USNA
THERMAL -FLUID SCIENCES
FIRST EDITION
WARZOHA , SMITH AND BROWNELL

PAGE 1
Introduction ....................................................................................................9
Chapter 1: Introduction to Fluid Mechanics ...................................................10
Fluid properties .......................................................................................................11
Dimensions and Units ...........................................................................................11
Pressure ..............................................................................................................13
Ideal Gas Law .....................................................................................................13
Pressure vs. Depth ...............................................................................................15
Example 1.1 - Surviving the ocean (depth-pressure problem) ................................................16
Example 1.2 - How high is too high? (depth-pressure problem) .............................................16
Measuring Pressure .............................................................................................17
Common Errors when Calculating Pressure ............................................................18
Example 1.3 - Differential U-tube manometer ......................................................................19
Example 1.4 - Extending the range of a manometer with multiple uids (U-tube) ...................21
Fluid Statics ............................................................................................................23
Forces on Planar, Stationary Surfaces ...................................................................23
Example 1.5 - Hydrostatic forces on the Three Gorges dam spillway gate .............................27
Pressure Prisms ....................................................................................................31
Example 1.6 - Force on a vertical wall with water/oil separation ..........................................37
Example 1.7 - Force on a levee .........................................................................................39
Buoyancy ............................................................................................................41
Example 1.8 - Iced water ...................................................................................................43
Stability ..............................................................................................................44
Fluid Dynamics........................................................................................................49
Conservation of Mass ..........................................................................................49
Bernoulli Equation ...............................................................................................51
Applications of the Bernoulli Equation...................................................................53
Example 1.9 - Squirt gun (free jet - Bernoulli) ......................................................................57
Example 1.10 - Pitot-static tube (Bernoulli) ...........................................................................59
Conservation of Momentum .................................................................................60
Example 1.11 - Vane-water Jet ............................................................................................63






























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PAGE 2
Example 1.12 - Nozzle/elbow combination .........................................................................66
Dimensional Analysis ...........................................................................................68
Example 1.13 - Rise of capillary uid ..................................................................................80
Example 1.14 - Contraction in a pipe ..................................................................................83
Example 1.15 - Explosion from an atomic bomb ..................................................................86
Modeling ............................................................................................................89
Example 1.16 - Scaling a spillway .......................................................................................91
Fluid Kinematics and the Navier-Stokes Equation ...................................................93
Example 1.17 - Fluid ow between two stationary plates (Poiselle ow) ................................98
Example 1.18 - Fluid ow between a stationary plate and a moving plate ...........................101
Introduction to Laminar and Turbulent Flow .........................................................104
Internal ow .........................................................................................................104
Pipe Flow ..........................................................................................................104
Example 1.19 - Calculating major head loss in a pipe ........................................................108
Example 1.20 - Pumping pressurized water .......................................................................115
Example 1.21 - Sizing a pool pump ...................................................................................117
Multi-path Flow in Piping Systems ........................................................................119
Pump Curves and Net Positive Suction Head (NPSH) ...........................................120
Example 1.22 - Using pump curves ...................................................................................126
External ow.........................................................................................................129
Boundary Layers ...............................................................................................129
Drag Forces ......................................................................................................136
Example 1.23 - External ow parallel to a at plate (drag force) .......................................139
Example 1.24 - Drag force acting on a parachute .............................................................145
Lift Forces..........................................................................................................147
Example 1.25 - Lift Force acting on a commercial aircraft ...................................................151
Chapter 2: Introduction to Heat Transfer ....................................................154
Fundamental Modes ..............................................................................................155
Conduction Heat Transfer ...................................................................................156
Example 2.1. - Heat transfer through a copper rod ............................................................159
Example 2.2. - Conduction heat transfer through a composite wall .....................................160































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PAGE 3
Convection Heat Transfer ...................................................................................162
Example 2.3. - Cartridge heater operating at steady-state .................................................165
Radiation Heat Transfer......................................................................................166
Thermal Resistor Analysis .......................................................................................169
Example 2.4 - Plane wall, series resistor ............................................................................171
Example 2.5 - Plane wall, parallel resistor.........................................................................173
Example 2.6 - Thermal transport in a pipe with moving uids, series resistor .......................175
Example 2.7 - Heat ow in a multi-layered cable ...............................................................177
Example 2.8 - Heat transfer in a pipe with moving uids, parallel resistor ...........................179
Example 2.9 - Boiler screen tube (energy balance and heat transfer rate per unit length) ....183
Extended Surfaces and Heat Sinks ......................................................................185
Example 2.10 - Rectangular n .........................................................................................191
Example 2.11 - Pin ns and parallel paths for heat ow .....................................................193
Example 2.12 - Circular annular ns ..................................................................................196
Contact Thermal Resistance ................................................................................199
Example 2.13 - Contact thermal resistance in an electronic component ...............................201
External Convection Heat Transfer..........................................................................204
Laminar Flow over a Flat Plate............................................................................206
Example 2.14 - Convection heat transfer over a at plate ...................................................212
Cylinders in Cross Flow ......................................................................................215
Example 2.15 - Cylindrical heater in an oil bath.................................................................217
Chapter 3: Introduction to Thermodynamics ...............................................219
Introduction and Systems .......................................................................................220
Introduction to Systems ......................................................................................220
Properties, Units, and Temperature .........................................................................223
Terminology ......................................................................................................223
Properties .........................................................................................................225
Temperature and the 0 Law of Thermodynamics ..................................................226
Example 3.1 - Finding gas pressure in a spring-piston system .............................................229
Energy and the First Law of Thermodynamics ..........................................................231
Heat and Work Energy..........................................................................................233































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PAGE 4
Heat Energy, Q .................................................................................................233
Work Energy, W ...............................................................................................233
Power ...............................................................................................................234
Types of Work ...................................................................................................234
Energy Transfers Between states and boundary work ..............................................236
States, Paths, and Boundary Work .....................................................................236
Path Types ........................................................................................................242
First Law for Closed Systems ..............................................................................243
Example 3.2 - Calculating Boundary Work .......................................................................245
Example 3.3 - First Law Energy Balance and Boundary Work .............................................246
Finding Thermodynamic Properties .........................................................................247
Ideal Gases ......................................................................................................247
Finding Properties for Ideal Gases in Underspeci ed Cases..................................250
Calculating Boundary Work for Ideal Gases .......................................................252
Example 3.4 - Air Processes (Ideal Gases and Paths) .........................................................253
Pure Substances ................................................................................................255
Phases of Pure Substances and Corresponding Phase Diagrams ...........................255
The Vapor Dome ...............................................................................................258
Property Tables (Pure Substances): Determining Phase .........................................259
Property Tables (Pure Substances): Extracting Properties ......................................264
Example 3.5 - Cooled Refrigerant ....................................................................................270
Thermodynamic Properties for Incompressible Substances (Oil, Sea Water, etc.) ...273
Example 3.6 - Cooling of Copper (Incompressible Substance) ............................................274
Open Systems .......................................................................................................276
Conservation of Mass ........................................................................................276
Conservation of Energy .....................................................................................277
Open Systems Analysis .........................................................................................280
Nozzles and Diffusers........................................................................................280
Example 3.7 - Analyzing a Nozzle ...................................................................................282
Turbines ............................................................................................................283






























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PAGE 5
Example 3.8 - Analyzing a Steam Turbine .........................................................................284
Pumps and Compressors ....................................................................................287
Throttles ............................................................................................................288
Heat Exchangers ...............................................................................................288
Example 3.9 - Combination Heat Exchanger .....................................................................292
Heat Exchanger Analysis: Effectiveness-NTU ...........................................................296
Heat Exchanger Types .......................................................................................296
Overall Heat Transfer Coef cient ........................................................................300
Example 3.10 - Calculating an Overall Heat Transfer Coef cient ........................................302
Effectiveness-NTU Method .................................................................................305
Example 3.11 - Effectiveness-NTU: Determining Outlet Temperatures (CASE A) ....................309
Example 3.12 - Effectiveness-NTU: Sizing a Heat Exchanger (CASE B) .................................312
Introduction to Thermodynamic Cycles ....................................................................314
Second Law of Thermodynamics .............................................................................318
Carnot Cycle .....................................................................................................318
Entropy and the Clausius Inequality.....................................................................319
Finding Entropy .................................................................................................321
Isentropic Processes ...........................................................................................322
Isentropic Ef ciencies .........................................................................................324
Example 3.13 - Isentropic Ef ciency of a Steam Turbine .....................................................327
Example 3.14 - Isentropic Ef ciency of a Pump ..................................................................330
Chapter 4: Thermodynamic Systems ..........................................................332
Introduction to Combustion Processes .....................................................................333
Fuels .................................................................................................................333
Combustion Air .................................................................................................334
Combustion Reaction .........................................................................................334
Air-Fuel Ratio ....................................................................................................335
Heating Values ..................................................................................................336
Example 4.1 - Combustion in a Gas Turbine Engine ...........................................................338
Otto Cycle ............................................................................................................340






























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PAGE 6
Thermodynamic Analysis ....................................................................................343
Example 4.2 - Otto Cycle .................................................................................................345
Diesel Cycle ..........................................................................................................348
Thermodynamic Analysis ....................................................................................351
Example 4.3 - Diesel Cycle...............................................................................................353
Gas Turbine Engines (Brayton Cycle) ......................................................................356
Ideal Brayton Cycle ...........................................................................................358
Non-ideal Brayton Cycle ....................................................................................359
Example 4.4 - Non-ideal Brayton Cycle .............................................................................361
Regenerative Brayton Cycle ...............................................................................363
Example 4.5 - Regenerative Brayton Cycle........................................................................366
Split-Shaft Gas Turbines .........................................................................................368
Example 4.6 - Split-shaft Gas Turbine ................................................................................369
Jet Propulsion Cycles .............................................................................................372
Turbojet Engines ................................................................................................372
Example 4.7 - Turbojet Engine ..........................................................................................374
Vapor Power Cycles ..............................................................................................377
Ideal Rankine Cycle ...........................................................................................377
Example 4.8 - Ideal Rankine Cycle ...................................................................................381
Real Rankine Cycle ............................................................................................384
Example 4.9 - Real Rankine Cycle ....................................................................................385
Reheat Rankine Cycle ........................................................................................389
Example 4.10 - Reheat Rankine Cycle ...............................................................................391
Nuclear Power ......................................................................................................395
Principles of Nuclear Fission ...............................................................................395
Types of Nuclear Reactors .................................................................................397
Example 4.11 - Pressurized Water Reactor ........................................................................402
Vapor Compression Refrigeration/Heat Pumps ........................................................407
Thermodynamic Analysis ....................................................................................407
Example 4.12 - Vapor Compression Refrigeration ..............................................................409
Appendix ....................................................................................................412
































PAGE 7
Unit conversions ....................................................................................................413
Values of Surface Roughness for pipe ow ..............................................................415
Additional Drag coef cients over 3-D Bodies ...........................................................416
Additional Drag coef cients over Real Bodies .........................................................417
Fluid Properties (SI Units) .......................................................................................418
Fluid Properties (BG Units) .....................................................................................422
Thermal Properties of Solids...................................................................................424
Thermal Properties of Fluids ...................................................................................429
Ideal Gas Speci c Heats (SI) .................................................................................435
Ideal Gas Speci c Heats (EEU) ..............................................................................436
Saturated Water - Temperature Table (SI) ...............................................................437
Saturated Water - Pressure Table (SI) .....................................................................440
Superheated Water Table (SI)................................................................................443
Saturated R-134a - Temperature Table (SI) ..............................................................449
Saturated R-134a - Pressure Table (SI) ....................................................................451
Superheated R-134a Tables (SI) .............................................................................453
Saturated Water - Temperature Table (EEU) ............................................................456
Saturated Water - Pressure Table (EEU) ..................................................................459
Superheated Water Table (EEU) .............................................................................461
Compressed Liquid Water Table (EEU) ...................................................................467
Saturated R-134a - Temperature Table (EEU)...........................................................468
Saturated R-134a - Pressure Table (EEU) .................................................................470
Superheated R-134a Tables (EEU) ..........................................................................472























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PAGE 8
INTRODUCTION
A practical understanding of uid mechanics, thermodynamics and heat transfer is critical
for the design and implementation of propulsive and energy systems that are themselves
vital to national infrastructure and the operation of defense systems. This textbook is
designed to provide undergraduate-level engineering students with su cient knowledge to
understand, analyze and design practical engineering systems that (1) produce power and
(2) cool or heat an environment.
This text begins with a rudimentary treatment of
insight into topics that include
uid mechanics. The student will gain
uid properties and units, pressure measurements,
uid
statics, uid dynamics and dimensional similitude. We place a speci c emphasis on Naval
applications throughout the text, given the unique background of the instructors. However,
we note that this textbook can serve as a valuable reference for any engineering discipline.
Likewise, we introduce students to basic concepts in thermodynamics and heat transfer.
Because of the unique nature of the course, we introduce topics in heat transfer prior to
those introduced for thermodynamics. To overcome issues associated with the sequence of
instruction, we provide students with a basic formulation of the rst law of thermodynamics
in our introduction of heat transfer. Topics covered in heat transfer include the basic modes
of thermal transport (Conduction, Convection and Radiation), thermal resistor networks,
heat sink design and analysis, coupled heat transfer and
uid
ow and heat exchanger
design. It is important to understand that this material is not intended as a substitute for a
standard undergraduate course in heat transfer.
Finally, we cover both the foundational and practical elements of thermodynamics for
undergraduate engineers. Topics include conventional thermodynamic properties, energy
conservation, second law analyses (Carnot cycle and device e ciencies), closed and open
system analyses and the fundamental aspects of thermodynamic cycles (Otto, Diesel,
Brayton, split-shaft gas turbines, aircraft engines, Rankine and vapor-compression).
This book is meant to provide students with an adequate resource for the treatment of uid
mechanics, thermodynamics and heat transfer in courses that are not tethered to Mechanical
Engineering curricula.
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PAGE 9
CHAPTER 1: INTRODUCTION TO
FLUID MECHANICS
OVING
WATER
COURSE
PURSUANT
OCCASIONS
OBSTACLE
THE
IN
COURSE
ITS
IT
STRIVES
IT
PATH,
HAS
TO
MAINTAIN
TO
THE
POWER
AND,
IF
IT
COMPLETES
COMMENCED
BY
WHICH
FINDS
THE
A
THE
SPAN
AN
OF
CIRCULAR
A N D R E VO LV I N G M OV E M E N T .
- Leonardo Da Vinci
What Causes the Eddies of Water, Da Vinci’s Notebooks

PAGE 10
luid properties are fundamental to
our understanding of
uid
movement and for the design of
energy systems.
You are likely to have seen many of these
properties while studying Mechanics and
Electricity and Magnetism in your
primary physics courses. In fact, much of
the work you’ll do as part of this course
will draw on the concepts you learned in
Physics and Chemistry. Before we get
ahead of ourselves, it is important to get a
sense for the dimensions, units and
properties that you’ll use extensively in this textbook and in the homework problems you
solve.
Dimensions and Units
In our study of the thermal- uid sciences (and, coincidentally, all areas of Mechanics), there
is some basic terminology that is useful to become familiar with. In particular, properties are
typically distinguished by whether they are intensive or extensive.
It is useful to rst visualize the di erence between the two before we de ne them. Examine
the image below, which represents your classroom. We will consider the air inside of the
classroom, which has some volume, mass, temperature, pressure and density. Now consider
what happens to each parameter when you draw an imaginary line down the middle of the
room.
Mass, m
1
V
2
1
m
2
1
V
2
1
m
2
Temperature, T
T
T
Pressure, P
P
P
Density, ρ
ρ
ρ
Classroom
Volume, V
Figure 1.1. Two-dimensional schematic representation of a classroom that is lled with air. Consider what happens to the quantities of volume (V),
mass (m), temperature (T), pressure (P) and density (ρ) when you look at two halves of the class independently.
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P A G E 11







FLUID PROPERTIES
and the mass are both halved when the room itself is split into two equal portions, while the
temperature, pressure and density remain the same on each side. In other words, some
properties depend on the size or extent of the system. These properties are termed Extensive
Properties. Conversely, properties that do not depend on the size or extent of the system are called
Intensive Properties.
In many of the problems we will solve in both EM316 and EM317, we will focus exclusively
on speci c properties. These properties are extensive properties that are normalized (i.e. divided
by) mass. A few examples of speci c properties are included in the table below.
Table 1.1. The "total" and "speci c" forms of the extensive uid properties discussed above. Units are provided for both SI (Standard Imperial, or
metric) and EEU (English Engineering Units), which are included in brackets and separated by a comma.
Forms of Properties
Property Name
Total Value [Units]
Speci c Value [Units]
Volume
V [m 3, f t 3]
V
m3 f t 3
,
m * [ kg lbm ]
Internal Energy
U [k J, Bt u]
Heat Capacity
Cp
U k J Bt u
,
m * [ kg lbm ]
k J Bt u
,
[K R ]
cp
kJ
Bt u
,
[ kg ⋅ K lbm ⋅ R ]
*Note: m in the denominator is mass, and not meters as in the corresponding units in the brackets. Note that cp in the bottom right cell is lower
case to indicate Cp /m * .
Speci c properties are useful in that they allow us to compare systems independent of mass.
Other important properties that we will utilize throughout the next few chapters are
included in Table 1.2., below.
Table 1.2. Important properties that will be relevant to those discussed in this chapter.
Types of Properties
Property
Name
Symbols
Relationship
Density
ρ
Viscosity
μ
Speci c
Gravity
SG
ρH2O
Speci c
Weight
γs
ρ⋅g
Pressure
P
F
A
Units
Values (Water)**
Values (Air)**
SI*
EEU*
SI
EEU
SI
EEU
m
V
kg
m3
lbm
f t3
1000
62.3
1.20
0.0752
Experiment
kg
m⋅s
lbm
ft ⋅ s
1.00 ⋅ 10−3
6.73 ⋅ 10−4
1.83 ⋅ 10−5
1.23 ⋅ 10−5
-
-
11.8
0.0752
ρf
Unitless
N
m3
N
m2
1.00
lbf
f t3
lbf
in 2
9810
62.3
Pressure varies with altitude/depth!!
*Note: SI = System International, EEU = English engineering units; **These are properties taken at standard pressure and temperature.

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PA G E 12
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Our question, now, is what happened to each of these properties? Notice that the volume
One of the important concepts covered in this chapter is termed Hydrostatics. In
uid
mechanics, we are often concerned with forces that are acting on solid objects by a
uid.
Typically, we concern ourselves with either the force distribution along the surface of the
object, or the resultant force acting on the object. Before we can calculate either, we must
familiarize ourselves with the general concept of Pressure.
Mathematically, we recognize pressure as an applied force from a uid that is normal to a
solid surface per unit area,
F
P=
A
(1.1)
We relate pressure to either an absolute zero reference or a local atmospheric pressure.
Our observation and measurement of pressure deans on knowledge of both, and how they
relate to one another. The diagram below is a useful reference for de ning each type.
P
Pabs,2
Pgauge = Pabs,2 − Patm
Patm
Pabs = 0
Pvac = Patm − Pabs,1
Pabs,1
Figure 1.2. Schematic of the relationships between atmospheric pressure (Patm), absolute pressure (Pabs), gage pressure (Pgage) and vacuum
pressure (Pvac).
Often, Patm is referenced to year-round average conditions at mid-latitude and sea-level conditions.
These values can be assumed when no atmospheric pressure is de ned, where,
1 atm ≈ 14.7 psia
1 atm = 101.325 kPa
1 atm = 760 mm Hg
1 atm = 29.92 in Hg
N
lbf
5
,
1
Pa
=
1
,
1
bar
=
10
Pa, 1 mm Hg = 1 torr. Also note that 1 kPa =
m2
in 2
1000 Pa and 1 MPa = 106 Pa.
Note: 1 psia =
Ideal Gas Law
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PAG E 13








Pressure
When the uid of interest is a gas that is far from condensing, the relationship between its
pressure, volume and temperature can be described by the Ideal Gas Law as,
p⋅V=m⋅R⋅T
(1.2)
where both the temperature and the pressure are written in absolute units. Thus, the
pressure is an absolute pressure and the temperature is written in either K (SI units) or R
(EEU).
Additionally, m represents the mass of the gas (kg or lbm) and R is the speci c gas constant.
The speci c gas constant for several common gases is provided in Table 1.3.
Table 1.3. Speci c gas constants for some common gases.
Speci c Gas Constants
Ideal Gas
R
kJ
( kg ⋅ K )
R
psi a ⋅ f t 3
( lbm ⋅ R )
R
psi a ⋅ f t 3
( slug ⋅ R )
Air
0.2870
0.3704
11.92
Oxygen (O2)
0.2598
0.3353
10.79
Nitrogen (N2)
0.2968
0.3830
12.32
Argon (Ar)
0.2081
0.2686
8.64
Helium (He)
2.0769
2.6809
86.26
Methane (CH4)
0.5182
0.6688
21.52
Propane (C3H8)
0.1885
0.2433
7.83
Hydrogen (H2)
4.1240
5.3324
171.53
Xenon (Xe)
0.06332
0.08172
2.63
Neon (Ne)
0.4119
0.5316
17.10
It should also be noted that Eqn. 1.2 can be rewritten in terms of density (ρ) or speci c
volume (ν, where ν = 1/ρ) as,
p =ρ⋅R⋅T
(1.3)
p⋅ν =R⋅T
(1.4)
and,
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PAG E 14
Pressure vs. Depth
It is quite likely that, provided you have travelled by air or been submerged under water, you
have experienced a variation in pressure with altitude (or depth) that accompanies a
relatively static uid under the in uence of a gravitational eld.
Let’s take a column of uid and perform a straightforward equilibrium analysis,
p
Cross-sectional
area, Ac
h
y
W
p + Δp
If you look at the forces acting in the - y direction, it is obvious that some additional
pressure (Δp) is required to keep the uid at rest. Thus, constructing a force balance (recall,
F = P⋅Ac) in y yields,
∑
Fy = 0 = (p + Δp) ⋅ Ac − p ⋅ Ac − W
(1.5)
W = m ⋅ g = ρ ⋅ V ⋅ g = ρ ⋅ Ac ⋅ h ⋅ g
(1.6)
where,
After distributing Ac in the rst term (p+Δp) and cancelling pressure terms, we obtain,
Δp = ρ ⋅ g ⋅ h
(1.7)
Provided with Eqn. 1.7, we can say the pressure in a static uid changes with height!
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PAG E 15
Humans can withstand a maximum of 58 psia of pressure under water. If the pressure at the
top of the ocean is approximately atmospheric pressure, what is the maximum depth you
can dive to without sustaining bodily harm?
Solution
The pressure at the top of the ocean is 14.7 psia, while the pressure at our maximum depth
is 58 psia. Thus,
Δp = 58
lbf
lbf
lbf
−
14.7
=
43
.
3
in 2
in 2
in2
Now, we manipulate Eqn. 1.4 to solve for h as,
h=
lbf
43.3
144in 2
1ft 2
⋅
2
Δp
in
=
ft
lbm
ρ⋅g
62.3
⋅ 32.2 ⋅
ft 3
s2
≈ 100ft
1lbf
32.2 lbm2⋅ ft
s
Example 1.2 - How high is too high? (depth-pressure problem)
The highest altitude having an urban settlement exists in La Rinconada Puno, Peru, which is
5100 m above sea level. How much does the air pressure change at this height? Assume the
gas density does not change, despite the large elevation change.
Solution
To determine the change in pressure above sea level, Δ p, we can again utilize equation 1.4
in its native form,
Δp = ρ ⋅ g ⋅ h = 1.20
kg
m
⋅
9.8
⋅ 5100m ⋅
m3
s2
1N
= 59,976
1kg ⋅ m
N
1Pa
1kPa
⋅
⋅
≈ 59 . 9kPa
N
m2
1000Pa
1 2
m
s2
This means that for a standard atmospheric pressure of Patm = 101.325 kPa, the air pressure
in La Rinconda Puno is ≈ 40 kPa. Consider that climbers often refer to elevations that result
in anything less than 40 kPa a "Death Zone" and it becomes clear that living in these
conditions for long periods of time can have signi cant negative consequences on long-term
health.


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PAG E 16


Example 1.1 - Surviving the ocean (depth-pressure problem)
Pressure measurements can be made using a variety of devices. Several standard devices,
shown below, can be used to measure particular pressures (or relative pressures). Each device
takes advantage of the relationship between pressure and depth (or height) that was
outlined in the previous section.
A
Patm
h
h
Patm
1
A
Pabs, A = Patm + ρ ⋅ g ⋅ h
Patm = ρ ⋅ g ⋅ h
Manometer
Barometer
!B
Patm
A
B
h3
A
h2
!A
h1
h2
!A
h1
!m
!m
PA = Patm + ρm ⋅ g ⋅ h 2 − ρA ⋅ g ⋅ h1
PB − PA = ρA ⋅ g ⋅ h1 − ρm ⋅ g ⋅ h 2 − ρB ⋅ g ⋅ h 3
U-tube Manometer
Di erential U-tube Manometer
Figure 1.3. Devices used to measure atmospheric (Barometer), absolute (Manometer and U-tube Manometer), and di erential pressure
(Di erential U-tube Manometer)
A barometer is used to measure atmospheric pressure by reading the height from point 1 to
point A. As the atmospheric pressure increases, so too does the height of the column. A
manometer is used to measure absolute pressure when the atmospheric pressure is known
(or can be measured). A U-tube manometer utilizes a set of uids to improve the accuracy
and/or range of absolute pressures that can be measured. Finally, a di erential U-tube
manometer provides its user with a relative pressure. Note that there could be more than 3
uids used in this type of manometer.
Note that the equations for each type of manometer have been directly provided in the text,
but are unlikely to be the expressions you’ll develop for more complex manometers.
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PA G E 17
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Measuring Pressure
Although Eqn. 1.7 appears straightforward at
rst, there are several common errors that
novel uid dynamists should look out for.
• The pressure at the deeper point is always greater than the pressure at the shallower
point. This seems obvious at
rst glance, but at times the de nition of ΔP within the
equation ΔP = ρgh is not trivial. If h is de ned as a length and always positive, ΔP must
be the pressure at the deeper point minus the pressure at the shallower point,
ΔP = Pdeep − Pshallow. This structure enforces a sign convention that ensures consistency
regardless of which pressure is unknown. In this manner, Eqn. 1.7 may be rewritten as,
Pdeep = Pshallow + ρgh.
• You cannot calculate a pressure somewhere within a uid without rst knowing the
pressure somewhere else within the uid. Looking at the barometer in Fig. 1.3, it appears
like this is exactly what happens: the barometric pressure is ρgh , with no other pressure
appearing in the equation. Note, however, the barometer creates a vacuum as a reference
and the absolute pressure of a vacuum is zero by de nition. Using the form of Eqn. 1.7 in
the bullet above, the barometric pressure is actually calculated via, P1 = PA + ρgh where
PA = 0 (abs).
• You may use either absolute or gage pressures, but you need to be consistent. If you use a
gage pressure as your reference, you will calculate a gage pressure. If you use an absolute
pressure as reference, you will get an absolute pressure.
• This tool was derived for a single uid, so the pressures examined must be contained
within a single uid with a single density. At any horizontal uid- uid interface (either
liquid-liquid or liquid-gas) we can assume that pressure is continuous, i.e. that the pressure
in the lower uid at the interface is the same as the pressure in the upper uid at some
very small distance from the interface.

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PAG E 18
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Common Errors when Calculating Pressure
Find the pressure di erence between points A and B shown in the di erential U-tube
manometer schematic, below. Air, water, benzene, glycerin and mercury are all used within
the manometer. The speci c weight (γs = ρ ⋅ g ) of benzene is 8643 N/m3, the density of
glycerin is 1.26 g/cm3 and the speci c gravity (SG) of mercury is 13.6, while the air is at
standard conditions. Assume that the pressure does not vary with height for any of the
uids in the manometer. Find the pressure di erence between points A and B in kPa.
Solution
The simplest way to solve this problem is to "start" at point A and trace your way through
each
uid until you reach point B. Recall that ΔP = ρ ⋅ g ⋅ Δh , where we use Δh here to
represent a change in height. We choose the direction of gravity to be positive as we trace.
Thus, we end up with,
PA + ρM ⋅ g ⋅ (0.15m − (0.23m − 0.05m − 0.08m)) − ρG ⋅ g ⋅ (0.15m) − ρB ⋅ g ⋅ (0.05m) + ρW ⋅ g ⋅ (0m) − ρair ⋅ g ⋅ (0.08m) = PB
where subscripts M, G, B and W represent the uids mercury, glycerin, benzene and water,
respectively. Now we must obtain the density for each relevant uid. Note that before we
proceed with the computation of densities, the water does not have a change in height, so
the corresponding ρ ⋅ g ⋅ Δh term is 0. Beginning with mercury,
ρM
kg
kg
SG =
→ ρM = SG ⋅ ρW = 13.6 ⋅ 1000 3 = 13,600 3
ρW
m
m


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PAG E 19
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Example 1.3 - Di erential U-tube manometer
those used to solve for ρM. As such,
g
1kg
106cm 3
kg
ρG = 1.26 3 ⋅
⋅
= 1,260 3
cm
1000g
1m 3
m
Now, we compute ρB using the speci c weight (γs) provided in the problem statement,
γS,B
γS,B = ρB ⋅ g → ρB =
=
g
8643 N3
m
m
9.81 2
s
= 881.04
kg
1m⋅s
2
N⋅s
⋅
m4
1 N ⋅2s
= 881.04
m
kg
m3
With air at standard conditions, we can simply extract its value from Table 1.2 as,
ρair = 1.20
kg
m3
With all values of density, we solve for PB - PA as,
PB − PA = 13,600
kg
kg
kg
kg
⋅
g
⋅
0.05m
−
1,260
⋅
g
⋅
0.15m
−
881.04
⋅
g
⋅
0.05m
−
1.20
⋅ g ⋅ 0.08m
m3
m3
m3
m3
Now we substitute the gravitational constant and perform some simple algebra to obtain,
PB − PA = 4,384
kg
1N
⋅
kg ⋅ m
m ⋅ s2
1 2
= 4,384
N
1Pa
1kPa
⋅
⋅
= 4 . 384kPa
N
2
m
1000Pa
1 2
m
s
Instructor’s Note: As you can see in this problem, there are a signi cant number of unit
conversions. You will need to nd a way to keep track of the units in problems. Write them
down next to the values they are associated with. This is not just important for receiving
partial credit on assessments. There have been some infamous (and expensive!) mistakes
made with unit conversions, including the $125M Mars Climate Orbiter spacecraft, which
sent the craft roughly 60 miles o
course after engineers failed to convert the required
thrust from EEU to SI. You can read more about some famous errors with unit conversions
by clicking here.
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PAGE 20


We were provided with the density of glycerin, but must convert it to the the same units as
Example 1.4 - Extending the range of a manometer with multiple uids (U-tube)
One of the advantages of a U-tube manometer is its ability to extend the range of a pressure
measurement. Consider both the ordinary manometer (left) and the U-tube manometer
(right) shown in the gure below.
In both images, uid A is water, while the image to the right also contains a secondary uid
(" uid m”), in this case, mercury. The ordinary manometer has a maximum height h = 20
cm, while the uid on the right has a height h1 = 5 cm and h2 = 20 cm. Water is considered
at standard conditions while the mercury has a density of 13,600 kg/m3. Determine the
maximum possible range in pressure di erence that the device can read.
Solution
The maximum change in pressure is represented as PA - Patm, which can be found for the
ordinary manometer using,
PA − Patm = ρW ⋅ g ⋅ h = 1,000
kg
m
1N
⋅
9.81
⋅
0.2m
⋅
kg ⋅ m
m3
s2
1 2
⋅
s
1Pa
N
m
1 2
⋅
1k Pa
= 1 . 962kPa
103Pa
where the minimum pressure di erence is 0 kPa when h = 0 m.
On the other hand, the maximum pressure di erence established by the U-tube manometer
is,
PA − Patm = ρm ⋅ g ⋅ h2 − ρW ⋅ g ⋅ h1 = 13,600
= 26,912.7
kg
1N
⋅
kg ⋅ m
m ⋅ s2
1 2
s
⋅
kg
m
kg
m
kg
⋅
9.81
⋅
0.2m
−
1,000
⋅
9.81
⋅
0.05m
=
26,192.7
m3
s2
m3
s2
m ⋅ s2
1Pa
N
m
1 2
⋅
1k Pa
≈ 26 . 9kPa
103Pa
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PAG E 21
PA − Patm = ρm ⋅ g ⋅ h2 − ρW ⋅ g ⋅ h1 = 13,600
kg
m
kg
1N
⋅
9.81
⋅
0.15m
=
20,012.4
⋅
kg ⋅ m
m3
s2
m ⋅ s2
1 2
s
⋅
1Pa
1 N2
m
⋅
1kPa
≈ 20kPa
103Pa
Thus, we have changed the measurable range of the manometer by using a combination of
uids and orientations. Below is a table that summarizes the range covered by each type of
manometer.
Bound
ΔPordinary = PA − Patm
ΔPU−tube = PA − Patm
Lower
0 kPa
20 kPa
Upper
1.962 kPa
26.9 kPa

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PAGE 22



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and the minimum pressure di erence is found when h1 = 0 m as,
FLUID STATICS
As you might imagine, the impact of depth on pressure has important consequences for
objects that are immersed in a
uid, including aircraft, submarines and underwater
structures. In this section, we examine the forces exerted on objects by pressure.
Forces on Planar, Stationary Surfaces
In this section, we are principally interested in how a uid at rest imparts forces on a solid
object from one point to the next. To construct the expression that represent a variation in
pressure with depth, we can visualize a small “element" of that uid and examine how the
pressure changes in each direction (x, y and z, where +z is the direction opposite
gravity). We then sum such forces in all directions to obtain the relationship between
pressure and location in each direction.
In the case of a static
uid, the pressure on each side of the element balance in x and y.
However, the weight of the
uid imparts a negative force in z, resulting in the following
pressure gradients,
∂p
=0
∂x
∂p
=0
∂y
∂p
=−γ
∂z
Because pressure is now only a function of z (or depth), it is no longer a partial di erential
and becomes,
dp
=−γ
dz
This result is the same as that o ered in the previous section when the
uid is
incompressible (i.e. its density does not vary with depth), that is,
P2
∫P
dp = − γ ⋅
1
h2
∫h
dz → Δp = ρ ⋅ g ⋅ Δh
1
What we ultimately care about is the resultant force and the pressure distribution acting
on a submerged surface and the. Let’s examine the expected distribution of forces acting on
di erent surfaces in a water tank, including at the bottom of the tank and its side walls, in
Fig. 1.4.
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PAGE 23
Bottom of Tank
Sides of Tank
Free surface (p = 0)
h
FR
Free surface (p = 0)
p = !h
p = !h
Figure 1.4. Pressure distribution along the bottom of a tank (left) and the sides of a tank (right) for a stationary uid with a speci c weight γ.
As we already know that pressure varies linearly with depth, it should come as no surprise
that at the bottom of the tank, the pressure is uniform across its surface, and that at the
sides of the tank, the magnitude of the pressure acting on the side walls increases linearly
with h (where the free surface represents h = 0).
The term FR in Fig. 1.4 is particularly important and represents the force acting on a surface
due to the hydrostatic pressure distribution (i.e. the resultant force). The resultant force
has both a magnitude and a location on a planar surface. To determine the resultant force
acting on a submerged surface, we manipulate the relationship F = P⋅A by integrating over
the area and substituting the depth-dependent relationship P = γ ⋅ h,
FR =
∫A
γh d A
(1.8)
More speci cally, if the surface is inclined relative to the free surface, as shown in Fig. 1.5,
below, then we substitute and integrate across the exposed surface area (As) to yield,
FR =
∫A
γ ⋅ (y ⋅ sinθ) d A = γ ⋅ As ⋅ yc ⋅ sinθ = γ ⋅ hc ⋅ As
(1.9)
Figure 1.5. Submerged surface (green line) at an incline angle θ. Note that the x-axis is considered to be the depth into the page.
Be careful to consider the axes we are looking at in this problem!

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We also want to determine where the resultant force acts on a submerged plate. Although
the resultant force is clearly located at the centroid when considering the pressure
distribution along a submerged horizontal plane, this is not the case when the pressure
acting on a surface changes with depth. In this case, the resultant force is weighted toward the
higher magnitudes of pressure acting on the surface.
To account for this mathematically, we sum the moments around the x-axis, where
FR ⋅ yR =
∫A
y dF =
∫A
γ ⋅ sinθ ⋅ y 2 d A
(1.10)
We can substitute the expression for resultant force that we obtained in Eqn. 1.8
(FR = γ ⋅ yc ⋅ sinθ ⋅ As), where,
yR =
∫A y 2 d A
(1.11)
yc ⋅ As
In Eqn. 1.11, the numerator represents the moment of inertia about x, or Ix. When combined
with the parallel axis theorem (Ix = Ixc + As ⋅ yc2), we obtain,
yR =
Ixc
+ yc
yc ⋅ As
(1.12)
where Ixc is the second moment of inertia with respect to an axis passing through its
centroid.
A similar mathematical approach can be taken to obtain the x-coordinate of the resultant
force, which yields,
xR =
Ixyc
yc ⋅ As
+ xc
(1.13)
where Ixyc is the moment of inertia about an orthogonal coordinate system passing through
the centroid when the x-y coordinate system is translated.
The corresponding moments of inertia (Ixc and Ixyc) can be found for di erent geometries
using the gures provided on the following page.
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PAGE 25
Rectangle
centroid
a
x
As = a ⋅ b
Ixc =
1
⋅ b ⋅ a3
12
Ixyc = 0
y
b
Circle
As = π ⋅ r
centroid
x
R
πR 4
Ixc =
4
2
Ixyc = 0
y
Semicircle
centroid
y
R
x
4"#
3"%
π ⋅ R2
As =
2
Ixc = 0.1098 ⋅ R 4
Ixyc = 0
R
Triangle
a⋅b
As =
2
b
d
a
centroid
y
!+#
3
x
%
3
b ⋅ a3
Ixc =
36
b ⋅ a2
Ixyc =
⋅ (b − 2 ⋅ d )
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PAGE 26
Example 1.5 - Hydrostatic forces on the Three Gorges dam spillway gate
The Three Gorges Dam facility is one of the largest power-generating facilities in the world.
The damn contains a number of sluice gates to prevent the water level from rising too
rapidly. The sluice gates are located at the bottom of the dam, which has a water level that
reaches 175 m below the surface. The sluice gates are 483 m long and rest against the
riverbank oor at an angle of θ = 60∘C. The height of the sluice gates from the bottom-up
are 10 m. On the other side of the gates, the river itself has a height of 35 m. Determine the
resultant force on each side of the gate, the location of the resultant forces on each side of
the gate structure, and the net mass of the gate required to keep it closed under these
conditions.
Solution
It is useful to rst visualize the distribution of pressure acting on each side of the gate as,


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PAGE 27
We treat the gate as a rectangular plate that is angled at θ = 60∘ relative to the
uid
surfaces. If the centroid of the rectangular plate is directly in the center of its area, then the
height to the centroid on each side of the gate is:
hc,dam = hdam −
1
1
⋅ hgate = 175m − ⋅ (20m ⋅ sin(60∘)) = 166.34m
2
2
and,
1
1
hc,river = hriver − ⋅ hgate = 35m − ⋅ (20m ⋅ sin(60∘)) = 26.34m
2
2
To nd the resultant force, then,
FR = γ ⋅ As ⋅ hc
Substituting on each side of the gate, we compute the resultant force as,
FR,dam = 9810
N
10
⋅
20m
⋅
483m
⋅
166.34m
=
1.58
⋅
10
N
m3
and,
FR,river = 9810
N
⋅ 20m ⋅ 483m ⋅ 26.34m = 2.50 ⋅ 109 N
3
m
Now we need to nd the location of each resultant force on both sides of the plate. In this
case, we are only concerned with the location in y. For the dam,
yR,dam =
Ixc
yc,dam ⋅ As
+ yc,dam =
1
3
⋅
b
⋅
a
12
+ yc,dam
yc,dam ⋅ As
where a and b are the height (20 m) and length (483 m) of the gate, respectively. Likewise,
we know that yc,dam = hc,dam / sin(θ) = 192.07 m. As a result,
1
⋅ 483m ⋅ (20m)3
12
yR,dam =
192.07m ⋅ 483m ⋅ 20m
+ 192.07m = 192.24m
And performing a similar computation for yR,river, we obtain:
yR,river =
1
⋅ 483m ⋅ (20m)3
12
26.34m ⋅ 483m ⋅ 20m
+ 30.41m = 31.67m
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Now, we can solve for the weight of the gate required to keep it shut under these conditions.
We know that for the gait to remain stationary when subjected to forces that have both an xand y-component, the sum of the moments about the point where it moves (i.e. a "hinge")
must be equal to zero,
∑
MB = 0 = W ⋅ 10m + FR,river ⋅ 14.54m − FR,dam ⋅ 11.60m
Therefore, the weight required is,
W = 1.469 ⋅ 1010 N
Before we calculate the mass, let’s think about how many people it would take to hold this
gate shut. If a single person can reasonably exert a force of 100 N, it would take,
1.469 ⋅ 1010 N
# of people =
=
= 146,900,000 people
Fperson
100N
W
Or roughly 2% of the world’s population!
Now, let’s calculate the mass of the gate required,
W
m=
=
g
1.469 ⋅ 1010 N ⋅
9.81 m2
s
1 kg 2⋅ m
s
1N
= 1 . 497 ⋅ 109 kg





PAGE 29
As we mostly relate mass to pounds and tons in the US,
m = 1.497 ⋅ 109 kg ⋅
2.2046lbm
1ton
= 3 . 30 ⋅ 109 lbm ⋅
= 1, 650, 143 tons
kg
2000lbm
And, if this door were constructed from stainless steel (ρ = 7700 kg/m3), the thickness of
the door would need to be,
m
m
1.497 ⋅ 109 kg
V=
→t=
=
= 20 . 12 m
kg
ρ
ρ ⋅ Ac
7700 ⋅ 483 m ⋅ 20m
m3





PAGE 30
Pressure Prisms
There exists a simpler, more intuitive method to compute the resultant force on rectangular
and circular planar surfaces. This method relies on the development of a "pressure prism".
Pressure prisms are quite simple to construct for equally simple geometries, but can not be
used for the wide variety of geometries described in the previous section.
Instead, we focus speci cally on rectangular and circular planes of interest due to the ease
with which we can determine the location of their centroids. To construct a pressure prism,
we consider the surface of interest. For simplicity, let’s consider the tank below and the
three walls labeled A, B and C.
Figure 1.5. Tank (thin black lines) lled to the rim with water. Sections of di erent types of walls relevant to this section are labeled A, B and C to
represent vertical, horizontal and inclined walls, respectively.
CASE A - VERTICAL WALL AT THE TOP OF THE FLUID
Let’s take a look at how we expect the pressure to vary along wall A. Starting with the
equation P = ρ ⋅ g ⋅ h , we recognize that pressure (P) is a linear function of height (h),
provided that density varies negligibly with height. Then,
Figure 1.6. (left) Vertical wall with the pressure distribution measured in gage pressure (Pg). H is the total height of the partial wall shown in Fig.
1.5. (right) Vertical wall with the pressured distribution measured in absolute pressure (Pabs).


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PAG E 31
The left-most schematic in Fig. 1.6 shows the expected pressure distribution (red lines)
acting on the wall from the uid to the right of it. We use gage pressure to measure the
pressure distribution as a function of depth. This is because the top of the uid is open to
the atmosphere, and the atmospheric pressure that acts on the left hand side of the wall
(from the air) is cancelled out by that acting on the right, as shown below:
The schematic on the right-hand side of Fig. 1.6 is useful if we wish to nd the absolute
pressure on either side of the wall, but most often we are looking for the net pressure
acting on the wall, which we measure using gage pressure (as in the schematic on the left).
Ultimately, we want to compute the resultant force (FR) acting on the wall, which we
de ned in the previous section as,
FR = ρ ⋅ g ⋅ hc ⋅ As
where hc is the distance from the top of the wall to its centroid. Because the pressure is
distributed linearly as a function of height, h, the average pressure results in an equivalent
force that occurs at the centroid hc, as shown in the gure below.
Figure 1.7. (left) Actual pressure distribution, which varies linearly with height (red) and is imposed across the area outlined in yellow in the gure
on the right, and the average pressure distribution acting on the wall (green), (right) 3-dimensional schematic of the wall with yellow highlighting
the area across which pressure acts on.
Note that in this course we will restrict ourselves to rectangular vertical walls to
accommodate the prism method and the simple identi cation of the centroid. Of course, the
method described in the previous sub-section can be used for alternate geometries.
In addition to the resultant force, we want to know where it acts on the wall and the
distance from the center of pressure to the resultant force. Imagine for a moment that the
schematic to the left in Fig. 1.6 is rotated to the left 90∘, and the shape of the pressure
distribution is replaced by a solid object of equivalent size.
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PAGE 32
Figure 1.8. (left) Vertical wall with the pressure distribution measured in gage pressure (Pg). H is the total height of the partial wall shown in Fig.
1.5. (right) Vertical wall with the pressured distribution measured in absolute pressure (Pabs).
For a triangular solid (assuming a uniform density), we would expect there to be a center of
gravity where the weight acts to apply a force downward. To counteract this, we would need
an opposing force in the same location as the gravitational force. For a triangular solid
speci cally, we calculate the center of gravity as being 1/3 of H from the tallest side (as
shown in the left-most image of Fig. 1.8).
An analogous statement can be made about the pressure distribution acting on a vertical
surface, where the resultant force emanates from the center of pressure (rather than the
center of gravity). Instead of a distributed weight acting to force a solid downward at some
location, a distributed pressure is providing this force on a solid surface.
To determine the distance between the center of pressure and the centroid, we use,
Figure 1.9. Location of the centroid and the center of pressure for the rectangular, vertical wall shown in Fig. 1.5 (wall A).
Mathematically, then, the distance between the centroid and the center of pressure can be
expressed as,
Δhc−cp = hc − hcp =
1
1
3
2
1
⋅H− ⋅H= ⋅H− ⋅H= ⋅H
2
3
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PAGE 33
CASE B - HORIZONTAL SURFACE
The case of the horizontal surface is straightforward relative to its vertical counterpart.
Recall that pressure varies linearly with depth, but at a single depth, the pressure is uniform.
Within the context of the horizontal surface in Fig. 1.5, the pressure distribution can be
visualized as,
Figure 1.10. (left) Horizontal wall with the pressure distribution measured in gage pressure (Pg). The horizontal wall could be presented in
rectangular or circular form, in which case the corresponding areas for each are highlighted in yellow in the schematics on the right.
For the horizontal, submerged surface, the center of pressure is in the same location as
the centroid. Therefore, the resultant force acts at both the center of pressure/centroid.
Again, the resultant force can be computed as,
FR = ρ ⋅ g ⋅ hc ⋅ As
where hc is the distance to the centroid, or the depth of the wall.
CASE C - INCLINED SURFACES AND VERTICAL SURFACES BELOW THE WATER LINE
For the case where surfaces are below the water line, the concept of a pressure prism is
uniquely suited to compute the resultant force and its location acting on the wall relative to
the centroid. Let’s consider the following:
Figure 1.11. Vertical (A) and inclined (C) walls that are submerged with upper values below the top of the uid that’s open to the atmosphere.
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PAGE 3 4
uid that is open to the
atmosphere, the the gage pressure at the top of each wall is not 0. The pressure
distribution can then be visualized as:
θ
Figure 1.12. Vertical (A) and inclined (C) walls with corresponding pressure distributions for the case where both. uids are submerged and the
top of each wall does not reach the surface of the uid open to the atmosphere.
Note that for these cases, we will restrict ourselves to rectangular walls only, so that the
centroid is directly in the center of the wall and the shape of the pressure distribution does
not change across the area. To help compute the resultant force, it is easiest to view each
wall in three dimensions as,
Figure 1.13. Vertical (A) and inclined (C) walls with corresponding pressure prisms, which are broken up into rectangular and triangular prisms.
The pressure distribution is broken up into rectangular and triangular prisms, where each
prism has a representative "volume", though not with the traditional units of volume.
Because this is a pressure prism, we are e ectively multiplying the magnitude of the pressure
by the area that the pressure is acting over.
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PAGE 35


If the top of either type of wall does not reach the top of the
We draw two distinct pressure prisms to nd the resultant force in each individual pressure
prism, and sum them to obtain the total resultant force, FR, as,
FR = F1 + F2
(1.14)
where F1 and F2 are represented by,
Figure 1.14. Individual resultant forces acting on each of the rectangular and triangular prisms, where the center of pressure for F1 exists at the
centroid for the rectangular prism (or H/2) and the center of pressure for F2 exists H/6 below the centroid (or 1/3⋅H from the bottom of the prism
upward).
Using wall A as an example, F1 and F2 are mathematically computed according to,
F1 = Ptop,wall ⋅ As = ρ ⋅ g ⋅ htop,A ⋅ HA ⋅ W
and,
Pbottom,wall − Ptop,wall
F2 =
⋅ As = ρ ⋅ g ⋅
2
(
hbottom,A − htop,A
2
)
⋅ HA ⋅ W
For the inclined wall, the basic form for both F1 and F2 remains the same (i.e. Pavg⋅As).
F1 = Ptop,wall ⋅ As = ρ ⋅ g ⋅ htop,C ⋅ Hc ⋅ W
and,
F2 =
Pbottom,wall − Ptop,wall
⋅ As = ρ ⋅ g ⋅
2
(
hbottom,C − htop,C
2
)
⋅ HC ⋅ W









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PAGE 36
Example 1.6 - Force on a vertical wall with water/oil separation
A tank is lled with water (ρw = 1.940 slugs/ft3) to a height Hw = 70 ft. Oil (ρo = 1.643
slugs/ft3) is added to the water. Because the oil is immiscible in water, it creates a layer on
top with a height Ho = 20 ft. What is the resultant force acting on the left wall, and where is
it located? The tank is 10 ft wide.
Solution
This problem can easily be solved using the prism method. First, let’s examine the expected
pressure distribution acting along the wall. Remember, we draw the pressure gradient in
gage pressure.





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PAGE 37
resultant forces after each prism is drawn. We compute each force as,
F1 = Pavg,o ⋅ As = ρo ⋅ g ⋅
F2 = Pavg,w,1 ⋅ As,w =
F3 = Pavg,w,2 ⋅ As,w =
Ho
slugs
ft
⋅ As = 1.643 3 ⋅ 32.174 2 ⋅
2
ft
s
1 lbf
1 slug ⋅ ft
slugs
ft
ρo ⋅ g ⋅ Ho ⋅ As = 1.643 3 ⋅ 32.174 2 ⋅
(
)
ft
s
20ft
⋅ 20ft ⋅ 10ft = 1.06 ⋅ 105 lbf
2
⋅
s2
1 lbf
⋅ 20ft ⋅ 70ft ⋅ 10ft = 7.40 ⋅ 105 lbf
1 slug ⋅ ft
s2
slugs
ft
ρw ⋅ g ⋅ Hw ⋅ As = 1.94 3 ⋅ 32.174 2 ⋅ 70ft ⋅
(
)
(
)
ft
s
1 lbf
1 slug ⋅ ft
⋅ 70ft ⋅ 10ft = 3.06 ⋅ 106 lbf
s2
We recognize that the force above is essentially calculated using the equivalent "volume" of
the prism. More accurately, we compute the Pressure-Area prism value, which is the force.
Now, the total force acting on the wall is,
FR = F1 + F2 + F3 = 3.92 ⋅ 106 lbf
To
nd the location of the resultant force, we simply sum the moments acting on both
surfaces due to each individual force.
Fr ⋅ hcp = F1 ⋅ hcp,1 + F2 ⋅ hcp,2 + F3 ⋅ hcp,3
where,
hcp,1 =
2
⋅ Ho = 13.33 ft
3
hcp,2 = Ho +
1
⋅ Hw = 55 ft
2
2
hcp,3 = Ho + ⋅ Hw = 66.67 ft
3
Therefore,
hcp =
1.06 ⋅ 105lbf ⋅ 13.33ft + 7.40 ⋅ 105lbf ⋅ 55ft + 3.06 ⋅ 106lbf ⋅ 66.67ft)
(
)
3.92 ⋅ 106lbf
= 62.79ft






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Note that the image above is not drawn to scale. In this problem, there are 3 individual
Example 1.7 - Force on a levee
Levees are used to prevent hazardous
ood waters from damaging low-lying towns and
cities that are close to water. In 2005, 50 levees failed during hurricane Katrina due to
inadequate assumptions about the strength of the soil in which they were embedded.
Extensive damage was done to the lowest lying areas of New Orleans, highlighted in the
map, below (Attribution: http://en.wikipedia.org/wiki/File:New_Orleans_Levee_System.svg).
Provided the levee below, what is the force the levee should be designed for under nonhurricane conditions? Image not drawn to scale. The levee is has a width of W = 20 ft.
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PAGE 39
We are only concerned with the resultant force acting on the wall. Constructing individual
pressure prisms, we obtain,
We focus on two di erent surfaces for the levee: (a) the vertical surface and (b) the inclined
surface. For the vertical surface, we compute the resultant force F1 as,
F1 = Pavg ⋅ As,1 = ρw ⋅ g ⋅
H1
slugs
ft
1lbf
⋅ H1 ⋅ W = 1.94 3 ⋅ 32.174 2 ⋅
slug ⋅ ft
2
ft
s
1 2
8.45ft
⋅ 8.45ft ⋅ 20ft = 89,135.4lbf
2
⋅
s
Now we nd the individual resultant forces for each prism acting on the inclined surface as:
F2 = Pavg,2 ⋅ As,2 = ρw ⋅ g ⋅ (H1) ⋅ H2 ⋅ W = 1.94
slugs
ft
1lbf
⋅
32.174
⋅
slug ⋅ ft
ft 3
s2
1 2
⋅ 8.45ft ⋅ 8ft ⋅ 20ft = 84,388.0lbf
s
F3 = Pavg,3 ⋅ As,2 = ρw ⋅ g ⋅
1
slugs
ft
1lbf
⋅ H2 ⋅ sin(θ) ⋅ H2 ⋅ W = 1.94 3 ⋅ 32.174 2 ⋅
slug ⋅ ft
(2
)
ft
s
1 2
s
⋅
1
⋅ 8ft ⋅ sin(55∘) ⋅ 8ft ⋅ 20ft = 32,722.9lbf
2
Therefore, the total resultant force is,
FR = F1 + F2 + F3 = 206,139.4 lbf


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Solution
Buoyancy is a physical concept that is critical for the design of a number of Naval platforms,
including ships and submarines, and features prominently in the assessment of ship
stability, which is covered in the next section.
Archimedes Principle
We consider buoyancy from the perspective of an object that is submerged in a column of
uid. Using what we just learned about the variation of pressure with depth, and the
corresponding resultant force(s) acting on each side of the object, we construct a force
balance to account for buoyant force acting on the object. Consider the schematic below,
which shows a submerged object with equal side lengths.
Figure 1.15. Pressure distributions surrounding a submerged cube and corresponding resultant forces on the top, bottom, left and right of the
cube.
Provided that the object is not misaligned and/or asymmetric, the resultant forces on the
sides of the cube (i.e. where the applied pressure varies with depth) are equal and opposite
one another.
Fleft = Fright
There is, however, a di erence between the resultant forces acting on the top and bottom of
the cube due to their di erence in depth below the surface. This di erence is known as the
buoyant force, and is expressed as,
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PAG E 41

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Buoyancy
FB = Fbottom − Ftop
Assuming the object is submerged in a single uid whose density does not vary signi cantly
with depth, we can simplify the above expression and re-write it as,
FB = ρ ⋅ g ⋅ (hbottom − htop) ⋅ As
The surface area term represents the area of the top or bottom of the submerged cube (both
being equal in area). When multiplied by the change in height (Δh = hbottom − htop), this
term itself becomes a volume. We call this volume the displacement volume, and rewrite
the buoyant force as,
FB = ρ ⋅ g ⋅ Vdisp
(1.15)
To be clear, the displacement volume is actually the volume of uid that is displaced by the solid.
Thus, for partially submerged objects, the displacement volume is only the volume of uid
that is displaced by the solid (and not the entire volume of the solid object).
The buoyant force is often considered alongside other forces acting on the object. For
instance, the buoyant force often counteracts the weight of the object itself.
Another important note is that buoyancy is not restricted to a solid object in a uid. In fact,
two or more di erent
uids that are mixed will often experience buoyant forces. Fluid
buoyancy is also induced by changes in its density, which is impacted strongly by changes in
temperature (which is why hot air will make its way to the very highest level of your home).
We will consider this aspect of buoyancy in our discussion of Convection Heat Transfer in
Chapter 3.
For now, we will restrict our focus to the case of a solid object that is immersed (wholly or
partially) in a uid.
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Example 1.8 - Iced water
A small cube of ice is 12.5 cm on each side. The density of the ice in its solid state is 910
kg/m3 and is put into a cup with liquid water. What fraction of the cube is submerged in the
water?
Solution
The weight of the ice cube must be balanced by the buoyant force applied by the
surrounding liquid water. Thus,
FB = FW
where FB is the buoyant force applied by the liquid water and FW is the force due to gravity
acting on the mass of the ice cube. Substituting the equation for each force, we obtain,
ρwater ⋅ g ⋅ Vdisp = mice ⋅ g
As we do not have the mass of the ice, we recast this problem in terms of its density and
volume,
ρwater ⋅ g ⋅ Vdisp = ρice ⋅ Vice ⋅ g
Now, solving for Vdisp, we obtain:
910
ρice ⋅ Vice
Vdisp =
=
ρwater
kg
m
⋅ 12.5cm 3 ⋅
3
1,000
1 ⋅ 10−6 m 3
1cm 3
kg
= 0.00178m 3
m3
Finally, we compare the volume of water displaced (i.e. the volume of the cube that’s
submerged) to the volume of the cube itself in order to nd the fraction of the cube that’s
submerged below the top of the water,
S( % ) =
Vdisp
Vice
0.00178m 3
=
(12.5cm)3 ⋅
1 ⋅ 10−6 m 3
1cm 3
= 0.911 ⋅ 100 % = 91.1 %

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PAGE 43
Stability
When considering submerged and oating bodies, it is important to also consider whether
such a body is stable under a variety of conditions. Broadly speaking, an object is considered
to be stable when, after some displacement, it returns to its initial condition. On the other
hand, an object is unstable if it does not return back to its original position. The illustration
below serves as a general example of these principles.
Figure 1.16. (left) When the ball is moved to the right on the concave surface, it will eventually return to its equilibrium position, (right) when the
ball is pushed to the right on the convex surface, it rolls down the surface and does not recover to its initial position.
Though the above schematic seems trivial, its importance for the stability of a oating body
can not be overstated. Consider that a ship at sea might be exposed to a rotational force due
to the presence of extremely large waves. In this case, the relative positions for the center of
gravity and the center of buoyancy will govern whether the ship is overturned by this force.
For simplicity, let’s consider the case where an object is fully submerged within a uid, as
shown in the schematic below.
Figure 1.17. (left) Case where the center of gravity is below the centroid and, (right) case where the center of gravity is above the centroid.
In the case where the the center of gravity is below the centroid, the buoyant force creates a
restorative moment once the body is tilted by some external means (e.g. when a wave hits
the side of ship), as shown in the left-most schematic of Fig. 1.17. Conversely, when the
center of gravity is above the centroid, the weight acts to provide an overturning moment.
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Let’s consider the case when an object is partially submerged. For the sake of this text (and the
curriculum speci c to USNA), we exclusively utilize ships to demonstrate stability in this
context. The image below illustrates the case when an external (upsetting) force is applied
to the side of a vessel, which shifts the centroid in the displaced uid volume.
Figure 1.18. (left) Ship in equilibrium and not yet exposed to upsetting force, with center of buoyancy (the centroid in the displaced uid volume)
immediately below the center of gravity) and (right) ship that is lilted due to exposure to the upsetting force, where the location of the buoyancy
force (i.e. the centroid of the displaced volume) has shifted to the right.
Although the centroid is below the center of gravity here, the object is partially submerged
and the buoyancy force acts to counteract the upsetting force. The angle that forms with the
free surface as the boat lilts is called the heel angle, ϕ , which is shown in the diagram
below (along with other relevant geometries that can be used to compute the restoring
moment, RM, that counteracts the upsetting force and which ultimately provides stability).
Figure 1.19. Schematic of forces acting on a vessel in response to an upsetting force (caused by wind or waves) and the resulting geometric
relationships between the heel angle (ϕ) and the forces (displacement, Δ , and buoyancy, FB) acting on the vessel. MT represents the transverse
metacenter, which is an imaginary point of intersection between the centerline of the vessel and the line of action for FB.
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PAGE 45
the buoyant force and the weight of the displaced uid (Δ, which is equivalent to the weight
¯ )
of the vessel and everything on it). This is critical to determining the righting arm (GZ
and righting moment (RM) as a function of heel angle. For small heel angles (0∘ to ~ 10∘),
¯ = GM
¯ ⋅ sin(ϕ). More generally, we compute the righting arm based on a projected
GZ
upsetting force applied across an equivalent radial dimension (or characteristic dimension)
of the ship.
The righting arm can be found by summing the moments about the centroid of the entire
vessel (in other words, not limited to the displaced volume only) as,
¯
0 = FU ⋅ rU − FB ⋅ GZ
where FU is the upsetting force, rU is the equivalent radius of the ship (from the point of
impact to the centroid of the entire vessel), and FB is the buoyant force. Recall from our
previous discussion of buoyancy that FB = W (where W is the weight of the object, or
displacement, Δ, as it is referred to when dealing with stability). Therefore,
¯ =
GZ
FU ⋅ rU
Δ
¯ to calculate the righting moment as,
We can use GZ
¯ = Δ ⋅ GZ
¯
RM = FB ⋅ GZ
(1.16)
The righting moment is a true measure of the ship’s stability. In general, we consider the
righting moment to be a description of a ship’s ability to return to an equilibrium position.
Typically, we plot the righting moment and/or the righting arm as a function of heel angle
in order to determine the maximum heel angle that can sustain upright stability.
Notice that in Fig. 1.20, an overturning moment is formed when the center of gravity is
above the transverse metacenter, or the point where the line of action for the buoyant
force and the centerline for the object intersect. The vessel is close to capsizing at point D,
and will overturn at point E. We note that the value of GZ as a function of heel angle is
di erent for di erent vessel designs.
Di erent types of vessels and their advantages and disadvantages are illustrated and
described in Fig. 1.22.






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The gure above (Fig. 1.19) reveals the way in which the heel angle (ϕ) is related to both
angle to demonstrate its impact on the type of moment that is formed (righting vs. overturning). Positive values of ϕ represent the starboard
side of the vessel while negative values of ϕ indicate heel angles for the port heel side.
There does exist a maximum heel angle prior to our being in danger of capsizing. This heel
¯ max (in the above curve, this equates to ϕmax ≈ 44∘).
angle can be determined by locating GZ
¯ approaches 0 ft is termed the angle of vanishing stability, and
The angle at which GZ
represents the angle at which the vessel will capsize.
Practically, we can account for changes in the height of the center of gravity by comparing
¯ values for di erent weight distributions. Here, if the center of gravity rises, so too does
GZ
¯ , as shown below.
GZ
Figure 1.21. (left) Ship with original weight distribution (denoted with subscript "o"), (right) ship with a vertical shift in weight distribution with
¯ due to a vertical rise in the center of gravity (GoGv).
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¯ ) as a function of heel angle (often termed an intact stability curve) with corresponding schematics of the heel
Figure 1.20. Righting arm (GZ
To compute the positional shift in the center of gravity (GoGv), or the new righting arm due
to a change in the center of gravity (GvZv), we use,
Gv Zv = Go Zo − Go P = Go Zo − GoGv ⋅ sin(θ)
(1.17)
If we know both GvZv and GoZo we can calculate the corresponding shift in the center of
gravity, GoGv, as a function of heel angle (ϕ). Let’s evaluate the following scenario, where
the center of gravity is shifted 2 ft in the vertical direction.
Figure 1.22. Ship with a rise in the center of gravity of 10 ft., and where the original righting arm, GoZo has a de ned distribution of lengths as a
3
⋅ϕ .
(2
)
function of heel angle, where GoZo = 6 ft. ⋅ si n
Here we plot GoZo and GvZv simultaneously to illustrate the impact of the rise in the center
of gravity.
Figure 1.23. GZ versus heel angle for a ship whose original (at GoZo) is increased by 2 ft (at GvZv). Maximum ϕ included for comparison.
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Thus far, we have only concerned ourselves with stationary
uids. However, we routinely
encounter uids that are moving, both in nature and in the engineering systems we use on a
daily basis. The movement of a uid results in signi cantly di erent physics that govern the
forces exerted on objects, for instance.
Consider the case where people in villages in rural areas must travel several miles to the
nearest clean water source (which often begins at some higher elevation). In many such
communities, women and children bear the brunt of this labor, often carrying several gallons
of water back several miles over the span of an entire day in order to provide clean water to
the rest of the community. Engineers Without Borders, a non-pro t engineering group, has
resolved to build a variety of watery delivery systems (including both gravity-driven systems
and pump-driven systems) to eliminate the need for this type of labor. In order to design
these systems, one must have a fundamental understanding of fluid dynamics. In this
section, we cover the basic fundamental aspects of uid dynamics, including the conservation
of mass, the Bernoulli relationship, dimensional analysis, conservation of momentum, the NavierStokes equation and laminar and turbulent ow conditions.
Conservation of Mass
Mass conservation provides us with a mechanism to account for the rate at which mass
ows into or out of an open system (recall that an open system implies that mass can cross
the system’s boundaries).
Our discussion of mass conservation begins with our understanding of mass
ow rate, m· .
A mass ow rate describes the rate at which a mass of uid (or gas) is moving, and as you
might imagine is coupled to a uid’s velocity. We can write the mass ow rate for a moving
uid as,
m· = ρf ⋅ V̄ ⋅ Ac
(1.18)
where ρf is the dentist of the uid, V̄ is the uid’s velocity and Ac is the cross-sectional area
of the
uid through which the mass ows (in other words, it is the area perpendicular to the
direction of uid ow). Combining V̄ and Ac, we can rewrite Eqn. 1.18 as,
·
m· = ρf ⋅ V
(1.19)
·
where V is a volume ow rate.

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F LU I D DY N A M I C S
following scenario, where two rivers each having velocity V̄1 and V̄2 meet to form a third
river having a di erent velocity V̄3 . If we want to predict V̄3 based on the conditions of the
rst two rivers, we would do this using the Conservation of Mass.
Figure 1.23. Two rivers converging to form a third river. Rivers have widths W1, W2 and W3, respectively, and each has a di erent velocity. In this
problem, we’ll assume that all three rivers also have the same depth (i.e. the dimension "into the page".)
To solve a problem of this nature, we use the principle of Conservation of Mass, which we
write in its steady-state form as,
∑
m· i =
∑
m· e
(1.20)
Equation 1.20 mathematically tells us that all of the mass ow rates coming into a system must
equal all of the mass ow rates going out of a system.
When applied to the case outlined in Fig. 1.18, we account for the inlets (rivers 1 and 2) and
exits (river 3) as,
m· 1 + m· 2 = m· 3
When dealing with velocity, we can substitute Eqn. 1.16 into the above expression, which
yields,
ρf 1 ⋅ V̄1 ⋅ Ac,1 + ρf 2 ⋅ V̄2 ⋅ Ac,2 = ρf 3 ⋅ V̄3 ⋅ Ac,3
This problem is greatly simpli ed by virtue of the fact that all three rivers contain the same
uid (water) and remain at the same depth (where Ac in this case is generally represented
as Ac = W⋅depth). Simplifying, we obtain the following expression,
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We are particularly interested in how mass is conserved in an open system. Consider the
V̄1 ⋅ W1 + V̄2 ⋅ W2 = V̄3 ⋅ W3
Now, solving for V̄3, we are left with,
V̄1 ⋅ W1 + V̄2 ⋅ W2
V̄3 =
W3
Note that the above equations are not general forms for the conservation of mass. Only Eqn.
1.18 should be applied broadly to problems concerning mass in ows and out ows.
Likewise, the conservation of mass can also be used to in tandem with the expression that
relates mass ow rate to volume ow rate (i.e. one can substitute Eqn. 1.19 into Eqn. 1.20
rather than substituting Eqn. 1.18 into Eqn. 1.20).
Conservation of Mass - Unsteady Problems
A more general form of the conservation of mass accounts for changes in the mass within a
system as a function of time.
dMsys
dt
=
∑
m· i −
∑
m· e
(1.21)
In the vast majority of problems we encounter, systems will operate under steady-state
conditions.
Bernoulli Equation
We begin our discussion of
uid
ow by applying Newton’s Second Law to describe the
steady-state (i.e. time-independent) motion of a
uid. Recall that Newton’s Second Law
states,
F = m ⋅ ā
In order to simplify our consideration of the way in which a uid ows across a surface, we
start by assuming that the uid itself is inviscid. In other words, we make the assumption
that the
uid viscosity is negligible. Of course, no
uid has a viscosity which is zero, but
viscous forces are considered to be secondary e ects in many cases where forces due to
pressure or gravity are signi cant. Without considering viscous forces, we can rewrite
Newton’s second law as,
Pnet ⋅ A + mf ⋅ g = mf ⋅ ā
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PA G E 51
uid particle and ā is the
the mass of the
uid particle’s acceleration. The inviscid
assumption used to construct this equation is generally applicable for
uids having low
viscosity and for those uid particles traveling far from a wall (where frictional forces are not
negligible close to a wall, regardless of uid viscosity).
To further simplify the problem, we also consider the
assumes that the
uid to be incompressible. This
uid density does not vary with pressure (or temperature), which is a
reasonable approach for uids and gases that are not moving too fast.
This treatment leads us to the development of the Bernoulli equation (for the full
derivation, see additional reference1). For an inviscid, incompressible uid traveling along a
streamline at steady-state, the Bernoulli equation reads,
P V¯2
+
+g⋅z =c
ρ
2
(1.22)
where z is the height of the uid (and g⋅z is representative of the potential energy of the uid),
V̄ is the uid velocity (and V¯2 /2 is the kinetic energy of the uid), and c is a constant. A
streamline is de ned as a set of curves that are tangent to the velocity vector for the ow.
You can think of it as a way to track an individual
uid particle as it moves in a two-
dimensional plane (for the purposes of this course). Visually, we describe this in the
schematics below.
Figure 1.24. Schematics showing two
“irrotational".
So long as the
uid
uid particles separated by height and/or width in the same
uid stream. We assume that the
ows are
ow is irrotational and meets the constraints mentioned above, Eqn.
1.24 can be used to relate pressure, density, velocity and height of a uid at points 1 and 2.
1 Cengel, Cimbala, and Ghajar. Fundamentals of Thermal-Fluid Sciences, 5th ed., p. 467-468.

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where Pnet is the net pressure applied to an individual uid particle across an area, A, mf is
P1 V12
P2 V22
+
+ g ⋅ z1 = c =
+
+ g ⋅ z2
ρ
2
ρ
2
There are a variety of forms that the Bernoulli equation can take. In the above expression,
each term represents a speci c energy (or an energy per unit mass). Occasionally, it is useful
to de ne each term as a length, or what we refer to as a “head".
P
V¯2
+
+z =c
ρ⋅g 2⋅g
where the
rst term (P/ρ ⋅ g) is called the pressure head, the second term (V¯2/2 ⋅ g) is
termed the velocity head and the nal term (z) is called the elevation head.
Applications of the Bernoulli Equation
It is useful to discuss some conventional ways that the Bernoulli equation is applied in uid
mechanics. The following types of problems are solved routinely by engineers to characterize
the impact of owing uids on their surroundings (or vice versa). We speci cally focus on
free jets, Venturi meters, and pitot-static tubes. Several examples are provided to give
context for each type of problem discussed herein.
Free Jets
The "free jets" discussed in this section are speci cally characterized as free
owing
uids
whose motion is governed by a change in potential energy. Consider the tank in the image
below, whose uid is open to the atmosphere at height h and which has an opening at the
bottom of the tank.
Figure 1.25. Conditions for the free jets discussed in this section, including the tank lled with uid at some height h above the mid-point of an
opening which allows a the uid to ow freely out.
Here, we are often most interested in determining the uid velocity at the exit (2) of Fig.
1.25. Assuming a steady ow and an inviscid, incompressible uid, we use the following,
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2
We can simplify the above expression using assumptions that are readily made by
examination of Fig. 1.25. Here, both
uids are exposed to the atmosphere. As a result,
ΔP = P2 − P1 = 0. Likewise, we can assume that the velocity of the uid at position 1 is
much less than that of position 2 (V̄1 < < V̄2, for proof of this, consider that m· 1 = m· 2 and
therefore ρ ⋅ V̄1 ⋅ Ac,1 = ρ ⋅ V̄2 ⋅ Ac,2; if Ac,2 << Ac,1, then V̄1 < < V̄2). Consequently, we can
rewrite the above expression as,
V¯22
g ⋅ z1 =
+ g ⋅ z2
2
Solving for V2, we are left with:
V̄2 =
2 ⋅ g ⋅ (z1 − z2)
As you might expect, the velocity at the exit (2) is based entirely on a speci c potential
energy.
Venturi Meters
Venturi meters are used to measure the volume
ow rate of a moving
uid. To do this, a
converging section of the pipe is used to create a low pressure region and induce a pressure
drop (and an increase in velocity) across the device. A schematic of a Venturi Meter is
provided below.
Figure 1.26. Schematic of a Venturi meter with cross sections located in regions (1) and (2).
In the schematic shown in Fig. 1.26, region (1) represents an area of high pressure and low
velocity, while region (2) represents an area of low pressure and high velocity. By measuring
the pressure di erence between points (1) and (2), the volume ow rate can be found via
the simpli ed Bernoulli equation. We start with the general Bernoulli expression, below, as,
2
2
P1 V̄1
P
V̄
+
+ g ⋅ z1 = 2 + 2 + g ⋅ z2
ρ
2
ρ
2
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2
P1 V̄1
P
V̄
+
+ g ⋅ z1 = 2 + 2 + g ⋅ z2
ρ
2
ρ
2
Δz = z1 − z2 = 0. Because the purpose of a Venturi Meter is to measure the volume ow
rate, we rewrite velocity as,
·
V
·
V̄ =
→ V = V̄ ⋅ Ac
Ac
Conservation of mass also tells us that,
·
·
m· 1 = m· 2 → ρ1 ⋅ V1 = ρ2 ⋅ V2
Since the uid is air at both points (1) and (2), ρ1 = ρ2, so,
·
·
·
V1 = V2 = V
·
Now we substitute V in for both V̄1 and V̄2, which yields:
P1
+
ρ
·
2
·
V
( A1 )
2
P2
=
+
ρ
2
·
V
( A2 )
2
Solving for V,
·
V = A2 ⋅
2⋅
p1 − p2
1
⋅
A2
ρ
1− 2
A12
Some images of a pair of Venturi tubes attached to a small aircraft are shown below.
Figure 1.27. Venturi tubes attached to the side of an aircraft for air-driven, precision gyroscopic measurements of ow rate.2
2Attribution: https://upload.wikimedia.org/wikipedia/commons/2/2c/Aircraft_venturi_3.JPG
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PAGE 55


In this case, we assume that the average height at (1) and (2) is equivalent, so
Pitot-static Tube
A Pitot-static tube is conventionally used as a speedometer on aircraft. In other words, it is
principally used to as a method to measure an aircraft’s velocity. The image below provides
both a schematic for an ordinary Pitot tube and an actual Pitot tube mounted on a Cessna
aircraft.
Figure 1.28.(left) Schematic of Pitot-static tube used to measure aircraft speed and (right) Pitot-static tube mounted to a Cessna aircraft.
The schematic shown on the left of Fig. 1.28 describes the basic working principle for a
manometer. The green regions of the pitot tube in this schematic represent air that enters
the system through a series of small holes and represents the atmospheric pressure at the
altitude of the aircraft (i.e. the static pressure, or pS. The total pressure (static pressure +
dynamic pressure, pT) is represented in the left chamber (white space with label pT). The
pressure di erence is measured by a pressure transducer that sits between the two
chambers in the pitot tube.
For a pitot tube, the end of the tube is a stagnation point, where the uid hits the tube and
the velocity V̄2 = 0. This is where the total pressure pT is measured. Since Δz = 0 , the
Bernoulli Equation simpli es to,
2
p1 V̄1
p
+
= 2
ρ
2
ρ
Solving for V̄1, we have,
V̄1 =
2⋅
p2 − p1
=
ρ
2⋅
pT − pS
ρ


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Example 1.9 - Squirt gun (free jet - Bernoulli)
A squirt gun uses a build up in pressure inside of its reservoir (these types of squirt guns
are often referred to as "Air Pressurized Reservoir" guns). To do this, a pump is used to
bring additional air into the reservoir, which is only partially lled with water. The opening
where the water exits the squirt gun is open to the atmosphere, and can therefore be
modeled as a "free jet". If a velocity of the water exiting the squirt gun must be 50 ft/s to
achieve a range of 8 ft, what must the minimum pressure of the reservoir be? What if you
replaced the water with ethanol (ρglycerin = 49.3
lbm
and pumped to the same pressure inside of
f t3 )
the reservoir, what would its exit velocity be? Use the dimensions provided in the diagram
below, and assume the
uid
ow is inviscid and the
uid is incompressible. Also assume
that the device is being operated at steady-state.
Solution
Here, we are looking for a pressure in the reservoir of the squirt gun such that we achieve
some velocity at its exit. Thus, we can use the Bernoulli equation to solve for Preservoir as,
2
2
pR V¯R
pE V¯E
+
+ g ⋅ zR =
+
+ g ⋅ zE
ρ
2
ρ
2
where the subscripts R and E represent the Reservoir and Exit, respectively. Given that the
cross-sectional area of the opening at the exit is much smaller than the average crosssectional area of the reservoir, we can use the Conservation of Mass to say:
ρR ⋅ V¯R ⋅ Ac,R = ρE ⋅ V¯E ⋅ Ac,E
Since water is the uid at both points, ρR = ρE, and if Ac,R >> Ac,E, then,
V¯R < < V¯E


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PAGE 57
also assume that pE = 0 psig. Rearranging our
obtain,
rst expression to solve for pR, then, we
2
V¯E
pR = ρ ⋅
+ g ⋅ (zE − zR)
( 2
)
Substituting values,
1 lbm
(50 s )
lbm
ft
3
pR = 62.3 3 ⋅
+ 32.174 2 ⋅ 0ft −
ft
⋅
ft 2
( 2
( 12 ) ))
ft
s (
25,037 2
ft 2
Btu
s
5.40395psia ⋅ ft 3
⋅
1Btu
And,
pR = 16.7 psig + 14.7 psi = 31.4 psia
Now, if we replace the water with glycerin, the exit velocity at a gage pressure of pR = 16.7
psig is:
V¯E =
2⋅
pR
+ g ⋅ (zR − zE )
(ρ
)
Substituting the values provided (and working in gage pressure), we obtain:
V¯E =
16.7
ft 2
lbf
25,037 2
1Btu
ft
3
s
⋅
⋅
+
32.174
⋅
ft − 0ft
Btu
2 (( 12 )
)
)
5.40395psia ⋅ ft 3
s
1 lbm
in 2
2⋅
( 49.3 lbm
ft 3
Therefore,
V¯E = 56.16
ft
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PAGE 58



And we can therefore neglect V¯R . If we solve for the gage pressure in the reservoir, we can
Example 1.10 - Pitot-static tube (Bernoulli)
The density of air at a speci c altitude is 1.1 kg/m3. A Pitot-static tube is used to measure
the velocity of an aircraft by measuring the di erence in pressure between the atmosphere
and that being applied to the front of the Pitot-static tube. A manometer is used to measure
the pressure di erence, as shown below. If the uid in the manometer is mercury (SGHg =
13.6), height h1 = 4 cm and height h2 = 1 cm, what is the indicated velocity of the aircraft?
Solution
Beginning with the Bernoulli equation,
2
2
pS V̄S
pT V̄T
+
+ g ⋅ zT =
+
+ g ⋅ zS
ρ
2
ρ
2
Although there is a change in height in the manometer, there is no signi cant change in
height for the air at static pressure and that at the total pressure (i.e. green vs. white uids).
Likewise, we can assume that the velocity of the static uid is V̄S = 0. Simplifying,
pS − pT
2⋅
=
( ρair )
V̄T =
2⋅
(
)
ρHg ⋅ g ⋅ Δh
ρair
So,
V̄T =
2⋅
(
13,600
kg
m
m
s
⋅ 9.81 2 ⋅ (4cm − 1cm) ⋅
3
1m
100cm
)
kg
1.1 3
m
= 85.3
m
s

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PAGE 59
Conservation of Momentum
Just as we accounted for forces acting on submerged surfaces that are surrounded by
quiescent uids, we must also account for forces impacting surfaces due to interactions with
moving uids. In this section, we use the Conservation of Linear Momentum to account for such
forces. As a disclaimer, this section includes content that can be di cult to absorb from a
fundamental perspective, but (like most other subjects) can be mastered with practice
working through problems.
We begin our discussion of momentum conservation by examining an arbitrary open system.
Figure 1.29. Arbitrary control volume with one inlet and one exit having mass ow rates m· i and m· e, respectively, and corresponding velocities of V̄i
and V̄e, respectively.
In order to determine the force(s) acting on a surface due to a moving uid, we must rst
recall the method used to compute momentum,
M = m ⋅ V̄
For an open system, the momentum is computed according to the product of a mass ow
rate and the uid’s velocity. Accounting for each inlet and exit, we can use Newton’s 2nd law
to obtain the time rate of change of linear momentum within a control volume as,
d(m ⋅ V̄ )cv
=
F̄cv +
m· i ⋅ V̄i −
m· e ⋅ V̄e
∑
∑
∑
dt
i
e
(1.23)
In this course, we will restrict our study of momentum conservation to steady-state
problems. Thus,
0=
∑
F̄cv +
∑
i
m· i ⋅ V̄i −
∑
e
m· e ⋅ V̄e
(1.24)

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PAGE 60
Also note that F̄cv , V̄i , and V̄e are all vector quantities, which means they have some
directionality to them!
rst turn our attention to the external forces acting on the control volume, F̄cv .
Let’s
These forces are either body forces that act on the control volume or surface forces that act on
the surface of the control volume. Such forces include:
Body forces
1. Gravity, g
2. Electric eld, Ē
3. Magnetic eld, B̄
Surface forces
1. Pressure (F = P⋅As)
2. Viscosity, (F = μ ⋅ V̄ ⋅ As / S, where S is a characteristic length for the ow)
3. Walls on the uid (Resultant force!)
We note that when we eventually apply momentum conservation, we will be working in
gage pressure given our assumption that the atmospheric pressure acts on all sides of our
object or surface. Thus, Patm will not impact the solution when looking at the net forces in
any one direction.
Typically, we wish to
nd forces acting on a surface (or, conversely, the force needed to
support an object). When we apply linear conservation to the problem, we generally follow
the procedure as outlined below:
1. Determine the relevant control volume.
2. Provide the coordinate system you will use (note that velocity and force are
directional).
3. Sketch the free body diagram showing the relevant body and surface forces.
4. Divide the linear momentum equation into each of its constituent coordinate systems.
5. Use the resulting equations to solve for the unknown forces exerted by the cv.

