First law of thermodynamics

1. First Law of
Thermodynamics
Dr. Rohit Singh Lather

2. The First Law of Thermodynamics
ü The quantity (Q – W) is the same for all processes
ü It depends only on the initial and final states of the system
ü Does not depend at all on how the system gets from one to the other
ü This is simply conservation of energy
(Q is the heat absorbed and W is the work done by the system)
The internal energy E of a system tends to increase, if energy is added as heat Q and tends to
decrease if energy is lost as work W done by system
dE = dQ – dW ( first law)

3. Internal Energy
• Internal energy (u): that portion of total energy E which is not kinetic or potential
energy. It includes thermal, chemical, electric, magnetic, and other forms of energy
• Change in the specific internal energy
du = CdT
• In case of gases internal energy is given by u = CvdT
• Specific heat changes with temperature is given by
C = Co (a+bT), a and b are constants and Co is specific heat at 0oC
• The total energy in the mass m is the sum of internal energy as well as PE and KE in the mass
E = U + PE + KE = m .u + mg. Z +
!
"
m V2
For unit mass E = em = u +
$
%
+
!
"
V2 (p = 𝜌gZ)

4. Enthalpy
1Q2 =U2 − U1 + P2V2 − P1V1
= (U2 + P2V2) − (U1 + P1V1)
u = h - pv
The enthalpy is especially valuable for analyzing isobaric processes
Enthalpy h = u + pvs
• The heat transfer in a constant-pressure quasi-equilibrium process is equal to the change in
enthalpy, which includes both the change in internal energy and the work for this particular
process
• This is by no means a general result
• It is valid for this special case only because the work done during the process is equal to the
difference in the PV product for the final and initial states
• This would not be true if the pressure had not remained constant during the process

5. Source: http://www4.ncsu.edu/~kimler/hi322/Rumford-expt.gif
Sir Benjamin Thompson
Count Rumford • Thompson’s theory of heat was demonstrated by a test tube full
of water within wooden paddles
• Water boiled due to friction
• The heat of friction is unlimited

6. The Joules Experiment
Rise in
temperature
One calorie corresponds to the amount
of heat that is needed to get one gram
of water from 14.5 C to 15.5 C
1 Cal = 4.1840 J
𝑱 ∮ 𝝏𝑸 = ∮ 𝝏𝑾
A paddle wheel turns
in liquid water
proportionality constant - mechanical equivalent of heat
W
W
H
Insulating walls
prevent heat transfer from the
enclosed water to the surroundings
As the weight falls at
constant speed, they
turn a paddle wheel,
which does work on
water
If friction in mechanism is negligible, the work done by the paddle wheel on the water equals the
change of potential energy of the weights

7. Joule showed that the same temperature rise could be obtained using an electrical resistor heated
by an electric current
Work can be transformed into heat
Heat and work are of the same nature and constitute different forms of energy

8. First Law of Thermodynamics
“If a system executes a cycle transferring work and heat through its boundaries,
then net work transfer is equal to net heat transfer”
E2 – E1 = Q - W
dE = Q - W
dE = E2 – E1 = ∫ 𝑑(𝑄 − 𝑊)
"
!
For a closed system
dU = U2 – U1 = ∫ 𝑑 𝑄 − 𝑊
"
!
dU = U2 – U1 = Q - W
“During any cycle, the cyclic integral of heat added to a system is proportional to the cyclic
integral of work done by the system”
∮ 𝝏𝑸 = ∮ 𝝏𝑾

9. Adiabatic processes
Process that occurs so rapidly or occurs in a system that is so well insulated that no transfer of
thermal energy occurs between the system and its environment
Free expansion
Adiabatic processes in which no heat transfer occurs between the system and its environment and no
work is done on or by the system
Cyclical Processes
Processes in which, after certain interchanges of heat and work, the system is restored to its initial
state. No intrinsic property of the system—including its internal energy—can possibly change
Constant-volume processes
If the volume of a system (such as a gas) is held constant, so that system can do no work

