Fundamentals of Thermodynamics
SENG 205
School of Engineering Sciences
University of Ghana
Lecturer: Prof. Emmanuel Nyankson
Email: enyankson@ug.edu.gh
Teaching Assistant: Daniel, Christian, Petrina, Mary
Credit Hours: Three (3)
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Chapter 3 (Robert DeHoff)
Chapters 2, 6 (section 6.6) and 7 (section
7.2, 7.3) (Van Ness)
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Objectives
1. Introduce the laws of thermodynamics with emphasis on the first
law.
2. Introduce the concept of internal energy
3. Introduce enthalpy
4. Introduce the concept of reversible processes/entropy
generation/dissipation
5. Introduce state function and process variables
6. Introduce extensive and intensive properties
Fundamentals of Thermodynamics 2022-2023 Academic Year
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Laws of Thermodynamics
➢The laws of thermodynamics are highly condensed expressions of a broad
body of experimental evidence.
➢The three laws of thermodynamics are empirical: derived from experimental
observations of how matter behaves.
➢The laws of thermodynamics are:
1st law: there exists a property of the universe, called its energy, which cannot
change no matter what processes occur.
2nd law: there exists a property of the universe, called its entropy, which always
changes in the same direction no matter what processes occur.
Fundamentals of Thermodynamics 2022-2023 Academic Year
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Laws of Thermodynamics
1st law: there exists a property of the universe, called its energy, which cannot
change no matter what processes occur.
2nd law: there exists a property of the universe, called its entropy, which always
changes in the same direction no matter what processes occur.
3rd law: there exists a lower limit to the temperature that can be attained by
matter, called the absolute zero of temperature (0 K , -273.15℃), and the entropy
of all substances is the same at that temperature.
Zeroth law: this law acknowledges that a temperature scale exists for all
substances in nature and provides an absolute measure of their tendencies to
exchange heat.
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The 1st Law of Thermodynamics
Fundamentals of Thermodynamics 2022-2023 Academic Year
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The 1st Law of Thermodynamics
1st law: there exists a property of the universe, called its energy, which cannot
change no matter what processes occur.
➢Energy can be defined both physically and mathematically and can be
measured with precision.
➢Three broad categories of energy have been identified in scientific experience:
1. Kinetic energy
2. Potential energy
3. Internal energy
The first law requires that:
∆ 𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 + ∆ 𝐸𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = 0
Fundamentals of Thermodynamics 2022-2023 Academic Year
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The 1st Law of Thermodynamics: Closed Systems
➢Except for a system that is isolated, changes that occur inside the system are
always accompanied by changes in the condition of the matter in the vicinity
of the system (surroundings).
The sum of the changes that
occur in the system and the
surroundings includes all of
the changes in the universe
associated with that process
Surroundings
System
By the first law the total
energy of the universe cannot
change for any process.
The only way that the internal
energy of a system can change
is by transferring energy across
its boundary.
What are some of the ways that energy
can be transferred across the boundary of a system?
Fundamentals of Thermodynamics 2022-2023 Academic Year
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The 1st Law of Thermodynamics: Closed System
➢A mathematical statement of the first law is formulated from the statement: the
change in internal energy of a system for a process must be equal to the sum of
all energy transfers across the boundary of the system during the process
Possible Energy Transfer
• Q-the quantity of heat
flows into the system
during the process.
Surroundings
Possible Energy Transfer
• W-the mechanical work
done on the system by the
force exerted by the
external pressure in the
surroundings.
Q
System
Possible Energy Transfer
• W1-all other kinds of work
done on the system during
the process.
W1
W
The First Law of Thermodynamics
ΔU = q + w
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The 1st Law of Thermodynamics
Relationship Between
Heat and Work
Benjamin Thompson, Count Rumford
Cannon Boring Experiment
The heat produced during the boring is proportional to the work performed during the boring
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Prof Nyankson
The 1st Law of Thermodynamics
James Joules
The First Law of Thermodynamics
ΔU = q + w
Fundamentals of Thermodynamics 2022-2023 Academic Year
Same effect if water
is placed
in thermal contact with a source of heat
Prof Nyankson
Laws of Thermodynamics
Joule’s Experiment
• Energy is added to the fluid as work, but is
transferred from the fluid as heat.
• What happens to this energy between its
addition to and transfer from the fluid?
• It is contained in the fluid in another form
called INTERNAL ENERGY.
• Kinetic energy of translation
• Kinetic energy of rotation
• Kinetic energy of internal vibration
The First Law of Thermodynamics
ΔU = q + w
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https://www.youtube.com/watch?v=ZOynwQ75pms
Prof Nyankson
The 1st Law of Thermodynamics: Closed System
➢The increase in the internal energy of the system during a process is the sum of
the transfers across the boundary.
➢Mathematical statement of the first law:
∆𝑈 = 𝑄 + 𝑊 + 𝑊 1
For infinitesimal change in the condition of the system:
𝑑𝑈 = 𝛿𝑄 + 𝛿𝑊 + 𝛿𝑊 1
Why 𝑑𝑈 𝑎𝑛𝑑 𝑛𝑜𝑡 𝛿𝑈?
The first law applies to any
Why 𝛿𝑄 𝑎𝑛𝑑 𝑛𝑜𝑡 𝑑𝑄?
system taken through any
process.
Why 𝛿𝑊 𝑎𝑛𝑑 𝑛𝑜𝑡 𝑑𝑊?
Fundamentals of Thermodynamics 2022-2023 Academic Year
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The 1st Law of Thermodynamics: Closed System
WORKING EXAMPLE 1
When a system is taken from state a to
state b along path acb, 100 J of heat
flows into the system and the system
does 40 J of work.
a. How much heat flows into the
system along path aeb if the work
done by the system is 20 J?
b. The system returns from b to a along
path bda. If the work done on the
system is 30 J, does the system
of Thermodynamics 2022-2023 Academic Year
absorb or liberate heat?Fundamentals
How much?
Prof Nyankson
The 1st Law of Thermodynamics: Closed System
WORKING EXAMPLE 1
When a system is taken from state a to
state b along path acb, 100 J of heat
flows into the system and the system
does 40 J of work.