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PA G E 61
Vanes: Vanes alter the direction of a uid that is bound only on one side.
Nozzles: Nozzles constrict the ow of uid by reducing the area between inlet(s) and exit(s).
Y-joint: A y-joint splits the uid into two separate (typically smaller diameter) paths.
Pipe Bend: A pipe bend is meant to redirect uid ow while fully bounding the uid.
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PAGE 62
Example 1.11 - Vane-water Jet
A vane redirects water having a cross-section of 4 in x 2 in. The vane redirects the water at
an angle of 35∘ and the velocity of the uid in the jet is 145 ft/s. If the weight of the uid is
negligible, determine the force of the uid acting on the vane.
Solution
Note rst that the two ends of the vane (positions 1 and 2) are exposed to the atmosphere.
Thus,
p1 = p2 = patm
Remember that when solving conservation of momentum problems, we use gage pressure.
As a result,
p1,g = p2,g = 0
By conservation of mass, we know that,
ρ ⋅ V̄1 ⋅ A1 = ρ ⋅ V̄2 ⋅ A2
Because the uid is the same at the inlet and exit, and because the areas at each location are
also equivalent, we can say that V̄1 = V̄2.
Knowing this, we can apply the conservation of momentum as outlined in Eqn. 1.24 as,
∑
0=
F̄cv +
∑
i
m· i ⋅ V̄i −
∑
e
m· e ⋅ V̄e
We break the above equation down into its respective x- and y-components according to the
coordinate system outline in the image above.
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PAGE 63
Looking at the forces applied in only the x-direction, we have,
0 = FR,x + m· ⋅ V̄1 − m· ⋅ V̄2,x
While V̄1 only has an x-component, V̄2 has both an x- and y-component. We can resolve the
force due to the momentum in the x-direction using trigonometry. An illustration of each
component is provided below.
As a result, the conservation of momentum in the x-direction (with V̄1 = V̄2) becomes,
0 = FR,x + m· ⋅ V̄1 − m· ⋅ V̄1 ⋅ cos(35∘)
We can solve for the resultant force in the x-direction as,
FR,x = m· ⋅ V̄1 ⋅ (cos(35∘) − 1) = ρ ⋅ A1 ⋅ V1 ⋅ V1 ⋅ (cos(35∘) − 1)
and,
lbm
FR,x = ρ ⋅ A1 ⋅ V12 ⋅ (cos(35∘) − 1) = 62.3 3 ⋅ (4 in ⋅ 1 in) ⋅
ft
1ft 2
ft
1lbf
∘
⋅
145
⋅
cos(35
)
−
1
⋅
(
)
lbm ⋅ ft
144in 2 (
s)
32.174 2
2
s
Therefore,
FR,x = − 204.4 lbf
Because it is negative, this represents the force of the vane on the uid. Now, applying the
conservation of momentum in the y-direction,
0 = FR,y − m· ⋅ V̄2,y

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PAGE 6 4


x-momentum
Recall that there is no y-component of the velocity at point (1) to generate a force in the ydirection at the entrance. Using the same diagram that is laid out on the previous page, we
can obtain the y-component of the velocity at point (2) as V̄2,y = V̄1 ⋅ sin(35∘). Therefore,
FR,y = m· ⋅ V̄1 ⋅ (sin(35∘)) = ρ ⋅ A2 ⋅ V1 ⋅ (sin(35∘))
and,
lbm
1ft 2
ft
1lbf
∘
FR,y = 62.3 3 ⋅ (4 in ⋅ 1 in) ⋅
⋅
145
⋅
sin(35
)
⋅
(
)
lbm ⋅ ft
ft
144in 2 (
s)
32.174 2
2
s
Therefore,
FR,y = 648.5 lbf
Again, this is the force of the vane on the uid. To compute the total force of the uid on the
vane, we use,
| FR | =
(204.4 lbf )2 + (−648.5 lbf )2 = 679 lbf
We can also calculate the angle that the force is acting on using trigonometry. A visual of the
resultant force and its x- and y-components from the perspective of the uid acting on the vane
is provided rst.
Here, we’ve arranged the direction of F̄R to represent the resultant force of the vane on the
water. The force of the water on the vane (F̄w) is equal and opposite that of F̄R. Now, we nd the
angle as,
θw = tan
( Fx,y )
Fw,y
−1
= tan −1
−648.8 lbf
= − 72∘
( 204.4 lbf )
which is 72∘ below the horizontal line of action (Fw,x).
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PAGE 65
Example 1.12 - Nozzle/elbow combination
An elbow joint is reduced at its end (section 2) from D1 = 90 cm to D2 = 30 cm. Water
enters the elbow at section 1 with a velocity V̄1 = 7
m
and a gage pressure of 100 kPa at
s
standard atmospheric conditions. The water exits to the atmosphere at section 2. What is
the force needed for the apparatus to remain stationary, and what is the direction of this
force?
Solution
As before, we start with the conservation of momentum. Unlike the previous problem,
however, we do not have the same gage pressure at both inlet and exit. Therefore, the
expression for momentum conservation becomes,
0=
∑
F̄cv +
∑
i
m· i ⋅ V̄i −
∑
e
m· e ⋅ V̄e
Considering the resultant force (F̄R), the pressure force at the inlet (F̄1 = p1,g ⋅ A1), and
the pressure force at the exit (F̄2 = p2,g ⋅ A2) as the forces acting on the control volume
(F̄cv), we write the x-momentum conservation equation as,









PAGE 66
0 = p1x,g ⋅ A1 + p2x,g ⋅ A2 + FR,x + m· 1 ⋅ V̄1x − m· 2 ⋅ V̄2x
and,
FR,x = − p2x,g ⋅ A2 − p1x,g ⋅ A1 + m· ⋅ (V̄2x − V̄1x)
We know p2,g = 0 and p1,g = 100 kPa, but need V̄2 . To obtain this, we can use the
Conservation of Mass as,
m· 1 = m· 2 → ρ ⋅ V̄1 ⋅ A1 = ρ ⋅ V̄2 ⋅ A2
∴ m· = 1000
kg
m
kg
2
⋅
7
⋅
π
⋅
(0.45
m)
=
4,453
m3
s
s
and,
A1
m π ⋅ (0.45m)2
m
V̄2 = V̄1 ⋅
=7 ⋅
=
63
A2
s π ⋅ (0.15m)2
s
The above velocity is the magnitude of velocity. We know that V̄2x = − V̄2, thus,
1 N2
1000Pa
kg
m
m
1N
m
⋅
+ 4,453 ⋅ − 63 − 7
1 kPa
1Pa
s (
s
s ) 1 kg ⋅ m
FR,x = − 100 kPa ⋅ π ⋅ (0.45m)2 ⋅
s2
FR,x = − 375,327 N = − 375.33 kN
Because there is no pressure force or velocity in the y-direction at either opening,
FR,y = 0 N
Therefore,
F̄R = FR,x = − 375.33 kN
This is the force of the vane on the water, so the resultant force needs to be applied in the
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PAGE 67


x-momentum
Dimensional Analysis
As engineers, it is often not possible to analyze full scale designs due to economic
constraints. Instead, we turn to dimensional analysis and "scaling". We begin our discussion
of dimensional analysis with a detailed look at Similtude, and proceed with the use of the
so-called Buckingham Pi Theorem, which follows with a discussion on scaling analysis.
Similitude
Similitude refers to the similarity of two problems. This principle is extremely useful when
designing an experiment, or when generalizing the results of a model or a set of
experiments. If the similarity of two problems can be established, we may use the behavior
of a model system to predict the behavior of a di erent, related system.
important when conducting experiments.
This is very
Often we are unable to make the required
measurements on the actual system of interest, but we can conduct an experiment on a
similar, smaller system. For example, we may not be able to measure the lift on a full-size
airplane wing. However, we can put a smaller model wing into a wind tunnel, measure the
lift, and relate the measured force to the force that would be expected on the full-size wing.
The process by which we determine the variables and parameters that govern the similarity
of two systems is called Dimensional Analysis. Every system is characterized by some number
of variables and parameters that are required in order to fully describe the system. Each
variable or parameter can be be expressed in terms of primary dimensions, typically mass
[M], length [L], time [t], and temperature [θ] (sometimes force [F] is used instead of mass.)
Because a large number of variables are all constructed from a small set of primary
dimensions, there are often only a few ways in which variables may be arranged in an
equation to produce a dimensionally correct result. By eliminating the dimensional aspects
of the system, we can reduce the number of required quantities needed to solve for a given
variable.
Again, the primary dimensions of a problem are typically mass [M], length [L], time [t]
and/or temperature [θ]. However, we often deal with secondary dimensions in a problem
that are a combination of these primary dimensions.
Examples of these secondary
dimensions include Area [L2], Force [ML/t2], Pressure [M/Lt2], and Energy [ML2/t2]. MLt
dimension for some common units are provided in Appendix A.
As an example, consider the classic equation for the position of an object subject to a
constant linear acceleration.
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PAGE 68
X = Xo + V̄o ⋅ t +
1
⋅ ao ⋅ t 2
2
The problem is to nd X as a function of the other variables, or X = f(Xo, Vo, ao, t) where f is
some function. In total, this system contains 5 quantities: Xo, Vo, ao, X, t. In this example,
the rst three quantities (Xo, Vo, ao) are usually xed for a particular problem and are called
dimensional constants. Variables X and t will vary and are called dimensional variables. In
general, the objective is to non-dimensionalize the variables in terms of the dimensional
constants. Constants such as (½) that do not have dimensions are not part of the process.
Position X has dimensions of length [L] and t has dimensions of time [t].
Notice that the
following dimensional constants (or combination of dimensional constants) have the same
dimensions as the variables we wish non-dimensionalize.
X* =
t* =
X
Xo
V̄o ⋅ t
Xo
Solving for X and t, substituting those expressions back into the original equation, and
dividing by Xo results in the following equation:
X* = 1 + t* +
X
1
⋅ a o ⋅ t *2
2
V̄ 2o
Where before each term in the equation had dimension [L], now each term is
dimensionless.
The remaining dimensional constants, when grouped together, form a
dimensionless quantity (a nondimensional acceleration) that we will call α. Substituting
this yields,
X* = 1 + t* +
α=
1
⋅ α ⋅ t *2
2
ao ⋅ Xo
V̄ 2o
The expression for position X* is now only a function of two quantities, t* and α, which we
can write as X* = f(t*,α). The original equation [1] was in terms of 5 quantities; we reduced
the number of quantities in the equation by 2, by going from 5 to 3.
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PAGE 69
Note that all of the quantities in the original expression use only two primary dimensions,
[L] and [t]. This leads us to the Buckingham Pi Theorem.
Buckingham Pi Theorem
The Buckingham Pi Theorem is introduced by rst examining the relevant unit systems we
use in Fluid Mechanics. Speci cally, we refer to these as the FLt and MLt unit systems. The
MLt unit system is used to construct Π groupings (dimensionless terms) and represents units
of mass, length, and time. The FLt unit system provides us with a mechanism to con rm Π
groupings and represents units of force, length, and time. Formally, the Buckingham Pi
Theorem is understood as,
I F
A N
E Q U A T I O N
I N V O LV I N G
N
VA R I A B L E S
I S
D I M E N S I O N A L LY H O M O G E N E O U S , I T C A N B E R E D U C E D T O
A
R E L AT I O N S H I P
DIMENSIONLESS
NUMBER
OF
A M O N G
PRODUCTS
BASIC
( N - R )
WHERE
DIMENSIONS
R
I N D E P E N D E N T
IS
THE
REQUIRED
TO
MINIMUM
DESCRIBE
T H E VA R I A B L E S .
We will combine the above unit systems to form Π groupings, but before we do that we
discuss the relevance of the Buckingham Pi Theorem and how we use Π groupings to
implement it. A widely used introductory example is the force of drag acting on a stationary,
smooth sphere that is immersed in a
uid having a uniform velocity, as shown in the
illustration below.
Figure 1.30. Sphere suspended in a uid with uniform ow. Gray lines represent streamlines for uid, where the uid initially meets a stagnation
point at the sphere’s leading point, and then ows around the sphere and produces some rotational ow at the trailing edge of the sphere.
It is useful to be able to measure the drag around the sphere; however, in some cases (e.g.
when the sphere is very large or the
uid velocity is very high) it is di cult to make

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PAGE 70
between such cases, it is useful to non-dimensionalize the expression for drag force. To do this,
we must specify the parameters that we think will impact the drag force.
Those with su cient experience in engineering design can typically use intuition to
determine the parameters that most impact a computation like this. In the relatively simple
case of drag force on the sphere above, one such engineer would point to its likely
dependence on the size of the sphere (or characteristic dimension, D), the uid viscosity (μ),
the uid velocity (V̄) and the uid density (ρf). Thus, we can say that the drag force (FD)
takes the following functional form,
FD = f (D, μ, V̄, ρ)
It is important to note that we are examining the drag force on a smooth sphere in this case,
and that we have not factored in parameters like surface roughness as a result. We also note
that the above parameters are all tunable in an experiment. For instance, the size of the
sphere can be controlled in the fabrication process, while the density and velocity can be
tuned via the choice of
uid, and the
uid velocity can be controlled by varying the
electronic signal to a fan or blower. Let’s illustrate the use of Π groupings to set up a nondimensional relationship between drag force and the functional parameters on the right of
the above equation.
First, let’s count the number of dimensional parameters:
FD
D
V̄
μ
ρ
→
n = 5 dimensional parameters
Now we select the primary dimensions in MLt units as,
FD =
M⋅L
[ t2 ]
D = [L]
V̄ =
L
[t]
ρ=
M
[ L3 ]
μ=
M
[L ⋅ t ]
→
r = 3 primar y dimensions
Next, we select the number of repeating parameters (m), which must be equal to the number of
primary dimensions (r). If m = r = 3, we select 3 of the above parameters (any 3). In this
illustrative example, we choose repeating parameters ρ, V̄, and D.
The number of dimensionless groups is then calculated as,
n − m = 2 dimensionless groups


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PAGE 71
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experimental measurements in laboratory conditions. In order to make comparisons
groups as ρ a ⋅ V̄ b ⋅ D c ⋅ P, where P is replaced by each non-repeating parameter. Thus, the two
Π groups here are,
Π1 = ρ a ⋅ V̄ b ⋅ D c ⋅ FD
and,
Π2 = ρ d ⋅ V̄ e ⋅ D f ⋅ μ
We solve for the exponents (a, b, c, d, e, and f) by substituting the primary dimensions into
the above equations for each parameter (ρ, V̄, D, FD, μ) and setting this equal to MLt, we
obtain (for the rst Π group, Π1),
M
L
M⋅L
ρ a ⋅ V̄ b ⋅ D c ⋅ FD → 3 ⋅
⋅ [L]c ⋅
= [M ]0 ⋅ [L]0 ⋅ [t]0
2
[L ] [ t ]
[ t ]
a
b
Where we equate the exponents of M, L, and t, as,
M:
L:
a+1=0
∴ a=−1
−3⋅a+b+c+1=0
t:
−b−2=0
∴ c=−2
∴ b=−2
Note that we solved for the coe cients a, b and c by manipulating the system of
simultaneous equations. Now, working with our second Π group, Π2, we obtain,
M
L
M
ρ d ⋅ V̄ e ⋅ D f ⋅ μ → 3 ⋅
⋅ [L] f ⋅
= [M ]0 ⋅ [L]0 ⋅ [t]0
[L ] [ t ]
[L ⋅ t ]
d
e
Again equating exponents of M, L, and t, we obtain,
M:
L:
d+1=0
∴ d=−1
−3⋅d+e+f−1=0
t:
−e−1=0
∴ f=−1
∴ e=−1
With the coe cients above, our Π groups become,

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PAGE 72


So we now know that we will need to develop 2 Π groups. To do this, we de ne each Π
ρ ⋅ V̄ 2 ⋅ D 2
and,
μ
ρ ⋅ V̄ ⋅ D
Π2 =
We can double-check the resulting dimensionality by applying the FLt units (i.e. force-lengthtime) to our Π groups. If the result is equivalent to unity (1), then we have properly
determined our Π groups.
L3
t
1
L4
t
1
Π
=
→
[F
]
⋅
⋅
⋅
=
1
→
[F
]
⋅
⋅
⋅
=1
[ 1] [
[ M ] [( L ) ] [ L 2 ]
[ F ⋅ t 2 ] [( L ) ] [ L 2 ]
ρ ⋅ V̄ 2 ⋅ D 2 ]
2
FD
2
Note that in the above expression, we use Newton’s second law to relate mass, M, and force,
F via F = m ⋅ ā , where ā is acceleration and has units of [L/t2]. Therefore, converting from
MLt to FLt yields [M] = [F⋅t2/L]. For the second Π group, we obtain,
M
L3
t
1
F⋅t
L4
t
1
Π
=
→
⋅
⋅
⋅
=
1
→
⋅
⋅
⋅
=1
[ 2] [
2
2
2
2
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
[
]
L
⋅
t
M
L
L
L
F
⋅
t
L
L
ρ ⋅ V̄ ⋅ D
FD
where we’ve used the same conversion between M and F as with Π1.
The functional relationship between Π1 and Π2 is then Π1 = f (Π2),
μ
=f
( ρ ⋅ V̄ ⋅ D )
ρ ⋅ V̄ 2 ⋅ D 2
FD
The actual functional relationship between the two sides of the above expression is found via
experiment. Notice that this relationship allows us to vary any of the above parameters to
create a range of values for Π1 and Π2, making it easier to conduct an experiment to predict
the drag force based on parameters that are more di cult to measure.




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PAGE 73



FD
Π1 =
Figure 1.31. The actual (experimentally obtained), non-dimensionalized functional form of the drag force acting over a smooth sphere submerged
in a uid of uniform velocity.
The real power of the Buckingham Pi Theorem is that you do not need to know the
functional form in order to determine the non-dimensional parameters.
There are many
problems that exist for which we cannot easily determine the functional form of the
governing equations.

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PAG E 74
Provided the example above, we can formalize the procedure for determining Π groups as
follows:
1. Establish a list of the parameters that are likely to impact the parameter of interest.
• Let "n" be the number of parameters established above.
• The establishment of such parameters takes practice, but unimportant parameters
can be identi ed using this procedure; therefore, it is useful to include more
parameters than less.
2. Select a set of primary dimensions (MLt or FLt).
• In heat transfer problems, you will also need T for temperature, which can be
implemented in either unit system.
3. Write the dimensions of each parameter provided in Step 1 in terms of the primary
dimensions (MLt or FLt).
• Let "r" be the number of primary dimensions used.
- If just one of M, L or t are used, for instance, then r = 1 (if two of M, L or t are
used, r = 2, and so on).
4. Select a set of "r" dimensional parameters (from Step 1) that contain all of the primary
dimensions identi ed in Step 3.
• These parameters will be referred to as "repeating parameters”.
• None of the repeating parameters should have dimensions that could be a power of
the dimensions of another repeating parameter.
- e.g. Do not use both Area (L2) and the second moment of inertia (L4)
• Do not include the parameter of interest as a repeating parameter.
• When combined, the dimensions of the repeating parameters cannot cancel to unity.
5. Establish your dimensional equations by combining the parameters identi ed in Step 4
with the remaining parameters (one at a time) and form your Π groups.
6. Check to see that the primary dimensions (MLt or FLt; the opposite of what was used in
Step 3) multiply to unity.

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PAGE 75
Alternative Method of Forming Π Groups
Step 5 in the above procedure, involving a system of equations derived from the exponents
of M, L, and t, may alternatively be replaced by an inspection process that is less structured but
nonetheless provides the correct form for the relevant Π terms.
The starting ingredients for forming dimensionless groups by inspection are the same as
using the exponent method. One dimensionless parameter will be formed for each nonrepeating variable, while the repeating variables will appear in any group where they are
necessary. To begin, write two fraction bars - one for the variables that will be assembled,
and one for the dimensions in each of the variables. Put the non-repeating variable you wish
to non-dimensionalize in the numerator on the variables line, and put the dimensions of
that variable on the dimension line. In the example above, FD was a non-repeating variable
that forms a term with ρ, V̄, and D as repeating variables. The two fractions would then be:
FD
VARIABLES:
1
ML
DIMENSIONS:
t2
Because the objective is to cancel out all the dimensions in the term, the next step is to
multiply or divide the working term in the variables line with a repeating variable such that
it will remove at least one of the existing dimensions, as seen in the dimensions line.
Because the dimension mass is present in the numerator, we can divide by the repeating
variable ρ , which was chosen to represent mass. We divide by the variable in the variables
line, and we divide by the dimensions of the variable in the dimensions line. This leaves,
FD 1
1 ρ
ML L 3
DIMENSIONS:
t2 M
VARIABLES:
This achieved the goal of removing the dimension mass from the working term, although it
did add additional dimensions of length at the same time. The process then continues with
each of the other dimensions. Because the existing variable group has t2 in the denominator,
the speed V̄ (which also has time in the denominator) can be divided through twice:











PAG E 76
Just as before, we were able to remove the dimension of interest represented by the
repeating variable. At this point it can be seen that the only dimension remaining in the
term is length, as both the mass (from ρ ) and the time (from V̄ ) have been removed.
Because there is a net L 2 dimension in the working term, we can remove it by dividing by
diameter D twice.
FD 1 1 1
1 ρ V̄ 2 D 2
ML L 3 t 2 1
DIMENSIONS:
t2 M L2 L2
VARIABLES:
Once we can see as above that all of the dimensions have been canceled, then what remains
in the variable line must be the dimensionless parameter we seek.
FD 1 1 1
FD
= Π1 =
1 ρ V̄ 2 D 2
ρ ⋅ V̄ 2 ⋅ D 2
Because this is a less formal process than the exponent method, there are times where
additional steps may be required before all dimensions are removed. Even though the force
FD in this example was a non-repeating variable, it is permissible to add additional force
variables (or take the root of the one you start with) if that helps remove the dimensions
from the working term. (The term “non-repeating” means that it cannot show up in other Π
terms. It’s welcome to “repeat” within the term it anchors.) When working with thermouid variables, it also tends to be easiest to remove the dimensions in the order illustrated
above. Mass M was removed
rst, followed by time t , with the removal of length L last.
Because a length variable, e.g. a height, diameter, etc., is often available as a repeating
variable, it can be used last to remove the dimension L without inserting any additional
dimensions in the process.






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PAGE 7 7



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FD 1 1
1 ρ V̄ 2
ML L 3 t 2
DIMENSIONS:
t2 M L2
VARIABLES:
Notes: 1. Ev = bulk modulus of elasticity, 2. Either of Ca or Ma can be used to assess uid compressibility; c is the speed of sound in the uid, 3. ω
is the frequency of vortex shedding, 4. h here is called the heat transfer coe cient and Cp is the uid’s heat capacity, 5. Lc is a characteristic length
and kf is the thermal conductivity of the uid, 6. ks is the thermal conductivity of the solid object, not the uid!, 7. β is the thermal expansion
coe cient of the uid (β = 1/Tabs for gasses, and can be found in the Appendix for liquids), α is the uid’s thermal di usivity, represented by α =
kf / (ρ ⋅ Cp)
Dimensionless Numbers
Name
Expression
Meaning
Reynolds Number, Re
ρ ⋅ V̄ ⋅ L
μ
Ratio of inertial force to viscous force
V̄
Froude Number, Fr
Ratio of inertial force to gravitational force
g⋅L
Euler Number, Eu
p
ρ ⋅ V̄ 2
Ratio of pressure force to inertial force
Cauchy Number, Ca1
ρ ⋅ V̄ 2
Ev
Ratio of inertial force to compressibility force
Mach Number, Ma2
V̄
c
Ratio of inertial force to compressibility force
Strouhal Number, St3
ω⋅L
V̄
Ratio of inertial (local) force and inertial (convective) force
Weber Number, We
ρ ⋅ V̄ 2 ⋅ L
σ
Ratio of inertial force to surface tension force
Stanton Number, St4
h
Nu
=
Re ⋅ Pr
ρ ⋅ V̄ ⋅ Cp
Ratio of heat transfer into a uid to uid thermal capacity
Nusselt Number, Nu5
h ⋅ Lc
kf
Ratio of convective heat transfer to conductive heat transfer
Cp ⋅ μ
Prandtl Number, Pr
Ratio of momentum di usivity to thermal di usivity
kf
Biot Number, Bi6
h ⋅ Lc
ks
Ratio of convection heat transfer to conduction heat transfer
Rayleigh Number, Ra7
ρ ⋅ β ⋅ ΔT ⋅ Lc3 ⋅ g
μ⋅α
Ratio of di usive heat transfer to convective heat transfer


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PAGE 78













Table 1.4. Common dimensionless numbers
Name
FLt System
MLt System
Acceleration, ā
L
t2
L
t2
Angle, α
F 0 ⋅ L0 ⋅ t0 = 1
M 0 ⋅ L0 ⋅ t0 = 1
Angular Velocity, ωV
1
t
1
t
Area, A
L2
L2
Density, ρ
F ⋅ t2
L4
M
L3
Energy, E
F⋅L
Force, F
F
Heat Energy, Q
F⋅L
Length, L
L
L
Mass, m
F ⋅ t2
L
M
Momentum, M
F⋅t
M⋅L
t
Power, P
F⋅L
t
M ⋅ L2
t3
F
L2
L2
t2 ⋅ θ
F
L3
F
L
M
L ⋅ t2
L2
t2 ⋅ θ
M
2
L ⋅ t2
M
t2
Pressure, p
Speci c Heat, Cp
Speci c Weight, γ
Surface Tension, σ
M ⋅ L2
t2
M⋅L
t2
M ⋅ L2
t2
Torque, T
F⋅L
M ⋅ L2
t2
Velocity, V̄
L
t
L
t
Viscosity (dynamic), μ
F⋅t
L2
M
L ⋅t
Viscosity (kinematic), ν
L2
t
L2
t
Work, W
F⋅L
M ⋅ L2
t2

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PAGE 79


















Primary Dimensions for Common Parameters


















Table 1.5. Primary dimensions for common uid dynamics and heat transfer parameters. Note θ is the primary dimension of temperature.
The rise of a capillary
uid’s surface tension (σ),
uid (h) can be determined based on a
tube diameter (d), interfacial angle of the uid and adjacent wall (α), and the speci c weight
of the uid (γf). We want to determine the dimensionless, functional form of the expression
that allows us to determine h based on the above parameters.
Solution
For this problem, we will follow the steps as outlined on page 59.
Step 1: Identify the parameters likely to impact h.
In this problem, we have been told which parameters are likely to impact the height, h.
These include:
h
σ
d
α
γ
→
n = 5 dimensional parameters
Step 2: Select a set of primary dimensions (MLt or FLt).
For this problem (as with the subsequent examples provided in this text), we will use the
MLt primary dimensions.
Step 3: Write the parameters chosen in Step 1 in terms of their primary dimensions.
h = [L]
σ=
M
[ t2 ]
d = [L]
α = [1]
γf =
M
[ L2 ⋅ t2 ]
→
r = 3 primar y dimensions
Step 4: Select a set of "r" dimensional parameters as repeating parameters
Here, the number of r dimensional parameters is set to 3 (and thus we have m = 3
repeating parameters). We choose our repeating parameters to be σ, d and γ.

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PAGE 80
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Example 1.13 - Rise of capillary uid
However, we must check that these 3 variables can not be combined such that their units
cancel out. Here we have:
M
[ t2 ]
σ=
d = [L]
γf =
M
[ L2 ⋅ t2 ]
Problematically, there is a way in which these parameters can be arranged such that the
units cancel to unity:
σ
=1
2
d ⋅ γf
Therefore, we must reduce the number of repeating parameters by 1, where now m =
2 repeating parameters. We choose parameters σ and d to satisfy the requirement for m.
Step 5: Establish Π groups
In this case, we have 2 repeating parameters, though we still will end up with 3 Π terms.
Π1 = σ a ⋅ d b ⋅ h
Π2 = σ c ⋅ d d ⋅ α
Π3 = σ e ⋅ d f ⋅ γf
First Π Group
M
σ ⋅ d ⋅ h → 2 ⋅ [L]b ⋅ [L] = [M ]0 ⋅ [L]0 ⋅ [t]0
[t ]
a
a
b
M:
a=0
∴ a=0
L:
b+1=0
∴ b=−1
t:
−2⋅a =0
∴ a=0
Π1 =
h
d















PA G E 81
Second Π Group
M
σ c ⋅ d d ⋅ α → 2 ⋅ [L]d ⋅ [1] = [M ]0 ⋅ [L]0 ⋅ [t]0
[t ]
c
M:
c=0
∴ c=0
L:
d=0
∴ d=0
As with the previous Π group, we don’t need to use t to compute c or d, and,
Π2 = α
This should make sense, as α is an angle and therefore already dimensionless.
Third Π Group
M
M
e
f
f
σ ⋅ d ⋅ γf → 2 ⋅ [L] ⋅ 2 2 = [M ]0 ⋅ [L]0 ⋅ [t]0
[t ]
[L ⋅ t ]
e
M:
e+1=0
∴ e=−1
L:
f−2=0
∴ f=2
t:
−2⋅e−2=0
∴ e=−1
Like the rst Π group, the primary dimension t serves to con rm the calculation of one of
our unknown exponents. Now,
Π3 =
γf ⋅ d 2
σ
With Π1 = f (Π2, Π3):
γf ⋅ d
h
= f α,
(
d
σ )
2
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PAGE 82
A pipe with radius R1 converges to a radius R2, as in a Venturi meter. According to
Bernoulli’s equation, a pressure drop (Δp) occurs from point (1) to point (2), as well as an
increase in velocity (ΔV̄). The pressure drop is known to be some function of the
uid
density (ρ) and dynamic viscosity (μ), in addition to the velocity at the inlet (V̄1).
Solution
To solve this problem, we apply the Buckingham Pi theorem in the step-by-step manner
outlined on the page 59.
Step 1: Identify the parameters likely to impact Δp.
In this problem, we are told which parameters are likely to impact Δp. These include,
Δp
R1
R2
V̄1
μ
ρ
→
n = 6 dimensional parameters
Step 2: Select a set of primary dimensions (MLt or FLt).
In this example problem, we choose to use MLt primary dimensions.
Step 3: Write the parameters chosen in Step 1 in terms of their primary dimensions.
Δp =
M
[ L ⋅ t2 ]
R1 = [L]
R2 = [L]
V̄ =
L
[t]
μ=
M
[L ⋅ t ]
ρ=
M
[ L3 ]
→
r = 3 primar y dimensions
Step 4: Select a set of "r" dimensional parameters as repeating parameters
The number of repeating parameters (m) is set equal to the number of dimensional
parameters (r).
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PAGE 83


Example 1.14 - Contraction in a pipe
we choose R1, V̄, and μ as repeating parameters (m).
Step 5: Establish Π groups
In this case, we have n - m = 6 - 3 = 3 repeating parameters. Thus, we will end up with 3
Π groups.
Π1 = R1a ⋅ V̄ b ⋅ μ c ⋅ Δp
Π2 = R1d ⋅ V̄ e ⋅ μ f ⋅ R2
Π3 = R1g ⋅ V̄ h ⋅ μ i ⋅ ρ
Now we must solve for each of the exponents in the above Π groups.
First Π Group
L
M
M
0
0
0
R1a ⋅ V̄ b ⋅ μ c ⋅ Δp → [L]a ⋅
⋅
⋅
=
[M
]
⋅
[L]
⋅
[t]
[ t ] [ L ⋅ t ] [ L ⋅ t2 ]
b
M:
L:
t:
c
c+1=0
∴ c=−1
a+b−c−1=0
∴ a=1
c+2
∴ b=
=−1
−1
−b−c−2=0
Π1 =
Δp ⋅ R1
V̄ ⋅ μ
Second Π Group
L
M
R1d ⋅ V̄ e ⋅ μ f ⋅ R2 → [L]d ⋅
⋅
⋅ [L] = [M ]0 ⋅ [L]0 ⋅ [t]0
[ t ] [L ⋅ t ]
e
M:
L:
f
f =0
∴ f=0
d+e−f+1=0
∴ d=−1

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PAGE 8 4


Recall that this step requires that we use a set of well-de ned rules. To satisfy these rules,
t:
−e−f =0
Π2 =
∴ e=0
R2
R1
Third Π Group
L
M
M
R1g ⋅ V̄ h ⋅ μ i ⋅ ρ → [L]g ⋅
⋅
⋅ 3 = [M ]0 ⋅ [L]0 ⋅ [t]0
[ t ] [L ⋅ t ] [L ]
h
M:
L:
i
i+1=0
∴ i=−1
g+h−i−3=0
t:
−h−i =0
Π3 =
∴ g=1
∴ h=1
ρ ⋅ V̄ ⋅ R1
μ
Now, the functional form of Δp in terms of all other parameters becomes,
Π1 = f (Π1, Π2)
and,
Δp ⋅ R1
V̄ ⋅ μ
=f
R2 ρ ⋅ V̄ ⋅ R1
,
( R1
)
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PAGE 85
Example 1.15 - Explosion from an atomic bomb
Below is an image of the rst atomic bomb ever detonated, taken 0.016 s after detonation.
From an image like this, the radius of the cloud can be related to the energy of the bomb
itself. In fact, these types of images are routinely used to calculate the energy released.
Using the Buckingham Pi theorem, determine the relationship between the radius of the
cloud and the energy of the bomb, and determine the energy of the rst atomic bomb from
the image and data found below. (Image attribution: https://medium.com/imaginary-papers/
leaving-trinity-ten-ground-zero-swerves-74d02d7bfbbf). The density of air was measured to
be 1 kg/m3 and that the ratio of speci c heats for air is 1.038.
Solution
To solve this problem, we need to rst understand which parameters impact the size of the
hemispherical shape shown above. Let’s proceed in alignment with the steps outlined on
page 59.
Step 1: Identify the parameters likely to impact the size of the blast (i.e. its radius, rblast )
rblast
Eblast
t
ρ
→
n = 4 dimensional parameters

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PAGE 86
For this problem, we elect to use MLt to reinforce the concepts outlined above.
Step 3: Write out the dimensions of each parameter chosen in Step 1
rblast = [L]
M ⋅ L2
Eblast =
[ t2 ]
t = [t]
M
[ L3 ]
ρ=
→
r = 3 primar y dimensions
Step 4: Select a set of "r" dimensional parameters
The only three parameters we can use are those other than the radius, which include Eblast, t
and ρ.
Step 5: Establish Π groups
In this case, we only have a single Π group:
Π1 = E a ⋅ t b ⋅ ρ c ⋅ rblast
Now we substitute our primary dimensions for each dimensional parameter,
M ⋅ L2
M
a
b
c
b
0
0
0
E ⋅ t ⋅ ρ ⋅ rblast →
⋅
[t]
⋅
⋅
[L]
=
[M
]
⋅
[L]
⋅
[t]
[ t2 ]
[ L3 ]
a
M:
c
a+c =0
∴ c=−a=
L:
2⋅a−3⋅c+1=0
t:
− 2a + b = 0
1
5
1
∴ a=−
5
∴ b=2⋅a=−
2
5
Therefore,
1
1
2
Π1 = rblast ⋅ ρ 5 E − 5 ⋅ t − 5
Because there is only one Π group, we set it equal to a constant (γ) as,
1
1
2
rblast ⋅ ρ 5 E − 5 ⋅ t − 5 = γ













PAGE 87


Step 2: Select a set of primary dimensions (MLt or FLt)
As it turns out, γ was experimentally determined to be the ratio of speci c heats (Cp / Cv),
which you will learn more about in your thermodynamics course. For air at the temperature
and pressure associated with the blast, γ = 1.038.
Now, with an approximate radius of 120 m, and a time of 0.016 s, the energy released by
this explosion was,
kg
5
5
2
120m
⋅
1
(
) ( m3 )
rblast ⋅ ρ
13 kg ⋅ m
Eblast = 5 2 =
= 8.07 ⋅ 10
γ ⋅t
1.0385 ⋅ (0.016s)2
s2
This is equivalent to 19 ktons (kilo-tons!) of energy, which is close to the measured energy
in this time range (between 18 and 22 ktons!)


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PAGE 88
Modeling
When conducting an experiment, it is obviously advantageous to vary as few parameters as
possible. Let’s say that we wish to determine the drag of a new hydrofoil, and we know that
the drag on a submerged object is a function of the velocity (V̄) and size (L) of the object
along with the viscosity (µ) and density (ρ) of the uid.
FD = f (V̄, L, μ, ρ)
The logical set of experiments would include foils of di erent size tested at a range of
speeds in both fresh and salt water. The problem is that we are talking about a lot of tests,
and depending on the size of the foil it could be very di cult and expensive to conduct these
experiments at the desired scale. However, by using the Buckingham Pi Theorem it is
possible reduce the number of quantities to two non-dimensional variables.
FD
=f
ρ ⋅ V̄ 2 ⋅ L 2
ρ ⋅ V̄ ⋅ L
( μ )
These two non-dimensional parameters are the non-dimensional drag force and the nondimensional viscosity since those are the two quantities that we used to form the Pi terms.
The non-dimensional drag force is called the drag coe cient.
FD
CD = 1
⋅ ρ ⋅ V̄ 2 ⋅ As
2
The non-dimensional viscosity is the Reynolds Number, a very important non-dimensional
parameter in uid mechanics.
ρ ⋅ V̄ ⋅ Lc
μ
Re =
Now we only need to conduct experiments where we vary a single parameter, such as the
velocity, using a water tunnel; and we can determine the coe cient of drag as a function of
Reynolds number.
Let’s say we want to understand the performance of the same hydrofoil, but the foil is too
large to t in our water tunnel. The solution is to test a sub-scale model that will t in the
tunnel. First, it is necessary to determine the range of Reynolds numbers over which we
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PAGE 89
expect the foil will operate. Then we can determine the range of velocities at which we need
to test the model. These velocities must produce the same Reynolds numbers.
Rem = Rep
CD,m = CD,p
where the subscripts m and p represent model and prototype, respectively.
In the experiment, the uid properties are known along with the dimensions of the model.
The drag force is then measured for the model and used to calculate the coe cient of drag
for the prototype. Since the model was tested at the same Reynolds number, the coe cient
of drag is the same for both the model and the foil. The coe cient of drag can then be used
to calculate the expected drag force on the actual foil.
Most problems are not dependent on a single non-dimensional parameter, and therefore
complete similitude is not possible. For example, when using a wind tunnel it is sometimes
necessary to match both the Reynolds number and the Mach number, which is the ratio of
the velocity to the speed of sound. In ows that involve a free surface, the Reynolds number
(along with the Froude number) must be matched, and again this is di cult if not
impossible. The subject of incomplete similitude will be discussed in the Tow Tank Lab.
If the length scale of the model is proportional to the problem, the problem is said to be
geometrically similar. In order to be geometrically similar all linear distances must be
reduced by the same ratio; therefore all angles will be preserved. If the model is related to
the problem by some length scale ratio and also by some time scale ratio, it follows that
there is a velocity scale ratio. If there is a velocity scale ratio, the problem is said to be
kinematically similar. Finally, if the model and problem are related by a length scale ratio,
time scale ratio and mass scale ratio; the problem is said to be dynamically similar. These
distinctions are important when we are conducting an experiment where complete
similitude is not possible.
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PAGE 90
Example 1.16 - Scaling a spillway
A spillway model is constructed at 1:50 the size of its full-scale counterpart and discharges
water at a rate of 1.25 m3/s. Using dimensionless groups, what is the corresponding
prototype discharge rate? Use the following general function to establish Π groups and
answer the questions above. In this problem, we assume that the viscosity is negligible
relative to gravity.
·
V = f (h, g, V̄ )
Solution
First we must establish our dimensionless Π groups. Here we know that n = 4 dimensional
parameters. We now establish the primary dimensions for each parameter and choose to use
MLt analysis,
L3
L
L
·
V→
, h → [L] , g → 2 , V̄ →
[ t ]
[t ]
[t]
∴ r = 2 primar y dimensions
If r = 2 primary dimensions, then there should be m = r = 2 repeating parameters and NΠ =
n - m = 4 - 2 = 2 Π groups. We choose our repeating parameters to be h and g such that,
L
L3
a
b ·
a
h ⋅ g ⋅ V → [L] ⋅ 2 ⋅
= L0 ⋅ t0
[t ] [ t ]
b
and,
L
L
·
h c ⋅ g d ⋅ V → [L]c ⋅ 2 ⋅
= L0 ⋅ t0
[t ] [ t ]
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PAG E 91
L:a+b+3=0
∴a=−
5
2
t :−2⋅b−1=0
∴b=−
1
2
Therefore,
Π1 =
·
V
5
1
h2 ⋅ g2
For the second grouping we have,
L:c+d+1=0
1
∴c=−
2
t :−2⋅d−1=0
∴d=−
1
2
And,
Π2 =
V̄
1
1
h2 ⋅ g2
Now we can answer the modeling questions posed in the original problem. If the model is
1:50 the scale of the full-sized spillway, then:
1
hm =
⋅ hp
50
·
If we want the corresponding ow rate (V) then we should use the Π1 group with,
Π1m = Π1p →
·
Vm
5
2
hm ⋅ g
1
2
=
·
Vp
5
1
hp2 ⋅ g 2
We know the gravitational constant should remain the same as both the model test and the
prototype will operate on earth. Now we cans substitute the height relationship as,
5
1
2
( 50 ⋅ hp)
·
·
Vm = Vp ⋅
m3
·
3
1.25
V
m
·
·
s
m
= 5.66 ⋅ 10−5 ⋅ Vp → Vp =
=
= 22,084.8
−5
−5
5.66 ⋅ 10
5.66 ⋅ 10
s
5
hp2





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Starting with the rst grouping,
Fluid Kinematics and the Navier-Stokes Equation
We will now transition to the study of uid ow that is bounded in one or more directions,
with speci c considerations for the way in which uid velocity varies spatially. We will also
cover energy considerations that account for as yet to be discussed forces (e.g. the frictional
force between a moving uid and a stationary or moving surface).
To aid in our understanding of the former (i.e. the spatially varying velocity pro le within a
pipe, between two
at plates, or across a single surface), we examine the expressions for
continuity and momentum across an individual uid particle in cartesian coordinates (i.e. x, y,
and z). This allows us to resolve the spatial distribution of velocity within the
uid. In this
course, we will restrict our understanding of this phenomenon to the x- and y-directions,
though we start with the full 3-D solution to interpret these expressions.
We’ll begin our discussion by examining
volume of
uid continuity. Let’s take a look at a really small
uid. In fact, we assume volume is so small that we call it a di erential uid
element. We’ll assume for the moment that the di erential
uid element is a cube with
dimensions dx, dy, dz.
Figure 1.32. Di erential uid element with mass ow rates into and out of each face (xyz coordinates provided in the schematic above.
We understand that we can write the mass
ow rate in terms of a velocity. We have
colloquially referred to velocity as V̄ , but will use the variables u, v, and w here to convey
directionality (where u, v, and w represent velocities in the x-, y- and z-directions,
respectively). This allows us to rewrite the mass ow rate in the x-direction (at location x),
for instance, as:
m· x = (ρ ⋅ u)x ⋅ dy ⋅ dz
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PAGE 93
and the mass ow rate at location x + dx becomes,
m· x+dx = (ρ ⋅ u)x+dx ⋅ dy ⋅ dz
Likewise, the equations above can be rewritten in the y- and z-directions. We recognize that
the conservation of mass for the di erential control volume suggests,
dMcv
=
m· in −
m· out
∑
∑
dt
The sum of mass ow rates entering the control volume becomes,
m· in = (ρ ⋅ u)x ⋅ dy ⋅ dz + (ρ ⋅ v)y ⋅ d x ⋅ dz + (ρ ⋅ w)z ⋅ d x ⋅ dy
∑
while the sum of mass ow rates leaving the control volume becomes,
∑
m· out = (ρ ⋅ u)x+dx ⋅ dy ⋅ dz + (ρ ⋅ v)y+dy ⋅ d x ⋅ dz + (ρ ⋅ w)z+dz ⋅ d x ⋅ dy
Finally, the mass in the control volume (Mcv) can be rewritten as the product of density (ρ)
and volume (V),
Mcv = ρ ⋅ d x ⋅ dy ⋅ dz
such that,
dMcv
dρ
=
⋅ d x ⋅ dy ⋅ dz
dt
dt
Now we substitute back into the expression for the conservation of mass and divide through
by d x ⋅ dy ⋅ dz,
∂ρ (ρ ⋅ u)x+dx − (ρ ⋅ u)x (ρ ⋅ v)y+dy − (ρ ⋅ v)y (ρ ⋅ w)z+dz − (ρ ⋅ w)z
0=
+
+
+
∂t
∂x
∂y
∂z
This equation can be rewritten in di erential form as,
0=
∂ρ ∂(ρ ⋅ u) ∂(ρ ⋅ v) ∂(ρ ⋅ w)
+
+
+
∂t
∂x
∂y
∂z
And, if we assume that the uid is incompressible and that the ow is steady, we have,
0=
∂u ∂v ∂w
+
+
∂x ∂y
∂z
(1.25)

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PAGE 94
The previous equation is often called the Continuity Equation, and helps us to satisfy the
conservation of mass across the di erential element’s control volume.
To fully satisfy the kinematic aspects of particle motion in view of the di erential element,
we must also conserve momentum across the control volume. We begin this analysis by
analyzing the forces acting on the control volume at all inlets and exits. Generally, we can
satisfy momentum conservation by considering Newton’s Second Law,
∑
F̄ = m ⋅ ā
Thus, the forces acting on the surfaces of the control volume should be equivalent to the
momentum of the uid entering and exiting the control volume.
The forces acting on the control volume include both shear stress and normal stress. Shear
stress is principally the result of frictional forces that exist between uid particles. Referring
to a di erential uid element, we visualize shear stress as,
Figure 1.33. Di erential uid element an applied shear stress in one dimension. Note that shear stress does not lead to deformation of the uid
element when it is stationary (one of the core characteristics that distinguishes a uid from a solid). However, when the uid is moving, viscous
forces result in some deformation.
The normal stresses act on the faces of the
uid element and are conventionally due to
pressure forces. We label these forces using the variable σ . The
nal force acting on the
di erential element (assuming it isn’t negligible) is due to the weight of the uid. We can
now break up the individual terms that act in each direction (x, y, and z) and insert them
into separate momentum equations, which collectively make up the Navier-Stokes equations.
We can separate the shear terms and the momentum terms into a tensor that acts on each
face of the
uid element. The subsequent
gure shows the forces acting on each of the
forward-facing planes on the di erential uid element (forces on the remaining 3 planes are
not shown to limit the complexity of the visual aid).
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PAGE 95
Now, we account for each of the forces acting in the x-direction as,
∑
F̄x = ρ ⋅ gx ⋅ d x ⋅ dy ⋅ dz + (σxx(x + d x) − σxx(x)) ⋅ dy ⋅ dz + (τyx(y + dy) − τyx(y)) ⋅ d x ⋅ dz + (τzx(z + dz) − τzx(z)) ⋅ d x ⋅ dy
Note that for the above, τxx = 0 as it is not acting to shear the
uid on its face. We can
divide each term by the di erential volume of the element to simplify as,
∂σxx ∂τyx ∂τzx
F̄ = ρ ⋅ gx +
+
+
∑ x
∂x
∂y
∂z
These represent the forces acing in the x-direction. We can now account for the acceleration of
the particles (i.e. the term on the right-hand side of Newton’s 2nd Law). Recall that the
acceleration term can be written as,
āx =
Du
∂u ∂u d x ∂u dy ∂u d x
=
+
⋅
+
⋅
+
⋅
Dt
∂t
∂x dt
∂y dt
∂z dt
where the resulting equivalency (the sum of the terms on the right-hand side of the above
expression) is the result of u being a function of u(x,y,z,t). Notice that the original
equivalency is a total derivative (d/dt) whereas the terms on the right hand side are partial
derivatives (∂/∂x, ∂/∂y, ∂/∂z). Note also that the terms dx/dt, dy/dt, and dz/dt are velocities.
We previously labeled these as,
u=
dx
dy
dz
,v =
,w =
dt
dt
dt

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PAGE 96


Figure 1.34. Di erential uid element with each of the normal stresses (left) acting on the forward facing planes, and each of the shear stresses
(right) acting on the forward facing planes. (Gravitational force not shown).
Thus, the acceleration term becomes,
āx =
Du
∂u
∂u
∂u
∂u
=
+u⋅
+v⋅
+w⋅
Dt
∂t
∂x
∂y
∂z
Now we can multiply by the mass to obtain the full term on the right-hand side of Newton’s
2nd Law (or m = ρ ⋅ d x ⋅ dy ⋅ dz), and the full expression describing the motion of the uid
element in the x-direction becomes,
∂P
∂ 2u ∂ 2u ∂ 2u
∂u
∂u
∂u
∂u
(1.26)
ρ ⋅ gx −
+μ⋅
+
+
=
ρ
⋅
+
u
⋅
+
v
⋅
+
w
⋅
2
2
2
(
)
(
)
∂x
∂x
∂y
∂z
∂t
∂x
∂y
∂z
where the stress term is replaced with a pressure (P) and the shear stress term is replaced
du
du
du
(or μ ⋅
, as appropriate).
,μ ⋅
dx
dy
dz
by τ = μ ⋅
The y-momentum and z-momentum forms of the Navier-Stokes equations can be derived by
analyzing the same forces in the y- and z-directions, which produces,
∂P
∂ 2v ∂ 2v ∂ 2v
∂v
∂v
∂v
∂v
ρ ⋅ gy −
+μ⋅
+
+
=
ρ
⋅
+
u
⋅
+
v
⋅
+
w
⋅
(1.27)
2
2
2
(
)
(
)
∂y
∂x
∂y
∂z
∂t
∂x
∂y
∂z
∂P
∂ 2w ∂ 2w ∂ 2w
∂w
∂w
∂w
∂w
ρ ⋅ gz −
+μ⋅
+
+
=
ρ
⋅
+
u
⋅
+
v
⋅
+
w
⋅
(1.28)
( ∂x 2
( ∂t
∂z
∂y 2
∂z 2 )
∂x
∂y
∂z )
Equations 1.26, 1.27, and 1.28 represent the Navier-Stokes equations. In tandem with
the Continuity Equation (1.25), we can solve a variety of useful (albeit simple) problems
in
uid mechanics. Particularly useful are the set of solutions to the above equations that
describe the velocity distribution in the uid (as demonstrated in the following examples).
We note, however, that solving the Navier-Stokes equations is not a trivial endeavor. In fact,
as the uid ow becomes more complex in nature, these equations become di cult (if not
impossible) to solve analytically, and we turn to numerical analyses ( nite element or nite
volume) to solve for the resulting velocity distribution.
We also note that, even for basic solutions to the Navier-Stokes equations, a background in
integral calculus is required (in particular, solutions to ordinary and partial di erential
equations) to obtain solutions to practical problems. In this course, we will not be tasked
with solving the Navier-Stokes equations (and we will avoid two-dimensional problems such
that knowledge of PDEs is not at all required). Nevertheless, we provide the following
(simple, widely used) examples to demonstrate the utility of the Navier-Stokes equations.
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PAGE 97
Let’s consider a uid owing between two at plates. It is useful for us to know where the
location of the maximum velocity is in this case, and to solve for the shear stress at di erent
locations within the uid. Note that the uid is owing in the x-direction in the schematic
below, and that the ow is fully developed (i.e. it is not varying at all in the x-direction; we
will learn more about this in the coming sections),
Solution
If we want to solve for the location of the maximum velocity, we will need to know how the
velocity varies in each direction. Since we know the uid is moving in the x-direction and the
ow is fully developed, the velocity distribution will not change as a function of x ( ∴ u ≠ f (x)). We
also don’t consider it to vary in the "z" direction for the at plates. Thus, the velocity must only be
a function of y ( ∴ u = f (y)). Because the
uid is
owing in the x-direction, we start with the
general form of the x-momentum equation as,
∂P
∂ 2u ∂ 2u ∂ 2u
∂u
∂u
∂u
∂u
ρ ⋅ gx −
+μ⋅
+
+
=
ρ
⋅
+
u
⋅
+
v
⋅
+
w
⋅
( ∂x 2 ∂y 2 ∂z 2 )
( ∂t
∂x
∂x
∂y
∂z )
As is often the case with the Navier-Stokes equation, there are a host of simpli cations that
can be made. These include:
1. No gravitational constant in the x-direction ( ∴ gx = 0)
2. The uid ow is steady
(
∴
∂u
=0
)
∂t
∂ 2u
∂ 2u
∂u
∂u
3. The velocity u is not a function of x or z ∴
=
0
,
=
0
,
=
0
,
=0
2
( ∂x 2
)
∂z
∂x
∂z
4. The velocity in the y-direction (v) is zero ( ∴ v = 0)
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PAGE 98
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Example 1.17 - Fluid ow between two stationary plates (Poiselle ow)
The x-momentum expression for the Navier-Stokes equation becomes,
dP
d 2u
d 2u
1 dP
−
+μ⋅ 2 =0→ 2 = ⋅
dx
dy
dy
μ dx
Now we integrate twice to obtain u(y),
du
1 dP
= ⋅
⋅ y + C1
dy
μ dx
and,
u(y) =
1
dP 2
⋅
⋅ y + C1 ⋅ y + C2
2 ⋅ μ dx
There are two ways to solve for C1 and C2. If we recognize that the problem is symmetric
about the center line shown in the gure on the previous page (and locate the center line at
y = 0), our boundary conditions become,
du
dy
=0
y=0
and,
u(y = W ) = 0
The second boundary condition is due to the fact that there is a no slip condition at the uid
walls, meaning uid particles "stick" to the wall and do not move in that " rst layer".
Therefore,
1
dP
⋅
⋅ y + C1
2 ⋅ μ dx
0=
→ ∴ C1 = 0
and,
1
dP
0=
⋅
⋅ (W 2) + C2
2 ⋅ μ dx
1
dP
→ ∴ C2 = −
⋅
⋅ ( W 2)
2 ⋅ μ dx
so that the expression describing the velocity distribution becomes,
u(y) =
1
dP 2
1
dP
⋅
⋅y −
⋅
⋅ W2
2 ⋅ μ dx
2 ⋅ μ dx
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PAGE 99
from the center line to the top plate for both octane and water when the pressure drop is
Δp = 80 kPa and the length of the pipe L = 50 m. The width of the pipe W = 4 cm and the
dynamic viscosities of the oil and water are μoct = 0.0005 Pa⋅s and μwater = 0.001 Pa⋅s.
An example calculation for the velocity at the center line for each uid reveals that,
1
dP
1
uoct(y = 0) = −
⋅
⋅ W2 = −
⋅
2 ⋅ μ dx
2 ⋅ 0.0005Pa ⋅ s
1
dP
1
uwater(y = 0) = −
⋅
⋅ W2 = −
⋅
2 ⋅ μ dx
2 ⋅ 0.001Pa ⋅ s
−80kPa 1000Pa
1kPa
50m
−80kPa 1000Pa
1kPa
50m
1m
m
⋅ 4cm
= 2,560
(
100cm )
s
2
1m
m
⋅ 4cm
= 1,280
(
100cm )
s
2
Clearly there are going to be signi cant di erences between the velocities, which are a direct
function of the
uid viscosities in this case (μ). A plot of
uid velocity versus height is
provided below for context.
The plot above highlights the distinction between the velocity pro les at a single point (really,
any single point) in the x-direction. The velocity is plotted on the x-axis such that the
variation in y matches the perspective of the gure provided for this problem.


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PAGE 100




What exactly does this look like for ow in a pipe? Let’s plot the velocity as a function of y
Example 1.18 - Fluid ow between a stationary plate and a moving plate
Let’s now look at the case when the top plate is moving and there is no pressure gradient
driving the uid ow (in this case, the top plate will be driving the uid ow). The top plate
will be moving with velocity Uplate. Provided the no-slip condition at each plate, the velocity
of the
uid in contact with the top plate must also be Uplate. Solve for the velocity
distribution u(y).
Solution
Again, we know the uid is owing in the x-direction because the plate itself is moving in the
+ x-direction. Starting with the x-momentum expression,
∂P
∂ 2u ∂ 2u ∂ 2u
∂u
∂u
∂u
∂u
ρ ⋅ gx −
+μ⋅
+
+
=
ρ
⋅
+
u
⋅
+
v
⋅
+
w
⋅
( ∂x 2 ∂y 2 ∂z 2 )
( ∂t
∂x
∂x
∂y
∂z )
We simplify the above expression using the following assumptions:
1. No gravitational constant in the x-direction ( ∴ gx = 0)
2. The uid ow is steady
(
∴
∂u
=0
)
∂t
∂ 2u
∂ 2u
∂u
∂u
3. The velocity u is not a function of x or z ∴
=
0
,
=
0
,
=
0
,
=0
( ∂x 2
)
∂z 2
∂x
∂z
4. The velocity in the y-direction (v) is zero ( ∴ v = 0)
dP
5. No pressure gradient driving the uid ow ∴
=0
(
)
dx
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PAG E 101
Thus, the simpli ed x-momentum equation becomes,
d 2u
d 2u
μ⋅ 2 =0→ 2 =0
dy
dy
Integrating twice yields,
du
= C1
dy
and,
u(y) = C1 ⋅ y + C2
Using the boundary conditions for this problem, we solve for u(y),
u(y = − W ) = 0
→
0 = C1 ⋅ (−W ) + C2
→ ∴ C2 = C1 ⋅ W
and,
u(y = W ) = Uplate
→
Uplate = C1 ⋅ W + C1 ⋅ W
→ ∴ C1 =
Uplate
2⋅W
Finally,
u(y) =
(2 ⋅ W )
Uplate
⋅y+
Uplate
2
Given the same conditions as those provided in the previous problem, with Uplate = Umax ,
we see the following velocity distributions.






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PAGE 102
of the two types of ows. For the case of Poiselle ow, the maximum velocity occurs at the
center-line, while for the case where the top plate is moving with velocity Uplate, the
maximum velocity occurs at W.
Neither of these results are surprising mathematically. To nd the location of the maximum
point on each curve, we can simply take the derivative of the velocity distribution and set it
equal to zero. For Poiselle ow,
du
=0
dy
→
1 dP
⋅
⋅ ymax = 0
μ dx
∴ ymax = 0
and for the case where the top plate is moving, the maximum velocity occurs where
u(y) = Uplate.
Uplate =
Uplate
2⋅W
⋅ ymax +
Uplate
2
∴ ymax = W
We can also solve for the shear stress, τ, at the walls for the case with the moving plate via,
du
τ
=μ⋅
dy
y=W
=μ⋅
y=W
Uplate
2⋅W
This does not change for the case of τ(y = − W ) . To nd the shear stress at the wall in the
case of Poiselle ow,
τ
y=W
=μ⋅
du
dy
=μ⋅
y=W
1 dP
dP
⋅
⋅W =
⋅W
(μ dx
) dx
In the latter, we prove to ourselves that a pressure gradient (likely the result of pumping a
uid) must be established in order to overcome frictional forces associated with shear stress
acting on the wall.
Though it results in larger pump powers, larger shear stresses are also associated with
reduced fouling (i.e. the buildup of particles on a surface) in heat exchangers, which
improves their lifespan.

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PAGE 103


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As shown in the gure above, the maximum velocity occurs at di erent values of y for each
In the previous section, we discuss methods to solve the Navier-Stokes equations for
di erent types of uid ow. However, we restricted ourselves to a series of assumptions in
order to simplify our analyses. One assumption that was made, though not discussed, was
that the uid ow was laminar.
In the following sections, we will examine head losses (akin to energy losses) in bound uids. In
particular, we intend to categorize losses as either major or minor. Major losses are a strong
function of the uid ow type, which is generally characterized as being laminar or turbulent.
Laminar ow is conventionally understood as a uid whose streamlines do not interrupt one
another. Colloquially, we say that the
turbulent
ow pattern is "smooth". We contrast this with
ow, which is chaotic by nature and whose streamlines can be entangled in
rotational ows (called "eddies" at the local scale, and a "wake" on a semi-global scale). With
any characteristic
ow, we can determine whether the
ow conditions are laminar or
turbulent by examining the Reynolds number, which is generally de ned as,
ρ ⋅ V̄ ⋅ Lc
V̄ ⋅ Lc
Re =
=
μ
ν
where Lc is a characteristic length based on the geometry of the surface that the
uid is
owing over. It is important to note that the Reynolds number can also be a function of the
roughness of this surface. We can determine the e ect of surface roughness on Re through
the use of an engineering chart called the Moody Diagram, developed in 1944 by Lewis Ferry
Moody and which is still widely used today.
INTERNAL FLOW
Pipe Flow
The physical mechanisms that impact uid ow in piping are critical to understand from an
engineering design perspective. Piping systems are ubiquitous in industrial and energy
technologies. From a uid mechanics perspective, we must understand the impact of shear
stress on the
uid at the boundaries of the pipes in order to determine the pressure
di erence that must be achieved across some length, size and type of pipe. We will also
learn how sudden changes in direction can result in additional pressure drops that must be
overcome. That is to say, the size of the pump (or fan/blower in the case of gases) and the
power required for its operation are governed by the frictional losses through a particular
piping system.
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Introduction to Laminar and Turbulent Flow
To compute a pressure drop through a system with frictional losses, we utilize the
Conservation of Energy, which says that,
∑
·
Ein =
·
Eout
∑
We account for the ow work, speci c kinetic energy, and speci c potential energy on each side of
the equation. In an actual energy system, we also incorporate the energy consumed by a
pump and the energy put out by a device like a turbine. Finally, we account for energy losses
in our system, which we must overcome with a pump. Such an equation looks like,
P V̄ 2
P V̄ 2
+
+g⋅z
+ g ⋅ hpump =
+
+g⋅z
+ g ⋅ hturbine + g ⋅ hlosses
(ρ
)
(ρ
)
2
2
in
out
It is often convenient to express each term as a length (or head), such that,
P
V̄ 2
P
V̄ 2
+
+z
+ hpump =
+
+z
+ hturbine + hlosses
(ρ ⋅ g 2 ⋅ g
)
(ρ ⋅ g 2 ⋅ g
)
in
out
(1.29)
We often rewrite hlosses as hL, where the head loss is the sum of both major and minor losses
in the system. We write this as,
hL = hL,major +
h
∑ L,minor
Before we address the utility of Eqn. 1.29, it is useful to rst determine (a) what governs
each component of head loss and (b) how to calculate each type of head loss.
Major Losses
We can determine the pressure drop across a circular pipe of length L and diameter D as,
L ρV̄ 2
L V̄ 2
ΔpL,major = ρ ⋅ g ⋅ hL,major = f ⋅ ⋅
→ hL,major = f ⋅ ⋅
D
2
D 2⋅g
where f is known as the friction factor. In circular pipes, we can use,
64
Re
laminar ow (Re ≤ 2,300)
ϵ
f = g Re,
(
D)
turbulent ow (Re > 4,000)
f=
(1.30)
where g indicates that f is a "function of" Reynolds number and hydraulic roughness (ϵ).


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PAGE 105
Figure 1.34. Moody diagram. (Attribution to: S Beck and R Collins, University of She eld (Donebythesecondlaw at English Wikipedia) Conversion
to SVG: Marc.derumaux - File:Moody_diagram.jpg, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=52681200). Note:
additional values of ε are included in the Appendix for both SI and EEU unit systems (with additional materials). Note also that ϵ is the
equivalent sandgrain roughness, or the height of sandgrain that results in the same friction drag as the given roughness.
We can use the above diagram to determine the friction factor for a number of di erent
conditions ( uid
ows and hydraulic roughnesses). For non-laminar
ows, a decent
approximation for the friction factor, f, is provided in the Colebrook equation,
1
ϵ
D
2.51
= − 2 ⋅ log
+
( 3.7 Re ⋅ f )
f
(1.31)
For non-circular cross-sections, we can use the following table when the ow is laminar. To
determine the Reyndolds number for ow in a circular tube, we use,
Re =
ρ ⋅ V̄ ⋅ D
μ
(1.32)
where, for non-circular cross-sections, we replace D (diameter) with a hydraulic diameter,
4 ⋅ Ac
Dh =
(Ac = cross-sectional area, P = perimeter)
P
(1.33)

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PAGE 106
Friction Factors for Laminar Flow
Friction Factor, f
Geometry
a
59.92
=1 → f =
b
Re
a
62.20
=2 → f =
b
Re
a
68.36
=3 → f =
b
Re
a
72.92
=4 → f =
b
Re
a
78.80
=6 → f =
b
Re
a
82.32
=8 → f =
b
Re
a
96.00
=∞ → f =
b
Re
a
64.00
=1 → f =
b
Re
a
67.28
=2 → f =
b
Re
a
72.96
=4 → f =
b
Re
a
76.60
=8 → f =
b
Re
a
78.16
= 16 → f =
b
Re
θ = 10∘ → f =
50.80
Re
θ = 30∘ → f =
52.28
Re
θ = 60∘ → f =
53.32
Re
θ = 90∘ → f =
52.60
Re
θ = 120∘ → f =
50.96
Re
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PAGE 107


















Table 1.5. Friction factors for common geometries and channel ow under laminar ow conditions.
·
Water at 20∘C ows at a volume ow rate of V = 1.2 m3/s through a 200 m long, 0.35 m
diameter cast iron pipe. Determine the friction factor, f, and the major head loss through the
pipe. The kinematic viscosity of water at 20∘C is 1.004⋅10−6 m2/s.
Solution
In order to determine the friction factor, we rst need to know the Reynolds number for the
uid ow in the pipe, which can be determined via,
ReD =
·
V⋅D
V̄ ⋅ D
=
=
D 2
ν
π( 2 ) ⋅ ν
π⋅(
m3
1.2 s ⋅ 0.35m
0.35m 2
) ⋅ 1.004 ⋅ 10−6 s
2
At the above value of ReD, we have fully turbulent
m
6
=
4.35
⋅
10
2
ow. Thus, we obtain f on the Moody
diagram. From the Moody diagram, we nd that ϵ = 0.15 mm for a cast iron pipe such that,
ϵ
=
D
0.15mm
0.35m ⋅
1,000mm
1m
= 0.00043
Using the Moody diagram, we obtain a friction factor of f ≈ 0.017.
The major head loss is then calculated as,
L V̄ 2
200m
V̄ 2
hL,major = f ⋅ ⋅
= 0.017 ⋅
⋅
D 2⋅g
0.35m 2 ⋅ 9.81 m2
s
Where V̄ is calculated as,
m3
·
1.2
V
m
s
V̄ =
=
= 12.47
0.35m 2
Ac
s
π( s )
Therefore, the major head loss is,
hL,major = 76.99 m







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PAGE 108
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Example 1.19 - Calculating major head loss in a pipe
Minor Losses
Pressure drops also occur across devices where the uid direction is abruptly changed. This
creates a separation in the uid that resembles those ow patterns most closely associated
with "wake" regions.
Figure 1.35. Schematic of the rotational ows that result in losses as a uid abruptly changes direction as it passes through a 90∘ bend (left) and as
it expands into a larger cavity (right).
Minor losses are most often determined via experiment, and we must therefore turn to
empirical correlations for di erent types of devices. The general form of the empirical
correlation we will use to calculate minor losses in piping systems is provided by,
V̄ 2
hL,minor =
KL ⋅
∑
2⋅g
Components
(1.34)
The empirical part of Eqn. 1.34 is termed the "loss coe cient" and is represented by the
variable KL. The loss coe cient is determined for a variety of di erent piping components
via experiment, and their values are provided in the gures below.
Figure 1.36. Loss coe cients (KL) for (left) 90∘ smooth pipe bend, (middle) 90∘ miter bend and (right) 90∘ miter bend with vanes.

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PAGE 109
Figure 1.37. Loss coe cients (KL) for (left) 45∘ threaded elbow and (right) 180∘ return bend for both anged and threaded piping.
Figure 1.38. Loss coe cients (KL) for (left) tee with branch ow, (middle) tee with line ow, and (right) a threaded union between two pipes.
Figure 1.39. Loss coe cients (KL) for (left) pipe inlet with reentrant, (middle) sharp-edged pipe inlet, and (right) a rounded pipe inlet with varying
degrees of "roundness".
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P A G E 11 0
Figure 1.40. Loss coe cients (KL) for (left) pipe exit with reentrant, (middle) sharp-edged pipe exit, and (right) a rounded pipe exit. The value of KL
is the same for all three exit conditions and only depends on whether the ow is laminar or turbulent (though it must be fully developed in both
cases).
Figure 1.41. Loss coe cients (KL) for (left) a pipe with a sudden expansion, (right) a pipe with a sudden contraction (KL on plot below).


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P A G E 111
Valves:
Table 1.6. Loss coe cients (KL) for di erent types of valves that are inserted within piping systems.
Loss Coe cients (KL) for di erent Valves
Valve Type (and Condition)
KL
Globe Valve (Fully Open)
KL = 10
Angle Valve (Fully Open)
KL = 5
Ball Valve (Fully Open)
KL = 0.05
Swing Check Valve
KL = 2
Gate Valve:
Fully Open
KL = 0.2
1
Closed
4
KL = 0.3
1
Closed
2
KL = 2.1
3
Closed
4
KL = 17


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P A G E 11 2












Figure 1.42. Plot of loss coe cient (KL) versus d2/D2 for a pipe that experiences a sudden contraction. Note that the diameter d is the smaller of
the two diameters.
Figure 1.43. Loss coe cients (KL) for (left) angled expansions at di erent values of θ and (right) angled contractions at di erent values of d/D
(when θ = 20∘).
Given the above values of KL, we can obtain the total head loss in our system using,
L
V̄ 2
hL = f ⋅ +
KL ⋅
∑
( D
) 2⋅g
Components
(1.35)
The above equation can be used to determine the pumping power required to overcome the
frictional forces due to
uid
ow through a piping system with components that abruptly
change the characteristics of the uid ow.
Several examples are provided hereafter in order to solidify your understanding of how we
use Eqn. 1.35 in practice.
·
One question we might have is how this impacts pump power, Wpump or the rotational power
·
delivered to a turbine, Wturbine. To compute pump power, we use,
·
·
Wpump = m· ⋅ g ⋅ hp = ρ ⋅ V ⋅ g ⋅ hp
(1.36)
and to compute turbine power, we use,
·
·
Wturbine = m· ⋅ g ⋅ hT = ρ ⋅ V ⋅ g ⋅ hT
(1.37)
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P A G E 11 3
A discussion of e ciency is also prudent at this point. Your intuition might already tell you
that devices like pumps and turbines do not operate with 100% e ciency (and your intuition
would be correct!). Although we will not yet work through the physics that dictate the
e ciency of such devices (we will touch on this in EM317), it is nevertheless important to
understand that they typically do not operate under ideal conditions. For both a pump and a
turbine, we de ne e ciency as:
·
Wp, ideal
ηpump = ·
Wp, actual
·
∴ Wp, actual =
·
Wp, ideal
ηpump
(1.38)
We recognize that the pump e ciency as outlined above must yield a value between 0 and 1
(corresponding to 0% and 100%, respectively). As a result, in the ideal case, we would need
to put less power in than the actual case, where a reduced e ciency (relative to 100%) acts
to penalize the magnitude of energy consumption required by the pump.
For a turbine, the e ciency is de ned as:
·
WT, actual
ηturbine = ·
WT, ideal
·
·
∴ WT, actual = ηturbine ⋅ WT,ideal
(1.39)
Here, the turbine e ciency is also a value that must be between 0 and 1 (again,
corresponding to 0% and 100%, respectively). For a turbine, we expect that the ideal power
output (what we gain from an energy system) is greater than what we actually get out if the
e ciency is less than 100%.
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P A G E 11 4
Example 1.20 - Pumping pressurized water
A pump is used to deliver water from a pressurized storage tank to a tank that is open to the
atmosphere. The piping system, depicted in the
gure, is (in total) 50 m long and has a
diameter of 10 cm with f = 0.02. There are (3) 90∘ threaded bends (smooth edged), (4) fully
opened gate valves, (1) fully opened globe valve and reentrant inlets/exits on the tanks.
With a
ow rate of 35 L/s, the pump head is 40 m. What is the gage pressure in the
pressurized tank?
Solution
We begin with the general form of the energy equation, where,
P
V̄ 2
P
V̄ 2
+
+z
+ hpump =
+
+z
+ hturbine + hL
(ρ ⋅ g 2 ⋅ g
)
(ρ ⋅ g 2 ⋅ g
)
in
out
Now we make the following assumptions:
1. The velocities at points (1) and (2) are very slow relative to that of the uid in the
pipe.
2. There is no turbine (or any other energy-generating device) in the system, so there is
no energy coming out
3. The pressure at point (2) is P2 = Patm, so that P2,gage = 0.
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P A G E 11 5
P1
P
+ z1 + hpump = 2 + z2 + hL
ρ⋅g
ρ⋅g
We have the height di erence, the uid type and the pump head, and also know that P2 = 0
(assuming that we are working in gage pressure). Thus, to solve for P1 we must compute
the total head loss, hL, where,
L
V̄ 2
hL = f ⋅ +
KL ⋅
∑
( D
) 2⋅g
Components
Substituting our given values of f, D, and L, and inserting values of KL from Figs. 1.36 - 1.43
(or Table 1.6), we can solve for hL. First, we solve for the velocity (V̄) as,
V̄ =
·
V
=
2
π ⋅ D2
( )
35 Ls ⋅
1m 3
1,000L
= 4.46
2
0.1m
π⋅
( 2 )
m
s
Thus,
hL =
4.46 ms
(
)
(
2
50m
+ (3 ⋅ 0.9 + 4 ⋅ 0.2 + 1 ⋅ 10 + 0.8 + 2) ⋅
m = 26.66m
)
0.1m
2 ⋅ 9.81 2
0.02 ⋅
s
We note that KL,exit = 2 because the Reynolds number is 4.44⋅106 (with νwater = 1.004
⋅ 10−6 m2/s) and it’s a reentrant opening.
Now,
P1 = P2 + ρ ⋅ g ⋅ (hL − hpump) + ρ ⋅ g ⋅ (z2 − z1)
P1 =
(
1,000
kg
m
kg
m
1N
⋅
9.81
⋅
40m
−
26.66m
+
1,000
⋅
9.81
⋅
20m)
⋅
(
)
(
kg ⋅ m
)
m3
s2
m3
s2
1 2
s
P1 = 327,065Pa ⋅
⋅
1Pa
1 N2
m
1 kPa
≈ 327.1 kPa
1,000Pa







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P A G E 11 6



The energy equation then simpli es to:
The ltration system used in a swimming pool is shown in the diagram below and consists
of 20 ft of 2 in diameter piping. The pipe has a friction factor of f = 0.02 and loss
coe cients through the inlet and exit are both KL,inlet = KL,exit = 1.0. There are also (2) fully
opened gate valves, (3) 90∘ threaded elbows (smooth-edged), a lter, and a pump. When the
·
ow rate (V) through the
lter is clean, its loss coe cient is KL, filter = 15. The volume
·
system is V = 0.25 ft3/s. Determine the shaft horsepower (hp) required to run the pump.
Solution
In the above problem, several components result in frictional forces that need to be
overcome in order to sustain uid motion. We can calculate the required pump power via
the energy equation,
P
V̄ 2
P
V̄ 2
+
+z
+ hpump =
+
+z
+ hturbine + hL
(ρ ⋅ g 2 ⋅ g
)
(ρ ⋅ g 2 ⋅ g
)
in
out
We choose points (1) and (2) (inlet/exit) for the control volume based on the fact that the
uid has to make its way around the entire loop (in other words, it must start and end at
the same place). In this way, we take into account all of the major and minor losses that are
required to pump the uid.


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Example 1.21 - Sizing a pool pump
words,
1. The pressure, velocity and height do not change from inlet to exit (i.e.
P1 = P2, V̄1 = V̄2, z1 = z2).
2. There is no power being generated by the system, therefore hturbine = 0.
Using the above two assumptions, it becomes clear that:
hpump = hL
where,
L
V̄ 2
hL = f ⋅ +
KL ⋅
∑
( D
) 2⋅g
Components
To calculate the velocity of the uid in the pipe, we have,
ft 3
·
V
V̄ =
π⋅
=
2
D
(2)
0.25 s
π⋅
2in
⋅
(2
1ft
12in
)
= 11.45
2
ft
s
Thus, the head loss is,
hL =
(
20ft .
0.02 ⋅
1ft
2in ⋅
+ (2 ⋅ 0.2 + 3 ⋅ 0.9 + 1 ⋅ 15 + 1 + 1) ⋅
)
12in
(
ft
11.45 s
)
ft
2 ⋅ 32.2 2
2
= 49.47 ft
s
The power required by the pump is expressed as the energy required by the pump
·
(ρ ⋅ g ⋅ hpump) multiplied by the volume ow rate (V),
·
·
Wpump = ρ ⋅ g ⋅ hpump → Wpump = V ⋅ ρ ⋅ g ⋅ hpump
Therefore,
lbm
ft
1lbf
·
Wpump = 0.25 ft 3 ⋅ 62.3 3 ⋅ 32.2 2 ⋅
lbm ⋅ ft
ft
s
32.2 2
⋅ 49.47 ft = 637.33 ft ⋅ lbf ⋅
s
1hp
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P A G E 11 8




Our assumptions are based on the fact that we start and end at the same location. In other
Multi-path Flow in Piping Systems
Thus far, we have restricted ourselves to piping system in which there exists only one route
that the
uid can take. In many practical applications, of course, the
uid can take more
than one route. As a result, it is important to consider what happens when the uid takes
multiple paths to reach some destination. The schematic below is a representation of a multipath piping system.
Figure 1.44. Multiple path piping system, where the uid can take more than one path to get from the inlet to the exit. Diagram includes purple
dashed lines to indicate the location of piping components (tees and elbows in the above schematic). Lengths indicate the sections where circular
piping exists.
In the above case, the entire piping system can be considered an open system. Assuming the
system is operating at steady-state and that the uid is incompressible, we must satisfy the
Conservation of Mass for a single-inlet/single-exit case as,
m· in = m· out
→
·
·
Vin = Vout
If we consider the entrance tee and the exit tee to be "nodes", we can examine the
Conservation of Energy across each branch from the inlet node to the exit node, we obtain the
following:
Pin
V̄ 2in
Pout
V̄ 2out
+
+ zin = hL,1 +
+
+ zout
ρ⋅g 2⋅g
ρ⋅g 2⋅g
and,
Pin
V̄ 2in
Pout
V̄ 2out
+
+ zin = hL,2 +
+
+ zout
ρ⋅g 2⋅g
ρ⋅g 2⋅g
Thus, hL,1 = hL,2 as is mathematically shown above. This is true between any parallel nodes.
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P A G E 11 9
Though we have talked brie y about pumps in engineering systems that require uid ow,
we have not explored their operational characteristics. Here we provide a brief overview of
pump design in order to better understand how we utilize power to create a pressure
di erence such that we can move a
uid with some volume
·
ow rate V . We provide a
particular focus on centrifugal pumps, which are the most common pumps used in
industry and are used extensively across the Naval eet.
Pump Types
We generally classify pumps in two categories: positive displacement pumps and nonpositive displacement pumps (the latter of which is occasionally referred to as a dynamic
pressure or roto-dynamic pump).
Positive Displacement Pumps
There are a wide variety of positive displacement pump types (only one of which is covered
here), but they all share some common features. In principle, positive displacement pumps
take advantage of rotating machinery (or reciprocating components, like a piston) to move a
uid. This machinery generally results in a large build-up of pressure, but does not typically
result in large uid ow rates. This makes them good candidates for high-viscosity liquids
(which are often more di cult to "push", or rotate in the case of a non-positive displacement
centrifugal pump), or when a
uid does not need to be transported quickly from one
location to another. An example of a positive displacement pump (in this case, a gear pump)
is provided in the schematic below.
Figure 1.45. (left) Schematic of a gear pump, which provides for positive displacement of a uid and (right) a cutaway image of a gear pump used in an
industrial application.
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PA G E 12 0
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Pump Curves and Net Positive Suction Head (NPSH)
ow rate remains
constant and independent of system pressure. This is principally due to the fact that there
remains a relatively xed volume of uid in the pump, and is re ected in the pump curve
below.
Figure 1.46. (top) Plot of pump power versus discharge pressure (red line represents a higher motor RPM than the blue line) and (bottom) plot of
·
volume ow rate (V) versus discharge pressure (red line represents a higher motor RPM than the blue line, each of which correspond to the same
color lines on the top plot).
Notice that in the plot above, the volume ow rate of the uid doesn’t change signi cantly
as a function of discharge pressure. The volume ow rate often changes so little that many
uid mechanics textbooks consider it to be constant and plot it as a vertical line on a
traditional pump curve (see the curve in the following section), though we note that this is
not strictly correct (as shown in Fig. 1.46). Now let’s turn our attention to centrifugal
pumps (i.e. non-positive displacement pumps) as they are ubiquitous in standard
engineering systems, particularly within the wider Naval eet.
Non-Positive Displacement Pumps
Centrifugal pumps are an important type of non-positive displacement pump, and as such
will constitute the majority of our discussion for this particular pump classi cation. The
centrifugal pump relies on its ability to create a suction pressure at the inlet (also referred to as
the "eye" of the pump).
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PA G E 121
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One unique feature of positive displacement pumps is that the volume
shown below.
Figure 1.47. (left) Schematic of centrifugal-type, non-positive displacement pump with important features and (right)Schematic of the impeller and
its operational characteristics (e.g. suction and discharge mechanisms). Figure on the right adapted from: McMahon, Glenda; Chatterton, Ken
(2019): Centrifugal pump impeller arrangements. Loughborough University. Figure. https://doi.org/10.17028/rd.lboro.7981925.v1.
The schematics shown in Fig. 1.47 represent cutaway images of a centrifugal pump (left)
and the impeller used in a centrifugal pump (right) to create suction, an increase in
uid
pressure, and a subsequent discharge of the uid into a piping system.
Generally speaking, a suction pressure (Ps) is created in the impeller eye, which is represented
by,
Ps = P +
1
⋅ ρf ⋅ V̄ 2
2
(1.40)
where P in the above expression represents the static pressure of the
1
⋅ ρf ⋅ V̄ 2 is the dynamic pressure of the
2
"stagnation pressure”). To
uid and the term
uid (together, these are also termed the
nd the discharge pressure (and ultimately the pressure
di erence across the pump, Pp and/or the pump head, hp), we can use the energy equation
as,
P
V̄ 2
P
V̄ 2
+
+z
+ hpump =
+
+z
+ hturbine + hL
(ρ ⋅ g 2 ⋅ g
)
(ρ ⋅ g 2 ⋅ g
)
in
out
Critically, we must also make sure that the suction pressure (which is the lowest pressure in
the system), does not get so low that it is below the saturation pressure of the uid (Psat).
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PA G E 12 2
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The basic working mechanism for this type of pump is best visualized using the schematic
Though we’ll talk much more about saturation pressure in the second part of this course
(Thermal- uid Sciences II), it is a simple endeavor to understand the saturation pressure as the
pressure at which a
uid begins to boil. We wish to avoid exceeding this threshold such
that we can prevent cavitation on the pump blades. Cavitation is perhaps best understood as
the rapid collapsing of bubbles that form in low pressure regions on a surface; the resulting
force imposed on the surface due to the collapse of this bubble is so large that, over time, it
can cause signi cant erosion. A wide variety of freely accessible videos exist across the
internet that depict this phenomenon, and you are strongly encouraged to view a handful of
them to familiarize yourself with an issue that is critical to naval propulsion systems.
The primary reason we are concerned with the saturation pressure is because we know that
as the pressure of a substance decreases, so too does the temperature at which it begins to
boil! Given that there are losses in our piping systems, we expect the pressure to drop ahead
of the pump. You might understand this relationship best by examining the saturation
pressures and temperatures of water as one or the other is varied. A (brief) table describes
this relationship below.
Table 1.7. Saturation pressures and temperatures for water
Saturation Pressure (Psat) vs. Saturation Temperature (Tsat) for Water
Psat [kPa]
Tsat [∘C]
100
99.6
10
45.8
1
6.79
It is relatively clear from the table above that as the pressure of a
uid (like water)
decreases, so does its saturation temperature! Thus, as we experience pressure drops in our
system, we must ensure that we do not reach a pressure below the saturation pressure of
the uid just prior to the pump’s entrance.
In practice, we compare the suction head to the vapor head and make sure that our net positive
suction head is itself a positive value. These values are de ned as follows.
Ps
ρf ⋅ g
(1.41)
Psat
hv =
ρf ⋅ g
(1.42)
hs =
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PA G E 12 3
Ps − Psat
= hs − hv
ρf ⋅ g
(1.43)
Equation 1.43 represents the net positive suction head required by the pump at its inlet. As
engineers, we must ensure that we meet the constraints of Eqn. 1.43 by comparing this
value to the net positive suction head required (NPSHR), which is speci ed by the pump
manufacturer.
As a
nal way to think about this issue, let’s look at how the pressure changes within a
system that uses a pump. Consider the following system:
Figure 1.48. (top) A large tank lled with water up to height S results in a discharge of uid ow at the exit (point F), which is at atmospheric
pressure. Points A, B, C, D, E, and F are provided to illustrate the pressure drops (or increases) at important points in the uid path, as shown in
(bottom) a plot of system pressure head versus location along the length of a pipe which has a uid owing through it and a pump at some length
down the pipe from its entrance.
As shown in the gure above, the pressure head at point A is constant along the direction of
ow (green dashed line) until it gets close to the pipe entrance (location B), at which point
the uid accelerates and develops some losses. As it enters the pipe at point C (points B and
C are, in reality, adjacent to each other at individual points across the openings boundary), a
minor loss across the entry length occurs, after which point major head losses induce a
pressure drop in the uid to the inlet of the pump. Provided the pump inlet (point D) has a
pressure head that exceeds hv, there will (ideally) be no cavitation that occurs on the
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PA G E 12 4
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NPSH =
impeller blades. The pump then supplies some pressure head from the power put into
rotating the impeller, which is represented by the length hp in the diagram above. Finally,
the remaining major losses through the pipe result in a pressure drop until the
uid
discharges at atmospheric pressure (point F).
To nd the pressure head, one can simply use the energy equation we started with as,
V22
hp = hL +
− z1
2⋅g
Here, we’ve assumed that the pressure at points (1) and (2) (which correspond to points "S"
and "F" on the schematic shown in Fig. 1.48) are both atmospheric pressures. As a result the
gage pressures at both locations are p1,gage = p2,gage = 0 . Likewise, we assume that the
velocity at point (1) is V̄1 = 0 due to the fact that the tank is very large. Finally, there is no
turbine in our system and we assume that the height at point (2) is normalized to z2 = 0.
If we substitute the expression for total head loss into the previous expression, we obtain,
L
V̄ 2
hp = f ⋅ +
K ⋅
− z1
( D ∑ L) 2 ⋅ g
SystemCurve
·
Since we know that V̄ = V/A, we can plot the pressure head versus a volume ow rate. This
is, in fact, how pump curves are plotted by manufacturers for speci c individual pumps. On
the right-hand side of the above equation, we relate the pump head versus volume ow rate
for a particular piping system. We can use this to help us select a pump that achieves a speci c
volume ow rate for a particular system. When we plot both the pump curve and the system
curve, the operational volume ow rate is considered to be the point at which they intersect.
Let’s take a look at a brief example in order to better assess how we choose a pump to meet
the needs of a full piping system. We note that in the above analysis, the pressure head will
be both a function of the uid’s velocity and its friction factor (which itself is a function of
velocity). This can get somewhat cumbersome to compute without a calculation. For a
simple problem where the uid is laminar, we can assume that the friction factor, f , can be
expressed as,
64 ⋅ μ
L
V̄ 2
→ ∴ hL =
⋅ +
K ⋅
( ρ ⋅ V̄ ⋅ D D ∑ L) 2 ⋅ g
64
f=
Re
The following problem will utilize the Moody Chart and the equation used to develop f for a
smooth pipe.
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PA G E 12 5
Let’s assume that we have a system that can be represented by the schematic below. The
system has two tanks whose water levels are separated by a height of 15 ft. In both the
lower tank and the upper tank, the inlets and exits are reentrant type openings. There is also
a globe valve that sits between the pump and the upper tank that can be used to restrict the
ow rate, and (2) 90∘ rounded elbows (threaded) in the pump path. The diameter of the
pipe is 2 ft. and the total length of the pipe is 35 ft. Provided the pump curve next to the
schematic, what will the volume ow rate of the uid be for the water when the globe valve
is fully-opened? We’ll assume the pipe is smooth.
Solution
In the above schematic, we are provided with a system having several components and a
change in height. We also have a pump curve on the right, and our goal is to nd the point
where it intersects the system for 3 di erent conditions. First, let’s solve for the system
curve using the information provided above. We start with the general form of the energy
equation, which is,
P
V̄ 2
P
V̄ 2
+
+z
+ hp =
+
+z
+ hT + hL
(ρ ⋅ g 2 ⋅ g
)
(ρ ⋅ g 2 ⋅ g
)
in
out
We can now make several assumptions to simplify the energy equation as shown above,
including:
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Example 1.22 - Using pump curves
Assumptions:
1. The reservoirs (A and B) are both open to the atmosphere, and therefore negate one
another in the energy equation.
2. There is no turbine in the system, so hT = 0.
3. The velocity at both points (1) and (2) are V̄1 = V̄2 ≈ 0 due to the size of the tank
relative to that of the piping.
4. We assume that the water level at point (1) is z1 = 0.
Thus, our system curve can be represented by the following expression:
hp = hL + z2
And, substituting the expression for hL, we obtain,
L
V̄ 2
hp = f ⋅ +
K ⋅
+ z2
( D ∑ L) 2 ⋅ g
where V̄ in the above expression is the velocity of the uid moving through the pipe. To nd f, we
must know if we have laminar or turbulent ow. We use the ReD to determine the maximum
Re we expect for the maximum volumetric ow rate used in the pump curve (V· = 2,000
62.3 lbm3 ⋅
gal
2,000 min
ft
ReD,max =
⋅
π ⋅ (1ft)2
2.037 ⋅ 10−5
lbf ⋅ s
ft 2
3
1 fts
⋅ 2ft
gal
448.83 min
⋅
gal
).
min
= 269,444.8
32.2 lbm2⋅ ft
s
1lbf
which is clearly turbulent. We also know that the ow is laminar when,
2300 ⋅ 2.037 ⋅ 10
·
V≤
−5 lbf ⋅ s
ft 2
62.3
32.2 lbm2⋅ ft
⋅
⋅ π ⋅ (1ft)2
s
1lbf
gal
⋅
lbm
⋅ 2ft
ft 3
448.83 min
ft 3
1s
≤ 17
gal
min
when the ow is turbulent, we can use a modi ed form of the Colebrook equation to solve
for f (note that the Colebrook equation must be solved implicitly rather than explicitly, which
is quite a bit more intensive than required for this course).

