=
=
=
∮
δW
=
………………………………………………………………………………………………………………
…………………………………………
Enthalpy
=
internal energy of the system + pressure * volume.
Internal energy
= sum of all energies in the system
Internal energy
= kinetic energy+ potential energy +nuclear energy+ chemical
energy .
………………………………………………………………………………………………………………
………
Compressible substances
= substances which can be compressed.
Incompressible substances
are those which cannot be compressed.
Difference between
Internal energy and enthalpy in incompressible and
com
pressible fluids
Increasing
T
=temperature, increases
molecular vibrations, rotation and translation
Therefore,
U
= internal energy
increases as well.
Changes in
P
=pressure
do not affect
U
because the
molar volume
of an
incompressible
substance
is not affected by changes in
pressure
.
Consequently, there is
no change
in the extent or nature of molecular interactions and
therefore
U
does not change
e
ither.
OK, so then it shouldn’t be a surprise that U increases strongly as the temperature increases,
right ?
The molecules substances at higher temperatures are in a higher energy state and as a result
their molecules are vibrating, rotating and translating more
energetically
Specific Heat and Heat Capacity
Specific heat is another physical property of matter. All matter has a temperature
associated with it. The temperature of matter is a direct measure of the motion of the
molecules: The greater the motion
the higher the temperature:
Motion requires energy: The more energy matter has the higher temperature it will also
have. Typicall this energy is supplied by heat. Heat loss or gain by matter is equivalent
energy loss or gain.
With the observation above u
nderstood we con now ask the following question: by how
much will the temperature of an object increase or decrease by the gain or loss of heat
energy? The answer is given by the specific heat (S) of the object. The specific heat of
an object is defined in
the following way: Take an object of mass m, put in x amount of
heat and carefully note the temperature rise, then S is given by
In this definition mass is usually in either grams or kilograms and temperatture is either
in kelvin or degres Celcius. Note
that the specific heat is "per unit mass". Thus, the
specific heat of a gallon of milk is equal to the specific heat of a quart of milk. A related
quantity is called the heat capacity (C). of an object. The relation between S and C is C =
(mass of obect)
x (specific heat of object). A table of some common specific heats and
heat capacities is given below:
Some common specific
heats and heat capacities:
Substance
S
(J/g0C)
C
(J/0C)
for
100 g
Air
1.01
101
Aluminum
0.902
90.2
Copper
0.385
38.5
Gold
0.129
12.9
Iron
0.450
45.0
Mercury
0.140
14.0
NaCl
0.864
86.4
Ice
2..03
203
Water
4.179
41.79
Consider the specific heat of copper ,
0.385 J/g 0C. What this means is that it takes
0.385 Joules of heat to raise 1 gram of copper 1 degree celcius. Thus, if we take 1 gram
of copper at 25 0C and add 1 Joule of heat to it, we will find that the temperature of the
copper will have risen to 26
0C. We can then ask: How much heat wil it take to raise by
1 0C 2g of copper?. Clearly the answer is 0.385 J for each gram or 2x0.385 J = 0.770 J.
What about a pound of copper? A simple way of dealing with different masses of matter
is to dtermine the hea
t capacity C as defined above. Note that C depends upon the size
of the object as opposed to S that does not.
We are not in position to do some calculations with S and C.
Example 1
: How much energy does it take to raise the temperature of 50 g of copper by
10 0C?
Example 2
: If we add 30 J of heat to 10 g of aluminum, by how much will its temperature
increase?
Thus, if the initial temperture of the aluminum was 20 0C then after the heat is added the
temperature will be 28.3 0C.
Representation of first
law of thermodynamics as rate equation.
2
.
5
Control volume form of the conservation laws
The thermodynamic laws (as well as Newton's laws) are for a system, a specific quantity of
matter. More often, in propulsion and power problems, we are interested in
what happens in a
fixed volume, for example a rocket motor or a jet engine through which mass is flowing at a
certain
rate
. We may also be interested in the
rates
of heat and work into and out of a system.
For this reason, the control volume form of the sy
stem laws is of great importance. A schematic
of the difference is shown in Figure
2.8
. Rather than focus on a particle of mass which moves
through the engine, it is more convenient to focus on the volume occupied by the engine. This
requires us to use the control volume form of the thermodynamic laws, developed below.
Figure 2.8:
Control volume and system for flow through a propulsion
device
2
.
5
.
1
Conservation of mass
For the control volume shown, the rate of change of mass inside the volume is given by the
difference between the mass flow rate in and the mass flow rate out. For a single flow coming in
and a single flow coming out this
is
If the mass inside the control volume changes with time it is because some mass is added or
some is taken out. In the special case of a steady flow,
, therefore
Figure 2.9:
A control volume used to track mass flows
2
.
5
.
2
Conservation of energy
The first law of thermodynamics can be written as a rate equation:
where
To derive the first law as a rate equation for a
control volume
we proceed as with the mass
conservation equation. The physical idea is
that any rate of change of energy in the control
volume must be caused by the rates of energy flow into or out of the volume. The heat transfer
and the work are already included and the only other contribution must be associated with the
mass flow in and o
ut, which carries energy with it. Figure
2.10
shows two schematics of this
idea. The desired form of the equation will be
[Simple]
[More
General]
Figure 2.10:
Schematic diagrams illustrating terms in the energy equation for a simple and a
more general control volume
The fluid that enters or leaves has an amount of energy per unit mass given by
where
is the fluid
velocity relative to some coordinate system, and we have neglected
chemical energy. In addition, whenever fluid enters or leaves a control volume there is a work
term associated with the entry or exit. We saw this in Section
2.3
, example 1, and the present
derivation is essentially an application of the ideas presented there. Flow exiting at station ``e''
must push back the surrounding fluid, doing work
on it. Flow entering the volume at station ``i'' is
pushed on by, and receives work from the surrounding air. The rate of flow work at exit is given
by the product of the pressure times the exit area times the rate at which the external flow is
``pushed b
ack.'' The latter, however, is equal to the volume per unit mass times the rate of mass
flow. Put another way, in a time
, the work done on the surroundings by the flow at the exit
station is
The net rate of flow work is
Including all possible energy
flows (heat, shaft work, shear work, piston work, etc.), the first law
can then be written as:
where
includes the sign associated with the energy flow. If heat is added or work is
done
on
the system then the sign is positive, if work or heat are
extracted
from
the system then
the sign is negative. NOTE: this is consistent with
, where
is the work
done
by
the system on the environment, thus work is flowing out of the system.
We can then combine the specific internal energy term,
, in
and
the specific flow work
term,
, to make the enthalpy appear:
Thus, the first law can be written as:
For most of the applications in this course, there will be no shear work and no piston work.
Hence, the first law for a control volume will be most
often used as:
(
2
..
10
)
Note how our use of enthalpy has simplified the rate of work term. In writing the control volume
form of the equation we have assumed only one entering and one leaving stream, but this could
be generalized to any number of inlet
and exit streams.
In the special case of a steady