Data:
𝑄𝑎𝑐𝑏 = 100 𝐽
𝑊𝑎𝑐𝑏 = −40 𝐽
∆𝑈𝑎𝑐𝑏 = 𝑄𝑎𝑐𝑏 + 𝑊
𝑎𝑐𝑏
Fundamentals of Thermodynamics 2022-2023 Academic Year
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The 1st Law of Thermodynamics: Closed System
Data:
𝑄𝑎𝑐𝑏 = 100 𝐽
𝑊𝑎𝑐𝑏 = −40 𝐽
∆𝑈𝑎𝑐𝑏 = 𝑄𝑎𝑐𝑏 + 𝑊𝑎𝑐𝑏
∆𝑈𝑎𝑐𝑏 = 100 − 40 = 60 𝐽
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics: Closed System
WORKING EXAMPLE 1
a. How much heat flows into the
system along path aeb if the work
done by the system is 20 J?
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics: Closed System
WORKING EXAMPLE 1
a. How much heat flows into the
system along path aeb if the work
done by the system is 20 J?
∆𝑈𝑎𝑒𝑏 = 𝑄𝑎𝑒𝑏 + 𝑊𝑎𝑒𝑏
∆𝑈𝑎𝑒𝑏 = 𝑄𝑎𝑒𝑏 − 20 = ∆𝑈𝑎𝑐𝑏 = 60
𝑄𝑎𝑒𝑏 = 80 𝐽
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics: Closed System
WORKING EXAMPLE 1
b. The system returns from b to a along
path bda. If the work done on the system
is 30 J, does the system absorb or
liberate heat? How much?
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics: Closed System
WORKING EXAMPLE 1
b. The system returns from b to a along
path bda. If the work done on the system
is 30 J, does the system absorb or
liberate heat? How much?
∆𝑈𝑏𝑑𝑎 = 𝑄𝑏𝑑𝑎 + 𝑊𝑏𝑑𝑎
∆𝑈𝑏𝑑𝑎 = 𝑄𝑏𝑑𝑎 + 30 = −60
𝑄𝑏𝑑𝑎 = −60 − 30 = −90J
Heat is liberated
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Equilibrium
➢Equilibrium is a word denoting a static condition, the absence of change.
➢In thermodynamics, it also means the absence of any tendency toward change on
a macroscopic scale.
A system is in equilibrium if it is in
1. Thermal equilibrium
2. Mechanical equilibrium
3. Chemical equilibrium
Fundamentals of Thermodynamics 2022-2023 Academic Year
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Equilibrium
➢Equilibrium is a word denoting a static condition, the absence of change.
➢In thermodynamics, it also means the absence of any tendency toward change on
a macroscopic scale.
A system is in equilibrium if it is in
1. Thermal equilibrium
2. Mechanical equilibrium
3. Chemical equilibrium
Fundamentals of Thermodynamics 2022-2023 Academic Year
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Equilibrium
➢Equilibrium is a word denoting a static condition, the absence of change.
➢In thermodynamics, it also means the absence of any tendency toward change on
a macroscopic scale.
A system is in equilibrium if it is in
1. Thermal equilibrium
2. Mechanical equilibrium
3. Chemical equilibrium
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Equilibrium
➢Equilibrium is a word denoting a static condition, the absence of change.
➢In thermodynamics, it also means the absence of any tendency toward change on
a macroscopic scale.
➢Phase Rule:
Phase is a homogenous region of matter.
𝐹 =2−𝜋+𝑁
N=number of components, F=degree of freedom, 𝜋 is the number of phases
2=non compositional variables (P,T).
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Equilibrium
𝐹 =2−𝜋+𝑁
Fundamentals of Thermodynamics 2022-2023 Academic Year
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Working Example
How many phase-rule variables must be specified to fix the thermodynamic state of
each of the following systems?
(a) Liquid water in equilibrium with its vapor.
(b) Liquid water in equilibrium with a mixture of water vapor and nitrogen.
(c) A three-phase system of a saturated aqueous salt solution at its boiling point
with excess salt crystals present.
𝐹 =2−𝜋+𝑁
Fundamentals of Thermodynamics 2022-2023 Academic Year
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Working Example
How many phase-rule variables must be specified to fix the thermodynamic state of
each of the following systems?
(a) Liquid water in equilibrium with its vapor.
(b) Liquid water in equilibrium with a mixture of water vapor and nitrogen.
(c) A three-phase system of a saturated aqueous salt solution at its boiling point
with excess salt crystals present.
𝐹 =2−𝜋+𝑁
a. 𝐹 = 2 − 𝜋 + 𝑁 = 2 − 2 + 1 = 1
b. 𝐹 = 2 − 𝜋 + 𝑁 = 2 − 2 + 2 = 2
c. 𝐹 = 2 − 𝜋 + 𝑁 = 2 − 3 + 2 = 1
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Reversible Process
• The gas pressure is sufficient to balance the weight
of the piston and all that it carries.
• In an equilibrium condition, the system has no
tendency to change.
• Mass m is removed
• Oscillation
• Chaotic molecular motion
• Dissipation of work into internal energy
• Frictionless Piston (dissipation)
Irreversible
process.
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Reversible Process
Reversible Process: its direction can be reversed at any point by an infinitesimal change in
external conditions
Gas
Gas
b
a
Gas
Gas
c
1. PROCESS 1: The pan and its contents are
lifted off the top of the piston.
➢ The system experiences a sudden drop in
the force applied to it.
➢ Process accompanied with dissipation,
finite entropy production and permanent
changes in the universe.
2. PROCESS 2: The particles are removed
with a pair of tweezers one particle at a time
and letting the system come to rest before the
next particle is removed
➢ No dissipation, no entropy production and
no permanent changes.
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Reversible Chemical Reaction
➢The concept of a reversible chemical reaction is illustrated by the decomposition
of calcium oxide carbonate, which when heated forms calcium oxide and carbon
dioxide gas.
𝐶𝑎𝐶𝑂3 ↔ 𝐶𝑎𝑂 + 𝐶𝑂2
• Increasing the pressure of CO2 above the
equilibrium CO2 pressure, CO2 combines with
CaO to form CaCO3
• Decreasing the pressure of CO2 below the
equilibrium CO2 pressure, CaCO3
decomposes to form more CO2 and CaO.
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Reversible Chemical Reaction
• Summary Remarks on Reversible Processes
A reversible process:
➢Is frictionless
➢Is never more than differentially removed from equilibrium
➢Traverses a succession of equilibrium states
➢Can be reversed at any point by a differential change in external conditions
➢When reversed, retracts its forward path, and restores the initial state of the system and
surroundings.