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The modi ed form of the Colebrook equation can be written as,
f=
[
2.51
+ 1.1513 ⋅ δ
ReD
δ−
( 3.71 )
ε
D
]
2
− 2.3026 ⋅ δ ⋅ log(δ)
with,
δ=
6.0173
0.109
ε
Re ⋅ (0.07 ⋅ ( D + Re −0.885)
ε
D
+
3.71
·
If we then plot f as a function of volume ow rate, V, we obtain (when ε = 0),
which is consistent with the Moody Diagram. Our system curve can now be plotted as,


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Our previous discussion of internal ow was principally focused on the analysis of pipe ow
and pumps, as such devices are ubiquitous in engineering systems. Now we turn our
attention to external ow, which is also an important engineering problem, but requires a
wider range of analyses (particularly with respect to geometry).
Boundary Layers
Our foray into external ow begins with a discussion of a boundary layer. Despite the long
history associated with the study of
uid mechanics, the recognition that viscous e ects
cause a drag force on an object is a relatively recent phenomenon. In 1904, Ludwig Prandtl3
developed a mathematical construct to describe a boundary layer, or a thin layer of
uid
where viscous forces were not negligible. This paved the way for analysis of shear stress
within the uid (and at the surface) and, ultimately, our understanding of drag force.
Though perhaps an oversimpli cation, the development of a boundary layer on the surface
of an object is the result of a "no-slip" condition, whereby the uid particles closest to the
surface are considered motionless, and act to slow down the layer of
uid particles above
them. As the uid moves parallel to the direction of motion, the impact of each individual
uid layer’s shear stress increases in the direction perpendicular to the wall. A visualization
of this phenomenon is provided in the gure below.
Figure 1.49. Laminar ow over a at plate with a uniform velocity U∞ imposed at the leading edge. The distinction between laminar and turbulent
ow is absent in the above gure and addressed later.
3 Prandtl, L., 1905. Uber Flüssigkeitsbewegung bei sehr kleiner Reibung, Verhandlungen des dritten internationalen Mathematiker-Kongresses
(Heidelberg 1904), pag. 484-491, 1905. English translation available from http://digital. library. unt. edu/ark:/67531/metadc65275.
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PA G E 12 9

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EXTERNAL FLOW
uid moves down the plate. We should note that in most applications, this layer is
extremely thin relative to the length-scale of the bulk uid that surrounds it. At the same
time, several practical applications exist where this thickness is on the order of the system’s
characteristic length. In some computer hard drives, for instance, the boundary layer is on
the order of the mean-free path of energy carriers, which has important consequences for
heat dissipation. However, we don’t cover the relevant physics that govern uid ow in such
cases here.
Within a boundary layer, the change in the velocity in the direction perpendicular to the
plate can be quite steep; in other words,
du
is large in close proximity to the surface. As a
dy
result, boundary layers develop for most uids owing over a surface that permits the uid
to pass over it in such a way that streamlines can be sustained, even if the uid’s viscosity is
relatively low.
The size of the boundary layer depends on whether the ow is laminar or turbulent. For a
smooth at plate, the transition from laminar to turbulent ow occurs at a critical Reynolds
number of,
Recr =
ρ ⋅ V̄ ⋅ L
= 5 ⋅ 105
μ
where L is the length of the plate in the direction of
internal
ow, the transition fro laminar to turbulent
uid
ow. We note that, as with
ow can occur at di erent Recr (for
instance, if turbulence is "tripped" - or activated - in the presence of some obstruction close
to the leading edge of the plate). However, use the above value as a general "rule of thumb".
There is a transition region where the ow characteristics do not precisely resemble those of
laminar or turbulent ow. This occurs in the following range of Re,
5 ⋅ 105 ≤ Re < 107
Beyond a value of Re = 107, the
ow is considered to be completely turbulent. The
following schematic illustrates the transition between laminar and turbulent
ow as the
uid moves down the length of the plate.
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As shown above, the boundary layer "grows" (i.e. its thickness increases) the further the
Figure 1.50. Flow transitions over a at plate, from laminar to transitional to turbulent ow. Note that the boundary layer will keep growing across
the length of the at plate.
The thickness of the boundary layer was an open question for several years after the
explanation of its existence by Prandtl. In 1908, Heinrich Blasius (one of Prandtl’s
rst
students!) used a similarity solution to solve the momentum integral equation and
developed a correlation for the thickness of a boundary layer over a smooth, at plate.
In a broader context, we consider the boundary layer to "end" where the velocity in the
direction away from the plate is 99% of the free-stream velocity, U∞. We call the resulting
thickness (δ) the displacement thickness for the boundary layer. To determine the
displacement thickness, we turn to a physical representation of both the mass de cit and
momentum de cit formed at a point along the length of the boundary layer. Let’s examine the
mass de cit rst, where,
Figure 1.51. (left) The "mass de cit" that results from the shear force generated at the surface of a smooth at plate exposed to a uniform ow and
(right) the equivalent displacement (δ*) the surface would need to be displaced from the uid in a frictionless ow to yield the same mass de cit.
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PAG E 131
from the surface as a result of the reduction in velocity from the boundary layer to the
surface itself. We can equate this to the distance (δ*) that a surface must be displaced from
the uid in the absence of frictional forces (i.e. where no boundary layer is formed) and the
velocity pro le remains uniform in the direction perpendicular to the ow. Thus we can say
that,
ρ ⋅ U∞ ⋅ δ* =
δ
∫0
ρ ⋅ (U∞ − u) dy
and when the uid is incompressible (i.e density remains constant), we obtain,
δ* =
∫0 (
δ
1−
u
dy
)
U∞
Now let’s examine the momentum thickness (θ). Recall that momentum is the product of mass
(mass ow rate on a rate basis) and velocity. Just as we experience a mass de cit due to the
velocity gradient within the boundary layer, we also experience a momentum de cit.
Figure 1.52. (left) The "momentum de cit" that results from the shear force generated at the surface of a smooth at plate exposed to a uniform
ow and (right) the equivalent displacement (θ) the surface would need to be displaced from the
uid in a frictionless
ow to yield the same
2
momentum de cit. Note that the units above result in kg⋅m/s for both the integral and the expression in the gure on the right. These units are
the same as those obtained when we compute m· ⋅ V̄.
Equating the two, we obtain,
δ
∫0
ρ ⋅ U∞ ⋅ θ =
ρ ⋅ u ⋅ (U∞ − u) dy
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PAG E 13 2
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In the above gure, the mass de cit is de ned as the mass of uid that is displaced vertically
and, assuming that the uid is incompressible,
θ=
For laminar
ow over a
u
u
⋅ 1−
dy
)
∫0 U∞ (
U∞
δ
at plate, let’s assume that the velocity distribution within the
boundary layer is parabolic and takes the following form,
u(y) = a + b ⋅ y + c ⋅ y 2
Reexamining the boundary layer and its corresponding boundary conditions, we have:
Figure 1.53. Boundary conditions for laminar ow over a at plate at the physical boundaries of a boundary layer. Three boundary conditions are
provided to solve for coe cients a, b, and c.
Mathematically, we use the boundary conditions listed in the previous
gure to solve for
constants a, b, and c, which yields,
u
y
y
=2⋅
−
(δ ) (δ )
U∞
2
Now, let’s de ne y/δ = η, such that,
u
= 2 ⋅ η − η2
U∞
To determine how the boundary layer grows, we relate the momentum de cit to the wall
shear stress (as the shear stress directly causes the momentum de cit and therefore controls
the thickness of the boundary layer). This relationship is represented by,
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PAG E 13 3
(1.44)
We insert the velocity distribution obtained after solving for constants a, b, and c into Eqn.
1.44 such that,
2
τw = ρ ⋅ U∞
⋅
d δ u
u
⋅ 1−
dy
d x ∫0 U∞ (
U∞ )
To be consistent with our simpli cation of the velocity distribution, we use the de nition y/
δ = η, such that,
dy = δ ⋅ dη
and the shear stress at the wall can be de ned as,
1
dδ
u
u
2
τw = ρ ⋅ U∞ ⋅
⋅ 1−
dη
d x ∫0 U∞ (
U∞ )
(1.45)
At the same time, we also know that,
∂u
τw = μ ⋅
∂y
U∞
=μ⋅
⋅
δ
y=0
∂ U
( ∞)
u
∂ δ
( )
y
y
δ =0
U∞
=μ⋅
⋅
δ
∂ U
( ∞)
u
∂η
η=0
Knowing that u/U∞ = 2 ⋅ η − η 2, then d(u/U∞)/dη = 2 - 2 ⋅ η and,
U∞
τw = μ ⋅
⋅ 2 − 2 ⋅ η)
( δ ) (
η=0
U∞
=2⋅μ⋅
( δ )
(1.46)
Therefore, we can equate Eqns. 1.45 and 1.46 to yield,
1
U∞
u
u
2 dδ
2⋅μ⋅
= ρ ⋅ U∞ ⋅
⋅ 1−
dη
( δ )
d x ∫0 U∞ (
U∞ )
Substituting the velocity distribution, we obtain,
1
U∞
2 dδ
2⋅μ⋅
= ρ ⋅ U∞ ⋅
2 ⋅ η − η 2) ⋅ 1 − 2 ⋅ η + η 2 dη
(
( δ )
(
)
d x ∫0
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PAG E 13 4


dδ
dx
2
τw = ρ ⋅ U∞
⋅
δ ⋅ dδ =
15 ⋅ μ
dx
ρ ⋅ U∞
Integrating each side from 0 to δ and 0 to x, we obtain,
δ2
15 ⋅ μ
=
+C
2
ρ ⋅ U∞
To solve for the constant c above, we recognize that the boundary layer has no thickness at
the leading edge (i.e. δ(x = 0) = 0). Therefore,
δ=
30 ⋅ μ ⋅ x
ρ ⋅ U∞
Finally, since we recognize that Rex (the local Reynolds number at any point x along the
length of the plate) is Rex = ρ ⋅ U∞ ⋅ x / μ, we can say that,
δ
5.48
=
x
Rex
The above expression is an approximation for the boundary layer thickness along a smooth
at plate (recall that we approximated the shape of the boundary layer). Blasius obtained the
"exact" mathematical solution via similarity, where,
δ
5.00
=
x
Rex
(1.47)
For turbulent ow, a similar mathematical derivation provides,
δ
0.382
=
1
x
Re 5
(1.48)
x
You will utilize equations 1.47 and 1.48 to solve for the thickness of the boundary layer at
any point x along the length of a at plate. Of course, these expressions are only valid for
the simple case of
uid
ow over a smooth
at plate. We do not cover boundary layer
development over other geometries in this textbook as it is beyond the scope of the course.
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Evaluating the integral and rearranging, we obtain,
Drag Forces
Given what we now know about the boundary layer - that it represents a region across
which shear stress induces a resistive force across the surface of an object - it is desirable to
understand the consequences of forces that act within the boundary layer and on the surface
of an object. In this section, we formalize those consequences using the terms lift and drag.
Let’s consider what each term represents physically, and why we require two terms for a
complete understanding of the forces acting on bodies that are immersed in a uid where
there is some relative motion between the two. Consider the case of a stationary sphere that
is immersed in moving uid,
Figure 1.54. Sphere that is suspended in a uid that is moving with a uniform incoming velocity U∞. Terms FL and FD represent the lift and drag
forces acting on the surface of the solid sphere.
In the above diagram we recognize that the lift and drag forces are perpendicular and
parallel to the direction of relative motion between the object and the uid, respectively. We
note here (and expand on this further later) that the lift force can be oriented in either the
positive or negative vertical direction.
We will rst consider the drag force acting over the surface of an object that is immersed in
a uid. There are two classi cations of drag force that are considered, which make up the
total drag force when summed. These include:
1. Friction Drag and,
2. Pressure Drag
The friction drag acting on an object is principally due to the shear stress at the wall, and is
analogous to the losses caused by wall shear stress in pipes. We also refer to this type of
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PAG E 13 6
drag force as surface drag due to the fact that the drag is caused by viscous uid ow over the
wall itself.
On the other hand, pressure drag (also referred to as form drag) is the drag force due to the
formation of an adverse pressure gradient at the back-end of an object due to the object’s
shape (and how that shape impacts the acceleration of a uid as it ows around the object).
These two types of drag can be visualized below.
Figure 1.55. (left) Horizontal at plate that experiences shear stress (and thus friction drag) as a result of the boundary layer that develops along
the length of the plate and (right) vertical at plate that does not shear in the direction of the ow, but results in an adverse pressure gradient that
forms as a wake behind the vertical plate.
In general, the total drag force acting on a surface is a combination of the shear drag and the
pressure drag, where,
FD = p ⋅ sin(θ) d A + τw ⋅ cos(θ) d A
∫
∫
(1.49)
The angle θ is described in the gure above. For the horizontal at plate, the angle θ = 0∘
while for the vertical plate the angle θ = 90∘. We know that sin(0∘) = 0 and cos(0∘) = 1, and
the drag force on the horizontal plate is therefore completely dominated by friction drag. We
also know that sin(90∘) = 1 and cos(90∘) = 0 such that the pressure drag completely
governs the drag force acting on the vertical plate.
Though Eqn. 1.49 is useful for conceptualizing the mechanisms that most impact the drag
force acting over a surface, we generally turn to a dimensionless relationship and empirical
correlations to solve for the total drag force that is acting over a surface. In this case, we
relate a drag coe cient, CD, to the total drag force, FD, where CD is experimentally
determined by a variation of parameters and modeling.


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PAG E 137
CD = 1
2
FD
(1.50)
⋅ ρ ⋅ V̄ 2 ⋅ Afrontal
where Afrontal is the frontal area of the object, which is the area you would see if you were
facing the object and could not render it in 3 dimensions. For example, the frontal area of a
sphere is the area of a circle (viewed straight on, a sphere looks like it is a circle). We will
practice using this in several example problems momentarily and within the context of
individual geometries.
Flow over a Flat Plate Parallel to the Direction of Flow
The drag coe cient on a horizontal at plate (left-most gure in Fig. 1.55) that is immersed
within a uid is described by the relationships shown in the table below for di erent ow
conditions.
Table 1.8. Drag coe cients (CD) for ow over a at plate that is parallel to the direction of the uid ow.
Drag Coe cients over a Horizontal Flat Plate
Condition
Drag Coe cient, CD
Laminar Flow, ReL ≤ 5⋅105
CD =
Turbulent Flow, 5 ⋅ 10 5 < ReL < 107
(when most of the ow is turbulent)
CD =
Turbulent Flow, ReL < 109
(when most of the ow is turbulent)
CD =
Turbulent Flow, 5 ⋅ 10 5 < ReL < 107
(when neither laminar nor turbulent ow dominates and there exists a transition)*
CD =
1.33
ReL
0.0742
1
ReL5
0.455
(log(ReL ))
0.0742
1
ReL5
−
2.58
1742
ReL
*This is valid when the critical Reynolds number is Recr = 5⋅105. If Recr ≠ 5⋅105, e.g. when a wire is placed close to the leading edge to "trip"
turbulence, then a correction factor (B) must be used, whereby the quantity B/ReL is subtracted from the drag coe cient (CD) for fully turbulent
ow. The correction factor B = Recr ⋅ (CD,turbulent − CD,laminar ).
The equations in Table 1.8 are valid when the the plate is completely parallel to the direction
of uid ow. In such a case, the only relevant component of drag is the friction drag caused
by the shear stress at the wall. If the plate itself is angled, the above expressions would not
accurately predict the drag force acting on the plate, as form drag would surely result due to
the presence of an adverse pressure gradient acting behind the plate.
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We express the aforementioned relationship as,
Example 1.23 - External ow parallel to a at plate (drag force)
Air at 20∘C, ρ = 1.204 kg/m3, and μ = 1.825⋅10−5
kg
m⋅s
ows over a very thin at plate at a
velocity of V̄ = 3.5 m/s in the directions shown in the diagrams below.
For each of the ow con gurations, determine:
1. The Reynolds number (ReL) and the boundary layer thickness (δ) at the trailing edge
of the plate.
2. The drag force (FD) acting on both sides of the plate.
Solution
First we compute the Reynolds number for the ow on the left as,
ReL,left =
ρ ⋅ V̄ ⋅ L
=
μ
1.204
kg
m
m
⋅ 3.5 s ⋅ 0.5m
3
kg
1.825 ⋅ 10−5 m ⋅ s
= 1.15 ⋅ 105
We calculate the Reynolds number for the ow on the right schematic as,
ReL,right =
1.204
ρ ⋅ V̄ ⋅ L
=
μ
kg
m
m
⋅
3.5
⋅ 2m
3
s
kg
1.825 ⋅ 10−5 m ⋅ s
= 4.62 ⋅ 105
In both cases, the uid ow is laminar (i.e. ReL < 5⋅105). Now, we solve for the boundary
layer thickness in each case as,
δleft =
5.00 ⋅ x
Rex
=
5.00 ⋅ 0.5m
= 0.0074 m = 7.4 mm
1.15 ⋅ 105
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PAG E 13 9
and,
δright =
5.00 ⋅ x
Rex
=
5.00 ⋅ 2m
= 0.0147 m = 14.7 mm
4.62 ⋅ 105
To nd the drag force imposed in each scenario, we use,
FD =
CD ⋅ ρ ⋅ V̄ 2 ⋅ Afrontal
2
For laminar ow, we nd CD in Table 1.8 is,
CD =
1.33
ReL
For the drag force on the left-most schematic, we recognize that boundary layers will
develop on each side of the plate, and their magnitude impact on FD will be equivalent. For
one side of the left-most schematic, the drag force is solved as,
2
m
⋅ 1.204 3 ⋅ 3.5 s
⋅ 2m ⋅ 0.5m
m
(
)
1.15 ⋅ 10 5
kg
1.33
FD, top, left =
2
⋅
1N
1
kg ⋅ m
= 0.0289N
s2
Therefore the total drag force acting on the plate in the left-most image of the schematic is,
FD, left = 2 ⋅ FD, top, left = 0.058 N
Similarly, the drag force acting across one side (we’ll say the top here, arbitrarily) of the
image in the right-most image of the schematic is,
1.33
4.62 ⋅ 10 5
FD, top, right =
⋅ 1.204
kg
m3
⋅ 3.5 ms
⋅ 2m ⋅ 0.5m
(
)
2
2
⋅
1N
1
kg ⋅ m
= 0.0145N
s2
and,
FD, right = 2 ⋅ FD, top, right = 0.029 N






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PAG E 14 0
As with the case of ow over a at plate that is oriented parallel to the direction of ow, a
at plate that is vertical relative to the uid ow direction experiences only one type of drag.
In this case, however, it is pressure drag that governs the drag force on a vertical at plate
(as visualized by the right-most schematic in Fig. 1.55) rather than friction drag. The table
below is compiled to list the drag coe cient for di erent object shapes that are oriented
perpendicular to the direction of uid ow.
Table 1.9. Drag coe cients (CD) for ow over vertically oriented geometries (relative to the direction of uid ow). These drag coe cients are valid
for Re ≥ 103. Other drag coe cients can be found in the Appendix of this textbook.
Drag Coe cients over a Series of Objects Oriented Vertically to the Direction of Fluid Flow
Object Geometry
Schematic
Drag Coe cient, CD
Rectangular Prism
L
=0
t
→
CD = 1.9
L
= 0.5
t
→
CD = 2.5
L
=2
t
→
CD = 1.7
Circular Disk
CD = 1.1
Hemisphere
(open-end facing ow)
CD = 1.17
Hemisphere
(open-end facing downstream)
CD = 0.38
The table above provides drag coe cients for a variety of shapes in cross- ow. This is an
important type of ow and we will see it again when we discuss convection heat transfer.
For now, we will use Table 1.9 as a reference when we encounter relevant problems.
The remaining geometries that we will encounter for external ow-type problems explored
in this course include both a sphere and a cylinder. Each is detailed below.
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PAG E 141

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Flow over Geometries that are Perpendicular to the Direction of Flow
Flow over Spheres and Cylinders
When a sphere or cylinder is immersed within a uid and relative uid motion exists, both
friction drag and pressure drag are relevant to the drag force, as shown in the schematic
below.
Figure 1.56. Cross-section of a cylinder or sphere in cross- ow, with corresponding shear stress and adverse pressure gradient that produce drag on
the surface .
The drag coe cient acting over a sphere is provided in the plot below, which is the same as
that shown in Fig. 1.31 and recreated here in Fig. 1.57 for ease of use.
Figure 1.57. Drag coe cient
FD
CD = 1
(
⋅ ρV̄ 2 ⋅ A
2
frontal
)
versus Reynolds number
Re =
( D
ρ ⋅ V̄ ⋅ D
for uniform ow over a smooth sphere.
)
μ
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PAG E 14 2
Similarly, we can obtain the coe cient of drag for a cylinder in cross- ow using a plot of CD
vs. ReD, as shown below.
Figure 1.58. Drag coe cient
FD
CD = 1
(
⋅ ρV̄ 2 ⋅ A
2
frontal
)
versus Reynolds number
Re =
( D
ρ ⋅ V̄ ⋅ D
for uniform ow over a smooth cylinder.
)
μ
The gures above can be used to determine the drag force acting over the surface of a sphere
or a cylinder in cross- ow, provided that the surface is smooth. Combined, we have,
1.59. Drag coe cient
FD
C = 1
( D
⋅ ρV̄ 2 ⋅ A
2
frontal
)
versus Reynolds number
Re =
( D
ρ ⋅ V̄ ⋅ D
for uniform ow over a sphere/cylinder.
)
μ
Figure

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PAG E 14 3
Drag over Airfoils
Airfoils are shapes that are speci cally designed to reduce drag and increase lift based upon
their "teardrop" shape. Here, the uid is streamlined such that it doesn’t create the adverse
pressure gradient that forms downstream of the uid ow, like we see with the vertical at
plate, sphere and cylinder in cross- ow. The streamlining nature of
uid
ow around an
airfoil is shown in the schematic below.
Figure 1.60. Streamlined nature of uid ow around an airfoil, where the boundary layer does not diverge from the surface and result in a wake in
the same way it does with a sphere or a cylinder.
The decrease in the contribution from pressure drag is clearly shown in the gure below; as
the the thickness of the airfoil is reduced relative to the chord length, we enhance the
streamlining shape and thus the pressure drag acting on the airfoil.
Figure 1.60. Drag coe cient acting on an ordinary airfoil with relative contributions from friction drag and pressure drag.
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PAG E 14 4
Example 1.24 - Drag force acting on a parachute
In order to land safely, a paratrooper having a mass of 170 lbm can fall no faster than 15
mph. Determine the minimum diameter of the parachute that allows for a safe landing.
Solution
The mass of the person above is represented by the darkened circle attached to the
parachute chords. We are told that the terminal velocity (i.e. the maximum velocity that is
reached during a free-fall) is 15 mph. The rst thing we’ll do is examine the forces acting on
the parachute (we will assume the drag coe cient on the human is negligible relative to
that acting on the parachute itself). We use Newton’s 2nd Law to account for the forces
acting on the parachute, which is in motion.
∑
Fy = m ⋅ ā
Since the acceleration is ā = 0 (due to the fact that we are at terminal velocity), we write the
force balance as,
FB + FD − W = 0
where FB is the buoyant force acting on the parachute, FD is the drag force and W is the
weight acting on the parachute (we will neglect the weight of the parachute itself). Because
we are not displacing much air (and because the density of air is low relative to the mass
density of a human), we can neglect the buoyant force altogether. We can now rewrite this
expression as,
CD ⋅
ρair ⋅ V̄ 2 ⋅ Afrontal
2
=m⋅g


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PAG E 14 5
Here, we consider the frontal area to be Afrontal = π ⋅ D 2 /4 (i.e. the area of a circle) due to
the assumption that the parachute itself is hemispherical. Solving for the required diameter,
we obtain,
D=
8⋅W
CD ⋅ ρair ⋅ V̄ 2 ⋅ π
According to Table 1.9, CD = 1.42 for a hemisphere that is open-ended and facing the
direction of uid ow. As a result,
ft
D=
8 ⋅ 170 lbm ⋅ 32.2 2
1.42 ⋅ 0.07518 lbm ⋅ 15 mi
⋅
( hr
ft 3
s
5280ft
1mi
⋅
1hr
3600s
)
2
= 16.42 ft
⋅π




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PAG E 14 6
Lift Forces
A discussion of drag forces naturally leads us to a corresponding discussion of lift forces that
act on objects. We note that for most objects, drag is the dominant force acting on its
surface. However, some geometries (like airfoils) have been developed to achieve lift. We
previously de ned drag as a force that is resistive and parallel to the direction of
uid
motion across the surface of the object. Lift, on the other hand, is broadly de ned as the
force exerted by the uid perpendicular to the direction of uid motion.
Like the drag force, we can likewise de ne lift force as,
FL = CL ⋅
1
⋅ ρ ⋅ V̄ 2 ⋅ Ap
(2 )
(1.51)
We’ll focus our discussion of lift on airfoils, which are widely used in the aerospace industry
for ight. In particular, we wish to understand the mechanisms that govern the design of an
airfoil such that it can achieve some desired magnitude of lift force. Before we do that, it is
worth indicating the geometric components of an airfoil via the schematic below.
Figure 1.61. Individual geometric features of a standard airfoil, including the chord, span, angle of attack (α), and the directionality of the forces
(lift, FL, and drag, FD) acting on the surface of the airfoil.
In Eqn. 1.51, the term Ap represents a "planform area". The planform area is calculated by
projecting the area where the uid is acting on the wing as if you were looking down at the
wing (i.e. a "top-down" view of the wing’s area).
We reconstruct several basic wing shapes from this perspective in the gure below. The area
is e ectively calculated as the two-dimensional space that the surface occupies.
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PAG E 147
Figure 1.62. Planform geometries (left = rectangular, middle = trapezoidal, right = triangular) with corresponding chord and span values. For chord
lengths, we indicate the root of the wing (closest to the body of the aircraft) with the subscript r, while we indicate the tip of the wing with the
subscript t. Reconstructed from portions of https://www.grc.nasa.gov/WWW/K-12/airplane/area.html.
Much of the robust airfoil data available within the wider literature comes directly from
experiments conducted in a wind tunnel on airfoils with varying section thickness, camber
location (the distance from the leading edge of the airfoil to the maximum height along the
upper surface), and the lift coe cient that the airfoil is designed for. These tests were
(remarkably) conducted in the 1930’s and 1940’s by the National Advisory Committee for
Aeronautics (the precursor agency for NASA). We now refer to the airfoils as having NACA
shapes, which are designated as follows,
In the above expression, the numerical values in the NACA designation (i.e. 23015) are
separated into segments that represent the lift coe cient, the location of maximum camber
and the section thickness, respectively. In the case of the lift coe cient, the leading value is
divided by 10 and multiplied by 1/2 to
nd CL. The next two numbers in the sequence
represent the location of maximum camber in % of the chord length when multiplied by
1/2. Finally, the last two digits represent the section thickness as a % of chord length.
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PAG E 14 8
The airfoil characteristics mentioned on the previous page are shown in the schematic below
such that the reader can visualize the meaning of the geometric parameters de ned by the
NACA numerical designator.
Figure 1.63. Schematic of an airfoil with a negative camber and the geometric implications of the camber within the context of the NACA airfoil
designation.
The lift and drag coe cients for di erent NACA airfoils are provided in the plots below.
Figure 1.64. Plots of lift coe cients (top plots) and drag coe cients (bottom plots) for NACA 23015, 2412 and 0012 airfoils verses angle of attack.
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PAG E 14 9
angles of attack. For the NACA 23015 and 2412 airfoils4, we see that there is some lift at a
0∘ angle of attack (α).
Though we have discussed how to calculate the lift force, we have not actually provided a
description of how lift is achieved. In principle, aerodynamic lift is due to the fact that there
exists a pressure decrease (and corresponding velocity increase) across the top of the wing,
while there is a pressure increase (and a corresponding velocity decrease) across the bottom
of the wing. This net di erent in pressure results in a lift force being generated in the
positive vertical direction. We notice that as the angle of attach increases, so too does CL,
which is representative of an increase in the pressure di erence (Δp) between the top and
bottom surfaces.
However, at some point the lift coe cient begins to decrease with increasing angle of attack.
This phenomenon is known as a "stall" and results in a loss of lift force. At some point, the
angle of attack becomes so large that the boundary layer can no longer "stick" to the upper
surface (i.e. it is no longer streamlined). This is colloquially referred to as ow separation.
Finally, the lift coe cient for a NACA 23015 airfoil is plotted against α and CD for a
comparison of the lift and drag coe cients achieved with the aps that are often attached at
the trailing edge of a wing, which can produce increased lift (but also result in increased
drag).
Figure 1.65. Lift coe cients for a NACA 23015 airfoil having di erent types of aps at the trail-end of the wing.
4 Note that for a 4-digit NACA code, the rst digit represents the maximum camber value as a % of chord length, the second digit represents the
location of the maximum chord length in % of chord length when multiplied by 10 (i.e. the "4" in 2412 represents 40% of the chord length), and the
nal two digits represent the maximum thickness of the airfoil as a % of the chord length.



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PAG E 15 0

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The lift coe cients in the previous plot can be used to calculate the lift force at various
Example 1.25 - Lift Force acting on a commercial aircraft
A commercial aircraft has a total mass of 175,000 lbm and uses wings with a planform area
of 1,800 ft2. The plane has a cruising speed of 550 mph and a cruising altitude of 38,000 ft,
where the air density is ρ = 0.0205 lbm/ft3. The plane has double-slotted
aps for use
during takeo and landing, but it cruises with all aps retracted. Assuming the lift and drag
characteristics of the wings can be approximated by a NACA 23012 airfoil, determine:
1. The minimum speed for takeo , assuming a ground-level air density of ρ = 0.075
lbm/ft3.
2. The angle of attack to cruise steadily at cruising altitude.
3. The power needed to overcome drag at cruising altitude [in hp].
Solution
To solve the rst part of this problem, we must understand that to take o ,
∑
Fy > 0
In this case, the forces acting in the y-direction are lift and weight (due to the mass of the
aircraft that must itself be lifted). There is no buoyancy force acting on the wing. A freebody diagram reveals that,
Therefore, we have,
FL − W > 0
→
FL > W
→
FL > m ⋅ g
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PA G E 151
CL ⋅
1
⋅ ρ ⋅ V̄ 2 ⋅ Ap > m ⋅ g
(2 )
We solve for the minimum speed necessary for take o as,
2⋅m⋅g
=
CL ⋅ ρ ⋅ Ap
V̄ =
ft
2 ⋅ 175,000lbm ⋅ 32.2 2
s
ft 3
3.48 ⋅ 0.07518 lbm ⋅ 1800ft 2
= 154.7
ft
3600s
1mi
⋅
⋅
= 105.5 mph
s
1hr
5,280ft
where we obtained the lift coe cient as CL = 3.48 by virtue of knowing that the minimum
velocity required to take o corresponds to the maximum coe cient of lift. This value was then taken
from the top distribution in the left-most plot in Fig. 1.65.
To nd the angle of attack required to cruise steadily when traveling 550 mph, we must nd
the lift coe cient produced when traveling at this velocity for this particular wing. In order
to cruise, we know that the forces must balance (so that there is not relative motion in the ydirection), or,
FL = W
Therefore,
CL =
ft
2 ⋅ 175,000lbm ⋅ 32.2 2
2⋅m⋅g
=
ρ ⋅ V̄ 2 ⋅ Ap
0.0205 lbm3 ⋅ 550 mi
⋅
hr
ft
(
s
5280ft
1mi
⋅
1hr
3600s
)
2
= 0.47
⋅ 1800ft 2
With a lift coe cient of 0.47 and a fully retracted ap, we can use the bottom distribution in
the left-most plot in Fig. 1.65 to nd that,
α ≈ 3.5∘
Finally, we nd the power required to overcome drag at cruising altitude via,
·
W = Fd ⋅ V̄
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PAG E 15 2
fi

where we know that the lift force can be rewritten via Eqn. 1.51 such that,
FD = CD ⋅
1
⋅ ρ ⋅ V̄ 2 ⋅ Afrontal
(2 )
Thus,
·
W = CD ⋅
1
⋅ ρ ⋅ V̄3 ⋅ Afrontal
(2 )
Because we have CL for this airfoil at cruising speed (CL = 0.47 from the previous part of
the problem), we can nd CD using the right-most plot in Fig. 1.65, which plots CL vs. CD
for this airfoil. At CL = 0.47, we nd that CD ≈ 0.01. Therefore,
·
W = 0.01 ⋅
1
lbm
mi
5,280ft
1hr
1hp
1lbf
⋅ 0.0205 3 ⋅ 550
⋅
⋅
⋅ 1800 ft 2 ⋅
⋅
lbm ⋅ ft
(2 )
(
ft
hr
1mi
3600s )
550lbf ⋅ ft
32.2 2
3
= 5,468 hp
s





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PAG E 15 3


where,
CHAPTER 2: INTRODUCTION TO
HEAT TRANSFER
I HEATED A THIC K PIECE OF IRON TILL IT WAS
RED
HOT, AND TAKING
W I T H
A
H O T
I M M E D I AT E LY
WHERE
THE
IT
P A I R
PUT
WIND
OUT
O F
IT
IN
BLEW
OF
THE
FIRE
P I N C E R S ,
A
COLD
I
PLACE,
C O N S T A N T LY. . .
THE
A I R H E AT E D BY T H E I R O N M I G H T A LWAYS B E
C A R R I E D AWAY BY T H E W I N D , A N D T H E C O O L
AIR MIGHT SUCCEED IN ITS PL ACE WITH [A]
UNIFORM
MOTION.
FOR
THUS
EQUAL
PARTS
OF THE AIR WOULD BE MADE HOT IN EQUAL
TIMES,
AND
WOULD
PROPORTIONAL
-SIR
ISAAC
TO
NEWTON,
TRANSACTIONS,
"A
CONCEIVE
THE
THE
SCALE
HEAT
A
HEAT
OF
IRON.
PHILOSOPHICAL
OF
THE
DEGREES
O F H E A T " , 17 01.
eat transfer is a physical and relatable science that has important consequences for
our daily lives. In this course, we concern ourselves with fundamental empirical
phenomena that allow us to analyze and design devices and energy systems,
including heat sinks and heat exchangers. We couple these to our knowledge of
uid
mechanics and thermodynamics.
Heating and cooling are phenomena that we regularly encounter in our daily lives. Heating,
ventilating and air conditioning systems have been ubiquitous in the United States since
their introduction into homes in the late 1940's. More tangible, perhaps, is the experience of
a hot cup of co ee slowly cooling throughout the day, or conversely, a cold bottle of water
slowly heating up while at the beach. In both cases, recent advancements in vacuum
insulation technology have led to the design of beverage containers that can keep drinks
warmer (or cooler) for longer durations of time (in some cases, for 8+ hours!).
In this course, we will immerse ourselves in the development of simple tools that allow us
to design or analyze speci c heat transfer equipment, including: (1) heat sinks and (2) heat
exchangers. As with all other applications, this will require a rudimentary understanding of
the basic modes of heat transfer, which include Conduction, Convection and Radiation.
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PAG E 15 4
In the remainder of this chapter, we discuss
the fundamental modes that govern heat ow
within stationary objects and moving
uids,
the use of thermal resistor networks for
Vacuum
Insulation
simpli ed analysis and design of thermal
transport in engineering applications, the
design and analysis of heat sinks for active and
passive cooling of devices, the fundamentals of
Inner
Container
convection heat transfer and the use of an
e ectiveness-NTU methodology to design and
analyze industrial-scale heat exchangers.
We note that this text is not meant to provide
an exhaustive overview of heat transfer and is
not suitable for a standalone course in heat
transfer o ered in many Mechanical
Engineering departments across the country.
For such courses, other texts are likely to be
more appropriate. This textbook is suitable for
Figure 2.1. Cut-plane of a typical vacuum insulated beverage
container. The design of this container is based on an understanding
of the fundamental modes of heat transfer, including conduction,
convection and radiation (described in the following section). In
particular, the vacuum insulation eliminates thermal transport by
conduction and convection, reducing the amount of heat that can
enter (or escape) from (or to) the environment. Photo taken at
USNA.
a practical, systems-based approach to heat transfer and is taught to engineers that are not in
the Department of Mechanical Engineering at the United States Naval Academy.
FUNDAMENTAL MODES
In this text, we provide a fundamental understanding of the predominant modes of heat
transfer, which includes: conduction, convection, and radiation. It is worth mentioning
that more that one mode is often present in any application where heat is being transferred.
That said, it is often the case that one mode governs thermal transport in a device or system
and the others can be neglected. It will be important, then, to be able to calculate the
magnitude contribution of each mode for a speci c problem. To do this, we can analyze the
rate equation for each mode.
Before discussing each individual mode of heat transfer, it is instructive to consider how and
when heat moves into or out of a system. Heat transfer is explicitly related to an object’s
relative temperature. Consider the case discussed brie y in the introduction of this section: if
you put a hot cup of co ee in a space with a colder environment, the co ee itself will
eventually come into thermal equilibrium with the surroundings. While the co ee is
cooling, heat is being transferred from the co ee into the surroundings. Thus, we can say that heat
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PAG E 15 5
The physical reason for this is best explained by the second law of thermodynamics and is
therefore beyond the scope of what we cover in this section.
We can account for heat transfer to or from a system by setting up an energy balance for a
system. Figure 6.2 represents a simple schematic for a system that is exposed to some
thermal reservoir at a di erent temperature.
qout
System
dU
dt
qin
Figure 2.2. Schematic of a system of interest and the corresponding changes to that system (dU/dt) that result due to heat transfer across the
boundary (dashed line).
In the schematic above, dU/dt is the rate of change of the internal energy in a system. From
a thermodynamics perspective, the rate of change of internal energy is best thought of as
either a change in temperature within a substance or a change in a substance’s phase (e.g. a
liquid-vapor phase change). If the substance is incompressible (most solids and liquids can
be approximated this way), and does not undergo a phase transition, the resulting energy
balance yields,
qnet = m ⋅ Cp ⋅
dT
dt
(2.1)
Problematically, this equation can not be used to directly calculate q independent of context.
This is because we rarely know how the temperature of an object or a
uid changes with
time without measuring it directly. Consequently, we turn our attention to a series of rate
equations that can be used to predict a heat transfer rate based on the properties and
conditions of a given solid, uid or gas.
Conduction Heat Transfer
Conduction heat transfer occurs when there exists a thermal gradient across a solid or
stationary object or
uid. The rate equation for conduction heat transfer is in some sense
empirical. In other words, this equation was developed through a parametric experimental
analysis.
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PAG E 15 6




is transferred from a hot object (or environment) to a cold object (or environment).
First, consider an object that is exposed to a
xed (hot) temperature on one end a
xed
(cold) temperature on the opposing end, with all other sides insulated.
Hot
Reservoir
Cold
Reservoir
Figure 2.3. Schematic of a material between hot and cold reservoirs at temperatures TH and TC , respectively. Gray shaded regions represent areas
where the material is thermally insulated from the surrounding environment.
When heat is restricted to owing in one dimension (in the above case, from left to right),
we nd that the rate of heat ow, q, is directly proportional to the temperature gradient in
that direction and the area that the heat passes through, expressed as,
q ∝ A⋅
By changing the material type, we
dT
dx
(2.2)
nd that this proportionality can be written as a direct
relationship in the form,
q =κ⋅A⋅
dT
dx
(2.3)
where κ is the thermal conductivity of the material. Thermal conductivity is therefore
considered to be a material property that governs the rate of heat that can pass through an
incompressible substance due to an imposed temperature di erence. Since the formulation
of Eqn. 2.3, we have since discovered that a material’s thermal conductivity is governed by
the fundamental characteristics of its microstructure and electrical conductivity (which is
well beyond the scope of this text).
We note that the heat transfer rate itself can alternatively be expressed as a rate per unit
area or a rate per unit length, as shown below (where Lc is some speci ed distance):
q
Lc
q
q′′ =
A
q′ =
A variety of material thermal conductivities are found in Figs. 2.4 and 2.5. Figure 2.4 shows
the thermal conductivity of various solid materials as a function of temperature, while Fig.
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PAG E 157
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2.5 shows the thermal conductivity of materials in di erent phases
Figure 2.4. Thermal conductivity of various solids as a function of temperature in (K). Note that metals tend to have a higher thermal conductivity
than non-metals due to the availability of electrons as heat energy carriers.
Figure 2.5. Thermal conductivity of di erent materials in di erent phases of matter (gases, liquids, and solids). Note that gases typically have lower
thermal conductivity than liquids, which typically have lower thermal conductivity than solids.
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PAG E 15 8
A copper rod has a diameter of d = 1 cm and a length of L = 0.6 m and is exposed to a
temperature of 600∘C on one of its circular faces and a temperature of 20∘C on the other.
Assuming the outer surface of the rod is perfectly insulated and that the thermal
ux [W/m 2] and heat
conductivity of copper is κCu = 393W/m ⋅ K , determine the heat
transfer rate [W] across and through the rod, respectively.
TH = 600∘C
TC = 20∘C
Solution
Note that the heat
ux is the heat transfer rate divided by the cross-sectional area (or the
area perpendicular to the direction of heat ow). To nd the heat ux, then, we can use the
rate equation that describes conduction heat transfer.
q
dT
ΔT
=κ⋅
= κCu ⋅
A
dx
L
q′′ =
In the equation above, q'' is the heat ux across the length of the copper rod. Note that we
integrate from TC to TH and from x = 0 to x = L in order to arrive at the expression on the
right. Now we simply substitute the values from the original problem statement:
W
600∘C − 20∘C
W
q′′ = 393
⋅
= 379, 900 2
m⋅K
0.6m
m
You might notice that the units of temperature cancelled directly, despite the presence of ∘C
in the numerator and K in the denominator. Because the numerator represents a temperature
di erence, and we know that ΔT(∘C ) = ΔT(K ) (for example: 600∘C − 20∘C = 580∘C and
873.15K − 293.15K = 580K), the units cancel.
To nd the total heat transfer rate, we simply multiply the heat ux by the cross-sectional
area that is perpendicular to the direction of heat ow.
πd 2
W 3.1415 ⋅ (0.01m)2
q = q′′ ⋅ Ac = q′′ ⋅
= 379,900 2
= 29 . 83W
4
m
4


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PAG E 15 9











fi

This is the rate of heat that is transferred from one end of the rod to the other.
ff



Example 2.1. - Heat transfer through a copper rod
A composite window is shown in the gure below and contains a 5.08 cm thick stainless
steel strut that sits between two pieces of glass that are each 10 mm thick. The temperature
on the outer part of the window is 33∘C and the temperature on the inner part of the
window is 15∘C. Determine the temperature distribution and the heat
ux through each
part of the composite window.
T(∘C )
33∘C
ΔTglass
ΔTsteel
ΔTglass
15∘C
x
Lglass
Lsteel
Lglass
To determine the temperature distribution through each component of the composite
window, we can use knowledge that:
ΔTtotal = ΔTglass + ΔTsteel + ΔTglass = ΔTsteel + 2 ⋅ ΔTglass
where the temperature di erence across the glass on the left is equivalent to that on the
right because they are made of the same material and have the same thickness. In other
words, the conservation of energy (i.e. the energy leaving one material = the energy entering
the next) requires that,
q′′ = κglass,left ⋅
ΔTglass,left
Lwindow,left
= κglass,right ⋅
ΔTglass,right
Lwindow,right
PAG E 16 0
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As a result, we can say that ΔTglass,left = ΔTglass,right = ΔTglass.







Example 2.2. - Conduction heat transfer through a composite wall
q′′ = (κ ⋅
ΔT
ΔT
=
κ
⋅
)
(
)
L glass
L steel
Thus, to determine the temperature drop across each material layer, we can say that,
ΔTtotal = ΔTsteel + 2 ⋅ ΔTglass = ΔTsteel ⋅ (1 + 2 ⋅
κsteel /Lsteel
)
κglass /Lsteel
Note that in the above, we multiplied the 2 ⋅ ΔTglass term by ΔTsteel /ΔTsteel (which is a
mathematical trick where we are e ectively multiplying a term by 1). This helps us to use
the conser vation of energy expression at the top of the page to become
2⋅
ΔTglass
ΔTsteel
⋅ ΔTsteel = 2 ⋅
κsteel /Lsteel
⋅ ΔTsteel.
κglass /Lglasss
Solving for ΔTsteel, we obtain,
κsteel /Lsteel −1
16.3W/m ⋅ K /0.0508m −1
∘
∘
∘
) = (33 C − 15 C ) ⋅ (1 + 2 ⋅
) ≈ 3 . 22 C
κglass /Lglass
1.4W/m ⋅ K /0.01m
ΔTsteel = ΔTtotal ⋅ (1 + 2 ⋅
The remaining temperature di erence is then split evenly across the two glass layers, or
ΔTglass ≈ 7 . 39∘C. The heat
ux can then be obtained with the use of any individual layer’s
material characteristics and its individual temperature gradient,
q′′ = κglass ⋅
ΔTglass
Lglass
W
7.39∘C
= 1.4
⋅
= 1034 . 6W/m2
m ⋅ K 0.01m
Recall that the temperature di erence in the numerator in ∘C is equivalent to a temperature



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PA G E 161





di erence in K (see previous problem).
ff



In the same way, we can also use the conservation of energy to say that,
The term "convection" is used to describe the
ow of heat from a solid surface into an
adjacent, moving uid. Like conduction heat transfer, convection is present in many of the
applications that are familiar to us. Perhaps the most tangible example of convection is the
movement of cold air over the human body. In fact, this formed the basis for our early
understanding of convection when, in 1701, Sir Isaac Newton considered how long it takes
for a body to cool down in an environment whose temperature is T∞,
dTbody
∝ Tbody − T∞
dt
(2.4)
However, it became quickly apparent that the energy needed to maintain a body temperature
was constantly being replenished, and that this speci c case represented a steady-state
problem (i.e. your body temperature does not cool to the temperature of the outside air over
time, else in many cases you would freeze!). Rather, the steady-state formulation becomes,
q′′ ∝ Tbody − T∞
(2.5)
As with conduction heat transfer, an empirical relationship is developed to provide a direct
correlation between an applied heat transfer rate and the temperature di erence between a
surface and its surroundings, Ts − T∞, as,
q′′ = h̄ ⋅ (T̄s − T∞)
(2.6)
where h̄ is the average heat transfer coe cient that acts on the body and describes the
magnitude with which heat can be carried away from the surface by a moving uid.
Consider the relatively simple case where a owing uid moves along a at plate,
Velocity pro le within
the boundary layer
U∞, T∞
T̄s
Free-stream
Velocity and
Temperature
q’’
x

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PAG E 16 2



Figure 2.6. Laminar ow over a at plate with a constant, applied heat ux at the surface of the at plate. The terms Ts and T∞ represent the
surface temperature and the free-stream temperature of the uid, respectively.



Convection Heat Transfer
there is also a thermal boundary layer that will be covered in a separate section on external
ow). To determine the heat ux acting along the entire length of the plate, an average heat
transfer coe cient is required. This is because the velocity distribution along the length (i.e.
in the x-direction) of the plate changes. As a result, the heat ux, surface temperature and
heat transfer coe cient also change in the x-direction when examined locally. The local heat
ux (i.e. the heat ux acting at any point in the x-direction) can be determined as,
q′′(x) = h(x) ⋅ (Ts(x) − T∞)
(2.7)
The above two equations are referenced as forms of the convection rate equation and represent
what is now called Newton’s Law of Cooling (though it should be noted that Newton never
actually wrote these equations in this form or made mention of a heat transfer coe cient).
Classi cations
A number of di erent thermal phenomena can technically be classi ed under what has
become the umbrella term "convection". These include:
1. Free Convection - The term "free convection" is used to describe a uid that is being
advected over a surface due to a di erence in its buoyancy near the surface relative to
the surrounding quiescent
uid. A tangible example of this occurs in homes with
more than one oor, where often the hottest space in the house is the highest oor
due to the lower density associated with hot air.
2. Forced Convection - This type of convection occurs when a uid is physically moved
over a surface in order to cool it or heat it. Typically this is done with the use of a fan,
blower or pump. When you sit in front of a fan and "feel cool", this is forced
convection in action.
3. Boiling - When a uid boils, the energy that is otherwise used to sensibly heat it is
instead used to form a vapor, keeping the temperature at the surface relatively stable.
As vapor bubbles depart (either from natural uid movement or forced advection), the
hot vapor bubble is replenished with a cooler liquid from the surrounding volume.
4. Condensation - Condensation occurs when a surface is cooler than its surroundings.
Consequently, a uid forms on the surface due to the presence of that uid as a vapor
in the surrounding environment. The energy used to do this again maintains a
relatively stable temperature, and heat dissipation is aided by the uid moving along
PAG E 16 3
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the surface.
fi
fl
fl
Figure 2.6 shows the velocity boundary layer along that develops along a at plate (note that
The heat transfer coe cient itself represents the magnitude "strength" of convection heat
transfer and depends on a variety of parameters, including:
1. Fluid properties like:
• Viscosity and density
• Thermal conductivity
2. Fluid velocity
3. The geometry of the surface
Some examples of representative heat transfer coe cients for speci c applications are
included in the table below.
Table 2.1. Range of convection heat transfer coe cients based on classi cation.
Type
Typical range of h
W
( m2 ⋅ K )
Natural Convection
4 - 4,000
Forced Convection
40 - 100,000
Boiling & Condensation
300 - 10,000,000
We note that generally, each of these types can occur either internally or externally, meaning
the
uid is either bound on all sides or is left free on one or more sides. An extensive
discussion of both internal
ow and external
ow is provided in a later section titled
"Convection Heat Transfer".
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PAG E 16 4
Example 2.3. - Cartridge heater operating at steady-state
A cylindrical cartridge heater emits a heat ux of 5,000 W/m2 at its outer surface and its
measured temperature is 80∘C when placed in water with a temperature of 20∘C. What is
the convection coe cient acting on the outer part of the heater? If the heater power is
suddenly increased to 8,000 W/m2, and assuming the convection coe cient acting on the
heater does not change much, what is the approximate temperature increase at the heater
surface?
Solution
When a hot object is immersed in a cold uid, natural convection occurs due to the presence
of buoyancy forces that are generated within the uid. This is analogous to hot air rising in
your home. The corresponding movement of the
uid then acts to remove heat from the
cartridge heater. To solve for the convection heat transfer coe cient we use Newton’s Law
of Cooling,
q′′
q′′ = h̄ ⋅ (Ts − T∞) → h̄ =
(Ts − T∞)
Substituting, we have,
h̄ =
5,000 W2
m
(80∘C − 20∘C)
≈ 83 . 3
W
m2 ⋅ K
Assuming h̄ does not change with temperature (which may not be a good assumption in
reality), we obtain the following surface temperature when q′′ increases to 8,000 W/m2,
q′′
h̄
+ T∞ =
m
83.3 2W
m ⋅K
+ 20∘C ≈ 116∘C
PAG E 16 5


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Ts =
8,000 W2
Radiation Heat Transfer
Radiant thermal energy is perhaps the most tangible of the three fundamental modes of heat
transfer. Most (if not all) of us have had the experience of stepping outside on a sunny day
and feeling the "warmth" of the sun. This type of heating is achieved principally through the
emission of electromagnetic radiation. Thus, radiant heat is emitted from any surface that
has a non-zero absolute temperature.
To understand this, one must examine the nature of the electromagnetic spectrum, as
shown below.
Figure 2.8. Electromagnetic spectrum showing the relative wavelength, size, frequency and maximum temperature of the seven most commonly
utilized parts, including radio, microwave, infrared, visible, ultraviolet, x-ray, and gamma ray. Attribution to: Inductiveload, NASA / CC BY-SA (http://
creativecommons.org/licenses/by-sa/3.0/).
Electromagnetic waves are emitted from a surface due to atomic excitation and the
subsequent elevation of electrons to a higher energy level. When this happens, a photon of
light is emitted at the surface of the material at a frequency that corresponds to the energy
di erence between the electrons in their ground state and the electrons in their excited
state.
Practically, the way that this manifests at larger scales is through temperature. As the
temperature of an object (or surface) is increased, there is a corresponding increase in
"atomic excitation", or the rate at which atoms vibrate around their equilibrium positions.

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PAG E 16 6
whose temperature Ts is greater than 0 K. With respect to the electromagnetic spectrum
referenced in Fig. 2.8, thermal radiation occurs at wavelengths ranging from 100 nm to 1000
μm.
Although radiation transport mechanisms are rooted in quantum mechanics, we account for
thermal radiation e ects using a rate equation similar to those developed for conduction and
convection (although this rate equation is derived from physics, where the convection rate
equation is largely empirical while the conduction rate equation has an alternate set of
physics not explained in this text that allow us to derive the rate equation).
In this book we restrict our study of radiation heat transfer to that which occurs between a
surface at temperature Ts and surroundings with an ambient temperature Tsurr. We can also
impart a solar ux from the sun via an energy balance. It is critical to know that each of these
temperatures must be in absolute units when we deal with thermal radiation. A schematic
of the radiative transport mechanisms dealt with in this course is outlined below.
Figure 2.9. Schematic representation of the radiative emissive terms that interact with the surface and balance to produce a nite, steady-state
temperature. Terms in red highlight heat energy interacting with the surface. The term “G” is known as irradiance (i.e. a heat ux from a separate
source that interacts with the surface), and the term E is known as the radiative emissive power (i.e. the heat ux leaving the surface).
The relationship between the heat transfer rate from a surface and its temperature scales to
the fourth power as,
(2.8)
PAG E 167
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Es = q′s′ = σ ⋅ ε ⋅ Ts4


As a result, there is always some amount of thermal radiation being emitted from a surface
W
) and ε is a measure of
m2 ⋅ K4
the surface emissivity. The surface emissivity itself is a property of the surface conditions
(e.g. color, surface roughness, etc.).
The heat
ux that is emitted by the surroundings (Esurr) is absorbed on the surface as
αsurr ⋅ Gsurr (where αsurr is the absorptivity of the surface) such that,
4
Esurr = σ ⋅ ε ⋅ Tsurr
and, at the surface, this energy is absorbed at a rate represented by,
4
αsurr ⋅ Gsurr = αsurr ⋅ σ ⋅ εsurr ⋅ Tsurr
Because the surroundings are typically much larger than the surface, we consider ε = 1 (i.e. a
blackbody). Likewise, since the surface and surroundings are likely to be at around the
same temperature, we consider the surface to be gray (i.e. its absorptivity to the energy
being emitted by the surroundings is equivalent to its emissivity across all wavelengths, or
αsurr,λ = εs,λ, where λ is the spectrum of light where thermal radiation occurs).
Thus, if we want to compute the net heat ux leaving the surface, we obtain,
4
4
q′net
′ = Es − αsurr ⋅ Gsurr = εs ⋅ σ ⋅ Ts4 − εs ⋅ σ ⋅ Tsurr
= εs ⋅ σ ⋅ (Ts4 − Tsurr
)
(2.9)
If the sun is also present (and its irradiance, G, is provided), and assuming the surface
remains gray (i.e. αsun,λ = εs,λ), we obtain,
4
4
q′net
′ = Es − αsurr ⋅ Gsurr − αsun ⋅ Gsun = εs ⋅ σ ⋅ Ts4 − εs ⋅ σ ⋅ Tsurr
− εs ⋅ Gsun = εs ⋅ σ ⋅ (Ts4 − Tsurr
) − εs ⋅ Gsun
It is not always the case that a surface is gray, or that we have a single emissivity across all
wavelengths where thermal radiation matters. However, this text only provides a brief
introduction to radiation heat transfer. For additional insight into radiation heat transfer




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PAG E 16 8



problems, it is best to consult standard heat transfer textbooks.




where σ represents the Stefan-Boltmann constant (σ = 5.67 ⋅ 10−8
T H E R M A L R E S I S T O R A N A LY S I S
In complex systems, there is often more than one mode of heat transfer present and/or
more than one path that the heat can take to escape or enter into a surface. As a result, it
becomes a matter of convenience to work with thermal resistor networks.
You are likely familiar with electrical resistor networks, where a change in voltage drives a
current through a series of resistive elements. For example, consider the case where an
electrical circuit passes current through a series resistor network where the voltage drops
from V1 = 400 V to V3 = 300 V, as shown in the diagram below. You know that the rst
resistor, R1 = 5 Ω and the second resistor R2 = 10 Ω.
Now say we want to determine the current, I, and the voltage V2. To do this, we use
knowledge of the relationship between voltage, current and resistance as,
I=
ΔV
∑R
or,
I=
V1 − V3
400V − 300V
=
= 6.67 A
R1 + R2
5Ω + 10Ω
Likewise, we can nd the voltage V2 via,
I = 6.67 A =
V1 − V2
400V − V2
=
R1
5Ω
∴ V2 = 400V − 6.67 A ⋅ 5Ω = 350V
We take an analogous approach to solving heat transfer problems (where heat ow can be
approximated as being "one-dimensional").
Rather than a voltage drop, the ow of heat is governed by the temperature drop across an
object (and, as you may have guessed, the energy that Δ T regulates is heat rather than
current).
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PAG E 16 9
To nd the heat ow rate, q, we can therefore use,
ΔT
∑R
q=
(2.10)
Here, though, each resistance we encounter depends on the mode of heat transfer that
occurs between two endpoints.
For Conduction Heat Transfer, we use,
Rcd =
L
κ ⋅ Ac
(Cartesian, 2.11)
for a cartesian coordinate system (where Ac is the area perpendicular to heat
ow). For a
cylindrical coordinate system, we use,
Rcd =
ln
( di )
do
(Cylindrical, 2.12)
2 ⋅ π ⋅ kw ⋅ L
For Convection Heat Transfer, we use,
Rcv =
1
h ⋅ As
(2.13)
which is valid for both cartesian and cylindrical coordinates (though you should be careful to
understand that As is the surface area for heat transfer by convection, in this case).
We often encounter the cylindrical coordinate system in heat transfer for uids that move
through pipes, or when wires generate heat, and an example problem is provided within this
text to this end.
We also note that when heat can move in more than one direction, we no longer have only a
series resistor. In this case, we must recognize that a parallel resistor has to be used to model
thermal transport.
Finally, we note that this text does not cover the spherical coordinate system to any extent.
For spherical coordinates, you should consult a standard heat transfer textbook.
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PA G E 17 0
Example 2.4 - Plane wall, series resistor
A plane wall is held at a constant temperature of Ts,i = 400 K on the left side. Heat moves
through the wall, which has a thermal conductivity of κ = 0.2 W/m⋅K, and is convected on
the outer part of the wall to the ambient environment, which has a temperature of T∞ =
300 K. The convection coe cient acting on the outer wall is h = 10 W/m2 ⋅ K. Determine
the rate of heat transfer through the wall and the temperature on the outer surface of the
wall, Ts,o. The thickness of the wall L = 0.01 m and the cross-sectional area is Ac = 0.01 m2.
Solution
In this problem, heat ows through the wall by conduction from left to right, and then out
into the ambient environment by convection. Therefore, we set up a series resistor network
as follows:
The heat ow rate can then be computed using,
ΔT
∑ Rth
q=








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PA G E 17 1
ow rate
using,
q=
Ts,i − Ts,o
Rcd,wall + Rcv,o
where,
Rcd,wall =
L
0.01m
K
=
=
5
κ ⋅ Ac
W
0.2 W ⋅ 0.01m 2
m⋅K
and,
1
Rcv,o =
=
h ⋅ As
10
W
1
m2 ⋅ K
K
= 10
2
W
⋅ 0.01m
Now,
q=
400K − 300K
K
K
5 W + 10 W
= 6.67 W
To solve for the outer wall temperature, we also recognize that,
Ts,i − T∞
Ts,i − Ts,o
Ts,o − T∞
ΔT
q=
=
=
=
∑ Rth
Rcd,wall + Rcv,o
Rcd,wall
Rcv,o
We can use either of the above two quotients on the right to solve for Ts,o. Here we use the
rst as,
400K − Ts,o
q = 6.67 W =
K
5W
→
Ts,o = 400K − 6.67W ⋅ 5
K
= 350 K
W
Note that this problem is directly analogous to the electrical resistor network described at
the beginning of this section.
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PA G E 17 2

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Using the two temperatures we have in this problem, we can calculate the heat
Example 2.5 - Plane wall, parallel resistor
A plane wall is heated on one side at a rate of 30 W over a 0.01 m2 area. The wall
experiences convection on both sides with air at T∞ = 300 K. The convection heat transfer
acting on both sides is h = 20 W/m2 ⋅ K. The wall has a thickness of 0.01 m and a thermal
conductivity of κ = 0.2 W/m⋅K. Calculate the temperature on both sides of the wall.
Solution
In this case, the heat can move either through the wall by conduction and to the outside by
convection or by convection to the inside space. In other words, the heat has an opportunity to
move in two directions. As a result, we use a parallel resistor network to model heat ow.
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PA G E 17 3
In a parallel circuit, the heat is split proportionally through the upper and lower branches (i.e.
its values are proportional to the resistance in each branch). For this type of circuit, we can
write,
Ts,i − T∞
q = q1 + q2 =
Rcd,wall + Rcv,o
+
Ts,i − T∞
Rcv,i
To solve for the inner surface temperature, then, we need each of the thermal resistances,
Rcd,wall =
L
0.01m
K
=
=
5
κ ⋅ Ac
W
0.2 W ⋅ 0.01m 2
m⋅K
and,
1
Rcv,o =
=
ho ⋅ As,o
20
1
W
m2 ⋅ K
K
=5
2
W
⋅ 0.01m
and,
1
Rcv,i =
=
hi ⋅ As,i
20
1
W
m2 ⋅ K
K
=5
2
W
⋅ 0.01m
Therefore,
Ts,i = 30W ⋅
[5 K + 5 K
1
W
W
+
K
5W ]
1
−1
+ 300K = 400K
Now we can use Ts,i to determine Ts,o provided we know q1, where,
Ts,i − T∞
q1 =
Rcd,wall + Rcv,o
=
400K − 300K
K
10 W
= 10 W
Thus,
q1 =
Ts,i − Ts,o
Rcd,wall
→
Ts,o = 400K − 10W ⋅ 5
K
= 350 K
W











PA G E 174
Example 2.6 - Thermal transport in a pipe with moving uids, series resistor
A cast iron pipe (κ = 52 W/m⋅K) moving hot oil with a temperature of T∞,oil = 60∘C is used
to heat a room with air to a temperature of T∞,air = 25∘C. The convection heat transfer
coe cient acting on the inner part of the pipe is hi = 15 W/m2⋅K, while the convection heat
transfer coe cient acting on the outer part of the pipe is ho = 35 W/m2 ⋅ K. Determine the
heat transfer rate, q, through the pipe and the inner and outer surface temperatures (Ts,i and
Ts,o, respectively) of the pipe walls. The inner and outer radii of the pipe are ri = 25.4 mm
and ro = 29 mm and the length of the pipe is 1 m.
Solution
In this problem, we know that heat has only one direction it can move in (the positive radial
direction, from the center of the inner uid to the uid convecting over the outer part of the
pipe). Thus, the problem requires the use of a series resistor.
In general, the heat ow rate can be determined using,
ΔT
q=
∑ Rth









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Rth = Rcv,i + Rcd,wall + Rcv,o
Let’s now compute each individual thermal resistance.
Rcv,i =
1
=
hi ⋅ As,i
1
1m
15 2W ⋅ π ⋅ 25.4mm 1,000mm
⋅ 1m
= 0.84
K
W
m ⋅K
Rcd,wall =
Rcv,o =
( di )
ln
do
0.029m
2 ⋅ π ⋅ kw ⋅ L
1
=
ho ⋅ As,o
ln 0.0254m
(
)
=
2 ⋅ π ⋅ 52 mW⋅ K ⋅ 1m
= 0.00041
1
1m
35 2W ⋅ π ⋅ 29mm 1000mm
⋅ 1m
K
W
= 0.31
K
W
m ⋅K
Now,
q=
60∘C − 25∘C
K
K
K
0.84 W + 0.00041 W ⋅ 0.31 W
= 30.33 W
We can now use q to solve for the inner and outer temperatures of the pipe wall.
T∞,i − Ts,i
q=
Rcv,o
Ts,i = T∞,i − q ⋅ Rcv,o = 60∘C − 30.33W ⋅ 0.83
→
K
= 34.83∘C
W
and,
q=
Ts,i − Ts,o
→
Rcd,wall
Ts,o = Ts,i − q ⋅ Rcd,wall = 34.83∘C − 30.33W ⋅ 0.00041
K
= 34.82∘C
W
Note here that the thermal conductivity and wall thickness of the pipe result in a very low
thermal resistance. This results in a very low temperature drop across the pipe wall
(ΔTpipe = 0.01∘C). Our largest source of thermal resistance is due to the convection acting
on the inner part of the pipe.








PA G E 176


For the series resistor on the previous page, Rth is,
Example 2.7 - Heat ow in a multi-layered cable
A 2 m long coaxial cable generates heat at a rate of q = 20 W. The heat generating cable has
a radius of 12 mm and is surrounded by a 5 mm thick layer of epoxy (κepoxy = 1.2 W/m⋅K),
a 10 mm thick layer of aluminum to provide rigidity (κAl = 205 W/m⋅K) and an outer
rubber (κrubber = 0.25 W/m⋅K)insulating layer that is 8 mm thick. Natural convection on
the outer radius provides a convection coe cient of h = 10 W/m2 ⋅ K at an ambient
temperature of T∞ = 21∘C. Determine the temperature at each surface.
Solution
Here, heat ows from the core of the coaxial cable (red circle) outward toward the ambient
environment. Because heat has only one direction to move in, we can model the system
using a series resistor, as shown below.
Again, we can compute the heat transfer rate, q, via,
q=
ΔT
∑ Rth





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PA G E 17 7
where the resistances can be added in series as,
∑
Rth = Rcd,epoxy + Rcd,Al + Rcd,rubber + Rcv,o
Solving for each individual thermal resistance,
ln do
( i)
d
Rcd,epoxy =
Rcd,Al =
2 ⋅ π ⋅ kw ⋅ L
( di )
=
=
2 ⋅ π ⋅ kw ⋅ L
ln do
( i)
Rcv,o =
K
W
ln 0.027m
(
)
K
W
0.035m
2 ⋅ π ⋅ kw ⋅ L
1
=
h ⋅ As
10
= 0.023
= 0.00018
2 ⋅ π ⋅ 205 mW⋅ K ⋅ 2m
d
Rcd,rubber =
W
2 ⋅ π ⋅ 1.2 m ⋅ K ⋅ 2m
ln 0.027m
( 0.017m )
do
ln
0.017m
ln 0.012m
(
)
=
2 ⋅ π ⋅ 0.25 mW⋅ K ⋅ 2m
1
W
⋅ π ⋅ 0.037m ⋅ 2m
m2 ⋅ K
= 0.103
= 0.43
K
W
K
W
Therefore,
Ts,cable − T∞
q=
∑ Rth
→
Ts,epoxy − T∞
q=
→
Rcd,Al + Rcd,rubber + Rcv,o
q=
q=
Ts,Al − T∞
Rcd,rubber + Rcv,o
Ts,rubber − T∞
Rcd,rubber + Rcv,o
K
Ts,cable = 20W ⋅ 0.56 + 21∘C = 32.2∘C
W
Ts,epoxy = 20W ⋅ 0.533
K
+ 21∘C = 31.7∘C
W
→
K
Ts,Al = 20W ⋅ 0.533 + 21∘C = 31.66∘C
W
→
Ts,rubber = 20W ⋅ 0.43
K
+ 21∘C = 29.6∘C
W
Here, the thermal conductivity of the Al is so high that, despite it being the thickest layer in
the resistor network, it represents the lowest thermal resistance (by far!).










PA G E 17 8
Example 2.8 - Heat transfer in a pipe with moving uids, parallel resistor
Heat is generated in the wall of a thin copper pipe at a rate of 1 kW. The pipe has a length of
3 m and is surrounded on each side by a 10 mm thick coating of bronze (κCuSn = 26 W/m⋅
ows through the inside part of the pipe at a temperature T∞,i = 15∘C and a
K). Water
convection coe cient of hi = 40 W/m2 ⋅ K. Air
ows on the outside part of the pipe at a
temperature of 22∘C and has a convection coe cient of 20 W/m2 ⋅ K. Determine the rate of
heat ow into each uid. The outermost diameter of the pipe is 54 mm.
Solution
In this problem, heat can escape the copper pipe and move either into the
through the pipe or into the
uid moving
uid convecting on the outer part of the pipe. The relevant
thermal resistance network can be drawn as,


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PA G E 17 9
the total thermal resistance as,
TCu − T∞,i
TCu − T∞,o
ΔT
q=
=
+
∑ Rth
Rcd,bronze,i + Rcv,i Rcd,bronze,o + Rcv,o
Thus, we can solve for TCu as,
q ⋅ (Rcd,bronze,o + Rcv,o) =
Rcd,bronze,o + Rcv,o
Rcd,bronze,i + Rcv,i
⋅ (TCu − T∞,i) + (TCu − T∞,o)
and,
q ⋅ (Rcd,bronze,o + Rcv,o) = TCu ⋅ 1 +
− T∞,i ⋅
− T∞,o
(
( Rcd,bronze,i + Rcv,i )
Rcd,bronze,i + Rcv,i )
Rcd,bronze,o + Rcv,o
Rcd,bronze,o + Rcv,o
Finally,
TCu =
q ⋅ (Rcd,bronze,o + Rcv,o) + T∞,i ⋅
( Rcd,bronze,i + Rcv,i )
Rcd,bronze,o + Rcv,o
+ T∞,o
1+ R
(
cd,bronze,i + Rcv,i )
Rcd,bronze,o + Rcv,o
In order to calculate TCu, we need each of the thermal resistances, which can be found via,
ln do
( i)
d
Rcd,bronze,o =
=
2 ⋅ π ⋅ kCuSn ⋅ L
ln do
( i)
d
Rcd,bronze,i =
2 ⋅ π ⋅ kCuSn ⋅ L
Rcv,i =
1
=
hi ⋅ As,i
40
Rcv,o =
1
=
ho ⋅ As,o
20
=
ln 54mm
( 44mm )
2 ⋅ π ⋅ 26 mW⋅ K ⋅ 3m
ln
44mm
( 34mm )
2 ⋅ π ⋅ 26 mW⋅ K ⋅ 3m
1
W
⋅ π ⋅ 0.034m ⋅ 3m
m2 ⋅ K
W
m2 ⋅ K
1
⋅ π ⋅ 0.054m ⋅ 3m
= 0.00042
K
W
= 0.00053
K
W
= 0.078
K
W
= 0.098
K
W
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PAG E 18 0


We know that the total heat ow rate is q = 30 W, and that the heat ow rate is related to
Now we can nd the temperature of the copper pipe, which is,
K
K
0.00042 W
+ 0.098 W
K
K
∘
1,000W ⋅ (0.00042 W + 0.098 W ) + 15 C ⋅
+ 22∘C
K
K
( 0.00053 + 0.078 )
TCu =
(
1+
K
K
)
0.00053 W
+ 0.078 W
W
W
K
K
0.00042 W
+ 0.098 W
= 61.78∘C
Now we can solve for the rate of heat transfer in each direction as,
qo =
TCu − T∞,o
Rcd,bronze,o + Rcv,o
=
61.78∘C − 22∘C
K
K
0.00042 W
+ 0.098 W
= 404.18W
and,
qi =
TCu − T∞,i
Rcd,bronze,o + Rcv,o
=
61.78∘C − 15∘C
K
K
0.00053 W
+ 0.078 W
= 595.82W
As a check for the above two equations, we know that,
q = qi + qo = 404.18 W + 595.82 W = 1,000 W
which is equal to the power dissipated from the heater.



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PA G E 181
It is worth pointing out that when we wish to normalize the heat transfer rate (e.g. we want to
compare the heat transfer rate per unit length or per unit area for pipes of di erent length or
surface area), we must recast thermal resistance in these same terms. For instance, let’s
consider the case of a plane wall whose area we do not know, with convection on each side
of the wall.
Figure 2.10. A plane wall with thickness L subjected to convection on each side of the wall, where Ti and To are the inner and outer wall
temperatures, and T∞,i and T∞,o are the temperatures of the surroundings for the inside and outside spaces, respectively.
If we want to normalize this problem to the area that each individual mode of heat transfer
acts along or across, we begin with,
q′′ =
q
ΔT
=
∑ Rth ⋅ A
A
In the case of the the plane wall shown in Fig. 2.10, we can write q'' as,
ΔT
Rcv,i ⋅ As,i + Rcd,wall ⋅ Ac + Rcv,o ⋅ As,o
q′′ =
Note that for the plane wall discussed here, As,i = As,o = Ac. Therefore, we can rewrite the
above expression as,
q′′ =
ΔT
= 1
1
L
1
⋅
A
+
⋅
A
+
⋅A
( hi ⋅ As,i s,i) ( κw ⋅ Ac c) ( ho ⋅ As,o s,o)
ΔT
L
1
w
o
+κ +h
h
i
In the case above, the areas are nulli ed in the term that represents
∑
Rth. Let’s now look

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PAG E 18 2




at an example where we normalize with respect to some characteristic distance, Lc.
Example 2.9 - Boiler screen tube (energy balance and heat transfer rate per unit length)
High pressure water enters an array of 30 mm diameter screen tubes at a pressure of 4 MPa
(which corresponds to a saturation temperature of ~ 250∘C). The convection heat transfer
coe cient for the forced convection boiling inside of the tubes is 100,000 W/m2 ⋅ K.
Assume that the walls of the tube are very thin and the tube is at a uniform temperature.
Determine: (1) the temperature of the copper tubes, (2) the heat transfer per unit length
due to radiation, convection from the combustion gases, and boiling inside the tubes.
Solution
In a boiler screen tube, a ame produces heat by radiation and warms up the surrounding
air (combustion gases) such that it moves over the cooler screen tubes via natural
convection.
An energy balance around the screen tube can be drawn as,


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PAG E 18 3
·
·
Ein = Eout
→
qconv,o + qrad = qconv,i
To determine the temperature of the copper tubes, we have,
4
hi ⋅ As,i = ε ⋅ σ ⋅ As,o ⋅ (Tsurr
− Ts4) + ho ⋅ As,o
Since As,o = As,i, we have,
W
W
W
−8
4
4
⋅
T
−
523.15
K
)
=
0.8
⋅
5.67
⋅
10
⋅
(1500
K
)
−
T
+
100
⋅ (1000 K − Ts)
( s
(
s)
m2 ⋅ K
m2 ⋅ K4
m2 ⋅ K
100,000
We solve for Ts numerically as Ts = 525.7 K.
Now, we solve for the heat transfer rate per unit length of pipe as,
q′rad =
As,o
qrad
W
4
4
=ε⋅σ⋅
⋅ (Tsurr
− Ts4) = ε ⋅ σ ⋅ π ⋅ D ⋅ (Tsurr
− Ts4) = 0.8 ⋅ 5.67 ⋅ 10−8 2 4 ⋅ π ⋅ 0.03m ⋅ ((1500 K )4 − (525.7 K )4)
L
L
m ⋅K
∴
qconv,o
q′conv,o =
L
= ho ⋅
As,o
L
qconv,i
q′conv,i =
L
As,i
= hi ⋅
L
q′rad = 21.3
kW
m
⋅ (T∞,o − Ts) = ho ⋅ π ⋅ D ⋅ (T∞,o − Ts) = 100
∴
q′conv,o = 4.5
W
⋅ π ⋅ 0.03m ⋅ (1,000 K − 525.7 K )
m2 ⋅ K
kW
m
⋅ (T∞,i − Ts) = hi ⋅ π ⋅ D ⋅ (T∞,i − Ts) = 100,000
∴
q′conv,i = 25.8
W
⋅ π ⋅ 0.03m ⋅ (525.7 K − 523.15 K )
m2 ⋅ K
kW
m
Note that the above values satisfy the energy balance (i.e. q′conv,o + q′rad = q′conv,i).

















PAG E 18 4







Mathematically, we construct an energy balance as,
Extended Surfaces and Heat Sinks
In many cases, the thermal resistance that most limits heat ow is convection. When this is
true, there are several options one can choose from to reduce the resistance due to
convection. From the previous section, we know that convection thermal resistance can be
expressed as,
1
Rcv =
h ⋅ As
To lower the resistance above, the two logical options we have are to either increase h or
increase As. In order to increase h, we can choose a uid with a lower viscosity or increased
thermal conductivity, or we can increase the
uid’s velocity. Often, it is much simpler to
increase the surface area for heat dissipation by convection.
A surface can be "extended" in a variety of ways. The most common types of extended
surfaces include those shown in Table 2.1 on the following page. The extension of di erent
surfaces can be produced by extrusion, electrical discharge machining, water jet cutting, and
can even be welded/brazed/soldered.
For the case of an extended surface, we approximate heat ow as being one-dimensional. Of
course, this is not strictly the case, as heat is transferred across a n (or an array of ns) by
conduction and convection, as shown in the schematic below.
Figure 2.11. An individual n (part of a larger array of ns) where heat ows through the base by conduction, and out from the upper surface of
the base by convection. Heat also ows by conduction through the n, and out to the environment in a direction perpendicular to the ow of heat
from below.
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PAG E 18 5
surrounding environment by convection across the area of the exposed portion of the base,
or can move into the n, and out into the surrounding environment on each face of the n.
Clearly, the heat moves in more than one direction here. Despite the multi-directional
nature of heat ow in this case, we still endeavor to model thermal transport through this
system using a 1-D resistor network.
To account for the multi-directional aspect of thermal transport, we de ne a n e ciency,
which is described as the actual rate of heat transfer through a particular n relative to the
maximum possible rate of heat transfer through an optimized
n of the same type and
geometry. Mathematically, we write this as,
ηf =
qf
qmax
=
qf
(2.14)
h ⋅ Af ⋅ θb
where Af is the surface area of a n (provided for di erent n types in Table 2.1) and,
θb = Tb − T∞
(2.15)
The heat transfer rate through the n depends on the boundary condition at the n tip (i.e. at
the edge of the n, which is exposed to the surroundings).
Table 2.2. Heat transfer rate through a n with uniform cross section (e.g. longitudinal and pin- ns) for di erent boundary conditions.
Heat Flow Rate through Di erent Fin Tip Boundary Conditions
Case
Condition
Fin Heat Transfer Rate, qf
Convection
sin h(m ⋅ L) + m ⋅hκ
dθ
h ⋅ θ (L) = − κ ⋅
dx
1
M⋅
x=L
f in
h
cosh(m ⋅ L) + m ⋅ κ
f in
⋅ cosh(m ⋅ L)
⋅ sin h(m ⋅ L)
Adiabatic (Insulated)
dθ
dx
2
M ⋅ ta n h(m ⋅ L)
=0
x=L
θ
Constant Temperature
3
M⋅
θ (L) = θs
cosh(m ⋅ L) − θs
sin h(m ⋅ L
In nite Fin (L → ∞)
4
b
M
θ (L) = 0
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PAG E 18 6
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In Fig. 2.11, heat ows through the base of a heat sink, and can either move directly into the


shown are the schematics that describe each n type (with relevant geometries) and equations that can be used to compute n e ciencies.