state flow,
Applying this to Equation
2.10
produces a form of the ``Steady
Flow Energy Equation'' (SFEE),
(
2
..
11
)
which has units of Joules per second. We could also divide by the mass flow to produce
which
has units of Joules per second per kilogram. For problems of interest in aerospace
applications the velocities are high and the term that is associated with changes in the elevation
is small. From now on, we will neglect the
terms unless explicitly stat
ed.
Free or Unresisted Expansion
Consider two vessels
A
and
B
which are connected to each other by a pipe and a valve.
Vessel
A
is initially filled with a fluid at a certain pressure and
B
is completely evacuated. By
opening the valve, the fluid in
the vessel
A
will expand until it fills both vessels. This process is
known as free or unresisted expansion. It is an
irreversible
process because it needs external
work to be done to restore
the fluid to its initial condition. Consider a
system
, consisting of both
vessels which is perfectly thermally insulated. Apply
the first law of thermodynamics
to the
system, i.e.
Q + W = U2

U1
where indices 1 and 2 represent initial and final states.
Q = 0
, because the thermal insulation will not allow any heat transfer between the system and
the surroundings.
W=0
because
the boundaries of the system are not moved. The result will
then be:
U2=U1
The free expansion process is
adiabatic
but irreversible. If the
working fluid
is a
perfect gas
,
then
U2=U1
is equivalent to
T2=T
1
open systems
.
For
closed systems
we considered a
control
mass
.
The
mass
remains
constant
throughout the
process
.
Thus
.
Closed
System
For
open systems
we consider a
control volume
.
The
mass
does
not necessarily
remain
constant
.
Thus, we must keep track of the
mass flow rate
in
,
, and the
mass flow rate
out
,
, of
the
control volume
.
Recall that mass crosses the boundary of open systems.
In an integral mass balance, mass crossing a system boundary is expressed as an
amount.
For example during a process 32 kg of water entered a system.
In this course,
however, a differential mass balance is more common.
A differential mass balance is used when mass flows continuously across the system
boundary.
For example, watered entered that system at a RATE of 12 kg/min during a process.
The symbol for a mass flow r
ate is an m with a dot over it and I will sometimes call this
symbol m

dot
Systems can also have mass leaving the system during a process as well.
In that case, we can use the subscripts in and out on the symbol m

dot, so we don’t get
the two streams confu
sed.
Internal energy in common processes.
Adiabatic process= no heat is transferred
∆
U=∆W (dU=

dW)
Isobaric process
–
constant volume process
∆
U=∆Q (dU=dQ)
Isothermal process

constant temperature.
∆
U=∆Q

∆W
(dU=dQ

dW )
(for an ideal gas dU = 0)
Isobaric process
–
constant pressure
∆
U=∆Q

∆W (dU=dQ

dW )
Free expansion
–
adiabatic with no work done.
∆
U=0 (dU=o)
Analysis of flow/no flow process for a control mass underg
oing a constant
volume, constant temperature, and constant pressure, adiabatic and polytrophic
processes
.
Work done in an
Isothermal process
= p(v) = nRT/v
dW= P(v).dV
w=
∫
∫
∫
= nRT log V
∫
Work done in an isobaric process
P(v)= Pa
W
∫
∫
∫
Work done in an isochoric process
W
∫
∫
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