10. First Law Corollaries
Corollary 3
“The perpetual motion machine (PMM) of first Kind is impossible”
Corollary 2
“The internal energy of a closed system remains unchanged if the system is isolated from its
surroundings”
Corollary 1
“There exists a property of a closed system such that the change in its value is equal to the
difference between heat supplied and work done during the change of state”

11. Corollary 1
“There exists a property of a closed system such that the change in its value is equal to the
difference between heat supplied and work done during the change of state”
∮ 𝜕𝑄 = ∮ 𝜕𝑊
∫ 𝜕𝑄
"
! A + ∫ 𝜕𝑄
"
! B = ∫ 𝜕𝑊
"
! A + ∫ 𝜕𝑊
"
! B
∫ 𝜕𝑄
"
! C +∫ 𝜕𝑄
"
! B = ∫ 𝜕𝑊
"
! C + ∫ 𝜕𝑊
"
! B
∫ 𝜕𝑄
"
! A - ∫ 𝜕𝑄
"
! C =∫ 𝜕𝑊
"
! A - ∫ 𝜕𝑊
"
! C
Subtracting and rearranging
∫ (𝜕𝑄
"
!
- 𝜕𝑊) 𝐴 = ∫ (𝜕𝑄
"
!
- 𝜕𝑊) 𝐶
Volume
Pressure
1
2
C
B
A
A, B, & C are arbitrary processes
between state 1 and state 2

12. 𝝏𝑸 − 𝝏𝑾 𝒊𝒔 𝒔𝒂𝒎𝒆 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒕𝒉𝒆 𝒑𝒓𝒐𝒄𝒆𝒔𝒔𝒆𝒔 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒔𝒕𝒂𝒕𝒆 𝟏 𝒂𝒏𝒅 𝒔𝒕𝒂𝒕𝒆 𝟐
Therefore 𝝏𝑸 − 𝝏𝑾 depends only on the initial and final states and not the path followed
between the two states
dE = 𝜕𝑄 − 𝜕𝑊
E2 – E1 = 1Q2 – 1W2
• The physical significance of the property E is that it represents all the energy of the system in the given
state
• In thermodynamics, it is convenient to consider the bulk kinetic and potential energy separately and then to
consider all the other energy of the control mass in a single property that we call the internal energy and to
which we give the symbol U
E = Internal energy + kinetic energy + potential energy
E = U + KE + PE

13. The perpetual motion machine (PMM)
HYPOTHETICAL SYSTEM
produce useful work
indefinitely
produce more work or
energy than they
consume
There is undisputed scientific consensus that “Perpetual motion would violate the Laws of Thermodynamics”

14. What is “Perpetual Motion”?
• Describes a theoretical machine that, without any losses due to friction or other forms of
dissipation of energy, would continue to operate indefinitely at the same rate without any external
energy being applied to it
• Machines which comply with both the 1st & 2nd Laws of Thermodynamics but access energy from
obscure sources are sometimes referred to as “Perpetual Motion” machines

15. Perpetual Machine of First Kind (PMFK)
• A “perpetual motion” machine of the first kind produces work without the input of energy
It thus violates the 1st Law of Thermodynamics: the Law of conservation of energy
• First law states “that the total amount of energy in an isolated system remains constant over”
• A consequence of this law is that energy can neither be created nor destroyed: it can only be
transformed from one state to another
• So, It is clearly impossible for a machine to do the work infinitely without consuming energy
Examples of the 1st Kind of “Perpetual Motion” Machine
The Overbalanced Wheel The Float Belt The Capillary Bowl

16. Laws of Conservation
Law of Conservation of Mass: Mass can neither be created nor be destroyed, but may be converted
from one form to another form
A1V1 𝜌1 = A2V2 𝜌2 (One dimensional continuity equation)
Law of Conservation of Momentum: If no external force acts on a system, linear momentum is
conserved in both direction and magnitude
m1V1 + m2V2 = (m1+m2)V
Law of Conservation of Energy: Energy can neither be created nor be destroyed