Table 2.1. Extended surfaces ( n) types with corresponding n surface areas (Af), corrected lengths (Lc), and corrected n pro le areas (Ap). Also
t
2
ηf =
ta n h(m ⋅ Lc )
m ⋅ Lc
ηf =
ta n h(m ⋅ Lc )
m ⋅ Lc
Ap = t ⋅ L
Rectangular Pin Fin
Af = π ⋅ D ⋅ Lc
Lc = L +
D
4
Triangular Fin
2 1
2
t
Af = 2 ⋅ W ⋅ L +
[
(2 ) ]
2
Ap =
ηf =
t
⋅L
2
I ⋅ (2 ⋅ m ⋅ L)
1
⋅ 1
m ⋅ L I0 ⋅ (2 ⋅ m ⋅ L)
Lc = L
Parabolic Fin
Af = ⋅ W ⋅
1
2 1
2
2
t
L2
t
t
1+
⋅L +
⋅ ln
+ 1+
[(
(L ) )
( t
(L [
( L ) ] ))
2
t
Ap = ⋅ L
3
ηf =
2
1
2
4 ⋅ (m ⋅ L)2 + 1 + 1
[
]
Lc = L

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PAG E 187



Lc = L +


Fin E ciency
Af = 2 ⋅ W ⋅ Lc


Schematic
Longitudinal Rectangular Fin


Fin Geometry


Extended Surface (Fin) Types and Characteristics
Heat Flow Rate through Di erent Fin Tip Boundary Conditions
m=
θs = Ts − T∞
θb = Tb − T∞
M=
h⋅P
κ ⋅ Ac
h ⋅ P ⋅ κ ⋅ Ac ⋅ θb
The e ciency of the n, then, governs the rate of heat transfer through the n as,
qf = ηf ⋅ h ⋅ Af ⋅ θb
For an individual n, as shown below, we can model thermal transport via a resistor network
with heat ow acting in parallel through the base and out into the ambient environment, as
well as through the n and out through the ambient environment.
Figure 2.12. Schematic of a longitudinal n with a uniform heat ux applied at the bottom of the n base. Convection acts along the upper base
surface and all surfaces of the n itself. We assume that the temperature at the upper surface of the base and at the n base is constant at T = Tb.
The thermal resistor network can then be drawn as,

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PAG E 18 8
For the above resistor network, we can
nd the e ciency of a single
n using either the
expressions for n e ciency in Table 2.1, or we can use the charts on the following page. We
note also that the heat ow rate splits across the parallel resistors on the right (i.e. via (1)
convection across the exposed base area and (2) conduction through the ns to convection
across the exposed n area.
Figure 2.13. Plot of n e ciencies for longitudinal rectangular, parabolic, and triangular ns. Lc is the corrected n length and is listed in Table 2.1
for each n type listed in the legend.
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PAG E 18 9
Figure 2.14 Plot of n e ciency for annular ns with rectangular pro le for di erent n aspect ratios.
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PAG E 19 0
A rectangular n has a thickness of t = 2 mm, a length of L = 50 mm, and a width of W =
100 mm. The n is made of aluminum (κAl = 175 W/m⋅K) and is used in an application
where h = 10 W/m2⋅K and T∞ = 20∘C. The base temperature is Tb = 100∘C. Assume that
there is convection at the n tip. Determine: (1) the rate of heat transfer through the n, qf,
using Table 2.2, (2) the n e ciency using the corrected length (Lc) approximation, (3) the
rate of heat transfer, qf, using the corrected length (Lc) approximation, and (4) the thermal
resistance across the n (Rfin).
Solution
To nd the rate of heat transfer through the n (qf), one option we have is to use Table 2.2,
where (for a rectangular n with convection acting at the n tip),
sinh(m ⋅ L) + m ⋅hκ ⋅ cosh(m ⋅ L)
fin
qf = M ⋅
h
cosh(m ⋅ L) + m ⋅ κ ⋅ sinh(m ⋅ L)
fin
where sinh and cosh are hyperbolic trigonometric functions. First, let’s calculate parameters M
and m, which are provided at the bottom of Table 2.2.
M = θb ⋅
∴
M = (100∘C − 20∘C ) ⋅
h ⋅ P ⋅ κfin ⋅ Ac = (Tb − T∞) ⋅
10
h ⋅ P ⋅ κfin ⋅ Ac
W
W
⋅
(2
⋅
0.002
m
+
2
⋅
0.1
m)
⋅
170
⋅ 0.002 m ⋅ 0.1 m
m2 ⋅ K
m⋅K
∴
M = 21.1 W

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PAG E 191
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Example 2.10 - Rectangular n
h⋅P
=
κfin ⋅ Ac
m=
10
W
⋅ (2 ⋅ 0.002 m + 2 ⋅ 0.1 m)
m2 ⋅ K
170 mW⋅ K ⋅ 0.002 m ⋅ 0.1 m
= 7.75
1
m
Therefore,
10 2
1
m ⋅K
sinh 7.75 m ⋅ 0.05 m +
(
) 7.75 1 ⋅ 170 W
W
qf = 21.1 W ⋅
m
⋅ cosh 7.75 m1 ⋅ 0.05 m
(
)
m⋅K
1
cosh 7.75 m1 ⋅ 0.05 m +
⋅
sinh
7.75
⋅ 0.05 m
m
(
) 7.75 m1 ⋅ 170 mW⋅ K
(
)
10
W
m2 ⋅ K
= 7.92 W
Now, solving for the n e ciency using the corrected length, Lc, we have,
tanh(m ⋅ Lc)
ηf =
m ⋅ Lc
where Lc = L + t/2 = 0.05 m + (0.002 m)/2 = 0.051 m. Thus,
ηf =
tanh 7.75 m1 ⋅ 0.051 m
(
)
1
7.75 m ⋅ 0.051m
= 95.1 %
As a result, qf can also be calculated as,
qf = ηf ⋅ h ⋅ Af ⋅ θb = 0.951 ⋅ 10
W
∘
∘
⋅
(2
⋅
0.051
m
⋅
0.1
m
+
0.1
m
⋅
0.002
m)
⋅
(100
C
−
20
C)
2
m ⋅K
∴
qf = 7.91 W
which is the same as that obtained with the expression in Table 2.2. Now, the thermal
resistance across the n can be computed as,
Rfin =
1
=
ηfin ⋅ h ⋅ Afin
0.951 ⋅ 10
1
W
⋅ (2 ⋅ 0.05 m ⋅ 0.1 m + 0.002 m ⋅ 0.1 m)
m2 ⋅ K
= 10.3
K
W




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PAG E 19 2


and,
Example 2.11 - Pin ns and parallel paths for heat ow
A set of 25 aluminum pin ns, each having diameter dfin = 5 mm, make up a heat sink with
a base thickness tbase = 2 mm. The ns are Lfin = 15 mm tall, and are in contact with air
that is forced over the heat sink with a convection coe cient hR = 400 W/m2 ⋅ K and an
ambient temperature T∞ = 22∘C. A very thin heater is attached to the base of the heat sink
and its temperature can not exceed Theater = 150∘C. A 1 cm thick copper heat spreader is
attached to the back of the heater to help maintain a uniform temperature, and is also
exposed to forced convection across its outer surface with a convection coe cient hL = 650
W/m2 ⋅ K. Determine the maximum allowable heater power, qheater. The dimensions of the
base are Lbase = 50.8 mm and Wbase = 50.8 mm. Assume the
n tip is exposed to
convection.
Solution
To determine the maximum allowable heater power, we can use,
ΔT
qheater =
∑ Rth
To model the heat ow in this system, we use the following thermal resistor network.


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PAG E 19 3
The thermal resistor network above is representative of how heat ows through the system
shown in the schematic on the previous page. In this network, the heat
ow rate splits
through the upper and lower branches such that,
q = qupper + qlower =
ΔTupper
∑ Rth,upper
+
ΔTlower
∑ Rth,lower
Thus,
q=
Theater − T∞
Theater − T∞
+
−1
Rcd,Cu + Rcv,base
−1
−1 )
Rcd,b + ((Rcv,base
+ Rfins
)
In order to obtain q, we need to solve for each individual thermal resistance.
Rcd,b =
Rcd,Cu =
Rcv,Cu =
0.002 m
237 mW⋅ K ⋅ 0.0508 m ⋅ 0.0508 m
401 mW⋅ K
= 0.0033
0.01 m
K
= 0.0097
W
⋅ 0.0508 m ⋅ 0.0508 m
1
650
K
W
= 0.596
W
⋅ 0.0508 m ⋅ 0.0508 m
m2 ⋅ K
K
W
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PAG E 19 4
For the thermal resistance through the ns, we need to nd the n e ciency. For a pin n,
we can use,
tanh(m ⋅ Lc)
ηf =
m ⋅ Lc
where,
h ⋅ Pfin
m=
W
⋅ π ⋅ 0.005 m
m2 ⋅ K
237 mW⋅ K ⋅ π ⋅ (0.0025 m)2
400
=
κfin ⋅ Ac, fin
= 36.75
1
m
and,
D
0.005 m
Lc = L + = 0.015 m +
= 0.01625 m
4
4
Thus, the e ciency of the n is calculated as ηf = 89.5 % . The thermal resistance through
the array of ns is therefore,
Rfins =
1
K
= 0.41
ηf ⋅ hR ⋅ Afin ⋅ Nfins
W
Finally, we calculate the Rcv,base as,
Rcv,base =
1
400
W
⋅ (0.0508 m ⋅ 0.0508 m − 25 ⋅ π ⋅ (0.0025 m)2)
2
m ⋅K
= 1.196
K
W
Now, we calculate qheater as,
(150 − 20) K
qheater =
K
K
0.0033 W +
1.196 W
((
)
∴
−1
K
+ 0.41 W
(
)
−1
+
−1
)
(150 − 20) K
K
K
0.0097 W + 0.596 W
qheater = 635.8 W
The above heater power is the maximum allowable heater power to keep the heater
temperature at or below 150∘C.
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PAG E 19 5
Example 2.12 - Circular annular ns
A 20 m long electrical wire with d = 12.5 mm generates 1 kW of heat. The wire is coated in
a thin (t = 1 mm) dielectric material with κ = 0.1 W/m⋅K. Copper tubing (t = 3 mm) is
pressed against the dielectric material and contains a series of annular ns with Lfin = 10
mm and a n thickness of tfin = 1 mm. In total, there are 200 copper ns along the length of
the wire. Determine the temperature of the electrical wire. The ambient temperature is T∞
= 20∘C and the convection coe cient acting over the outside of the base and ns is h = 200
W/m2 ⋅ K.
Solution
In this problem, heat is generated within the wire and
ows outward toward the
surrounding environment. The heat ow can be modeled via the following thermal resistor
network,
To determine the temperature of the wire, we use,
q=
ΔT
∑ Rth


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PAG E 19 6
Rth by rst determining each individual thermal resistance,
0.0145 m
( 0.0125 m )
ln
Rcd,d =
Rcd,b =
W
2 ⋅ π ⋅ 0.1 m ⋅ K ⋅ 20 m
m
ln 0.0205
( 0.0145 m )
= 0.012
K
W
−6 K
= 6.9 ⋅ 10
2 ⋅ π ⋅ 401 mW⋅ K ⋅ 20 m
W
To nd Rfins, we must rst nd ηf. For the case of an annular n, we turn to Fig. 2.14, which
requires the calculation of,
h
κ ⋅ Ap
3
Lc2 ⋅
where,
Lc = L +
t
0.001 m
= 0.01 m +
= 0.0105 m
2
2
and,
Ap = Lc ⋅ t = 0.0105 m ⋅ 0.001 m = 1.05 ⋅ 10−5 m 2
Thus,
200
h
3
= (0.0105 m) 2 ⋅
κ ⋅ Ap
3
Lc2 ⋅
W
m2 ⋅ K
401 mW⋅ K ⋅ 1.05 ⋅ 10−5m 2
= 0.2334
t
= 20.25 mm and r1 = 10.25 mm. Thus,
2
r2,c ≈ 2. Using Fig. 2.14, we obtain a n e ciency of ηf ≈ 93 % . Thus,
We also need the quantity r2c/r1. Here, r2c = r2 +
Rfins =
1
0.93 ⋅ 200
(0.02025 m)2
(0.01025 m)2
W
⋅
π
⋅
2
⋅
−
+ 0.02025 m ⋅ 0.001 m ⋅ 200
4
4
m2 ⋅ K
( (
)
)

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PAG E 19 7


∑
We calculate
or,
Rfins = 0.05
K
W
Finally, the thermal resistance by convection over the base is,
Rcv,b =
1
200
W
⋅ π ⋅ 0.01025 m ⋅ 20m − 200 ⋅ π ⋅ 0.01025 m ⋅ 0.001 m)
m2 ⋅ K (
Now we calculate
∑
= 0.0078
K
W
Rth as,
∑
Rth = Rcd,d + Rcd,b + ((Rcv,b)−1 + (Rfins)−1)
−1
Thus,
K
K
−6 K
R = 0.0012 + 6.9 ⋅ 10
+
0.05
∑ th
W
W ((
W)
−1
K
+ 0.0078
(
W) )
−1
−1
and,
∑
Rth = 0.0188
K
W
Now,
Twire − T∞
ΔT
q=
=
∑ Rth
0.0188 K
W
→
K
Twire = 1,000W ⋅ 0.0188 + 20∘C = 38.8∘C
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PAG E 19 8
Thus far, we have discussed the development of thermal resistor networks to model heat
ow along a variety of thermal pathways, including heat sinks. However, we have not yet
touched on the issues that arise due to imperfect contact between adjacent surfaces. For
instance, two machined surfaces that are held together by an applied force are likely to form
an interface that looks like that shown in the gure below.
Figure 2.13. Heat sink above a heated electronic component. The heat sink is clamped to a heat spreader. Inset shows a magni ed view of the
interface between the heat sink and heat spreader.
The inset above the heat sink apparatus shown in Fig. 2.13 demonstrates the imperfect
contact made between adjacent surfaces due to surface asperities that are the byproduct of
machining the components (i.e. there is some roughness to each surface).
This results in a temperature drop across the interface, as shown in the plot below.
Figure 2.14. Temperature distribution from the bottom of the copper heat spreader to the top of the heat sink base.
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Contact Thermal Resistance
show in the next example problem.
Typically, a contact thermal resistance is provided in such a way that it is normalized across
an area. This allows for an extension of the contact resistances measured in experiment to
devices with the same materials (and same surface roughness) but di erent areas across
which heat is conducted. Some common values of interfacial thermal resistance (or contact
thermal resistance) between two surfaces with di erent
uids occupying the gap between
surfaces are provided in the table below.
Table 2.3. Some common values of thermal contact resistance, in units of m2 ⋅ K/W, for di erent surfaces in contact with one another, and for a
variety of gap llers or applied pressures.
Common Values of Contact Thermal Resistance
Material Types
Thermal Contact Resistance, R′c′
Gap Filler
Aluminum / Aluminum
Aluminum / Aluminum
m2 ⋅ K
0.000105
W
Helium
σ = 10 μm
m2 ⋅ K
W
0.000275
Air
σ = 10 μm
Aluminum / Aluminum
0.0000265
Silicone Oil
σ = 10 μm
Aluminum / Aluminum
Vacuum
Stainless Steel / Stainless Steel
Vacuum
Copper / Copper
Vacuum
m2 ⋅ K
W
m2 ⋅ K
0.002
W
→
F = 10
m2 ⋅ K
0.00003
W
→
F = 100
m2 ⋅ K
0.01
W
→
F = 10
m2 ⋅ K
0.00007
W
→
F = 100
m2 ⋅ K
0.005
W
→
F = 10
m2 ⋅ K
W
→
F = 100
0.000025
kN
m2
kN
m2
kN
m2
kN
m2
kN
m2
kN
m2
The values in the table above help illustrate that displacing the air in the gap with a highly
conformable, high thermal conductivity material (silicone oil) reduces the thermal contact
resistance (and thus ΔTinterface ). In fact, most of your CPUs inside of your own computers
have a "thermal paste" where the heat sink is attached in order to reduce the contact thermal
resistance and maintain a lower operating temperature.
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PAGE 200













The temperature di erence across the interface generally results in a hotter device, as we will
Example 2.13 - Contact thermal resistance in an electronic component
An electronic component generates 100 W of heat and is put in contact with a copper heat
spreader. The heat spreader has a thickness of 3 mm and an area (discounting the anged
area for the spring-loaded screws) is 50.8 mm x 50.8 mm. A heat sink is attached to the
copper heat spreader and a pressure is applied to maintain a low thermal contact resistance
m2 ⋅ K
at the interface (R′c′ = 0.001
). The heat sink is made of copper with a base thickness of 5
W
mm. The ns on the heat sink are also copper, and each has a thickness and n height of tfin
= 2 mm and Lfin = 12 mm, respectively. The convection coe cient acting across the nned
surfaces is h = 80 W/m2 ⋅ K, while the ambient air temperature is 21∘C. Assume that there
is no contact thermal resistance between the heater and the heat spreader. Determine the
temperature of the electronic component. What happens if you apply a thermal grease at the
heat spreader/heat sink interface (such that R′c′ = 0.00004
m2 ⋅ K
)?
W
Solution

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PAG E 2 01


We begin by drawing the thermal resistor network that models heat ow in this problem,
q=
T
− T∞
ΔT
= heater
∑ Rth
∑ Rth
where each of the individual thermal resistance values must be determined.
Rcd,hs =
Lhs
0.003 m
K
=
=
0.0029
W
κCu ⋅ Ac
W
401 m ⋅ K ⋅ 0.0508 m ⋅ 0.0508 m
Rinterface =
m2 ⋅ K
0.001 W
R′c′
K
=
= 0.388
Ac
0.0508 m ⋅ 0.0508 m
W
Lbase
0.005 m
K
Rcd,base =
=
= 0.0048
W
κCu ⋅ Ac
W
401 m ⋅ K ⋅ 0.0508 m ⋅ 0.0508 m
To nd Rfins, we must rst nd the e ciency of each n, which can be expressed as,
tanh(m ⋅ Lc)
ηf =
m ⋅ Lc
where,
h ⋅ Pfin
m=
κfin ⋅ Ac, fin
80
=
W
⋅ (2 ⋅ 0.002 m + 2 ⋅ 0.0508 m)
m2 ⋅ K
W
401 m ⋅ K ⋅ 0.002 m ⋅ 0.508 m
= 14.4
1
m
and,
t
0.002 m
Lc = L + = 0.012 m +
= 0.013 m
2
2
tanh 14.4 m ⋅ 0.013 m
(
)
1
∴
ηf =
14.4 m1 ⋅ 0.013 m
= 98.8 %
Thus,
1
W
⋅ (2 ⋅ 0.013 m ⋅ 0.0508 m + 0.002 m ⋅ 0.0508 m) ⋅ 5
m2 ⋅ K
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PAGE 202






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1
Rfins =
=
ηfin ⋅ h ⋅ Afin ⋅ Nfins
0.988 ⋅ 80
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To nd the temperature of the heater, we can use,
Rfins = 1.78
K
W
Finally, the thermal resistance by convection across the exposed heat sink base area is,
1
=
h ⋅ As,base
80
Rcv,base =
W
m2 ⋅ K
1
⋅ (0.0508 m ⋅ 0.0508 m − 5 ⋅ 0.002 m ⋅ 0.0508 m
= 6.03
K
W
Now, the total thermal resistance can be expressed as,
∑
−1
−1
Rth = Rcd,hs + Rinterface + Rcd,base + (Rcv,base
+ Rfins
)
−1
Therefore,
K
K
K
K
Rth = 0.0029 + 0.388 + 0.0048 +
6.03
∑
W
W
W ((
W)
−1
K
+ 1.78
(
W) )
−1
−1
= 1.77
K
W
= 1.39
K
W
Now we can calculate the temperature of the electronic component as,
Theater = q ⋅
∑
Rth + T∞ = 100 W ⋅ 1.77
K
+ 21∘C = 198∘C
W
When a thermal paste is applied at the contacting region, we now have,
Rinteface =
m2 ⋅ K
0.00004 W
0.0508 m ⋅ 0.0508 m
= 0.016
K
W
such that,
K
K
K
K
Rth = 0.0029 + 0.016 + 0.0048 +
6.03
∑
W
W
W ((
W)
−1
K
+ 1.78
(
W) )
−1
−1
and,
K
Theater = q ⋅
Rth + T∞ = 100 W ⋅ 1.39 + 21∘C = 160∘C
∑
W
which represents a decrease in operating temperature of 38∘C!






PAGE 203




∴
EXTERNAL CONVECTION HEAT TRANSFER
We have previously discussed that the strength of natural convection is some function of
di erent parameters (e.g. uid speed, uid type, and object geometry). However, convection
coe cients have simply been provided to us in our assessment of heat transfer problems to
this point. In this section, we will learn how to calculate the convection coe cient for two
basic geometries:
at plates and cylinders. We will also restrict our analysis to forced
convection. Let’s rst examine how we have historically determined convection coe cients by
experiment.
Figure 2.15. A at plate with a sharp leading edge that is internally heated. The heat is uniformly applied across its area in such a way that either
the surface temperature (Ts) is constant, or the surface heat ux (q′′) is constant.
As shown in the
gure above, a
at plate having some length Lplate has heat applied
internally via Joule heating. By measuring the power applied to the electrical circuit and
equating it to the heat dissipated in the plate, in addition to monitoring either the average
or the local temperature at the sample surface, an average (h̄) or local (hx) convection
coe cient can be obtained via,
q = h̄ ⋅ As ⋅ (T̄s − T∞)
or,
q = hx ⋅ As ⋅ (T(x) − T∞)
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PAGE 20 4
In practice, we develop a set of Nusselt number expressions for di erent geometries and ow
conditions such that the cooling and heating performance of di erent uids that advect over
a surface can be compared to one another. The general form of each Nusselt number
expression is given as,
NuL = C ⋅ ReLm ⋅ Pr n
(2.16)
where NuL is the average Nusselt number, C, m, and n are constants, ReL is the total
Reynolds number active over the plate, and Pr is the Prandtl number for the uid. Both the
total Reynolds number and the Prandtl number can be expressed as,
ReL =
ρ ⋅ V̄ ⋅ Lplate
μ
and,
Pr =
ν
α
where Lplate is the total length of the plate (as shown in Fig. 2.15), ν is the kinematic viscosity
of the uid, and α is the thermal di usivity of the uid.
To establish a correlation for a particular geometry and/or
ow condition, we generally
conduct the experiment shown in Fig. 2.15 for di erent uids (i.e. di erent Pr) and di erent
free-stream velocities (i.e. di erent Re) and plot Nu vs. Re and Nu/Pr vs. Re as,
Figure 2.16. (left) log-log plot of NuL vs. ReL for uids with di erent Pr and (right) log-log plot of NuL /Pr n vs. ReL. Comparing left and right plots
allow us to determine C, m and n that make all functions linear.
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PAGE 205
Note that all uid parameters in ReL and Pr are evaluated at the uid’s lm temperature, Tf as,
Tf =
Ts + T∞
2
(2.17)
where Ts is the surface temperature of the plate and T∞ is the
uid’s free-stream
temperature.
Nearly all of the Nusselt number correlations presented in this section are empirical in
nature, meaning that they were obtained via experiment (similar to the way we just
discussed here).
Laminar Flow over a Flat Plate
We have already discussed the establishment of a boundary layer and its in uence on drag
across a at plate. Our development of the friction coe cient acting over the at plate can
now be used to solve the energy equation that describes heat ow into a boundary layer that
forms over a at plate. To begin, we’ll assume the following conditions can be met:
1. The uid ow is steady.
2. The uid is incompressible.
3. The uid ow is laminar.
4. Fluid properties are constant (e.g. not a function of temperature)
5. There is negligible viscous dissipation within the uid.
We have already solved the continuity and momentum equations for uid ow over a at
plate. The energy equation for two-dimensional ow is expressed as,
∂T
∂T
∂ 2T ∂ 2T
u⋅
+v⋅
=α⋅
+ 2
2
(
∂x
∂y
∂x
∂y )
Convection of Energy
Conduction of Energy
If we consider the surface temperature along the length of the plate to be constant, we can also
say that,
∂ 2T
∂ 2T
<< 2
∂x 2
∂y
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PAGE 206
We also know that the boundary conditions for this problem are:
(BC 1)
T(y = 0) = Ts
(BC 2)
T(y → ∞) = T∞
and,
Before solving, let’s also non-dimensionalize the governing equation and boundary
conditions using:
θ=
T(y) − Ts
T∞ − Ts
such that,
∂θ
∂θ
∂ 2θ
u⋅
+v⋅
=α⋅ 2
∂x
∂y
∂y
and,
(BC 1)
θ(y = 0) = 0
(BC 2)
θ(y → ∞) = 1
Also note the presence of a third boundary condition, whereby the slope of the temperature
gradient does not change once you are outside of the boundary layer. As a result, we have,
(BC 3)
∂θ
∂y
=0
y→∞
In the case where α ≠ ν (i.e. the thermal di usivity does not equal the kinematic viscosity),
the thermal boundary layer that forms over the surface is di erent than the hydrodynamic
boundary layer (it will either be thicker or thinner depending on the relationship between α
and ν ). To solve for the non-dimensional form of the temperature distribution within the
boundary layer, we assume that θ takes the following functional form,
θ = f (η)
→
η =y⋅
V
ν⋅x

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PAGE 207
which is the same functional form of the velocity we used to solve the governing momentum
expressions for the hydrodynamic case. Now, each of the terms in the governing equation
for heat di usion becomes,
∂θ
∂θ ∂η
=
⋅
∂x
∂η ∂x
∂θ
∂θ ∂η
=
⋅
∂y
∂η ∂y
∂ 2θ
∂ 2θ
∂η
=
⋅
∂y 2
∂η 2 ( ∂y )
2
such that the governing equation now becomes,
∂θ ∂η
∂θ ∂η
∂ 2θ
∂η
u⋅
⋅
+v⋅
⋅
=α⋅ 2 ⋅
∂η ∂x
∂η ∂y
∂η ( ∂y )
2
Finally, we can use a similarity solution (as with the Blasius solution earlier) and algebra to
produce,
∂ 2θ 1
∂θ
+
⋅
F
⋅
Pr
⋅
=0
2
∂η
2
∂η
Mathematically, this expression has a non-trivial solution. To nd an analytical solution, we
assume that the product of F and Pr (in this case, Pr2/3) is a separate in nite series function,
where the governing equation becomes,
∂ 2θ
1
∂θ
+ ⋅ F(η*) ⋅
=0
∂η *2
2
∂η *
where η * = η ⋅ Pr 2/3. One parameter that the solution to the above expression yields is the
thickness of the thermal boundary layer, which is expressed as a function of the
hydrodynamic boundary layer,
δ
= Pr 1/3
δth
(2.18)
Now the relevant question is: how do we use this information to determine the Nusselt
number correlation for laminar
ow over a
at plate? Let’s take a look at what’s actually
happening inside if the boundary layer.
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PAGE 208
Figure 2.17. Schematic of uid ow over a at plate having a constant surface temperature Ts. Inset shows an enlarged image of the surface, in
which heat is transferred by conduction into a stationary layer of uid particles and out by convection into an adjacent layer of uid particles that
move inside of the boundary layer.
The schematic above describes the mechanisms of heat transfer from the surface of the plate
into the moving
uid above it. Because we have a no-slip condition at the boundary, the
bottom layer of
uid particles is stationary, and thus we conduct heat into them. Thermal
energy is then convected into the adjacent uid layer. Thus, we can say that,
q′cd′
y=0
= q′cv′
y=0
or,
dT
−κ
dy
= h ⋅ (T(y = 0) − T∞)
y=0
Let’s recall that we’ve de ned the following:
θ=
and,
1/3
V
⋅ Pr 1/3
ν⋅x
=y⋅
PAGE 209
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η * = η ⋅ Pr
T(y) − Ts
T∞ − Ts
where,
dθ =
dT
T∞ − Ts
and,
V
⋅ Pr 1/3
ν⋅x
dη * = dy ⋅
Using
dT
q′′
=−κ⋅
dy
y=0
and substituting for dy and dt using the expressions
y=0
developed above, we obtain,
q′′
y=0
=−κ
dθ ⋅ (T∞ − Ts)
(
dη *
V̄
ν⋅x
)
⋅ Pr 1/3
= − κ ⋅ (T∞ − Ts) ⋅
V̄
dθ
1/3
⋅ Pr ⋅
ν⋅x
dη *
η*=0
We can rearrange this algebraically and multiply through by x/x (i.e. a value of 1) to
produce,
q′′
=κ⋅
y=0
Ts − T∞
⋅
x
V̄ ⋅ x
⋅ Pr 1/3 ⋅ F′′(0)
ν
Rex1/2
Notice that by multiplying through by x/x, we have produced the Reynolds number in our
expression for surface heat ux. Likewise, the same similarity solution used by Blasius tells
us that F′′(0) = 0.332. Additionally, since q′cd
′
h ⋅ (Ts − T∞) = κ ⋅
y=0
= q′cv′
y=0
, we can say that,
Ts − T∞
⋅ Rex1/2 ⋅ Pr 1/3 ⋅ 0.332
x
Rearranging, and knowing that Nux = h⋅x/κf (where κ from above represents the thermal
conductivity of the uid, κf), we obtain,















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(2.19)
PAG E 210


h⋅x
= 0.332 ⋅ Rex1/2 ⋅ Pr 1/3
κf





Nux =
Recall that we speci ed a number of assumptions when we began to derive this expression,
ow needs to be laminar (i.e. ReL ≤ 5 ⋅ 105 ). We also derived the local
including that the
correlation for the Nu with laminar ow over a at plate. For other conditions, the following
table should be used to compute Nu and h.
Table 2.4. Nusselt number correlations for uid owing over a at plate when subjected to either a constant surface temperature (Ts) or constant
surface heat ux (q′′) when the uid ow is laminar, turbulent, or contains a mixture of the two. Mixed boundary layer Nu correlation is valid when
the critical Reynolds number is Recr = 5 ⋅ 10 5.
Nusselt Number Correlations for Fluid Flow over a Flat Plate
Condition
Nu Correlation
Constraints
Laminar Flow, Constant Ts, Local Nu (Nux)
Nux = 0.332 ⋅ Rex1/2 ⋅ Pr 1/3
Rex ≤ 5 ⋅ 105, Pr ≥ 0.6
Laminar Flow, Constant Ts, Average Nu (NuL)
NuL = 0.664 ⋅ ReL1/2 ⋅ Pr 1/3
ReL ≤ 5 ⋅ 105, Pr ≥ 0.6
Laminar Flow, Constant q′′, Local Nu (Nux)
Nux = 0.453 ⋅ Rex1/2 ⋅ Pr 1/3
Rex ≤ 5 ⋅ 105, Pr ≥ 0.6
Laminar Flow, Constant q′′, Average Nu (NuL)
NuL = 0.680 ⋅ ReL1/2 ⋅ Pr 1/3
ReL ≤ 5 ⋅ 105, Pr ≥ 0.6
Mixed Boundary Layer
(Part Turbulent/Part Laminar)
NuL = (0.037 ⋅ ReL4/5 − 871) ⋅ Pr 1/3
Constant Ts
Average Nu (NuL)
Turbulent Flow, Constant Ts, Local Nu (Nux)
Nux = 0.0296 ⋅ Rex4/5 ⋅ Pr 1/3
Turbulent Flow, Constant Ts, Average Nu (NuL)
NuL = 0.037 ⋅ ReL4/5 ⋅ Pr 1/3
Turbulent Flow, Constant q′′, Local Nu (Nux)
Nux = 0.0308 ⋅ Rex4/5 ⋅ Pr 1/3
5 ⋅ 105 < ReL ≤ 108
0.6 ≤ Pr ≤ 60
ReL > 108, 0.6 ≤ Pr ≤ 60
0.6 ≤ Pr ≤ 60





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P A G E 2 11


The above also represents one of a series of Nu correlations for uid ow over a at plate.


ow in the system), and should not be thought of as solely an empirical
relation.


describe heat


number, which indicates that Nu is a physical quantity (i.e. it represents the physics that


The expression developed on the previous page is an analytical solution for the Nusselt
Example 2.14 - Convection heat transfer over a at plate
Air ows over a at plate with a velocity of 5 m/s and a free-stream temperature of T∞ =
300 K. The plate is held at a constant temperature of 400 K, and the Reynolds number for
the plate is ReL = 4 ⋅ 105. Assume Prair = 0.71 and κair = 0.028
W
. Determine (1) the
m⋅K
length of the plate [m], (2) the average convection coe cient acting over the plate
[W/m2 ⋅ K], (3) the rate of heat transfer from the plate per unit width [W/m], (4) the local
Reynolds number (Rex) and heat ux (q′x′) [W/m2] at the midpoint of the plate, and (5) the
hydrodynamic and thermal boundary layer thicknesses at the end of the plate [m]. What
happens to the value in parts (2) and (3) when the velocity is increased by a factor of 3?
Solution
In the rst part of this problem we are asked to solve for the length of the plate, L. Because
we are provided with a total Reynolds number (ReL) and the velocity of the air moving over
it, we can determine the uid properties for the air and use ReL to calculate the length of the
plate as,
ReL =
ρ⋅V⋅L
μ
→
L=
ReL ⋅ μ
ρ⋅V
In order to nd the uid properties for the air (ρ and μ), we rst need to determine the lm
temperature, Tf,
Ts + T∞
400 K + 300 K
=
= 350 K
2
2
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PA G E 212
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Tf =
At a lm temperature of 350 K, we nd that ρ ≈ 0.9996
N⋅s
kg
and μ ≈ 2.07 ⋅ 10−5
. Thus,
3
2
m
m
kg
L=
4 ⋅ 105 ⋅ 2.07 ⋅ 10−5 m ⋅ s
kg
0.9996 3 ⋅ 5 ms
m
= 1.66 m
Because ReL < 5 ⋅ 105, we know the uid is laminar. To nd the convection coe cient (h),
we use the following Nu correlation,
NuL = 0.664 ⋅ ReL1/2 ⋅ Pr 1/3 = 0.664 ⋅ (4 ⋅ 105)1/2 ⋅ 0.711/3 = 374.7
and,
h=
NuL ⋅ κf
L
374.7 ⋅ 0.028 mW⋅ K
=
1.66 m
W
= 6.32 2
m ⋅K
Now we can solve for the heat transfer rate per unit width across the plate as,
q′ =
q
W
= h ⋅ L ⋅ (Ts − T∞) = 6.32 2
⋅ 1.66 m ⋅ (400 K − 300 K ) = 1049.1 W/m
W
m ⋅K
At the midpoint of the plate, the local Reynold’s number is computed to be,
Rex=L/2 =
ρ ⋅ V ⋅ L2
=
μ
0.9996
kg
m
m 1.66 m
⋅
5
⋅ 2
3
s
kg
2.07 ⋅ 10−5 m ⋅ s
= 2 ⋅ 105
and the local heat ux at the midpoint can be found via determination of Nux,
Nux = 0.332 ⋅ Rex1/2 ⋅ Pr 1/3 = 0.332 ⋅ (2 ⋅ 105)1/2 ⋅ (0.71)1/3 = 132.5
and thus hx is,
hx =
Nux ⋅ kf
L
2
132.5 ⋅ 0.028 mW⋅ K
=
1.66 m
2
= 4.47
W
m2 ⋅ K
such that the heat ux is,
PAG E 213
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W
W
q′′
= hx=L/2 ⋅ (Ts − T∞) = 4.47 2
⋅ (400 K − 300 K ) = 447 2
m ⋅K
m
x=L/2
Finally, we determine both the hydrodynamic and thermal boundary layer thicknesses as,
δ=
5⋅L
ReL
=
5 ⋅ 1.66 m
4 ⋅ 105
= 0.013 m = 13 mm
δth = Pr 1/3 ⋅ δ = (0.71)1/3 ⋅ 0.013 m = 0.0115 m = 11.5 mm
Now we ask ourselves what happens when the velocity (V) is increased by a factor of 3, such
that V = 15 m/s. In this case, the length of the plate is still L = 1.66 m, but the value of ReL
changes. Here,
ReL =
ρ⋅V⋅L
=
μ
0.9996
kg
m
m
⋅
15
⋅ 1.66 m
3
s
kg
2.07 ⋅ 10−5 m ⋅ s
= 1.2 ⋅ 106
Since ReL > 5 ⋅ 105 but < 1 ⋅ 108, we have a mixed boundary layer. In this case,
NuL = (0.037 ⋅ ReL4/5 − 871) ⋅ Pr 1/3 = (0.037 ⋅ (1.2 ⋅ 106)4/5 − 871) ⋅ (0.71)1/3 = 1632.7
and,
h=
NuL ⋅ κf
L
=
1632.7 ⋅ 0.028 mW⋅ K
1.66 m
= 27.54
W
m2 ⋅ K
such that,
q
W
W
= h ⋅ L ⋅ (Ts − T∞) = 27.54 2
⋅ 1.66 m ⋅ (400 K − 300 K ) = 4571.6
W
m ⋅K
m
q′ =













PAG E 214
Cylinders in Cross Flow
Cylindrical geometries appear in a wide variety of practical problems that are important in
the wider eld of heat transfer. We’ve seen cylinders used in both pin- n heat sinks and in
piping systems. We have also discussed the details of uid ow around cylinders, and will
further restrict our analysis of convection heat transfer to that caused by external
ow
around the outer diameter of a cylinder.
For a smooth cylinder in cross ow (as shown in the schematic below), the transition from
laminar to turbulent ow generally occurs at a critical Reynolds number of ReD,cr = 2 ⋅ 105.
Figure 2.18. Fluid ow over a cylinder (in cross- ow orientation) for laminar (left) and turbulent (right) ow conditions. Note that the separation
point is signi cantly delayed for the case of turbulent ow due to the increase in uid momentum across the surface. Because the boundary layer
sticks across more of the surface for the turbulent case than for the laminar case, heat is transferred more e ectively at ReD > 2 ⋅ 10 5.
Remember also that for a cylinder in cross ow we de ne Re for ow across the cylinder as,
ρ⋅V⋅D
μ
ReD =
We note that the value of the Nu reveals convection heat transfer to be stronger in the case
ow (i.e. ReD > 2 ⋅ 105), which is principally due to an increase in
of turbulent
uid
momentum with an increase in ReD.
Figure 2.19. Local Nu as a function of θ around the cylinder for di erent values of ReD.
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PAG E 215
The previous gure suggests that Nuθ is higher for larger ReD, as was suggested.
In most cases, we care about the average Nu (NuD) for convection heat transfer across the
cylinder. There are two sets of correlations that we can use to obtain the average Nu (and
therefore the average convection coe cient, h). The rst uses the following general form of
the Nusselt number,
NuD =
h⋅D
= C ⋅ ReDm ⋅ Pr 1/3
κf
(2.20)
where C and m are again constants that have been obtained via experiment. For the case of a
cylinder subjected to cross ow, values for C and m are a function of ReD and can be found in
the table below.
Table 2.5. Constants C and m as a function of ReD for use in the NuD correlation shown in Eqn. 2.20.
Constants C and m for Eqn. 2.20
ReD
C
m
0.4 - 4
0.989
0.330
4 - 40
0.911
0.385
40 - 4000
0.683
0.466
4000 - 40,000
0.193
0.618
40,000 - 400,000
0.027
0.805
An all-encompassing equation has also been developed by Churchill and Bernstein for a
wide range of ReD and Pr (and is valid when ReD⋅Pr ≥ 0.2).
ReD
NuD = 0.3 +
⋅ 1+
1/4 [
( 282,000 ) ]
2/3
[1 + (0.4/Pr) ]
5/8 4/5
0.620 ⋅ ReD1/2 ⋅ Pr 1/3
In the above expressions, the properties found in ReD and Pr must be determined at the lm
temperature, Tf.
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PAG E 216
A heater element is being used to heat an oil bath. The heater is cylindrical with an outer
diameter of 15 mm. The element generates 1 kW/m of power. Oil
ows over the heating
element in cross- ow at a rate of 2 m/s. The oil is at a temperature of 20∘C. Calculate the
surface temperature of the heater.
Solution
To determine the surface temperature, we can use,
q′ =
q
= h ⋅ π ⋅ D ⋅ (Ts − T∞)
L
Thus, we need to determine h to solve for Ts. For a cylinder in cross- ow, we can use the
general form of the NuD correlation as,
NuD = C ⋅ ReDm ⋅ Pr 1/3
where C and m depend on ReD. Problematically, in order to determine ReD and Pr, we must
rst have Tf = (Ts - T∞)/2. Since we don’t have Ts, we make an initial guess and will check
to ensure that our solution does not change as we iterate to converge on the correct surface
temperature. Let’s guess that the surface temperature Ts = 120∘C such that Tf = 50∘C ≈ 323
K. At this lm temperature, we nd that,
ReD =
ρ⋅V⋅D
=
μ
kg
m
871.8 3 ⋅ 2 s ⋅ 0.015m
m
kg
0.141 m ⋅ s
= 185.5
Using Table 2.5, we nd that C = 0.683 and m = 0.466. Thus,





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PA G E 217

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Example 2.15 - Cylindrical heater in an oil bath
NuD = 0.683 ⋅ (185.5)0.466 ⋅ (1,965)1/3 = 97.56
Now we can solve for h as,
NuD ⋅ κf
h=
=
D
97.56 ⋅ 0.143 mW⋅ K
0.015 m
= 930.07
W
m2 ⋅ K
With this value of h̄, we obtain a surface temperature Ts of,
Ts =
W
q′
h⋅π⋅D
+ T∞ =
1000 m
W
⋅ π ⋅ 0.015 m
m ⋅K
930.07 2
+ 20∘C = 42.81∘C
This surface temperature is not near our original guess value, so we use this new surface
temperature to iterate a second time. Our new lm temperature becomes,
Ts + T∞
42.81∘C + 20∘C
Tf =
=
= 31.4∘C = 304.6K
2
2
At this temperature, our Reynolds number becomes,
ReD =
kg
881.4
m
m
⋅
2
⋅ 0.015 m
3
s
kg
0.486 m ⋅ s
= 54.41
Using Table 2.5 again, we obtain C = 0.683 and m = 0.466 again. Thus,
NuD = 0.683 ⋅ (54.1)0.466 ⋅ (6,400)1/3 = 81.44
And h is calculated to be,
h=
NuD ⋅ κf
D
=
81.44 ⋅ 0.145 mW⋅ K
0.015 m
= 787.3
W
m2 ⋅ K
With this average heat transfer coe cient, our surface temperature is found to be,
q′
Ts =
h⋅π⋅D
+ T∞ =
1000 W
m
787.3 2W ⋅ π ⋅ 0.015 m
+ 20∘C = 46.95∘C
m ⋅K
This is close to our previous computation of Ts, and more iterations would bring us closer to
a nal value of Ts. Typically, we would iterate until we are within 1% of the previous value.

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PAG E 218
CHAPTER 3: INTRODUCTION TO
T H E R M O DY N A M I C S
E LIVE SUBMERGED AT THE BOTTOM OF AN
OCEAN
OF
THE
UNQUESTIONED
ELEMENT
AIR,
EXPERIMENTS
WHICH
IS
BY
KNOWN
T O H AV E W E I G H T.
- Evangelista Torricelli
Letter to Michelangelo Ricci, June 11th, 1644

PAG E 219
INTRODUCTION AND SYSTEMS
The subject of thermodynamics has played an enormous role in the development of energy
systems, environmental practices, and improvements in human comfort and health. Modern
technological advancements in biotechnology and nanotechnology, as well as improvements
in the e ciency of electronics, internet technologies, and advanced hypersonic also rely on
fundamental aspects of thermodynamics. In this Chapter, we will examine the
rst and
second law of thermodynamics in detail, and use them to analyze and design power and
refrigeration systems for propulsion, heating, and cooling technologies.
Introduction to Systems
It is useful to begin our discussion of thermodynamics with the etymology of the word
itself. The word thermodynamics can be broken down into its constituent parts, where
therme means heat and dynamis means power. Not coincidentally, each of these terms is a
form of energy. Thus, in thermodynamics we study the way in which we:
1. Transfer energy
2. Obtain energy
and,
3. Apply energy
to or between power or heating/cooling systems. Some examples that you may be familiar
with include automotive engines, jet engines, and nuclear power plants. These examples
are all forms of systems. For now, let’s consider a system to be a kind of “black box” across
which energy can ow, as in the gure below.
Figure 3.1. A schematic of a generalized system, which is enclosed by a boundary and exposed to its surroundings
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PAGE 220
Before we take a look at more speci c energy systems, we need to understand how energy
interacts with a system more generally. Thus, we need to de ne the components in Fig. 3.1 in
order to understand how energy can interact between a system and its environment.
1. System: A region of space (volume) or a quantity of matter (mass)
2. Surroundings: The mass or region outside of the system
3. Boundary: The real or imaginary surface that separates the system and surroundings
Note: the boundary can be xed or moveable
One obvious question from our de nition of a boundary is: what is meant by real and
imaginary boundaries?
To answer the above question, we need to de ne the di erence between a closed versus and
open system:
1. Closed System (Control Mass System):
• Has a xed mass
• No mass can cross its boundaries!
• Energy (Heat and Work) can cross the boundaries!
Figure 3.2. Schematic of a piston-cylinder system, which we will see a lot. We consider the “system” to be everything inside of the dashed line
because, generally, we care about how the substance inside of it expands and contracts to give us useful work (more on this later). This system is
assumed to be closed, meaning none of the substance inside can get out, and vise versa. But, energy can clearly move into or out of the system.
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PAG E 2 21
2.
Isolated System
• Special case of closed system
• Energy can not cross boundaries! (e.g., thermally insulated)
3. Open System (Control Volume System)
• Represents a region in space
• Both energy and mass can cross the boundaries!
Figure 3.3. Schematic of a nozzle. We consider the “system” to be everything inside the dashed line. On the left and right sides, ow is allowed in
and out of the system. Energy can cross any part of the system boundary.
The study of closed and open systems in this course is a foundational component for
understanding energy transport and generation in larger thermodynamic systems. In fact,
the core systems we will learn about in this book are categorized by whether they are open or
closed. As you will see, these classi cations dictate how energy moves into and out of a
system, and also drive the utility of power, propulsion, and heating, ventilating, and air
conditioning (HVAC) technologies.
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PAGE 222
PROPERTIES, UNITS, AND TEMPERATURE
In order to design and analyze thermodynamic systems, we must be familiar with the
properties of substances. The property of a substance will often govern how it behaves in
application. For instance, you may know that materials with a large electrical conductivity
are able to transfer a lot more energy than those that are electrically insulating. In the rst
chapter of this book, you also learned about how a uid’s viscosity can alter the frictional
stresses imposed on its surface. As in these applications, the uids used in thermodynamic
systems (namely air and refrigerants) will govern their performance. Here we will de ne a
number of terms that are critical to understanding thermodynamic properties, and will
subsequently learn how to determine them for various states and phases of matter.
Terminology
Here we de ne the terms used to classify properties of substances that are commonly used
in thermodynamic systems. In this book, we focus strictly on air and refrigerants
(R-134a), though there are many others that are important in applications that are beyond
the scope of this course.
First, let’s de ne what a property is:
A PROPERT Y IS ANY C HARACTERISTIC OF A SYSTEM
It is important to understand that a “characteristic” is something that can be used to
distinguish between the system and its surroundings. For instance, temperature is referred
to as a characteristic. If your home is being cooled, and the indoor space is your system,
then its temperature is a characteristic that can be reference and contrasted to the
temperature of the surroundings, or even the temperature at another time of day.
We can also classify a property as being either intensive or extensive, which is important to
understand if we want to compare systems.
1. Intensive Properties
• Independent of system mass!
• Examples: Temperature, Pressure, and Density
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PAGE 223
2. Extensive Properties
• Depend on the size or extent of the system
• Examples: Total mass, total volume, and total momentum
For visual learners, it may be best to start with a practical example to determine how to tell
whether a property is intensive or extensive. Let’s say we have a classroom and assume that
the classroom can be modeled as an isolated system.
Figure 3.4. Fictitious classroom with air inside of the system boundary having a particular temperature, pressure, mass, volume, and density. The
system is assumed to be isolated.
The dashed line is the system (the inside of the room), and the substance in the system is air,
which has a particular temperature, pressure, mass, volume, and density. Now divide the
room in half with a ctitious plane (blue line in Fig. 3.5). If the property does not have
the same value after splitting the system, we call it extensive (red properties).
Otherwise, it is intensive (green properties).
Figure 3.5. Fictitious classroom from Fig. 3.4 split into two halves, represented by the blue line. Extensive properties in red, intensive properties in
green.
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PAGE 22 4
Speci c Properties are extensive properties divided by mass. When we divide a property by
mass, we can normalize the performance of systems to compare them to one another
independent of the amount of a substance in that system. One example is speci c volume,
which is the total volume of a substance divided by its mass. We will talk more about this
property below and in the coming chapters.
Properties
Let’s take a look at some of the properties we’ll use in this class. These are properties that
you will need to nd or calculate in order to analyze the performance of a system operating
between two characteristic states (later, we’ll call these “state points”).
A. Density (ρ) and Speci c Volume (ν)
Density:
ρ=
m
V
“Mass per unit volume”
Speci c Volume:
ν=
V
m
“Volume per unit mass”
B. Speci c Gravity (SG) and Speci c Weight (γ)
ρ
Speci c Gravity:
SG =
Speci c Weight:
γ =ρ⋅g
ρH2O → density of water at 4∘C
ρH2O
g → acceleration due to gravity
C. Pressure (General)
Pressure de ned: A normal force that’s exerted by a substance per unit area (Eqn.
1.1).
P=
F
A
Units (SI):
N
= Pa
2
m
Note: In practice, the unit of a Pa is too small! Instead, we turn to kPa or MPa.
For a reminder on the types of pressure we can encounter in a system, please refer
to page 10 in your textbook.
For the thermodynamic property tables in this textbook, all pressures are listed in absolute
units.

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PAGE 225
Temperature and the 0th Law of Thermodynamics
Generally, we think of temperature in relative terms - e.g., how much hotter is the weather
in Florida versus Annapolis? To understand the answer to this question, one must have
knowledge of a particular temperature scale, and a sense for what a measurement of relative
temperatures entails. Let’s start by addressing di erent temperature scales. In this course,
we will principally deal with the Fahrenheit (∘F), Celsius (∘C), Rankine, and Kelvin
scales.
Both the Fahrenheit (∘F) and Celsius (∘C) scales are “zeroed” using the freezing point of
di erent substances (i.e., 0∘ was established as the freezing point of a particular liquid). The
Fahrenheit scale was proposed in 1724 by Dutch scientist Daniel Fahrenheit and its 0∘
reference point was de ned by the freezing point of a water and ammonium chloride
mixture. We now de ne the scale by two
xed points, the melting and boiling points of
water (32∘F and 212∘F, respectively). Likewise, the Celsius scale was developed by Swedish
scientist Anders Celsius and established 0∘C and 100∘C reference points, again based on the
melting and boiling points of water (though this scale was not proposed by Celsius himself in fact, it was reversed; it was not until 1743 that the modern scale was proposed, and
eventually adopted, by French scientist Jean-Pierre Christen).
Though useful, these scales do not de ne an absolute zero reference point (note that in
colder parts of the U.S. and Canada, negative temperatures are common!). The Rankine and
Kelvin scales, on the other hand, do provide for an absolute temperature and are necessary
to de ne the characteristic temperature of a system. We can convert between the various
systems as follows:
Celsius to Kelvin:
K
T(K ) = 1 ∘ ⋅ T(∘C ) + 273.15K
C
(3.1)
R
T(R) = 1 ∘ ⋅ T(∘F ) + 459.67R
F
(3.2)
Fahrenheit to Rankine:
Celsius to Fahrenheit:
∘
F
T( F ) = 1 ∘ ⋅ T(∘C ) + 32∘F
C
∘
(3.3)




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PAGE 226
continues to remain a somewhat abstract concept. A standard interpretation of temperature
can instead be made with the 0th Law of Thermodynamics, which de nes temperature based
on the concept of equilibrium.
Consider two blocks held at di erent temperatures, as shown below.
We know from experience (and, as will be discussed later in this chapter, the 2nd Law of
Thermodynamics) that when blocks A and B are put in contact with each other, heat will
ow from the hot block (A) to the cold block (B). If we now let this new system rest for
enough time, the contacting blocks will eventually come to some intermediate temperature.
Here, the system comes to an equilibrium where TA > T > TB , where T is the temperature
of the combined system A and B. Now consider a 3-object system with TA > TB > TC.


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PAGE 227

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While our perception of relative temperatures is grounded in everyday experiences, it
Let us again put these three blocks into contact with one another and let them rest for some
amount of time. If we nd there to be no heat transfer from block A to B, and also nd there
to be no heat transfer from block B to block C, then we indeed have no heat transfer from
block A to block C. In this way, we can use block B as a reference point, or a thermometer.
As a brief example, consider two objects, one at Annapolis and one at West Point.
If we take the thermometer (block B) to West Point and put it in contact with block (C)
until it reaches equilibrium, then insulate it and bring it down to Annapolis and nd that
there is no heat
ow between the thermometer and block A, we can say that the
temperature in Annapolis is the same as the temperature at West Point. Thus, the 0th Law of
Thermodynamics de nes both the concept and the measurement of temperature!
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PAGE 228
Example 3.1 - Finding gas pressure in a spring-piston system
Air is contained in a piston-cylinder assembly, as shown in the gure below. The piston has
a mass of 5 kg. The spring has a spring constant (k) of 130 N/cm and pushes down on the
piston over a distance of 350 mm. The piston diameter is 6 cm. If the atmospheric pressure
is 100 kPa, determine the pressure of the gas inside of the cylinder to hold the piston in
place.
Solution
Here, we consider the piston to be the system in order to determine the pressure of the air
in the cylinder. We do this because we can produce a free body diagram with all relevant
forces acting on both sides of it; we do not have enough information if we consider the air to
be the system. Let’s start by establishing a free body diagram, as shown below.
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PAGE 229
∑
Fy = 0 = Fp,atm + Fspring + Wpiston − Fp,air
Knowing that p = F/A, Fspring = k⋅y, and W = m⋅g, the above equation become,
∑
Fy = 0 = patm ⋅ Ap + k ⋅ y + mpiston ⋅ g − pair ⋅ Ap
Now we solve for pair as,
pair =
patm ⋅ Ap + k ⋅ y + mpiston ⋅ g
Ap
Finally, we substitute and solve as,
3
10 Pa
1kPa
100kPa ⋅
pair =
1 N2
⋅
m
1Pa
⋅π⋅
π ⋅ 0.06m
( 2 )
π⋅
2
N
+ 130 cm
⋅ 35cm + 5kg ⋅ 9.81 m2 ⋅
π ⋅ 0.06m
( 2 )
s
1N
1 kg 2⋅ m
s
2
where,
pgas = 1,728,000
N
1Pa
1kPa
⋅
⋅
= 1,728kPa
N
3Pa
m2
10
1 2
m








PAGE 230


Given the free body diagram above, a force balance yields,
E N E R GY A N D T H E F I R S T L AW O F T H E R M O DY N A M I C S
Provided with an understanding of thermodynamic properties and the concept of
temperature, we can now examine the way in which energy is transferred across the
boundaries of a system, or exchanged with an environment. In particular, we are interested
in how this exchange of energy impacts our system and its thermodynamic characteristics
(e.g., its temperature or pressure). To quantify these impacts, we must perform an accounting
of energy in ows and out ows across the boundaries of a system, as illustrated by a
standard production bookkeeping method.
Let’s consider an arbitrary system which has (for now) unspeci ed in ows and out ows
across its boundaries, as well as productions and destructions that occur within it.
Figure 3.6. Schematic of a system with arbitrary in ows and out ows crossing the boundaries, and productions and destructions within the
system. Common examples of each are provided below.
We can use the terms in Fig. 3.6 to determine the accumulation of something within the
system. Mathematically, we write this as,
Accumulation = In ows - Out ows + Productions - Destructions
There are some common applications of production bookkeeping that you may be familiar
with. For instance, in nance we can track the balance change in our account via,
Balance Change = Deposits - Withdrawals + Interest - Fees
In the example above, deposits represent an in ow of cash into our system (in this case, our
bank account), withdrawals are an out ow of cash from our system, the interest accrued in
our bank account produces additional funds, and the fees we owe to maintain the account
act to destroy funds we have in it.
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PAG E 2 31
We can use the same process to account for changes in the energy internal to our system.
However, the 1st Law of Thermodynamics restricts how we can apply this concept. Taken
in its modern context, the 1st Law states that:
E N E RGY C A N B E T R A N S F E R R E D F RO M O N E F O R M TO
ANOTHER, BUT IT C AN NEITHER BE CREATED NOR
D E S T ROY E D
It took roughly half a century to validate this statement via empirical methods, and the
statement itself has appeared in many di erent forms since. However, the form presented
above is useful to our application of production bookkeeping to track changes in the energy
internal to our system. In particular, the 1st Law tells us that the production and destruction
terms must be negated in our energy accounting. Thus, a resulting energy balance across
our system is represented by,
ΔEsys = Ein − Eout
(3.4)
The change in energy within our system (ΔEsys) is a combination of three forms of energy
that should be familiar from an undergraduate level course in Newtonian physics. These
include kinetic energy, potential energy, and internal energy, each described below.
mv 2
1. Kinetic Energy (KE) =
, describes the relative motion of a system
2
2. Potential Energy (PE) = m ⋅ g ⋅ h, describes the gravitational potential of a system
3. Internal Energy (U), describes sensible and latent heat storage, and chemical and
nuclear reactions.
In this course, we will assume that most systems are static. In this case, both the kinetic
energy and potential energy terms do not contribute to ΔEsys. Thus, much of our discussion
of the rst law will focus on changes in the internal energy, U of a system.
For the energy in ow (Ein) and energy out ow (Eout) terms, we will consider both energy
transfer across the boundaries (heat and work) as well as mass transfer across the
boundaries. As we will initially focus on closed systems, we will rst examine the heat and
work energies that can cross system boundaries.
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PAGE 232
H E AT A N D WO R K E N E RGY
In this section, di erent forms of heat and work energy are described in detail. Some of
these forms will then be used to describe the heat transfer into and out of a system, as well
as the work done to or by a system. These concepts are critical to power and HVAC systems.
For example, we will soon be tasked with calculating the amount of power that can be
produced by an internal combustion engine. In order to do that, we must understand how
much heat we put into that system in order to produce the mechanical power that we
require by the engine.
Heat Energy, Q
In 1843, James Prescott Joule (initially a brewery operator in England, but now recognized
as one of the greatest physicists of the 19th century) proposed that heat can be converted to
work and vise versa. Eventually, he presented evidence that this was indeed true in a paper
titled On the Mechanical Equivalent of Heat, though it was met with great skepticism given his
lack of a scienti c background and due to the fact that it was contrary to much of the work
done in the early 19th century by famous physicist Emile Clapeyron. However, it is now
widely accepted that heat can be converted to work, and vise versa.
For the purposes of energy in ow and out ow accounting, we de ne a sign convention for
Q, which is de ned as the amount of energy transferred across a system boundary by heat.
This sign convention is denoted by,
Q > 0 : Heat is transferred to the system
Q < 0 : Heat is transferred from the system
Work Energy, W
In this course, we will spend a great deal of time discussing, analyzing, and designing energy
systems. For instance, we will discuss the working principles of nuclear power plants in
some detail later in this chapter. In order to analyze this type of system, we must
understand portions of the system that generate some amount of energy (or what we will
·
call work, W; when we discuss energy output on a rate basis, we will call this power, W).
For the purposes of this course, we will broadly de ne work as the force, F, acting on an
object through some distance, s. This is generally de ned by the following expression,

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PAGE 233
W=
s2
∫s
F ⋅ ds
(3.5)
1
As with heat energy, we must also assign a convention for work energy to denote its
entrance or exit into or out of a system, respectively. This convention is as follows,
W > 0 : Work done by the system
W < 0 : Work done on the system
Power
In some instances, we are more concerned with the time rate of change of energy transfer
·
across the boundary. For heat transfer, we simply describe this as a heat transfer rate, Q ,
which is the total heat transferred relative to the total amount of time that it’s been
·
transferred (i.e., Q = Q /t). However, it is useful to consider the concept of power in greater
detail. We can de ne the power as the rate of energy transferred by work, which is
mathematically represented by,
·
W = F ⋅ V̄
(3.6)
·
where W is power, F is an applied force, and V̄ is the velocity at the point of applied force.
Note that the dot above both W and Q is used to denote a rate of energy transfer.
Types of Work
There are many di erent forms of work (or power) that we might consider in
thermodynamic system. Let’s focus on four di erent types that we might encounter in this
text.
Electrical Work
Electricity can produce either work or power (though we are likely more comfortable with
the latter, given its ubiquity in our everyday life). You are likely familiar with the way in
which we calculate power or work that is driven by an electrical current from an Electricity
and Magnetism undergraduate physics course, which can be written as a function of voltage
(V) and current (I),
·
We = V ⋅ I
(3.7)
We = V ⋅ I ⋅ Δt
(3.8)

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PAGE 23 4
Shaft Work
Shaft work is one way that we can produce energy
in thermodynamic systems with rotating
machinery. We can calculate the power and work
imposed by this rotating shaft as,
·
Wsh = 2 ⋅ π ⋅ n· ⋅
(3.9)
Wsh = 2 ⋅ π ⋅ n ⋅
(3.10)
In the above expressions,
is the torque applied
by the shaft, n· is the number of revolutions per unit
time of the shaft (Eqn. 3.9), and n is the total number
Fig. 3.7. Rotating shaft inside of a system with an arbitrary
substance.
of shaft revolutions (Eqn. 3.10).
Spring Work
Like electrical work, spring work should be familiar
from your undergraduate physics course (in
Newtonian Mechanics). Here, a spring can absorb or
impart energy when it is compressed or allowed to
expand through a distance, Δy.
·
Wspr =
1
⋅ k ⋅ (y22 − y12)
2 ⋅ Δt
(3.11)
1
⋅ k ⋅ (y22 − y12)
2
(3.12)
Wspr =
Fig. 3.8. Displacement of a spring by an applied for F
through a distance Δy = y2 − y1.
In the above expressions, k represents the spring
constant, y1 is the rest position of the spring, and y2 is the nal position of the spring after
displacement.
Boundary Work
The
nal type of work we will discuss is termed Boundary Work, which we de ne as the
work done to or by a system due to a moving boundary. As this is the most important
type of work for closed systems (and is the principal mechanism that drives work in a
piston-cylinder device, itself the featured component of an internal combustion engine), we
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will discuss this in detail in the following section.
E N E RGY T R A N S F E R S B E T W E E N S TAT E S A N D BO U N DA RY
WORK
In thermodynamic systems, energy transfers between so-called “states”, which we de ne
here as the current physical condition of a substance. The physical condition of a system is
de ned by a system’s characteristics (e.g., its temperature, pressure, volume, entropy, etc.).
In this section, we will take a closer look at states and how they impact heat and work
transfer from the system to the environment, or vise versa. In particular, we will focus much
of our attention on changes in system volume (i.e., expansion and compression) as we will
focus rst on closed systems. However, energy transfer can occur any time a system changes
one or more of its characteristics.
The rst concept we need to understand is how we “ x” a state, or how we know what all of
a system’s properties are. To do this, we follow the procedure outline by the so-called State
Postulate, which says that you must know at least 2 characteristics of a system to x its
state (i.e., to
nd all other characteristic properties of that system). From this point
forward, we will refer to these characteristic properties as State Variables. Recall that the
state variables of a system are properties like pressure (p), temperature (T), volume (V),
mass (m), and as we will learn in subsequent sections, internal energy (U), enthalpy (H),
and entropy (S). Note that in this book, we will focus on systems that contain substances
which are homogeneous or heterogeneous (i.e., substances that exist in one or more
phases), systems that are in equilibrium (or approximated as such at individual state
points), and systems that contain only one component (i.e., a single substance like air,
water, or refrigerant).
States, Paths, and Boundary Work
Let’s rst take a closer look at a system that has moving boundaries. In order to maintain
consistency and relate this content to the internal combustion engines we will eventually
study, we will utilize a piston cylinder
apparatus to represent closed systems
with moving boundaries for the
remainder of this text.
Consider the piston-cylinder apparatus
shown in Fig. 3.9. Initially the gas in the
cylinder is kept at a high pressure and
temperature. Then, heat is released and
the gas is allowed to expand.
Fig. 3.9. Piston-cylinder apparatus at an initial state with a pressure p1 and
temperature T1 (left) and a nal state with pressure p2 and temperature T2 (right).

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PAGE 236
In order to know how much work, W, is involved in this process, we need to know how we get
from the initial state point to the nal state point. Another way to express this is to say we need to
know the path taken between states. Let’s consider a plot of pressure versus temperature
for the scenario above. One way we can achieve the reduction in pressure and temperature
shown in Fig. 3.9 is to rst cool the car, and then lower its pressure (how we achieve each of
these processes will be the subject of further study in this text; for now, let’s assume we can
do this in the order we just speci ed).
Fig. 3.10. One possible path from the initial state point to the nal state point described in Fig. 3.9.
Note that the blue and pink lines in Fig. 3.10 represent the combined path between state
points 1 and 2 (also labeled in Fig. 3.10). In theory, there are an in nite number of paths
that can be drawn between state points 1 and 2 in Fig. 3.10, though there are only a limited
number of realistic paths for the thermodynamic processes we will study. The path types we
will study are termed as follows.
1. Adiabatic - No heat transfer between the system and surroundings; both pressure
and volume change simultaneously from one state point to the next.
2. Isobaric - Constant pressure path.
3. Isothermal - Constant temperature path.
4. Isometric - Constant volume path.
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PAGE 237
In problems where the system has a moving boundary that changes due to changes in the
system’s state variables, the path you take between state points has a signi cant impact on
the work done to or by the system.
Let’s take a closer look at how di erent paths can impact the work produced or consumed by
a system. Here we will compress an arbitrary gas, where a compression process is the result
of transitioning between the following state variables:
State Point 1: p1, V1
State Point 2: p2, V2
where:
V1 > V2
p1 < p2
and
On a p − V diagram, we can draw this as:
Figure 3.11. p − V diagram showing two obvious paths between state points 1 and 2. Arrows show the direction of each path (A and B).
In Fig. 3.11, two obvious paths to take between points are drawn in dashed lines, and are
labeled path A and path B. Our focus in this brief example will be to determine the in uence
of taking one path versus the other on how much work we need to impart to the system in
order to compress the gas in the cylinder.
Before we illustrate the impact of path choice on the calculation of work into the system,
let’s re-examine the nature of work and determine how to calculate it for a moving
boundary.
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Recall that work can be calculated as the product of an applied force multiplied by a
distance.
W = F ⃗⋅ d
(3.13)
Within the context of compression or expansion work, we can rewrite our expression for the
work done to or by the system as a function of the pressure being applied on either side of
the piston, as shown in Fig. 3.12, below.
Figure 3.12. Schematic of a piston undergoing a compression process due to a pressure applied uniformly over the top surface of a piston. The
piston is compressed a distance d toward the bottom of the cylinder.
The force applied on the piston due to pressure can be expressed as,
F ⃗ = pext ⋅ Ap
(3.14)
and the work done on the system (to compress the gas) is now rewritten by substituting
Eqn. 3.14 into 3.13,
W = pext ⋅ Ap ⋅ d = pext ⋅ dV
(3.15)
Thus, boundary work can be expressed as the product or pressure and the change in system
volume, d ∀ .





PAGE 239
It is critical to understand that the pressure of the gas inside of the system is itself a
function of volume. Recall that there are theoretically an in nite number of paths between
state points, as shown in Fig. 3.13.
Figure 3.13. p − V diagram showing multiple paths between state points 1 and 2. Arrows show the direction of each path (A, B, C, and D).
Because we can take multiple paths between state points, we can generalize our expression
for the work done between any two state points (in this case, state points 1 and 2), as,
W12 =
2
∫1
ðW =
2
∫1
pdV
(3.16)
Equation 3.16 reveals that work can be computed by nding the area under a p − V curve
bounded by two state points, like those found between state points 1 and 2 in Fig. 3.13.
Note that the di erential symbol ð is meant to imply that the integral in Eqn. 3.16 is path
dependent.
Now let’s refocus our attention on the original problem of a gas that is compressed by
taking one of paths A or B. If we evaluate the work done to compress the gas using each
path, we obtain,

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Path A
2
∫1
W12,A =
pdV
Here we can integrate the path in two segments as,
W12,A =
V2
∫V
p1dV +
1
V2
∫V
pdV
2
Note that the second integral in the above expression reduces to 0. Knowing that p1 is
constant, we can simplify and solve for work as,
W12,A = p1 ⋅ (V2 − V1)
Path B
W12,B =
2
∫1
pdV
Here we can integrate the path in two segments as,
W12,B =
V1
∫V
1
pdV +
V2
∫V
p2dV
2
In this case, we note that the rst integral in the above expression reduces to 0. Knowing
that p2 is constant, we can simplify and solve for work as,
W12,B = p2 ⋅ (V2 − V1)
If we compare W12,A to W12,B, we quickly realize that we need to put more work in to achieve
compression via path B!







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PAG E 2 41
Path Types
The types of paths that we will encounter in this course are illustrated in Fig. 3.14, below.
Figure 3.14. p − V diagrams for isometric (left), isobaric (middle), and polytropic (right) thermodynamic processes (paths). Arrows indicate the
direction of each path.
In Fig. 3.14, three di erent types of paths are highlighted: isometric, isobaric, and
polytropic. Below, we discuss each of these paths within the context of calculating work.
Isometric - Constant Volume Process
An isometric process is one in which the pressure inside of a system changes, but not its
volume. When volume does not change, then no work can be done to or by the system (if
the piston isn’t moving, then no work is being done!). Mathematically, we express this as,
W12 =
V2
∫V
pdV
1
Since there is no change in volume, then dV = 0, and,
W12 = 0
Isobaric - Constant Pressure Process
An isobaric process is one in which the pressure inside of a system remains constant, but its
volume changes. A general form of the expression for work can be mathematically resolved
as,



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PAGE 2 42
W12 =
V2
∫V
pdV = p ⋅
1
V2
∫V
dV
1
or,
W12 = p ⋅ (V2 − V1)
Polytropic - Pressure varies with volume as p ⋅ ∀n = Constant
In a polytropic process, the pressure in a system is a function of its volume. Speci cally,
pressure and volume are related via the following function,
p ⋅ Vn = C
(3.17)
where C is an arbitrary constant and n is a constant exponent. Note that we can rearrange
the above expression and substitute for p in Eqn. 3.16 such that,
W12 =
2
C
dV
∫1 V n
Two solutions exist for this expression, one for n = 1 and the other for n ≠ 1, as shown
below.
n=1
W12 = p1 ⋅ V1 ⋅ ln
V2
V
= p2 ⋅ V2 ⋅ ln 2
( V1 )
( V1 )
Note that in the above expression, the product of pressure and volume should be equal
to the same constant, C (when n = 1 in Eqn. 3.17).
n≠1
p2 ⋅ V2 − p1 ⋅ V1
W12 =
1−n
Note here that if volume is replaced with speci c volume, we solve for speci c work,
which we de ne as,
w12 =
W12
m
(3.18)
First Law for Closed Systems
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We revisit the 1st Law of Thermodynamics for a closed system. Because our analyses will be
restricted to rigid tanks and piston-cylinder systems, we will neglect changes in kinetic and
potential energy. Recall that we generally de ned the 1st Law of Thermodynamics as,
ΔEsys = Ein − Eout
Substituting the energy terms detailed on pages 229 and 230, we expand the expression
above as,
ΔU12 = Q12 − W12
(3.19)
In Eqn. 3.19, whether energy comes into the system or goes out of the system is identi ed by
the sign convention we de ned previously, namely that heat into the system is positive and
work into the system is negative. Thus, if work is going into the system, W12 itself is
negative and its contribution in Eqn. 3.19 is re ected by a positive work into the system.
For example, if I compress a gas using 100 kJ of energy, and add 100 kJ of heat into the
system, the change in the internal energy of the gas in this system is,
ΔU12 = 100 kJ − (−100 kJ ) = 200 kJ
If instead I expand a gas and gain 100 kJ of energy from the system, and remove 100 kJ of
heat, the change in the internal energy of the gas in this system is,
ΔU12 = − 100 kJ − (100 kJ ) = − 200 kJ
Note that Eqn. 3.19 represents the 1st Law of Thermodynamics for a closed system. We will
discuss the 1st Law within the context of open systems later in this course.
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Air in a piston-cylinder apparatus undergoes an expansion process for which the
relationship between pressure and volume is given by p ⋅ V n = C, where n = 1.4. The initial
pressure of the gas in the cylinder is 50 psia, the initial volume is 6 ft3 and the nal volume
is 12 ft3. What is the work done by the system during this process [Btu]?
Solution
In this problem, we identify the path as polytropic by virtue of the relationship p ⋅ V n = C .
In order to calculate work with n = 1.4, we use the expression,
W12 =
p2 ⋅ V2 − p1 ⋅ V1
1−n
However, the problem statement does not provide us with V2 . In order to nd V2 , we can
use our knowledge of the fact that,
p1 ⋅ V1n = C = p2 ⋅ V2n
Rearranging to solve for p2 provides us with,
V2
lbf
6ft 3
p2 = p1 ⋅
= 50 2 ⋅
( V1 )
in ( 12ft 3 )
n
1.4
= 19
lbf
in 2
Now we can solve for W12 using p2 from above,
19
W12 =
lbf
in 2
3
⋅ 12ft ⋅
144in 2
ft 2
− 50
lbf
in 2
3
⋅ 6ft ⋅
144in 2
ft 2
1 − 1.4
= 25,920 ft ⋅ lbf ⋅
1Btu
= 33Btu
778 ft ⋅ lbf


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Example 3.2 - Calculating Boundary Work
Example 3.3 - First Law Energy Balance and Boundary Work
Air in a piston-cylinder apparatus is compressed isobarically at a pressure of p = 300 kPa
with a displacement volume (i.e., the di erence between the initial and
nal volumes) of
1100 cm3. There is also an electric heater embedded within the cylinder that adds 1.4 kJ of
heat into the system. What is the change in internal energy (ΔU) during this process? Note
that there is no change in the kinetic or potential energy of a piston-cylinder system.
Solution
For a closed system that is not itself moving, the rst law of thermodynamics can be written
as,
ΔU12 = Q12 − W12
Here, we know Q12 from the problem statement, but we need to solve for W12. Since it is an
isobaric process, we solve for W12 as,
W12 = p ⋅ (V2 − V1) = 300 kPa ⋅ − 1100 cm
(
3
1m 3
1kJ
⋅
= − 0.33 kJ
6
3
3
)
10 cm
1kPa ⋅ m
Thus, the change in the internal energy within the system is calculated to be,
ΔU12 = 1.4 kJ − (−0.33 kJ ) = 1.73 kJ
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PAGE 2 46
F I N D I N G T H E R M O DY N A M I C P RO P E R T I E S
In examples 3.2 and 3.3, we had everything we needed to provide answers to speci c
thermodynamics questions related to boundary work and the 1st Law of Thermodynamics.
However, we often do not have su cient information to use the expressions we’ve already
developed. Speci cally, we do not always have the properties we need to solve problems like
those in the aforementioned example problems. In this section, perhaps the most important
in this chapter, we will discuss methods to nd thermodynamic properties at an individual
state point.
This section is organized by the type of substance that we have in a given system. These
include:
1. Ideal Gases (e.g., Air, Oxygen, Nitrogen, Methane, etc.)
2. Pure Substances (e.g., Water, R-134a)
3. Incompressible Substances (e.g., Seawater, Oil, Copper)
In each section, you will learn how to nd all thermodynamic properties knowing any two
other characteristics of a system.
Ideal Gases
Though the term ideal gases represents a theoretical construct, many real gases do behave as
if they are ideal under particular conditions. An ideal gas is one in which the gas molecules
behave as randomly moving, individual particles that do not interact with one another. In
general, gases behave this way at high temperatures and low pressures, where
intermolecular forces and particle size contributes negligibly to energy transport relative to
the speed at which they move.
Ideal Gas Law
The characteristics of an ideal gas can be determined using the Ideal Gas Law, expressed
below (and identical to Eqns. 1.2 and 1.4) as,
p⋅V=m⋅R⋅T
(3.20)
or, we can divide through by mass and rewrite Eqn. 3.20 in speci c form as,
p⋅ν =R⋅T
(3.21)
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Here, ν is a speci c volume, and R is a speci c gas constant (i.e., speci c to the gas in our
system).
A table of speci c gas constants can be found below (note that the table below is identical to
Table 1.3).
Table 3.1. Speci c gas constants for some common gases.
Speci c Gas Constants
Ideal Gas
R
kJ
( kg ⋅ K )
R
psi a ⋅ f t 3
( lbm ⋅ R )
R
psi a ⋅ f t 3
( slug ⋅ R )
Air
0.2870
0.3704
11.92
Oxygen (O2)
0.2598
0.3353
10.79
Nitrogen (N2)
0.2968
0.3830
12.32
Argon (Ar)
0.2081
0.2686
8.64
Helium (He)
2.0769
2.6809
86.26
Methane (CH4)
0.5182
0.6688
21.52
Propane (C3H8)
0.1885
0.2433
7.83
Hydrogen (H2)
4.1240
5.3324
171.53
Xenon (Xe)
0.06332
0.08172
2.63
Neon (Ne)
0.4119
0.5316
17.10
In addition to p, V, and T, we also need to be able to calculate a change in internal energy (for
closed systems) or a change in enthalpy (for open systems) to use an energy balance in
thermodynamic problems.
Internal Energy (ΔU)
To nd the change in internal energy between states for an ideal gas, we use,
ΔU = m ⋅ cv ⋅ ΔT
(3.22)
In Eqn. 3.22, the change in internal energy (ΔU) and the change in temperature (ΔT)
represent changes in these properties between states. As with work, each side of Eqn. 3.22 can
be divided through by mass, m, to produce a speci c internal energy, u.
Enthalpy (ΔH)
To nd the change in the enthalpy between states for an ideal gas, we use,
ΔH = m ⋅ cp ⋅ ΔT
(3.23)

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PAGE 2 48
Equations 3.22 and 3.23 di er only in the type of heat capacity used in each expression. It is
important to understand that these two heat capacities are di erent, in the same way that
both enthalpy and internal energy are di erent.
The heat capacity at constant volume, cv, is a measure of the energy it takes to increase the
temperature of a substance with unit mass when the substance itself is held at constant
volume. Because this measure of heat capacity is considered when the volume is held
constant, there can be no boundary work considered between states during this measurement. Note
that this does not mean that the thermodynamic process in our analysis of Eqn. 3.22
itself is isometric. A measurement of cv requires that heat be added to a gas at constant
volume, which yields,
Qv = m ⋅ cv ⋅ ΔT = ΔU + W = ΔU
(3.24)
Recall that at constant volume, W = 0. Thus, ΔU can be found according to m ⋅ cv ⋅ ΔT.
Because we de ne cv as a measure of heat capacity based on a small increase in temperature
(e.g., 1∘C), we can look at small changes in internal energy, i.e., dU , and small changes in
temperature, i.e., dT. Thus, we can rewrite the above expression as,
dU = Cv ⋅ dT
dU
dT
Cv =
and
where Cv = m ⋅ cv . If instead heat is added at constant pressure, we obtain the following
relation,
Qp = m ⋅ cp ⋅ ΔT = ΔU + W = ΔU + P ⋅ ΔV
(3.25)
For very small changes in the characteristics of a system at an individual state point (with
the exception of pressure, which is held constant), we obtain,
m ⋅ cp ⋅ dT = dU + p ⋅ dV = m ⋅ cv ⋅ dT + p ⋅ dV
(3.26)
In the above expression, we de ne the term m ⋅ cp ⋅ ΔT as enthalpy, ΔH , which we will
de ne in detail later in this chapter. For the moment, we will generally suggest that ΔU
will be applicable to closed systems, and ΔH to open systems, and provide clearer
explanations and restrictions later.
For an ideal gas, we can utilize the IDG Law to construct a relationship between cv and cp .
Recall that one de nition of the IDG Law reads,







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PAGE 2 49
or, through use of the chain rule, we can say that p ⋅ dV = m ⋅ R ⋅ dT. Substituting into Eqn.
3.26 yields,
m ⋅ cp ⋅ dT = m ⋅ cv ⋅ dT + m ⋅ R ⋅ dT
(3.27)
cp = cv + R
(3.28)
Simplifying Eqn. 3.27 yields,
Useful to some thermodynamic problems is the de nition of a speci c heat ratio, k, where,
k=
cp
(3.29)
cv
For an ideal gas, k replaces the exponent n in Eqn. 3.17. This will be covered
extensively when we introduce the compression and expansion processes in 4-stroke
Otto and Diesel cycles.
Finding Properties for Ideal Gases in Underspeci ed Cases
The State Postulate tells us that we need two properties at any given state in order to nd all
other properties. For an illustration of this, let’s look at the Ideal Gas Law in speci c form
(Eqn. 3.21), which says that p ⋅ ν = R ⋅ T. Given that R is the speci c gas constant and depends
only on the type of gas in your system, it becomes obvious that if we know two properties
(e.g., p and T), we can nd the remaining property (in the example mentioned, ν).
It should also be clear that if you know two properties at each of two states, you can nd
ΔU or ΔH. However, in many practical cases, we only know 3 properties between 2 states.
In these cases, we are missing information (i.e., 1 property at one of the states) and would
not be able to nd the change in internal energy or enthalpy between state points. In these
cases, we must instead look at the path we take between states to solve for the
missing property.
Let’s say, for example, that you have an isometric process for which you know p1 and T1 at
state point 1, but only p2 at state point 2. If in this case we are looking for T2, we could not
simply use the Ideal Gas Law at state point 2 to solve for it. Instead, we can take advantage
of our knowledge of the path we take, which in this case is isometric (i.e., constant volume).
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p⋅V=m⋅R⋅T
For an Ideal Gas, we can relate the Ideal Gas Law at each state through the constant volume
process. Given our knowledge of p1 and T1, we can solve for ν1 as,
ν1 =
R ⋅ T1
p1
And, knowing that ν2 = ν1 for an isometric process, we can say that,
T2 =
p2 ⋅ ν1
R
Notice that in the above expression we directly substituted ν1 in for ν2 within the Ideal Gas
Law, which provides us with a set of known values to solve directly for T2.
It should likewise be clear that this process can be used in the same way when the path is
isobaric or (albeit less direct) when the path is polytropic. In this case, we can use a modi ed
relationship which suggests that, between states, we can rewrite the Ideal Gas Law as,
p1 ⋅ ν1
p2 ⋅ ν2
=R=
T1
T2
(3.30)
Note that in the above expression, the speci c gas constant, R, is the same in a closed system
where the gas does not change, and can be cancelled. Recall also that a polytropic process
takes the following form,
p1 ⋅ V1k = C = p2 ⋅ V2k
(3.31)
We can also rewrite this relationship more conveniently in three forms that provide us with
an opportunity to solve for a given property as,
T2
p2
=
T1 ( p1 )
k−1
k
(3.32)
T2
V1
=
T1 ( V2 )
(3.33)
p2
V1
=
p1 ( V2 )
(3.34)
k−1
k
Note that Eqn. 3.34 can be extracted directly from a rearrangement of Eqn. 3.31. The
relationship in Eqn. 3.31 can also be inserted into Eqn. 3.30 to obtain Eqns. 3.32 and 3.33.
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PA G E 2 51
absolute units (i.e., K or R)!
Calculating Boundary Work for Ideal Gases
When calculating boundary work for an Ideal Gas, we can again take advantage of the Ideal
Gas Law to simplify the problem. Recall that the full form of the Ideal Gas Law tells us that
p ⋅ V = m ⋅ R ⋅ T. Thus, for polytropic processes, we can write,
k=1
V2
V2
W12 = p1 ⋅ V1 ⋅ ln
= m ⋅ R ⋅ T1 ⋅ ln
( V1 )
( V1 )
(3.35)
and,
k≠1
W12 =
p2 ⋅ V2 − p1 ⋅ V1
m ⋅ R ⋅ (T2 − T1)
=
1−k
1−k
(3.36)
As we will see, these relationships will be extremely helpful for the compression and
expansion processes that occur in 4-stroke Otto and Diesel cycles.



