Fundamentals_of_thermodynamics_and_refrigeration_cycles.pptx

Important concepts in thermodynamics
(Based on chapters 1 to 5 of Van Wylen)
Process engineering course
Supervisor: Dr. Avami
TA: Alireza Ghader Tootoonchi
Department of energy engineering – Sharif university of technology- Fall 2021

Control volume
Control mass
Flows
Mass
Energy
2
Alireza Ghader Tootoonchi – Sharif university of technology- Fall 2021

𝑃𝑉 = 𝑛𝑅𝑇
𝐸 =
1
2
𝑚𝑉2 =
3
2
𝐾𝐵𝑇
Macro
Micro
MACROSCOPIC VERSUS MICROSCOPIC POINTS OF VIEW
3

A phase is defined as a quantity of matter that is homogeneous
throughout.
State, which is determined by some parameters called thermodynamic properties
(e.g. Temperature, pressure, quality, enthalpy, entropy, specific volume, density and
so on)
thermodynami
c
properties
Intensiv
e
Extensiv
e
Temp,
density…
Heat, internal
energy…
PROPERTIE
S
4

Whenever one or more of the properties of a system change, we
say that a change in state has occurred. For example, when one of
the weights on the piston in Fig. 2.3 is removed, the piston rises and
a change in state occurs, for the pressure decreases and the
specific volume increases. The path of the succession of states
through which the system passes is
called the process.
Several processes are described by the fact that one property remains constant.
The prefix iso- is used to describe such a process. An isothermal process is a
constant-temperature process, an isobaric (sometimes called isopiestic) process is
a constant-pressure process, and an isochoric process is a constant-volume
process
PROPERTIE
S
5

When a system in a given initial state goes through a number of different
changes of state or processes and finally returns to its initial state, the system
has undergone a cycle.
Energy
One very important concept in a study of thermodynamics is energy.
Energy is a fundamental concept, such as mass or force, and, as is
often the case with such concepts, it is very difficult to define. Energy
has been defined as the capability to produce an effect. Fortunately
the word energy and the basic concept that this word represents are
familiar to us in everyday usage, and a precise definition is not
essential at this point. Energy can be stored within a system and can
be transferred (as heat, for example) from one system to another.
CYCLE
6

The zeroth law of thermodynamics states that when two bodies have equality of
temperature with a third body, they in turn have equality of temperature with each
other. This seems obvious to us because we are so familiar with this experiment.
Because the principle is not derivable from other laws, and because it precedes
the first and second laws of thermodynamics in the logical presentation of
thermodynamics, it is called the zeroth law of thermodynamics. This law is really
the basis of temperature measurement. Every time a body has equality of
temperature with the thermometer, we can say that the body has the temperature
we read on the thermometer.
Zeroth law of
thermodynamics
7

8

Triple
point
9

A pure substance is one that has a homogeneous and invariable chemical
composition. It may exist in more than one phase, but the chemical composition is
the same in all phases. Thus, liquid water, a mixture of liquid water and water vapor
(steam), and a mixture of ice and liquid water are all pure substances; every phase
has the same chemical composition.
THE PURE SUBSTANCE
The term saturation temperature designates the temperature at which vaporization
takes place at a given pressure. This pressure is called the saturation pressure for
the given temperature. Thus, for water at 99.6◦C the saturation pressure is 0.1
MPa, and for water at 0.1 MPa the saturation temperature is 99.6◦C.
10

saturated
liquid
subcooled liquid
compressed
liquid
saturated
vapor
quality superheated
vapor
States of a pure
substance
11

Critical
point
12

𝑠 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔 ℎ = ℎ𝑓 + 𝑥ℎ𝑓𝑔 𝑢 = 𝑢𝑓 + 𝑥𝑢𝑓𝑔
QUALITY
13

INDEPENDENT PROPERTIES OF A PURE SUBSTANCE
To understand the significance of the term independent property, consider the
saturated-liquid and saturated-vapor states of a pure substance. These two states
have the same pressure and the same temperature, but they are definitely not the
same state. In a saturation state, therefore, pressure and temperature are not
independent properties. Two independent properties, such as pressure and specific
volume or pressure and quality, are required to specify a saturation state of a pure
substance.
the state of a simple compressible pure substance (that is, a pure substance in the
absence of motion, gravity, and surface, magnetic, or electrical effects) is defined by
two independent properties.
14