Minimum work input required or maximum work output
attainable from a specified process
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Constant Volume and Constant Pressure Processes
• The energy balance for a homogenous closed system of n moles is:
𝑑 𝑛𝑈 = 𝛿𝑄 + 𝛿𝑊
𝑑𝑊 = −𝑃𝑑(𝑛𝑉)
Hence
𝑑 𝑛𝑈 = 𝛿𝑄 − 𝑃𝑑(𝑛𝑉)
This is the general first law equation for a mechanically reversible, closed system
process.
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Constant Volume and Constant Pressure Processes
• Constant Volume Process
➢If the process occurs at constant volume, the work is zero. Thus
𝛿𝑄 = 𝑑(𝑛𝑈) integration yields 𝑄 = 𝑛∆𝑈
Thus for a mechanically reversible, constant volume, closed system process, the
heat transferred is equal to the internal energy of the system.
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Constant Volume and Constant Pressure Processes
• Constant Pressure Process
𝛿𝑄 = 𝑑 𝑛𝑈 + 𝑃𝑑(𝑛𝑉)
For a constant pressure change of state
𝛿𝑄 = 𝑑 𝑛𝑈 + 𝑑 𝑛𝑃𝑉 = 𝑑 𝑛(𝑈 + 𝑃𝑉)
A new thermodynamic property needs to be defined
𝐻 = 𝑈 + 𝑃𝑉
H,U and V are molar volume or unit mass values. Hence
𝛿𝑄 = 𝑑 𝑛𝐻
Integration yields
𝑄 = 𝑛∆𝐻
For a mechanically reversible, constant pressure, closed-system process, the heat
Fundamentals
of Thermodynamics
2022-2023
Academic Year
transferred equals the enthalpy
change
of
the
system.
Prof Nyankson
Constant Volume and Constant Pressure Processes
Calculate ΔU and ΔH for 1 kg of water when it is vaporized at the constant
temperature of 100°C and the constant pressure of 101.33 kPa. The specific
volumes of liquid and vapor water at these conditions are 0.00104 and 1.673
m3·kg−1, respectively. For this change, heat in the amount of 2256.9 kJ is added
to the water.
Data
T = 100°C
P = 101.33 kPa
Find
VL = 0.00104 m3·kg−1
ΔU and ΔH
VV = 1.673 m3·kg−1
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Q = 2256.9 kJ
Prof Nyankson
Constant Volume and Constant Pressure Processes
Data
At constant pressure ΔH = Q = 2256.9 kJ
T = 100°C
P = 101.33 kPa
∆𝐻 = ∆𝑈 + 𝑃∆𝑉
3
−1
VL = 0.00104 m ·kg
VV = 1.673 m3·kg−1
𝑁
𝑚3
2256.9𝑘𝐽 = ∆𝑈 + 101.33𝑘 2 (1.673 − 0.00104)
Q = 2256.9 kJ
𝑚
𝑘𝑔
∆𝑈 = 2256.9 − 169.4 = 2087.5 𝑘𝐽
Fundamentals of Thermodynamics 2022-2023 Academic Year
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Constant Volume and Constant Pressure Processes
• Heat capacity: the heat capacity is defined as 𝐶 ≡
Two heat capacities can be defined
➢Heat Capacity at Constant Volume
𝜕𝑈
𝐶𝑣 =
𝜕𝑇
Integration yields
𝛿𝑄
𝑑𝑇
, 𝑑𝑈 = 𝐶𝑣 𝑑𝑇
𝑣
𝑇2
∆𝑈 = න 𝐶𝑣 𝑑𝑇
𝑇1
Hence, 𝑄 = 𝑛∆𝑈 =
𝑇2
𝑛 ‫𝑇𝑑 𝑣𝐶 𝑇׬‬
1
Fundamentals of Thermodynamics 2022-2023 Academic Year
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Constant Volume and Constant Pressure Processes
➢Heat Capacity at Constant Pressure
𝜕𝐻
𝐶𝑃 ≡
𝜕𝑇
Integration yields
Reversible adiabatic Process
Isothermal process
, 𝑑𝐻 = 𝐶𝑃 𝑑𝑇
𝑃
𝐶𝑃 > 𝐶𝑉 , 𝑊ℎ𝑦?
𝑇2
∆𝐻 = න 𝐶𝑃 𝑑𝑇
𝑇1
Hence, 𝑄 = 𝑛∆𝐻 =
For an ideal gas
𝐶𝑃 − 𝐶𝑉 = 𝑅
𝑇2
𝑛 ‫𝑇𝑑 𝑃𝐶 𝑇׬‬
1
Above room temperature, the temperature variation of Cp follows a relatively simple empirical
Fundamentals of Thermodynamics 2022-2023 Academic Year
relation:
Prof Nyankson
𝑐
𝐶𝑃 𝑇 = 𝑎 + 𝑏𝑇 + 2 + 𝑑𝑇 2
𝑇
The 1st Law of Thermodynamics
𝛿𝑄 = 𝑑𝐻 = 𝐶𝑝 𝑑𝑇
𝐶𝑝 = 𝐴 + 𝐵𝑇 +
𝑑𝑄 = 𝑑𝑈 = 𝐶𝑉 𝑑𝑇
𝐶
+ 𝐷𝑇 2
2
𝑇
Metal
C / (J mol-1 K-1)
selenium
24.9
gold
25.2
zinc
25.0
titanium
25.1
tungsten
25.0
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The 1st Law of Thermodynamics
𝐶𝑝 = 𝑎 + 𝑏𝑇 +
𝑐
2
+
𝑑𝑇
𝑇2
If CuO is heated from 50 ℃ to 150
℃ at constant pressure of 1 atm.
Calculate the enthalpy change.
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
Air at 1 bar and 298.15 K is compressed to 3 bar and 298.15 K by two different
closed-system mechanically reversible processes:
(a) Cooling at constant pressure followed by heating at constant volume.
(b) Heating at constant volume followed by cooling at constant pressure.
Calculate the heat and work requirements and ΔU and ΔH of the air for each path.
The following heat capacities for air may be assumed independent of temperature:
C V = 20.785 and C P = 29.100 J· mol −1 ·K −1
Assume also that air remains a gas for which PV/T is a constant, regardless of the
changes it undergoes. At 298.15 K and 1 bar the molar volume of air is 0.02479 m3·
Fundamentals of Thermodynamics 2022-2023 Academic Year
mol−1.