PAGE 252


Note that the temperatures used in Eqns. 3.20, 3.21, 3.30, and 3.31-3.34 must be in
A piston-cylinder initially contains 1 kg of air at 27∘C and 100 kPa. Heat is added to the air
with the piston stationary until the air temperature is 527∘C. The air is then expanded
according to the relationship p ⋅ V 1.2 = C (where C is a constant), until the volume is 5
times the original. What is the heat transferred and work done during each of the two
individual processes outlined here?
Solution
Process 1 → 2:
In order to nd the heat transferred during each process, we must make use of our rst law
energy balance. For the heat transferred between states 1 and 2 (Q12) for example, we have,
ΔU12 = Q12 − W12
Thus, in order to solve for Q12 we need both ΔU12 and W12 . For an ideal gas, we can nd
ΔU12 via,
ΔU12 = m ⋅ cv ⋅ ΔT12
or,
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Example 3.4 - Air Processes (Ideal Gases and Paths)
kJ
⋅ 500 K = 359.0 kJ
kg ⋅ K
Now we turn our attention to the work, W12 . For an isometric process, the boundary work
done to or by the system must be 0! Recall that if the piston does not move, the volume does
not change, and thus there can be no boundary work. As a result, we solve for Q12 as,
Q12 = ΔU12 + W12 = 359.0 kJ
Process 2 → 3:
We being with a
rst law analysis, which yields ΔU23 = Q23 − W23 . Because air can be
approximated as an ideal gas at these pressures and temperatures, we can solve for ΔU23 as,
ΔU23 = m ⋅ cv ⋅ ΔT23
However, we do not know the temperature at state point 3. Because we know that process 2
→ 3 is polytropic, we can use the following relationship to solve for the unknown
temperature T3,
k−1
T3
V2
=
T2 ( V3 )
Recall that V2 = V1 and that V3 = 5 ⋅ V1 . Substituting each of these relationships into the
above expression provides,
k−1
1.2−1
T3
V1
1
=
→ T3 = 800 K ⋅
= 579.82 K = 306.82∘C
(5 )
T2 ( 5 ⋅ V1 )
With T3 in hand, we can now solve for both W23 and ΔU23,
W23 =
1 kg ⋅ 0.2870 kgkJ⋅ K ⋅ (579.82 K − 800 K
1 − 1.2
= 316.0 kJ
and,
ΔU23 = m ⋅ cv ⋅ (T3 − T2) = 1 kg ⋅ 0.7180
kJ
⋅ (579.82 K − 800 K ) = − 158.1 kJ
kg ⋅ K
∴ Q23 = ΔU23 + W23 = − 158.1 kJ + 316.0 kJ = 157.9 kJ











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PAGE 25 4


ΔU12 = 1 kg ⋅ 0.718
Pure Substances
In a variety of power and refrigeration systems, so-called “pure” substances are used to drive
energy transfer. We restrict our study of these substances to include water and refrigerant
R-134a, which is su cient to gain a deep understanding of the operating principles of
thermodynamic power and HVAC systems.
As with Ideal Gases, the state postulate governs our ability to
nd thermodynamic
properties; namely, one needs to know any two thermodynamic characteristics of a system
in order to nd any other unknown property of it. However, we must also know the phase of
the substance in our system, or at least be able to determine what it is based on a set of
known thermodynamic properties.
You are likely familiar with the three principal phases of matter from core courses in
Chemistry. These include solid, liquid, and vapor. In particular, we will focus on how a
substance transitions from a liquid to a vapor and vise versa. As the substance transitions
between these phases, there is also an intermediate phase where both liquid and vapor exist.
This is analogous to a pot of water that is boiling on your stove; in this case, vapor bubbles
are generated as energy is put into the system. The system remains at the same
temperature, but an increasing fraction of the substance is vapor as energy continues to be
put into the system, until the substance is entirely vapor. There are a few key points to be
made about the conditions of such a system within the context of thermodynamic state
points. We will introduce this system with an example that you may be familiar with: water
boiling at atmospheric pressure.
Let us start with liquid water at 20∘C and 1 atm (standard atmospheric pressure) and
add energy. We will “pause” to examine a set of important points along the way. The above
condition is termed a Compressed Liquid.
Phases of Pure Substances and Corresponding Phase Diagrams
Compressed Liquids
A substance that is entirely liquid is termed a Compressed Liquid.
The expandable tank to the left is lled with water. At 1 atm, water
exists in liquid form at T = 20∘C.
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PAGE 255
Saturated Liquid
As energy continues to enter the system, the water heats up until it reaches its boiling point
at a given pressure (in this case, the boiling point at 1 atm is T = 100∘C ). We label the
substance a Saturated Liquid when in the instant just prior to boiling.
The expandable tank to the left remains lled with water. At 1 atm,
water exists in liquid form at the instant that T reaches 100∘C, and
the volume of water expands.
Saturated Mixture
Once the water reaches its boiling point, the energy going into the system no longer causes
the substance to heat up; instead, the energy going into the system causes the bonds
between uid molecules to weaken until they no longer interact. We label the substance a
Saturated Mixture when it exists in both liquid and vapor phases.
The expandable tank to the left now contains both liquid and vapor.
The amount of vapor inside of the system is quanti ed by a property
called Quality, χ, which is a value between 0 and 1. Gray coloring
inside of the circles indicates a that the substance is vapor. Note that
the volume continues to increase.
Saturated Vapor
At the instant when all of the uid just becomes vapor, we have a substance that exists as a
Saturated Vapor.
The expandable tank to the left is now full of water vapor. The
amount of vapor inside of the system is quanti ed by a property
called Quality, χ, which is now a value of 1. Note that the volume
of the substance continues to increase.


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PAGE 256
Superheated Vapor
Finally, energy continues to heat up the substance beyond the saturated vapor point. Once
this happens, we have a substance that exists as a Superheated Vapor.
The expandable tank to the left is now full of water vapor, whose
temperature continues to increase as energy continues to ow into
the system. Note that the volume of the substance continues to
increase.
We can plot all of this on a phase diagram to visualize this process, which can happen both
in the sequence shown above as well as in the reverse order (when energy is removed from a
system). Two common diagrams, temperature vs. speci c volume (T − ν) and pressure vs.
speci c volume (p − ν) are illustrated in Fig. 3.15, below, for the case where volume is
allowed to vary (such as in a piston-cylinder apparatus).
Figure 3.15. T − ν and p − ν phase diagrams that illustrate phase transitions as a function of the system’s temperature (left) and pressure (right)
with corresponding impacts on speci c volume. Phases are labeled on the diagrams.
Note that the labels for saturation temperature and saturation pressure indicate the
phase transition temperature and phase transition pressure, respectively. The p − ν diagram
also indicates that a liquid can be brought to a boiling state when the its pressure is


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PAGE 257
reduced. Finally, the distribution shown in the T − ν diagram is a constant pressure line, while
the distribution shown in the p − ν diagram is a constant temperature line.
The Vapor Dome
One helpful way to visualize the e ects of pressure and temperature on volume and phase is
to examine multiple constant temperature or constant pressure lines on the above diagrams.
Let’s take a look at the p − ν diagram for water, shown in Fig. 3.16, below.
Figure 3.16. p − ν diagram for water. The Vapor Dome is shown as a dash-dot line, and three constant temperature lines are plotted on the gure.
The vapor dome is plotted as a dash-dot line in Fig. 3.16. Its inclusion provides us with a
way to distinguish between the di erent phases of water at di erent pressures,
temperatures, and volumes. Our reference points in this diagram are: (1) the bounds set by
the vapor dome, and (2) the critical point, which is the apex of the vapor dome. At high
enough pressures and temperatures, the addition of energy into a liquid will eventually
cause a phase transition from liquid directly to vapor, without transitioning through the
saturated mixture region. If a state point lies within the vapor dome, then we have a
saturated mixture. If it lies to the left of the vapor dome and the imaginary line drawn
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PAGE 258
upward from the critical point, then the phase of the substance is a compressed liquid.
Likewise, if our state point lies to the right of the vapor dome and the imaginary point
drawn upward from the critical point, we have a superheated vapor. Finally, if a state point
lies on the vapor dome line and to the left of the critical point, we have a saturated liquid,
whereas if it is on the vapor dome but to the right of the critical point, we have a
superheated vapor.
For pure substances, we need to use the thermodynamic properties that we already know at
a given state in order to rst determine the phase of a substance. This is because the physical
properties of a pure substance are listed in tables that are separated by phase. Thus, we
must
rst examine how to determine the phase of a substance given any two
thermodynamic properties.
Property Tables (Pure Substances): Determining Phase
In order to
nd thermodynamic properties of a pure substance at a given state and solve
problems like the one in Example 3.4, we must
rst determine its phase using known
thermodynamic properties. In general, you will encounter two possible scenarios for which
you will need to be able to x a state. These are detailed below.
Scenario 1: Both the pressure, p and the temperature, T of a substance are given for an
individual state.
Scenario 2: Either the pressure, p, or the temperature, T will be given, along with one of
speci c internal energy, u, speci c enthalpy, h, speci c volume, v, or speci c entropy, s
(we will discuss entropy later). Recall that speci c properties are properties per unit mass!
Let us rst have a discussion about Scenario 1, in which both pressure and temperature are
known at an individual state. In this case, the following diagram will be helpful.
Figure 3.17. p − ν and T − ν diagram for water at T = 100∘C (red line) and p = 101 kPa (blue line). Note that the lines for Tsat and Psat overlap.

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PAGE 259
use the saturation temperature or pressure to determine whether the phase of the substance
is a compressed liquid, saturated mixture, or superheated vapor.
Below are the “rules” you should use to determine which of the above phases you have for a
particular state with a given pressure and temperature. Note that pgiven and Tgiven are the
known values of temperature and pressure at a state. To nd psat or Tsat , you must look at
at the second column in the saturation pressure and temperature tables provided in the
Appendix.
1. If Tgiven < Tsat at pgiven (OR if pgiven > psat at Tgiven): Compressed Liquid
2. If Tgiven > Tsat at pgiven (OR if pgiven < psat at Tgiven): Superheated Vapor
3. If Tgiven = Tsat at pgiven (OR if pgiven = psat at Tgiven): Saturated Mixture
Figure 3.17 can be used to verify the rules written above. For example, the second rule
suggests that you have a superheated vapor if Tgiven > Tsat at pgiven. Note that the red line is
our constant pressure line (where the constant pressure is pgiven ). At this pressure, if
Tgiven > Tsat, and the only part of the red line that is above Tsat is to the right of the vapor
dome, then you must have a superheated vapor! This exercise can be done for all three
rules and using either the constant pressure (solid red) or constant temperature (solid blue)
lines.
Let’s take a look at this in practice. Below are two examples where p and T are known, and
we are attempting to determine the phase of the substance.
Example 1: Determine the phase of water at T = 14∘C and p = 1.5989 kPa.
In this example, we are able to choose whether we use the saturated temperature table or
saturated pressure table for water. Since temperature is presented as a whole number, it is
simpler to use this table. The relevant portion of this table is shown on the following page.
The speci ed temperature of the water (T = 14∘C) is outlined with a blue box. Note that
the corresponding saturation pressure is shown in the column to the right of the
temperature. In the example, our given pressure is p = 1.5989 kPa , which matches the
saturation pressure listed on the table. Therefore, according to the rules outlined above,
we must have a saturated mixture!







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PAGE 260


Scenario 1: In the case where you are given both temperature and pressure, it is easiest to
Example 2: Determine the phase of water at p = 7,000 kPa and T = 200∘C.
For this example, we choose to use the saturated water - pressure table. In theory, we can use
either saturation table given the use of two whole numbers that are each located on their
respective saturation tables. In order to illustrate the di erence between the two tables,
though, we highlight the use of the saturated pressure table, below.
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PA G E 2 61
where Tgiven = 200∘C and Tsat = 285.83∘C, the substance exists as a compressed liquid.
Scenario 2: As a reminder, Scenario 2 describes the case in which either the pressure, p, or
the temperature, T will be given, along with one of speci c internal energy, u, speci c
enthalpy, h, speci c volume, v, or speci c entropy, s.
In general, the approach you should take for this type of problem is to locate the appropriate
saturation table (i.e., if T is given, use the saturated temperature table, and if p is given, use
the saturated pressure table), and then use your other given property (i.e., u, h, v, or s) to
determine the phase of the substance you have. For this second part, you will need to
determine where your other given property falls with respect to the saturated liquid and
saturated vapor values at your given pressure or temperature. We can visualize this using
Fig. 3.16. Let’s say we are provided with a temperature, T, of Tgiven = 100∘C , and a speci c
volume of νgiven = 0.5 m 3 /kg . On a p − ν diagram, constant temperature line is plotted for
Tgiven = 100∘C. Our given speci c volume is then plotted on this line, and is represented by
the yellow star in the gure below.
Figure 3.18. p − ν diagram for water with Tgiven = 100∘C line (purple) and our given speci c volume plotted along that line.
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PAGE 262


Following the rules provided for Scenario 1, we nd that for pgiven = 7000 kPa, Tgiven < Tsat,
νg at this temperature, which puts the state point squarely in the saturated mixture region.
Moreover, this
gure can help to visualize the “rules” for this scenario. Let’s consider an
arbitrary property (u, v, h, or s) to be labeled as “Z”. In this case, the rules suggest,
1. If Zgiven > Zg at pgiven or Tgiven: Superheated Vapor
2. If Zgiven < Zf at pgiven or Tgiven: Compressed Liquid
3. If Zgiven = Zf at pgiven or Tgiven: Saturated Liquid
4. If Zgiven = Zg at pgiven or Tgiven: Saturated Vapor
5. If Zgiven > Zf AND Zgiven < Zg at pgiven or Tgiven: Saturated Mixture
As before, let’s look at some examples to reinforce our understanding of the above rules.
Example 1: Determine the phase of R-134a at T = 25∘F and ν = 1 ft 3 /lbm.
Note that in this example, the temperature, T, and speci c volume, v are provided in English
Engineering Units (EEU), and that the substance is R-134a, so be sure to use the appropriate
tables. An illustration of the table in the appendix of this text is provided for context, below.

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On this gure, it should be clear that our given value of speci c volume falls between νf and
Using the table above, we
saturated mixture.
nd that at T = 25∘F , νf < νgiven < νg , so we must have a
Example 2: Determine the phase of R-134a at p = 15 psia and s = 0.5 Btu /lbm /R.
For this example, we must use the pressure table for R-134a and examine our value of
entropy, s, relative to the saturated liquid (sf) and saturated vapor (sg) values at the pressure
we’re given. An illustration of the table in the appendix of this text is provided for context,
below.
According to the rules outlined for this scenario, we
nd that the substance must be a
superheated vapor due to the fact that sgiven > sg at pgiven.
Property Tables (Pure Substances): Extracting Properties
Now that we have established how to determine the phase of a pure substance, we can
determine all of its unknown properties. Extracting these properties is described in detail
below.


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PAGE 26 4
Superheated Vapor
If you nd that you have a superheated vapor, you can simply nd other properties using the
superheated vapor tables located in the appendix. These tables are organized in such a way
that individual pressures have “blocks”, and the temperature, speci c volume (v), speci c
internal energy (u), speci c enthalpy (h), and speci c entropy (s) are arranged in columns. A
portion of the superheated water table is provided for illustration, below.
The table above re ects the rst row of the pressure “blocks” in the superheated tables in
your Appendix. If a pressure is given with any other property, it is a simple enough exercise
to nd the appropriate pressure block, locate the value of the property you have, and nd all
other properties (including the temperature). If instead the temperature is provided, one
must locate the temperature and work through each pressure box to nd the property value
closest to the one you are given. Two examples are provided below.
Example: Determine the speci c internal energy of R-134a at T = 200∘F and ν = 1 ft 3 /lbm.
At T = 200∘F , we nd that νf = 0.02010 ft 3 /lbm and νg = 0.06441 ft 3 /lbm . Since our given
speci c volume is ν = 1 ft 3 /lbm, we must have a superheated vapor (i.e., νgiven > νg). Thus,
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PAGE 265
the speci c internal energy, u), we must use the superheated vapor tables.
Shown below is the portion of the superheated vapor table that is relevant to this problem.
In this case, we are given a temperature and a speci c volume. Thus, we must
nd the
pressure block in which the speci c volume is equivalent to ν = 1 ft 3 /lbm at a temperature
of T = 200∘F.
From the table above, the speci c volume is equivalent to 1 ft 3 /lbm at a temperature of
T = 200∘F when the pressure is between 60 psia and 70 psia. In order to nd an exact value
of the speci c internal energy (which, as you may recall, is what the example problem asks
for), you must interpolate between the two pressure blocks. The speci c internal energy, u,
is calculated as:
ft 3
ft 3
1 lbm − 0.9447 lbm
u=
ft 3
ft 3
1.1101 lbm − 0.9447 lbm
Btu
Btu
Btu
Btu
⋅ 131.64
− 131.40
+ 131.40
= 131.48
(
lbm
lbm )
lbm
lbm
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PAGE 266



to nd the remaining thermodynamic properties of the substance in this state (in particular,
If you nd that your substance is a saturated mixture at a particular state point, your rst
objective will be to determine the amount of the substance that exists as a liquid versus how
much exists as a vapor. The transition from liquid to vapor occurs along the horizontal line
in Figure 3.18, or between the saturated liquid and saturated vapor points. We de ne the
distance away from the saturated liquid phase as the Quality of a substance, χ .
Mathematically, we de ne quality as:
χ=
mvapor
(3.37)
mvapor + mliquid
Thus, the quality of a saturated liquid is 0, and the quality of a saturated vapor is 1. We can
therefore rewrite Eqn. 3.37 as an interpolation between the saturated liquid and saturated
vapor points using any given thermodynamic property, as shown below.
χ=
χ=
χ=
χ=
νgiven − νf
(3.38)
νg − νf
ugiven − uf
(3.39)
ug − uf
hgiven − hf
(3.40)
hg − hf
sgiven − sf
(3.41)
sg − sf
Note that we can use the quality found in any one of the above equations to
nd any
unknown value in the remaining expressions (i.e., you can rearrange any of Eqns. 3.38-3.41
to solve for unknown properties). Consider the following example.
Example: Determine the speci c enthalpy of water having a temperature of T = 16∘C and a
speci c internal energy, u = 1,000 kJ/kg . In order to determine the phase of the substance,
we examine the saturated temperature table for water (SI), which reads as:
In the previous table, we nd that uf = 67.169 kJ/kg and ug = 2396.9 kJ/kg . Our value of
u = 1,000 kJ/kg is between the saturated liquid and saturated vapor values of speci c
internal energy at our given temperature. Thus, we have a saturated mixture. In order to

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PAGE 267
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Saturated Mixture
this case, we use Eqn. 3.39 to calculate quality as:
χ=
kJ
1,000 kJ
−
67.169
kg
kg
kJ
2396.9 kJ
−
67.169
kg
kg
= 0.4
This indicates that 40% of the substance exists in vapor form!
Now, in order to calculate the speci c enthalpy, we rearrange Eqn. 3.40 (where our “given”
value is actually our unknown value in this case) as,
h = hf + χ ⋅ (hg − hf ) = 67.170
kJ
kJ
kJ
kJ
+ 0.4 ⋅ 2530.2
− 67.170
= 1052.38
(
kg
kg
kg )
kg
Compressed Liquid
Finally, if you
nd that you have a compressed liquid at some state point, you can
approximate most thermodynamic properties as their saturated liquid values. In principle,
this is because properties in the compressed liquid region deviate little from the properties
found in the saturated liquid phase (see Fig. 3.18). These properties should be found on the
saturated temperature table. In other words,
ν = νf (T )
(3.42)
u = uf (T )
(3.43)
s = sf (T )
(3.44)
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PAGE 268

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nd other thermodynamic properties, we must rst calculate the quality of the substance. In
The single exception is for the case of speci c enthalpy when the given pressure is relatively
large. In this case, the following expression must be used:
h = hf (T ) + νf (T ) ⋅ (Pgiven − Psat(T ))
(3.45)
Note that the T in the parenthesis in Eqns. 3.42-3.45 is used to indicate that these
properties should be found on the saturated temperature table.
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PAGE 269
Example 3.5 - Cooled Refrigerant
A rigid tank contains 5 kg of refrigerant 134a with an initial pressure of 0.5 MPa and an
initial temperature of 90∘C. The refrigerant is then cooled until it becomes a saturated vapor.
What is the heat that is transferred during the process?
Solution
Ultimately, this questions asks us to solve for the heat transferred during a process, which
we typically solve for using the rst law of thermodynamics as written for a closed system.
ΔU12 = Q12 − W12
Since the tank is rigid (i.e., it does not contain a moving boundary), there can be no
boundary work. Likewise, we know that ΔU12 can be rewritten as ΔU12 = m ⋅ (u2 − u1). As
a result, the closed system, rst law expression can be rewritten as,
Q12 = m ⋅ (u2 − u1)
Thus, we need to determine the speci c internal energy at each state. We know that State 2
is a saturated vapor and must have the same speci c volume as State 1 (since the tank is
rigid and neither the mass nor the volume of the substance inside of it change).
Consequently, we must rst identify the phase of the substance at State 1.

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Using a temperature of T1 = 90∘C and a pressure of p1 = 0.5 MPa = 500 kPa , we can
determine the phase of the R-134a at State 1 with either the saturated temperature or
saturated pressure table. Let’s use the pressure table for this example. Here we nd that,
At p1 = 500 kPa , Tgiven = 90∘C > Tsat . Therefore, we must have a superheated vapor.
Recall that to nd the properties of a superheated vapor, we use superheated vapor tables.
Using the superheated vapor table, shown above, we nd that the speci c internal energy, u,
at State 1 is u1 = 302.51 kJ/kg. Now we must nd u2.
At State point 2, we know that the substance is a saturated vapor. Since there is no moving
boundary, we also know that ν2 = ν1 = 0.056205 m 3 /kg . To
nd u2 , we can go back to
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PAGE 271
nd where νg = ν2 , and extract u2 (which in this case
happens to be equal to ug because we know the substance is a saturated vapor at this state
point). Here we use the saturated pressure table, but either saturation table can be used.
We nd that ν2 is between the saturated vapor speci c volumes (νg) for pressures of 360 kPa
and 400 kPa. Using interpolation, we nd that u2 is,
m3
m3
0.056205 kg − 0.056738 kg
kJ
kJ
kJ
kJ
u2 =
⋅ 235.07
− 233.38
+ 233.38
= 233.54
( 0.051201 m3 − 0.056738 m3 ) (
kg
kg )
kg
kg
kg
kg
Thus, the heat transferred during this process can be calculated as,
kJ
kJ
Q12 = 5 kg ⋅ 233.54
− 302.51
= − 344.85 kJ
(
kg
kg )
This should make sense, as heat must be removed from the system in order to transition
from a superheated vapor to a saturated vapor during a constant volume process. Because
there is no phase transition, we also see a corresponding drop in temperature (from
T1 = 90∘C to T2 = 6.12∘C).


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PAGE 272
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either of the saturation tables,
Thermodynamic Properties for Incompressible Substances (Oil, Sea Water, etc.)
Many practical thermodynamic systems (e.g., heat exchangers) contain one or more
incompressible substances (e.g., engine oil, sea water, metals, etc.). In order to design and
analyze these systems, we must be able to solve for changes in speci c internal energy or
speci c enthalpy. One can calculate these properties via the following expressions,
ΔU = m ⋅ c ⋅ ΔT
(3.46)
ΔH = m ⋅ c ⋅ ΔT + m ⋅ νf (Tavg) ⋅ ΔP
(3.47)
and,
In the above expressions, c represents the speci c heat capacity of the substances, which is
most often referenced at constant pressure (or cp ). In Eqn. 3.47, νf (Tavg) is the speci c
volume (approximated to be a saturated liquid) taken at the average temperature between
states.
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PAGE 273
4 lbm of copper (c = 0.095 Btu/lbm ⋅∘ F), initially at 350∘F , is cooled in a well-insulated, 1
ft 3 bath of oil (c = 0.43 Btu/lbm ⋅∘ F, ρ = 57 lbm / ft 3). What is the nal temperature of the
copper if both the copper and the bath are allowed to come to equilibrium? What is the heat
transferred from the copper to the bath during this process?
Solution:
If we consider everything inside of the oil tank to be our “system”, then we can write the
rst law of thermodynamics for a thermally insulated, rigid system as,
ΔU12 = 0
Recall that when there is no moving boundary, W12 = 0 . Here we have two systems whose
internal energy changes. Thus, we can rewrite the rst law as,
mo ⋅ (uo,2 − uo,1) + mCu ⋅ (uCu,2 − uCu,1) = 0
Now, using the relationship for Δu that we de ned for incompressible substances, we can
rewrite the above expression as,
mo ⋅ co ⋅ (To,2 − To,1) + mCu ⋅ cCu ⋅ (TCu,2 − TCu,1) = 0
Given that To,2 = TCu,2 = T2, we can solve for the nal temperature (T2) as,
T2 =
mCu ⋅ cCu ⋅ TCu,1 + mo ⋅ co ⋅ To,1
mCu ⋅ cCu + mo ⋅ co
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PAG E 2 74


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Example 3.6 - Cooling of Copper (Incompressible Substance)
T2 =
Btu
lbm
Btu
3
4 lbm ⋅ 0.095 lbm
⋅
810
R
+
57
⋅
1
ft
⋅
0.43
⋅ 530R
⋅R)
lbm ⋅ R
ft 3
(
Btu
lbm
3 ⋅ 0.43 Btu
4 lbm ⋅ 0.095 lbm
+
57
⋅
1
ft
⋅R
lbm ⋅ R
ft 3
= 534.3 R = 74.3∘F
Now, we calculate the heat transferred from the block by considering the block as the
system. In this case, heat is leaving the block and is therefore negative. Thus, the rst law of
thermodynamics reads,
ΔU12 = Q12
where W12 = 0 because the boundary of the system (the copper block) is not moving.
Hence, we solve for Q12 as,
Q12 = mCu ⋅ (uCu,2 − uCu,1) = mCu ⋅ cCu ⋅ (TCu,2 − TCu,1)
and,
Q12 = 4 lbm ⋅ 0.095
Btu
⋅ (534.3 R − 806.9 R) = − 103.59 Btu
lbm ⋅ R
Here, the value of Q12 is negative, indicating that heat is leaving the system. This must be
the case given that the block cools from State 1 to State 2.
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PAGE 275


Substituting the known values provided in the problem statement, we obtain,
OPEN SYSTEMS
We have thus far restricted our study of energy transfers to those that occur in closed
systems, such as the basic piston-cylinder apparatus and rigid tanks. While these systems
form the basis for internal combustion engines, most of the thermodynamic systems that we
will analyze include so-called open systems.
Conservation of Mass
An open system is one that permits mass to cross its boundaries. Contrary to closed
systems, open systems have inlets and exits that allow uids to ow through the system. As
you might imagine, the amount of uid (or the rate of uid ow) can impact the transfer of
energy between a system and its surroundings. One common example is a sink faucet; while
both the hot and cold water lines are open, the rate at which one uid ows relative to the
other governs the temperature of the
uid at the outlet. Thus, we must understand how
much uid ows into and out of a system.
All open systems must satisfy the Conservation of Mass principle. The Conservation of
Mass can be formally expressed as,
Written mathematically, we obtain,
dMcv
=
m· in −
m· out
∑
∑
dt
inlets
exits
(3.48)
In this course, we will only focus on thermodynamic systems that operate at steady-state,
which implies that the time rate of change of a property is equivalent to zero. Thus, the
Conservation of Mass at steady-state reads,
∑
inlets
m· in =
∑
exits
m· out
(3.49)
Note that we discussed the Conservation of Mass in grate detail in Chapter 1 (pages 48-49);
in fact, Eqn. 3.49 is identical to Eqn. 1.20. As a result, the reader is encouraged to re-read
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PAG E 2 76
this section. In particular, we develop a relationship between mass ow rate and velocity (as
well as mass ow rate and volume ow rate) that will be important for the analysis of the
thermodynamic systems we will analyze in subsequent sections. These are rewritten below
for convenience.
V̄ ⋅ Ac
m· = ρ ⋅ V̄ ⋅ Ac =
ν
(3.50)
where ν is the speci c volume of the uid, and Ac is the cross-sectional area of the inlet or
exit (i.e., the area perpendicular to the direction of uid ow).
·
V = ν ⋅ m·
(3.51)
where ν is again the speci c volume of the uid.
Conservation of Energy
In addition to abiding by the Conservation of Mass, open systems must also obey the
Conservation of Energy. This is particularly important in the analysis of power,
refrigeration, and heat exchanger systems.
The Conservation of Energy is formally de ned as,
The energy that ows into or out of the system can be divided into three categories:
·
1. Net rate of energy transferred in by heat, Qcv
·
2. Net rate of energy transferred out by work, Wnet
·
3. Net rate of energy transferred in by mass ow, Em
Note that the subscript cv indicates that we are analyzing a control volume, or open system.
The
·
rst type of energy transfer, Qcv , remains consistent with our de nition of the heat
transfer rate in closed systems. That is, we examine heat exchange between the system (in
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PAGE 27 7
this case, our control volume) and its surroundings. When the heat leaves the system, it is
negative, and when it enters the system, it is positive.
The net rate of energy transferred in by work can be divided into two separate terms. These
include:
·
1. The work done by a rotating shaft, Wcv, and,
2. The “ ow work” done by the system to move the
·
uid mass through system inlets
and exits, Wf low.
Thus,
·
·
·
Wnet = Wcv + Wf low
·
In order to provide a conceptual framework to understand the rate of ow work, Wf low, let’s
examine the following control volume, which uses a piston to “push” a uid into its inlet.
Figure 3.19. Control volume with one inlet and one exit. The uid is being pushed by a piston with force F across a distance d.
In this case, the force applied to the uid by the piston can be written as,
F = p ⋅ Ac
Recall also that work can be de ned as,
W=F⋅d

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PAGE 278
Wf low = p ⋅ Ac ⋅ d = p ⋅ V
where V = Ac ⋅ d is the volume of the uid. Now, we can write the ow work in time rate
form as,
d
·
Wf low = p ⋅ Ac ⋅ = p ⋅ Ac ⋅ V̄ = p ⋅ m· ⋅ ν
t
where m· is the uid’s mass ow rate and ν is the speci c volume of the uid.
Now, we can rewrite the net ow rate out of the system as,
·
·
Wnet = Wcv +
∑
exits
m· e ⋅ pe ⋅ νe −
Finally, the net rate of energy transferred by mass
∑
inlets
m· i ⋅ pi ⋅ νi
ow into the system can be written in
terms of a time rate form of the sum of the internal, kinetic, and potential energy of the
uid within the control volume, or,
V̄ 2i
V̄ 2e
·
·
m ⋅ u+
+ g ⋅ zi −
m ⋅ u +
+ g ⋅ ze
∑ i ( i
) ∑ e ( e
)
2
2
·
Em =
inlets
exits
We can now substitute each of these terms into our de nition for the Conservation of
Energy, which yields,
dEcv
V̄ 2i
V̄ 2e
·
·
·
·
·
·
= Qcv − Wcv +
m ⋅p ⋅ν −
m ⋅p ⋅ν +
m ⋅ u+
+ g ⋅ zi −
m ⋅ u +
+ g ⋅ ze
∑ i i i ∑ e e e ∑ i ( i
) ∑ e ( e
)
dt
2
2
inlets
exits
inlets
exits
In thermodynamics, the sum of a substance’s internal energy with the product of pressure
and volume is termed enthalpy. We can therefore write the speci c enthalpy, h as,
h =u+p⋅ν
(3.52)
We can use the de nition of enthalpy to simplify the conservation of energy. In tandem with
our assertion that only steady-state thermodynamic systems will be considered in this
course, the expression for the Conservation of Energy becomes,
·
·
0 = Qcv − Wcv +
2
2
V̄
V̄
i
m· i ⋅ hi +
+ g ⋅ zi −
m· e ⋅ he + e + g ⋅ ze (3.53)
∑
(
) ∑
(
)
2
2
inlets
exits
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PAGE 279

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Thus, the ow work can be written as,
thermodynamic system!
·
We will use Equation 3.53 to analyze the net rate of heat transferred (Qcv) into or out of (or
·
the power produced or consumed by, Wcv) a variety of open systems.
O P E N S Y S T E M S A N A LY S I S
In this section, we discuss common approaches used to analyze engineerings devices that
are critical to the operation of thermodynamic systems, including power and refrigeration
systems. These devices are considered in individual sub-sections.
Nozzles and Di users
Nozzles and di users are used to accelerate and decelerate uid ows, respectively, resulting
in a change in velocity between their inlet and exit ports. A thermodynamic schematic of
each device is provided in the gure below.
Figure 3.20. (Left) Nozzle with inlet (1) and exit (2), where V̄1 < V̄2, p1 > p2, and h1 > h 2, and (Right) Di user with inlet (1) and exit (2), where
V̄1 > V̄2, p1 < p2, and h1 < h 2.
Mathematically, we make several assumptions to reduce the energy equation to a useful
form, such that we can su ciently analyze its performance in application. These
assumptions include:
1. A nozzle/di user has only one inlet and one exit.
2. Only ow work contributes to the net rate of work being done inside of the system.
3. The di erence in the average height of the inlet and exit is negligible.

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PAGE 280
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The above expression is the general form of the Conservation of Energy for a steady-state
Recall that we will also assume that all devices operate at steady-state. These assumptions
result in the following reduced form of the energy equation:
V̄ − V̄ e
·
0 = Qcv + m· ⋅ (hi − he) + i
(
)
2
2
2
(3.54)
For a nozzle or a di user, there must always be a change in kinetic energy (this is the point
of using such a device!). However, if the nozzle is insulated, then no heat exchange can
occur between the system (the uid inside of the device) and the surrounding environment,
·
which results in Qcv = 0.
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PA G E 2 81
Example 3.7 - Analyzing a Nozzle
Air enters an insulated nozzle with negligible velocity and a temperature of Ti = 400∘F. Air
exits the nozzle with a velocity of 1,000 ft/s. What is the exit temperature? Assume constant
speci c heats for the inlet and exit for a temperature of 100∘F.
Solution:
Note that the energy equation, which governs the temperatures at the inlet and exit of the
device, can be simpli ed to,
V̄ 2i − V̄ 2e
0 = (hi − he) +
2
·
since Qcv = 0 due to the surrounding insulation. Also note that the entire solution can be
divided through by m· in Eqn. 3.54.
For an ideal gas, we know that,
hi − he = cp ⋅ (Ti − Te)
Substituting into the reduced form of the energy equation above, and taking advantage of
the fact that V̄i = 0, we solve for Te as,
V̄ 2i
Te = Ti −
2 ⋅ cp
Now we compute Te by substituting our known values into the above expression as,
1000 s
(
)
ft
Te = 860 R −
2
Btu
2 ⋅ 0.240 lbm
⋅R
1Btu
1lbf ⋅ s 2
⋅
⋅
= 776.8 R = 316.8∘F
778ft ⋅ lbf
32.2lbm ⋅ ft






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PAGE 282
Turbines
A turbine is a device that generates power as a result of a liquid or a gas passing through it
and rotating a shaft. These devices have been used for fossil fuel-free power generation in
wind turbines and conventional power plants, and to achieve thrust in modern aircraft.
Thermodynamically, we consider a turbine to be a device that converts
uid enthalpy to
useful power. A thermodynamic schematic is provided in the gure below.
Figure 3.20. Turbine with one inlet (1) and one exit (2). A rotating shaft is drawn to help visualize the way in which power is produced by the
turbine.
With respect to mathematical analysis, our only simplifying assumptions are that the
turbine operates at steady-state and that the change in height between inlet and exit is
negligible. Thus, the general form of the energy equation for a turbine reads as,
·
·
0 = Qcv − Wcv +
2
2
V̄
V̄
i
m· i ⋅ hi +
−
m· e ⋅ he + e
∑
∑
(
(
2 ) exits
2 )
inlets
(3.55)
For the common case where there is only a single inlet and single exit, Eqn. 3.55 simpli es
to,
V̄ − V̄ e
·
·
0 = Qcv − Wcv + m· ⋅ (hi − he) + i
(
)
2
2
2
Finally, if the turbine is insulated and the change in kinetic energy is negligible, we nd,
·
Wcv = m· ⋅ (hi − he)
Note that you should always start your analysis of a turbine with Eqn. 3.55!
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PAGE 283
A well-insulated steam turbine operates at steady-state with a mass
ow rate of
·
m· = 10,000 lbm /hr and develops a power output of Wcv = 4.5 ⋅ 106 Btu /hr . The inlet
pressure is p1 = 500 psia and the inlet temperature is Ti = 1,000∘F . The turbine exhaust
pressure is pe = 5 psia . What is the quality at the exhaust? Assume there is no change in the
velocity from inlet to exit.
Solution
In order to
nd the quality at the exhaust of the turbine, χe , we must know one other
thermodynamic property at the exit. However, there simply is not enough information
provided in the problem to do that. Instead, we turn to the energy equation for a turbine.
Given our assumptions, the energy equation reduces to,
·
Wcv = m· ⋅ (hi − he)
Using the above expression, we can solve for he and use it to
nd χe . First, we must
determine hi so that we can rearrange the above expression and, ultimately, solve for he.
In order to nd hi , we must rst determine the phase of the water entering the turbine. To
do this, we use our saturated pressure table and nd Tsat at pgiven , after which point we

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PAGE 28 4


Example 3.8 - Analyzing a Steam Turbine
compare it to Tgiven to determine phase. The saturated pressure table is shown in the image
below.
With Tgiven = 1,000∘F > Tsat , our phase at the inlet is clearly a superheated vapor. Thus,
we must use the superheated vapor tables to determine hi . Below is a portion of the
superheated vapor table that can be used to extract hi.
At Tgiven = 1,000∘F and pgiven = 500 psia, hi = 1521
Btu
.
lbm
Now we can use hi to solve for he via,
6 Btu
·
Wcv
Btu 4.5 ⋅ 10 hr
Btu
he = hi − · = 1521
−
= 1071
lbm
m
lbm
lbm
10,000
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PAGE 285
At the exit, we now know that pe = 5 psia and he = 1071
Btu
. Although we are asked what
lbm
the quality is at the exit, and can therefore safely assume that the phase of the steam at the
exit is a saturated mixture, it remains good practice to determine phase the usual way. That
is, we determine he relative to hf and hg at pe = 5 psia on the saturated pressure table for
water.
Btu
, hf ≤ he ≤ hg . Thus, we do indeed have a saturated
lbm
At pe = 5 psia and he = 1071
mixture. We can therefore nd the quality at the exit via,
χe =
he − hf
hg − hf
Btu
=
Btu
1071 lbm − 130.18 lbm
Btu
Btu
1130.7 lbm
− 130.18 lbm
= 0.94
Here we nd that 6% of the substance is liquid at the exit. Typically, we attempt to design a
steam turbine such that χe > 0.9 in order to avoid corrosion of the turbine blades over time.






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PAGE 286
Pumps and Compressors
Pumps and compressors are integral components in many power and refrigeration systems.
These devices use (i.e., consume) power to increase the pressure of a
uid. Pumps are
designed to increase the pressure of liquids, while compressors are designed to increase the
pressure of gases.
·
Figure 3.21. Pump (left) and compressor (right) with one inlet (1) and one exit (2). A rotating shaft is drawn to help visualize Wcompressor
The general form of the energy equation reads as follows,
·
·
0 = Qcv − Wcv +
2
2
V̄
V̄
i
m· i ⋅ hi +
−
m· e ⋅ he + e
∑
∑
(
(
2 ) exits
2 )
inlets
(3.56)
which is equivalent to Eqn. 3.55. And, if there is only a single inlet and single exit,
V̄ − V̄ e
·
·
0 = Qcv − Wcv + m· ⋅ (hi − he) + i
(
)
2
2
2
Finally, if the device is insulated and the change in kinetic energy is negligible, we nd,
·
Wcv = m· ⋅ (hi − he)
For incompressible liquids in a pump, the change in speci c enthalpy can be simpli ed as,
hi − he = cp ⋅ (Ti − Te) + ν ⋅ (pi − pe)
Normally, we assume that ΔT across the pump is negligible, such that,
·
Wpump = m· ⋅ νf (T ) ⋅ (pi − pe)
Where νf (T ) is the saturated liquid speci c volume of the uid at T = Ti = Te.
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PAGE 287
Throttles
A throttle controls the ow rate of a uid by changing the open area through which the uid
can pass. From a thermodynamics perspective, the analysis of a throttle is relatively
straightforward. For the thermodynamic systems we will study, one can neglect the change
in kinetic and potential energies between the inlet and the exit of the throttle. Likewise,
there is no input power to the throttle during steady-state operation (when it is not
moving), and it is unlikely that heat is lost to (or gained from) the surroundings. Thus, the
simpli ed form of the energy equation becomes,
hi = he
(3.58)
Note, however, that if the nozzle gains or loses heat, one must instead begin their energy
analysis with Eqn. 3.57.
Heat Exchangers
Devices that produce power or cool or heat spaces often rely on heat exchangers to transfer
heat into a separate uid for rejection into a particular space. In general, heat exchangers are
devices that transfer heat between uid streams, and we can classify them into two distinct
types:
1. Non-mixing
2. Mixing
We will analyze each individual type of heat exchanger and examine how their physical
aspects lead to mathematical assumptions that simplify the energy equation.
Non-mixing Heat Exchangers
A non-mixing heat exchanger exchanges heat between two
uids that are completely
separated from one another by a solid surface. These types of heat exchangers are commonly
found in car radiators, baseboard home radiators, and condensers in common air
conditioning units. An engineering schematic is provided in the gure below.
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PAGE 288
Figure 3.22. Schematic of a non-mixing heat exchanger with
·
·
uids A (hot) and B (cold). Q is shown to indicate the direction of heat transfer
between the uids. Note that Qcv is the heat exchange between the heat exchanger and the surrounding environment, demarcated by the dashed
line.
In this case, we have a device that has multiple inlets and exits that do not interact with one
another. Therefore, the conservation of mass tells us that,
m· Ai = m· Ae = m· A
and,
m· Bi = m· Be = m· B
However, energy can transfer between uid streams. Assuming that there are no changes in
potential and kinetic energy, and that no power is required to transfer heat from the hot stream to the
cold stream, the energy equation is initially written as,
·
0 = Qcv +
∑
inlets
m· i ⋅ hi −
∑
exits
m· e ⋅ he
Simplifying with the inlets and exits of the heat exchanger, we have,
·
0 = Qcv + m· Ai ⋅ hAi + m· Bi ⋅ hB,i − m· Ae ⋅ hA,e − m· B,e ⋅ hB,e
We can further simplify the above expression using the relationship between the mass ow
rates that we developed by applying the conservation of mass to each uid stream,
·
0 = Qcv + m· A ⋅ (hA,i − hA,e) + m· B ⋅ (hB,i − hB,e)
(3.57)
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PAGE 289
·
Note here that Qcv is NOT the heat exchange between the two uids, but the heat transfer
rate between the heat exchanger and the surrounding environment (see Fig. 3.22). Thus, if
the heat exchanger is insulated on the outside (i.e., between the environment and the uids
inside of the heat exchanger), Eqn. 3.57 simpli es to,
0 = m· A ⋅ (hA,i − hA,e) + m· B ⋅ (hB,i − hB,e)
Thus, the change in uid enthalpies controls the rate at which heat is exchanged between
the two uids.
Mixing Heat Exchangers
A number of applications use mixing heat exchangers to achieve an outlet temperature
intermediate to the temperatures of the inlet uids. For instance, the faucet in your sink or
shower uses one hot and one cold stream of water to produce a uid that exits the faucet at
some intermediate temperature. The temperature of the water at the exit is controlled by
adjusting the mass ow rate of one or both uids with a valve. An engineering schematic is
provided in the gure below.
·
Figure 3.23. Schematic of a mixing heat exchanger with uids A (hot), B (cold) and C (intermediate). Note that Qcv is the heat exchange between
the heat exchanger and the surrounding environment, demarcated by the dashed line.
In this case, the uids in the heat exchanger do interact. Thus, the conservation of mass reads,
m· Ai + m· B,i = m· C,e
Again, we assume that the potential and kinetic energy terms are negligible in the energy
equation, and that no power is required to drive heat from the hot uid to the cold uid.
Thus, the energy equation reads,
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PAGE 290
·
0 = Qcv +
∑
inlets
m· i ⋅ hi −
Considering all inlets and exits in Fig. 3.22, we nd,
∑
exits
m· e ⋅ he
(3.58)
·
0 = Qcv + m· Ai ⋅ hA,i + m· B,i ⋅ hB,i − m· C,e ⋅ hC,e
Note that the number of inlets and exits may change between heat exchangers. Thus, you
should always begin with Eqn. 3.58 when analyzing mixing heat exchangers. Note also that
if the heat exchanger is insulated on the outside, no heat can cross the system boundaries
·
and Qcv = 0 in Eqn. 3.58.
Now let’s look at an example problem that combines these two di erent types of heat
exchangers.
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PAG E 2 91
A well-insulated feedwater heater is also used to pre-heat oil. The oil and water are kept
separate. Water exits the heat exchanger as a saturated liquid at 20 psia and with a mass
ow rate of 40 lbm/s. Of the water entering the heat exchanger, 85% enters at 20 psia and
80 ∘F and the rest enters at 20 psia and 250 ∘F . The oil ( ρ = 57 lbm /ft 3 ,
c = 0.4 Btu /lbm ⋅∘ F) enters at 70∘F and exits at 150∘F. What is the volumetric ow rate of
the oil [in gpm]?
Solution
Since we are not given any information about the ow rate (e.g., mass ow rate) of the oil,
·
there is no way to directly solve for Voil using the conservation of mass for the oil stream.
Instead, we must turn to the energy equation. Recall that the system energy (and ultimately
the temperature changes within the system) is controlled by the mass ow rate. Because the
heat exchanger is well insulated, the energy equation is simpli ed to,
0=
∑
inlets
m· i ⋅ hi −
∑
exits
m· e ⋅ he
Substituting for all inlets and exits, the above expression becomes,
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PAGE 292

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Example 3.9 - Combination Heat Exchanger
Likewise, the conservation of mass for each independent ow stream yields,
m· w1,i + m· w2,i = m· w,e
and,
m· oil,i = m· oil,e = m· oil
·
Ultimately, we are looking for Voil, which is related to the mass ow rate via,
·
m· oil = Voil ⋅ ρoil
Thus, we can ultimately rearrange the energy equation to solve for m· oil as,
m· w,e ⋅ hw,e − m· w1,i ⋅ hw1,i − m· w2,i ⋅ hw2,i
m· oil =
ho,i − ho,e
To solve for m· oil, we need to nd all mass ow rates and enthalpies on the right-hand side of
the above expression. To nd the mass ow rates, we can use the conservation of mass as
applied to the water stream. We know that m· e = 40 lbm /s, and that 85% of the water enters
the
rst inlet and the remaining 15% enters the second inlet. Thus, the conservation of
mass for the water stream becomes,
lbm
lbm
lbm
·
·
·
mw2,i = mw,e − 0.85 ⋅ mw,e = 40
− 0.85 ⋅ 40
=6
s
s
s
lbm
.
s
This must mean that m· w1,i = 34
Our nal task is to determine the enthalpy at each inlet and exit. For the oil, which is an
incompressible substance, we can solve for ho,i − ho,e as,
ho,i − ho,e = coil ⋅ (Toil,i − Toil,e) = 0.4
Btu
Btu
∘
∘
⋅
(70
F
−
150
F
)
=
−
32
lbm ⋅∘ F
lbm
Because our other uid is water, we must use the appropriate tables to nd each enthalpy.
For the
rst entrance, we determine the phase of our substance at pw1,i = 20 psia and
Tw1,i = 80∘F using the saturated pressure table for water.
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PAGE 293
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0 = m· w1,i ⋅ hw1,i + m· w2,i ⋅ hw2,i + m· oil,i ⋅ ho,i − m· w,e ⋅ hw,e − m· oil,e ⋅ hoil,e
liquid. In this case,
hw1,i = hf (Tw1,i) + νf (Tw1,i) ⋅ (Pgiven − Psat(Tw1,i))
Btu
ft 3
lbf
lbf
144in 2
1Btu
hw1,i = 48.07
+ 0.01607
⋅ 20 2 − 0.507 2 ⋅
⋅
lbm
lbm ( in
in )
ft 2
778ft ⋅ lbf
hw1,i = 48.13
Btu
lbm
where we use the saturated temperature table to nd hf, νf, and Psat.
In the same vein, we use the saturated pressure table to determine the phase of the water
entering the second inlet. Here, pw2,i = 20 psia and Tw2,i = 250∘F . Using the saturated
water pressure table, we nd that Tw2,i = 250∘F > Tsat = 227.92∘F , and we must therefore
have a superheated vapor.
Using the table above, we interpolate to nd hw2,i,
250∘F − 240∘F
Btu
Btu
Btu
Btu
hw2,i =
⋅ 1181.9
− 1162.3
+ 1162.3
= 1167.2
( 280∘F − 240∘F ) (
lbm
lbm )
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PAGE 294



At 20 psia, Tw1,i = 80∘F < Tsat = 227.92∘F . Thus, the water at entrance 1 is a compressed
nd hw,e by
determining hf at pw,e = 20 psia. This value is determined to be hw,e = 196.27 Btu /lbm.
Now we solve for m· oil as,
m· oil =
Btu
lbm
Btu
lbm
Btu
40 lbm
⋅
196.27
−
34
⋅
48.13
−
6
⋅
1167.2
s
s
s
lbm
lbm
lbm
lbm
= 24.65
s
Btu
−32 lbm
Now we solve for the volume ow rate of the oil as,
lbm
·
24.65
m
1gal
60s
gal
·
s
Voil = oil =
⋅
⋅
=
194.1
ρoil
0.1337ft 3
1min
min
57 lbm
ft 3

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PAGE 295



Finally, we know that the exit is a saturated liquid at 20 psia. Thus, we
H E A T E X C H A N G E R A N A LY S I S : E F F E C T I V E N E S S - N T U
In the previous section, our discussion of heat exchangers focused on the use of enthalpy to
determine inlet and outlet temperatures or the heat transfer rate between uids. While this
is generally appropriate for our analysis of the performance of power and refrigeration
systems, it typically isn’t a su cient method to design or size heat exchangers for these
systems. In fact, it is common not to know the outlet temperatures of the heat exchanger apriori. In
such a case, it would not be possible to assess the performance of a power or refrigeration
system as there would be not mechanism to nd the enthalpy at the appropriate outlet.
In this section, we will use information about the inner workings of speci c types of heat
exchangers in order to determine heat exchanger performance, design heat exchangers, or
solve for unknown temperatures. The method used here, and described in other heat
transfer textbooks, is widely referred to as ε -NTU, where ε represents the e ectiveness of
the heat exchanger and NTU is referred to as the “Number of Transfer Units”. We will
describe each term in detail as we learn to use this method for heat exchanger analysis.
Heat Exchanger Types
As we will see, the type of heat exchanger we have will in part dictate its performance.
Details of di erent heat exchanger types are provided below.
Double-pipe Heat Exchangers
A double-pipe heat exchanger is one in which heat is transferred between two
di erent temperatures that
uids at
ow through concentric pipes. Heat is transferred across the
Figure 3.24. Parallel (left) and counter- ow (right) heat exchangers, in which uids ow through concentric tubes in speci c directions.
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PAGE 296
uids, as shown in the schematics in Fig. 3.24. Note that we
specify two con gurations according to the direction that each uid ows: (1) parallel ow,
in which both uids ow the same direction, and (2) counter- ow, in which the uids move
in opposite directions.
Figure 3.25. Parallel (left) and counter- ow (right) heat exchangers, with corresponding
length of the heat exchanger.
uid temperatures as a function of distance along the
Note that as the length of the heat exchanger, L, approaches ∞, the relative e ectiveness of
each heat exchanger will be di erent. In the case of parallel
ow, there is some critical
distance at which the uids reach an equilibrium temperature, and beyond which there is
negligible heat transfer between the uids. On the other hand, as L is extended in the case
of the counter- ow heat exchanger, the outlet temperatures begin to approach the inlet
temperatures of the opposite uid (e.g., the outlet temperature of the cold uid approaches
the inlet temperature of the hot uid). Thus, beyond some critical distance L and prior to
L → ∞, the e ectiveness of the counter- ow heat exchanger is larger than that of the
parallel ow heat exchanger. The selection of which should be used, then, depends on the
required outlet temperature and the allowable size of the heat exchanger.
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PAGE 297


thin wall that separates the
One typical constraint when designing a heat exchanger is an upper-limit to its size. For
some applications (e.g., car radiators), there is a limit to the amount of space that the heat
exchanger can possibly occupy. In such cases, compact heat exchangers are required to achieve
reasonable heat transfer rates while also maintaining small form factors.
In general, compact heat exchangers are also used when heat must be transferred between a
liquid and a gas. This is primarily due to the fact that gases generally produce low values of
the convection coe cient, h, acting over a surface (for a full description of convection heat
transfer, please refer to Chapter 2).
In order to achieve higher heat transfer rates and smaller design footprints, compact heat
exchangers are designed to provide ultra-high surface area-to-volume ratios, where,
β=
As
V
(3.59)
In Eqn. 3.59, As is the surface area for heat transfer and V is the volume of the heat
exchanger. For a heat exchanger to be considered compact, its surface area to volume ratio
must be greater than 700 m 2 /m 3 (or 200 ft 2 /ft 3 ). Some examples of compact heat
exchangers include car radiators (~1,000 m 2 /m 3 ), gas turbine heat exchangers (~6,000
m 2 /m 3), and the human lung (~ 20,000 m 2 /m 3).
For applications in which heat must be transferred between a gas and a liquid, a
uid is
usually passed over a tube (or series of tubes) in a parallel or perpendicular orientation
(usually guided by
ns that increase the surface area between the
uids). Likewise,
orientations that promote uid mixing have very di erent heat transfer characteristics from
those that do not, as we will see in the ε − NTU charts in the following sections.
Shell-and-Tube Heat Exchangers
Industrial systems that require heat exchange with the ambient environment most often
utilize shell-and-tube heat exchangers. This is principally due to their capacity, or ability to
discharge/absorb heat at very high rates.
In these types of heat exchangers, one uid ows through the outer “shell” side, and one
ows through an inner series of tubes. It is important to understand that the shell-and-tube
con guration can be run in parallel ow or counter ow, and its e ectiveness mimics that of
the counter ow and parallel ow descriptions provided above. For that reason, we generally
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PAGE 298
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Compact Heat Exchangers
use the ε − NTU correlations associated with those ow types. A schematic of a shell-andtube heat exchanger is provided below.
Figure 3.26. Schematic of a shell-and-tube heat exchanger showing the system’s ba es, shell, and series of tubes. Inlets and outlets are marked
with arrows and colored for the appropriate uid inlet temperature.
Note that we have indicated ba es in Fig. 3.26, which are used to ensure that all of the uid
on the shell side comes into contact with the surface containing the uid in the tubes. This
maximizes the heat transfer rate between the two uids because it maximizes the surface
area for heat transfer.
We classify shell-and-tube heat exchangers by the number of passes the uid makes on each
of the shell side and tube side. A “pass” is considered to be when the uid makes its way
from one side of the heat exchanger to the other. As an example, a one shell pass, two tube
pass shell-and-tube heat exchanger is shown in the image below.
Figure 3.27. One shell pass, two tube pass shell-and-tube heat exchanger con guration. Tube passes from one end to the other twice.
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PAGE 299
The performance (i.e., e ectiveness) of a heat exchanger is governed by the heat transfer
rate between
uids. Thus, we must understand how to determine the heat transfer rate
between the
uids in order to ultimately gauge the e ectiveness of any particular heat
exchanger. To do this, we de ne an overall heat transfer coe cient, which combines the e ects
of multiple modes of heat transfer.
In the heat exchangers we’ll encounter, heat must transfer from one moving uid, through a
solid wall, and to another moving
uid. The solid wall may or may not include extended
surfaces ( ns) to improve the heat transfer rate into one or both of the uids. We therefore
have both conduction and convection heat transfer between the two
uids in a heat
exchanger. In order to calculate a heat transfer rate between them, we use combine these
modes into an overall heat transfer coe cient, U (W/m 2 ⋅ K). We can relate the overall
heat transfer coe cient to the heat transfer rate via,
Q = U ⋅ As ⋅ ΔT
(3.60)
To calculate the overall heat transfer coe cient, we utilize the concepts we developed in our
discussion of thermal resistor networks. Consider the case of a concentric tube, as shown
below.
Figure 3.28. Fluid ow in a concentric tube. Outer uid is hot and inner uid is cold. Right image shows the equivalent resistor network that
describes heat transfer from the hot uid to the cold uid at a speci c point along the length of the heat exchanger.
The resistor network in Fig. 3.28 describes the resistance to heat ow from the hot (outer)
uid to the cold (inner) uid. In this case, heat ows by convection from the outer uid to
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PAGE 300

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Overall Heat Transfer Coe cient
the wall, by conduction through the wall, and nally by convection to the inner uid. The
total thermal resistance is therefore,
Rtot = Rcv,o + Rcd,w + Rcv,i
(3.61)
where the convection resistance is described by,
Rcv =
1
h ⋅ As
and As is the surface area of the part of the tube that a uid comes in contact with,
As = π ⋅ d ⋅ L
In the above expression, L is the length of the tube. Likewise, the conduction resistance
through a wall bounded by concentric cylinders is de ned as,
ln do
( i)
d
Rcd =
2 ⋅ π ⋅ κw ⋅ L
where κw is the thermal conductivity of the wall. Finally, we relate the overall heat transfer
coe cient to the total thermal resistance as,
U ⋅ As =
1
Rtot
(3.62)
In Eqn. 3.62, the product U ⋅ As requires that an area be speci ed if the inner and outer
diameters of the solid wall are di erent. Thus, we can either nd the overall heat transfer
coe cient based on the inner area of the wall (Ui ⋅ As,i) or the outer area of the wall (Uo ⋅ As,o). In
other words,
U ⋅ As = Ui ⋅ As,i = Uo ⋅ As,o
(3.63)
In order to determine the e ectiveness of a heat exchanger, you will need to be able to
compute the overall heat transfer coe cient according to Eqn. 3.62. Note that Eqn. 3.61 is
more complex for the case where
separating the two
ns are attached to either (or both) sides of the wall
uids. When designing a heat exchanger, keep in mind that
ns are
typically used to transfer heat to or from a gas.
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PAG E 3 01
A double-pipe shell-and-tube heat exchanger is constructed with AISI 1010 Carbon Steel
and contains 50 tubes each making a single pass through the heat exchanger. The tubes have
an inner diameter of di = 15 mm and an outer diameter of do = 30 mm. Water ows inside
of the tubes with a mass ow rate of m· w = 0.5 kg/s, while oil ows through the shell side.
The oil on the shell side produces a convection coe cient of ho = 120 W/m 2 ⋅ K . Assume
that the properties of water can be found at an average temperature of 50∘C. The length of
the tube is L = 1 m and the kinematic viscosity of the water is measured to be
5.53 ⋅ 10−7 m 2 /s. Determine the overall heat transfer coe cient based on the inner area of
the tube. For reference, the Nusselt number correlation for turbulent ow through a pipe is,
Nu = 0.023 ⋅ Re 0.8 ⋅ Pr 1/3
and recall that,
Nu =
h⋅d
κf
where d is the diameter of the tube and κf is the thermal conductivity of the uid owing
through it.
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PAGE 302


Example 3.10 - Calculating an Overall Heat Transfer Coe cient
Solution
To calculate the overall heat transfer coe cient based on the inner area of the tube, we use,
Ui ⋅ As,i =
1
Rtot
or,
Ui =
1
Rtot ⋅ As,i
Thus, in order to nd Ui, we must rst compute Rtot as,
Rtot = Rcv,o + Rcd,w + Rcv,i
We determine Rcv,o using the information provided in the problem,
1
Rcv,o =
=
ho ⋅ As,o
120
1
K
= 0.177
W
⋅ π ⋅ 0.015m ⋅ 1m
W
m2 ⋅ K
Likewise, we nd Rcd,w via,
Rcd,w =
ln
( di )
0.03m
ln 0.015m
(
)
do
2 ⋅ π ⋅ κw ⋅ L
=
2 ⋅ π ⋅ 49.8 ⋅ mW⋅ K ⋅ 1m
Finally, we must determine Rcv,i . To do this, we must
requires knowledge of the
= 0.0022
K
W
nd the Nusselt number, which
ow regime (i.e., laminar, mixed, or turbulent
ow). The
ow
regime itself can be determined using the Reynolds number, which is calculated as,
kg
4 ⋅ 0.5 s
V̄ ⋅ d
4 ⋅ m·
ReD =
=
=
= 77,620
N
⋅
s
−4
ν
π⋅d⋅μ
π ⋅ 0.015m ⋅ 5.468 ⋅ 10
m2
Since ReD > 2300 , we clearly have turbulent
ow. Thus, we can use the Nu correlation
provided in the problem statement, where Prw = 3.77.
Nu = 0.023 ⋅ (77,620)0.8 ⋅ (3.77)1/3 = 292.29
Now we can solve for hi as,
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PAGE 303
hi =
Nu ⋅ κf
di
=
292.29 ⋅ 0.64 mW⋅ K
0.015m
W
= 12,471 2
m ⋅K
Thus,
Rcv,i =
1
=
hi ⋅ As,i
12,471
1
W
⋅ π ⋅ 0.03m ⋅ 1m
m2 ⋅ K
= 0.00085
K
W
Now,
Rtot = 0.177
K
K
K
K
+ 0.0022 + 0.00085 = 0.18
W
W
W
W
Finally,
Ui =
1
K
0.18 W
⋅ π ⋅ 0.015m ⋅ 1m
= 117.9
W
m2 ⋅ K
Note that the dominant thermal resistor in this network was the convection resistance on
the shell-side. In order to facilitate a lower thermal resistance, and therefore a higher
overall heat transfer coe cient, one could add ns to the outer surface of the wall.
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PAGE 30 4
In practice, we wish to either: (1) size a heat exchanger (e.g., limit or solve for As ), or (2)
determine a particular heat exchanger’s total capacity (Q). To do this, we turn to a
procedural method that relates heat exchanger e ectiveness to its total capacity, expressed
in terms of the heat exchanger’s “Number of Transfer Units”, or NTU.
Let’s consider a very basic heat exchanger, where two uids are adjacent to each other and
separated by a wall, as shown in the
gure below. In this particular case, the hot
uid is
above the cold uid, but the orientation of the uids is irrelevant to our present discussion.
Figure 3.29. Heat transfer rate, Q, from the hot uid to the cold uid. From the perspective of system A, heat is leaving the system, while from the
·
perspective of system B, heat is entering the system. Note that Qcv is the heat leaving or entering the system to or from the surroundings in this
case, as we have chosen our system such that it crosses its boundaries.
We examine Fig. 3.29 from the perspective of two individual systems. If we consider the hot
uid to be the system, then heat is leaving the system. On the other hand, if we consider the
cold uid to be the system, then heat is entering the system. We can write the heat transfer
rate according to an energy balance. Let’s consider the water in the system to be
incompressible. The energy balance reads,
·
0 = Qcv +
∑
inlets
m· i ⋅ hi −
∑
inlets
m· e ⋅ he
We assumed in this case that there are not changes in potential and kinetic energy. Given
that we have only one inlet and one exit, and that we are assuming that water is
incompressible (Δh = cp ⋅ ΔT), we obtain,
·
Qcv = m· ⋅ cp ⋅ (Te − Ti)
(3.64)
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PAGE 305
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E ectiveness-NTU Method
·
Using the above expression for Qcv, we can write the heat transfer rate from the perspective
of each uid system,
Fluid System A - Hot Fluid (subscript h)
·
Qcv = m· ⋅ cp,h ⋅ (Th,in − Th,out )
(3.65)
Fluid System B - Cold Fluid (subscript c)
·
Qcv = m· ⋅ cp,c ⋅ (Tc,out − Tc,in)
(3.66)
Note that in Eqn. 3.65, the temperature di erence does not re ect the order of the
temperatures shown in Eqn. 3.64. This is because heat leaves system A, and is therefore
negative.
When designing a heat exchanger, it is common not to know all of the temperatures in Eqns.
·
3.65 and 3.66, and we often can not nd the heat transfer rate Qcv (i.e., when designing a
heat exchanger, we are often left with the two equations 3.65 and 3.66, but 3 unknown
values between them). To solve such problems, we de ne the terms heat exchanger
e ectiveness, ε, and the number of transfer units, NTU.
·
Qcv
ε= ·
Qmax
(3.67)
·
where Qmax is the maximum possible heat transfer rate for a particular heat exchanger,
de ned as,
·
Qmax = Cmin ⋅ (Thi − Tc,i)
(3.68)
We de ne Cmin as the minimum capacity rate. Each uid has a capacity rate, C, de ned as,
Ch = m· h ⋅ cp,h
(3.69)
Cc = m· c ⋅ cp,c
(3.70)
for the hot uid, and,
where cp is the speci c heat capacity for each particular uid. Thus, the minimum capacity
rate is de ned as the minimum vale produced by Eqns. 3.69 and 3.70.
Now we de ne the Number of Transfer Units as,
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PAGE 306
U ⋅ As
Cmin
(3.71)
where NTU is a function of the overall heat transfer coe cient. A useful relationship to
remember when computing NTU is Eqn. 3.63, or U ⋅ As = Ui ⋅ As,i = Uo ⋅ As,o.
·
In practical design problems, we either don’t know the capacity of the heat exchanger (Qcv)
or we want to size the heat exchanger (i.e., solve for As ). Let’s call the former Case A and
the latter Case B. In both cases, we will not have enough information to directly solve for
·
Qcv or As directly. Instead, we use well-established relationships between ε and NTU for
speci c types of heat exchangers. These relationships are provided in the gures and tables
below. These relationships are shown as a function of the capacity ratio, cr, de ned as,
Cmax
Cmin
cr =
(3.72)
Figure 3.29. E ectiveness-NTU plots for concentric pipe parallel ow and counter ow heat exchangers.
Additional plots of ε − NTU are provided in the Appendix of this textbook. The charts in
the Appendix have
ner resolution and can better be used to solve practical engineering
problems. Note that cr = 0 when one uid undergoes a phase change!
Below are tables with relationships that are used to calculate heat exchanger e ectiveness.
For the shell-and-tube con guration, we calculate the total e ectiveness of the heat exchanger
by
rst computing the e ectiveness for one shell pass, ε1 . For a one shell pass heat
exchanger, then, ε = ε1 . Additionally, for a single shell pass we use the surface area
for only one shell, and label NTU as NTU1.

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PAGE 307
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NTU =
ε − N TU: Solving for ε when NTU can be calculated with known parameters
Heat Exchanger Type
E ectiveness Relation
Concentric Pipe, Parallel Flow
ε=
1 − ex p[−N T U ⋅ (1 + cr )]
1 + cr
Concentric Pipe, Counter Flow
ε=
1 − ex p[−N T U ⋅ (1 − cr )]
1 − cr ⋅ ex p[−N T U ⋅ (1 − cr )]
Shell-and-tube, One shell pass (2,4,… tube passes)
ε1 = 2 ⋅ 1 + cr + (1 + cr )
(
Shell-and-tube, n shell passes (2n, 4n,… tube passes)
ε=
Cross- ow (single pass): Both uids unmixed
1/2
1 + ex p[−N T U1 ⋅ (1 + cr2 )1/2]
⋅
1 − ex p[−N T U1 ⋅ (1 + cr2 )1/2] )
−1
1 − ε1 ⋅ cr
1 − ε1 ⋅ cr
−1 ⋅
− cr
[( 1 − ε1 )
] [( 1 − ε1 )
]
n
ε = 1 − ex p
n
−1
1
⋅ (N T U )0.22 ⋅ (ex p[−cr ⋅ (N T U )0.78] − 1)
[( cr )
]
1
⋅ 1 − ex p[−cr ⋅ (1 − ex p(−N T U ))])
( cr ) (
Cross- ow (single pass): Cma x (mixed), Cmin (unmixed)
ε=
Cross- ow (single pass): Cma x (unmixed), Cmin (mixed)
ε = 1 − ex p(−cr−1 ⋅ (1 − ex p[−cr ⋅ (N T U )]))
All Heat Exchangers when cr = 0
ε = 1 − ex p(−N T U )
Below is a table to determine the Number of Transfer units based on a computation of ε.
Table 3.3. E ectiveness-NTU Equations.
ε − N TU: Solving for NTU (or As) when ε can be calculated with known parameters
Heat Exchanger Type
Concentric Pipe, Parallel Flow
E ectiveness Relation
NTU = −
ln[1 − ε ⋅ (1 + cr )]
1 + cr
NTU =
1
ε −1
⋅ ln
( ε ⋅ cr − 1 )
cr − 1
NTU =
ε
1−ε
Concentric Pipe, Counter Flow
Shell-and-tube, One shell pass (2,4,… tube passes)
E=
(cr < 1)
(cr = 1)
(N T U )1 = − (1 + cr2 )−1/2 ⋅ ln
E−1
where,
(E+1)
2
− (1 + cr )
ε1
(1 + cr2 )1/2
N T U = n ⋅ (N T U )1 where,
Shell-and-tube, n shell passes (2n, 4n,… tube passes)
F−1
ε1 =
F − cr
ε ⋅ cr − 1
and F =
( ε −1 )
1/n
Cross- ow (single pass): Cma x (mixed), Cmin (unmixed)
1
N T U = − ln 1 +
⋅ ln(1 − ε ⋅ cr )
[
( cr )
]
Cross- ow (single pass): Cma x (unmixed), Cmin (mixed)
ε =−
All Heat Exchangers when cr = 0
1
⋅ ln[cr ⋅ ln(1 − ε) + 1]
( cr )
N T U = − ln(1 − ε)







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Table 3.2. E ectiveness-NTU Equations.
Hot engine oil requires cooling in a shell-and-tube heat exchanger that contains one shell
and 2 tubes. The tubes have very thin walls with an internal diameter of di = 2 cm. The
length of each tube pass in the heat exchanger is L = 26 m and the overall heat transfer
coe cient based on the inner part of the tube wall is Ui = 310 W/m2⋅K. Cooling water
ows through the tubes while oil ows through the shell. The cooling water ows through
the tube with a mass
ow rate of m· tube = 0.5 kg/s and oil
ows through the shell with a
mass ow rate of 0.4 kg/s. Determine the outlet temperatures for the water and the oil. The
water and oil enter at temperatures of 23∘C and 157∘C, respectively.
Solution
In principle, the outlet temperatures can be determined using the energy equation for either
of the water or oil systems,
·
Qcv,water = Cw ⋅ (Tw,e − Tw,i)
and,
·
Qcv,oil = Coil ⋅ (Toil,i − Toil,e)
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PAGE 309
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Example 3.11 - E ectiveness-NTU: Determining Outlet Temperatures (CASE A)
·
determine the e ectiveness of this heat exchanger,
·
Qcv
ε= ·
Qmax
·
which can be rearranged to solve for Qcv as,
·
Qcv = ε ⋅ Cmin ⋅ (Toil,i − Tw,i)
Thus, we must determine the e ectiveness of the heat exchanger using an ε − NTU
relation. For a one shell pass (2,4,… tube pass) heat exchanger, this relationship is de ned
as,
1 + exp[−NTU1 ⋅ (1 + cr2)1/2]
1/2
ε1 = 2 ⋅ 1 + cr + (1 + cr ) ⋅
(
1 − exp[−NTU1 ⋅ (1 + cr2)1/2] )
−1
where ε = ε1 because there is only one shell pass. To compute the e ectiveness, then, we
must determine cr and NTU. Here, cr is determined by solving for the capacity rate of each
uid,
kg
J
W
kW
Cw = m· water ⋅ cp,water = 0.5 ⋅ 4182
= 2091 = 2.091
s
kg ⋅ K
K
K
and,
kg
J
W
kW
Coil = m· oil ⋅ cp,oil = 0.4
⋅ 2471
= 988.4 = 0.988
s
kg ⋅ K
K
K
Thus, Cmin = Coil = 988.4 W/K and Cmax = Cwater = 2091 W/K. Now,
988.4 W
Cmin
K
cr =
=
= 0.47
W
Cmax
2091
K
Finally, we calculate NTU as,
NTU =
U ⋅ As
Cmin
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PAG E 310

fl
·
However, we do not know either of the exit temperatures or Qcv . To solve for Qcv we must
Here, we must calculate U ⋅ As . Since we are given an inner diameter, we will actually
compute Ui ⋅ As,i. The problem provides us with Ui and thus,
Ui ⋅ As,i = 310
W
W
⋅
2
⋅
π
⋅
0.02m
⋅
26m
=
1012.8
m2 ⋅ K
K
We included the factor 2 in the expression for the surface area because there are 2 tube
passes in this system. Now,
NTU =
Ui ⋅ As,i
Cmin
=
1012.8 W
K
988.4 W
K
= 1.02
We can substitute both cr and NTU into the expression for ε to yield,
2 1/2
2 1/2 1 + exp[−1.02 ⋅ (1 + 0.47 )
ε = 2 ⋅ 1 + 0.47 + (1 + 0.47 ) ⋅
[
1 − exp[−1.02 ⋅ (1 + 0.472)1/2 ]
−1
= 0.55
Using ε, we nd that,
W
·
·
Qcv = ε ⋅ Qcv,max = 0.55 ⋅ 988.4 ⋅ (157∘C − 23∘C) = 72,845 W
K
·
Using Qcv, we solve for Toil,e as,
·
Qcv
72,845 W
∘
Toil,e = Toil,i −
= 157∘C −
=
83.3
C
W
Coil
988.4
K
and for Twater,e as,
·
Qcv
72,845W
∘
∘
Twater,e = Twater,i +
= 23 C +
=
57.84
C
W
Cwater
2,091
K










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P A G E 3 11
Example 3.12 - E ectiveness-NTU: Sizing a Heat Exchanger (CASE B)
Let’s imagine that in problem 3.11, the oil temperature at the exit was required to be 70∘C,
and we wanted to know what the size of the heat exchanger should be to achieve the
proposed temperature drop for the oil. If all other parameters remain the same (and we no
longer know the exit temperature of the water), solve for L.
Solution
In this problem, we no longer know the e ectiveness as NTU will change. Instead, we solve
·
·
for e ectiveness using Qcv and Qcv,max,
·
Qcv
Coil ⋅ (Toil,i − Toil,e)
∘
∘
988.4 W
⋅
(157
C
−
70
C)
K
ε= ·
=
=
= 0.65
W
∘
∘
Cmin ⋅ (Toil,i − Twater,i)
Qcv,max
988.4 ⋅ (157 C − 23 C )
K
Thus, the required e ectiveness in order to achieve an oil exit temperature of 70∘ C is 0.65.
Now we can solve for the Number of Transfer Units using an ε − NTU relation.
NTU = − (1 + cr2)−1/2 ⋅ ln
E−1
(E + 1)
where,



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PA G E 312
2/ε − (1 + cr )
2/0.65 − (1 + 0.47)
=
= 1.45
2
1/2
2
1/2
(1 + cr )
(1 + 0.47 )
E=
Therefore,
NTU = (−1 + 0.472)−1/2 ⋅ ln
1.45 − 1
= 1.53
( 1.45 + 1 )
Now we can use NTU to solve for As, and ultimately L.
As = π ⋅ di ⋅ L ⋅ n =
NTU ⋅ Cmin
Ui
where n is the number of tube passes. Thus,
W
1.53 ⋅ 988.4 K
NTU ⋅ Cmin
L=
=
= 38.82 m
W
Ui ⋅ π ⋅ di ⋅ n
310
⋅ π ⋅ 0.02m ⋅ 2
m2 ⋅ K






PAG E 313
I N T RO D U C T I O N T O T H E R M O DY N A M I C C YC L E S
Thus far, we have discussed two important topics:
1. Closed system processes with moving boundaries and,
2. Open system components (pumps, turbines, heat exchangers, etc.)
We will now use these concepts to develop and analyze closed system cycles (piston-cylinderbased engines) and open system cycles (power and refrigeration systems).
Using the 1st Law of Thermodynamics will allow us to analyze the performance of an actual
cycle, while the 2nd Law of Thermodynamics will provide us with a means to compare the
actual performance of each system with the maximum possible e ciency of that system.
At this point, you might be wondering what we mean by the term “cycle”. While we may
have learned that an expansion process produces power, the process alone does not produce power
continuously. Let’s look at where the piston ends up after an expansion,
Figure 3.30. Expansion process in a piston-cylinder apparatus. After expansion, the piston rests near the top of the cylinder.
In order to repeat the expansion, the gas must eventually be compressed. However, we can
not repeat the same expansion process without
rst cooling the gas and returning to an
ambient condition. And, we must heat the gas to expand it in the rst place. These steps can
be put together to form the basis of a simple power cycle (using an ideal gas) as follows,
1. Isobaric compression (p = constant, V = ↓ )
2. Isometric heat addition (V = constant, p = ↑ )
3. Isobaric expansion (p = constant, V = ↓ )
4. Isometric heat removal (V = constant, p = ↓ )
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PAG E 314
This manifests in a series of process that appear in the following gure.
Figure 3.31. A series of thermodynamic processes that make up a cycle, which begins and ends at the same state to achieve continuous net power.
In the series of processes depicted in Fig. 3.31, we achieve a cycle, in which we begin and end
the series of processes at the same state point (state point 1). We also note that the area
bounded by this series of processes is the net work done by the cycle. While it takes some
work to compress the gas (i.e., the area under process 1-2), we gain much more energy from
the expansion process (i.e., the area bounded by all four processes). However, we do not
get this net energy for free! We do have to heat the substance with a fuel, which has some
cost associated with it. Likewise, we can not start the process over without discharging heat
(i.e., to get from state point 4 to state point 1). Thermodynamically, this implies that we
must be in contact with two thermal reservoirs: one hot thermal reservoir and one
cold thermal reservoir, as shown in the gure below.
Figure 3.32. Schematic of a power system in contact with two thermal reservoirs (hot: top, and cold: bottom) with energy transfer rates.
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PAG E 315
course, one can achieve some net power using an expansion process, but without releasing
the heat to a cold reservoir, you can not maintain a cycle.
If we analyze the closed system in Fig. 3.32 using a 1st Law analysis, we nd that,
·
·
·
·
ΔUcycle = Qh − Qc − Wcycle
·
Since the cycle begins and ends at the same state then ΔUcycle = 0 and,
·
·
·
Wcycle = Qh − Qc
Thus, the net power produced by the cycle can found by computing the di erence between
·
·
Qh and Qc. This is applicable to both closed and open system cycles.
From our construction of the 1st Law for a cycle, we can also obtain a system’s e ciency, ηth.
We generally de ne e ciency as,
get
ηth =
pay
or what you “get” from the cycle versus what you “pay” to get it. For a power cycle, this
amounts to,
·
·
·
Wcycle
Qh − Qc
get
ηth =
= ·
=
·
pay
Qh
Qh
(3.73)
where we “pay” with the rate of heat we put into the system to achieve some useful power
out.
There are two other types of cycles that we will examine in detail in this course, including
refrigeration and heat pump cycles. A schematic and the corresponding performance
metric for each system is provided below.
Refrigeration Cycle
A refrigeration system is often referred to as a backward heat engine, because we must
supply power to remove heat from the cold reservoir and transfer it into the hot reservoir.
While it seems counterintuitive, our means to achieve this are actually quite practical and
will be described in much greater detail toward the end of this course. For now, we restrict
our attention to the operation of a refrigeration system as a cycle.

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PAG E 316


Figure 3.32 depicts a simple schematic that shows what’s required to run a power cycle. Of
Figure 3.33. Schematic of a refrigeration system in contact with two thermal reservoirs (hot: top, and cold: bottom); we are trying to remove heat
from the cold reservoir to maintain it at some temperature, Tc.
A 1st Law analysis likewise provides us with,
·
·
·
·
ΔUcycle = Qc − Qh − (− Wcycle)
·
Again, ΔUcycle = 0 and,
·
·
·
Wcycle = Qh − Qc
For a refrigeration system, performance is de ned by a metric called “Coe cient of
Performance”,
·
·
Qc
Qc
get
COPR =
= ·
= ·
·
pay
Wcycle
Qh − Qc
(3.74)
Heat Pump Cycle
A heat pump uses the same working principle as a refrigerator,
Figure 3.34. Schematic of a heat pump system in contact with two thermal reservoirs (hot: top, and cold: bottom); we are trying to put heat into
the hot reservoir to maintain it at some temperature, Th.
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PA G E 317
As shown in Fig. 3.34, the point of the heat pump system is to maintain the hot reservoir at
some temperature, Th . Note that the basic operating principle remains the same as that
shown in Fig. 3.33.
Because the direction of each energy transport mechanism is identical to Fig. 3.33, the 1st
Law analysis remains the identical as well. The only signi cant di erence is our calculation
of the Coe cient of Performance, which reads as,
·
·
Qh
Qh
get
COPHP =
= ·
= ·
·
pay
Wcycle
Qh − Qc
(3.75)
S E C O N D L AW O F T H E R M O DY N A M I C S
The second law of thermodynamics is of fundamental importance to our characterization of
the performance of thermodynamic cycles. In particular, we will see that the 2nd Law
provides us with an understanding of irreversibility, which limits the maximum possible
performance we can achieve for a given cycle.
Carnot Cycle
Here we introduce the concept of a maximum theoretical e ciency and use the Carnot cycle to
provide context for the limitations of performance governed by the 2nd Law.
The Carnot cycle describes a power cycle that is completely reversible when in contact with
thermal reservoirs at two di erent temperatures (Th and Tc ). When we refer to a completely
reversible process, we are referring to an idealized process - e.g., a process in which everything
will return to its initial state. One very simple example of a completely reversible process is
a pendulum that swings and returns to its initial position, as shown in the gure below.
Figure 3.35. Schematic of a pendulum swinging; release occurs in state 1, swings out to state 2, and returns to state 3, which has the same position
as state 1.


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PAG E 318
In the case of the swinging pendulum, the absence of friction is required for the pendulum
to reach State 1 upon its return from State 2. Important to note is that the pendulum can
not swing beyond its initial position in State 1 or its nal position in State 2 without some
additional external force. Thus, an irreversible process represents the best possible
performance of a system.
When a power cycle operates between two reservoirs at Th and Tc , its performance is also
governed by these two bounding states. Thus, for a perfectly reversible cycle, we can say,
·
Qc
Tc
=
( Q· h )
Th
rev
(3.76)
which requires that there be no heat loss between the reservoirs and the system.
This relationship can be used to de ne a maximum possible e ciency, which we call the
Carnot E ciency, ηmax,
Tc
ηmax = 1 −
Th
(3.77)
Equation 3.76 can also be used to determine the maximum possible coe cient of
performance for both a refrigeration and heat pump cycle as,
COPR,max = T
h
Tc
1
(3.78)
−1
and,
1
COPHP,max =
(3.79)
Tc
1− T
h
Entropy and the Clausius Inequality
An irreversible process is one in which there is no change in entropy. Thus, it is useful to frame
the above discussion with respect to entropy. This will likewise inform our discussion of the
maximum e ciency of individual thermodynamic systems, like compressors, pumps, and
turbines.
Entropy is a quantitative measure of the state of disorder in a system. Practically, it
represents the amount of a system’s thermal energy that can not be converted into
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PAG E 319
mechanical work during an individual process. Consequently, the entropy of a system can
never decrease between state points. It can either (1) remain the same in a completely
reversible process, or (2) increase due to an irreversibility (e.g., heat loss, friction, etc.).
Such a statement ( rst stated by German physicist Rudolf Clausius, and now termed the
Clausius Inequality) allows us to determine whether a system is reversible, irreversible, or
impossible. The Clausius Inequality states,
δQ
dS =
≤0
∮ T
(3.80)
where the integral symbol represents a “cyclic integral”, which means that one must
integrate across the entire cycle. If the evaluation of Eqn. 3.80 produces a value greater than
0, it is impossible to generate the power or cooling loads indicated. If instead the result is
equal to 0, then the process is internally reversible. Finally, if the value is less than 0, then
the process is irreversible. Below is an example to demonstrate the way in which we
evaluate Eqn. 3.80 for the thermodynamic cycles we will consider in this course.
Example A
A power cycle is connected to two thermal reservoirs. The hot thermal reservoir has a
temperature of Th = 500 K and the cold thermal reservoir has a temperature of Tc = 300 K.
An inventor claims that she can operate a power cycle with corresponding heat transfer rates
of Qh = 1,000 kJ and Qc = 590 kJ. Is this cycle reversible, irreversible, or impossible?
Solution
To evaluate the integral in Eqn. 3.80, and assuming the thermal reservoirs are at a constant
temperature, we nd,
1,000 kJ −590 kJ
+
= 0.033 > 0
500 K
300 K
Thus, this cycle is impossible.
Example B
For the problem in Example A, evaluate whether the cycle is possible if it is instead used as a
refrigerator.




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PAGE 320
Solution
Now that we have a refrigeration system, the direction of heat
ow into and out of the
system is opposite that of a power cycle. Considering this, our expression becomes,
−1,000 kJ 590 kJ
+
= − 0.033 < 0
500 K
300 K
Thus, this cycle is irreversible.
Finding Entropy
In order to fully analyze the performance of thermodynamic cycles, we will also need to use
entropy. It is therefore useful to explain how we
nd entropy for di erent types of
substances.
Ideal Gases
In this course, we will evaluate all thermodynamic properties assuming constant speci c
heats. The assumption that the speci c heat is not a function of temperature as an ideal gas
changes states typically holds so long as the temperature di erence between the two states
is not greater than a few hundred degrees. The expression used to nd the entropy change
between states is then,
T
v
s2 − s1 = cv,avg ⋅ ln 2 + R ⋅ ln 2
( T1 )
( v1 )
(3.81)
or, given the relations between cp, cv, and R, we can rewrite Eqn. 3.81 as,
T
p
s2 − s1 = cp,avg ⋅ ln 2 − R ⋅ ln 2
( T1 )
( p1 )
(3.82)
where cv,avg and cp,avg are found at the average temperatures between state points. Although
we will focus exclusively on cases for which the constant speci c heat method will produce
reasonable results, practicing engineers should utilize a variable speci c heat approach to
analyze and design thermodynamic systems.
Pure Substances
The entropy of a pure substance can be found according to the same concepts that govern
speci c volume and internal energy.
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PAG E 3 21
If the substance is determined to be a saturated liquid, the entropy can be found according
to,
s(T, p) = sf (T )
(3.83)
If instead the substance is found to be a saturated mixture, then,
s = sf + χ ⋅ (sg − sf )
(3.84)
Finally, if the substance is determined to be a superheated vapor, then one must use the
superheated vapor tables in the usual way.
Isentropic Processes
An isentropic process is one in which the entropy does not change from one state to the
next; in other words, the process is reversible. As with the Carnot cycle, an isentropic
process represents a best case scenario. This is particularly useful for de ning individual
device e ciencies, which have isentropic e ciencies. As we will see, isentropic e ciencies
relate an actual change in entropy to an isentropic change in entropy.
For the moment, we will restrict our discussion of helpful relationships that are the
consequence isentropic processes.
Ideal Gases
When the constant speci c heat assumption is valid, a variety of useful relationships can be
used when a process is isentropic (i.e., Δs = 0). In this case, Eqn. 3.81 becomes,
T
v
R
ln 2 = −
⋅ ln 2
( T1 )
( v1 )
cv,avg
which is equivalent to,
T
v
ln 2 = ln 2
( T1 )
( v1 )
R
cv,avg
Knowing that R = cp − cv and k = cp/cv, R/cv = k − 1. Now,
v1
=
( T1 ) ( v2 )
T2,s
k−1
(3.85)
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PAGE 322
where the subscript s in T2,s is used to indicate that the process from state point 1 to
state point 2 is isentropic.
One can also rearrange Eqn. 3.82 with Δs = 0 as,
( T1 )
T2,s
=
p2
( p1 )
k−1
k
(3.86)
These two equations can also be set equal to one another to produce a relationship between
pressure and volume,
v1
=
( p1 ) ( v2 )
p2,s
k
(3.87)
Pure Substances
It is useful to examine an isentropic process for a pure substance on a T − s (temperatureentropy) diagram.
Figure 3.36. Temperature-entropy diagram for water showing a constant entropy process between states 1 and 2, where we refer to state point 2 as
state point “2s” to indicate that the process is isentropic. Dashed-dotted black line is the vapor dome.