17. Energy and System
Change in Total Energy
of the System
Total Energy Entering
the System
Total Energy Leaving
the System
= -
∆E = Ein - Eout
∆Esystem = Efinal state – Einitial state
Mechanism of Energy Transfer
Work TransferMass TransferHeat Transfer
Energy balance in rate form, Ein – Eout =
𝒅𝑬 𝒔𝒚𝒔𝒕𝒆𝒎
𝒅𝒕
0 for steady state
For a closed system undergoing a cycle , ∆Esystem = E2 – E1 = 0 -> Ein = Eout

18. • The conservation of mass and the conservation of energy principles for open systems or control
volumes apply to systems having mass crossing the system boundary or control surface
• Thermodynamic processes involving control volumes can be considered in two groups: steady-flow
processes and unsteady-flow processes
• Defining a Steady Flow Process: A process during which the fluid flows steadily through the
control volume (CV)
- Flow process à fluid flows through CV
- Steady à not changing with time
- During a steady flow process:
- Conditions (fluid properties, flow velocity, elevation) at any fixed point within the CV are unchanging with time
- Properties, flow velocity or elevation may change from point to point within CV
- Size, shape, mass and energy content of the CV do not change with time
- Rate at which heat and work interactions take place with surroundings do not change with time
- Devices/systems which undergo steady flow process:
Conservation of Energy for Control Volumes
compressors, pumps, turbines, water supply pipes, nozzles, heat exchangers, power plants, aircraft engines etc.

19. • During a steady-flow process, the fluid flows through the control volume steadily,
experiencing no change with time at a fixed position
Vcm
miVi
meVe
Wnet
Qnet
Control Surface
Zi
Zcm
Ze
The mass and energy content of the open system may change when mass enters or leaves the control volume
e – exit
i - inlet
Z – height
Q – Heat
W – Work
m – Mass
V – Velocity
For Steady-State, Steady-Flow Processes
• Most energy conversion devices operate steadily over long periods of time
• The rates of heat transfer and work crossing the control surface are constant with time
• The states of the mass streams crossing the control surface or boundary are constant with time
• Under these conditions the mass and energy content of the control volume are constant with time
MNOP
MQ
= ∆ṁcv = 0
MROP
MQ
= ∆Ė = 0

20. Steady-state, Steady-Flow Conservation of Mass: ! ! ( / )m m kg sin out∑ ∑=
Steady-state, steady-flow conservation of energy: The energy of the control volume is constant
with time during the steady-state, steady-flow process
! ! !E E E kWin out system− =
Rate of net energy transfer
by heat, work, and mass
Rate change in internal, kinetic,
potential, etc., energies
( )
" #$ %$ "#$ %$
Δ
0

21. • The conservation of energy principle for the control volume or open system has the same word
definition as the first law for the closed system
• Expressing the energy transfers on a rate basis, the control volume first law is
! ! !E E E kWin out system− =
Rate of net energy transfer
by heat, work, and mass
Rate change in internal, kinetic,
potential, etc., energies
( )
" #$ %$ "#$ %$
Δ
Considering that energy flows into and from the control volume with the mass, energy enters
because net heat is transferred to the control volume, and energy leaves because the control volume
does net work on its surroundings, the open system, or control volume, applying the first law of
thermodynamics
Conservation of Energy for General Control Volume
Energy balance in differential form dE = 𝜕Q - 𝜕W
Time rate form of energy balance
MR
MQ
= Q – W
. .