Prof Nyankson
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
Air at 1 bar and 298.15 K is compressed to 3 bar and 298.15 K by two different
closed-system mechanically reversible processes:
Data
(a) Cooling at constant pressure followed by heating at constant
volume
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
𝑷𝑽
= 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑻
Final
Pf = 3 bar
Tf = 298.15 K
Vf
(b) Heating at constant volume followed by cooling at constant
pressure
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
Air at 1 bar and 298.15 K is compressed to 3 bar and 298.15 K by two different
closed-system mechanically reversible processes:
Data
(a) Cooling at constant pressure followed by heating at constant
volume
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
𝑷𝑽
= 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝑻
Final
Pf = 3 bar
Tf = 298.15 K
Vf
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
(a) Cooling at constant pressure followed by heating at constant volume.
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
Intermediate
P* = P1= 1 bar
T* =
V* = Vf
Final
Pf = 3 bar
Tf = 298.15 K
Vf =V*
V
T
initial
initial
V1
T1
V*
T*
final
P*=P1
Pf
final
P
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
P1
Pf
P
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
(a) Cooling at constant pressure followed by heating at constant volume.
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
Intermediate
P* = P1= 1 bar
T* =
V* = Vf
Final
Pf = 3 bar
Tf = 298.15 K
Vf =V*
V
T
initial
initial
V1
T1
V*
T*
final
P1
Pf
P
𝑃𝑓 𝑉𝑓 𝑃∗ 𝑉 ∗
3 ∗ 𝑉𝑓
1 ∗ 𝑉∗
= ∗
=
𝑇
𝑇of Thermodynamics
298.15
𝑇 ∗ Year
𝑓
Fundamentals
2022-2023 Academic
Prof Nyankson
final
P1
3
1
= ∗
298.15 𝑇
Pf
P
𝑇 ∗ = 99.4 𝐾
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
(a) Cooling at constant pressure followed by heating at constant volume.
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
Intermediate
P* = P1= 1 bar
T* =99.4 K
V* = Vf
Final
Pf = 3 bar
Tf = 298.15 K
Vf =V*
V
T
initial
initial
V1
T1
V*
T*
final
P1
Pf
P
𝑃1 𝑉𝐼 𝑃∗ 𝑉 ∗
1 ∗ 0.02479 1 ∗ 𝑉 ∗
= ∗
=
𝑇
𝑇of Thermodynamics
298.15
99.4
1
Fundamentals
2022-2023 Academic
Year
Prof Nyankson
final
P1
Pf
𝑉 ∗ = 0.00826 𝑚3 = Vf
P
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
(a) Cooling at constant pressure followed by heating at constant volume.
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
V
initial
V1
𝑄𝑎 = 𝑄1−𝑖 + 𝑄𝑖−𝑓
𝑊𝑎 = 𝑊1−𝑖 + 𝑊𝑖−𝑓
∆𝐻 = ∆𝐻1−𝑖 + ∆𝐻𝑖−𝑓
∆𝑈 = ∆𝑈1−𝑖 + ∆𝑈𝑖−𝑓
𝑄1−𝑖 , 𝑊1−𝑖 , ∆𝐻1−𝑖 , ∆𝑈1−𝑖
Intermediate
P* = P1= 1 bar
T* =99.4 K
V* = 0.00826 𝑚3
Final
Pf = 3 bar
Tf = 298.15 K
Vf =V*= 0.00826 𝑚3
V*
final
P1
Pf
P
𝑄𝑖−𝑓 , 𝑊𝑖−𝑓 , ∆𝐻𝑖−𝑓 , ∆𝑈𝑖−𝑓
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
(a) Cooling at constant pressure followed by heating at constant volume.
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
V
𝑄1−𝑖 , 𝑊1−𝑖 , ∆𝐻1−𝑖 , ∆𝑈1−𝑖
𝑄1−𝑖 = ∆𝐻1−𝑖 = 𝐶𝑝 ∆𝑇 = 29.1 99.4 − 298.15
initial
V1
𝑄1−𝑖 = ∆𝐻1−𝑖 = −5784 𝐽
𝑄1−𝑖 , 𝑊1−𝑖 , ∆𝐻1−𝑖 , ∆𝑈1−𝑖
Intermediate
P* = P1= 1 bar
T* =99.4 K
V* = 0.00826 𝑚3
Final
Pf = 3 bar
Tf = 298.15 K
Vf =V*= 0.00826 𝑚3
𝑊 = න −𝑃𝑑𝑉 = −105 (0.00826 − 0.02479)
𝑊1−𝑖 = 1653 𝐽
V*
final
∆𝑈1−𝑖 = ∆𝐻1−𝑖 − 𝑃1 ∆𝑉
P1
Pf
P
∆𝑈1−𝑖 = −5784 − 1 ∗ 105 (0.00826 − 0.02479)
∆𝑈1−𝑖 = −4131 𝐽
𝑄𝑖−𝑓 , 𝑊𝑖−𝑓 , ∆𝐻𝑖−𝑓 , ∆𝑈𝑖−𝑓
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
(a) Cooling at constant pressure followed by heating at constant volume.
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
𝑄𝑖−𝑓 , 𝑊𝑖−𝑓 , ∆𝐻𝑖−𝑓 , ∆𝑈𝑖−𝑓
V
𝑊𝑖−𝑓 = න −𝑃𝑑𝑉 = 0 𝐽
initial
V1
𝑄𝑖−𝑓 = ∆𝑈𝑖−𝑓 = 𝐶𝑉 ∆𝑇 = 20.785 298.15 − 99.4
𝑄1−𝑖 , 𝑊1−𝑖 , ∆𝐻1−𝑖 , ∆𝑈1−𝑖
Intermediate
P* = P1= 1 bar
T* =99.4 K
V* = 0.00826 𝑚3
Final
Pf = 3 bar
Tf = 298.15 K
Vf =V*= 0.00826 𝑚3
𝑄𝑖−𝑓 = ∆𝑈𝑖−𝑓 = 4131 𝐽
V*
final
P1
∆𝐻𝑖−𝑓 = ∆𝑈𝑖−𝑓 + 𝑉∆𝑃𝑖−𝑓
∆𝐻𝑖−𝑓 = 4131 + 0.00826 3 − 1 ∗ 105
Pf
P
∆𝐻𝑖−𝑓 = 5784 𝐽
𝑄𝑖−𝑓 , 𝑊𝑖−𝑓 , ∆𝐻𝑖−𝑓 , ∆𝑈𝑖−𝑓
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics
WORKING EXAMPLE 2
(a) Cooling at constant pressure followed by heating at constant volume.