PAGE 323
Figure 3.36 shows an isentropic process between two arbitrarily chosen state points. The
second state point is labeled with the subscript “s” to indicate that the process is isentropic.
With knowledge that a process is isentropic, one can determine unknown properties
knowing that s2s = s1.
Isentropic E ciencies
We can also use isentropic processes to nd the true e ciency of engineering devices (e.g.,
pumps, compressors, turbines, etc.). To do this, recall that our de nition of a reversible
process is one that represents the optimal performance, or when Δs = 0 . Let us look at a
Mollier (h-s) diagram for water as an example,
Figure 3.37. Enthalpy-entropy diagram for water showing a constant entropy process between states 1 and 2s and an irreversible process between
states 1 and 2. Shown for a pressure drop.
Shown in Fig. 3.37 are both a constant entropy process across a particular pressure drop
(i.e., from state point 1 to state point 2s between pressures of 10 MPa and 1 MPa) and an
irreversible process between state points 1 and 2 across the same pressure drop.
The above process might describe the pressure drop through a turbine, which produces
power. In its simplest form, the energy equation for a turbine can be expressed as,
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PAGE 32 4
assuming that there kinetic energy changes are negligible and that the turbine is wellinsulated. If we examine the isentropic process for this pressure drop, we
nd that its
change in enthalpy is greater than the change in enthalpy for an irreversible process (i.e.,
Δhs > Δh). Therefore, the isentropic process yields the largest possible power we might
expect under these conditions.
Turbines
The concept described above helps us to de ne an isentropic e ciency for a turbine, which we
de ne as,
·
WT
h − h2
η= ·
= 1
h1 − h2s
WT,s
·
(3.88)
·
where WT is the actual power produces by the turbine, and WT,s is the power that the turbine
would produce if its operating conditions were isentropic.
Compressors and Pumps
As compressors and pumps are designed to increase the pressure of a uid from inlet to exit,
the state points in Fig. 3.37 must be revered, as shown in the gure below.
In this gure, the change in entropy for the irreversible process is now greater than that for
the isentropic process. This should make sense based on our de nition of power. Again, in
the simplest case, we can express the power required by a pump or compressor as,
·
WT = m· ⋅ (Δh)
In this case, though, we wish to minimize the power required to operate a pump or
compressor. As a result, the isentropic process again represents our optimal case.
Our isentropic e ciency can therefore be de ned as,
·
WT,s
h − h2s
η= · = 1
h1 − h2
WT
(3.89)
For a pump, we can express the isentropic process as,
h1 − h2s = νf (T ) ⋅ (p1 − p2)
(3.90)
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PAGE 325
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·
WT = m· ⋅ (Δh)
Figure 3.38. Enthalpy-entropy diagram for water showing a constant entropy process between states 1 and 2s and an irreversible process between
states 1 and 2. Shown for an increase in pressure.
Nozzles
Nozzles are generally very e cient, often surpassing a 95% isentropic e ciency. Nozzle
e ciencies are based on the relationship between the actual and isentropic kinetic energies
at the nozzle exit, and can be calculated using,
η=
V̄ 22
(3.91)
V̄ 22s
The nozzle e ciency can also be expressed in terms of enthalpies as,
η=
h1 − h2
h1 − h2s
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PAGE 326
Steam enters an adiabatic turbine at 1100∘F and 1000 psia and exits as a saturated vapor at 5
psia. What is the isentropic e ciency of the turbine?
Solution
The isentropic e ciency of a turbine can be expressed as,
·
WT
h − h2
η= ·
= 1
h1 − h2s
WT,s
In this problem, we are told that steam is our substance. We consider steam to be a pure
substance and must therefore use property tables to nd the values of enthalpy required for
us to compute the turbine’s isentropic e ciency.
We begin with the inlet, which has a temperature and pressure of 1100∘F and 1000 psia,
respectively. First, we must determine the phase of the water at the inlet, which required we
use the saturation tables for water. We use the saturated pressure table and locate a pressure
of 1000 psia.
Because Tgiven = 1100∘F > Tsat = 544.65∘F at pgiven = 1000 psia, we know that we have a
superheated vapor. Thus, we must use our superheated vapor tables to obtain h1.
Btu
From the superheated vapor table, we obtain h1 = 1563.1
.
lbm
Now we must nd h2s and h2. In the problem statement, we are provided with information
about the actual conditions at state point 2. We can simply use the saturated pressure table
to determine h2 and locate the saturated vapor condition. However, we will focus on nding
h2s rst.
To determine h2s , we know that p2s = p2 = 5 psia and that s2s = s1 . From the superheated
vapor table, we nd that s2s = s1 = 1.6911
Btu
.
lbm ⋅ R
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PAGE 327
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Example 3.13 - Isentropic E ciency of a Steam Turbine
We now examine the saturated pressure table at p2s = p2 = 5 psia and determine the phase
of state 2s,
Btu
is between sf and sg . Thus, we
lbm ⋅ R
have a saturated mixture. To determine h2s, then, we must nd χ2s,
We nd that at p2s = p2 = 5 psia, s2s = s1 = 1.6911



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PAGE 328
sg − sf
=
Btu
Btu
1.6911 lbm
−
0.23488
⋅R
lbm ⋅ R
Btu
Btu
1.8438 lbm ⋅ R − 0.23488 lbm ⋅ R
= 0.905
and h2s can be calculated as,
Btu
Btu
Btu
Btu
h2s = hf + χ2s ⋅ (hg − hf ) = 130.18
+ 0.905 ⋅ 1130.7
− 130.18
= 1035.7
(
lbm
lbm
lbm )
lbm
Finally, h2 can be found directly from the saturated pressure table. We are told that we have
Btu
a saturated vapor, and h2 is therefore 1130.7
.
lbm
Now we calculate the turbine e ciency as,
Btu
Btu
·
1563.1 lbm − 1130.7 lbm
WT
h1 − h2
η= ·
=
=
= 0.82 = 82 %
Btu
Btu
h1 − h2s
WT,s
1563.1
− 1035.7
lbm
lbm

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PAGE 329


χ2s =
s2s − sf
Saturated liquid water enters a 90% e cient pump at p1 = 200 kPa and discharges at
kg
p2 = 4000 kPa. If the mass ow rate of the water owing through the pump is m· = 1.2 ,
s
what is the power required to operate the pump?
Solution
Ultimately, we must nd the power required to operate the pump, which can be expressed
as,
·
Wp = m· ⋅ (h1 − h2)
We do not have enough information at state 2 to determine h2 . However, we are given the
pump’s isentropic e ciency, which does contain h2.
The isentropic e ciency of a pump can be written as,
·
ν1, f (T ) ⋅ (p1 − p2)
WT,s
h1 − h2s
η= · =
=
h
−
h
h1 − h2
WT
1
2
In this problem, we must
nd enthalpies h1 and h2 . We know that state 1 is a saturated
liquid, so that h1 = hf. Since we are provided with p1, we use the saturated pressure table to
nd h1.
kJ
Therefore, h1 = hf = 504.71
. To nd h2 we can rearrange the expression for isentropic
kg
e ciency and solve for it directly as,
ν1, f (T ) ⋅ (p1 − p2)
h2 = h1 −
η
= 504.71
kJ
−
kg
3
0.00106 mkg ⋅ (200 kPa − 4000 kPa)
0.9
⋅
1 kJ
kJ
=
509.18
1 kPa ⋅ m 3
kg




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PAGE 330
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Example 3.14 - Isentropic E ciency of a Pump
Note that the speci c volume was taken from the saturated temperature table (not shown
here) at T1 = 120.21∘C (which is the saturation temperature; in this case, the uid enters as
a saturated liquid, so it must be at the saturation temperature).
Finally, the power to operate the pump is calculated to be,
kg
kJ
kJ
·
Wp = 1.2
⋅ 504.71
− 509.18
= − 5.36 kW
s (
kg
kg )
Note here that the power is negative because the pump requires power to be put in to the
system.
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PAG E 3 31
C H A P T E R 4 : T H E R M O DY N A M I C
SYSTEMS
T I S A R G UA B L E W H E T H E R T H E H U M A N R AC E H AV E B E E N GA I N E R S BY T H E
M A RC H O F S C I E N C E B E YO N D T H E S T E A M E N G I N E . E L E C T R I C I T Y O P E N S A
FIELD
THEY
OF
M AY
INFINITE
WELL
CONVENIENCES
H AV E
TO
PAY
TO
D E A R LY
EVER
FOR
GREATER
THEM.
BUT
NUMBERS,
ANYHOW
BUT
IN
MY
THOUGHT I STOP SHORT OF THE INTERNAL COMBUSTION ENGINE WHICH
HAS MADE THE WORLD SO MUCH SMALLER. STILL MORE MUST WE FEAR THE
CONSEQUENCES
FROM
THEIR
OF
ENTRUSTING
PREDECESSORS
OF
A
THE
HUMAN
RACE
SO-CALLED
SO
LITTLE
BARBAROUS
DIFFERENT
AGES
SUCH
AWFUL AGENCIES AS THE ATOMIC BOMB. GIVE ME THE HORSE.
- S I R W I N S T O N C H U R C H I L L , A D D R E S S T O T H E R OYA L C O L L E G E O F
S U R G E O N S ( 19 5 2 )
Unknown authorUnknown author, Public domain, via Wikimedia Commons, Downloaded from: https://commons.wikimedia.org/wiki/File:PSM_V18_D500_An_american_internal_combustion_otto_engine.jpg

PAGE 332
INTRODUCTION TO COMBUSTION PROCESSES
There is plenty of controversy surrounding the development of the internal combustion
engine, particularly with respect to its impact on the environment. Nevertheless, it has
“made the world so much smaller”, as Churchill mentions, and has therefore provided us
with an opportunity to engage with other nations that might lack access to modern
healthcare, clean drinking water, and the like. While we will not discuss the ethics or virtues
of the modern internal combustion engine beyond what we’ve already identi ed here,
students are encouraged to keep these issues in mind while reading through the remainder
of this textbook. In particular, pay careful attention to the chemical reaction processes that
are described in the subsequent section, as their impact on the environment is currently a
major subject of climatology.
Fuels
Combustion is an oxidation reduction reaction where a fuel is broken down into
combustible elements that react with the oxygen and release chemical energy,
Fuel + Oxidizer → Products
(4.1)
The typical fuels that we use are hydrocarbons, which contain hydrogen and carbon as the
combustible elements. In a hydrocarbon, the carbon atom forms four covalent bonds with
either another carbon atom or a hydrogen atom. Hydrocarbons can be either saturated,
meaning that it contains the maximum possible amount of hydrogen, or unsaturated.
Ethane C H is an example of a saturated hydrocarbon, and acetylene C H is an example of
2
6
2
2
an unsaturated hydrocarbon due to the multiple bonds between the carbon atoms.
Figure 4.1. Diagrams of ethane and acetylene, two hydrocarbons that are saturated (ethane) or unsaturated (acetylene)
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PAGE 333
Most common fuels are actually a mixture of many di erent hydrocarbons. Natural gas is
principally methane, CH , ~95%; but it also contains ethane (C H ), propane (C H ), butane
4
2
6
3
8
(C H ) and other gaseous hydrocarbons. The composition of natural gas varies depending on
4
10
the source, but, for example it might be represented as C H , meaning that on average
1.1
3.8
there are 1.1 carbon atoms and 3.8 hydrogen atoms in each molecule of fuel.
Combustion Air
While it is possible to burn a fuel with pure oxygen, this is only done in special applications
that are trying to achieve very high temperatures such as cutting and welding. The classic
example of combustion with pure oxygen is an oxyacetylene torch. The thermodynamic
engines that we have discussed in this course use air as their oxygen source.
Air is a mixture of gases that primarily contains diatomic oxygen, O , and diatomic nitrogen,
2
N . Since oxygen and nitrogen can be treated as ideal gases near room temperature, it
2
follows that a mixture of ideal gases can also be treated as an ideal gas. For simplicity, all of
the gases other than oxygen are lumped in with the nitrogen, and we assume that air is 21%
O and 79% N . These percentages are mole fractions, and can be thought of as the
2
2
percentage of the volume or pressure that is contributed by each gas.
yO2 =
yN2 =
NO2
Ntot
NN2
Ntot
=
1 kmol
= 0.21
4.76 kmol
=
3.76 kmol
= 0.79
4.76 kmol
Based on the assumption that air is 21% oxygen and 79% nitrogen, it is evident that in
order to get 1 kmol of O it is necessary to bring in 4.76 kmol of Air.
2
4.76 kmol Air = 1 kmol O2 + 3.76 kmol N2
(4.2)
Combustion Reaction
Fuel and air are the reactants of the combustion reaction, while the products depend on the
relative proportions of the reactants along with the kinetics of the reaction. Complete
combustion occurs when the combustible elements are completely oxidized; all of the C
becomes CO and all of the H becomes H O. Complete combustion requires a certain
2
2
amount of oxygen, and can only occur when there is su cient oxygen present. If there is
just enough oxygen present to completely combust all the carbon and all the hydrogen, that
reaction is referred to as stoichiometric or theoretical combustion. The products of

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PAGE 33 4
stoichiometric combustion will be carbon dioxide, CO ; water, H O; and nitrogen, N . The
2
2
2
combustion reaction is balanced on a molar basis, and while the N does not participate in
2
the reaction it is typically included in the reaction since the N is present and will absorb
2
energy and in uence the product temperature.
Theoretical/Stoichiometric Combustion of Methane
CH4 + 2 ⋅ (O2 + 3.76 ⋅ N2) → CO2 + 2 ⋅ H2O + 7.52 ⋅ N2
Complete combustion becomes more probable when there is excess air. The products for
complete combustion with excess air include CO , H O, O and N
2
2
2
2.
Complete Combustion of Methane with Excess Air (n > 2)
CH4 + n ⋅ (O2 + 3.76 ⋅ N2) → CO2 + 2 ⋅ H2O + (n − 2) ⋅ O2 + (n ⋅ 3.76) ⋅ N2
If there is insu cient air, incomplete combustion occurs and the combustible elements are
not completely oxidized. The products of incomplete combustion are likely to include
carbon monoxide, CO, and unburnt hydrocarbons.
Air-Fuel Ratio
The air fuel ratio is one method for reporting the proportion of air and fuel in the
combustion reaction. Chemical reactions are balanced on a molar basis, which lends to the
calculation of a molar air fuel ratio:
Nair
AFmolar =
Nfuel
Fuel consumption and the
(4.3)
ow rate of air are usually measured as mass
ow rates. To
convert a molar air fuel ratio to a mass air fuel ratio requires the molar mass of the air and
the fuel,
AFmass = AFmolar ⋅
Mair
Mfuel
(4.4)
The amount of air can also be reported as the percent theoretical air, where 100% theoretical
would represent the air required for stoichiometric combustion.
% TA =
AF
(4.5)
AFstoichiometric
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PAGE 335
Heating Values
The heating value is essentially the chemical energy that is released by the reaction.
However, since at least some of the energy leaves the reaction as enthalpy in the products,
the phase of the products, speci cally the water, can in uence the heating value. Assuming
that the reactants enter at standard temperature and pressure (298 K and 101.3 kPa - or 1
atm), complete combustion occurs, and the products leave at standard temperature and
pressure, the heating value is the amount of heat removed in the combustion chamber. If the
heat is not removed, the thermal energy would leave as enthalpy in the products resulting in
an elevated product temperature.
Figure 4.2. Combustion event causes heating, which is subsequently removed from the system and used to drive a power cycle.
The Lower Heating Value (LHV) assumes that the water leaves as a vapor, and the Higher
Heating Value (HHV) assumes the water leaves as a liquid. The LHV is typically used for
rating engine performance, while the HHV is typically used to determine furnace or boiler
e ciencies. Rarely does the water in the combustion products leave an engine as a liquid.
However, high e ciency furnaces in fact condense some of the water vapor. Draining the
condensate that occurs in the exhaust is an important part of the design. Table I provides
heating values for common fuels.
When using the air standard assumptions, the combustion process is removed and
notionally replaced by heat addition. The amount of heat addition can be determined based
on the heating value, typically the lower heating value, and the mass ow rate of the fuel.
Since the air standard assumption assumes that the air is heated, the mass ow rate of air is
the same at the inlet and exit. The mass ow rate of the fuel is therefore not included in the
exit mass
ow rate. The higher the air fuel ratio the more reasonable this assumption
becomes.
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PAGE 336
We can model the combustion process as a heat exchanger in a thermodynamic system, and
can use fuel-based heating values to determine the amount of heat that is inserted into the
system due to combustion using the following expressions.
Combustion in Air: Energy Balance for Open Systems and Diesel Cycle
·
Q = m· fuel ⋅ LHV = m· air ⋅ (h2 − h1) = m· air ⋅ cp ⋅ (T2 − T1)
(4.6)
Note that the values for cp, T1, and T2 are those for air.
Combustion in Air: Energy Balance for Otto Cycle
·
Q = m· fuel ⋅ LHV = m· air ⋅ (u2 − u1) = m· air ⋅ cv ⋅ (T2 − T1)
(4.7)
Below are some common heating values for di erent types of fuels.
Table 4.1. Heating values for some common types of fuels used in combustion processes.
Heating Values of Common Fuels
Fuel Type
Chemical
Formula
Molar Mass
HHV
(Btu/lbm)
LHV
(Btu/lbm)
HHV
(kJ/kg)
LHV
(kJ/kg)
Methane (g)
CH4
16.043
23,880
21,520
55,530
50,050
Ethanol (l)
C2 H6O
46.069
12,760
11,530
29,670
26,810
Propane (g)
C3 H8
44.097
21,640
19,930
50,330
46,340
Butane (l)
C4 H10
58.123
21,130
19,510
49,150
45,370
Octane (l)
C8 H18
114.23
20,590
19,100
47,890
44,430
Gasoline
C7.2 H13.5
100
20,300
18,900
47,300
44,000
Diesel
C12.3 H22.1
170
19,800
18,600
46,100
43,200
Air
O2 + 3.76N2
28.97
—
—
—
—
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PAGE 337









Figure 4.3. Schematic representing the way that we model combustion (i.e., as a heat exchanger) in a thermodynamic system.
Example 4.1 - Combustion in a Gas Turbine Engine
In a gas turbine engine, air enters an adiabatic combustion chamber at a rate of 10 lbm/s
with a temperature of 1000 R. The air is mixed with propane gas and complete combustion
occurs. The maximum allowable temperature in the engine is 2200 R. Calculate the
minimum Air-Fuel Ratio and %TA.
Solution
Part (a): The Air-Fuel ratio can be found according to,
AFmass =
LHV
cp,air ⋅ (T2 − T1)
To calculate the value of AFmass, we need to nd the LHV and cp,air. The LHV can be pulled
directly from Table 4.1,
LHVpropane = 19,930
Btu
lbmfuel
while cp,air can be found in the Ideal Gas Speci c Heats table in your appendix,

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PAGE 338
Interpolating, we nd,
Btu
1140∘F − 1000∘F
Btu
Btu
Btu
cp,avg = 0.263
+
⋅
0.276
−
0.263
=
0.267
lbm ⋅ R 1500∘F − 1000∘F (
lbm ⋅ R
lbm ⋅ R )
lbm ⋅ R
and nally,
Btu
19,930 lbm ⋅ R
lbmair
LHV
AFmass =
=
=
62.2
Btu
cp,air ⋅ (T2 − T1)
lbmfuel
0.267 lbm
⋅ (2200 R − 1000 R)
⋅R
Part (b): The %TA can be found according to % TA =
AFmass, we nd AFstoich via,
AFmass
. Since we already found
AFstoich
C3H8 + a ⋅ (O2 + 3.76 ⋅ N2) → b ⋅ CO2 + c ⋅ H2O + d ⋅ N2
We solve for coe cients a, b, c, and d by identifying the atomic multiplier of each
constituent element and balancing them on each side of the expression above,
C:
3=b
H:
8=2⋅c
O:
2⋅a =2⋅b+c
→
∴a=5
N:
3.76 ⋅ 2 ⋅ a = d
→
∴ d = 37.6
→
∴c=4
Now,
5 ⋅ (1 + 3.76) lbmolair
lbmolair
AFmolar,stoich =
= 23.8
1 lbmolfuel
lbmolfuel
lbm
AFstoich = AFmolar,stoich ⋅
28.97 lbmolair
Mair
lbmolair
lbmair
air
= 23.8
⋅
=
15.6
Mfuel
lbmolfuel 44.097 lbmfuel
lbmfuel
lbmolfuel
lbm
∴ % TA =
62.2 lbm air
fuel
lbmair
15.6 lbm
= 3.99 = 399 %
fuel



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PAGE 339
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nd the speci c heat at the average temperature, which is Tavg = 1600 R ≈ 1140∘F .
We
O T T O C YC L E
The
rst four-stroke engine was built by the German inventor Nicolaus Otto in 1866 in
collaboration with a technician named Eugen Langen, landing them the Gold Medal at the
Paris World Exhibition in 1867. Nearly a decade later, he built a compressed charge, four
cycle engine with fellow inventors Francis and William Crossley; this iteration of the fourstroke engine is what is now commonly referred to as the “Otto Cycle”. This remarkable
achievement quickly accelerated technologies after the Industrial Revolution.
The Otto Cycle uses four primary strokes to create energy continuously. These include,
1. Compression
2. Power
3. Exhaust
4. Intake
Before we describe the process outlined in the list above, let’s take a closer look at the
components that make up a piston-cylinder system,
Figure 4.4. Schematic of a piston-cylinder with critical components used to achieve a four process Otto Cycle.
Shown above is a schematic of a piston-cylinder apparatus with corresponding intake and
exhaust valves, as well as a spark plug to ignite an air-fuel mixtur

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PAGE 3 40

PAG E 3 41
The schematic on the previous page shows the series of processes used to complete a fourstroke cycle. First, the fuel/air mixture is compressed such that it reaches a higher pressure
and temperature. Next, the fuel/air mixture is ignited using a spark plug; the pressure and
temperature increase almost instantaneously. In fact, they increase so fast that the piston
has very little time to move before the gas reaches its peak temperature. After ignition, the
gas expands and produces useful work. In order to begin the cycle again, the spent fuel/air
mixture must be exhausted to the atmosphere, after which point new cold air is taken into
the cylinder. A pressure-volume diagram is provided below in order to highlight each
thermodynamic process.
Figure 4.5. A pressure-volume diagram which shows the processes used to complete the four-stroke Otto Cycle.
Note that both the compression and expansion processes are modeled as polytropic
processes, while the ignition and exhaust/intake processes are isometric. Recall that a
polytropic process takes the following form,
p ⋅ Vn = C
where C is a constant.


PAGE 3 42
We base our thermodynamic analysis on the following p-V and T-s diagrams.
Figure 4.6. Pressure-volume and temperature-entropy diagrams which show the processes used to complete the four-stroke Otto Cycle. Note that
Qnet can be calculated using the area within the temperature-entropy diagram, where Qx y =
y
∫x
Td s.
Let’s now look at the energy transferred in each process using the above two plots in
tandem with the 1st Law of Thermodynamics (i.e., ΔUxy = Qxy − Wxy , where the subscript
“xy” represents the state points that bound the process).
Otto Cycle Energy Transport Processes
Process
Heat Transferred
Work Done
1 → 2: Isentropic Compression
Q12 = 0
W12 = U1 − U2
2 → 3: Isometric Heat Addition
Q23 = U3 − U2
W23 = 0
3 → 4: Isentropic Expansion
Q34 = 0
W34 = U3 − U4
4 → 1: Isometric Heat Removal
Q41 = U1 − U4
W41 = 0
Because we have an ideal gas, we also know that ΔU = m ⋅ cv ⋅ ΔT . We can use this to
determine the heat transferred and work done during each process.
We can use these computations to determine the e ciency of an Otto Cycle-based engine,
where,
η=
W + W34
Wnet
= 12
qin
q23
(4.8)



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Thermodynamic Analysis
There are also several important terms we will use when dealing with the Otto Cycle (and
eventually the Diesel Cycle). These are listed and de ned below.
Top Dead Center (TDC): Piston position when it forms the smallest volume in the cylinder.
Bottom Dead Center (BDC): Piston position when it forms the largest volume in the
cylinder.
Stroke: Distance that the piston travels from TDC to BDC.
Bore: Diameter of the piston
Clearance Volume: Smallest possible volume in the cylinder (i.e., the volume at TDC).
Compression Ratio (r): The ratio between the maximum and minimum volumes,
VBDC
Vmax
V1
V4
r=
=
=
=
Vmin
VTDC
V2
V3
(4.9)
Mean E ective Pressure (MEP): A ctitious pressure applied by the gas on the cylinder which
is equivalent to the work produced by the cycle, or,
Wcycle = MEP ⋅ Vdisp
(4.10)
Cold Air Standard: The speci c heat for each state point is evaluated at the point in which
we take in cold, fresh air (i.e., state point 1).
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PAGE 3 4 4
Example 4.2 - Otto Cycle
The compression ratio of a cold air standard ideal Otto Cycle is 9. Prior to the isentropic
compression, air is 40∘F and 15 psia. The maximum temperature of the cycle is 2100∘F.
Determine the speci c heat (qnet) and speci c work (wnet) transferred per unit mass for each
process, as well as the cycle thermal e ciency. Assume constant speci c heats evaluated at 40∘F.
Solution
Shown in the schematic above are general p-V and T-S diagrams. We will focus on the
isometric and isentropic processes, which provide useful relationships to calculate
thermodynamic properties and energy transfer rates. These computations are provided for
each process, below.
Process 1 → 2
Since the process is isentropic, q12 = 0 . The speci c work done to the system during the
compression process is therefore calculated as,
w12 = u1 − u2 = cv ⋅ (T1 − T2)
However, we do not know T2 from the information provided. But, we do know that the
process is isentropic, and that we have an ideal gas. Thus, we can use the isentropic ideal
gas relationships we developed in the section on Entropy. In this case, we are given a


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PAGE 3 45
nd T2,
v1
. Thus, we can use the following relationship to
v2
T2
v1
=
T1 ( v2 )
k−1
= (r)k−1
Now, we solve for T2 as,
T2 = T1 ⋅ r k−1 = 500 R ⋅ (9)1.4−1 = 1206.8 R
Based on T1, we nd that cv = 0.171
Btu
and,
lbm ⋅ R
Btu
Btu
w12 = 0.171
⋅ (500 R − 1206.8 R) = − 120.9
lbm ⋅ R
lbm
Note that the value for speci c work is negative because the gas is being compressed.
Process 2 → 3
Here, the process is isometric. Thus, w23 = 0. We can therefore nd q23 according to,
q23 = u3 − u2 = cv ⋅ (T3 − T2)
In this case, we are provided with T3 (i.e., the maximum temperature in our system) and,
Btu
Btu
⋅ (2560 R − 1206.8 R) = 231.4
lbm ⋅ R
lbm
q23 = 0.171
Process 3 → 4
This process is also isentropic, which means that q34 = 0 . Thus, the speci c work can be
found according to,
w34 = u3 − u4 = cv ⋅ (T3 − T4)
However, we are not provided with T4 . Fortunately, the isentropic process provides us with
the following relationship,
v3 k−1
T4
=
T3 ( v4 )
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PAGE 3 46

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compression ratio, which is de ned as r =
Note that the fraction v3/v4 is the reciprocal of v1 /v2. Consequently,
1
T4 = T3 ⋅
(9 )
k−1
1
= 2560 R ⋅
(9 )
1.4−1
= 1060.7 R
and,
w34 = 0.171
Btu
Btu
⋅ (2560 R − 1060.7 R) = 256.4
lbm ⋅ R
lbm
Process 4 → 1
Our nal process is isometric. Therefore, w41 = 0. Now, q41 can be calculated as,
q41 = u1 − u4 = cv ⋅ (T1 − T4) = 0.171
Btu
Btu
⋅ (500 R − 1060.7 R) = − 95.9
lbm ⋅ R
lbm
Cycle E ciency
Finally, the cycle e ciency can be calculated as,
η=
Btu
Btu
−120.9 lbm
+ 256.4 lbm
w + w23
w
get
= net = 12
=
pay
qin
q12
Btu
231.4 lbm
= 58.6 %
To put this in context, we must calculate the Carnot e ciency, which is,
Tc
500 R
ηC = 1 −
=1−
= 79.3 %
Th
2560 R
Thus, we have achieved a 58.6% e ciency out of a possible 79.3% Carnot e ciency.
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PAGE 3 47
D I E S E L C YC L E
The
rst successful Diesel engine was built and tested by Rudolf Diesel in 1897, several
decades after the
rst prototype four-stroke spark-ignition engine. On his suspicion that
additional compression could improve the e ciency of the engine, Diesel demonstrated that
his engine could operate at a nearly 17% higher e ciency than the steam engine. Further
modi cations, including the use of a turbocharger, substantially improved the e ciency to
the point that entire eets of tractor-trailers now use them for long commutes. They are also
widely used in the Navy, including on some larger carriers like the LSD-41 (Whidbey Island)
class and in auxiliary equipment for redundancy. However, they also tend to emit more
harmful gases into the atmosphere and can therefore contribute signi cantly to climate
change. While new regulations have limited their ability to produce harmful emissions, they
are at times economically unfavorable due to fuel and maintenance issues.
The Diesel Cycle uses four primary strokes to create energy continuously. These include,
1. Compression
2. Power
3. Exhaust
4. Intake
Note that these four strokes are the same as those listed for the Otto Cycle. However, the
process by which the power stroke happen is di erent between cycles.
Figure 4.7. Schematic of a piston-cylinder with critical components used to achieve a four process Diesel Cycle.
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PAGE 3 48

PAGE 3 49
The schematic on the previous page describes the thermodynamic processes involved in
each stroke of a Diesel cycle. The major processes are similar to those that an Otto cycle
works through. The one major di erence is the way in which we achieve combustion. Here,
we compress air only (i.e., we do not take in a fuel-air mixture during our intake process) to
such high pressures and temperatures that we can achieve combustion when we put fuel
into the system. While we inject fuel into the cylinder, there is some constant pressure
expansion that occurs; therefore, we also produce useful work during the power stroke.
When we are
nished injecting fuel into the system, the air-fuel mixture undergoes a
polytropic expansion. The remaining processes are the same as those described by the Otto
cycle. A pressure-volume diagram is provided below in order to highlight each
thermodynamic process.
Figure 4.8. A pressure-volume diagram which shows the processes used to complete the four-stroke Diesel Cycle.
Note that both the compression and expansion processes are modeled as polytropic
processes, while the ignition and exhaust/intake processes are isometric. Recall that a
polytropic process takes the following form,
p ⋅ Vn = C

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PAGE 350
process indicates p2 = p3 in this cycle.
Thermodynamic Analysis
We base our thermodynamic analysis on the following p-V and T-s diagrams.
Figure 4.9. Pressure-volume and temperature-entropy diagrams which show the processes used to complete the four-stroke Diesel Cycle. Note that
Qnet can be calculated using the area within the temperature-entropy diagram, where Qx y =
y
∫x
Td s.
We can use the above two plots in tandem with the 1st Law of Thermodynamics (i.e.,
ΔUxy = Qxy − Wxy).
Diesel Cycle Energy Transport Processes
Process
Heat Transferred
1 → 2: Isentropic Compression
Q12 = 0
2 → 3: Isobaric Heat Addition
Q23 = U3 − U2 + p ⋅ (V3 − V2)
3 → 4: Isentropic Expansion
Q34 = 0
4 → 1: Isometric Heat Removal
Q41 = U1 − U4
Work Done
W12 = U1 − U2
W23 = p ⋅ (V3 − V2)
W34 = U3 − U4
W41 = 0
Because we have an ideal gas, we also know that ΔU = m ⋅ cv ⋅ ΔT . We can use this to
determine the heat transferred and work done during each process.
Note that for Q23, we have Q23 = ΔU23 + p ⋅ ΔV23. Considering that Δh = Δu + p ⋅ Δν,
Q23 = m ⋅ (h3 − h2) = m ⋅ cp ⋅ (T3 − T2)
(4.11)








PA G E 3 51















where C is a constant. Also note that the ignition process is isobaric. Recall that an isobaric
since we have an ideal gas. Also note that we can utilize ideal gas relations to solve for the
work done between states 2 and 3 as,
W23 = p ⋅ (V3 − V2) = m ⋅ R ⋅ (T3 − T2)
(4.12)
where R is the speci c gas constant for air.
We can use these computations to determine the e ciency of a Diesel Cycle-based engine,
where,
W + W23 + W34
Wnet
= 12
Qin
Q23
η=
(4.13)
The Diesel cycle uses much of the same terminology that’s used to describe the equipment
used in the Otto cycle, with one addition: the cuto ratio. The cuto ratio is de ned as,
V3
V2
(4.14)
V2
r=
V1
(4.15)
rc =
The compression ratio, r, is also limited to,
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PAGE 352
A Diesel cycle has a compression ratio of r = 15 and a cuto
ratio of rc = 2. At the
beginning of the compression process, air is 100 kPa and 27∘C. Determine the thermal
e ciency of the cycle. Assume constant speci c heats taken at 23∘C.
Solution
The thermal e ciency of a Diesel cycle can be expressed as,
η=
W12 + W23 + W34
Q23
Thus, we need to solve for each of the individual terms in the above expression. These
computations are provided for each process, below.
Process 1 → 2
Since the process is isentropic, q12 = 0 . The speci c work done to the system during the
compression process is therefore calculated as,
w12 = u1 − u2 = cv ⋅ (T1 − T2)
However, we do not know T2 from the information provided. But, we do know that the
process is isentropic, and that we have an ideal gas. Thus, we can use the isentropic ideal
gas relationships we developed in the section on Entropy. In this case, we are given a
compression ratio, which is de ned as r =
nd T2,
v1
. Thus, we can use the following relationship to
v2
T2
v1
=
T1 ( v2 )
k−1
= (r)k−1
Now, we solve for T2 as,
T2 = T1 ⋅ r k−1 = 300 K ⋅ (15)1.4−1 = 886.3 K
Now we can solve for the speci c work done (since we do not have the mass of the air)
during compression as,

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PAGE 353
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Example 4.3 - Diesel Cycle
w12 = cv ⋅ (T1 − T2) = 0.718
kJ
kJ
⋅ (300 K − 886.3 K ) = − 421.0
kg ⋅ K
kg
Note that we obtained cv at 23∘C, as speci c in the problem, using the Appendix.
Process 2 → 3
During this process we have isobaric combustion. The speci c work and heat transfer are
determined via,
q23 = cp ⋅ (T3 − T2)
and,
w23 = R ⋅ (T3 − T2)
To solve for both q23 and w23, we must determine T3. Since the process between states 2 and
3 is isobaric, we can solve for T3 using the Ideal Gas Law,
p ⋅v
p2 ⋅ v2
=R= 3 3
T2
T3
where we know that the gas constant, R, is equivalent between states 2 and 3 (and, really,
all states in the cycle). Because p3 = p2, we can rearrange to solve for T3 as,
T3 = T2 ⋅
V3
= T2 ⋅ (rc) = 886.3 K ⋅ 2 = 1772.6 K
V2
Now,
q23 = 1.005
kJ
kJ
⋅ (1772.6 K − 886.3 K ) = 890.7
kg ⋅ K
kg
and,
kJ
kJ
w23 = 0.287
⋅ (1772.6 K − 886.3 K ) = 254.4
kg ⋅ K
kg
Process 3 → 4
We solve for the work done by the piston during the expansion process as,
w34 = cv ⋅ (T3 − T4)

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PAGE 35 4
and we also know that q34 = 0 due to the isentropic nature of the process. We therefore
require T4 in order to solve for w34 . Unlike with the Otto cycle, we cannot simply use the
isentropic ideal gas equations with the reciprocal of the compression ratio, r. Here, we have,
v3 k−1
T4
=
T3 ( v4 )
but
v3
1
≠ . Instead, we can use the following mathematical equivalency,
v4
r
v3
v3 v2 v1
=
⋅ ⋅
v4
v2 v1 v4
where
v3
v
r
v
v
1
= rc, 2 = , and 1 = 1. As a result, we can say that 3 = c and,
v2
v4
r
v1
r
v4
rc k−1
T4
=
T3 ( r )
→
2
T4 = 1772.6 K ⋅
( 15 )
0.4
= 791.7 K
We now solve for w34 as,
kJ
kJ
w34 = 0.718
⋅ (1772.6 K − 791.7 K ) = 704.3
kg ⋅ K
kg
Process 4 → 1
Although we don’t need q41 to solve for η (and we know that w41 = 0 because the exhaust/
intake process is isometric), we still show you how to solve for it here.
kJ
kJ
q41 = cv ⋅ (T1 − T4) = 0.718
⋅ (300 K − 791.7 K ) = − 353
kg ⋅ K
kg
Finally, we solve for the thermal e ciency of the cycle as,
537.7 kJ
kg
w12 + w23 + w34
=
= 60.4 %
kJ
q23
890.7
η=
kg
and,
ηcarnot = 1 −
Tc
300 K
=1−
= 83.1 %
Th
1772.6 K





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PAGE 355
G A S T U R B I N E E N G I N E S ( B R AY T O N C YC L E )
Gas turbine engines are principally used for
aircraft propulsion and electric power
generation. In fact, the majority of the
world’s Naval
eets currently use gas
turbine engines for both of these purposes.
The GE LM2500 gas turbine engine (shown
on the right), for instance, is used to power
many of the large warships in operation
today. Likewise, The LM2500+, which
delivers nearly 30,000 kW of power (when
connected to an electrical generator), now
serves as the propulsion system for many
Figure 4.10. GE LM2500 gas turbine engine. By Camera Operator: PH2
JEFFREY ELLIOTT - ID:DN-SN-88-03869 / Service Depicted: Navy,
Public Domain, https://commons.wikimedia.org/w/index.php?
curid=2979155
modern cruise ships.
One important complication of a typical gas turbine system is that it consumes a great deal
of fuel. In order to provide for e cient cruising speeds, propulsion systems are often
out tted with Diesel cycles, while the gas turbine engines mentioned above are reserved for
high speeds.
The basic operating principle of a gas turbine system is outlined in the schematic featured in
Fig. 4.11.
Figure 4.11. Schematic of a gas turbine system; major components include a compressor, combustion chamber, and turbine.
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PAGE 356
As shown in Fig. 4.11, ambient air is rst drawn into the compressor, after which point the
(now) high pressure air is sent through the combustion chamber. With the addition of fuel
and an ignition element, the air-fuel mixture burns to a higher temperature ahead of
entering the turbine. As the hot air passes through the turbine, it expands back to ambient
pressure. Note that some of the power produced by the turbine is used to power the
compressor. The amount of power required to operate the compressor is called the back
work ratio, and is de ned as,
·
WC
BWR = ·
WT
(4.16)
or the compressor power divided by the turbine power.
Thermodynamically, we can model this as a closed cycle (even though we intake cold, fresh
air and exhaust our spend air/fuel mixture) with the following assumptions:
1. The combustion chamber is modeled as a heat exchanger.
2. The exhaust/intake process is modeled as a heat exchanger.
3. The uid is modeled as air only (typically, the air-fuel ratio is very high, which means
we can neglect the thermodynamic impacts of the fuel).
A schematic, which includes these assumptions, is provided below.
Figure 4.12. Thermodynamic schematic of a gas turbine system; major components include a compressor, combustion chamber, and turbine.

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PAGE 357
Ideal Brayton Cycle
In the ideal case, both the compressor and the turbine have isentropic e ciencies of 100%.
We will learn how to account for these in the non-ideal case. For now, let’s focus on the
thermodynamic operating principles of a Brayton cycle. Using the state points outlined in
Fig. 4.12., we outline the thermodynamic processes as follows,
Process 1 → 2: Isentropic Compression
Process 2 → 3: Constant Pressure Heat Addition
Process 3 → 4: Isentropic Expansion
Process 4 → 1: Constant Pressure Heat Removal
Recall that these are open systems. When an open system has an isentropic e ciency of
100%, we call the process itself isentropic. Likewise, remember that we neglect the pressure
drop through a heat exchanger. As a result, we consider the process to be isobaric. Given
these processes, we can draw the p-v and T-s diagrams as shown in Fig. 4.13.
Figure 4.13. p-v (left) and T-s (right) diagrams for an ideal Brayton cycle. Note that heat enters and leaves the system between states 2 and 3 and
states 4 and 1, respectively.
For this cycle, we will follow the cold air standard analysis that we used in both the Otto
and Diesel cycles, meaning that our thermodynamic properties are taken at the temperature
of the cold air at the intake of the system. For each individual open system (i.e., heat
exchangers, compressor, and turbine) we will also assume that there are negligible changes
in kinetic and potential energy between inlet and exit. The heat transferred and work done
to/by each device is described in the following table.
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PAGE 358
Process
Device
1 → 2: Isentropic
Compression
2 → 3: Isobaric Heat
Addition
Heat Transferred
4 → 1: Isobaric Heat
Removal
Power Produced/
Consumed
Compressor
·
Q12 = 0
·
W12 = m· ⋅ (h1 − h2)
Heat Exchanger
·
Q23 = m· ⋅ (h3 − h2)
·
W23 = 0
·
Q34 = 0
·
W34 = m· ⋅ (h3 − h4)
·
Q41 = m· ⋅ (h1 − h4)
·
W41 = 0
3 → 4: Isentropic Expansion Turbine
Heat Exchanger
Recall that for an ideal gas, Δh = cp ⋅ ΔT.
To solve for the thermal e ciency of the cycle, we use,
·
·
Wnet
Qout
T − T4
η = · =1− · =1− 1
T3 − T2
Qin
Qin
(4.17)
For the ideal Brayton cycle, also remember that the following relationships hold for the
isentropic processes:
k−1
k−1
p3 k
T3
T2
p2 k
=
=
=
( p4 )
T1 ( p1 )
T4
(4.18)
We de ne the ratio p2 /p1 as the pressure ratio. In application, typical pressure ratios are
between 5 and 20, and the optimal pressure ratio depends on the minimum and maximum
temperatures of the cycle.
Non-ideal Brayton Cycle
In the non-ideal case, the compressor and turbine are not isentropic. Instead, we rely on the
isentropic e ciency of each device, represented by,
·
Ws
h − h2s
T − T2s
ηC = · = 1
= 1
h1 − h2
T1 − T2
Wa
(4.19)
for the compressor, and,




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PAGE 359
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Brayton Cycle Energy Transport Processes
·
h − h4
T − T4
Wa
ηT = · = 3
= 3
h3 − h4s
T3 − T4s
Ws
(4.20)
·
for the turbine. As a reminder, Ws is the power produced or consumed if the process were
·
isentropic, while Wa is the actual power produced or consumed by the device. As shown in
Eqns. 4.19 and 4.20, we can use temperatures to determine each individual device e ciency.
In Eqn. 4.19, T2s is the isentropic temperature at the exit of the compressor, and in Eqn.
4.20, T4s is the isentropic temperature at the exit of the turbine. These temperatures can be
determined using the isentropic ideal gas relationship provided in Eqn. 4.18,
k−1
k−1
p3 k
T
T2s
p2 k
=
=
= 3
( p1 )
( p4 )
T1
T4s
(4.21)
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PAGE 360
Example 4.4 - Non-ideal Brayton Cycle
A gas turbine power plant operates with a Brayton cycle and has a pressure ratio of rp = 12.
The ambient air temperature is 300 K and the maximum temperature of the cycle is 1400 K.
The isentropic e ciency of the compressor is 80% and the isentropic e ciency of the
turbine is 90%. Assume the ambient air pressure is 100 kPa and that there is negligible
pressure drop during the heat exchange processes. Determine:
1. The temperature at the exits of the compressor and the turbine [K]
2. The back work ratio (BWR) of the cycle.
3. The cycle thermal e ciency, η [%]
Solution
To nd the temperature at the exit of the turbine (T2), we can use the isentropic e ciency of
the compressor,
T1 − T2s
ηC =
T1 − T2
In order to rearrange the above expression and nd T2, we must rst nd T2s. We can do this
with the isentropic ideal gas relationship described by Eqn. 4.21,
T2s
p2
=
( p1 )
T1
k−1
k
and,
0.4
T2s = 300K ⋅ (12) 1.4 = 610.18 K
Now we can determine T2 via,
T2 = T1 −
T1 − T2s
300K − 610.18K
= 300K −
= 687.72 K
ηC
0.8
We can use the same process to nd the temperature at the exit of the turbine, T4,
T3 − T4
ηT =
T3 − T4s

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PA G E 3 61
Again, we must rst nd T4s prior to computing T4.
T4s
p1
=
( p2 )
T3
k−1
k
and,
T4s = 1400K ⋅
1
( 12 )
0.4
1.4
= 688.32 K
Now we solve for T4 as,
T4 = T3 − ηT ⋅ (T3 − T4s) = 1400K − 0.9 ⋅ (1400K − 688.32K ) = 759.48 K
To calculate the back work ratio, we use,
·
| WC |
T − T2
| (300K − 687.72K ) |
BWR = ·
= 1
=
= 61 %
T3 − T4
1400K − 759.48K
WT
Finally, we calculate the thermal e ciency of the cycle as,
·
Wnet
(300K − 687.72K ) + (1400K − 759.48K )
η= · =
= 35.4 %
1400K
−
687.72K
Qin


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PAGE 362
Regenerative Brayton Cycle
As mentioned previously, the e ciency of a standard Brayton cycle is low enough to require
the use of a Diesel engine during periods when warships operate at cruising (i.e., lower)
speeds. To make the Brayton cycle more e cient, we can implement a regenerating system.
This system takes the hot exhaust gases and feeds them back through the system to preheat
the air before it enters the combustion chamber. In this way, less heat energy is required by
the system to achieve some nominal power or thrust from the turbine. To achieve this
preheating scheme, a heat exchanger is used between the compressor and the combustion
chamber to transfer heat from the hot exhaust gas to the air entering the combustion
chamber, as pictured in Fig. 4.14, below.
Figure 4.14. Schematic of a regenerative Brayton cycle; major components include a compressor, combustion chamber, turbine, and non-mixing
heat exchanger for pre-heating the air ahead of the combustion chamber.
Note that state point 5 is in-between state points 2 and 3. This is done in order to keep the
former state points consistent with their previous locations in the standard Brayton cycle.
Because the temperature at state point 4 remains relatively high, we can use it to put heat
into the system prior to the entrance of the combustion chamber, as shown above. This
shifts the temperature at the entrance of the combustion chamber, as shown in the T-s
diagram provided in Fig. 4.15.
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PAGE 363
Figure 4.15. Temperature-entropy diagram for regenerative Brayton cycle. Note that point 5’ represents the maximum theoretical temperature that
can be reached if there were no losses in the system and if the heat exchanger e ectiveness were equivalent to 1.
In Fig. 4.15, the amount of heat that is regenerated is represented by the area under the
curve between points 2 and 5. Note that point 5’ represents the maximum theoretical
temperature that can be achieved for the air ahead of the combustion chamber; this assumes
that there are no losses between the exit of the turbine and the entrance of the non-mixing
heat exchanger, and that the e ectiveness of the heat exchanger is 1 with an in nite length.
The important feature of this system is that it reduces the amount of heat that has to be put
into the system to achieve the temperature at state point 3. This is re ected in the
·
·
calculation of e ciency, which now uses Qin = Q53.
The heat supplied during the regeneration process can be calculated as,
·
Qregen = m· ⋅ (h5 − h2) = m· ⋅ cp ⋅ (T5 − T2)
(4.22)
We also know that the maximum possible regeneration we can achieve can be expressed as,
·
Qregen,max = m· ⋅ (h5′ − h2) = m· ⋅ cp ⋅ (T5′ − T2) − h2) = m· ⋅ cp ⋅ (T4 − T2)
(4.23)
The ratio of these two values is actually equal to the heat exchanger e ectiveness! Assuming
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PAGE 36 4

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a cold air standard analysis is satisfactory, this can be written as,
·
Qregen
T − T2
ε= ·
= 5
T4 − T2
Qregen,max
(4.24)
Now, our cycle thermal e ciency can be expressed as,
·
(T1 − T2) + (T3 − T4)
Wnet
η= · =
T3 − T5
Qin
(4.25)


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PAGE 365
A gas-turbine power plant operates on an ideal Brayton cycle and has a pressure ratio of
rp = 8. The gas temperature is 300 K at the compressor inlet and 1300 K at the turbine
inlet. Assume that the ambient air pressure is 100 kPa. The compressor has an isentropic
e ciency of 80% and the turbine has an isentropic e ciency of 85%. For this problem,
assume that the pressure drop during any heat transfer process is negligible. Determine:
1. The gas temperature at the exits of the compressor and turbine [K]
2. The back work ratio [%]
3. The cycle thermal e ciency [%]
4. Now assume that a regenerator has been attached to the exit of the turbine and is
used to preheat the uid entering the combustion chamber. If the regenerator has an
e ectiveness of 80%, determine the thermal e ciency of the cycle [%]
Solution
In order to nd the temperatures at the exit of each component, we must use the isentropic
e ciencies (note that if the devices were 100% e cient, we could use the isentropic ideal
gas relationships we developed to solve for exit temperatures). Let’s start with the
compressor:
ηc =
T1 − T2s
= 0.8
T1 − T2
To solve for T2, we must rst determine T2s via,
T2s
p2
=
( p1 )
T1
k−1
k
0.4
→
T2s = 300K ⋅ (8) 1.4 = 543.4 K
Now we solve for T2 as,
T2 = T1 −
T1 − T2s
300K − 543.4K
= 300K −
= 604.3 K
η
0.8
We can follow this same procedure to solve for T4 . We begin with the e ciency of the
turbine as,
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PAGE 366
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Example 4.5 - Regenerative Brayton Cycle
T3 − T4
= 0.85
T3 − T4s
Now we solve for T4s using,
k−1
T4s
p4 k
=
( p3 )
T3
Here, we know that p3 = p5 = p2 = 800 kPa (no heat loss during heat exchange processes).
We also nd that p4 = p6 = 100 kPa for the same reason. Therefore, p4 /p3 = 1/rp and,
0.4
1 1.4
T4s = 1300K ⋅
= 717.7 K
(8 )
and,
T4 = T3 − ηT ⋅ (T3 − T4s) = 1300K − 0.85 ⋅ (1300K − 717.7K ) = 805 K
Now we can calculate the back work ratio as,
·
| Wc |
| T1 − T2 |
| 300K − 604.3K |
BWR = ·
=
=
= 61.4 %
T
−
T
1300K
−
805K
WT
3
4
And the thermal e ciency of the cycle is,
(T1 − T2) + (T3 − T4)
(300K − 604.3K ) + (1300K − 805K )
=
= 27 %
T3 − T2
1300K − 604.3K
η=
Let’s now consider the regenerator, which has a heat exchanger with ε = 0.8. Here, we need
the temperature at state point 5. To solve for T5 , we can use the expression for heat
exchanger e ectiveness, or,
T5 − T2
T4 − T2
ε=
→
T5 = T2 + ε ⋅ (T4 − T2) = 604.3K + 0.8 ⋅ (805K − 604.3K ) = 764.8 K
And the new cycle e ciency becomes,
η=
(T1 − T2) + (T3 − T4)
(300K − 604.3K ) + (1300K − 764.8)
=
= 35.6 %
T3 − T5
1300K − 604.3K
which is an improvement of over 8%!







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PAGE 367


ηT =
SPLIT-SHAFT GAS TURBINES
A split-shaft gas turbine operates using the Brayton cycle, but has two turbines operating on
two separate shaft (hence the name split-shaft). The two turbines are referred to as the “gas
generator turbine” (or high pressure turbine) and the “power turbine” (or low pressure
turbine). One common split-shaft gas turbine is the GE T700 gas turbine, which is used to
drive the propulsion system in the Army’s Blackhawk helicopter. A basic schematic of a
split-shaft engine is shown in the gure below.
Figure 4.16. Schematic of a split-shaft gas turbine engine. Note that there are two separate shafts - a low pressure shaft to drive supply power to
the compressor, and a high pressure shaft to provide rotational power out of the system.
The purpose of the gas generator turbine is to supply power to the compressor. Therefore,
we know that,
·
·
WGGT = | WC |
(4.26)
Neglecting kinetic and potential energy changes, power produced or consumed by each
·
device can generally be expressed as W = m· ⋅ Δh = m· ⋅ cp ⋅ ΔT (where Δh = cp ⋅ ΔT for an
ideal gas). Thus, we can also say that,
T3 − T4 = T2 − T1
(4.27)
In this case, note that the pressure drop across the gas generator turbine will not be the
same as the pressure increase across the compressor (i.e., p4 /p3 ≠ 1/rp as in the standard
Brayton cycle). The remaining analysis is the same as that used by the Brayton cycle.





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PAGE 368
A GE T700 gas turbine operates with a compressor having an isentropic e ciency of 88%
and a turbine having an isentropic e ciency of 86%. Determine:
1. The actual temperature at the exit of the compressor, T2 [K]
2. The power consumed by the compressor [kW]
3. The heat transfer rate into the combustion chamber [kW]
4. The actual temperature and pressure at the exit of the high pressure turbine, T4 and
p4 [K, kPa]
5. The actual temperature at the exit of the low pressure turbine [K]
6. The thermal e ciency of the cycle [%]
Solution
Here we recognize that the compressor is not isentropic. However, we can use its isentropic
e ciency to solve for the temperature at its exit, T2, where,
T1 − T2s
= 0.88
T1 − T2
ηc =
We use the isentropic ideal gas equation to solve for T2s as,
k−1
0.4
T2s = T1 ⋅ (rp) k = 289K ⋅ (15) 1.4 = 626.5 K
And now,
T1 − T2s
289K − 626.5K
T2 = T1 −
= 289K −
= 672.5 K
ηc
0.88
We can now use this to nd the power consumed by the compressor,
kg
kJ
·
WC = m· ⋅ (h1 − h2) = m· ⋅ cp ⋅ (T1 − T2) = 4.6
⋅ 1.004
⋅ (289K − 672.5K ) = − 1771.2 kW
s
kg ⋅ K
In the above expression, we nd cp at 289K (i.e., we use the cold air standard analysis). Now
we nd the heat transferred into the system via the combustion chamber using,

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Example 4.6 - Split-shaft Gas Turbine
Next, we are tasked with solving for T4 and p4 . We note here that p4 /p3 ≠ 1/rp . Since we
don’t have p4 , we also can not use the device’s isentropic e ciency to solve for T4s to then
·
·
solve for T4. However, we do know that WGGT = | WC | and therefore,
·
·
| WC | = 1771.2kW = WGGT = m· ⋅ cp ⋅ (T3 − T4)
Rearranging, we solve for T4 as,
·
| WC |
1771.2kW
T4 = T3 − ·
= 1273K −
kg
m ⋅ cp
4.6 ⋅ 1.004 kJ
s
= 889.5 K
kg ⋅ K
Now we can use T4 to solve for p4 via the isentropic ideal gas relationship, but rst we must
solve for T4s using the isentropic e ciency of the high pressure turbine,
T3 − T4
T3 − T4s
ηGGT =
→
1273K − 889.5K
= 827.06 K
0.86
T4s = 1273K −
Now we use the isentropic ideal gas relationship between temperature and pressure to nd
p4,
k−1
T4s
p4 k
=
( p3 )
T3
k
1.4
T4s k − 1
827.06K 0.4
p4 = p3 ⋅
= 1500kPa ⋅
= 332 kPa
( T3 )
( 1273K )
→
Next, we are asked to solve for the temperature at the exit of the power turbine. Here, we
know both p4
and p5
(recall that we are discharging into the atmosphere, so
p5 = p1 = 100 kPa). We can do this with the e ciency of the power turbine,
T4 − T5
ηPWT =
= 0.86
T4 − T5s
To solve for T5, we need to compute T5s via the isentropic ideal gas expression as,
T5s
p5
=
( p4 )
T4
k−1
k
→
T5s = 889.5K ⋅
100kPa
( 332kPa )
0.4
1.4
= 633.11 K
Now,
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PAGE 370



kg
kJ
·
Q23 = m· ⋅ cp ⋅ (T1 − T2) = 4.6
⋅ 1.004
⋅ (1273K − 672.5K ) = 2773.3 kW
s
kg ⋅ K
T5 = T4 − η ⋅ (T4 − T5s) = 889.5K − 0.86 ⋅ (889.5K − 633.1K ) = 669 K
The power produced by the power turbine can now be calculated according to,
kg
kJ
·
WPWT = m· ⋅ cp ⋅ (T4 − T5) = 4.6
⋅ 1.004
⋅ (889.5K − 669K ) = 1018.35 kW
s
kg ⋅ K
Finally, the thermal e ciency of the cycle can be computed according to,
·
WPWT
1018.35kW
η= ·
=
= 36.7 %
2773.3kW
Q23

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PAGE 371
J E T P RO P U L S I O N C YC L E S
Jet engines utilize gas turbines to power aircraft given their high power to weight ratios. In
order to achieve thrust, a Brayton cycle is used with several additional components.
Speci cally, the use of a di user and nozzle allow for the deceleration of air ow into the
system, and the acceleration of air at the exit of the system, respectively. We will focus our
e orts on turbojet engines.
Turbojet Engines
A turbojet engine uses both a di user and a nozzle at the entrance/exit of a standard
Brayton cycle, respectively. The di user is used to slow air ow headed into the compressor.
Without the di user, high air speeds may damage the compressor blades. Likewise, the
nozzle is used to accelerate air ow and provide a thrust force to propel the aircraft. In this
case, we can determine the thrust supplied by the engine according to,
F̄ = m· ⋅ (V̄exit − V̄inlet )
(4.28)
Note that if the aircraft is cruising in still air, V̄inlet = V̄aircraft , where V̄aircraft is the cruising
speed of the aircraft. A schematic of a turbojet engine is shown in the gure below.
Figure 4.17. Schematic of a turbojet engine. A nozzle at the exit of the system provides thrust to propel an aircraft forward. Note that all power
generated by the turbine is used to spin the compressor.
In the absence of any auxiliary equipment to run, the turbine only supplies power to the
compressor, such that,
·
·
WT = | WC |
(4.29)
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PAGE 372
Also note that at the exit of the di user, we can typically assume that V̄2 < < V̄1 and at the
entrance of the nozzle, V̄5 < < V̄6 . As we will see in the following example problem, this
greatly simpli es the conservation of energy for each of these devices.
The basic thermodynamic cycle is represented by the following T-s diagram for an ideal
turbojet engine (i.e., all components have isentropic e ciencies of 100%),
Figure 4.17. T-s diagram for an ideal turbojet engine, where the isentropic e ciency of all devices is 100%.
Note that for this cycle, we de ne an overall pressure ratio as,
rp,o =
p3
p1
(4.30)
We can also de ne a propulsive e ciency as the propulsive power generated relative to the
heat energy we put into the system, or,
·
m· ⋅ (V̄exit − V̄inlet ) ⋅ V̄aircraft
Wp
Propulsive Power
η=
= · =
·
Heat In
Qin
Q34
(4.31)

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Example 4.7 - Turbojet Engine
An F-4 Phantom is cruising at 900 ft/s where the local atmospheric pressure is
approximately 10.8 psi and the air temperature is 60∘F. The engine operates with an 8:1
compressor pressure ratio. Air owing at 210 lbm/s enters the turbine at 1900∘F. The nozzle
exhaust pressure is 11 psia. Assume that all applicable components in the jet engine operate
isentropically and that there is no pressure drop across the combustion chamber. Also
assume that a cold air standard analysis is appropriate for this problem. Determine:
1. The overall pressure ratio, rp,o
2. The temperature and pressure of the air entering the nozzle [∘F, psia].
3. The exit velocity from the nozzle [ft/s]
4. The thrust produced by the engine [lbf]
5. The propulsive e ciency [%]
Solution
The overall pressure ratio can be determined via,
p3
rp,o =
p1
Thus, we must determine p3 in order to calculate rp,o . From the problem statement, we
know that p3 = 8 ⋅ p2 . We must therefore nd p2 . Since the di user is isentropic, we know
that,
T2
p2
=
T1 ( p1 )
k−1
k
→
p2 = p1 ⋅
T2
( T1 )
k
k−1
We can also use the conservation of energy for the di user to solve for T2. The conservation
of energy expression applied across the di user yields,
2
2
V̄
−
V̄
1
2
0 = m· ⋅ h1 − h2 +
(
)
2
V̄ 21
0 = cp ⋅ (T1 − T2) +
2
→
where Δh = cp ⋅ ΔT for an ideal gas. Now we rearrange to solve for T2,
900 s
(
)
V̄ 21
T2 = T1 +
= 520R +
2 ⋅ cp
2 ⋅ 0.240 Btu
ft
2
⋅
lbm ⋅ R
Btu
1 lbm
ft 2
25,037 2
= 587.4 R
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PAG E 374
Now,
p2 = 10.8
lbf
587.4R
⋅
in 2 ( 520R )
1.4
0.4
= 16.55
lbf
in 2
Now we can use the compressor pressure ratio to determine p3, or
p3 = 8 ⋅ p2 = 8 ⋅ 16.55
lbf
lbf
=
132.4
in 2
in 2
Thus, the overall pressure ratio is,
rp,o =
132.4
lbf
p3
in 2
=
= 12.26
lbf
p1
10.8
in 2
Now we must solve for the pressure and temperature at the entrance of the nozzle (or the
exit of the turbine), p5 and T5. We know that the power generated by the turbine is equal to
that consumed by the compressor, or,
| m· ⋅ cp ⋅ (T2 − T3) | = m· ⋅ cp ⋅ (T4 − T5)
Ultimately, we can use this to solve for T5 , but we just
compressor is also isentropic, we can say that,
T3
p3
=
T2 ( p2 )
k−1
k
→
T3 = 587.4R ⋅
(
rst determine T3 . Since the
132.4
16.55
lbf
in 2
lbf )
0.4
1.4
= 1064 R
in 2
and,
T5 = T4 − T3 + T2 = 2360R − 1064R + 587.4R = 1883 R
To solve for p5, we also know that the turbine is isentropic such that,
k−1
T5
p5 k
=
T4 ( p4 )
k
→
T5 k − 1
p5 = p4 ⋅
( T4 )
Since there is negligible pressure drop through the combustion chamber (relative to the
pressure drop in all major system components), we can assume that p4 = p3 = 132.4 psia.
Thus,
p5 = 132.4
lbf
1883R
⋅
in 2 ( 2360R )
0.4
1.4
= 60.07
lbf
in 2
In order to determine the velocity at the exit of the nozzle, we apply the conservation of
energy, which yields (in simpli ed form),


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PAGE 375
where we have assumed that V̄5 < < V̄6 and that Δh = cp ⋅ ΔT for an ideal gas. To solve for
the velocity at the exit, then, we rst need to compute T6. To do this, we use,
T6
p6
=
T5 ( p5 )
k−1
k
→
T6 = 1883R ⋅
(
11
lbf
in 2
lbf )
60.07
0.4
1.4
= 1159 R
in 2
Now,
ft 2
V̄6 =
25,037 2
Btu
ft
s
2 ⋅ 0.240
⋅ (1883R − 1159R) ⋅
= 2950
Btu
lbm ⋅ R
s
1
lbm
The thrust force is then calculated as,
lbm
ft
ft
1lbf
F̄T = m· ⋅ (V̄6 − V̄1) = 210
⋅ 2950 − 900
s (
s
s ) 32.17 lbm ⋅ ft
= 13,370 lbf
s2
and the propulsive power becomes,
ft
1hp
·
Wp = F̄T ⋅ V̄aircraft = 13,370 lbf ⋅ 900 ⋅
lbf ⋅ ft
s
550 s
= 21,900 hp
Finally, the cycle e ciency is determined via,
·
Wp
21,900hp
η= · = ·
=
m
⋅
c
⋅
(T
−
T
)
Q34
p
4
3
21,900hp
lbm
Btu
210 s ⋅ 0.240 lbm ⋅ R ⋅ (2360R − 1064R) ⋅
= 23.7 %
1hp
0.7068 Btu
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PAG E 376


V̄ 26
0 = cp ⋅ (T5 − T6) −
2
VA P O R P O W E R C YC L E S
Vapor power systems convert water to steam in order to produce power through changes in
enthalpy. The vapor power systems we will discuss, all derivatives of the Rankine cycle,
utilize one or more turbines to generate useful power. In contrast to the systems that we’ve
discussed thus far, vapor power systems utilize water as the working uid (whereas each of
the previous cycles have used air). Thus, all of our thermodynamic state point variables are
found in the steam tables in the Appendix.
Ideal Rankine Cycle
The simplest form of the rankine cycle utilizes a pump to pressurize our working uid and a
boiler to increase its temperature (and change its phase) to achieve a large value of enthalpy
at the turbine inlet. The
uid is allowed to expand and is condensed to a liquid prior to
entering the pump, at which point the cycle is reinitiated. A schematic of this process is
provided in the gure below.
Figure 4.18. Ideal rankine cycle diagram with corresponding state points and likely phases of water.
In an ideal Rankine cycle, both the pump and the turbine are isentropic. Note also that in an
ideal process, water enters the pump as a saturated liquid. Thus, we have the following
processes that occur between individual devices.
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PAGE 37 7
Process
1 → 2: Isentropic
Compression
2 → 3: Isobaric Heat
Addition
Device
4 → 1: Isobaric Heat
Removal
We
Heat Transferred
Power Produced/
Consumed
Pump
·
Q12 = 0
·
W12 = m· ⋅ (h1 − h2)
Heat Exchanger
·
Q23 = m· ⋅ (h3 − h2)
·
W23 = 0
·
Q34 = 0
·
W34 = m· ⋅ (h3 − h4)
·
Q41 = m· ⋅ (h1 − h4)
·
W41 = 0
3 → 4: Isentropic Expansion Turbine
Heat Exchanger
nd the value of enthalpy at each state point (shown in the expressions in the table
above) using the procedure we developed earlier in this course, namely: (1) identify the
phase of the substance, and (2) use the appropriate table to determine enthalpy.
A T-s diagram for this cycle is provided in the gure, below.
Figure 4.19. T-s diagram (SI units) for an ideal Rankine cycle operating between two arbitrary pressures.
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Rankine Cycle Energy Transport Processes
There are some subtleties that the reader should be aware of when analyzing some of the
components that are associated with the Rankine cycle. In particular, we highlight the
nuances associated with the pump and condenser below.
Pump
Recall that pump power can be calculated according to,
·
WP = m· ⋅ (hi − he) = m· ⋅ νf (Ti) ⋅ (pi − pe)
In many cases, we can use the equivalency on the right hand side of the above expression to
solve for he as,
he = hi − νf (Ti) ⋅ (pi − pe)
(4.32)
Condenser
The condenser is often modeled as a non-mixing heat exchanger, where a separate stream of
cooling water is used to condense the steam at the exit of the turbine. In fact, sea water is
often used as a cooling source on Naval platforms due to its abundance in the surroundings.
Figure 4.20. Schematic of condenser modeled as a non-mixing heat exchanger. Shown are the cooling water inlet and exit from the second uid
stream.
Mathematically, we can model this with the use of the conservation of energy principle, or,
0 = m· s ⋅ (h1 − h4) + m· cw ⋅ (hcw,i − hcw,e)
(4.33)
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PAGE 379
where the subscript “s” refers to the steam owing through the condenser and the subscript
“cw” refers to the cooling water. We consider the secondary uid (the cooling water) to be
an incompressible substance, which yields,
0 = m· s ⋅ (h1 − h4) + m· cw ⋅ ccw ⋅ (Tcw,i − Tcw,e)
(4.34)
Cycle Thermal E ciency
We solve for the cycle thermal e ciency in the usual way,
·
·
Wnet
Qout
h − h1
η = · =1− · =1− 4
h3 − h 2
Qin
Qin
(4.35)
Back Work Ratio
We can also compute the back work ratio for a Rankine cycle according to,
·
WP
h − h1
BWR = · = 2
h3 − h4
WT
(4.36)
Heat Ratio
Finally, we can de ne a heat ratio as,
·
Qin
HR = ·
WT
(4.37)
Note that the heat ratio has unfamiliar units. For example, in SI the units are kJ/kW/hr.
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PAGE 380
A steam power plant operates on a simple ideal Rankine cycle between pressure limits of
1000 lbf/in2 and 5 lbf/in2. The turbine inlet temperature is 800∘F. The mass
ow rate of
steam is 5⋅105 lbm/hr. Determine:
1. The power produced by the turbine [hp]
2. The heat transfer rate out of the steam in the condenser [Btu/hr]
3. The power required by the pump [hp]
4. The heat transfer rate supplied by the boiler [Btu/hr]
5. The cycle’s thermal e ciency [%]
Solution
First, we must determine the phase of the working uid (water) at the inlet and the exit of
the turbine. At the inlet, the temperature and pressure are T3 = 800∘F
and
p3 = 1000 lbf /in 2, respectively. The saturated water pressure table is shown below.
Since T3 > Tsat at p3 , we have a superheated vapor. We must therefore use the superheated
vapor table to determine h3.
At T3 = 800∘F and p3 = 1000 lbf /in 2, we nd that h3 = 1389
Btu
.
lbm
Now we must nd the phase at state point 4. Because the turbine is isentropic, we know that
lbf
to determine what the
in 2
phase of the substance is by examining s4 within the context of sf and sg at p4.
s4 = s3. We can use the saturated water pressure table at p4 = 5
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PA G E 3 81



Example 4.8 - Ideal Rankine Cycle
The relevant portion of the saturated water pressure table is provided below.
Btu
lbf
, which is between sf and sg at 5
. Therefore, the phase at
lbm ⋅ R
in 2
state point 4 is a saturated mixture. We nd the quality, χ4, via,
Here, s4 = s3 = 1.5670
Btu
χ4 =
Btu
1.5670 lbm ⋅ R − 0.23488 lbm ⋅ R
Btu
Btu
1.8438 lbm
−
0.23488
⋅R
lbm ⋅ R
= 0.83




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PAGE 382
Btu
Btu
Btu
Btu
+ 0.83 ⋅ 1130.7
− 130.18
= 960.6
(
lbm
lbm
lbm )
lbm
Thus, the power produced by the turbine is calculated to be,
lbm
1hr
Btu
Btu
1hp
·
WT = m· ⋅ (h3 − h4) = 5 ⋅ 105
⋅
⋅ 1389
− 960.6
⋅
hr
3600s (
lbm
lbm ) 0.7058 Btu
s
= 8.43 ⋅ 104 hp
In order to nd the heat transfer rate out of the condenser, we use,
·
Q41 = m· ⋅ (h1 − h4)
Because state point 1 is a saturated liquid, h1 = hf,5psia = 130.18
Btu
. Therefore,
lbm
lbm
Btu
Btu
Btu
·
Q41 = m· ⋅ (h1 − h4) = 5 ⋅ 105
⋅ 130.18
− 960.6
= − 4.11 ⋅ 108
hr (
lbm
lbm )
hr
The power required by the pump can be calculated according to,
lbm
1hr
ft 3
lbf
lbf
144in 2
1hp
·
WP = m· ⋅ (h1 − h2) = m· ⋅ νf (Ti) ⋅ (p1 − p2) = 5 ⋅ 105
⋅
⋅ 0.01641
⋅ 5 2 − 1000 2 ⋅
⋅
lbf ⋅ ft
hr
3600s
lbm ( in
in )
ft 2
550 s
·
WP = − 593 hp
To nd the heat transfer rate supplied by the boiler, we nd,
·
Q23 = m· ⋅ (h3 − h2)
·
To calculate Q23, then, we require h2, which can be found via,
Btu
ft 3
lbf
lbf
144in 2
1Btu
Btu
h2 = h1 − νf (T1) ⋅ (p1 − p2) = 130.18
− 0.01641
⋅ 5 2 − 1000 2 ⋅
⋅
= 133.20
2
lbm
lbm ( in
in )
1ft
778.169ft ⋅ lbf
lbm
Now,
lbm
Btu
Btu
Btu
·
Q23 = 5 ⋅ 105
⋅ 1389
− 133.20
= 6.28 ⋅ 108
hr (
lbm
lbm )
hr
Finally,
Btu
Btu
960.6 lbm
− 130.18 lbm
h4 − h1
η =1−
=1−
= 33.8 %
Btu
Btu
h3 − h 2
1389
− 133.20
lbm
lbm
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h4 = hf + χ4 ⋅ (hg − hf ) = 130.18
Real Rankine Cycle
A real Rankine cycle is identical in layout to the ideal Rankine cycle shown in Fig. 4.18. The
primary di erence between the two is that either or both of the pump and turbine are not
isentropic. That is, the pump and/or turbine will have some isentropic e ciency. The
isentropic e ciency for each device is rewritten for convenience below.
Pump Isentropic E ciency
The isentropic e ciency for a pump can be written as,
·
νf (T1) ⋅ (p1 − p2)
Ws
h1 − h2s
ηP = · =
=
h1 − h2
h1 − h2
Wa
(4.38)
where the subscripts “s” and “a” refer to “isentropic” and “actual”, respectively.
Turbine Isentropic E ciency
The isentropic e ciency of a turbine can be written as,
·
h3 − h4
Wa
ηT = · =
h3 − h4s
Ws
(4.39)
where the subscripts “s” and “a” refer to “isentropic” and “actual”, respectively.
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PAGE 38 4
The adiabatic turbine in a Rankine cycle is 95% e cient. Steam leaves the boiler and enters
the turbine at 700 psia and 1100∘F with a mass ow rate of 5 lbm/s. The steam leaves the
turbine at a pressure of 2 psia, and the pump is 100% e cient. Determine:
1. The net power produced by the cycle [hp]
2. The heat transfer rate produced by the boiler [Btu/s]
3. The cycle thermal e ciency [%]
4. The quality of the steam at the turbine’s exit, χ4
Solution
The net power produced by the cycle can be calculated as,
·
·
·
Wnet = WP + WT = m· ⋅ (h1 − h2) + m· ⋅ (h3 − h4)
Clearly, we need to nd the enthalpy at each state point.
State Point 1
Because we are not told otherwise, we consider the water to be a saturated liquid at the inlet
of the pump. Therefore, h1 = hf at p1 = 2 psia. We use the saturated water pressure table to
nd h1.
Thus, h1 = 94.02
Btu
.
lbm
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Example 4.9 - Real Rankine Cycle
Because the pump is isentropic, we nd h2 to be,
Btu
ft 3
lbf
lbf
144in 2
1Btu
Btu
h2 = h1 − νf (T1) ⋅ (p1 − p2) = 94.02
− 0.01623
⋅ 2 2 − 700 2 ⋅
⋅
=
96.12
lbm
lbm ( in
in )
ft 2
778ft ⋅ lbf
lbm
State Point 3
We are provided with the temperature (T3 = 1100∘F) and pressure (p3 = 700 psia) at state
point 3. In this case, we must rst determine what the phase of the substance is ahead of the
turbine. To do this, we use the saturated water pressure table, as shown below.
C l e a r l y, t h e w a t e r ex i s t s a s a s u p e r h e a t e d v a p o r a h e a d o f t h e t u r b i n e
(Tgiven = 1100∘F > Tsat = 503.13∘F). Therefore, we must
nd h3 using the superheated
vapor table.


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PAGE 386


State Point 2
Btu
.
lbm
State Point 4
To determine h4 , we must rst determine the substance’s phase at state point 4. We know
that the pressure p4 = 2
Btu
lbf
and s4s = s3 = 1.7341
. We use the saturated water
2
lbm ⋅ R
in
pressure table (the rst table provided in this example) to x our state, and nd that
sf = 0.17499
Btu
Btu
Btu
< s4s = 1.7341
< sg = 1.9194
lbm ⋅ R
lbm ⋅ R
lbm ⋅ R
Thus, state point 4s exists as a saturated liquid. We emphasize that this is state point 4 in
the case where the turbine is isentropic. We can now use this to nd h4 via the isentropic
e ciency of the turbine,
ηT =
h3 − h4
h3 − h4s
Thus, we need to calculate h4s. First, we must calculate χ4s,
s4s − sf
χ4s =
sg − sf
Btu
=
Btu
1.7341 lbm ⋅ R − 0.1749 lbm ⋅ R
Btu
Btu
1.9194 lbm
−
0.17499
⋅R
lbm ⋅ R
= 0.89
Now,
Btu
Btu
Btu
Btu
h4s = hf + χ4s ⋅ (hg − hf ) = 94.02
+ 0.89 ⋅ 1115.8
− 94.02
= 1007
(
lbm
lbm
lbm )
lbm
Finally, we compute h4 via,
h4 = h3 − ηT ⋅ (h3 − h4s) = 1570.4
Btu
Btu
Btu
Btu
− 0.95 ⋅ 1570.4
− 1007
= 1035
(
lbm
lbm
lbm )
lbm
Net Power Produced by Cycle
The net power produced by the cycle is calculated as,
lbm
Btu
Btu
3600s
1hp
·
Wnet = 5
⋅ (94.02 − 96.12)
+ (1570.4 − 1035)
⋅
⋅
s (
lbm
lbm )
1hr
2545 Btu
hr
= 3678 hp
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Here, we nd that h3 = 1570.4
Heat Transfer Rate Produced by Boiler
The heat transferred into the system by the boiler is calculated as,
lbm
Btu
Btu
Btu
·
Q23 = m· ⋅ (h3 − h2) = 5
⋅ 1570.4
− 96.12
= 7370.4
s (
lbm
lbm )
s
Cycle Thermal E ciency
The cycle’s thermal e ciency is computed via,
1hr
3600s
3678.3hp
⋅
·
Wnet
η= · =
Btu
Qin
7370
2545 Btu
hr
1hp
= 35.3 %
s
Actual Quality at the Exit of the Turbine
We determine the actual quality at the exit of the turbine, χ4, using the enthalpies provided
in the saturated water pressure table at p4 = 2
χ4 =
h4 − hf
hg − hf
=
lbf
,
2
in
Btu
Btu
1035 lbm
− 94.02 lbm
Btu
Btu
1115.8 lbm
− 94.02 lbm
= 0.91 = 91 %


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PAGE 388
Reheat Rankine Cycle
There exists several methods to improve the thermal e ciency of a Rankine cycle. In this
section, we discuss the so-called “reheat” method, in which the working uid is passed back
through the boiler after exiting the rst (high pressure) turbine, and subsequently passed
through a second (low pressure) turbine, as shown in the schematic below.
Figure 4.21. Schematic of reheat Rankine cycle. Shown are the high pressure (HPT) and low pressure (LPT) turbines used to produce power in this
cycle.
In order to achieve a higher system e ciency with a reheat cycle, a higher pressure is
required through the boiler. However, this risks excess liquid at the exit of the low pressure
turbine, which can corrode its blades. To avoid this, we also increase the temperature by
adding additional heat to the system (i.e., reheat through the boiler) such that the quality at
the exit of the low pressure turbine remains relatively high (i.e., χe,LPT > 90 % ).
Thus, the major changes that occur within this system are to the net heat into the system
and the net power produced by the two-stage turbine system,
·
·
·
Qin = Qprimary + Qreheat = m· ⋅ (h3 − h2 + h5 − h4)
(4.40)
·
·
·
Wnet = WHPT + WLPT = m· ⋅ (h3 − h4 + h5 − h6)
(4.41)
and,
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PAGE 389
The corresponding T-s diagram is provided for visualization of these processes below. Note
that the diagram itself re ects the case when all devices operate isentropically.
Figure 4.22. T-s diagram for water (SI Units) showing a typical Rankine cycle. Note: all devices are assumed to be isentropic for ease of viewing on
the T-s diagram.
In Fig. 4.22, all devices are assumed to be isentropic for ease of viewing. Note, however, that
the pump and turbines are unlikely to operate with e ciencies approaching 100%.
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PAGE 390
Water is the working uid in a Rankine cycle with reheat. Superheated vapor enters the high
pressure turbine at 10 MPa and 500∘C, and expands to a pressure of 0.7 MPa before it is
reheated back to 500∘C ahead of the low pressure turbine. It then expands to 5 kPa before
entering the condenser. Assume that the water is a saturated liquid at the inlet of the pump.
If the high pressure turbine and pump are isentropic, and the low pressure turbine has an
isentropic e ciency of 90%, determine:
1. The speci c heat transfer rate into the working uid [kJ/kg]
2. The thermal e ciency of the cycle [%]
Solution
Ultimately, we require knowledge of the enthalpy at each of the 6 state points in our reheat
Rankine cycle. As a result, we rst solve for h at each state point, below.
State Point 1
We are told in the problem to assume that the working uid enters the pump as a saturated
liquid, and we also know that the working uid exists at its lowest pressure ahead of the
pump (i.e., p1 = 5 kPa ). Thus, we can use the saturated water pressure table to determine
h1,
Here, we nd that h1 = hf, 5 kPa = 137.75
kJ
.
kg
State Point 2
Since the pump operates isentropically, we can determine h2 via,
kJ
m3
kJ
h2 = h1 − νf (T1) ⋅ (p1 − p2) = 137.75
− 0.001005
⋅ (5kPa − 10,000kPa) = 147.8
kg
kg
kg

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PAG E 3 91


Example 4.10 - Reheat Rankine Cycle
State Point 3
We are explicitly told that the working uid (water) is superheated at state point 3. Thus,
we can utilize the superheated vapor tables to determine h3.
kJ
The superheated vapor table indicates that h3 = 3375.1 .
kg
State Point 4
At state point 4, we know that p4 = 0.7 MPa = 700 kPa and that s4 = s3 = 6.5995
kJ
,
kg ⋅ K
since the high pressure turbine is isentropic. We therefore determine the phase at state
point 4 using the saturated water pressure table.



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PAGE 392
According to the table above, sf < s4 < sg, and we have a saturated mixture. We now have to
determine the quality at this state point in order to nd h4.
χ4 =
6.5996 kgkJ⋅ K − 1.9918 kgkJ⋅ K
6.7071 kgkJ⋅ K − 1.9918 kgkJ⋅ K
= 0.977
Now we nd,
h4 = hf + χ4 ⋅ (hg − hf ) = 697
kJ
kJ
kJ
kJ
+ 0.977 ⋅ 2762.8
− 697
= 2715.3
(
kg
kg
kg )
kg
State Point 5
At state point 5, we know that p5 = p4 = 0.7 MPa = 700 kPa and T5 = 500∘C , since the
working uid is reheated back to the temperature ahead of the high pressure turbine. We
use the saturated water pressure table to rst determine the phase of the water ahead of the
low pressure turbine.
Clearly, Tgiven = 500∘C > Tsat = 164.95∘C and we have a superheated vapor. Thus, we must
use the superheated vapor table to determine h5.