22. Time rate change of energy is given by
MR
MQ
=
MSR
MQ
+
MTR
MQ
+
MU
MQ
Time rate change of energy is given by Q – W =
MSR
MQ
+
MTR
MQ
+
MU
MQ
Where the time rate change of the energy of the control volume has been written as Δ !ECV
Considering that energy flows into and from the control volume with the mass, energy enters
because heat is transferred to the control volume, and energy leaves because the control volume
does work on its surroundings, the steady-state, steady-flow first law becomes

23. Total energy crossing
boundary per unit time
Total energy of mass
leaving CV per unit time
Total energy of mass
entering CV per unit time
! ! !
! ! !
Q Q Q
W W W
net in out
net out in
= −
= −
∑ ∑
∑ ∑
Mass flow rate (kg/s) =
V!W!
P!
=
V"W"
P"
Steady Flow process Involving one fluid stream at the inlet and exit of the control volume
Where, V = Velocity (m/s)
v = specific volume (m3/kg)

24. • A number of thermodynamic devices such as pumps, fans, compressors, turbines, nozzles,
diffusers, and heaters operate with one entrance and one exit
• The steady-state, steady-flow conservation of mass and first law of thermodynamics for these
systems reduce to
mmm ie
!!! ==∑∑since)
2
()
2
(
22
i
i
ie
e
e gzhmgzhmWQ ++−++=−
VV
!!!!
Using subscript 1 and subscript 2 for denoting
inlet and exit states)](
2
[ 12
2
1
2
2
12 zzghhmWQ −+
−
+−=−
VV
!!!
)(
2
12
2
1
2
2
12 zzghhwq −+
−
+−=−
VV Dividing the equation by m! yields
wzghq +Δ+
Δ
+Δ=
2
or
2
V
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+Δ+
Δ
+Δ+Δ= wzgpvuq
2
2
V
Steady-State, Steady-Flow for One Entrance and One Exit

25. 𝒎 𝟏(
𝒁 𝟏 𝒈
𝟏𝟎𝟎𝟎
+
𝑽 𝟏
𝟐
𝟐𝟎𝟎𝟎
+ 𝒉 𝟏) + 𝑸 𝒏𝒆𝒕 = 𝒎 𝟐
𝒁 𝟐 𝒈
𝟏𝟎𝟎𝟎
+
𝑽 𝟐
𝟐
𝟐𝟎𝟎𝟎
+ 𝒉 𝟐 + 𝑾 𝒏𝒆𝒕
Solving Steady Flow Energy Equation
[h, W, Q should be in kJ/kg and V in m/s and g in m/s2]
𝒎 𝟏(𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒉 𝟏) + 𝑸 𝒏𝒆𝒕 = 𝒎 𝟐
𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒉 𝟐 + 𝑾 𝒏𝒆𝒕
[h, W, Q should be in J/kg and V in m/s and g in m/s2]
Our aim is to give heat to the system and gain work output from it.
So heat input → +ive (positive) Work output → +ive (positive)
+ive (positive) +ive (positive)-ive (negative) -ive (negative)
Work Work Heat Heat

26. Applications of SFEE
• Nozzles and diffusers (e.g. jet propulsion)
• Turbines (e.g. power plant, turbofan/turbojet aircraft engine), compressors and pumps (power
plant)
• Heat exchangers (e.g. boilers and condensers in power plants, evaporator and condenser in
refrigeration, food and chemical processing)
• Mixing chambers (power plants)
• Throttling devices (e.g. refrigeration, steam quality measurement in power plants)
All elements of a simple power plant/ refrigeration cycle and more! In principle, you can take the
elements together to calculate power generated/required, heat removed/supplied.
Source: www.google.com
Throttling DevicesHeat ExchangersGas Turbines

27. Water Turbine
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 − 𝒘 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐 + 𝒘
𝒗 𝒔𝟏 = 𝒗 𝒔𝟐 = 𝒗 𝒔
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
− 𝒘 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒗 𝒔(p2 –p1)
𝒁 𝟏
−𝒁 𝟐
𝑾
Datum
Water Turbine

28. Steam / Gas Turbine
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒉 𝟏 − 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒉 𝟐
+𝒘
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐 +𝒘
𝒅𝒁 = 𝟎
𝑾
Steam/Gas
Turbine
𝑽 𝟏
𝟐
𝟐
+ 𝒉 𝟏 =
𝑽 𝟐
𝟐
𝟐
+ 𝒉 𝟐
+ 𝒘
Steam/Gas In
Steam/Gas Out