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
V
𝑄𝑎 = 𝑄1−𝑖 + 𝑄𝑖−𝑓 = −5784 + 4131 = −1653 𝐽
𝑊𝑎 = 𝑊1−𝑖 + 𝑊𝑖−𝑓 = 1653 + 0 = 1653 𝐽
∆𝐻 = ∆𝐻1−𝑖 + ∆𝐻𝑖−𝑓 = −5784 + 5784 = −0 𝐽
∆𝑈 = ∆𝑈1−𝑖 + ∆𝑈𝑖−𝑓 = −4131 + 4131 = 0 𝐽
initial
V1
𝑄1−𝑖 , 𝑊1−𝑖 , ∆𝐻1−𝑖 , ∆𝑈1−𝑖
Intermediate
P* = P1= 1 bar
T* =99.4 K
V* = 0.00826 𝑚3
Final
Pf = 3 bar
Tf = 298.15 K
Vf =V*= 0.00826 𝑚3
V*
final
P1
Pf
P
𝑄𝑖−𝑓 , 𝑊𝑖−𝑓 , ∆𝐻𝑖−𝑓 , ∆𝑈𝑖−𝑓
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics
WORKING EXAMPLE 2 (To be completed by students)
(b) Heating at constant volume followed by cooling at constant pressure.
Initial
P1 = 1 bar
T1 = 298.15 K
V1 = 0.02479 m3/mol
V
T
initial
T*,V*
initial
V1
T1
final
Intermediate
P* = Pf= 3 bar
T* =
V* =V1 = 0.02479 m3/mol
Final
Pf = 3 bar
Tf = 298.15 K
Vf =
T*,V*
T*
final
P1
Pf
P
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
P1
Pf
P
The 1st Law of Thermodynamics
PRACTICE PROBLEM
a. An ideal gas at 300 K has a volume of 15 L at a pressure of 15 atm. This system
undergoes a reversible isothermal expansion to a pressure of 10 atm. Calculate
1. The final volume of the system
R = 8.314 J/mol K
R = 0.08205 L atm/mol K
2. The work done by the system
Cv = 1.5 R
3. The heat entering or leaving the system
4. The change in the internal energy
5. The change in enthalpy
b. Repeat the calculation if the process is taken through a reversible adiabatic
expansion to a pressure of 10
atm
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics
Solution (a)
1.
2.
3.
4.
5.
The final volume of the system (22.5 L)
The work done by the system (-9244 J)
The heat entering or leaving the system (+9244 J)
The change in the internal energy ( 0 J)
The change in enthalpy (0 J)
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
The 1st Law of Thermodynamics
Solution (b)
1.
2.
3.
4.
5.
The final volume of the system (19.13 L)
The work done by the system (-5130 J)
The heat entering or leaving the system (0 J)
The change in the internal energy (-5130 J)
The change in enthalpy (-8529.5 J)
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Mass and Energy Balances for Open Systems
▪ Measure of Flow
Open systems are characterize by flowing streams, for which there are four
common measures of flow:
ሶ
➢Mass flowrate (𝑚)
ሶ
➢Molar flow rate (𝑛)
➢Volumetric flow rate 𝑉ሶ
➢Velocity (𝑢)
The measures of flow are interrelated:
𝑚ሶ = 𝑀𝑛ሶ 𝑎𝑛𝑑 𝑉ሶ = 𝑢𝐴
𝑢𝐴𝜌
𝑚ሶ = 𝑢𝐴𝜌 𝑎𝑛𝑑 𝑛ሶ =
𝑀
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Mass and Energy Balances for Open Systems
Working Example 3
Liquid n-hexane flows at a rate of 𝑚ሶ = 0.75 𝑘𝑔𝑠 −1 in a pipe with inside diameter
𝐷 = 5 𝑐𝑚. What are 𝑞, 𝑛,ሶ 𝑎𝑛𝑑 𝑢? What would these quantities be for the same 𝑚ሶ if
𝐷 = 2 𝑐𝑚 ? Assume for liquid n-hexane that ρ = 659 𝑘𝑔𝑚3 . The molar mass of nhexane is 86.177 g/mol.
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Mass and Energy Balances for Open Systems
Working Example 3
Liquid n-hexane flows at a rate of 𝑚ሶ = 0.75 𝑘𝑔𝑠 −1 in a pipe with inside diameter
𝐷 = 5 𝑐𝑚. What are 𝑞, 𝑛,ሶ 𝑎𝑛𝑑 𝑢? What would these quantities be for the same 𝑚ሶ if
𝐷 = 2 𝑐𝑚 ? Assume for liquid n-hexane that ρ = 659 𝑘𝑔𝑚3 . The molar mass of n𝑘𝑔
hexane is 86.177 g/mol.
0.75
𝑚ሶ
𝑚3
𝑠
𝑉ሶ = =
= 0.00114
𝑘𝑔
𝜌
𝑠
Data
659 3
𝑚
𝑚ሶ = 0.75 𝑘𝑔𝑠 −1
ρ = 659 𝑘𝑔/𝑚3
5 cm
M = 86.177 g/mol
𝑘𝑔 1000𝑔
0.75
𝑚ሶ
𝑚𝑜𝑙
𝑠 × 1 𝑘𝑔
𝑛ሶ = =
=
8.7
𝑔
𝑀
𝑠
86.177
𝑚𝑜𝑙
𝑉ሶ = 𝑢𝐴
2
0.05
0.00114𝑚3 /𝑠 = 𝑢 ∗ 𝜋(
)𝑚2
4
Fundamentals of Thermodynamics 2022-2023 Academic Year
𝑢 = 0.581 𝑚/𝑠
Prof Nyankson
Mass and Energy Balances for Open Systems
Working Example 3
Liquid n-hexane flows at a rate of 𝑚ሶ = 0.75 𝑘𝑔𝑠 −1 in a pipe with inside diameter
𝐷 = 5 𝑐𝑚. What are 𝑞, 𝑛,ሶ 𝑎𝑛𝑑 𝑢? What would these quantities be for the same 𝑚ሶ if
𝐷 = 2 𝑐𝑚 ? Assume for liquid n-hexane that ρ = 659 𝑘𝑔𝑚3 . The molar mass of n𝑘𝑔
hexane is 86.177 g/mol.