Thermodynamic properties of a pure substance and the phase boundaries for solid,
liquid, and vapor states are discussed. Phase equilibrium for vaporization (boiling
liquid to vapor), or the opposite, condensation (vapor to liquid); sublimation (solid to
vapor) or the opposite, solidification (vapor to solid); and melting (solid to liquid) or the
opposite, solidifying (liquid to solid), should be recognized. The three-dimensional P–
v–T surface and the two-dimensional representations in the (P, T), (T, v) and (P, v)
diagrams, and the vaporization, sublimation, and fusion lines, are related to the
printed tables in Appendix B. Properties from printed and computer tables covering a
number of substances are introduced, including two-phase mixtures, for which we use
the mass fraction of vapor (quality).
SUMMARY
15

v-u
16

h-s
17

v-u
18

h-s
19

Super heat vapor
20

Subcooled liquid
21

Water
T = 80 C and P = 500 KPa
T = 100 C and P = 10 KPa
P = 30 KPa and x = 0
T = 65 C and s = 6.5 => h=?
T = 57 C and x = 1 => h=?
How to use the table
22

T h
55 2600
57 y
60 2609
55−60
2600−2609
=
55−57
2600−𝑦
=> 𝑦 = 2603.6
How to use the table
Saturated vapor
Linear interpolation
23

THE FIRST LAW OF THERMODYNAMICS FOR A MASS VOLUME
conservation of
energy
𝐸2 − 𝐸1 = 𝑄 − 𝑊
𝑒 = 𝑢 +
𝑣2
2
+ 𝑔𝑧
24

ENTHALPY
25

Any question??
a.ghadertotonchi@energy.sharif.edu
aghadertootoonchi@gmail.com
26

27
Important concepts in thermodynamics
(Based on chapters 6 to 10 of Van Wylen)

First-Law Analysis for a Control Volume
continuity
equation
28

Example 1
Hot
Cold
29

Example 1
Flow P T m
R-134a (Hot, in) 1 MPa 60 C 0.2 Kg/s
R-134a (Hot, out) 0.95 MPa 35 C 0.2 Kg/s (liq)
Water (cold, in) 10
?
Water (cold, out) 20
Enthalpies can be calculated
based on the above table and
thermodynamic tables of the
reference book.
(R134a: p810, water: p777)
30

Turbine (Gas or Steam)
Pump (incompressible fluids) Compressor (compressible fluids)
• 𝑃𝑒 > 𝑃𝑖
• ℎ𝑒 > ℎ𝑖, ∆ℎ ≅ 0
• 𝑤𝑝 = 𝑣 𝑃𝑒 − 𝑃𝑖 = ℎ𝑒 − ℎ𝑖
• 𝑠𝑒 = 𝑠𝑖 (𝑖𝑓 𝜂𝑖𝑠𝑛 = 1)
• 𝑇𝑒 > 𝑇𝑖, Δ𝑇 ≅ 0
• 𝑞𝑝 = 0 (𝑖𝑓 𝑛𝑜𝑡ℎ𝑖𝑛𝑔 𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑒𝑑)
• 𝑃𝑖 > 𝑃𝑒
• ℎ𝑖 > ℎ𝑒
• 𝑤𝑇 = ℎ𝑖 − ℎ𝑒
• 𝑇𝑖 > 𝑇𝑒
• 𝑞𝑇 = 0 (𝑖𝑓 𝑛𝑜𝑡ℎ𝑖𝑛𝑔 𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑒𝑑)
• 𝑃𝑒 > 𝑃𝑖
• ℎ𝑒 > ℎ𝑖
• 𝑤𝑝 = ℎ𝑒 − ℎ𝑖
• 𝑇𝑒 > 𝑇𝑖
• 𝑞𝑝 = 0 (𝑖𝑓 𝑛𝑜𝑡ℎ𝑖𝑛𝑔 𝑚𝑒𝑛𝑡𝑖𝑜𝑛𝑒𝑑)
i
i
i
e
e
e
w
w w
Common equipment
31

Condenser Boiler Evaporator
𝑞𝑐 𝑞𝑐
𝑞ℎ
i e i e i e
• 𝑃𝑒 = 𝑃𝑖 (𝑎𝑠𝑠𝑢𝑚𝑒 𝑛𝑜
𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝)
• ℎ𝑖 > ℎ𝑒
• 𝑤𝑐 = 0
• 𝑠𝑒 ≤ 𝑠𝑖
• 𝑇𝑒 ≤ 𝑇𝑖
• ℎ𝑖 < ℎ𝑒
• 𝑤𝑏 = 0
• 𝑠𝑒 ≥ 𝑠𝑖
• 𝑇𝑒 ≥ 𝑇𝑖
• ℎ𝑖 < ℎ𝑒
• 𝑤𝑏 = 0
• 𝑠𝑒 ≥ 𝑠𝑖
• 𝑇𝑒 ≥ 𝑇𝑖
Common equipment
32