29. Steam Nozzle
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒉 𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒉 𝟐 + 𝒘
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
+𝒘
𝑽 𝟏
𝟐
𝟐
+ 𝒉 𝟏 =
𝑽 𝟐
𝟐
𝟐
+ 𝒉 𝟐
Steam In
Steam Out
𝑽 𝟐 = 𝑽 𝟏
𝟐 + 𝟐(𝒉 𝟏 − 𝒉 𝟐)
𝑽 𝟐 = 𝟐(𝒉 𝟏 − 𝒉 𝟐)

30. Boiler
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
+ 𝒘
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
+ 𝒘
h1 + q = h 𝟐
𝑺𝒕𝒆𝒂𝒎
𝑸
𝑾𝒂𝒕𝒆𝒓
Control Surface
Boiler
q = h 𝟐 - h 𝟏

31. Heat Exchanger
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 − 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐 + 𝒘
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
+ 𝒘
h 𝟏 - q = h 𝟐
𝑺𝒕𝒆𝒂𝒎
𝑸
𝑾𝒂𝒕𝒆𝒓
Control Surface
1 kg steam
q = (hw2 – hw1) mw = mw Cw (Tw2 – Tw1)
h 𝟏 - (hw2 – hw1) mw = h 𝟐
h 𝟏 - h 𝟐 = (hw2 – hw1) mw =mw Cw (Tw2 – Tw1)
q - heat gained by water by passing through condenser
Tw1
Tw2
mw – flow of water per kg steam
q - heat lost by 1 kg steam to water passing through condenser

32. Reciprocating Compressor
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 − 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟏 𝒗 𝒔𝟐
− 𝒘
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
+ 𝒘
𝑹𝒆𝒄𝒊𝒆𝒗𝒆𝒓
𝑸
𝑨𝒊𝒓 𝑰𝒏
Control Surface
𝑾
𝒉 𝟏 + 𝒘 − 𝒒 = 𝒉 𝟐
If Velocity changes are neglected and flow process is
treated as adiabatic
Due to large area in contact and low flow rates appreciable heat transfer can take place between the system
and the surroundings. Therefore water cooling is required

33. Rotary Compressor
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 − 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐 − 𝒘
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
+ 𝒘
𝑨𝒊𝒓 𝑶𝒖𝒕
𝑸𝑨𝒊𝒓 𝑰𝒏
Control Surface
𝑾

34. Centrifugal Pump
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
+ 𝒘
𝑾
Water Sump
𝑽 𝟏
𝟐
𝟐
+ 𝒁 𝟏 𝒈 + 𝒘 =
𝑽 𝟐
𝟐
𝟐
+ 𝒁 𝟐 𝒈 + 𝐯𝐬(𝒑 𝟐 − 𝒑 𝟏)
Control Surface
Steam/Gas Out
𝒁 𝟏
𝒁 𝟐
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
+ 𝒘

35. A blower handles 1 kg/s of air at 20°C. Find the exit air temperature, assuming adiabatic
conditions. Take cp of air is 1.005 kJ/kg-K.
𝑸 = 𝟎
100 m/s 150 m/s
W = 15 kW
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒖 𝟏 + 𝒑 𝟏 𝒗 𝒔𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒖 𝟐 + 𝒑 𝟐 𝒗 𝒔𝟐
+ 𝒘
𝒁 𝟏 𝒈 +
𝑽 𝟏
𝟐
𝟐
+ 𝒉 𝟏 + 𝒒 = 𝒁 𝟐 𝒈 +
𝑽 𝟐
𝟐
𝟐
+ 𝒉 𝟐
- 𝒘
m1 = m2 = kg /s