0.75
𝑚ሶ
𝑚3
𝑠
𝑉ሶ = =
= 0.00114
𝑘𝑔
𝜌
𝑠
Data
659 3
𝑚
𝑚ሶ = 0.75 𝑘𝑔𝑠 −1
ρ = 659 𝑘𝑔/𝑚3
2 cm
M = 86.177 g/mol
𝑘𝑔 1000𝑔
0.75
𝑚ሶ
𝑚𝑜𝑙
𝑠 × 𝑘𝑔
𝑛ሶ = =
=
8.7
𝑔
𝑀
𝑠
86.177
𝑚𝑜𝑙
𝑉ሶ = 𝑢𝐴
2
0.02
0.00114𝑚3 /𝑠 = 𝑢 ∗ 𝜋(
)𝑚2
4
Fundamentals of Thermodynamics 2022-2023 Academic Year
𝑢 = 3.63 𝑚/𝑠
Prof Nyankson
Mass and Energy Balances for Open Systems
Practice Problem
In a major human artery with an internal diameter of 5 mm, the flow of blood,
averaged over the cardiac cycle, is 5 cm3·s−1. The artery bifurcates (splits) into two
identical blood vessels that are each 3 mm in diameter. What are the average
velocity and the mass flow rate upstream and downstream of the bifurcation? The
density of blood is 1.06 g·cm−3.
3 mm
5 mm
3 mm
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Mass Balance for Open Systems
The rate of change of mass within the control
volume equals the net rate of flow of mass into
the control volume.
𝑑𝑚𝑐𝑣
+ ∆ 𝑚ሶ 𝑓𝑠 = 0
𝑑𝑡
∆ 𝑚ሶ 𝑓𝑠 = 𝑚ሶ 3 − 𝑚ሶ 2 − 𝑚ሶ 1
This can be written as
𝑑𝑚𝑐𝑣
+ ∆ 𝜌𝑢𝐴 𝑓𝑠 = 0
𝑑𝑡
This equation is called the continuity equation.
The control surface is extensible
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Mass Balance for Open Systems
• For a steady state process, the condition within the control volume does not
change with time.
• The accumulation terms is zero, the continuity equation reduces to
∆ 𝜌𝐴𝑢 𝑓𝑠 = 0
When there is a single entrance and a single exit:
𝜌2 𝑢2 𝐴2 − 𝜌1 𝑢1 𝐴1 = 0
Hence
𝜌2 𝑢2 𝐴2 = 𝜌1 𝑢1 𝐴1 = 𝑚ሶ = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Also
𝑢1 𝐴1 𝑢2 𝐴2 𝑢𝐴
𝑚ሶ =
=
=
𝑉Thermodynamics
𝑉
𝑉 Year
Fundamentals of
2022-2023
Academic
1
2
Prof Nyankson
General Energy Balance
• The rate of change of energy within a control volume equals the net rate of
energy transfer into the control volume.
• Streams flowing into and out of the control volume have associated with them
energy in its internal, potential and kinetic forms.
• The rate of energy accumulation within the control volume is
𝑑 𝑚𝑈
𝑑𝑡
𝑐𝑣
= −∆
1 2
𝑈 + 𝑢 + 𝑧𝑔 𝑚ሶ
2
−∆= 𝑖𝑛 − 𝑜𝑢𝑡
+ 𝑄ሶ + 𝑤𝑜𝑟𝑘 𝑟𝑎𝑡𝑒
𝑓𝑠
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
General Energy Balance
The work rate term includes
• −∆ 𝑃𝑉 𝑚ሶ 𝑓𝑠
ሶ shaft work, work due to expansion or contraction of the
• All other work (𝑊):
control volume
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
General Energy Balance
𝑑 𝑚𝑈
𝑑𝑡
𝑐𝑣
= −∆
1 2
𝑈 + 𝑢 + 𝑧𝑔 𝑚ሶ
2
+ 𝑄ሶ − ∆ 𝑃𝑉 ሶ 𝑚ሶ
𝑓𝑠
Considering the definition of enthalpy; H=U+PV
𝑑 𝑚𝑈
𝑑𝑡
𝑐𝑣
= −∆
1 2
𝐻 + 𝑢 + 𝑧𝑔 𝑚ሶ
2
+ 𝑄ሶ +ሶ 𝑊ሶ
𝑓𝑠
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
𝑓𝑠
+ 𝑊ሶ
General Energy Balance
Which is usually written as
𝑑 𝑚𝑈
𝑑𝑡
𝑐𝑣
+∆
1 2
𝐻 + 𝑢 + 𝑧𝑔 𝑚ሶ
2
= 𝑄ሶ + 𝑊ሶ
𝑓𝑠
For the case where the kinetic and potential energies of the fluid are negligible
𝑑 𝑚𝑈
𝑑𝑡
𝑐𝑣
+ ∆ 𝐻𝑚ሶ
𝑓𝑠
= 𝑄ሶ + 𝑊ሶ
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Energy Balances for Steady State Processes
• Flow processes for which the accumulation term is zero is said to occur at
steady state.
∆
1 2
𝐻 + 𝑢 + 𝑧𝑔 𝑚ሶ
2
= 𝑄ሶ + 𝑊ሶ 𝑠
𝑓𝑠
For one entrance and one exit
1 2
∆ 𝐻 + 𝑢 + 𝑧𝑔 𝑚ሶ = 𝑄ሶ + 𝑊ሶ 𝑠
2
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Energy Balances for Steady State Processes
• Dividing through by the mass flow rate gives
1 2
𝑄ሶ 𝑊ሶ 𝑠
∆ 𝐻 + 𝑢 + 𝑧𝑔 = +
= 𝑄 + 𝑊𝑠
2
𝑚ሶ
𝑚ሶ
For the case where the kinetic and potential energy terms are omitted because
they are negligible when compared to others
∆𝐻 = 𝑄 + 𝑊𝑠
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Energy Balances for Steady State Processes
Working Example 4
Steam flows at a steady state through a converging insulated nozzle, 25 cm long
and with inlet diameter of 5 cm. at the nozzle entrance (state 1), the temperature
and pressure are 325 C and 700 kPa, and the velocity is 30 m/s. at the nozzle
exit (state 2), the steam temperature and pressure are 240 C and 350 kPa. What
is the velocity of the steam at the nozzle exit, and what is the exit diameter?
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Energy Balances for Steady State Processes
Working Example 4
Steam flows at a steady state through a converging insulated nozzle, 25 cm long
and with inlet diameter of 5 cm. at the nozzle entrance (state 1), the temperature
and pressure are 325 C and 700 kPa, and the velocity is 30 m/s. at the nozzle
exit (state 2), the steam temperature and pressure are 240 C and 350 kPa. What
is the velocity of the steam at the nozzle exit, and what is the exit diameter?