Example 2
33

Example 2
Thermodynamic tables
34

Example 3
Appendix A, Table A.8, p765
35

The Second Law of Thermodynamic
37

The Kelvin–Planck statement: It is impossible to construct a device that will
operate in a cycle and produce no effect other than the raising of a weight and
the exchange of heat with a single reservoir.
The Clausius statement: It is impossible to construct a device that operates in a
cycle and produces no effect other than the transfer of heat from a cooler body to a
hotter body.
The question that can now logically be posed is this: If it is impossible to have a heat
engine of 100% efficiency, what is the maximum efficiency one can have? The first
step in the answer to this question is to define an ideal process, which is called a
reversible process.
The Second Law of Thermodynamic
38

IRREVERSIBILITY
Friction
Unrestrained Expansion
Heat Transfer (requires work to get back to the initial stage)
Mixing of Two Different Substances
39

THE CARNOT CYCLE
It is impossible to construct an engine that
operates between two given reservoirs and
is more efficient than a reversible engine
operating between the same two
reservoirs.
First Proposition
All engines that operate on the Carnot cycle
between two given constant-temperature
reservoirs have the same efficiency.
Second Proposition
Power cycle: 𝑊
𝑟𝑒𝑣 ≥ 𝑊𝑖𝑟𝑒𝑣 heat pumps: 𝑊𝑖𝑟𝑒𝑣 ≥ 𝑊
𝑟𝑒𝑣 40

THE CARNOT CYCLE
41

IDEAL VERSUS REAL MACHINES (Example 4)
REAL
42

FIRST AND SECOND LAWS
43

44

Entropy
Negative zero positive
𝑠 = 1 − 𝑥 𝑠𝑓 + 𝑥𝑠𝑔
𝑠 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔
Pure substance
Reversible
45

irreversible
entropy balance equation for a control mass
Entropy generation is always associated with the irreversibilities.
Entropy change = Heat transfer + disorder
ENTROPY IN IRREVERSIBLE PROCESSES
46

Energy is conserved but entropy is
not
Change = +in - out +
generation
principle of the increase of
entropy
Feasible processes are those
with nonnegative entropy
change
CONSERVATION
47

45 C
25 C First law:
∆𝐸𝑡𝑜𝑡𝑎𝑙= 0
∆𝐸𝑡𝑜𝑡𝑎𝑙= 0
10 J
Second law:
∆𝑆𝑡𝑜𝑡𝑎𝑙= ∆𝑆𝑠 + ∆𝑆𝑒 ⇒
10
45 + 273
−
10
25 + 273
= −0.00211
𝐽
𝐾
∆𝑆𝑡𝑜𝑡𝑎𝑙= ∆𝑆𝑠 + ∆𝑆𝑒 ⇒ −
10
45 + 273
+
10
25 + 273
= 0.00211
𝐽
𝐾
https://www.youtube.com/watch?v=WTtxlaeC9PY
PROCESS
DIRECTION
48

THERMODYNAMIC LAWS FOR A TYPICAL CONTROL VOLUME
49

THE SECOND LAW OF THERMODYNAMICS FOR A CONTROL
VOLUME
Mass volume
Control volume
steady-state process
steady-state single flow process
steady-state single flow process per unit mass
50

Example 5
51

For a thermodynamic cycle to be feasible, both first and second laws should be applicable.
FEASIBILITY OF A THERMODYNAMIC CYCLE
Violation of the first law of thermodynamics:
Violation of the second law of thermodynamics:
52

Any question??
53

54
Refrigeration cycles
(Based on chapters 11 & 12 of Van Wylen)

Saturated water
x = 0
Saturated vapor
x = 1
REGRIGERATION CYCLES
55

Example 6
Table of R134a: p810
56

Example 7
T1=-20, saturated vapor
P3=
Saturated liquid
saturated vapor
Table of R134a: p810
57

Example 7
58

DEVIATION OF THE ACTUAL VAPOR-COMPRESSION
REFRIGERATION CYCLE FROM THE IDEAL CYCLE
59

CASCADE CONFIGURATION
60

THE AIR-STANDARD REFRIGERATION CYCLE
If we consider the original ideal four-process refrigeration cycle with a
noncondensing (gaseous) working fluid, then the work output during the isentropic
expansion process is not negligibly small, as was the case with a condensing
working fluid.
61

THE AIR-STANDARD REFRIGERATION CYCLE
62

THE AIR-STANDARD REFRIGERATION CYCLE WITH HEAT
EXCHANGER
63

Example 8
P3=500KPa
T3=15C
P2=500KPa
P1=100KPa
T1=-20C=253K
P4=100KPa
64

Example 8
65

Any question??
66

Fundamentals_of_thermodynamics_and_refrigeration_cycles.pptx

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