36. Throttling process
• A flow is throttled when, for example, it flows through a partially open valve
• When it does so, we notice that there can be a significant pressure loss from one side of the
partially open valve to the other
So, we can say that such a throttling device is one in which pressure drops and enthalpy remains constant
In throttling devices there may be a change in velocity due to compressibility effects, but it is
observed to be small when the flow velocity is much less than the speed of sound.
We shall assume here the velocity is small relative to the speed of sound so as to recover v1 ∼ v2
Thus, h1 = h2
• A throttling process is modeled as steady device with one entrance and exit, with no control volume work or
heat transfer
• Changes in area as well as potential energy are neglected

37. This shows that enthalpy remains constant during adiabatic throttling process.
The throttling process is commonly used for the following purposes :
1. For determining the condition of steam (dryness fraction)
2. For controlling the speed of the turbine
3. Used in refrigeration plants
4. Liquefaction of gases

38. In this experiment gas is forced through a porous plug and is called a throttling process
iii TvP fff TvPpiston
porous plug adiabatic walls
The Joule-Thomson Experiment
• In an actual experiment, there are no pistons and there is a continuous flow of gas
• A pump is used to maintain the pressure difference between the two sides of the porous plug
In this experiment, as pressures are kept constant work is done
iiff
v
0
f
0
v
i vPvPvdPvdPw
f
i
−=+= ∫∫
𝜹𝒒 = 𝒅𝒖 + 𝜹𝒘
iiifffiiffif vPuvPuor)vPvP()uu(0 +=+−+−=
From the definition of enthalpy if hh =
Hence, in a throttling process, enthalpy is conserved

39. 39
• In the region where the atoms or molecules are very close together, then repulsive forces
dominate and as the volume expands, the energy goes down. Thus, for these conditions, pT is
negative
• In the region where the atoms or molecules are close enough that attractive forces dominate,
then as the volume expands, the energy goes up. Thus, for these conditions, pT is positive
• For most gases at not too large pressures, the molecules don't interact very much and so there is
little dependence of energy on volume so pT is very small. In the extreme of zero
interaction, pT is zero. This is the defining condition for an ideal (perfect) gas.

40. 40
Pump
Porous plug
Pi Ti Pf Tf
Throttling Process
(Joule-Thomson or Joule-Kelvin expansion widely used in refrigerators)
• The pump maintains the pressures Pi and Pf
• In the experiment Pi, Ti and Pf are set and Tf is
measured
• Consider a series of experiments in which Pi and Ti are
constant (hi constant) and the pumping speed is changed to
change Pf and hence Tf
• Since the final enthalpy does not change, we get points
of constant enthalpy
The enthalpy is the same on the
two sides of the porous plug i.e.,
hf = hi.
We plot Tf as a function of Pf
Pi > Pf

41. 41
Pressuref
Temperaturef
- A smooth curve is placed through the points yielding an isenthalpic curve
- Note that this is not a graph of the throttling process as it passes through irreversible states
•
•
• • •
•
•
•
Pf , Tf
Pi, , Ti
Isenthalpic Curve

42. 42
We now change Pi and Ti and obtain another isenthalpic curve
Pressuref
Temperaturef
Maximum Inversion T
Cooling
•
d
•
c
•
b
•
a
Heating
Inversion Curve
Inversion Curve
Ideal Gas
We are interested in the temperature change
due to the pressure change, therefore it is
useful to define the Joule-Thomson coefficient
𝝁
This is the slope of an isenthalpic curve and
hence varies from point to point on the graph
( )hP
T
∂
∂
=µ
• A point at which 𝝁 = 0 is called an inversion point
• Connecting all of these points produces the inversion curve

43. 43
Pressuref
Temperaturef
Maximum Inversion T
Cooling
•d
•c
•
b
•
a
Heating
Inversion Curve
Inversion Curve
Ideal Gas
If point a on the diagram
(𝝁 < 0) is a starting point
and point b is the final
point, then the T of the
gas will rise, i.e. we have
heatingIf we start at point c
(𝝁 > 0) and go to point
d, then the T of the
gas will drop, i.e. we
have cooling These curves are horizontal
lines for an ideal gas
As higher initial starting temperatures are used, the isenthalpic curves become flatter and more
closely horizontal
Maximum inversion T, the value of which depends on the gas.
For cooling to occur, the initial T must be less than the maximum inversion T, for such a T the
optimum initial P is on the inversion curve