State 2
State 1
d2 =
P2 = 350 kPa
T2 = 240 C
u2
d1 = 5 cm
P1 = 700 kPa
T1 = 350 C
u1= 30 m/s
Fundamentals of Thermodynamics 2022-2023 Academic Year
L Nyankson
= 25 cm
Prof
Energy Balances for Steady State Processes
Working Example 4
State 2
State 1
d2 =
P2 = 350 kPa
T2 = 240 C
u2
d1 = 5 cm
P1 = 700 kPa
T1 = 350 C
u1= 30 m/s
L = 25 cm
• Insulated so Q = 0
• ∆𝑧 is negligible
• No shaft work, Ws = 0
1 2
∆ 𝐻 + 𝑢 + 𝑧𝑔 = 𝑄 + 𝑊𝑠
2
1 2
∆ 𝐻+ 𝑢 =0
2
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Energy Balances for Steady State Processes
Working Example 4
State 2
State 1
d2 =
P2 = 350 kPa
T2 = 240 C
u2
d1 = 5 cm
P1 = 700 kPa
T1 = 350 C
u1= 30 m/s
L = 25 cm
1 2
1 2
∆ 𝐻 + 𝑢 = 0 = 𝐻2 − 𝐻1 + 𝑢2 − 𝑢12
2
2
𝑢22
−
𝑢12
= 2(𝐻1 − 𝐻2 )
From the steam table, at P1 = 700kPa and T1 = 350 C ,
H1 = 3112.1 kJ/kg, V1 = 388.61 cm3/g, U1 = 2840.1 kJ/kg
From the steam table, at P2 = 350kPa and T2 = 240 C ,
H2 = 2945.7 kJ/kg, V2 = 667.75 cm3/g, U2 = 2712.2 kJ/kg
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Read the enthalpy values from steam tables.
Energy Balances for Steady State Processes
Working Example 4
State 2
State 1
d2 =
P2 = 350 kPa
T2 = 240 C
u2
d1 = 5 cm
P1 = 700 kPa
T1 = 350 C
u1= 30 m/s
L = 25 cm
1 2
1 2
∆ 𝐻 + 𝑢 = 0 = 𝐻2 − 𝐻1 + 𝑢2 − 𝑢12
2
2
𝑢22
−
𝑢12
𝑢22 − 302 = 2(3112.1 − 2945.7) × 103
= 2(𝐻1 − 𝐻2 )
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Read the enthalpy values from steam tables.
𝑢2 = 577.7 𝑚/𝑠
Energy Balances for Steady State Processes
Working Example 4
State 2
State 1
d2 =
P2 = 350 kPa
T2 = 240 C
u2
d1 = 5 cm
P1 = 700 kPa
T1 = 350 C
u1= 30 m/s
L = 25 cm
1 2
1 2
∆ 𝐻 + 𝑢 = 0 = 𝐻2 − 𝐻1 + 𝑢2 − 𝑢12
2
2
𝑢22
−
𝑢12
𝑢22 − 302 = 2(3112.1 − 2945.7) × 103
= 2(𝐻1 − 𝐻2 )
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Read the enthalpy values from steam tables.
𝑢2 = 577.7 𝑚/𝑠
Energy Balances for Steady State Processes
Working Example 4
State 2
State 1
d2 =
P2 = 350 kPa V1 = 388.61 cm3/g, U1 =
2840.1 kJ/kg
T2 = 240 C
V2 = 667.75 cm3/g, U2 =
u2
2712.2 kJ/kg
d1 = 5 cm
P1 = 700 kPa
T1 = 350 C
u1= 30 m/s
L = 25 cm
Use the continuity equation (mass balance)
𝑚ሶ 1 = 𝑚ሶ 2
𝜌1 𝑉1ሶ = 𝜌2 𝑉ሶ2
𝜌1 𝑢1 𝐴1 = 𝜌2 𝑢2 𝐴2
𝑑2 = 𝑑1
𝜌1 𝑢1
𝑉2 𝑢1
= 𝑑1
𝜌2 𝑢2
𝑉1 𝑢2
667.75 × 30
𝑑2 = 5
= 1.49 𝑐𝑚
388.61 × 577.7
Fundamentals of Thermodynamics 2022-2023 Academic Year
Prof Nyankson
Tables of Thermodynamic Properties
• In many instances, thermodynamic properties are tabulated.
• Check appendix F in Van Ness.
Superheated steam originally at P1 and T1 expands through a nozzle to an
exhaust pressure P2. Assuming the process is reversible and adiabatic,
determine the down stream state of the steam and ΔH for the following
condition
P1=1000 kPa, t1 = 250 C and P2 = 200 kPa.
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Turbines (Expanders)
• The expansion of a gas through a nozzle produce a highvelocity stream in a process that converts internal energy
into kinetic energy.
• The kinetic energy is in turn converted to shaft work.
When the stream impinges on a blade attached to rotating
shaft.
1 2
∆ 𝐻 + 𝑢 + 𝑧𝑔 𝑚ሶ = 𝑄ሶ + 𝑊ሶ 𝑠
2
𝑚∆𝐻
ሶ
= 𝑊ሶ 𝑠
𝑚∆𝐻 = 𝑊𝑠
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The 2nd Law of Thermodynamics
➢Processes observed in nature have a natural direction of change
✓Motion picture ran backwards
➢ A wilted flower does not revive
✓Videotape ran backwards
to full bloom and then curl into
✓A pebble dropped into a pond
a bud.
➢ Time flows in one direction
➢ The 2nd law of thermodynamics distills
this aspect of experience and states it
succinctly and quantitatively.
➢ The 2nd law determines the direction and
extent of a process that brings a system
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The 2nd Law of Thermodynamics
➢Processes observed in nature have a natural direction of change
✓Motion picture ran backwards
➢ A wilted flower does not revive
✓Videotape ran backwards
to full bloom and then curl into
✓A pebble dropped into a pond
a bud.
➢ Time flows in one direction
➢ The 2nd law of thermodynamics distills
this aspect of experience and states it
succinctly and quantitatively.
➢ The 2nd law determines the direction and
extent of a process that brings a system
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The 2nd Law of Thermodynamics
The initial state of a system can be;
• Its equilibrium state (state of rest)
• No equilibrium state
A
Natural
or
Spontaneous
or
Irreversible Process
Irreversible
processes
are accompanied by
dissipation
or
degradation
B
Mixing
Once equilibrium state is reached, the capacity of the system for doing further work (to cause a change) is
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exhausted.
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The 2nd Law of Thermodynamics
There are two distinct spontaneous processes:
• Conversion of heat to work
• Flow of heat down temperature gradient
∆𝑆 =
1.
𝑄
𝑇
2.
3.