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PAGE 393
MPa. Because we are directly between these pressures at p5 = 0.7 MPa , we can simply
average the enthalpies to produce h5 = 3482.4
kJ
.
kg
State Point 6
State point 6 represents the exit of the low pressure turbine. Therefore, we are at a pressure
of p6 = 5 kPa . We also know that the low pressure turbine has an isentropic e ciency of
ηLPT = 0.9 and,
ηLPT =
h5 − h6
h5 − h6s
which we will use to calculate h6 . To do that, we must
s6s = s5 = 7.94
rst determine h6s , where
kJ
. At a pressure of 5 kPa (see the rst table in this example problem),
kg ⋅ K
sf < s6s < sg. Therefore, the isentropic state point 6s is a saturated mixture.
χ6s =
7.94 kgkJ⋅ K − 0.4762 kgkJ⋅ K
8.3938 kgkJ⋅ K − 0.4762 kgkJ⋅ K
= 0.943
Therefore,
kJ
kJ
kJ
kJ
h6s = hf + χ6s ⋅ (hg − hf ) = 137.75
+ 0.943 ⋅ 2560.7
− 137.75
= 2424.48
(
kg
kg
kg )
kg
Now we use ηLPT to nd h6,
h6 = h5 − ηLPT ⋅ (h5 − h6s) = 3482.4
kJ
kJ
kJ
kJ
− 0.9 ⋅ 3482.4
− 2424.48
= 2530.3
(
kg
kg
kg )
kg
Solving for q23
q23 = h3 − h2 = 3375.1
kJ
kJ
kJ
− 147.8
= 3227.3
kg
kg
kg
Solving for η
137.75 kJ
− 147.8 kJ
+ 3375.1 kJ
− 2715.3 kJ
+ 3482.4 kJ
− 2530.3 kJ
w12 + w34 + w56
h1 − h2 + h3 − h4 + h5 − h6
kg
kg
kg
kg
kg
kg
η=
=
=
= 49.6 %
kJ
q23
h3 − h 2
3227.3
kg
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PAGE 394





The enthalpy at state point 5 is found by interpolating between pressures of 0.6 MPa and 0.8
NUCLEAR POWER
On January 17, 1955, the USS Nautilus (SSN-571) signaled “Underway on nuclear power.”
This started the U.S. Navy’s operations with nuclear-powered vessels. The Nautilus initiated
many changes in the Navy, speci cally in the ways submarines operate, namely higher
speed, ability to maintain that speed almost inde nitely, and the ability to remain
submerged for long periods of time.
Nuclear power was later adapted to aircraft carriers and cruisers. The USS Enterprise
(CVN-65) was commissioned on November 25, 1961. This nuclear-powered platform had
eight nuclear reactors providing 280,000 HP divided equally between four shafts, plus all of
the electrical and steam-operated catapult needs of the ship. By the time the USS Nimitz
(CVN-68) was built, improvements in reactor design allowed the ships of this class to have
the same shaft power output provided by two nuclear reactors. The Navy also built a
number of surface combatants with nuclear propulsion. All of the U.S. Navy’s nuclearpowered vessels employ pressurized water reactors.
The overall propulsion e ciency of
current naval nuclear power plants is about 15%.
Nuclear power is the preferred power system for aircraft carriers as it allows more fuel to be
carried for aircraft and surface escorts, and dramatically reduces the number of tankers
required to support a Battle Group. Current aircraft carrier reactor designs can operate for
approximately 20 years before the nuclear fuel (or “core”) must be replaced.
The primary disadvantages of nuclear power compared to conventionally powered surface
vessels are: much higher initial cost, much higher maintenance costs, signi cantly more
personnel required for operation and maintenance, and it is politically unpopular with some
groups and countries, and thus some countries impose restrictive port visitation
requirements.
Principles of Nuclear Fission
Recall that a chemical element is de ned by the number of protons in its nucleus and that
neutrons are also present in the nucleus of most elements.
An element can also have
various isotopes due to di ering numbers of neutrons accompanying the same number of
protons.
The term,
ssionable, refers to the ability of an element to undergo a
reaction.
There are very few elements that are
ssion
ssionable and only certain isotopes are
readily ssionable. This is the case with the principle nuclear fuel used in ssion reactors uranium.
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PAGE 395
rare uranium isotope U-235 () which has 92 protons and molecular weight of 235. Only
approximately 0.7% of natural uranium, as it is mined from the earth, is U-235; the balance
is predominately U-238 which is more stable, and much more di cult to
U-235.
ssion than
Commercial reactors use uranium that has been enriched to approximately 3%
U-235; Navy reactors, since a higher power density is desired, use uranium with much
higher U-235 enrichment, thus reducing the weight of the undesired U-238 present in the
fuel.
Depleted uranium, used in projectiles, is the waste product U-238 from the enrichment
process. There are two enrichment processes in current use in the world: gaseous di usion
and gaseous centrifuging. Gaseous di usion micro- lters uranium that has been gasi ed into
uranium hexa uoride (UF6).
Centrifuge enrichment spins UF6 at high speeds and the UF6
with the heavier isotope (U-238) tends to migrate to the outer region of the centrifuge while
the lighter isotope tends to stay closer to the centrifuge axis. Many multiple centrifuges in
series are required to achieve signi cant enrichment by this technique.
Once the UF6 gas reaches the required degree of enrichment, it is reformulated to solid
uranium dioxide, UO2. The UO2 is shaped and then covered with a metal cladding to protect
the UO2 from corrosion. The cladding metal is selected based upon having high thermal
conductivity, low susceptibility to corrosion, and hardness to minimize erosion e ects from
water that is pumped through the reactor. The fuel is assembled into fuel elements that are
inserted into the reactor in geometric shapes referred to as fuel element assemblies that make
up the reactor core.
The U-235 ssion reaction is:
235
1
1
92 U +0 n → FP1 + FP2 + 2.430 n + E
(4.42)
where E is energy, n is a neutron and FP1 and FP2 are elemental ssion products (e.g. Iodine,
Barium, Cesium, etc) that are typically much more radioactive than the parent material.
Note that, on average, 2.43 neutrons are released from each
ssion of U-235.
These
neutrons are available to cause subsequent ssions in what is referred to as a chain reaction.
In a nuclear power reactor this chain reaction is controlled to maintain the desired power
level. In an atomic bomb, with approximately 80% U-235 enrichment, the chain reaction
runs unchecked which results in a massive release of energy in a fraction of a second. The
ssion products tend to decay to more stable elements over varying periods of time,
releasing energy referred to as decay heat. Decay heat can produce a signi cant amount of
heat for days and even weeks after a reactor is shutdown.
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PAGE 396

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U.S Navy nuclear propulsion plants, and most commercial reactor designs, use ssion of the
1010 BTU/lbm of U-235, which is many orders of magnitude greater than that released in a
typical chemical combustion reaction (e.g.
heating value of aircraft fuel JP-5 is
approximately 18,300 BTU/lbm).
The student may be familiar with the terms light water reactor, heavy water reactor, or graphitemoderated reactor. These terms relate to the way in which the neutrons that are produced by
ssion are slowed down (moderated) to the appropriate energy level to cause another
ssion. As noted in the equation above, there is an average of 2.43 neutrons produced per
ssion event. Neutrons are produced at an energy level that is too high to cause a ssion
and must be slowed down to the proper energy level without losing too many of them in the
process. Neutron moderation slows the neutrons and re-directs them to the core in order to
sustain the ssion chain reaction. Navy nuclear power uses light water reactors.
Control rods are devices made from materials that absorb neutrons.
Control rods are
inserted into spaces inside the core to shut it down. This may be done either by inserting
the rods slowly with their drive motors or dropping them quickly in what is called a scram.
A scram may be manually initiated by plant operators or automatically initiated by the
reactor safety control systems. Either way, the inserted control rods absorb neutrons and
prevent these neutrons from causing ssion in the fuel. To start up a reactor, the rods are
withdrawn to allow some of the otherwise absorbed neutrons to cause additional ssions.
This power level is raised until the chain reaction is self-sustaining and the reactor is
referred to as critical.
One additional factor in operating nuclear systems is that, once the fuel has been operated
to make steam, the reactor continues to emit heat even if it is shutdown. This decay heat
must be removed to prevent heating the water in the reactor to boiling conditions. The
approximate amount of decay heat is about 10% of what the power was immediately prior
to shutdown and it decays away exponentially.
Types of Nuclear Reactors
Current naval applications of nuclear power all use the pressurized water reactor (PWR)
con guration. In a PWR, high-pressure, subcooled water is pumped through the reactor to
carry the heat from the reactor core to a steam generator. The water ows around the fuel
elements and is referred to as primary water or coolant.
The steam generator is a heat
exchanger that separates the high-pressure primary water from secondary water, which is at a
lower pressure and allowed to boil. The steam generator is analogous to the boiler in a
conventional (hydrocarbon
red) steam plant.
The steam produced by a PWR’s steam
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PAGE 397
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The amount of energy released in ssion, or U-235 heating value, is approximately 3.51 x
otherwise similar to a conventional steam plant.
Pressurized Water Reactor
• •• ••• •••• •
• •• • • •• • • • •• •• •
•
• • •• •• •
• • • ••• •
•
• • •• • • • •• •
•• • •
• ••• • •• •
•• • •
•
•
•
•
• • •• •• • • •• • ••• • • • •
•• • • •
•
• • • • • • •• •
• • •• •• • • • •• • • • •
•• • • •
Figure 4.23. Schematic of a simpli ed pressurized water reactor.
Figure 4.23 shows a simpli ed pressurized water reactor, illustrating the two “loops” and
their components. The two loops are (1) the high-pressure, subcooled primary loop and (2)
the steam cycle secondary loop.
The principal components of the primary loop are the
reactor, the main coolant pump, and the high pressure side of the steam generator. The
principle components on the secondary loop include the low pressure side of the steam
generator, the main feed pump, steam turbine and the condenser.
described below.
Each component will be
The numbering of the state points as shown in Figure 1 neglects any
pressure drops due to major or minor losses in the piping connecting the components.
Otherwise states would be required at the inlet and exit of each component.
The primary loop’s working uid is water that is pressurized to such high levels (typically in
the range of 1500 to 2500 psi, depending upon the speci c design) that it will not boil in the
primary loop, even at the fairly high temperatures in the reactor. The state of the primary
coolant is thus a compressed or subcooled liquid in the owing portion of the system. The
reactor is basically a heat exchanger in which thermal energy generated by the nuclear
ssion process is removed by the primary coolant. The primary coolant then rejects the heat
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PAGE 398

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generator is saturated steam (not superheated). The steam portion of the overall system is
in the steam generator (S/G) to the steam in the secondary loop. Heat transfer in the steam
generator takes place across numerous small-diameter tubes.
The primary coolant
ows
through the inside of the tubes, while the secondary system’s water boils on the outside of
the tubes. After transferring heat, the primary coolant is returned to the reactor. The Main
Coolant Pumps (MCPs) providing the pressure rise necessary to overcome the frictional
head losses of the pipes, valves, reactor, and S/G. The pressurizer is a special component
that is discussed in greater detail later.
The pressurizer is a special primary system component. The pressurizer is connected to the
hot leg which runs from the reactor to the steam generator. The pressure in the hot leg is
essentially the pressurizer pressure. The pressurizer is
tted with electric heaters which
serve to boil the water in the pressurizer and create a steam bubble at the top of the
pressurizer.
If the primary pressure is lower than desired, the electric heaters in the
pressurizer will energize to raise the saturation temperature (and hence saturation
pressure). The pressurizer is also equipped with a spray nozzle connected to the cold leg
after the main coolant pumps. If pressure is too high, the spray valve is remotely opened to
allow water to spray in via the spray nozzle which lowers the saturation temperature and
hence the saturation pressure.
p7 = psat(Tpressurizer )
(4.43)
There will typically be a signi cant pressure drop in the primary loop across both the reactor
and the steam generator. In both cases the coolant must travel through small channels or
tubes to promote e ective heat transfer. There are also likely to be major and minor losses
in the piping that connect the components of the primary loop.
The pressure head of the
pump for a loop can be determined by summing all of the major and minor losses.
p6 − p5
Δhpump =
Δh
+
Δh
=
∑ minor ∑ major
ρ⋅g
Water is not incompressible at these pressures and temperature.
(4.44)
Therefore, the use of
compressed liquid tables is more appropriate, but generally requires a more complete set of
tables than what is provided in your textbook. However, across the steam generator and the
reactor the change in enthalpy with respect to temperature is much greater than the change
in enthalpy with respect to pressure; therefore it is reasonable to approximate the change in
enthalpy as Δh ≈ cp ⋅ ΔT. A more accurate result can be obtained by interpolation with the
compressed liquid tables.
·
QSG = m· ⋅ (h5 − h7) ≈ m· ⋅ cp ⋅ (T5 − T7)
(4.45)
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PAGE 399
The work of a reversible process can be determined by integrating the speci c volume over
the change in pressure. Speci c volume is not a strong function of pressure and remains
relative constant across the pump, therefore our standard equation for calculating the work
of an isentropic pump is still reasonable.
wrev = ν ⋅ dP ≈ ν5 ⋅ (p6 − p5)
∫
·
WMCP =
·
WMCP,isentropic
ηP
(4.46)
− m· ⋅ ν ⋅ (p6 − p5
=
ηP
(4.47)
where ηP is the isentropic e ciency of the pump.
Figure 4.24. Two-loop primary system with pressurizer shown.
Figure 4.24 illustrates the redundancy provided by having two primary loops connected to
the same reactor and also shows redundancy with multiple coolant pumps that can be
isolated in the event of system damage. This is generally characteristic of submarine
systems.
The steam generator is a heat exchanger that has high temperature hot water on the primary
loop side which is used to boil water at a lower pressure on the steam generation side.
Liquid water is supplied to the steam generator using the main feed pump, the water boils
and leaves as a saturated vapor, χ3 = 1 . Since the steam cannot be heated above the inlet
temperature of the coolant, T7, there is not much opportunity to superheat in a pressurized
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PAGE 400
water reactor. The rate of heat transfer in the steam generator can be determined from an
energy balance on the steam generator.
·
QSG = m· S ⋅ (h3 − h2) = m· P ⋅ (h7 − h5)
(4.48)
The rest of the secondary loop is the same as a conventional steam power plant. The steam
is expanded in a turbine that produces shaft power. The water leaving the turbine is directed
to a condenser. As with the conventional steam plant, the condenser is operated at a vacuum
(typically below 5 psia). The condenser is often a shell and tube heat exchanger with
numerous small-diameter tubes. The condenser’s function is to reject the heat necessary to
turn the secondary water from saturated mixture conditions to saturated liquid (SL)
conditions. If the system is modeled as ideal, the water leaving the condenser is a saturated
liquid. In an actual system the water leaving the condenser will be a subcooled liquid. The
amount of subcooling of the liquid leaving the condenser is referred to as condensate
depression and is reported in degrees of subcooling below saturation temperature at the
condenser pressure.
Figure 4.25 shows the states of the primary and secondary loops on a T-s diagram.
When
the Rankine cycle is sketched on a Ts diagram, state 2 is often shown as being above state 1
in order to illustrate the isentropic process. However, when drawn to scale, state 1 and state
2 are actually right on top of one another as shown in the Figure 3. Since the e ectiveness
of the steam generator will be less than 1, T3 must be less than T7 and in reality T3 will
likely be less than T5.
Figure 4.24. Representative T-s diagram from a pressurized water reactor.
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PAG E 4 01
Example 4.11 - Pressurized Water Reactor
Coolant enters the steam generator at 320∘C and leaves at 280∘C. The mass ow rate of the
coolant is 400 kg/s. A steam bubble is used to keep the pressure at 15 MPa inside the
pressurizer. Saturated vapor leaves the steam generator at 6 MPa and enters the turbine. The
pressure drop is 500 kPa in the reactor and 500 kPa in the steam generator on the primary
loop side. The isentropic e ciency of the turbine is 90% and both pumps are ideal.
Condensate enters the main feed pump at 10 kPa and 30∘C. Determine:
1. The temperature inside the pressurizer [∘C]
2. The rate of heat transfer in the steam generator [MW]
3. The power input to the main coolant pump [kW]
4. The rate of heat transfer in the reactor [MW]
5. The mass ow rate of the steam leaving the steam generator [kg/s]
6. The power output of the turbine [MW]
7. The power input to the main feed pump [kW]
8. The overall thermal e ciency of the system [%]
Solution
We determine the enthalpies at each state point (1-7) in order to answer 1-8 above.
State Point 1
We know that the temperature and pressure at state point 1 are 30∘ C and 10 kPa,
respectively. First, we use the saturated water pressure table to determine the phase of our
working uid.
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PAGE 402
h1 = hf (T1). We now use the saturated water temperature table to nd the enthalpy at state
point 1.
kJ
Thus, h1 = 125.74 .
kg
State Point 2
At state point 2, we only know that p2 = 6,000 kPa. Because the pump is isentropic,
however, we can use,
kJ
m3
kJ
h2 = h1 − νf (T1) ⋅ (p1 − p2) = 125.74
− 0.001004
⋅ (10kPa − 6,000kPa) = 131.74
kg
kg
kg
State Point 3
At state point 3 we have a saturated vapor at p3 = 6,000 kPa. Thus, we use the saturated
water pressure table to determine h3.
We nd that h3 = hg = 2784.6
kJ
.
kg
State Point 4
At state point 4, p4 = 10 kPa. We also know that,
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nd that Tgiven = 30∘C < Tsat = 45.81∘C . Therefore, we have a compressed liquid and
We
Therefore, to calculate h4 we must
h3 − h4
h3 − h4s
rst determine h4s . In this case, we know that
kJ
. At p4 = 10 kPa, we
kg ⋅ K
s4s = s3 = 5.8902
nd that sf < s4s = 5.8902
Therefore, the phase at state point 4 is a saturated mixture, where,
χ4s =
5.8902 kgkJ⋅ K − 0.6492 kgkJ⋅ K
8.1488 kgkJ⋅ K − 0.6492 kgkJ⋅ K
kJ
< sg .
kg ⋅ K
= 0.699
and,
kJ
kJ
kJ
kJ
h4s = hf + χ4s ⋅ (hg − hf ) = 191.81
+ 0.699 ⋅ 2583.9
− 191.81
= 1863.9
(
kg
kg
kg )
kg
Now,
kJ
kJ
kJ
kJ
h4 = h3 − ηT ⋅ (h3 − h4s) = 2784.6
− 0.9 ⋅ 2784.6
− 1863.9
= 1955.7
(
kg
kg
kg )
kg
State Point 5
The pressure at state point 5 is p5 = p7 − 0.5 MPa = 14.5 MPa (where p7 = ppressurizer = 15
MPa) and the temperature is T5 = 280∘ C. Here, we use the Compressed Liquid Water
Table to determine h5.



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ηT = 0.9 =
h5 = 1235.0
kJ
14.5MPa − 10MPa
kJ
kJ
kJ
+
⋅ 1233.0
− 1235.0
= 1233.2
kg ( 15MPa − 10MPa ) (
kg
kg )
kg
State Point 6
To determine the enthalpy at state point 6, we use,
m3
kJ
h6 = h5 − νf (T5) ⋅ (p5 − p6) = 0.001311
⋅ (14,500kPa − 15,500kPa) = 1234.5
kg
kg
because the pump is operating isentropically.
State Point 7
At state point 7, the pressure is p7 = ppressurizer = 15 MPa and the temperature is T7 = 320∘
C. Again, the substance is a compressed liquid, and we use the Compressed Liquid Water
Table to determine h7. Using the same table as the one shown for state point 5, we nd that
h7 = 1454
kJ
.
kg
Temperature Inside of Pressurizer
We know that Tpressurizer = Tsat,@ppressurizer = 342.16∘C (see the saturation temperature on the
compressed liquid water table).
Rate of Heat Transfer in the Steam Generator
kg
kJ
kJ
·
QSG = m· p ⋅ (h5 − h7) = 400
⋅ 1233.2
− 1454
= − 88,320 kW
(
)
s
kg
kg
Power Input to the Main Coolant Pump
kg
kJ
kJ
·
WMCP = m· p ⋅ (h5 − h6) = 400
⋅ 1233.2
− 1234.5
= − 520 kW
(
)
s
kg
kg
Rate of Heat Transfer in the Reactor
kg
kJ
kJ
·
QR = m· p ⋅ (h7 − h6) = 400
⋅ 1454
− 1234.5
= 87,800 kW
(
)
s
kg
kg
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PAGE 405


Here we interpolate between 10 MPa and 15 MPa to determine h5, where,
Mass Flow Rate of the Steam
·
Q
88,320kW
kg
SG
m· S =
=
=
33.29
kJ
h3 − h 2
s
2652.9
kg
Power Output of the Turbine
kg
kJ
kJ
·
WT = m· S ⋅ (h3 − h4) = 33.29
⋅ 2784.6
− 1955.0
= 27,617 kW
(
)
s
kg
kg
Power Required by the Main Feed Pump
kg
kJ
kJ
·
WMFP = m· S ⋅ (h1 − h2) = 33.29
⋅ 125.74
− 131.74
= − 199.7 kW
(
)
s
kg
kg
Overall Thermal E ciency of the Cycle
·
·
·
WT + WMCP + WMFP
26,897kW
η=
=
= 30.6 %
·
87,800kW
QR


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PAGE 406
VA P O R C O M P R E S S I O N R E F R I G E R AT I O N / H E AT P U M P S
Vapor compression refrigeration and heat pump systems are widely credited as one of the
most important technological developments of the 21st century, allowing for reasonable
living conditions and prosperity in climates that would otherwise be too harsh to live in
year-round. These systems also enabled the distribution of goods across the globe, and
advanced dilution refrigeration systems are expected to be critical components in future
quantum computers. Here, we will discuss the thermodynamic concepts associated with
conventional vapor compression refrigeration and heat pump systems.
Thermodynamic Analysis
Vapor compression systems can be used to achieve both cooling and heating of spaces,
where the working
uid’s
ow direction is often changed to achieve one or the other. A
simple schematic of a vapor compression system is shown below.
Figure 4.25. Schematic of a Vapor Compression system; in refrigeration-mode, we keep remove heat from a cold space to maintain Tc


fl
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PAGE 407
temperature of our refrigerant (Note: we will consider R-134a to be the working
uid
in vapor compression problems for this course, despite its adverse impacts on the
environment). The condenser is a heat exchanger used to keep a space warm at Th , while
the evaporator is a heat exchanger that is used to cool a space and maintain its temperature
at Tc. The expansion valve between state points 3 and 4 is a throttle that is used to lower the
pressure and temperature of the refrigerant.
Recall that for a throttle, hi = he; therefore, h3 = h4 according to the schematic in Fig. 4.25.
Refrigeration/Heat Pump Cycle Energy Transport Processes
Process
Device
1 → 2: Isentropic
Compression
2 → 3: Isobaric Heat
Addition
3 → 4: Isenthalpic
Expansion
4 → 1: Isobaric Heat
Removal
Heat Transferred
Power Produced/
Consumed
Compressor
·
Q12 = 0
·
W12 = m· ⋅ (h1 − h2)
Heat Exchanger
·
Q23 = m· ⋅ (h3 − h2)
·
W23 = 0
Expansion Valve
·
Q34 = 0
·
W34 = 0
Heat Exchanger
·
Q41 = m· ⋅ (h1 − h4)
·
W41 = 0
The process is shown below on a T-s diagram.
Figure 4.25. T-s diagram for an ideal vapor compression refrigeration/heat pump system. Note state point 1 must be a saturated or superheated
vapor in order to pass through the compressor. For a non-ideal system, the pump is no longer isentropic and state point 2 is shifted to the right on
its constant pressure line.
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PAGE 408













In the above schematic, the compressor is designed to increase the pressure and
Example 4.12 - Vapor Compression Refrigeration
R-134a is used in a vapor compression refrigeration system to cool a household refrigerator
with a cooling capacity of 1,000 Btu/hr. Refrigerant enters the evaporator at a temperature
of -10∘F and exits as a saturated vapor. The refrigerant exits the condenser as a saturated
liquid at a pressure of 120 psia. If the compressor has an isentropic e ciency of 80%,
determine:
1. The evaporator pressure [psia]
2. The mass ow rate of the refrigerant [lbm/min]
3. The compressor power input [hp]
4. The coe cient of performance for the refrigeration system (COP)
Solution
First, let’s nd the enthalpy at each state point so that we can answer the questions posed to
us above.
State Point 1
We are told that we have a saturated vapor at the exit of the evaporator, which maintains a
temperature of T1 = T4 = − 10∘F in this problem. Thus, we use the saturated R-134a
temperature table to obtain h1.
Using the table above, we nd that h1 = 101.61
Btu
.
lbm
State Point 2
At state point 2, we only know that the pressure p2 = 120 psia. However, we also know,
ηC = 0.8 =
h1 − h2s
h1 − h2
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PAGE 409
Btu
and that p2s = p2 = 120 psia. We must rst determine the phase
lbm ⋅ R
of the R-134a at state point 2s prior to determining h2s . To do this, we use the saturated
s2s = s1 = 0.2266
R-134a pressure table.
We nd that s2s > sg at p2s = p2 = 120 psia. Thus, we have a superheated vapor at state
point 2s. We therefore use the superheated vapor table to determine h2s.
Btu
Interpolating, we nd that h2s = 120
. Thus,
lbm
Btu
Btu
h1 − h2s
Btu 101 lbm − 120 lbm
Btu
h2 = h1 −
= 101
−
= 124.6
ηC
lbm
0.8
lbm

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PAG E 410
fi

Thus, we must determine h2s to then compute h2 . For state point 2s, we know that
At state point 3, we have a saturated liquid at p3 = p2 = 120 psia. We can use the saturated
R-134a pressure table to determine h3 , which is the
heading, above. Thus, h3 = 41.787
rst table under the State Point 2
Btu
.
lbm
State Point 4
Because the process between state points 3 and 4 is isenthalpic, we know that
h4 = h3 = 41.787
Btu
.
lbm
Evaporator Pressure
The evaporator pressure is psat at state point 1, which we nd to be 16.64 psia.
Refrigerant Mass Flow Rate
Here we can use the cooling capacity to solve for the mass ow rate of the refrigerant:
Btu
·
1,000
QC
1hr
lbm
hr
m· =
=
⋅
=
0.28
h1 − h4
60min
min
101 Btu − 41.789 Btu
lbm
lbm
Compressor Power Input
lbm
Btu
Btu
60min
1hp
·
WC = m· ⋅ (h1 − h2) = 0.28
⋅ 101
− 124.6
⋅
⋅
min (
lbm
lbm )
1hr
2545 Btu
hr
= − 0.16 hp
Cycle Coe cient of Performance (COP)
·
Qevap
COPR = ·
=
| WC |
1,000 Btu
hr
0.16hp ⋅
2545 Btu
hr
= 2.9
1hp
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P A G E 4 11



State Point 3
APPENDIX

PA G E 412
Dimension
SI Units
SI/English Units
Acceleration
1 m/s2= 100 cm/s2
1 m/s2 = 3.2808 ft/s2
1 m2 = 104 cm2
1 m2 = 1550 in2 = 10.764 ft2
Area
Density
1 ft/s2 = 0.3048 m/s2
… = 106 mm2 = 10−6 km2
1 ft2 = 144 in2 = 0.0929034 m2
1 g/cm3 = 1000 kg/m3 = 1 kg/L
1 g/cm3 = 62.428 lbm/ft3
1 kJ = 1000 J = 1000 N⋅m
1 kJ = 0.94782 Btu
1 Btu = 1.055056 kJ
Energy, heat, work,
and speci c energy
… = 1 kPa⋅m3
1 kJ/kg = 1000 m2/s2
1 kWh = 3600
… = 0.036127 lbm/in3
… = 5.40395 psia⋅ft3 = 778.169 lbf⋅ft
1 Btu/lbm = 25,037 ft2/s2 = 2.326 kJ/kg
1 kWh = 3412.14 Btu
Force
1 N = 1 kg⋅m/s2
Length
1 m = 100 cm = 1000 mm = 106 μm
1 km = 1000 m
1 ft = 12 in
1 mile = 5280 ft
1 in = 2.54 cm
1 kg = 2.2046226 lbm
Mass
1 kg = 1000 g
1 metric ton = 1000 kg
1 ounce = 28.3495 g
1 slug = 32.174 lbm
1 short ton = 2000 lbm
Power
1 W = 1 J/s
1 kW = 1 kJ/s = 1000 W
1 hp = 745.7 W
1 hp = 550 lbf⋅ft/s = 0.7058 Btu/s
… = 42.41 Btu/min = 2544.5 Btu/h
1 Pa = 1 N/m2
1 psi = 144 lbf/ft2 = 6.894757 kPa
1 atm = 14.696 psi
Pressure and
Pressure head
1 kPa = 103 Pa = 10−3 MPa
1 atm = 101.325 kPa
… = 760 mm Hg (at 0∘C)
1 lbf = 1 slug⋅ft/s2
1 lbf = 32.174 lbm⋅ft/s2 = 4.44822 N
… = 29.92 in Hg (at 30∘C)
Speci c heat
1 kJ/kg⋅∘C = 1 kJ/kg⋅K = 1 J/g⋅∘C
1 Btu/lbm⋅∘F = 4.1868 kJ/kg⋅∘C
Speci c volume
1 m3/kg = 1000 L/kg = 1000 cm3/g
1 m3/kg =16.02 ft3/lbm
Temperature
T(K) = T(∘C) + 273.15
T(R) = T(∘F) + 459.67 = 1.8⋅T(K)
ΔT(K) = ΔT(∘C)
Velocity
1 m/s = 3.60 km/hr
T(∘F) = 1.8⋅T(∘C) + 32
ΔT(∘F) = ΔT(R) = 1.8⋅ΔT(K)
1 m/s = 3.2808 ft/s
1 mi/hr = 1.46667 ft/s
























































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PAG E 413



UNIT CONVERSIONS
Dimension
SI Units
SI/English Units
Viscosity
(dynamic)
1 kg/m⋅s = 1 N⋅s/m2 = 1 Pa⋅s
1 kg/m⋅s = 2419.1 lbm/ft⋅hr
… = 0.020886 lbf⋅s/ft2
… = 0.67197 lbm/ft⋅s
Viscosity
(kinematic)
1 m2/s = 104 cm2/s
1 m2/s = 10.764 ft2/s
1 m3 = 1000 L = 106 cm3
1 in3 = 0.000579 ft3 = 0.00433 gal
1 stoke = 1 cm2/s
Volume
1 gal = 231 in3 = 3.7854 L
1
1 m3/s = 60,000 L/min = 106 cm3/s
Volume ow rate
ounce = 29.5735 cm3
1 gal = 128 ounces
1 ft3/s = 448.83 gal/min
Important Note: Values highlighted in red are used extremely often, while values
highlighted in blue and green are used very often.


















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fl








fl
PAG E 414
Material Type
(New Pipe)
ε (mm)
ε (ft)
Riveted steel
0.9 - 9.0
0.003 - 0.03
Concrete
0.3 - 3.0
0.001 - 0.01
Wood stave
0.18 - 0.9
0.0006 - 0.003
Cast iron
0.15
0.0005
Galvanized iron
0.15
0.0005
Asphalted cast iron
0.12
0.0004
Commercial steel or
wrought iron
0.045
0.00015
Drawn tubing
0.0015
0.000005
Carbon steel
0.035
0.00011
Ordinary concrete
0.3 - 1.0
0.001 - 0.0032
Smoothed cement
0.3
0.001
Flexible rubber
tubing
0.006 - 0.07
0.00002 - 0.00023
PVC and plastic
pipes
0.0015 - 0.007
0.000006 - 0.000023
Copper, lead, brass,
and aluminum
0.001 - 0.002
0.0000033 - 0.0000066
PAG E 415



VA L U E S O F S U R FAC E R O U G H N E S S F O R P I P E F L O W
Object Geometry
Schematic
Drag Coe cient, CD
Cube
CD = 1.05
Afrontal = D 2
Cone
(θ = 30∘)
Afrontal =
π ⋅D
4
2
CD = 0.5
Ellipsoid
π ⋅ D2
Afrontal =
4
Finite Cylinder
(vertical)
Afrontal = L ⋅ D
Finite Cylinder
(horizontal)
π ⋅ D2
Afrontal =
4

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PAG E 416





Drag Coe cients over a Series of 3-D Objects Oriented Vertically to the Direction of Fluid Flow




ADDITIONAL DRAG COEFFICIENTS OVER 3-D BODIES
Schematic
Drag Coe cient, CD
Standing
Person
(standing and sitting)
CD ⋅ A = 9 f t 2 = 0.84 m 2
Sitting
CD ⋅ A = 6 f t 2 = 0.56 m 2
Upright
Bike
Afrontal = 5.5 f t 2 = 0.51 m 2
CD = 1.1
Racing
Afrontal = 3.9 f t 2 = 0.36 m 2
CD = 0.9
Passenger Car
Frontal Area Varies
CD = 0.3 − 0.4
Tree
Afrontal = L ⋅ D

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PA G E 417


Object Geometry


Drag Coe cients over a Series of Real Objects Oriented Vertically to the Direction of Fluid Flow


ADDITIONAL DRAG COEFFICIENTS OVER REAL BODIES
FLUID PROPERTIES (SI UNITS)
Water (Liquid at STP)
( C)
Surface Tension, σ
(N/m)
Vapor Pressure, pv
(N/m2, absolute)
Sound Speed, c
(m/s)
0
0.0756
610.5
1403
5
0.0749
872.2
1427
10
0.0742
1,228
1447
20
0.0728
2,338
1481
30
0.0712
4,243
1507
40
0.0696
7,376
1526
50
0.0679
12,330
1541
60
0.0662
19,920
1552
70
0.0644
31,160
1555
80
0.0626
47,340
1555
90
0.0608
70,100
1550
99.99
0.0589
101,300
1543
Temperature
∘
Water (Liquid at STP)
Density, ρ
Speci c Weight, γ
Dynamic Viscosity, μ
Kinematic Viscosity, ν
0
999.9
9.806
1.787⋅10−3
1.787⋅10−6
5
1,000
9.807
1.519⋅10−3
1.519⋅10−6
10
999.7
9.804
1.307⋅10−3
1.307⋅10−6
20
998.2
9.789
1.002⋅10−3
1.004⋅10−6
30
995.7
9.765
7.975⋅10−4
8.009⋅10−7
40
992.2
9.731
6.529⋅10−4
6.580⋅10−7
50
988.1
9.690
5.468⋅10−4
5.534⋅10−7
60
983.2
9.642
4.665⋅10−4
4.745⋅10−7
70
977.8
9.589
4.042⋅10−4
4.134⋅10−7
80
971.8
9.530
3.547⋅10−4
3.650⋅10−7
90
965.3
9.467
3.147⋅10−4
3.260⋅10−7
99.99
958.4
9.399
2.818⋅10−4
2.940⋅10−7
Temperature
(∘C)
(kg/m3)
(kN/m3)
[(N⋅s)/m2]
(m2/s)











































fi




















PAG E 418
Air (Gas at Standard Atmospheric Pressure)
Temperature
Density, ρ
Speci c Weight, γ
Dynamic Viscosity, μ
Kinematic Viscosity, ν
-40
1.514
14.85
1.57⋅10−5
1.04⋅10−5
-20
1.395
13.68
1.63⋅10−5
1.17⋅10−5
0
1.292
12.67
1.71⋅10−5
1.32⋅10−5
5
1.269
12.45
1.73⋅10−5
1.36⋅10−5
10
1.247
12.23
1.76⋅10−5
1.41⋅10−5
15
1.225
12.01
1.80⋅10−5
1.47⋅10−5
20
1.204
11.81
1.82⋅10−5
1.51⋅10−5
25
1.184
11.61
1.85⋅10−5
1.56⋅10−5
30
1.165
11.43
1.86⋅10−5
1.60⋅10−5
40
1.127
11.05
1.87⋅10−5
1.66⋅10−5
50
1.109
10.88
1.95⋅10−5
1.76⋅10−5
60
1.060
10.40
1.97⋅10−5
1.86⋅10−5
70
1.029
10.09
2.03⋅10−5
1.97⋅10−5
80
0.9996
9.803
2.07⋅10−5
2.07⋅10−5
90
0.9721
9.533
2.14⋅10−5
2.20⋅10−5
100
0.9461
9.278
2.17⋅10−5
2.29⋅10−5
200
0.7461
7.317
2.53⋅10−5
3.39⋅10−5
300
0.6159
6.040
2.98⋅10−5
4.84⋅10−5
400
0.5243
5.142
3.32⋅10−5
6.34⋅10−5
500
0.4565
4.477
3.64⋅10−5
7.97⋅10−5
1,000
0.2772
2.719
5.04⋅10−5
1.82⋅10−4
(∘C)
(kg/m3)
(N/m3)
[(N⋅s)/m2]
(m2/s)


















































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PAG E 419
Other Common Fluids (Liquid at STP)
Density, ρ
Dynamic Viscosity, μ
Surface Tension, σ
Ammonia
608
2.20⋅10−4
2.13⋅10−2
Benzene
881
6.51⋅10−4
2.88⋅10−2
Carbon tetrachloride
1,590
9.67⋅10−4
2.70⋅10−2
Ethanol
789
1.20⋅10−3
2.28⋅10−2
Ethylene Glycol
1,117
2.14⋅10−2
4.84⋅10−2
Freon 12
1,327
2.62⋅10−4
-
Gasoline
680
2.92⋅10−4
2.16⋅10−2
Glycerin
1,260
1.49
6.33⋅10−2
Kerosene
804
1.92⋅10−3
2.80⋅10−2
Mercury
13,550
1.56⋅10−3
4.84⋅10−1
Methanol
791
5.98⋅10−4
2.25⋅10−2
SAE 10W oil
870
1.04⋅10−1
3.60⋅10−2
SAE 30W oil
891
2.90⋅10−1
3.50⋅10−2
Water
998
1.00⋅10−3
7.28⋅10−2
Seawater (30%)
1,025
1.07⋅10−3
7.28⋅10−2
Fluid
(kg/m3)
(kg/m⋅s)
(m2/s)































































PAGE 420
Gas
Molecular
Weight
Dynamic Viscosity, μ
γ =ρ⋅g
Hydrogen, H2
2.016
9.05⋅10−6
0.822
Helium, He
4.003
1.97⋅10−5
1.63
Water Vapor, H2O
18.02
1.02⋅10−5
7.35
Argon, Ar
39.944
2.24⋅10−5
16.3
Clean, Dry Air
28.96
1.80⋅10−5
11.8
Carbon Dioxide, CO2
44.01
1.48⋅10−5
17.9
Carbon Monoxide, CO
28.01
1.82⋅10−5
11.4
Nitrogen, N2
28.02
1.76⋅10−5
11.4
Oxygen, O2
32.00
2.00⋅10−5
13.1
Nitric Oxide, NO
30.01
1.90⋅10−5
12.1
Dinitrogen Monoxide, N2O
44.02
1.45⋅10−5
17.9
Chlorine Gas, Cl2
70.91
1.03⋅10−5
28.9
Methane, CH4
16.04
1.34⋅10−5
6.54
[(N⋅s)/m2]
(N/m3)







































PAG E 4 21


Other Common Gases (Vapor at STP)
FLUID PROPERTIES (BG UNITS)
Water (Liquid)
Temperature
Density, ρ
Speci c Weight, γ
Dynamic Viscosity, μ
Kinematic Viscosity, ν
32
1.940
62.42
3.732⋅10−5
1.924⋅10−5
40
1.940
62.43
3.228⋅10−5
1.664⋅10−5
50
1.940
62.41
2.730⋅10−5
1.407⋅10−5
60
1.938
62.37
2.344⋅10−5
1.210⋅10−5
70
1.936
62.30
2.037⋅10−5
1.052⋅10−5
80
1.934
62.22
1.791⋅10−5
9.262⋅10−6
90
1.931
62.11
1.500⋅10−5
8.233⋅10−6
100
1.927
62.00
1.423⋅10−5
7.383⋅10−6
120
1.918
61.71
1.164⋅10−5
6.067⋅10−6
140
1.908
61.38
9.743⋅10−6
5.106⋅10−6
160
1.896
61.00
8.315⋅10−6
4.385⋅10−6
180
1.883
60.58
7.207⋅10−6
3.827⋅10−6
200
1.869
60.12
6.432⋅10−6
3.393⋅10−6
212
1.860
59.83
5.886⋅10−6
3.165⋅10−6
(∘F)
(slug/ft3)
(lbf/ft3)
[(lbf⋅s)/ft2]
(ft2/s)


















































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PAGE 422
Air (Gas at Standard Atmospheric Pressure)
Temperature
Density, ρ
Speci c Weight, γ
Dynamic Viscosity, μ
Kinematic Viscosity, ν
-40
0.002939
0.09456
3.29⋅10−7
1.12⋅10−4
-20
0.002805
0.09026
3.34⋅10−7
1.19⋅10−4
0
0.002683
0.08633
3.38⋅10−7
1.26⋅10−4
10
0.002626
0.08449
3.44⋅10−7
1.31⋅10−4
20
0.002571
0.08273
3.50⋅10−7
1.36⋅10−4
30
0.002519
0.08104
3.58⋅10−7
1.42⋅10−4
40
0.002469
0.07942
3.60⋅10−7
1.46⋅10−4
50
0.002420
0.07786
3.68⋅10−7
1.52⋅10−4
60
0.002373
0.07636
3.75⋅10−7
1.58⋅10−4
70
0.002329
0.07492
3.82⋅10−7
1.64⋅10−4
80
0.002286
0.07353
3.86⋅10−7
1.69⋅10−4
90
0.002244
0.07219
3.90⋅10−7
1.74⋅10−4
100
0.002204
0.07090
3.94⋅10−7
1.79⋅10−4
120
0.002128
0.06846
4.02⋅10−7
1.89⋅10−4
140
0.002057
0.06617
4.13⋅10−7
2.01⋅10−4
160
0.001990
0.06404
4.22⋅10−7
2.12⋅10−4
180
0.001928
0.06204
4.34⋅10−7
2.25⋅10−4
200
0.001870
0.06016
4.49⋅10−7
2.40⋅10−4
300
0.001624
0.05224
4.97⋅10−7
3.06⋅10−4
400
0.001435
0.04616
5.24⋅10−7
3.65⋅10−4
500
0.001285
0.04135
5.80⋅10−7
4.51⋅10−4
750
0.001020
0.03280
6.81⋅10−7
6.68⋅10−4
1,000
0.0008445
0.02717
7.85⋅10−7
9.30⋅10−4
1,500
0.0006291
0.02024
9.50⋅10−7
1.51⋅10−3
(∘F)
(slug/ft3)
(lbf/ft3)
[(lbf⋅s)/ft2]
(ft2/s)
























































fi



















































PAGE 423
Thermal Properties of Non-metals at 300 K
Density
Thermal Conductivity
Speci c Heat Capacity
Aluminum Oxide, Al2O3
(Sapphire)
3970
46.0
765
Aluminum Oxide, Al2O3
(Polycrystalline)
3970
36.0
765
Boron
2500
27.6
1105
Diamond
3500
2,300
509
Pyroceram
(Corning 9606)
2600
3.98
808
Polypropylene
920
0.246
1800
Silicon Carbide, SiC
3160
490
675
Silicon Dioxide, SiO2
(Crystalline Quartz)
2650
Silicon Dioxide, SiO2
(Amorphous, Glass)
2220
1.38
745
Silicon Nitride, SiN
2400
16.0
691
Sulfur
2070
0.21
708
Thorium Dioxide
9110
13.0
235
Titanium Dioxide, TiO2
(Polycrystalline)
4157
8.4
710
Material
(kg/m3)
(W/m⋅K)
|| to c-axis: 10.4
⊥ to c-axis: 6.21
(J/kg⋅K)
745











PAGE 42 4
fi

THERMAL PROPERTIES OF SOLIDS
Thermal Properties of Metals at 300 K
Density
Thermal Conductivity
Speci c Heat Capacity
Aluminum
(Pure)
2702
237
903
Aluminum
(Alloy 2024-T6)
2770
177
875
Bismuth
9780
7.86
122
Boron
2500
27.0
1107
Chromium
7160
93.7
449
Cobalt
8862
99.2
421
Copper
(Pure)
8933
401
385
Copper - Commercial Bronze
(90% Cu, 10% Al)
8800
52
420
Copper - Brass
(70% Cu, 30% Zn)
8530
110
380
Germanium, Ge
5360
59.9
322
Gold, Au
19,320
317
129
Iridium
22,500
147
130
Iron
(Pure)
7870
80.2
447
Material
(kg/m3)
(W/m⋅K)
(J/kg⋅K)




fi
PAGE 425
Thermal Properties of Metals at 300 K (Continued)
Density
Thermal Conductivity
Speci c Heat Capacity
Stainless Steel
(AISI 302)
8055
15.1
480
Stainless Steel
(AISI 316)
8238
13.4
468
Lead, Pb
11,340
35.3
129
Magnesium
1740
156
1024
Molybdenum
10,240
138
251
Nickel
(Pure)
8900
90.7
444
Nickel
(Nichrome, 80% Ni 20% Cr)
8400
12.0
420
Nickel
(Inconel X-750)
8510
11.7
439
Palladium
12,020
71.8
244
Platinum, Pt
21,450
71.6
133
Silicon, Si
2330
148
712
Silver, Ag
10,500
429
235
Tantalum
16,600
57.5
140
Material
(kg/m3)
(W/m⋅K)
(J/kg⋅K)




fi
PAGE 426
Thermal Properties of Metals at 300 K (Continued)
Material
Density
(kg/m3)
Thermal Conductivity
Speci c Heat Capacity
Thorium
11,700
54.0
118
Tin
7310
66.6
227
Titanium
4500
21.9
522
Tungsten
19,300
174
132
Uranium
19,070
27.6
116
Vanadium
6100
30.7
489
Zinc
7140
116
389
Zirconium
6750
22.7
278
(W/m⋅K)
(J/kg⋅K)




fi
PAGE 427
Thermal Properties of Common Building Materials at 300 K
Material
Density
(kg/m3)
Thermal Conductivity
Speci c Heat Capacity
Brick
1750
0.77
1,000
Concrete
(Heavy)
2300
2.00
1,000
Concrete
(Light)
600
0.2
1,000
Mineral Wool
12
0.042
1,030
Plaster
(Dense)
1300
0.57
1,000
Plaster
(Light)
600
0.18
1,000
Polystyrene Foam
46
0.026
1,130
Rubber
930
0.138
2,092
Single-glass Window
2500
0.65
837
Thermalite Board
753
0.190
1,050
Wood - Oak
770
0.160
2,000
Wood - Pine
510
0.113
2,031
Wood - Pine Fiber Board
256
0.052
2,000
(W/m⋅K)
(J/kg⋅K)




fi
PAGE 428
THERMAL PROPERTIES OF FLUIDS
Thermal Properties of Saturated Liquid Water
Speci c Heat
Temperature (K)
Capacity, Cp
(kJ/kg)
Thermal Conductivity, κ
(W/m⋅K)
Prandtl Number, Pr
273.15
4.217
0.569
12.99
275
4.211
0.574
12.22
280
4.198
0.582
10.26
285
4.189
0.590
8.81
290
4.184
0.598
7.56
295
4.181
0.606
6.62
300
4.179
0.613
5.83
305
4.178
0.620
5.20
310
4.178
0.628
4.62
315
4.179
0.634
4.16
320
4.180
0.640
3.77
325
4.182
0.645
3.42
330
4.184
0.650
3.15
335
4.186
0.656
2.88
340
4.188
0.660
2.66
345
4.191
0.664
2.45
350
4.195
0.668
2.29
355
4.199
0.671
2.14
360
4.203
0.674
2.02
365
4.209
0.677
1.91




fi
PAGE 429
Thermal Properties of Saturated Liquid Water
Speci c Heat
Temperature (K)
Capacity, Cp
(kJ/kg)
Thermal Conductivity, κ
(W/m⋅K)
Prandtl Number, Pr
370
4.214
0.679
1.80
373.15
4.217
0.680
1.76
375
4.220
0.681
1.70
380
4.226
0.683
1.61
385
4.232
0.685
1.53
390
4.239
0.686
1.47
400
4.256
0.688
1.34
410
4.278
0.688
1.24
420
4.302
0.688
1.16
430
4.331
0.685
1.09
440
4.360
0.682
1.04
450
4.400
0.678
0.99
460
4.440
0.673
0.95
470
4.480
0.667
0.92
480
4.530
0.660
0.89
490
4.590
0.651
0.87
500
4.660
0.642
0.86
510
4.740
0.631
0.85
520
4.840
0.621
0.84
530
4.950
0.608
0.85
540
5.080
0.594
0.86




fi
PAGE 430
Thermal Properties of Saturated Liquid Water
Speci c Heat
Temperature (K)
Capacity, Cp
(kJ/kg)
Thermal Conductivity, κ
(W/m⋅K)
Prandtl Number, Pr
550
5.240
0.580
0.87
560
5.430
0.563
0.90
570
5.680
0.548
0.94
580
6.000
0.528
0.99
590
6.410
0.513
1.05
600
7.000
0.497
1.14
610
7.850
0.467
1.30
620
9.350
0.444
1.52
625
10.600
0.430
1.65
630
12.600
0.412
2.00
635
16.400
0.392
2.70
640
26.000
0.367
4.20
645
90.000
0.331
12.00




fi
PAG E 4 31
Thermal Properties of Air
Speci c Heat
Thermal Conductivity, κ
Temperature (K)
Capacity, Cp
100
1.032
0.00934
0.786
150
1.012
0.0138
0.758
200
1.007
0.0181
0.737
250
1.003
0.0223
0.720
300
1.005
0.0263
0.707
350
1.008
0.0300
0.700
400
1.013
0.0338
0.690
450
1.020
0.0373
0.686
500
1.029
0.0407
0.684
550
1.040
0.0439
0.683
600
1.051
0.0469
0.685
650
1.063
0.0497
0.690
700
1.075
0.0524
0.695
750
1.087
0.0549
0.702
800
1.099
0.0573
0.709
850
1.110
0.0596
0.716
900
1.121
0.0620
0.720
950
1.131
0.0643
0.723
1000
1.141
0.0667
0.726
1100
1.159
0.0715
0.728
1200
1.175
0.0763
0.728
(kJ/kg⋅K)
(W/m⋅K)
Prandtl Number, Pr





fi
PAGE 432
Thermal Properties of Air
Speci c Heat
Temperature (K)
Capacity, Cp
(kJ/kg⋅K)
Thermal Conductivity, κ
(W/m⋅K)
Prandtl Number, Pr
1300
1.189
0.0820
0.719
1400
1.207
0.0910
0.703
1500
1.230
0.1000
0.685
1600
1.248
0.1060
0.688
1700
1.267
0.1130
0.685
1800
1.286
0.1200
0.683
1900
1.307
0.1280
0.677
2000
1.337
0.1370
0.672
2100
1.372
0.1470
0.667
2200
1.417
0.1600
0.655
2300
1.478
0.1750
0.647
2400
1.558
0.1960
0.630
2500
1.665
0.2220
0.613
3000
2.726
0.4860
0.536





fi
PAGE 433
Temperature Density, ρ
(K)
(kg/m3)
Speci c Heat
Capacity, Cp
(kJ/kg⋅K)
Thermal
Conductivity,
κ
(W/m⋅K)
Dynamic
Viscosity, μ
Prandtl Number, Pr
(kg/m⋅s)
273.15
899.1
1.796
0.147
3.85
47,000
280
895.3
1.827
0.144
2.17
27,500
290
890.0
1.868
0.145
0.999
12,900
300
884.1
1.909
0.145
0.486
6,400
310
877.9
1.951
0.145
0.253
3,400
320
871.8
1.993
0.143
0.141
1,965
330
865.8
2.035
0.141
0.0836
1,205
340
859.9
2.076
0.139
0.0531
793
350
853.9
2.118
0.138
0.0356
546
360
847.8
2.161
0.138
0.0252
395
370
841.8
2.206
0.137
0.0186
300
380
836.0
2.250
0.136
0.0141
233
390
830.6
2.294
0.135
0.0110
187
400
825.1
2.337
0.134
0.00874
152
410
818.9
2.381
0.133
0.00698
125
420
812.1
2.427
0.133
0.00564
103
430
806.5
2.471
0.132
0.00470
88








PAGE 43 4
fi

Thermal Properties of (Engine) Oil (Unused)
Speci c Heats of Ideal Gases (SI)
cp
T
(K)
(kJ/kg/K)
cv
(kJ/kg/K)
k
cp
(kJ/kg/K)
cv
(kJ/kg/K)
k
Carbon Dioxide, CO2
Air
cp
(kJ/kg/K)
cv
(kJ/kg/K)
k
Carbon Monoxide, CO
250
1.003
0.716
1.401
0.791
0.602
1.314
1.039
0.743
1.400
300
1.005
0.718
1.400
0.846
0.657
1.288
1.040
0.744
1.399
350
1.008
0.721
1.398
0.895
0.706
1.268
1.043
0.746
1.398
400
1.013
0.726
1.395
0.939
0.750
1.252
1.047
0.751
1.395
450
1.020
0.733
1.391
0.978
0.790
1.239
1.054
0.757
1.392
500
1.029
0.742
1.387
1.014
0.825
1.229
1.063
0.767
1.387
550
1.040
0.753
1.381
1.046
0.857
1.220
1.075
0.778
1.382
600
1.051
0.764
1.376
1.075
0.886
1.213
1.087
0.790
1.376
650
1.063
0.776
1.370
1.102
0.913
1.207
1.100
0.803
1.370
700
1.075
0.788
1.364
1.126
0.937
1.202
1.113
0.816
1.364
750
1.087
0.800
1.359
1.148
0.959
1.197
1.126
0.829
1.358
800
1.099
0.812
1.354
1.169
0.980
1.193
1.139
0.842
1.353
900
1.121
0.834
1.344
1.204
1.015
1.186
1.163
0.866
1.343
1000
1.142
0.855
1.336
1.234
1.045
1.181
1.185
0.888
1.335
Hydrogen, H2
Nitrogen, N2
Oxygen, O2
250
14.051
9.927
1.416
1.039
0.742
1.400
0.913
0.653
1.398
300
14.307
10.183
1.405
1.039
0.743
1.400
0.918
0.658
1.395
350
14.427
10.302
1.400
1.041
0.744
1.399
0.928
0.668
1.389
400
14.476
10.352
1.398
1.044
0.747
1.397
0.941
0.681
1.382
450
14.501
10.377
1.398
1.049
0.752
1.395
0.956
0.696
1.373
500
14.513
10.389
1.397
1.056
0.759
1.391
0.972
0.712
1.365
550
14.530
10.405
1.396
1.065
0.768
1.387
0.988
0.728
1.358
600
14.546
10.422
1.396
1.075
0.778
1.382
1.003
0.743
1.350
650
14.571
10.447
1.395
1.086
0.789
1.376
1.017
0.758
1.343
700
14.604
10.480
1.394
1.098
0.801
1.371
1.031
0.771
1.337
750
14.645
10.521
1.392
1.110
0.813
1.365
1.043
0.783
1.332
800
14.695
10.570
1.390
1.121
0.825
1.360
1.054
0.794
1.327
900
14.822
10.698
1.385
1.145
0.849
1.349
1.074
0.814
1.319
1000
14.983
10.859
1.380
1.167
0.870
1.341
1.090
0.830
1.313






PAGE 435
fi









IDEAL GAS SPECIFIC HEATS (SI)
Speci c Heats of Ideal Gases (EEU)
cp
T
∘
( F)
(Btu/lbm/R)
cv
(Btu/lbm/R)
k
cp
(Btu/lbm/R)
cv
(Btu/lbm/R)
k
Carbon Dioxide, CO2
Air
cp
(Btu/lbm/R)
cv
(Btu/lbm/R)
k
Carbon Monoxide, CO
40
0.240
0.171
1.401
0.195
0.150
1.300
0.248
0.177
1.400
100
0.240
0.172
1.400
0.205
0.160
1.283
0.249
0.178
1.399
200
0.241
0.173
1.397
0.217
0.172
1.262
0.249
0.179
1.397
300
0.243
0.174
1.394
0.229
0.184
1.246
0.251
0.180
1.394
400
0.245
0.176
1.389
0.239
0.193
1.233
0.253
0.182
1.389
500
0.248
0.179
1.383
0.247
0.202
1.223
0.256
0.185
1.384
600
0.250
0.182
1.377
0.255
0.210
1.215
0.259
0.188
1.377
700
0.254
0.185
1.371
0.262
0.217
1.208
0.262
0.191
1.371
800
0.257
0.188
1.365
0.269
0.224
1.202
0.266
0.195
1.364
900
0.259
0.191
1.358
0.275
0.230
1.197
0.269
0.198
1.357
1000
0.263
0.195
1.353
0.280
0.235
1.192
0.273
0.202
1.351
1500
0.276
0.208
1.330
0.298
0.253
1.178
0.287
0.216
1.328
2000
0.286
0.217
1.312
0.312
0.267
1.169
0.297
0.226
1.314
Hydrogen, H2
Nitrogen, N2
Oxygen, O2
40
3.397
2.412
1.409
0.248
0.177
1.400
0.219
0.156
1.397
100
3.426
2.441
1.404
0.248
0.178
1.399
0.220
0.158
1.394
200
3.451
2.466
1.399
0.249
0.178
1.398
0.223
0.161
1.387
300
3.461
2.476
1.398
0.250
0.179
1.396
0.226
0.164
1.378
400
3.466
2.480
1.397
0.251
0.180
1.393
0.230
0.168
1.368
500
3.469
2.484
1.397
0.254
0.183
1.388
0.235
0.173
1.360
600
3.473
2.488
1.396
0.256
0.185
1.383
0.239
0.177
1.352
700
3.477
2.492
1.395
0.260
0.189
1.377
0.242
0.181
1.344
800
3.494
2.509
1.393
0.262
0.191
1.371
0.246
0.184
1.337
900
3.502
2.519
1.392
0.265
0.194
1.364
0.249
0.187
1.331
1000
3.513
2.528
1.390
0.269
0.198
1.359
0.252
0.190
1.326
1500
3.618
2.633
1.374
0.283
0.212
1.334
0.263
0.201
1.309
2000
3.758
2.773
1.355
0.293
0.222
1.319
0.270
0.208
1.298






fi
PAGE 436










IDEAL GAS SPECIFIC HEATS (EEU)
Saturated Water Table - Temperature (SI)
νf
(m 3 /kg)
νg
(m 3 /kg)
uf
(k J/kg)
ug
hf
(k J/kg) (k J/kg)
hg
(k J/kg)
sf
sg
(k J/kg/K ) (k J/kg/K )
0.01 0.6117 0.001000
206.00
0.000
2374.9
0.000
2500.9
0.0000
9.1556
5
0.8725 0.001000
147.03
21.019
2381.8
21.020
2510.1
0.0763
9.0249
10
1.2281 0.001000
106.32
42.020
2388.7
42.022
2519.2
0.1511
8.8999
12
1.4028 0.001001
93.732
50.408
2391.4
50.410
2522.9
0.1806
8.8514
14
1.5989 0.001001
82.804
58.791
2394.1
58.793
2526.5
0.2099
8.8038
15
1.7057 0.001001
77.885
62.980
2395.5
62.982
2528.3
0.2245
8.7803
16
1.8187 0.001001
73.295
67.169
2396.9
67.170
2530.2
0.2389
8.7571
20
2.3392 0.001002
57.762
83.913
2402.3
83.915
2537.4
0.2965
8.6661
25
3.1698 0.001003
43.340
104.83
2409.1
104.83
2546.5
0.3672
8.5567
30
4.2469 0.001004
32.879
125.73
2415.9
125.74
2555.6
0.4368
8.4520
35
5.6291 0.001006
25.205
146.63
2422.7
146.64
2564.6
0.5051
8.3517
40
7.3851 0.001008
19.515
167.53
2429.4
167.53
2573.5
0.5724
8.2556
45
9.5953 0.001010
15.251
188.43
2436.1
188.44
2582.4
0.6386
8.1633
50
12.352 0.001012
12.026
209.33
2442.7
209.34
2591.3
0.7038
8.0748
55
15.763 0.001015
9.5639
230.24
2449.3
230.26
2600.1
0.7680
7.9898
60
19.947 0.001017
7.6670
251.16
2455.9
251.18
2608.8
0.8313
7.9082
65
25.043 0.001020
6.1935
272.09
2462.4
272.12
2617.5
0.8937
7.8296
70
31.202 0.001023
5.0396
293.04
2468.9
293.07
2626.1
0.9551
7.7540
75
38.597 0.001026
4.1291
313.99
2475.3
314.03
2634.6
1.0158
7.6812
80
47.416 0.001029
3.4053
334.97
2481.6
335.02
2643.0
1.0756
7.6111
85
57.868 0.001032
2.8261
355.96
2487.8
356.02
2651.4
1.1346
7.5435
90
70.183 0.001036
2.3593
376.97
2494.0
377.04
2659.6
1.1929
7.4782
95
84.609 0.001040
1.9808
398.00
2500.1
398.09
2667.6
1.2504
7.4151
100
101.42 0.001043
1.6720
419.06
2506.0
419.17
2675.6
1.3072
7.3542
105
120.90 0.001047
1.4186
440.15
2511.9
440.28
2683.4
1.3634
7.2952
110
143.38 0.001052
1.2094
461.27
2517.7
461.42
2691.1
1.4188
7.2382
115
169.18 0.001056
1.0360
482.42
2523.3
482.59
2698.6
1.4737
7.1829
120
198.67 0.001060
0.89133
503.60
2528.9
503.81
2706.0
1.5279
7.1292
T
(∘C)
p
(kPa)

PAGE 437

















SATURATED WATER - TEMPERATURE TABLE (SI)
∘
( C)
p
(kPa)
νf
(m 3 /kg)
νg
(m 3 /kg)
uf
(k J/kg)
ug
hf
(k J/kg) (k J/kg)
hg
(k J/kg)
sf
sg
(k J/kg/K ) (k J/kg/K )
125
232.23 0.001065
0.77012
524.83
2534.3
525.07
2713.1
1.5816
7.0771
130
270.28 0.001070
0.66808
546.10
2539.5
546.38
2720.1
1.6346
7.0265
135
313.22 0.001075
0.58179
567.41
2544.7
567.75
2726.8
1.6872
6.9773
140
361.53 0.001080
0.50850
588.77
2549.6
589.16
2733.5
1.7392
6.9294
145
415.68 0.001085
0.44600
610.19
2554.4
610.64
2739.8
1.7908
6.8827
150
476.16 0.001091
0.39248
631.66
2559.1
632.18
2745.9
1.8418
6.8371
155
543.49 0.001096
0.34648
653.19
2563.5
653.79
2751.8
1.8924
6.7927
160
618.23 0.001102
0.30680
674.79
2567.8
675.47
2757.5
1.9426
6.7492
165
700.93 0.001108
0.27244
696.46
2571.9
697.24
2762.8
1.9923
6.7067
170
792.18 0.001114
0.24260
718.20
2575.7
719.08
2767.9
2.0417
6.6650
175
892.60 0.001121
0.21659
740.02
2579.4
741.02
2772.7
2.0906
6.6242
180
1002.8 0.001127
0.19384
761.92
2582.8
763.05
2777.2
2.1392
6.5841
185
1123.5 0.001134
0.17390
783.91
2586.0
785.19
2781.4
2.1875
6.5447
190
1255.2 0.001141
0.15636
806.00
2589.0
807.43
2785.3
2.2355
6.5059
195
1398.8 0.001149
0.14089
828.18
2591.7
829.78
2788.8
2.2831
6.4678
200
1554.9 0.001157
0.12721
850.46
2594.2
852.26
2792.0
2.3305
6.4302
205
1724.3 0.001164
0.11508
872.86
2596.4
874.87
2794.8
2.3776
6.3930
210
1907.7 0.001173
0.10429
895.38
2598.3
897.61
2797.3
2.4245
6.3563
215
2105.9 0.001181
0.094681
918.02
2599.9
920.50
2799.3
2.4712
6.3200
220
2319.6 0.001190
0.086094
940.79
2601.3
943.55
2801.0
2.5176
6.2840
225
2549.7 0.001199
0.078405
963.70
2602.3
966.76
2802.2
2.5639
6.2483
230
2797.1 0.001209
0.071505
986.76
2602.9
990.14
2802.9
2.6100
6.2128
235
3062.6 0.001219
0.065300
1010.0
2603.2
1013.7
2803.2
2.6560
6.1775
240
3347.0 0.001229
0.059707
1033.4
2603.1
1037.5
2803.0
2.7018
6.1424
245
3651.2 0.001240
0.054656
1056.9
2602.7
1061.5
2802.2
2.7476
6.1072
250
3976.2 0.001252
0.050085
1080.7
2601.8
1085.7
2801.0
2.7933
6.0721
255
4322.9 0.001263
0.045941
1104.7
2600.5
1110.1
2799.1
2.8390
6.0369
260
4692.3 0.001276
0.042175
1128.8
2598.7
1134.8
2796.6
2.8847
6.0017
265
5085.3 0.001289
0.038748
1153.3
2596.5
1159.8
2793.5
2.9304
5.9662
T

PAGE 438

















Saturated Water Table - Temperature (SI)
∘
( C)
p
(kPa)
νf
(m 3 /kg)
νg
(m 3 /kg)
uf
(k J/kg)
ug
hf
(k J/kg) (k J/kg)
hg
(k J/kg)
sf
sg
(k J/kg/K ) (k J/kg/K )
270
5503.0 0.001303
0.035622
1177.9
2593.7
1185.1
2789.7
2.9762
5.9305
275
5946.4 0.001317
0.032767
1202.9
2590.3
1210.7
2785.2
3.0221
5.8944
280
6416.6 0.001333
0.030153
1228.2
2586.4
1236.7
2779.9
3.0681
5.8579
285
6914.6 0.001349
0.027756
1253.7
2581.8
1263.1
2773.7
3.1144
5.8210
290
7441.8 0.001366
0.025554
1279.7
2576.5
1289.8
2766.7
3.1608
5.7834
295
7999.0 0.001384
0.023528
1306.0
2570.5
1317.1
2758.7
3.2076
5.7450
300
8587.9 0.001404
0.021659
1332.7
2563.6
1344.8
2749.6
3.2548
5.7059
305
9209.4 0.001425
0.019932
1360.0
2555.8
1373.1
2739.4
3.3024
5.6657
310
9865.0 0.001447
0.018333
1387.7
2547.1
1402.0
2727.9
3.3506
5.6243
315
10,556 0.001472
0.016849
1416.1
2537.2
1431.6
2715.0
3.3994
5.5816
320
11,284 0.001499
0.015470
1445.1
2526.0
1462.0
2700.6
3.4491
5.5372
325
12,051 0.001528
0.014183
1475.0
2513.4
1493.4
2684.3
3.4998
5.4908
330
12,858 0.001560
0.012979
1505.7
2499.2
1525.8
2666.0
3.5516
5.4422
335
13,707 0.001597
0.011848
1570.7
2464.5
1594.6
2622.0
3.6050
5.3907
340
14,061 0.001638
0.010783
1537.5
2483.0
1559.4
2645.4
3.6602
5.3358
345
15,541 0.001685
0.009772
1605.5
2443.2
1631.7
2595.1
3.7179
5.2765
350
16,529 0.001741
0.008806
1642.4
2418.3
1671.2
2563.9
3.7788
5.2114
355
17,570 0.001808
0.007872
1682.2
2388.6
1714.0
2526.9
3.8442
5.1384
360
18,666 0.001895
0.006950
1726.2
2351.9
1761.5
2481.6
3.9165
5.0537
365
19,822 0.002015
0.006009
1772.2
2303.6
1817.2
2422.7
4.0004
4.9493
370
21,044 0.002217
0.004953
1844.5
2230.1
1891.2
2334.3
4.1119
4.8009
373.95 22,064 0.003106
0.003106
2015.7
2015.7
2084.3
2084.3
4.4070
4.4070
T

PAGE 439

















Saturated Water Table - Temperature (SI)
Saturated Water Table - Pressure (SI)
(∘C)
νf
(m 3 /kg)
νg
(m 3 /kg)
uf
(k J/kg)
ug
hf
(k J/kg) (k J/kg)
hg
(k J/kg)
sf
sg
(k J/kg/K ) (k J/kg/K )
1.0
6.97
0.001000
129.19
29.302
2384.5
29.303
2513.7
0.1059
8.9749
1.5
13.02
0.001001
87.964
54.686
2392.8
54.688
2524.7
0.1956
8.8270
2.0
17.50
0.001001
66.990
73.431
2398.9
73.433
2532.9
0.2606
8.7227
2.5
21.08
0.001002
54.242
88.422
2403.8
88.424
2539.4
0.3118
8.6421
3.0
24.08
0.001003
45.654
100.98
2407.9
100.98
2544.8
0.3543
8.5765
4.0
28.96
0.001004
34.791
121.39
2414.5
121.39
2553.7
0.4224
8.4734
5.0
32.87
0.001005
28.185
137.75
2419.8
137.75
2560.7
0.4762
8.3938
7.5
40.29
0.001008
19.233
168.74
2429.8
168.75
2574.0
0.5763
8.2501
10
45.81
0.001010
14.670
191.79
2437.2
191.81
2583.9
0.6492
8.1488
15
53.97
0.001014
10.020
225.93
2448.0
225.94
2598.3
0.7549
8.0071
20
60.06
0.001017
7.6481
251.40
2456.0
251.42
2608.9
0.8320
7.9073
25
64.96
0.001020
6.2034
271.93
2462.4
271.96
2617.5
0.8932
7.8302
30
69.09
0.001022
5.2287
289.24
2467.7
289.27
2624.6
0.9441
7.7675
40
75.86
0.001026
3.9933
317.58
2476.3
317.62
2636.1
1.0261
7.6691
50
81.32
0.001030
3.2403
340.49
2483.2
340.54
2645.2
1.0912
7.5931
75
91.76
0.001037
2.2172
384.36
2496.1
384.44
2662.4
1.2132
7.4558
100
99.61
0.001043
1.6941
417.40
2505.6
417.51
2675.0
1.3028
7.3589
101.3 99.97
0.001043
1.6734
418.95
2506.0
419.06
2675.6
1.3069
7.3545
125
105.97 0.001048
1.3750
444.23
2513.0
444.36
2684.9
1.3741
7.2841
150
111.35 0.001053
1.1594
466.97
2519.2
467.13
2693.1
1.4337
7.2231
175
116.04 0.001057
1.0037
486.82
2524.5
487.01
2700.2
1.4850
7.1716
200
120.21 0.001061
0.88578
504.50
2529.1
504.71
2706.3
1.5302
7.1270
225
123.97 0.001064
0.79329
520.47
2533.2
520.71
2711.7
1.5706
7.0877
250
127.41 0.001067
0.71873
535.08
2536.8
535.35
2716.5
1.6072
7.0525
275
130.58 0.001070
0.65372
548.57
2540.1
548.86
2720.9
1.6408
7.0207
300
133.52 0.001073
0.60582
561.11
2543.2
561.43
2724.9
1.6717
6.9917
325
136.27 0.001076
0.56199
572.84
2545.9
573.19
2728.6
1.7005
6.9650
p
(kPa)
T

PAGE 4 40

















SATURATED WATER - PRESSURE TABLE (SI)
p
(kPa)
( C)
νf
(m 3 /kg)
νg
(m 3 /kg)
uf
(k J/kg)
ug
hf
(k J/kg) (k J/kg)
hg
(k J/kg)
sf
sg
(k J/kg/K ) (k J/kg/K )
350
138.86 0.001079
0.52422
583.89
2548.5
584.26
2732.0
1.7274
6.9402
375
141.30 0.001081
0.49133
594.32
2550.9
594.73
2735.1
1.7526
6.9171
400
143.61 0.001084
0.46242
604.22
2553.1
604.66
2738.1
1.7765
6.8955
450
147.90 0.001088
0.41392
622.65
2557.1
623.13
2743.4
1.8205
6.8561
500
151.73 0.001093
0.37483
639.54
2560.7
640.09
2748.1
1.8604
6.8207
550
155.46 0.001097
0.34261
655.16
2563.9
655.77
2756.2
1.8970
6.7886
600
158.83 0.001101
0.31560
669.72
2566.8
670.38
2756.2
1.9308
6.7593
650
161.98 0.001104
0.29260
683.37
2569.4
684.08
2759.6
1.9623
6.7322
700
164.95 0.001108
0.27278
696.23
2571.8
697.00
2762.8
1.9918
6.7071
750
167.75 0.001111
0.25552
708.40
2574.0
709.24
2765.7
2.0195
6.6837
800
170.41 0.001115
0.24035
719.97
2576.0
720.87
2768.3
2.0457
6.6616
850
172.94 0.001118
0.22690
731.00
2577.9
731.95
2770.8
2.0705
6.6409
900
175.35 0.001121
0.21489
741.55
2579.6
742.56
2773.0
2.0941
6.6213
950
177.66 0.001124
0.20411
751.67
2581.3
752.74
2775.2
2.1166
6.6027
1,000 179.88 0.001127
0.19436
761.39
2582.8
762.51
2777.1
2.1381
6.5850
1,100 184.06 0.001133
0.17745
779.78
2585.5
781.03
2780.7
2.1785
6.5520
1,200 187.96 0.001138
0.16326
796.96
2587.8
798.33
2783.8
2.2159
6.5217
1,300 191.60 0.001144
0.15119
813.10
2589.9
814.59
2786.5
2.2508
6.4936
1,400 195.04 0.001149
0.14078
828.35
2591.8
829.96
2788.9
2.2835
6.4675
1,500 198.29 0.001154
0.13171
842.82
2593.4
844.55
2791.0
2.3143
6.4430
1,750 205.72 0.001166
0.11344
876.12
2596.7
878.16
2795.2
2.3844
6.3877
2,000 212.38 0.001177
0.099587
906.12
2599.1
908.47
2798.3
2.4467
6.3390
2,250 218.41 0.001187
0.088717
933.54
2600.9
936.21
2800.5
2.5029
6.2954
2,500 223.95 0.001197
0.079952
958.87
2602.1
961.87
2801.9
2.5542
6.2558
3,000 233.85 0.001217
0.066667
1004.6
2603.2
1008.3
2803.2
2.6454
6.1856
3,500 242.56 0.001235
0.057061
1045.4
2603.0
1049.7
2802.7
2.7253
6.1244
4,000 250.35 0.001252
0.049779
1082.4
2601.7
1087.4
2800.8
2.7966
6.0696
5,000 263.94 0.001286
0.039448
1148.1
2597.0
1154.5
2794.2
2.9207
5.9737
6,000 275.59 0.001319
0.032499
1205.8
2589.9
1213.8
2784.6
3.0275
5.8902
T
∘

PAG E 4 41

















Saturated Water Table - Pressure (SI)
νf
(m 3 /kg)
νg
(m 3 /kg)
uf
(k J/kg)
ug
hf
(k J/kg) (k J/kg)
hg
(k J/kg)
sf
sg
(k J/kg/K ) (k J/kg/K )
7,000 285.83 0.001352
0.027378
1258.0
2581.0
1267.4
2772.6
3.1220
5.8148
8,000 295.01 0.001384
0.023525
1306.0
2570.5
1317.1
2578.7
3.2077
5.7450
9,000 303.35 0.001418
0.020489
1350.9
2558.5
1363.7
2742.9
3.2866
5.6791
10,000 311.00 0.001452
0.018028
1393.3
2545.2
1407.8
2725.5
3.3603
5.6159
11,000 318.08 0.001488
0.015988
1433.9
2530.4
1450.2
2706.3
3.4299
5.5544
12,000 324.68 0.001526
0.014264
1473.0
2514.3
1491.3
2685.4
3.4964
5.4939
13,000 330.85 0.001566
0.012781
1511.0
2496.6
1531.4
2662.7
3.5606
5.4336
14,000 336.67 0.001610
0.011487
1548.4
2477.1
1571.0
2637.9
3.6232
5.3728
15,000 342.16 0.001657
0.010341
1585.5
2455.7
1610.3
2610.8
3.6848
5.3108
16,000 347.36 0.001710
0.009312
1622.6
2432.0
1649.9
2581.0
3.7461
5.2466
17,000 352.29 0.001770
0.008374
1660.2
2405.4
1690.3
2547.7
3.8082
5.1791
18,000 356.99 0.001840
0.007504
1699.1
2375.0
1732.2
2510.0
3.8720
5.1064
19,000 361.74 0.001926
0.006677
1740.3
2339.2
1776.8
2466.0
3.9396
5.0256
20,000 365.75 0.002038
0.005862
1785.8
2294.8
1826.6
2412.1
4.0146
4.9310
21,000 369.83 0.002207
0.004994
1841.6
2233.5
1888.0
2338.4
4.1071
4.8076
22,000 373.71 0.002703
0.003664
1951.7
2092.4
2011.1
2172.6
4.242
4.5439
22,064 373.95 0.003106
0.003106
2015.7
2015.7
2084.3
2084.3
4.0470
4.4070
p
(kPa)
T
∘
( C)

PAGE 4 42

















Saturated Water Table - Pressure (SI)
Superheated Water (SI)
ν
u
h
( C) (m /kg) (k J/kg) (k J/kg)
T
∘
3
s
(k J/kg /K )
p = 0.01 MPa (Tsat = 45.81∘C)
ν
(m /kg)
3
u
(k J/kg)
h
s
ν
3
(k J/kg) (k J/kg /K ) (m /kg)
p = 0.05 MPa (Tsat = 81.32∘C)
u
(k J/kg)
h
s
(k J/kg) (k J/kg /K )
p = 0.1 MPa (Tsat = 99.61∘C)
Sat.
14.670
2437.2
2583.9
8.1488
3.2403
2483.2
2645.2
7.5931
1.6941
2505.6
2675.0
7.3589
50
14.867
2433.3
2592.0
8,1741
0.35212
2639.4
2850.6
6.9683
0.26088
2631.1
2839.8
6.8177
100
17.196
2515.5
2687.5
8.4489
3.4187
2511.5
2682.4
7.6953
1.6959
2506.2
2675.8
7.3611
150
19.513
2587.9
2783.0
8.6893
3.8897
2585.7
2780.2
7.9413
1.9367
2582.9
2776.6
7.6148
200
21.826
2661.4
2879.6
8.9049
4.3562
2660.0
2877.8
8.1592
2.1724
2658.2
2875.5
7.8356
250
24.136
2736.1
2977.5
9.1015
4.8206
2735.1
2976.2
8.3568
2.4062
2733.9
2974.5
8.0346
300
26.446
2812.3
3076.7
9.2827
5.2841
2811.6
3075.8
8.5387
2.6389
2810.7
3074.5
8.2172
400
31.063
2969.3
3280.0
9.6094
6.2094
2968.9
3279.3
8.8659
3.1027
2968.3
3278.6
8.5452
500
35.680
3132.9
3489.7
9.8998
7.1338
3132.6
3489.3
9.1566
3.5655
3132.2
3488.7
8.8362
600
40.296
3303.3
3706.3
10.1631
8.0577
3303.1
3706.0
9.4201
4.0279
3302.8
3705.6
9.0999
700
44.911
3480.8
3929.9
10.4056
8.9813
3480.6
3929.7
9.6626
4.4900
3480.4
3929.4
9.3424
800
49.527
3665.4
4160.6
10.6312
9.9047
3665.2
4160.4
9.8883
4.9519
3665.0
4160.2
9.5682
900
54.143
3856.9
4398.3
10.8429
10.8280
3856.8
4398.2
10.1000
5.4137
3856.7
4398.0
9.7800
1000
58.758
4055.3
4642.8
11.0429
11.7513
4055.2
4642.7
10.3000
5.8755
4055.0
4642.6
9.9800
1100
63.373
4260.0
4893.8
11.2326
12.6745
4259.9
4893.7
10.4897
6.3372
4259.8
4893.6
10.1698
1200
67.989
4470.9
5150.8
11.4132
13.5977
4470.8
5150.7
10.6704
6.7988
4470.7
5150.6
10.3504
1300
72.604
4687.3
5413.4
11.5857
14.5209
4687.3
5413.3
10.8429
7.2605
4687.2
5413.3
10.5229
p = 0.2 MPa (Tsat = 120.21∘C)
p = 0.3 MPa (Tsat = 133.52∘C)
p = 0.4 MPa (Tsat = 143.61∘C)
Sat.
0.886
2529.1
2706.2
7.127
0.606
2543.2
2724.9
6.992
0.462
2553.1
2738.1
6.896
150
0.960
2577.1
2769.1
7.281
0.634
2571.0
2761.2
7.079
0.471
2564.4
2752.8
6.931
200
1.081
2654.6
2870.7
7.508
0.716
2651.0
2865.9
7.313
0.534
2647.2
2860.9
7.172
250
1.199
2731.4
2971.2
7.710
0.796
2728.9
2967.9
7.518
0.595
2726.4
2964.5
7.380
300
1.316
2808.8
3072.1
7.894
0.875
2807.0
3069.6
7.704
0.655
2805.1
3067.1
7.568
350
1.433
2887.3
3173.9
8.064
0.954
2885.9
3172.0
7.875
0.714
2884.4
3170.0
7.740
400
1.549
2967.1
3277.0
8.224
1.032
2966.0
3275.5
8.035
0.773
2964.9
3273.9
7.900
450
1.666
3048.5
3381.6
8.373
1.109
3047.5
3380.3
8.185
0.831
3046.6
3379.0
8.051
500
1.781
3131.4
3487.7
8.515
1.187
3130.6
3486.6
8.327
0.889
3129.8
3485.5
8.193
600
2.013
3302.2
3704.8
8.779
1.341
3301.6
3704.0
8.591
1.006
3301.0
3703.2
8.458
700
2.244
3479.9
3928.8
9.022
1.496
3479.5
3928.2
8.834
1.122
3479.0
3927.6
8.701
800
2.476
3664.7
4159.8
9.248
1.650
3664.3
4159.3
9.060
1.237
3663.9
4158.8
8.927
900
2.707
3856.3
4397.6
9.460
1.804
3856.0
4397.3
9.272
1.353
3855.7
4396.9
9.139
1000
2.938
4054.8
4642.3
9.660
1.958
4054.5
4642.0
9.473
1.469
4054.3
4641.7
9.340
1300
3.630
4687.1
5413.1
10.203
2.420
4686.9
5413.0
10.016
1.815
4686.7
5412.8
9.882







PAGE 4 43

























SUPERHEATED WATER TABLE (SI)
T
∘
ν
u
h
s
ν
( C) (m 3 /kg) (k J/kg) (k J/kg) (k J/kg /K ) (m 3 /kg)
p = 0.5 MPa (Tsat = 151.83∘C)
u
(k J/kg)
h
s
ν
3
(k
J/kg
/K
)
(k J/kg)
(m /kg)
p = 0.6 MPa (Tsat = 158.83∘C)
u
(k J/kg)
h
s
(k
J/kg
/K )
(k J/kg)
p = 0.8 MPa (Tsat = 170.4∘C)
Sat.
0.37483
2560.7
2748.1
6.8207
0.31560
2566.8
2756.2
6.7593
0.24035
2576.0
2768.3
6.6616
200
0.42503
2643.3
2855.8
7.0610
0.35212
2639.4
2850.6
6.9683
0.26088
2631.1
2839.8
6.8177
250
0.47443
2723.8
2961.0
7.2725
0.39390
2721.2
2957.6
7.1833
0.29321
2715.9
2950.4
7.0402
300
0.52261
2803.3
3064.6
7.4614
0.43442
2801.4
3062.0
7.3740
0.32416
2797.5
3056.9
7.2345
350
0.57015
2883.0
3168.1
7.6346
0.47428
2881.6
3166.1
7.5481
0.35442
2878.6
3162.2
7.4107
400
0.61731
2963.7
3272.4
7.7956
0.51374
2962.5
3270.8
7.7097
0.38429
2960.2
3267.7
7.5735
500
0.71095
3120.0
3484.5
8.0893
0.59200
3128.2
3483.4
8.0041
0.44332
3126.6
3481.3
7.8692
600
0.80409
3300.4
3702.5
8.3544
0.66976
3299.8
3701.7
8.2695
0.50186
3298.7
3700.1
8.1354
700
0.89696
3478.6
3927.0
8.5978
0.74725
3478.1
3926.4
8.5132
0.56011
3477.2
3925.3
8.3794
800
0.98966
3663.6
4158.4
8.8240
0.82457
3663.2
4157.9
8.7395
0.61820
3662.5
4157.0
8.6061
900
1.08227
3855.4
4396.6
9.0362
0.90179
3855.1
4396.2
8.9518
0.67619
3854.5
4395.5
8.8185
1000 1.17480
4054.0
4641.4
9.2364
0.97893
4053.8
4641.1
9.1521
0.73411
4053.3
4640.5
9.0189
1100 1.26728
4259.0
4892.6
9.4263
1.05603
4258.8
4892.4
9.3420
0.79197
4258.3
4891.9
9.2090
1200 1.35972
4470.0
5149.8
9.6071
1.1330
4469.8
5149.6
9.5229
0.84980
4469.4
5149.3
9.3898
1300 1.45214
4686.6
5412.6
9.7797
1.21012
4686.4
5412.5
9.6955
0.90761
4686.1
5412.2
9.5625
p = 1.0 MPa (Tsat = 179.9∘C)
p = 1.2 MPa (Tsat = 188.0∘C)
p = 1.4 MPa (Tsat = 195.0∘C)
Sat.
0.1944
2582.7
2777.1
6.585
0.1633
2587.8
2783.7
6.522
0.1408
2591.8
2788.8
6.468
200
0.2060
2622.2
2828.3
6.696
0.1693
2612.9
2816.1
6.591
0.1430
2602.7
2803.0
6.498
250
0.2328
2710.4
2943.1
6.927
0.1924
2704.7
2935.6
6.831
0.1636
2698.9
2927.9
6.749
300
0.2580
2793.6
3051.6
7.125
0.2139
2789.7
3046.3
7.034
0.1823
2785.7
3040.9
6.955
350
0.2825
2875.7
3158.2
7.303
0.2346
2872.7
3154.2
7.214
0.2003
2869.7
3150.1
7.138
400
0.3066
2957.9
3264.5
7.467
0.2548
2955.5
3261.3
7.379
0.2178
2953.1
3258.1
7.305
450
0.3305
3040.9
3371.3
7.620
0.2748
3038.9
3368.7
7.533
0.2351
3037.0
3366.1
7.459
500
0.3541
3125.0
3479.1
7.764
0.2946
3123.4
3476.9
7.678
0.2522
3121.8
3474.8
7.605
600
0.4011
3297.5
3698.6
8.031
0.3339
3296.3
3697.0
7.946
0.2860
3295.1
3695.4
7.873
700
0.4478
3476.2
3924.1
8.276
0.3730
3475.3
3922.9
8.190
0.3195
3474.4
3921.7
8.118
800
0.4944
3661.7
4156.1
8.502
0.4118
3661.0
4155.2
8.418
0.3529
3660.2
4154.3
8.346
900
0.5408
3853.9
4394.8
8.715
0.4506
3853.3
4394.0
8.630
0.3861
3852.7
4393.3
8.559
1000
0.5872
4052.7
4639.9
8.916
0.4893
4052.2
4639.4
8.831
0.4193
4051.7
4638.8
8.759
1100
0.6335
4257.9
4891.4
9.106
0.5279
4257.5
4891.0
9.021
0.4525
4257.0
4890.5
8.950
1200
0.6798
4469.0
5148.9
9.287
0.5665
4468.7
5148.5
9.202
0.4856
4468.3
5148.1
9.131
1300
0.7261
4685.8
5411.9
9.459
0.6051
4685.5
5411.6
9.375
0.5187
4685.1
5411.3
9.304