44. • This also tells us that we cannot just use any gas at any set of pressures to make a refrigerator,
for example
- At a given pressure, some gases may be cooling (m > 0) but others may be heating (m < 0)
• The proper choice of refrigerant will depend on both the physical properties, esp. the Joule-
Thompson coefficient as well as the mechanical capacity of the equipment being used.
• Thus, we cannot just exchange our ozone-depleting freon in our car's air conditioner with any
other coolant unless the two gases behave similarly in the pressure - temperature ranges of the
mechanical device, i.e., they must have the same sign of m at the pressures the equipment is
capable of producing.
• Generally, to use a more environmentally friendly coolant, we need to replace the old equipment
with new equipment that will operate in the temperature range needed to make m positive.

45. http://faculty.chem.queensu.ca/people/faculty/mombourquette/Chem221/3_FirstLaw/ChangeFunctions.asp
• The sign of the Joule–Thomson coefficient, µ, depends on the conditions
Positive Negative
• The temperature corresponding to the boundary at a given pressure is the ‘inversion temperature’
of the gas at that pressure
The maximum inversion temperatures of
some gases are given below :
(i) He=24K
(ii) H2=195K
(iii) Air=603K
(iv) N2 =261K
(v) A=732K
(vi) CO2 =1500K

46. 46
To make the discussion clear, we have exaggerated the slopes in the above T-P diagram. In fact, for
most gases at reasonable T’s and P’s the isenthalpic curves are approximately flat and so 0≈µ
It can be shown that µP
h
P
T
c
P
T
c
P
h
−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
)(00 Thhsoand
P
h
then
T
==⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
=µ
We now have )1.5oblem(Pr0
P
h
v
u
TT
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
=⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
Constant temp. coefficient and can be
determined by Joules Thompson experiment
for an ideal gas
• For a given pressure, the temperature must be below a certain value if cooling is required
but, if it becomes too low, the boundary is crossed again and heating occurs
• Reduction of pressure under adiabatic conditions moves the system along one of the isenthalps, or
curves of constant enthalpy

47. 47
• Some gases can be liquefied in a simple process
- For example, carbon dioxide can be liquefied at room temperature by a simple isothermal
compression to about 60 bar
• To liquefy nitrogen or air is not so simple.
- At room temperature, regardless of any increase in pressure, these gases will not undergo a
phase transformation to the liquid state
• A method for these gases, using the throttling process, was invented in 1895 and is called the
Hampson-Linde Process
- The basis idea is to use the gas cooled in the throttling process to precool the gas going towards
the throttle
until the T is below the maximum inversion T
- Starting from room temperature, this cycle can be used to liquefy all gases except hydrogen and
helium
- To liquefy H by this process, it must first be cooled below 200K and to accomplish this liquid N
at 77K is used
- To liquefy He by this process, it must first be cooled below 43K and to accomplish this liquid H
can be used. (A device called the Collins helium liquifier is used to liquefy He.
Liquefaction of Gases

48. Hampson-Linde Process

49. 49
If a throttling process is used to liquefy a gas, the cooled gas is recycled through a heat exchanger to
precool the gas moving towards the throttle. The gas continues to cool and when a steady state is
reached a certain fraction, y, is liquefied and a fraction (1 - y) is returned by the pump.
Using the notation:
- hi = molar enthalpy of entering gas
- hf = molar enthalpy of emerging gas
- hL = molar enthalpy of emerging liquid
Since the enthalpy is constant we have
hi = y h L + (1-y)hf
Of course, as some of the gas liquefies, additional gas must be added to the system.
It should be mentioned that Joule-Thomson liquefaction of gases has these advantages:
• No moving parts that would be difficult to lubricate at low T.
• The lower the T , the greater the T drop for a given pressure drop.

50. Controlling the speed of the steam turbine by Throttling