𝑄
𝑇1
>
𝑄
𝑇2
Extent of degradation varies
from one process to another.
Heat reservoir is at T2, Work is performed
and the heat produced Q enters the heat
reservoir.
Heat reservoir at T2 is placed in thermal
contact with another heat reservoir at T1. Q
flows from T2 to T1.
Heat reservoir is at T1, Work is performed
and the heat produced Q enters the heat
reservoir.
Quantitative measure of degree of
irreversibility
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The 2nd Law of Thermodynamics
• Conduct the process in such a way that the degree of irreversibility is
minimal.
• The degree of irreversibility is zero (no degradation/dissipation).
• Reversible Process.
• If a process is reversible, then the concept of spontaneity is not
applicable.
• If spontaneity is removed, the system will be at equilibrium at all
times.
• The process will pass through continuum of equilibrium states.
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The 2nd Law of Thermodynamics
Entropy is a state function, which when summed up for both system
and surroundings for any process, always changes in the same
direction.
In every volume element of any system and surroundings that may be
experiencing change, at every instant in time, the entropy production
is positive.
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The 2nd Law of Thermodynamics
➢In every volume element of any system and surroundings that may be
experiencing a change, at every instant in time, the entropy production is
positive.
➢The change in entropy of a system is not restricted to entropy transferred
across its boundaries.
✓Production of entropy-the first law is conservative law but the 2nd law is not.
∆𝑆𝑠𝑦𝑠 = ∆𝑆𝑡 + ∆𝑆𝑃
The transfer term may be positive or negative depending upon the nature of the
process and the flows at the boundary of the system.
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The 2nd Law of Thermodynamics
➢The mathematical statement of the 2nd law:
∆𝑆𝑝 ≥ 0
Consider the entropy change experienced by the surroundings to a system during
a process. Treating the surroundings as the system:
∆𝑆𝑠𝑢𝑟 = ∆𝑆𝑡1 + ∆𝑆𝑝1
The entropy of the universe:
∆𝑆𝑢𝑛 = ∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟 = ∆𝑆𝑡1 + ∆𝑆𝑝1 + ∆𝑆𝑡 + ∆𝑆𝑝
This reduces to
∆𝑆𝑢𝑛 = ∆𝑆𝑝1 + ∆𝑆𝑝
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The 2nd Law of Thermodynamics
The entropy of the universe:
∆𝑆𝑢𝑛 = ∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟 = ∆𝑆𝑡1 + ∆𝑆𝑝1 + ∆𝑆𝑡 + ∆𝑆𝑝
This reduces to
∆𝑆𝑢𝑛 = ∆𝑆𝑝1 + ∆𝑆𝑝
Why?
Thus, while the entropy of a system may increase or decrease during a process, the
entropy of the universe, taken as system plus whatever surroundings are involved in
producing the changes within the system, can only increase.
For an infinitesimal change: 𝑑𝑆𝑠𝑦𝑠 = 𝑑𝑆𝑡 + 𝑑𝑆𝑝
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The 2nd Law of Thermodynamics
Reversible and Irreversible Process:
Gas
Gas
b
a
Gas
Gas
c
1. PROCESS 1: The pan and its contents are
lifted off the top of the piston.
➢ The system experiences a sudden drop in
the force applied to it.
➢ Process accompanied with dissipation,
finite entropy production and permanent
changes in the universe.
2. PROCESS 2: The particles are removed
with a pair of tweezers one particle at a time
and letting the system come to rest before the
next particle is removed
➢ No dissipation, no entropy production and
no permanent changes.
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Relationship Between Entropy Transfer and Heat
Absorbed
For a system taken through a reversible process, the heat absorbed by the system
during that step is:
𝛿𝑄𝑟𝑒𝑣
ර
𝑇
For any reversible cyclic path, the above expression is zero.
𝛿𝑄𝑟𝑒𝑣
𝑇
is a state function which is defined to be the entropy of the system
𝛿𝑄𝑟𝑒𝑣 = 𝑇𝑑𝑆
𝑄𝑟𝑒𝑣 = ර 𝑇𝑑𝑆
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Combined Statement of the 1st and 2nd Laws
𝛿𝑊 = 𝐹𝑑𝑥
𝐹
𝛿𝑊 = 𝐴𝑑𝑥,
𝐴
𝑑𝑉 = −𝐴𝑑𝑥
𝐹
𝛿𝑊 = − 𝑑𝑉
𝐴
𝛿𝑊 = 𝑃(−𝑑𝑉)
𝛿𝑊 = −𝑃𝑑𝑉
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Combined Statement of the 1st and 2nd Laws
First Law:
𝑑𝑈 = 𝛿𝑄 + 𝛿𝑊 + 𝛿𝑊 1
𝛿𝑄 = 𝑇𝑑𝑆,
𝛿𝑊 = −𝑃𝑑𝑉
𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 + 𝛿𝑊 1
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3rd Law of Thermodynamics
A system consisting of one mole of carbon and one mole of pure silicon initially
at absolute zero is heated from 0 to 1500 K and held at this temperature for Si to
react with C and form SiC. The compound is then cooled to absolute zero.
Si + C
SiC
Since entropy is a state function:
∆𝑆𝑠𝑦𝑠 = ∆𝑆1 + ∆𝑆11 + ∆𝑆111 + ∆𝑆𝐼𝑉 = 0
II
T(K)
I
Si + C
III
SiC
However, when individual entropy changes for
steps I,11,and III are calculated from experimental
measurements it is found that:
∆𝑆1 + ∆𝑆11 + ∆𝑆111 = 0
IV
0
Si + C
It can be deduced that
SiC
∆𝑆𝐼𝑉 = 0
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3rd Law of Thermodynamics
A system consisting of one mole of carbon and one mole of pure silicon initially
at absolute zero is heated from 0 to 1500 K and held at this temperature for Si to
react with C and form SiC. The compound is then cooled to absolute zero.
Si + C
SiC
➢ The entropy change of the compound SiC is the
same for the pure Si and C at 0 K.
II
T(K)
I
Si + C
III
There exists a lower limit to the temperature that can be
attained by matter, called the absolute zero of
temperature, and the entropy of all substances is the
same at that temperature.
IV
0
Si + C
SiC
➢ These empirical observations have established the
third law of thermodynamics:
SiC
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Example
Alumina can be formed from aluminum and oxygen as described by the reaction
below:
3
2𝐴𝑙 + 𝑂2 = 𝐴𝑙2 𝑂3
2
Calculate the change in entropy for this reaction at 298 K.
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