PAGE 4 4 4

























Superheated Water (SI)
T
∘
ν
u
h
s
ν
( C) (m 3 /kg) (k J/kg) (k J/kg) (k J/kg /K ) (m 3 /kg)
p = 1.6 MPa (Tsat = 201.37∘C)
u
(k J/kg)
h
s
ν
3
(k
J/kg
/K
)
(k J/kg)
(m /kg)
p = 1.8 MPa (Tsat = 207.11∘C)
u
(k J/kg)
h
s
(k
J/kg
/K )
(k J/kg)
p = 2.0 MPa (Tsat = 212.38∘C)
Sat.
0.1237
2594.8
2792.8
6.420
0.1104
2597.2
2795.9
6.378
0.0996
2599.1
2798.3
6.339
225
0.1329
2645.1
2857.8
6.554
0.1168
2637.0
2847.2
6.482
0.1038
2628.5
2836.1
6.416
250
0.1419
2692.9
2919.9
6.675
0.1250
2686.7
2911.7
6.609
0.1115
2680.2
2903.2
6.548
300
0.1587
2781.6
3035.4
6.886
0.1403
2777.4
3029.9
6.825
0.1255
2773.2
3024.2
6.768
350
0.1746
2866.6
3146.0
7.071
0.1546
2863.6
3141.8
7.012
0.1386
2860.5
3137.7
6.958
400
0.1901
2950.7
3254.9
7.239
0.1685
2948.3
3251.6
7.181
0.1512
2945.9
3248.3
7.129
450
0.2053
3035.0
3363.5
7.395
0.1821
3033.1
3360.9
7.338
0.1635
3031.1
3358.2
7.287
500
0.2203
3120.1
3472.6
7.541
0.1955
3118.5
3470.4
7.485
0.1757
3116.9
3468.2
7.434
600
0.2500
3293.9
3693.9
7.810
0.2220
3292.7
3692.3
7.754
0.1996
3291.5
3690.7
7.704
700
0.2794
3473.5
3920.5
8.056
0.2482
3472.6
3919.4
8.000
0.2233
3471.6
3918.2
7.951
800
0.3087
3659.5
4153.3
8.283
0.2743
3658.8
4152.4
8.228
0.2467
3658.0
4151.5
8.179
900
0.3378
3852.1
4392.6
8.497
0.3002
3851.5
4391.9
8.442
0.2701
3850.9
4391.1
8.393
1000
0.3669
4051.2
4638.2
8.697
0.3261
4050.7
4637.6
8.643
0.2934
4050.2
4637.0
8.594
1100
0.3959
4256.6
4890.0
8.888
0.3519
4256.2
5147.3
8.833
0.3167
4255.7
4889.1
8.784
1200
0.4249
4467.9
5147.7
9.069
0.3777
4467.6
5410.6
9.014
0.3399
4467.2
5147.0
8.965
p = 2.5 MPa (Tsat = 223.95∘C)
p = 3.0 MPa (Tsat = 233.85∘C)
p = 3.5 MPa (Tsat = 242.56∘C)
Sat.
0.0799
2602.1
2801.9
6.256
0.0667
2603.2
2803.2
6.186
0.0571
2602.9
2802.6
6.124
250
0.0871
2663.3
2880.9
6.411
0.0706
2644.7
2856.5
6.289
0.0588
2624.0
2829.7
6.176
300
0.0989
2762.2
3009.6
6.646
0.0812
2750.8
2994.3
6.541
0.0685
2738.8
2978.4
6.448
350
0.1098
2852.5
3127.0
6.842
0.0906
2844.4
3116.1
6.745
0.0768
2836.0
3104.8
6.660
400
0.1201
2939.8
3240.1
7.017
0.0994
2933.5
3231.7
6.923
0.0846
2927.2
3223.2
6.843
450
0.1302
3026.2
3351.6
7.177
0.1079
3021.2
3344.8
7.086
0.0920
3016.1
3338.0
7.007
500
0.1400
3112.8
3462.7
7.325
0.1162
3108.6
3457.2
7.236
0.0992
3104.5
3451.6
7.159
600
0.1593
3288.5
3686.8
7.598
0.1325
3285.5
3682.8
7.510
0.1133
3282.5
3678.9
7.436
700
0.1784
3469.3
3915.2
7.846
0.1484
3467.0
3912.2
7.759
0.1270
3464.7
3909.3
7.685
800
0.1972
3656.2
4149.2
8.074
0.1642
3654.3
4146.9
7.989
0.1406
3652.5
4144.6
7.916
900
0.2160
3849.4
4389.3
8.288
0.1799
3847.9
4387.5
8.203
0.1541
3846.4
4385.7
8.130
1000
0.2347
4048.9
4635.6
8.490
0.1955
4047.7
4634.1
8.405
0.1675
4046.4
4632.7
8.332
1100
0.2533
4254.8
4887.9
8.680
0.2111
4253.6
4886.7
8.596
0.1809
4252.5
4885.6
8.524
1200
0.2719
4466.3
5146.0
8.862
0.2266
4465.3
5145.1
8.777
0.1942
4464.4
5144.1
8.705







PAGE 4 45

























Superheated Water (SI)
T
∘
ν
u
h
s
ν
( C) (m 3 /kg) (k J/kg) (k J/kg) (k J/kg /K ) (m 3 /kg)
p = 4.0 MPa (Tsat = 250.35∘C)
u
(k J/kg)
h
s
ν
3
(k
J/kg
/K
)
(k J/kg)
(m /kg)
p = 4.5 MPa (Tsat = 257.44∘C)
u
(k J/kg)
h
s
(k
J/kg
/K )
(k J/kg)
p = 5.0 MPa (Tsat = 263.94∘C)
Sat.
0.0498
2601.7
2800.8
6.070
0.0441
2599.7
2797.9
6.020
0.0394
2597.0
2794.2
5.974
275
0.0546
2668.9
2887.3
6.231
0.0473
2651.3
2864.3
6.143
0.0414
2632.3
2839.5
6.057
300
0.0589
2726.2
2961.7
6.364
0.0514
2713.0
2944.2
6.285
0.0453
2699.0
2925.7
6.211
350
0.0665
2827.4
3093.3
6.584
0.0584
2818.6
3081.5
6.515
0.0520
2809.5
3069.3
6.452
400
0.0734
2920.7
3214.5
6.771
0.0648
2914.2
3205.6
6.707
0.0578
2907.5
3196.7
6.648
450
0.0800
3011.0
3331.2
6.939
0.0708
3005.8
3324.2
6.877
0.0633
3000.6
3317.2
6.821
500
0.0864
3100.3
3446.0
7.092
0.0765
3096.0
3440.4
7.032
0.0686
3091.7
3434.7
6.978
600
0.0989
3279.4
3674.9
7.371
0.0877
3276.4
3670.9
7.313
0.0787
3273.3
3666.8
7.261
700
0.1110
3462.4
3906.3
7.621
0.0985
3460.0
3903.3
7.565
0.0885
3457.7
3900.3
7.514
800
0.1229
3650.6
4142.3
7.852
0.1092
3648.8
4140.0
7.796
0.0982
3646.9
4137.7
7.746
900
0.1348
3844.8
4383.9
8.067
0.1197
3843.3
4382.1
8.012
0.1077
3841.8
4380.2
7.962
1000
0.1465
4045.1
4631.2
8.270
0.1302
4043.9
4629.8
8.214
0.1172
4042.6
4628.3
8.165
1100
0.1582
4251.4
4884.4
8.461
0.1406
4250.4
4883.2
8.406
0.1266
4249.3
4882.1
8.357
1200
0.1699
4463.5
5143.2
8.643
0.1510
4462.6
5142.2
8.588
0.1359
4461.6
5141.3
8.539
1300
0.1816
4680.9
5407.2
8.816
0.1614
4680.1
5406.5
8.761
0.1453
4679.3
5405.7
8.712
p = 6.0 MPa (Tsat = 275.59∘C)
p = 7.0 MPa (Tsat = 285.83∘C)
p = 8.0 MPa (Tsat = 295.01∘C)
Sat.
0.0324
2589.9
2784.6
5.890
0.0274
2581.0
2772.6
5.815
0.0235
2570.5
2758.7
5.745
300
0.0362
2668.4
2885.5
6.070
0.0295
2633.5
2839.9
5.934
0.0243
2592.3
2786.5
5.794
350
0.0423
2790.4
3043.9
6.336
0.0353
2770.1
3016.9
6.230
0.0300
2748.3
2988.1
6.132
400
0.0474
2893.7
3178.2
6.543
0.0400
2879.5
3159.2
6.450
0.0343
2864.6
3139.4
6.366
450
0.0522
2989.9
3302.9
6.722
0.0442
2979.0
3288.3
6.635
0.0382
2967.8
3273.3
6.558
500
0.0567
3083.1
3423.1
6.883
0.0482
3074.3
3411.4
6.800
0.0418
3065.4
3399.5
6.727
600
0.0653
3267.2
3658.7
7.169
0.0557
3260.9
3650.6
7.091
0.0485
3254.7
3642.4
7.022
700
0.0735
3453.0
3894.3
7.425
0.0629
3448.3
3888.2
7.349
0.0548
3443.6
3882.2
7.282
800
0.0816
3643.2
4133.1
7.658
0.0699
3639.5
4128.4
7.584
0.0610
3635.7
4123.8
7.518
900
0.0896
3838.8
4376.6
7.875
0.0768
3835.7
4373.0
7.801
0.0671
3832.6
4369.3
7.737
1000
0.0976
4040.1
4625.4
8.079
0.0836
4037.5
4622.5
8.006
0.0731
4035.0
4619.6
7.942
1100
0.1054
4247.1
4879.7
8.271
0.1955
4047.7
4634.1
8.405
0.0790
4242.8
4875.0
8.135
1200
0.1133
4459.8
5139.4
8.453
0.2111
4253.6
4886.7
8.596
0.0849
4456.1
5135.5
8.318
1300
0.1211
4677.7
5404.1
8.627
0.2266
4465.3
5145.1
8.777
0.0908
4674.5
5401.0
8.493







PAGE 4 46

























Superheated Water (SI)
T
∘
ν
u
h
s
ν
( C) (m 3 /kg) (k J/kg) (k J/kg) (k J/kg /K ) (m 3 /kg)
p = 9.0 MPa (Tsat = 303.35∘C)
u
(k J/kg)
h
s
ν
3
(k
J/kg
/K
)
(k J/kg)
(m /kg)
p = 10.0 MPa (Tsat = 311.00∘C)
u
(k J/kg)
h
s
(k
J/kg
/K )
(k J/kg)
p = 12.5 MPa (Tsat = 327.81∘C)
sat. 0.020489 2558.5
2742.9
5.6791
0.018028
2545.2
2725.5
5.6159
0.013496
2505.6
2674.3
5.4638
325 0.023284 2647.6
2857.1
5.8738
0.019877
2611.6
2810.3
5.7596
38.849
1074.5
1146.4
1.7926
350 0.025816 2725.0
2957.3
6.0380
0.022440
2699.6
2924.0
5.9460
0.016138
2624.9
2826.6
5.7130
400 0.029960 2849.2
3118.8
6.2876
0.026436
2833.1
3097.5
6.2141
0.020030
2789.6
3040.0
6.0433
450 0.033524 2956.3
3258.0
6.4872
0.029782
2944.5
3242.4
6.4219
0.023019
2913.7
3201.5
6.2749
500 0.036793 3056.3
3387.4
6.6603
0.032811
3047.0
3375.1
6.5995
0.025630
3023.2
3343.6
6.4651
550 0.039885 3153.0
3512.0
6.8164
0.035655
3145.4
3502.0
6.7585
0.028033
3126.1
3476.5
6.6317
600 0.042861 3248.4
3634.1
6.9605
0.038378
3242.0
3625.8
6.9045
0.030306
3225.8
3604.6
6.7828
650 0.045755 3343.4
3755.2
7.0954
0.041018
3338.0
3748.1
7.0408
0.032491
3324.1
3730.2
6.9227
700 0.048589 3438.8
3876.1
7.2229
0.043597
3434.0
3870.0
7.1693
0.034612
3422.0
3854.6
7.0540
800 0.054132 3632.0
4119.2
7.4606
0.048629
3628.2
4114.5
7.4085
0.038724
3618.8
4102.8
7.2967
900 0.059562 3829.6
4365.7
7.6802
0.053547
3826.5
4362.0
7.6290
0.042720
3818.9
4352.9
7.5195
1000 0.064919 4032.4
4616.7
7.8855
0.058391
4029.9
4613.8
7.8349
0.046641
4023.5
4606.5
7.7269
1100 0.070224 4240.7
4872.7
8.0791
0.063183
4238.5
4870.3
8.0289
0.050510
4233.1
4864.5
7.9220
1200 0.075492 4454.2
5133.6
8.6252
0.067938
4452.4
5131.7
8.2126
0.054342
4447.7
5127.0
8.1065
1300 0.080733 4672.9
5399.5
8.4371
0.072667
4671.3
5398.0
8.3874
0.058147
4667.3
5394.1
8.2819
p = 15.0 MPa (Tsat = 342.16∘C)
p = 17.5 MPa (Tsat = 354.67∘C)
p = 20 MPa (Tsat = 365.75∘C)
Sat.
0.01034
2455.6
2610.7
5.311
0.00793
2390.5
2529.3
5.143
0.00587
2295.0
2412.3
4.931
375
0.01390
2650.4
2858.9
5.705
0.01056
2567.5
2752.3
5.494
0.00768
2449.1
2602.6
5.228
400
0.01567
2740.6
2975.7
5.882
0.01246
2684.3
2902.4
5.721
0.00995
2617.9
2816.9
5.553
450
0.01848
2880.7
3157.9
6.143
0.01520
2845.4
3111.4
6.021
0.01272
2807.2
3061.7
5.904
500
0.02083
2998.4
3310.8
6.348
0.01739
2972.4
3276.7
6.242
0.01479
2945.3
3241.2
6.145
600
0.02492
3209.3
3583.1
6.680
0.02107
3192.5
3561.3
6.589
0.01819
3175.3
3539.0
6.508
700
0.02862
3409.8
3839.1
6.957
0.02434
3397.5
3823.5
6.873
0.02113
3385.1
3807.8
6.799
800
0.03212
3609.2
4091.1
7.204
0.02741
3599.7
4079.3
7.124
0.02387
3590.1
4067.5
7.053
900
0.03550
3811.2
4343.7
7.429
0.03035
3803.4
4334.5
7.351
0.02648
3795.7
4325.4
7.283
1000 0.03881
4017.1
4599.2
7.638
0.03322
4010.7
4592.0
7.562
0.02902
4004.3
4584.7
7.495
1100 0.04206
4227.7
4858.6
7.834
0.03603
4222.3
4852.8
7.759
0.03150
4216.9
4847.0
7.693
1200 0.04528
4443.1
5122.3
8.019
0.03881
4438.5
5117.6
7.945
0.03395
4433.8
5112.9
7.880
1300 0.04847
4663.3
5390.3
8.195
0.04156
4659.2
5386.5
8.122
0.03637
4655.2
5382.7
8.057







PAGE 4 47

























Superheated Water (SI)
ν
u
h
( C) (m /kg) (k J/kg) (k J/kg)
T
∘
3
s
(k J/kg /K )
p = 5.0 MPa (Tsat = 263.94∘C)
ν
(m /kg)
3
u
(k J/kg)
h
s
ν
3
(k J/kg) (k J/kg /K ) (m /kg)
p = 10.0 MPa (Tsat = 311.00∘C)
u
(k J/kg)
h
s
(k J/kg) (k J/kg /K )
p = 15 MPa (Tsat = 342.16∘C)
20
0.00100
83.609
88.607
0.29543
0.00100
83.308
93.281
0.29435
0.00100
83.007
97.934
0.29323
40
0.00101
166.92
171.95
0.57046
0.00100
166.33
176.36
0.56851
0.00100
165.75
180.77
0.56656
60
0.00101
250.29
255.36
0.82865
0.00101
249.42
259.55
0.82602
0.00101
248.58
263.74
0.8234
80
0.00103
333.82
338.95
1.0723
0.00102
332.69
342.94
1.0691
0.00102
331.59
346.92
1.0659
100
0.00104
417.64
422.85
1.3034
0.00104
416.23
426.62
1.2996
0.00104
414.85
430.39
1.2958
120
0.00106
501.90
507.19
1.5236
0.00105
500.18
510.73
1.5191
0.00105
498.49
514.28
1.5148
140
0.00108
586.79
592.18
1.7344
0.00107
584.71
595.45
1.7293
0.00107
582.69
598.75
1.7243
160
0.00110
672.55
678.04
1.9374
0.00110
670.06
681.01
1.9315
0.00109
667.63
684.01
1.9259
180
0.00112
759.46
765.08
2.1338
0.00112
756.48
767.68
2.1271
0.00112
753.58
770.32
2.1206
200
0.00115
847.91
853.68
2.3251
0.00115
844.31
855.8
2.3174
0.00114
840.84
857.99
2.3100
220
0.00119
938.39
944.32
2.5127
0.00118
934.00
945.81
2.5037
0.00118
929.80
947.43
2.4951
240
0.00123
1031.6
1037.7
2.6983
0.00122
1026.1
1038.3
2.6876
0.00121
1021.0
1039.2
2.6774
260
0.00128
1128.5
1134.9
2.8841
0.00127
1121.6
1134.3
2.8710
0.00126
1115.1
1134
2.8586
0.00132
1221.8
1235.0
3.0565
0.00131
1213.4
1233.0
3.0410
0.00147
1431.9
1454.0
3.4263
280
320
p = 20.0 MPa (Tsat = 365.75∘C)
p = 30 MPa
p = 50 MPa
20
0.00099
82.708
102.57
0.29207
0.00099
82.112
111.77
0.28968 0.0009805
80.93
129.95
0.2845
40
0.00100
165.17
185.16
0.56461
0.001
164.05
193.9
0.56069 0.0009872 161.90
211.25
0.5528
60
0.00101
247.75
267.92
0.8208
0.001
246.14
276.26
0.81564 0.0009962 243.08
292.88
0.8055
80
0.00102
330.50
350.9
1.0627
0.00102
328.40
358.86
1.0564
0.0010072 324.42
374.78
1.0442
100
0.00103
413.50
434.17
1.292
0.00103
410.87
441.74
1.2847
0.0010201 405.94
456.94
1.2705
120
0.00105
496.85
517.84
1.5105
0.00104
493.66
525
1.5020
0.0010349 487.69
539.43
1.4859
140
0.00107
580.71
602.07
1.7194
0.00106
576.89
608.76
1.7098
0.0010517 569.77
622.36
1.6916
160
0.00109
665.27
687.05
1.9203
0.00108
660.74
693.21
1.9094
0.0010704 652.33
705.85
1.8889
180
0.00111
750.77
773.02
2.1143
0.0011
745.40
778.54
2.1020
0.0010914 735.49
790.06
2.0790
200
0.00114
837.49
860.27
2.3027
0.00113
831.10
865.02
2.2888
0.0011149 819.45
875.19
2.2628
220
0.00117
925.77
949.16
2.4867
0.00116
918.14
952.93
2.4707
0.0011412 904.39
961.45
2.4414
240
0.00121
1016.1
1040.2
2.6676
0.00119
1006.9
1042.7
2.6491
0.0011708 990.55
1049.1
2.6156
260
0.00125
1109.0
1134
2.8469
0.00123
1097.8
1134.7
2.8250
0.0012044 1078.2
1138.4
2.7864
280
0.00130
1205.5
1231.5
3.0265
0.00128
1191.5
1229.8
3.0001
0.0012430 1167.7
1229.9
2.9547
300
0.00136
1307.1
1334.4
3.2091
0.00133
1288.9
1328.9
3.1760
0.0012879 1259.6
1324.0
3.1218
320
0.00144
1416.6
1445.5
3.3996
0.0014
1391.6
1433.7
3.3557
0.0013409 1354.3
1421.4
3.2888
340
0.00157
1540.2
1571.6
3.6086
0.00149
1502.3
1547.1
3.5438
0.0014049 1452.9
1523.1
3.4575
360
0.00182
1703.6
1740.1
3.8787
0.00163
1626.7
1675.6
3.7498
0.0014848 1556.5
1630.7
3.6301





PAGE 4 48

























C OMPRESSED LIQUID WATER (FOR VERY HIGH PRESS)
Saturated Refrigerant R-134a - Temperature (SI)
T
(∘C)
P
(kPa)
νf
(m 3 /kg)
νg
(m 3 /kg)
uf
(k J/kg)
ug
(k J/kg)
hf
(k J/kg)
hg
sf
sg
(k J/kg) (k J/kg/K ) (k J/kg/K )
-40
51.2 0.0007054 0.3611
-0.04
207.37
0.00
225.86
0.0000
0.9687
-36
62.9 0.0007112 0.2977
4.99
209.66
5.04
228.39
0.0214
0.9632
-32
76.7 0.0007172 0.2473
10.05
211.96
10.10
230.92
0.0425
0.9582
-28
92.7 0.0007234 0.2068
15.13
214.26
15.20
233.43
0.0634
0.9536
-26
101.7 0.0007265 0.1896
17.69
215.41
17.76
234.68
0.0738
0.9515
-24
111.3 0.0007297 0.1741
20.25
216.55
20.33
235.93
0.0841
0.9495
-22
121.7 0.0007329 0.1601
22.82
217.70
22.91
237.17
0.0944
0.9476
-20
132.7 0.0007362 0.1474
25.39
218.85
25.49
238.41
0.1046
0.9457
-18
144.6 0.0007396 0.1359
27.98
219.99
28.09
239.64
0.1148
0.9440
-16
157.3 0.0007430 0.1255
30.57
221.13
30.69
240.87
0.1249
0.9423
-14
170.8 0.0007464 0.1161
33.17
222.27
33.30
242.09
0.1350
0.9407
-12
185.2 0.0007499 0.1074
35.78
223.41
35.92
243.31
0.1451
0.9392
-10
200.6 0.0007535 0.0996
38.40
224.54
38.55
244.52
0.1550
0.9377
-8
216.9 0.0007571 0.0924
41.03
225.67
41.19
245.72
0.1650
0.9364
-6
234.3 0.0007608 0.0859
43.67
226.80
43.84
246.92
0.1749
0.9351
-4
252.7 0.0007646 0.0799
46.31
227.93
46.50
248.11
0.1848
0.9338
-2
272.2 0.0007684 0.0744
48.97
229.05
49.17
249.29
0.1946
0.9326
0
292.8 0.0007723 0.0693
51.63
230.17
51.86
250.46
0.2044
0.9315
2
314.6 0.0007763 0.0647
54.30
231.28
54.55
251.62
0.2142
0.9304
4
337.7 0.0007804 0.0604
56.99
232.39
57.25
252.78
0.2239
0.9294
6
362.0 0.0007845 0.0564
59.68
233.49
59.97
253.92
0.2336
0.9284
8
387.6 0.0007887 0.0528
62.39
234.58
62.69
255.05
0.2432
0.9274
12
443.0 0.0007975 0.0463
67.83
236.76
68.19
257.29
0.2625
0.9256
16
504.3 0.0008066 0.0408
73.32
238.91
73.73
259.47
0.2816
0.9240
20
571.7 0.0008161 0.0360
78.86
241.02
79.32
261.60
0.3006
0.9224
24
645.8 0.0008261 0.0319
84.45
243.11
84.98
263.68
0.3196
0.9210
26
685.4 0.0008313 0.0300
87.26
244.13
87.83
264.70
0.3290
0.9203
28
726.9 0.0008367 0.0283
90.09
245.15
90.70
265.69
0.3385
0.9196

PAGE 4 49

















S AT U R AT E D R - 13 4 A - T E M P E R AT U R E TA B L E ( S I )
∘
( C)
P
(kPa)
νg
(m 3 /kg)
uf
(k J/kg)
ug
(k J/kg)
hf
(k J/kg)
hg
sf
sg
(k J/kg) (k J/kg/K ) (k J/kg/K )
30
770.2 0.0008421 0.0266
92.93
246.16
93.58
266.67
0.3479
0.9189
32
815.4 0.0008478 0.0251
95.79
247.15
96.48
267.64
0.3573
0.9182
34
862.6 0.0008536 0.0237
98.66
248.13
99.40
268.58
0.3667
0.9175
36
911.9 0.0008595 0.0224
101.55
249.10
102.33
269.50
0.3761
0.9168
38
963.2 0.0008657 0.0211
104.46
250.05
105.29
270.41
0.3855
0.9162
40
1016.6 0.0008720 0.0200
107.38
250.99
108.27
271.28
0.3949
0.9155
42
1072.2 0.0008786 0.0189
110.32
251.91
111.26
272.14
0.4043
0.9147
44
1130.1 0.0008854 0.0178
113.28
252.81
114.28
272.97
0.4136
0.9140
48
1252.9 0.0008997 0.0160
119.26
254.56
120.39
274.55
0.4324
0.9125
52
1385.4 0.0009150 0.0143
125.33
256.23
126.60
276.01
0.4513
0.9108
56
1528.2 0.0009317 0.0128
131.49
257.79
132.92
277.32
0.4702
0.9089
60
1681.8 0.0009498 0.0114
137.76
259.24
139.36
278.49
0.4892
0.9068
70
2116.8 0.0010038 0.0087
154.01
262.19
156.14
280.51
0.5376
0.9000
80
2633.2 0.0010773 0.0064
171.41
263.69
174.25
280.67
0.5880
0.8894
90
3244.2 0.0011936 0.0046
190.91
262.30
194.78
277.27
0.6434
0.8706
100
3972.4 0.0015357 0.0027
219.05
248.89
225.15
259.54
0.7232
0.8153
101.06 4059.1 0.0019535 0.0020
233.57
233.57
241.49
241.49
0.7665
0.7665
T
νf
(m 3 /kg)

PAGE 450

















Saturated Refrigerant R-134a - Temperature (SI)
Saturated Refrigerant R-134a - Pressure (SI)
p
(kPa)
T
(∘C)
νf
(m 3 /kg)
νg
(m 3 /kg)
uf
(k J/kg)
ug
(k J/kg)
hf
(k J/kg)
hg
sf
sg
(k J/kg) (k J/kg/K ) (k J/kg/K )
60
-36.9 0.0007098 0.3112
3.8
209.1
3.9
227.8
0.0164
0.9645
80
-31.1 0.0007185 0.2376
11.2
212.5
11.2
231.5
0.0472
0.9572
100
-26.4 0.0007259 0.1926
17.2
215.2
17.3
234.5
0.0720
0.9519
120
-22.3 0.0007324 0.1621
22.4
217.5
22.5
237.0
0.0928
0.9478
140
-18.8 0.0007383 0.1402
27.0
219.6
27.1
239.2
0.1110
0.9446
160
-15.6 0.0007437 0.1235
31.1
221.4
31.2
241.1
0.1270
0.9420
180
-12.7 0.0007487 0.1104
34.9
223.0
35.0
242.9
0.1415
0.9397
200
-10.1 0.0007534 0.0999
38.3
224.5
38.5
244.5
0.1547
0.9378
220
-7.6
0.0007578 0.0912
41.5
225.9
41.7
245.9
0.1668
0.9361
240
-5.4
0.0007620 0.0839
44.5
227.2
44.7
247.3
0.1780
0.9347
260
-3.2
0.0007661 0.0777
47.3
228.4
47.5
248.6
0.1885
0.9333
280
-1.2
0.0007699 0.0724
50.0
229.5
50.2
249.7
0.1984
0.9322
300
0.7
0.0007737 0.0677
52.5
230.5
52.8
250.9
0.2077
0.9311
320
2.5
0.0007773 0.0636
54.9
231.5
55.2
251.9
0.2165
0.9301
340
4.2
0.0007808 0.0600
57.3
232.5
57.5
252.9
0.2248
0.9293
360
5.8
0.0007842 0.0567
59.5
233.4
59.8
253.8
0.2328
0.9284
400
8.9
0.0007907 0.0512
63.7
235.1
64.0
255.6
0.2477
0.9270
500
15.7 0.0008060 0.0411
73.0
238.8
73.4
259.3
0.2803
0.9241
600
21.6 0.0008200 0.0343
81.0
241.9
81.5
262.4
0.3081
0.9219
700
26.7 0.0008332 0.0294
88.3
244.5
88.8
265.1
0.3324
0.9200
800
31.3 0.0008459 0.0256
94.8
246.8
95.5
267.3
0.3541
0.9184
900
35.5 0.0008581 0.0227
100.9
248.9
101.6
269.3
0.3739
0.9170
1000
39.4 0.0008701 0.0203
106.5
250.7
107.4
271.0
0.3920
0.9157
1200
46.3 0.0008935 0.0167
116.7
253.8
117.8
273.9
0.4245
0.9131
1400
52.4 0.0009167 0.0141
126.0
256.4
127.3
276.2
0.4533
0.9106
1600
57.9 0.0009401 0.0121
134.5
258.5
136.0
277.9
0.4792
0.9080
1800
62.9 0.0009640 0.0106
142.4
260.2
144.1
279.2
0.5031
0.9051
2000
67.5 0.0009888 0.0093
149.8
261.6
151.8
280.1
0.5252
0.9020

PA G E 4 51

















S AT U R AT E D R - 13 4 A - P R E S S U R E TA B L E ( S I )
p
(kPa)
( C)
νg
(m 3 /kg)
uf
(k J/kg)
ug
(k J/kg)
hf
(k J/kg)
hg
sf
sg
(k J/kg) (k J/kg/K ) (k J/kg/K )
2500
77.6 0.0010569 0.0069
167.1
263.5
169.7
280.9
0.5755
0.8925
3000
86.2 0.0011413 0.0053
183.1
263.4
186.6
279.2
0.6215
0.8792
T
∘
νf
(m 3 /kg)

PAGE 452

















Saturated Refrigerant R-134a - Pressure (SI)
Superheated R-134a (SI)
ν
u
h
( C) (m /kg) (k J/kg) (k J/kg)
T
∘
3
s
(k J/kg /K )
p = 0.06 MPa (Tsat = − 36.95∘C)
ν
(m /kg)
3
u
(k J/kg)
h
s
ν
3
(k J/kg) (k J/kg /K ) (m /kg)
p = 0.1 MPa (Tsat = − 26.37∘C)
u
(k J/kg)
h
s
(k J/kg) (k J/kg /K )
p = 0.14 MPa (Tsat = − 18.77∘C)
Sat.
0.3112
209.1
227.8
0.964
0.1926
215.2
234.5
0.952
1.6941
2505.6
2675.0
7.3589
-20
0.3361
220.6
240.8
1.018
0.1984
219.7
239.5
0.972
0.1402
219.6
239.2
0.945
-10
0.3505
227.6
248.6
1.048
0.2074
226.8
247.5
1.003
0.1461
225.9
246.4
0.972
0
0.3648
234.7
256.5
1.077
0.2163
234.0
255.6
1.033
0.1526
233.2
254.6
1.003
10
0.3789
241.9
264.7
1.107
0.2251
241.3
263.8
1.063
0.1591
240.7
262.9
1.033
20
0.3930
249.4
272.9
1.135
0.2337
248.8
272.2
1.092
0.1654
248.2
271.4
1.062
30
0.4071
257.0
281.4
1.164
0.2423
256.5
280.7
1.120
0.1717
255.9
280.0
1.091
40
0.4210
264.7
290.0
1.192
0.2509
264.3
289.3
1.149
0.1780
263.8
288.7
1.120
50
0.4350
272.6
298.7
1.219
0.2594
272.2
298.2
1.176
0.1841
271.8
297.6
1.147
60
0.4488
280.7
307.7
1.246
0.2678
280.4
307.1
1.204
0.1903
280.0
306.6
1.175
70
0.4627
289.0
316.8
1.273
0.2763
288.6
316.3
1.231
0.1964
288.3
315.8
1.202
80
0.4765
297.4
326.0
1.300
0.2847
297.1
325.6
1.257
0.2024
296.8
325.1
1.229
90
0.4903
306.0
335.4
1.326
0.2930
305.7
335.0
1.284
0.2085
305.4
334.6
1.255
100
0.5041
314.8
345.0
1.352
0.3014
314.5
344.6
1.310
0.2145
314.2
344.2
1.282
p = 0.18 MPa (Tsat = − 12.73∘C)
Sat.
0.1104
223.0
242.9
0.940
-10
0.1119
225.0
245.2
0.948
0
0.1172
232.5
253.6
10
0.1224
240.0
20
0.1275
30
p = 0.2 MPa (Tsat = − 10.09∘C)
p = 0.24 MPa (Tsat = − 5.4∘C)
0.0999
224.5
244.5
0.938
0.0839
227.2
247.3
0.935
0.980
0.1048
232.1
253.1
0.970
0.0862
231.3
252.0
0.952
262.0
1.010
0.1096
239.7
261.6
1.001
0.0903
239.0
260.7
0.983
247.6
270.6
1.040
0.1142
247.4
270.2
1.030
0.0942
246.8
269.4
1.013
0.1325
255.4
279.3
1.069
0.1187
255.2
278.9
1.060
0.0981
254.6
278.2
1.043
40
0.1374
263.3
288.1
1.098
0.1232
263.1
287.7
1.088
0.1019
262.6
287.1
1.072
50
0.1423
271.4
297.0
1.126
0.1277
271.2
296.7
1.116
0.1057
270.7
296.1
1.100
60
0.1472
279.6
306.1
1.153
0.1321
279.4
305.8
1.144
0.1094
279.0
305.2
1.128
70
0.1520
287.9
315.3
1.181
0.1364
287.7
315.0
1.171
0.1131
287.4
314.5
1.156
80
0.1567
296.4
324.6
1.207
0.1407
296.3
324.4
1.198
0.1168
295.9
323.9
1.183
90
0.1615
305.1
334.1
1.234
0.1451
304.9
333.9
1.225
0.1204
304.6
333.5
1.209
100
0.1662
313.9
343.8
1.260
0.1493
313.7
343.6
1.251
0.1240
313.5
343.2
1.236







PAGE 453

























S U P E R H E AT E D R - 13 4 A TA B L E S ( S I )
T
∘
ν
u
h
s
ν
( C) (m 3 /kg) (k J/kg) (k J/kg) (k J/kg /K ) (m 3 /kg)
p = 0.28 MPa (Tsat = − 1.25∘C)
u
(k J/kg)
h
s
ν
3
(k
J/kg
/K
)
(k J/kg)
(m /kg)
p = 0.32 MPa (Tsat = 2.46∘C)
u
(k J/kg)
h
s
(k
J/kg
/K )
(k J/kg)
p = 0.40 MPa (Tsat = 8.91∘C)
Sat.
0.0724
229.5
249.7
0.932
0.0636
231.5
251.9
0.930
0.0512
235.1
255.6
0.927
10
0.0765
238.3
259.7
0.968
0.0661
237.5
258.7
0.954
0.0515
236.0
256.6
0.931
20
0.0800
246.1
268.5
0.999
0.0693
245.5
267.7
0.986
0.0542
244.2
265.9
0.963
30
0.0834
254.1
277.4
1.029
0.0723
253.5
276.7
1.016
0.0568
252.4
275.1
0.994
40
0.0867
262.1
286.4
1.058
0.0753
261.6
285.7
1.045
0.0593
260.6
284.3
1.024
50
0.0900
270.3
295.5
1.086
0.0782
269.8
294.9
1.074
0.0617
268.9
293.6
1.053
60
0.0932
278.6
304.7
1.114
0.0811
278.2
304.1
1.102
0.0641
277.3
303.0
1.081
70
0.0964
287.0
314.0
1.142
0.0839
286.6
313.5
1.130
0.0664
285.9
312.4
1.109
80
0.0996
295.6
323.5
1.169
0.0868
295.2
323.0
1.157
0.0687
294.5
322.0
1.137
90
0.1028
304.3
333.1
1.196
0.0895
304.0
332.6
1.184
0.0710
303.3
331.7
1.164
100
0.1059
313.2
342.8
1.222
0.0923
312.9
342.4
1.211
0.0733
312.3
341.6
1.191
110
0.1090
322.2
352.7
1.248
0.0950
321.9
352.3
1.237
0.0755
321.3
351.5
1.217
120
0.1121
331.3
362.7
1.274
0.0977
331.1
362.4
1.263
0.0777
330.6
361.6
1.243
p = 0.50 MPa (Tsat = 15.71∘C)
Sat.
0.0411
238.8
259.3
0.924
20
0.0421
242.4
263.5
0.938
30
0.0443
250.8
273.0
40
0.0465
259.3
50
0.0485
60
p = 0.60 MPa (Tsat = 21.55∘C)
p = 0.70 MPa (Tsat = 26.69∘C)
0.0343
241.9
262.4
0.922
0.0294
244.5
265.1
0.920
0.970
0.0360
249.2
270.8
0.950
0.0300
247.5
268.5
0.931
282.5
1.001
0.0379
257.9
280.6
0.982
0.0317
256.4
278.6
0.964
267.7
292.0
1.031
0.0397
266.5
290.3
1.012
0.0333
265.2
288.5
0.995
0.0505
276.3
301.5
1.060
0.0414
275.2
300.0
1.042
0.0349
274.0
298.4
1.026
70
0.0524
284.9
311.1
1.088
0.0431
283.9
309.7
1.071
0.0364
282.9
308.3
1.055
80
0.0543
293.6
320.8
1.116
0.0447
292.7
319.6
1.099
0.0378
291.8
318.3
1.084
90
0.0562
302.5
330.6
1.144
0.0463
301.7
329.5
1.126
0.0393
300.8
328.3
1.111
100
0.0581
311.5
340.5
1.171
0.0479
310.7
339.5
1.154
0.0406
310.0
338.4
1.139
110
0.0599
320.6
350.6
1.197
0.0495
319.9
349.6
1.180
0.0420
319.2
348.6
1.166
120
0.0617
329.9
360.7
1.223
0.0510
329.2
359.8
1.207
0.0434
328.6
358.9
1.192
130
0.0635
339.3
371.0
1.249
0.0525
338.7
370.2
1.233
0.0447
338.0
369.3
1.219
140
0.0653
348.8
381.5
1.275
0.0540
348.3
380.7
1.258
0.0460
347.7
379.9
1.244







PAGE 45 4

























Superheated R-134a (SI)
T
∘
ν
u
h
s
ν
( C) (m 3 /kg) (k J/kg) (k J/kg) (k J/kg /K ) (m 3 /kg)
p = 0.80 MPa (Tsat = 31.31∘C)
u
(k J/kg)
h
s
ν
3
(k
J/kg
/K
)
(k J/kg)
(m /kg)
p = 0.90 MPa (Tsat = 35.51∘C)
u
(k J/kg)
h
s
(k
J/kg
/K )
(k J/kg)
p = 1.00 MPa (Tsat = 39.37∘C)
Sat.
0.0256
246.8
267.3
0.918
0.0227
248.9
269.3
0.917
0.0203
250.7
271.0
0.916
40
0.0270
254.8
276.5
0.948
0.0234
253.1
274.2
0.933
0.0204
251.3
271.7
0.918
50
0.0285
263.9
286.7
0.980
0.0248
262.4
284.8
0.966
0.0218
260.9
282.7
0.953
60
0.0300
272.8
296.8
1.011
0.0261
271.6
295.1
0.998
0.0231
270.3
293.4
0.985
70
0.0313
281.8
306.9
1.041
0.0274
280.7
305.4
1.028
0.0243
279.6
303.9
1.016
80
0.0327
290.8
317.0
1.070
0.0286
289.9
315.6
1.057
0.0254
288.9
314.3
1.046
90
0.0339
300.0
327.1
1.098
0.0298
299.1
325.9
1.086
0.0265
298.2
324.7
1.075
100
0.0352
309.2
337.3
1.126
0.0310
308.3
336.2
1.114
0.0276
307.5
335.1
1.103
110
0.0364
318.5
347.6
1.153
0.0321
317.7
346.6
1.141
0.0286
316.9
345.5
1.131
120
0.0376
327.9
358.0
1.180
0.0332
327.2
357.0
1.168
0.0296
326.5
356.1
1.158
130
0.0388
337.4
368.5
1.206
0.0342
336.8
367.6
1.195
0.0306
336.1
366.7
1.185
140
0.0400
347.1
379.1
1.232
0.0353
346.5
378.2
1.221
0.0316
345.9
377.4
1.211
150
0.0411
356.9
389.8
1.258
0.0363
356.3
389.0
1.247
0.0325
355.7
388.2
1.237
160
0.0423
366.8
400.6
1.283
0.0374
366.2
399.9
1.272
0.0335
365.7
399.2
1.262
p = 1.20 MPa (Tsat = 46.29∘C)
Sat.
0.0167
253.8
273.9
0.913
50
0.0172
257.6
278.3
0.927
60
0.0184
267.6
289.6
70
0.0195
277.2
80
0.0205
90
p = 1.40 MPa (Tsat = 52.40∘C)
p = 1.60 MPa (Tsat = 57.88∘C)
0.0141
256.4
276.2
0.911
0.0121
258.5
277.9
0.908
0.961
0.0150
264.5
285.5
0.939
0.0124
260.9
280.7
0.916
300.6
0.994
0.0161
274.6
297.1
0.973
0.0134
271.8
293.3
0.954
286.8
311.4
1.025
0.0170
284.5
308.3
1.006
0.0144
282.1
305.1
0.987
0.0215
296.3
322.1
1.055
0.0179
294.3
319.4
1.036
0.0152
292.2
316.5
1.019
100
0.0224
305.8
332.7
1.084
0.0188
304.0
330.3
1.066
0.0160
302.1
327.8
1.050
110
0.0233
315.4
343.4
1.112
0.0196
313.8
341.2
1.095
0.0168
312.1
338.9
1.080
120
0.0242
325.0
354.1
1.139
0.0204
323.6
352.1
1.123
0.0175
322.0
350.0
1.108
130
0.0251
334.8
364.9
1.166
0.0212
333.4
363.0
1.150
0.0182
332.0
361.1
1.136
140
0.0259
344.6
375.7
1.193
0.0219
343.4
374.0
1.177
0.0189
342.1
372.3
1.163
150
0.0268
354.6
386.7
1.219
0.0226
353.4
385.1
1.204
0.0195
352.2
383.5
1.190
160
0.0276
364.6
397.7
1.245
0.0234
363.5
396.2
1.230
0.0202
362.4
394.7
1.216
170
0.0284
374.8
408.8
1.270
0.0241
373.8
407.4
1.255
0.0208
372.7
406.0
1.242
180
0.0292
385.1
420.1
1.295
0.0248
384.1
418.8
1.281
0.0215
383.1
417.4
1.268







PAGE 455

























Superheated R-134a (SI)
Saturated Water - Temperature (EEU)
T
(∘F)
p
νf
νg
uf
ug
hf
hg
sf
sg
(Psia) ( f t 3 /lbm) ( f t 3 /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm /R) (Btu /lbm /R)
32.018 0.08871 0.01602
3299.9
0
1021
0
1075.2
0
2.1867
35
0.09998 0.01602
2945.7
3.004
1022
3.004
1076.5
0.00609
2.1762
40
0.12173 0.01602
2443.6
8.032
1023.7
8.032
1078.7
0.0162
2.1589
45
0.14756 0.01602
2035.8
13.05
1025.3
13.05
1080.9
0.0262
2.1421
50
0.17812 0.01602
1703.1
18.07
1026.9
18.07
1083.1
0.03609
2.1256
55
0.21413 0.01603
1430.4
23.07
1028.6
23.07
1085.3
0.04586
2.1096
60
0.25638 0.01604
1206.1
28.08
1030.2
28.08
1087.4
0.05554
2.094
65
0.30578 0.01604
1020.8
33.08
1031.8
33.08
1089.6
0.06511
2.0788
70
0.36334 0.01605
867.18
38.08
1033.5
38.08
1091.8
0.07459
2.0639
75
0.43016 0.01606
739.27
43.07
1035.1
43.07
1093.9
0.08398
2.0494
80
0.50745 0.01607
632.41
48.06
1036.7
48.07
1096.1
0.09328
2.0352
85
0.59659 0.01609
542.8
53.06
1038.3
53.06
1098.3
0.10248
2.0214
90
0.69904 0.0161
467.4
58.05
1040
58.05
1100.4
0.11161
2.0079
95
0.81643 0.01612
403.74
63.04
1041.6
63.04
1102.6
0.12065
1.9947
100 0.95052 0.01613
349.83
68.03
1043.2
68.03
1104.7
0.12961
1.9819
110
1.2767 0.01617
264.96
78.01
1046.4
78.02
1109
0.14728
1.957
120
1.6951
0.0162
202.94
88
1049.6
88
1113.2
0.16466
1.9332
130
2.226
0.01625
157.09
97.99
1052.7
97.99
1117.4
0.18174
1.9105
140
2.8931 0.01629
122.81
107.98
1055.9
107.99
1121.6
0.19855
1.8888
150
3.7234 0.01634
96.929
117.98
1059
117.99
1125.7
0.21508
1.868
160
4.7474 0.01639
77.185
127.98
1062
128
1129.8
0.23136
1.8481
170
5.9999 0.01645
61.982
138
1065.1
138.02
1133.9
0.24739
1.8289
180
7.5197 0.01651
50.172
148.02
1068.1
148.04
1137.9
0.26318
1.8106
190
9.3497 0.01657
40.92
158.05
1071
158.08
1141.8
0.27874
1.793
200
11.538 0.01663
33.613
168.1
1074
168.13
1145.7
0.29409
1.776
210
14.136
0.0167
27.798
178.15
1076.8
178.2
1149.5
0.30922
1.7597
212
14.709 0.01671
26.782
180.16
1077.4
180.21
1150.3
0.31222
1.7565
220
17.201 0.01677
23.136
188.22
1079.6
188.28
1153.3
0.32414
1.7439

PAGE 456

















SATURATED WATER - TEMPERATURE TABLE (EEU)
∘
( F)
p
νf
νg
uf
ug
hf
hg
sf
sg
(Psia) ( f t 3 /lbm) ( f t 3 /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm /R) (Btu /lbm /R)
230
20.795 0.01684
19.374
198.31
1082.4
198.37
1157
0.33887
1.7288
240
24.985 0.01692
16.316
208.41
1085.1
208.49
1160.5
0.35342
1.7141
250
29.844
0.017
13.816
218.54
1087.7
218.63
1164
0.36779
1.6999
260
35.447 0.01708
11.76
228.68
1090.3
228.79
1167.4
0.38198
1.6862
270
41.877 0.01717
10.059
238.85
1092.8
238.98
1170.7
0.39601
1.673
280
49.222 0.01726
8.6439
249.04
1095.2
249.2
1173.9
0.40989
1.6601
290
57.573 0.01735
7.4607
259.26
1097.5
259.45
1177
0.42361
1.6475
300
67.028 0.01745
6.4663
269.51
1099.8
269.73
1180
0.4372
1.6354
310
77.691 0.01755
5.6266
279.79
1101.9
280.05
1182.8
0.45065
1.6235
320
89.667 0.01765
4.9144
290.11
1104
290.4
1185.5
0.46396
1.612
330
103.07 0.01776
4.3076
300.46
1105.9
300.8
1188.1
0.47716
1.6007
340
118.02 0.01787
3.7885
310.85
1107.7
311.24
1190.5
0.49024
1.5897
350
134.63 0.01799
3.3425
321.29
1109.4
321.73
1192.7
0.50321
1.5789
360
153.03 0.01811
2.958
331.76
1111
332.28
1194.8
0.51607
1.5683
370
173.36 0.01823
2.6252
342.29
1112.5
342.88
1196.7
0.52884
1.558
380
195.74 0.01836
2.3361
352.87
1113.9
353.53
1198.5
0.54152
1.5478
390
220.33
0.0185
2.0842
363.50
1115.1
364.25
1200.1
0.55411
1.5378
400
247.26 0.01864
1.86390
374.19
1116.2
375.04
1201.4
0.56663
1.5279
410
276.69 0.01878
1.67060
384.94
1117.1
385.90
1202.6
0.57907
1.5182
420
308.76 0.01894
1.50060
395.76
1117.8
396.84
1203.6
0.59145
1.5085
430
343.64 0.01910
1.35050
406.65
1118.4
407.86
1204.3
0.60377
1.4990
440
381.49 0.01926
1.21780
417.61
1118.9
418.97
1204.8
0.61603
1.4895
450
422.47 0.01944
1.09990
428.66
1119.1
430.18
1205.1
0.62826
1.4801
460
466.75 0.01962
0.99510
439.79
1119.2
441.48
1205.1
0.64044
1.4708
470
514.52 0.01981
0.90158
451.01
1119.0
452.90
1204.9
0.65260
1.4615
480
565.96 0.02001
0.81794
462.34
1118.7
464.43
1204.3
0.66474
1.4521
490
621.24 0.02022
0.74296
473.77
1118.1
476.09
1203.5
0.67686
1.4428
500
680.56 0.02044
0.67558
485.32
1117.3
487.89
1202.3
0.68899
1.4334
510
744.11 0.02067
0.61489
496.99
1116.2
499.84
1200.8
0.70112
1.4240
T

PAGE 457

















Saturated Water - Temperature (EEU)
∘
( F)
p
νf
νg
uf
ug
hf
hg
sf
sg
(Psia) ( f t 3 /lbm) ( f t 3 /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm /R) (Btu /lbm /R)
520
812.11 0.02092
0.56009
508.80
1114.8
511.94
1199.0
0.71327
1.4145
530
884.74 0.02118
0.51051
520.76
1113.1
524.23
1196.7
0.72546
1.4049
540
962.24 0.02146
0.46553
532.88
1111.1
536.70
1194.0
0.73770
1.3952
550
1044.8 0.02176
0.42465
545.18
1108.8
549.39
1190.9
0.75000
1.3853
560
1132.7 0.02207
0.38740
557.68
1106.0
562.31
1187.2
0.76238
1.3752
570
1226.2 0.02242
0.35339
570.40
1102.8
575.49
1183.0
0.77486
1.3649
580
1325.5 0.02279
0.32225
583.37
1099.2
588.95
1178.2
0.78748
1.3543
590
1430.8 0.02319
0.29367
596.61
1095.0
602.75
1172.8
0.80026
1.3433
600
1542.5 0.02362
0.26737
610.18
1090.3
616.92
1166.6
0.81323
1.3319
610
1660.9 0.02411
0.24309
624.11
1084.8
631.52
1159.5
0.82645
1.3201
620
1786.2 0.02464
0.22061
638.47
1078.6
646.62
1151.5
0.83998
1.3076
630
1918.9 0.02524
0.19972
653.35
1071.5
662.32
1142.4
0.85389
1.2944
640
2059.3 0.02593
0.18019
668.86
1063.2
678.74
1131.9
0.86828
1.2803
650
2207.8 0.02673
0.16184
685.16
1053.6
696.08
1119.7
0.88332
1.2651
660
2364.9 0.02767
0.14444
702.48
1042.2
714.59
1105.4
0.89922
1.2483
670
2531.2 0.02884
0.12774
721.23
1028.5
734.74
1088.3
0.91636
1.2293
680
2707.3 0.03035
0.11134
742.11
1011.1
757.32
1066.9
0.93541
1.2070
690
2894.1 0.03255
0.09451
766.81
987.6
784.24
1038.2
0.95797
1.1789
700
3093
0.03670
0.07482
801.75
948.3
822.76
991.1
0.99023
1.1354
705.10 3200.1 0.04975
0.04975
866.61
866.6
896.07
896.1
1.05257
1.0526
T

PAGE 458

















Saturated Water - Temperature (EEU)
Saturated Water - Pressure (EEU)
p
(Psia)
νf
νg
uf
ug
hf
hg
sf
sg
(∘F) ( f t 3 /lbm) ( f t 3 /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm /R) (Btu /lbm /R)
T
1
101.69 0.01614
333.49
69.72
1043.7
69.72
1105.4
0.13262
1.9776
2
126.02 0.01623
173.71
94.02
1051.5
94.02
1115.8
0.17499
1.9194
3
141.41 0.01630
118.70
109.39
1056.3
109.4
1122.2
0.20090
1.8858
4
152.91 0.01636
90.629
120.89
1059.9
120.9
1126.9
0.21985
1.8621
5
162.18 0.01641
73.525
130.17
1062.7
130.18
1130.7
0.23488
1.8438
6
170.00 0.01645
61.982
138.00
1065.1
138.02
1133.9
0.24739
1.8289
8
182.81 0.01652
47.347
150.83
1068.9
150.86
1139.0
0.26757
1.8056
10
193.16 0.01659
38.425
161.22
1072.0
161.25
1143.1
0.28362
1.7875
14.696 211.95 0.01671
26.805
180.12
1077.4
180.16
1150.3
0.31215
1.7566
15
212.99 0.01672
26.297
181.16
1077.7
181.21
1150.7
0.31370
1.7549
20
227.92 0.01683
20.093
196.21
1081.8
196.27
1156.2
0.33582
1.7319
25
240.03 0.01692
16.307
208.45
1085.1
208.52
1160.6
0.35347
1.7141
30
250.30 0.01700
13.749
218.84
1087.8
218.93
1164.1
0.36821
1.6995
35
259.25 0.01708
11.901
227.92
1090.1
228.03
1167.2
0.38093
1.6872
40
267.22 0.01715
10.501
236.02
1092.1
236.14
1169.8
0.39213
1.6766
45
274.41 0.01721
9.4028
243.34
1093.9
243.49
1172.2
0.40216
1.6672
50
280.99 0.01727
8.5175
250.05
1095.4
250.21
1174.2
0.41125
1.6588
55
287.05 0.01732
7.7882
256.25
1096.9
256.42
1176.1
0.41958
1.6512
60
292.69 0.01738
7.1766
262.01
1098.1
262.2
1177.8
0.42728
1.6442
65
297.95 0.01743
6.6560
267.41
1099.3
267.62
1179.4
0.43443
1.6378
70
302.91 0.01748
6.2075
272.50
1100.4
272.72
1180.8
0.44112
1.6319
75
307.59 0.01752
5.8167
277.31
1101.4
277.55
1182.1
0.44741
1.6264
80
312.02 0.01757
5.4733
281.87
1102.3
282.13
1183.4
0.45335
1.6212
85
316.24 0.01761
5.1689
286.22
1103.2
286.5
1184.5
0.45897
1.6163
90
320.26 0.01765
4.8972
290.38
1104.0
290.67
1185.6
0.46431
1.6117
95
324.11 0.01770
4.6532
294.36
1104.8
294.67
1186.6
0.46941
1.6073
100
327.81 0.01774
4.4327
298.19
1105.5
298.51
1187.5
0.47427
1.6032
110
334.77 0.01781
4.0410
305.41
1106.8
305.78
1189.2
0.48341
1.5954

PAGE 459

















SATURATED WATER - PRESSURE TABLE (EEU)
p
(Psia)
νf
νg
uf
ug
hf
hg
sf
sg
( F) ( f t 3 /lbm) ( f t 3 /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm /R) (Btu /lbm /R)
120
341.25 0.01789
3.7289
312.16
1107.9
312.55
1190.8
0.49187
1.5883
130
347.32 0.01796
3.4557
318.48
1109.0
318.92
1192.1
0.49974
1.5818
140
353.03 0.01802
3.2202
324.45
1109.9
324.92
1193.4
0.50711
1.5757
150
358.42 0.01809
3.0150
330.11
1110.8
330.61
1194.5
0.51405
1.5700
160
363.54 0.01815
2.8347
335.49
1111.6
336.02
1195.5
0.52061
1.5647
170
368.41 0.01821
2.6749
340.62
1112.3
341.19
1196.4
0.52682
1.5596
180
373.07 0.01827
2.5322
345.53
1113.0
346.14
1197.3
0.53274
1.5548
190
377.52 0.01833
2.4040
350.24
1113.6
350.89
1198.1
0.53839
1.5503
200
381.80 0.01839
2.2882
354.78
1114.1
355.46
1198.8
0.54379
1.5460
250
400.97 0.01865
1.8440
375.23
1116.3
376.09
1201.6
0.56784
1.5270
300
417.35 0.01890
1.5435
392.89
1117.7
393.94
1203.3
0.58818
1.5111
350
431.74 0.01912
1.3263
408.55
1118.5
409.79
1204.4
0.60590
1.4973
400
444.62 0.01934
1.1617
422.70
1119.0
424.13
1205.0
0.62168
1.4852
450
456.31 0.01955
1.0324
435.67
1119.2
437.3
1205.2
0.63595
1.4742
500
467.04 0.01975
0.92819
447.68
1119.1
449.51
1205.0
0.64900
1.4642
550
476.97 0.01995
0.84228
458.90
1118.8
460.93
1204.5
0.66107
1.4550
600
486.24 0.02014
0.77020
469.46
1118.3
471.7
1203.9
0.67231
1.4463
700
503.13 0.02051
0.65589
488.96
1116.9
491.62
1201.9
0.69279
1.4305
800
518.27 0.02087
0.56920
506.74
1115.0
509.83
1199.3
0.71117
1.4162
900
532.02 0.02124
0.50107
523.19
1112.7
526.73
1196.2
0.72793
1.4030
1000 544.65 0.02159
0.44604
538.58
1110.1
542.57
1192.6
0.74341
1.3906
1200 567.26 0.02232
0.36241
566.89
1103.8
571.85
1184.2
0.77143
1.3677
1400 587.14 0.02307
0.30161
592.79
1096.3
598.76
1174.4
0.79658
1.3465
1600 604.93 0.02386
0.25516
616.99
1087.7
624.06
1163.2
0.81972
1.3262
1800 621.07 0.02470
0.21831
640.03
1077.9
648.26
1150.6
0.84144
1.3063
2000 635.85 0.02563
0.18815
662.33
1066.8
671.82
1136.4
0.86224
1.2863
2500 668.17 0.02860
0.13076
717.67
1031.2
730.90
1091.7
0.91311
1.2330
3000 695.41 0.03433
0.08460
783.39
969.8
802.45
1016.8
0.97321
1.1587
3200.1 705.10 0.04975
0.04975
866.61
866.6
896.07
896.1
1.05257
1.0526
T
∘

PAGE 460

















Saturated Water - Pressure (EEU)
Superheated Water (EEU)
T
∘
ν
u
h
( F) ( f t /lbm) (Bt u /lbm) (Bt u /lbm)
3
s
(Bt u /lbm /R)
p = 1.0 psia (Tsat = 101.69∘ F)
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 5.0 psia (Tsat = 162.18∘ F)
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 10 psia (Tsat = 193.16∘ F)
sat.
333.49
1043.7
1105.4
1.9776
73.525
1062.7
1130.7
1.8438
38.425
1072.0
1143.1
1.7875
200
329.53
1077.5
1150.1
2.0509
78.153
1076.2
1148.4
1.8716
38.849
1074.5
1146.4
1.7926
240
416.44
1091.2
1168.3
2.0777
83.009
1090.3
1167.1
1.8989
41.326
1089.1
1165.5
1.8207
280
440.33
1105.0
1186.5
2.1030
87.838
1104.3
1185.6
1.9246
43.774
1103.4
1184.4
1.8469
320
464.20
1118.9
1204.8
2.1271
92.650
1118.4
1204.1
1.9490
46.205
1117.6
1203.1
1.8716
360
488.07
1132.9
1223.3
2.1502
97.452
1132.5
1222.6
1.9722
48.624
1131.9
1221.8
1.8950
400
511.92
1147.1
1241.8
2.1722
102.25
1146.7
1241.3
1.9944
51.035
1146.2
1240.6
1.9174
440
535.77
1161.3
1260.4
2.1934
107.03
1160.9
1260.0
2.0156
53.441
1160.5
1259.4
1.9388
500
571.54
1182.8
1288.6
2.2237
114.21
1182.6
1288.2
2.0461
57.041
1182.2
1287.8
1.9693
600
631.14
1219.4
1336.2
2.2709
126.15
1219.2
1335.9
2.0933
63.029
1219.0
1335.6
2.0167
700
690.73
1256.8
1384.6
2.3146
138.09
1256.7
1384.4
2.1371
69.007
1256.5
1384.2
2.0605
800
750.31
1295.1
1433.9
2.3553
150.02
1294.9
1433.7
2.1778
74.980
1294.8
1433.5
2.1013
1000
869.47
1374.2
1535.1
2.4299
173.86
1374.2
1535.0
2.2524
86.913
1374.1
1534.9
2.1760
1200
988.62
1457.1
1640.0
2.4972
197.70
1457.0
1640.0
2.3198
98.840
1457.0
1639.9
2.2433
1400
1107.8
1543.7
1748.7
2.5590
221.54
1543.7
1748.7
2.3816
110.762
1543.6
1748.6
2.3052
p = 15.0 psia (Tsat = 212.99∘ F)
p = 20.0 psia (Tsat = 227.92∘ F)
p = 40 psia (Tsat = 267.22∘ F)
sat.
26.297
1077.7
1150.7
1.7549
20.093
1081.8
1156.2
1.7319
10.501
1092.1
1169.8
1.6766
240
27.429
1087.8
1163.9
1.7742
20.478
1086.5
1162.3
1.7406
38.849
1074.5
1146.4
1.7926
280
29.085
1102.4
1183.2
1.8010
21.739
1101.4
1181.9
1.7679
10.713
1097.3
1176.6
1.6858
320
30.722
1116.9
1202.2
1.8260
22.980
1116.1
1201.2
1.7933
11.363
1112.9
1197.1
1.7128
360
32.348
1131.3
1221.2
1.8496
24.209
1130.7
1220.2
1.8171
11.999
1128.1
1216.9
1.7376
400
33.965
1145.7
1239.9
1.8721
25.429
1145.1
1239.3
1.8398
12.625
1143.1
1236.5
1.7610
440
35.576
1160.1
1258.8
1.8936
26.644
1159.7
1258.3
1.8614
13.244
1157.9
1256.0
1.7831
500
37.986
1181.9
1287.8
1.9243
28.458
1181.6
1286.9
1.8922
14.165
1180.2
1285.0
1.8143
600
41.988
1218.7
1335.3
1.9718
31.467
1218.5
1334.9
1.9398
15.686
1217.5
1333.6
1.8625
700
45.981
1256.3
1383.9
2.0156
34.467
1256.1
1383.7
1.9837
17.197
1255.3
1382.6
1.9067
800
49.967
1294.6
1433.3
2.0565
37.461
1294.5
1433.1
2.0247
18.702
1293.9
1432.3
1.9478
1000
57.930
1374.0
1534.8
2.1312
43.438
1373.8
1534.6
2.0994
21.700
1373.4
1534.1
2.0227
1200
65.885
1456.9
1639.8
2.1986
49.407
1456.8
1639.7
2.1668
24.691
1456.5
1639.3
2.0902
1400
73.836
1543.6
1748.5
2.2604
55.373
1543.5
1748.4
2.2287
27.678
1543.3
1748.1
2.1522
1600
81.784
1634.0
1861.0
2.3178
61.335
1633.9
1860.9
2.2861
30.662
1633.7
1860.7
2.2096







PA G E 4 61

























SUPERHEATED WATER TABLE (EEU)
ν
T
∘
u
h
( F) ( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 60 psia (Tsat = 292.69∘ F)
ν
u
h
( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 80 psia (Tsat = 312.02∘ F)
sat.
7.1766
1098.1
1177.8
1.6442
5.4733
1102.3
1183.4
1.6212
320
7.4863
1109.6
1192.7
1.6636
5.544
1105.9
1187.9
1.6271
360
7.9259
1125.5
1213.5
1.6897
5.8876
1122.7
1209.9
400
8.3548
1140.9
1233.7
1.7138
6.2187
1138.7
440
8.7766
1156.1
1253.6
1.7364
6.542
500
9.4005
1178.8
1283.1
1.7682
600
10.4256
1216.5
1332.2
700
11.4401
1254.5
800
12.4484
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 100 psia (Tsat = 327.81∘ F)
4.4327
1105.5
1187.5
1.6032
1.6545
4.6628
1119.8
1206.1
1.6263
1230.8
1.6794
4.9359
1136.4
1227.8
1.6521
1154.3
1251.2
1.7026
5.2006
1152.4
1248.7
1.6759
7.0177
1177.3
1281.2
1.735
5.5876
1175.9
1279.3
1.7088
1.8168
7.7951
1215.4
1330.8
1.7841
6.2167
1214.4
1329.4
1.7586
1381.6
1.8613
8.5616.
1253.8
1380.5
1.8289
6.8344
1253
1379.5
1.8037
1293.3
1431.5
1.9026
9.3218.
1292.6
1430.6
1.8704
7.4457
1292
1429.8
1.8453
1000 14.4543
1373
1533.5
1.9777
10.8313
.
1372.6
1532.9
1.9457
8.6575
1372.2
1532.4
1.9208
1200 16.4525
1456.2
1638.9
2.0454
12.3331
.
1455.9
1638.5
2.0135
9.8615
1455.6
1638.1
1.9887
1400 18.4464
1543
1747.8
2.1073
13.8306
.
1542.8
1747.5
2.0755
11.0612
1542.6
1747.2
2.0508
1600
20.438
1633.5
1860.5
2.1648
15.3257
1633.3
1860.2
2.133
12.2584
1633.2
1860
2.1083
1800
22.428
1727.6
1976.6
2.2187
16.8192
1727.5
1976.5
2.1869
13.4541
1727.3
1976.3
2.1622
2000
24.417
1825.2
2096.3
2.2694
18.3117
1825
2096.1
2.2376
14.6487
1824.9
2096
2.213
p = 120 psia (Tsat = 341.25∘ F)
p = 140.0 psia (Tsat = 353.03∘ F)
sat.
3.7289
1107.9
1190.8
1.5883
3.2202
1109.9
1193.4
1.5757
360
3.8446
1116.7
1202.1
1.6023
3.2584
1113.4
1197.8
1.5811
400
4.0799
1134.0
1224.6
1.6292
3.4676
1131.5
1221.4
450
4.3613
1154.5
1251.4
1.6594
3.7147
1152.6
500
4.6340
1174.4
1277.3
1.6872
3.9525
550
4.9010
1193.9
1302.8
1.7131
600
5.1642
1213.4
1328.0
700
5.6829
1252.2
800
6.1950
1000
p = 160 psia (Tsat = 363.54∘ F)
2.8347
1111.6
1195.5
1.5647
1.6092
3.0076
1129.0
1218.0
1.5914
1248.9
1.6403
3.2293
1150.7
1246.3
1.6234
1172.9
1275.3
1.6686
3.4412
1171.4
1273.2
1.6522
4.1845
1192.7
1301.1
1.6948
3.6469
1191.4
1299.4
1.6788
1.7375
4.4124
1212.3
1326.6
1.7195
3.8484
1211.3
1325.2
1.7037
1378.4
1.7829
4.8604
1251.4
1377.3
1.7652
4.2434
1250.6
1376.3
1.7498
1291.4
1429.0
1.8247
5.3017
1290.8
1428.1
1.8072
4.6316
1290.2
1427.3
1.7920
7.2083
1371.7
1531.8
1.9005
6.1732
1371.3
1531.3
1.8832
5.3968
1370.9
1530.7
1.8682
1200
8.2137
1455.3
1637.7
1.9684
7.0367
1455.0
1637.3
1.9512
6.1540
1454.7
1636.9
1.9363
1400
9.2149
1542.3
1746.9
2.0305
7.8961
1542.1
1746.6
2.0134
6.9070
1541.8
1746.3
1.9986
1600 10.2135
1633.0
1859.8
2.0881
8.7529
1632.8
1859.5
2.0711
7.6574
1632.6
1859.3
2.0563
1800 11.2106
1727.2
1976.1
2.1420
9.6082
1727.0
1975.9
2.1250
8.4063
1726.9
1975.7
2.1102
2000 12.2067
1824.8
2095.8
2.1928
10.4624
1824.6
2095.7
2.1758
9.1542
1824.5
2095.5
2.1610







PAGE 462

























Superheated Water (EEU)
T
∘
ν
u
h
( F) ( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 180 psia (Tsat = 373.07∘ F)
ν
u
h
( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 200 psia (Tsat = 381.80∘ F)
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 225 psia (Tsat = 391.80∘ F)
sat.
2.5322
1113.0
1197.3
1.5548
2.2882
1114.1
1198.8
1.5460
2.0423
1115.3
1200.3
1.5360
400
2.6490
1126.3
1214.5
1.5752
2.3615
1123.5
1210.9
1.5602
2.0728
1119.7
1206.0
1.5427
450
2.8514
1148.7
1243.7
1.6082
2.5488
1146.7
1241.0
1.5943
2.2457
1144.1
1237.6
1.5783
500
3.0433
1169.8
1271.2
1.6376
2.7247
1168.2
1269.0
1.6243
2.4059
1166.2
1266.3
1.6091
550
3.2286
1190.2
1297.7
1.6646
2.8939
1188.9
1296.0
1.6516
2.5590
1187.2
1293.8
1.6370
600
3.4097
1210.2
1323.8
1.6897
3.0586
1209.1
1322.3
1.6771
2.7075
1207.7
1320.5
1.6628
700
3.7635
1249.8
1375.2
1.7361
3.3796
1249.0
1374.1
1.7238
2.9956
1248.0
1372.7
1.7099
800
4.1104
1289.5
1426.5
1.7785
3.6934
1288.9
1425.6
1.7664
3.2765
1288.1
1424.5
1.7528
900
4.4531
1329.7
1478.0
1.8179
4.0031
1329.2
1477.3
1.8059
3.5530
1328.5
1476.5
1.7925
1000
4.7929
1370.5
1530.1
1.8549
4.3099
1370.1
1529.6
1.8430
3.8268
1369.5
1528.9
1.8296
1200
5.4674
1454.3
1636.5
1.9231
4.9182
1454.0
1636.1
1.9113
4.3689
1453.6
1635.6
1.8981
1400
6.1377
1541.6
1746.0
1.9855
5.5222
1541.4
1745.7
1.9737
4.9068
1541.1
1745.4
1.9606
1600
6.8054
1632.4
1859.1
2.0432
6.1238
1632.2
1858.8
2.0315
5.4422
1632.0
1858.6
2.0184
1800
7.4716
1726.7
1975.6
2.0971
6.7238
1726.5
1975.4
2.0855
5.9760
1726.4
1975.2
2.0724
2000
8.1367
1824.4
2095.4
2.1479
7.3227
1824.3
2095.3
2.1363
6.5087
1824.1
2095.1
2.1232
p = 250 psia (Tsat = 400.97∘ F)
p = 275 psia (Tsat = 409.45∘ F)
p = 300 psia (Tsat = 417.35∘ F)
sat.
1.8440
1116.3
1201.6
1.5270
1.6806
1117.0
1202.6
1.5187
1.5435
1117.7
1203.3
1.5111
450
2.0027
1141.3
1234.0
1.5636
1.8034
1138.5
1230.3
1.5499
1.6369
1135.6
1226.4
1.5369
500
2.1506
1164.1
1263.6
1.5953
1.9415
1162.0
1260.8
1.5825
1.7670
1159.8
1257.9
1.5706
550
2.2910
1185.6
1291.5
1.6237
2.0715
1183.9
1289.3
1.6115
1.8885
1182.1
1287.0
1.6001
600
2.4264
1206.3
1318.6
1.6499
2.1964
1204.9
1316.7
1.6380
2.0046
1203.5
1314.8
1.6270
650
2.5586
1226.8
1345.1
1.6743
2.3179
1225.6
1343.5
1.6627
2.1172
1224.4
1341.9
1.6520
700
2.6883
1247.0
1371.4
1.6974
2.4369
1246.0
1370.0
1.6860
2.2273
1244.9
1368.6
1.6755
800
2.9429
1287.3
1423.5
1.7406
2.6699
1286.5
1422.4
1.7294
2.4424
1285.7
1421.3
1.7192
900
3.1930
1327.9
1475.6
1.7804
2.8984
1327.3
1474.8
1.7694
2.6529
1326.6
1473.9
1.7593
1000
3.4403
1369.0
1528.2
1.8177
3.1241
1368.5
1527.4
1.8068
2.8605
1367.9
1526.7
1.7968
1200
3.9295
1453.3
1635.0
1.8863
3.5700
1452.9
1634.5
1.8755
3.2704
1452.5
1634.0
1.8657
1400
4.4144
1540.8
1745.0
1.9488
4.0116
1540.5
1744.6
1.9381
3.6759
1540.2
1744.2
1.9284
1600
4.8969
1631.7
1858.3
2.0066
4.4507
1631.5
1858.0
1.9960
4.0789
1631.3
1857.7
1.9863
1800
5.3777
1726.2
1974.9
2.0607
4.8882
1726.0
1974.7
2.0501
4.4803
1725.8
1974.5
2.0404
2000
5.8575
1823.9
2094.9
2.1116
5.3247
1823.8
2094.7
2.1010
4.8807
1823.6
2094.6
2.0913







PAGE 463

























Superheated Water (EEU)
T
∘
ν
u
h
( F) ( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 350 psia (Tsat = 431.74∘ F)
ν
u
h
( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 400 psia (Tsat = 442.62∘ F)
sat.
1.3263
1118.5
1204.4
1.4973
1.1617
1119
1205
1.4852
450
1.3739
1129.3
1218.3
1.5128
1.1747
1122.5
1209.4
1.4901
500
1.4921
1155.2
1251.9
1.5487
1.2851
1150.4
1245.6
550
1.6004
1178.6
1282.2
1.5795
1.384
1174.9
600
1.7030
1200.6
1310.9
1.6073
1.4765
650
1.8018
1221.9
1338.6
1.6328
700
1.8979
1242.8
1365.8
800
2.0848
1284.1
900
2.2671
1000
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 450 psia (Tsat = 456.31∘ F)
1.0324
1119.2
1205.2
1.4742
1.5288
1.1233
1145.4
1238.9
1.5103
1277.3
1.561
1.2152
1171.1
1272.3
1.5441
1197.6
1306.9
1.5897
1.3001
1194.6
1302.8
1.5737
1.565
1219.4
1335.3
1.6158
1.3807
1216.9
1331.9
1.6005
1.6567
1.6507
1240.7
1362.9
1.6401
1.4584
1238.5
1360.0
1.6253
1419.1
1.7009
1.8166
1282.5
1417
1.6849
1.6080
1280.8
1414.7
1.6706
1325.3
1472.2
1.7414
1.9777
1324
1470.4
1.7257
1.7526
1322.7
1468.6
1.7117
2.4464
1366.9
1525.3
1.7791
2.1358
1365.8
1523.9
1.7636
1.8942
1364.7
1522.4
1.7499
1200
2.7996
1451.7
1633.0
1.8483
2.4465
1450.9
1632
1.8331
2.1718
1450.1
1631.0
1.8196
1400
3.1484
1539.6
1743.5
1.9111
2.7527
1539
1742.7
1.896
2.4450
1538.4
1742.0
1.8827
1600
3.4947
1630.8
1857.1
1.9691
3.0565
1630.3
1856.5
1.9541
2.7157
1629.8
1856.0
1.9409
1800
3.8394
1725.4
1974.0
2.0233
3.3586
1725
1973.6
2.0084
2.9847
1724.6
1973.2
1.9952
2000
4.1830
1823.3
2094.2
2.0742
3.6597
1823
2093.9
2.0594
3.2527
1822.6
2093.5
2.0462
p = 500 psia (Tsat = 467.04∘ F)
p = 600 psia (Tsat = 486.24∘ F)
sat.
0.92815
1119.1
1205.0
1.4642
0.7702
1118.3
1203.9
1.4463
500
0.99304
1140.1
1231.9
1.4928
0.79526
1128.2
1216.5
1.4596
550
1.07974
1167.1
1267.0
1.5284
0.87542
1158.7
1255.9
600
1.15876
1191.4
1298.6
1.5590
0.94605
1184.9
650
1.23312
1214.3
1328.4
1.5865
1.01133
700
1.30440
1236.4
1357.0
1.6117
800
1.44097
1279.2
1412.5
900
1.57252
1321.4
1000 1.70094
p = 700 psia (Tsat = 503.13∘ F)
0.65589
1116.9
1201.9
1.4305
1.4996
0.72799
1149.5
1243.8
1.4730
1289.9
1.5325
0.79332
1177.9
1280.7
1.5087
1209
1321.3
1.5614
0.85242
1203.4
1313.8
1.5393
1.07316
1231.9
1351
1.5877
0.90769
1227.2
1344.8
1.5666
1.6576
1.19038
1275.8
1408
1.6348
1.01125
1272.4
1403.4
1.6150
1466.9
1.6992
1.3023
1318.7
1463.3
1.6771
1.10921
1316.0
1459.7
1.6581
1363.6
1521.0
1.7376
1.41097
1361.4
1518.1
1.716
1.20381
1359.2
1515.2
1.6974
1100 1.82726
1406.2
1575.3
1.7735
1.51749
1404.4
1572.9
1.7522
1.29621
1402.5
1570.4
1.7341
1200 1.95211
1449.4
1630.0
1.8075
1.62252
1447.8
1627.9
1.7865
1.38709
1446.2
1625.9
1.7685
1400
2.1988
1537.8
1741.2
1.8708
1.82957
1536.6
1739.7
1.8501
1.56580
1535.4
1738.2
1.8324
1600
2.4430
1629.4
1855.4
1.9291
2.034
1628.4
1854.2
1.9085
1.74192
1627.5
1853.1
1.8911
1800
2.6856
1724.2
1972.7
1.9834
2.2369
1723.4
1971.8
1.963
1.91643
1722.7
1970.9
1.9457
2000
2.9271
1822.3
2093.1
2.0345
2.4387
1821.7
2092.4
2.0141
2.08987
1821.0
2091.7
1.9969







PAGE 46 4

























Superheated Water (EEU)
ν
T
∘
u
h
( F) ( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 800 psia (Tsat = 518.27∘ F)
ν
u
h
( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 1000 psia (Tsat = 544.65∘ F)
sat.
0.56920
1115.0
1199.3
1.4162
0.44604
1110.1
1192.6
1.3906
550
0.61586
1139.4
1230.5
1.4476
0.45375
1115.2
1199.2
1.3972
600
0.67799
1170.5
1270.9
1.4866
0.51431
1154.1
1249.3
650
0.73279
1197.6
1306.0
1.5191
0.56411
1185.1
700
0.78330
1222.4
1338.4
1.5476
0.60844
750
0.83102
1246.0
1369.1
1.5735
800
0.87678
1268.9
1398.7
900
0.96434
1313.3
1000 1.04841
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 1250 psia (Tsat = 572.45∘ F)
0.34549
1102.0
1181.9
1.3623
1.4457
0.37894
1129.5
1217.2
1.3961
1289.5
1.4827
0.42703
1167.5
1266.3
1.4414
1212.4
1325
1.514
0.46735
1198.7
1306.8
1.4771
0.64944
1237.6
1357.8
1.5418
0.50344
1226.4
1342.9
1.5076
1.5975
0.68821
1261.7
1389
1.567
0.53687
1252.2
1376.4
1.5347
1456.0
1.6413
0.76136
1307.7
1448.6
1.6126
0.59876
1300.5
1439.0
1.5826
1357.0
1512.2
1.6812
0.83078
1352.5
1506.2
1.6535
0.65656
1346.7
1498.6
1.6249
1100 1.13024
1400.7
1568.0
1.7181
0.89783
1396.9
1563.1
1.6911
0.71184
1392.2
1556.8
1.6635
1200 1.21051
1444.6
1623.8
1.7528
0.96327
1441.4
1619.7
1.7263
0.76545
1437.4
1614.5
1.6993
1400 1.36797
1534.2
1736.7
1.8170
1.09101
1531.8
1733.7
1.7911
0.86944
1528.7
1729.8
1.7649
1600 1.52283
1626.5
1851.9
1.8759
1.2161
1624.6
1849.6
1.8504
0.97072
1622.2
1846.7
1.8246
1800 1.67606
1721.9
1970.0
1.9306
1.33956
1720.3
1968.2
1.9053
1.07036
1718.4
1966.0
1.8799
2000 1.82823
1820.4
2091.0
1.9819
1.46194
1819.1
2089.6
1.9568
1.16892
1817.5
2087.9
1.9315
p = 1500 psia (Tsat = 596.26∘ F)
sat.
0.27695
1092.1
1169.0
1.3362
600
0.28189
1097.2
1175.4
1.3423
650
0.33310
1147.2
1239.7
700
0.37198
1183.6
750
0.40535
800
p = 1750 psia (Tsat = 617.17∘ F)
p = 2000 psia (Tsat = 635.85∘ F)
0.22681
1080.5
1153.9
1.3112
0.18815
1066.8
1136.4
1.2863
1.4016
0.26292
1122.8
1207.9
1.3607
0.20586
1091.4
1167.6
1.3146
1286.9
1.4433
0.30252
1166.8
1264.7
1.4108
0.24894
1147.6
1239.8
1.3783
1214.4
1326.9
1.4771
0.33455
1201.5
1309.8
1.4489
0.28074
1187.4
1291.3
1.4218
0.43550
1242.2
1363.1
1.5064
0.36266
1231.7
1349.1
1.4807
0.30763
1220.5
1334.3
1.4567
850
0.46356
1268.2
1396.9
1.5328
0.38835
1259.3
1385.1
1.5088
0.33169
1250
1372.8
1.4867
900
0.49015
1293.1
1429.2
1.5569
0.41238
1285.4
1419
1.5341
0.3539
1277.5
1408.5
1.5134
1000 0.54031
1340.9
1490.8
1.6007
0.45719
1334.9
1482.9
1.5796
0.39479
1328.7
1474.9
1.5606
1100 0.58781
1387.3
1550.5
1.6402
0.49917
1382.4
1544.1
1.6201
0.43266
1377.5
1537.6
1.6021
1200 0.63355
1433.3
1609.2
1.6767
0.53932
1429.2
1603.9
1.6572
0.46864
1425.1
1598.5
1.64
1400 0.72172
1525.7
1726.0
1.7432
0.61621
1522.6
1722.1
1.7245
0.53708
1519.5
1718.3
1.7081
1600 0.80714
1619.8
1843.8
1.8033
0.69031
1617.4
1840.9
1.7852
0.60269
1615
1838
1.7693
1800 0.89090
1716.4
1963.7
1.8589
0.76273
1714.5
1961.5
1.841
0.6666
1712.5
1959.2
1.8255
2000 0.97358
1815.9
2086.1
1.9108
0.83406
1814.2
2084.3
1.8931
0.72942
1812.6
2082.6
1.8778







PAGE 465

























Superheated Water (EEU)
ν
T
∘
u
h
( F) ( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 2500 psia (Tsat = 668.17∘ F)
sat.
0.13076
1031.2
1091.7
1.2330
ν
u
h
( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
3
ν
0.0846
969.8
1016.8
1.1587
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
p = 3000 psia (Tsat = 695.41∘ F)
650
s
(Bt u /lbm /R)
p = 3500 psia
0.34549
1102.0
1181.9
1.3623
0.02492
663.7
679.9
0.8632
700
0.16849
1098.4
1176.3
1.3072
0.09838
1005.3
1059.9
1.196
0.03065
760
779.9
0.9511
750
0.20327
1154.9
1249.0
1.3686
0.1484
1114.1
1196.5
1.3118
0.1046
1057.6
1125.4
1.2434
800
0.22949
1195.9
1302.0
1.4116
0.17601
1167.5
1265.3
1.3676
0.13639
1134.3
1222.6
1.3224
850
0.25174
1230.1
1346.6
1.4463
0.19771
1208.2
1317.9
1.4086
0.15847
1183.8
1286.5
1.3721
900
0.27165
1260.7
1386.4
1.4761
0.2164
1242.8
1362.9
1.4423
0.17659
1223.4
1337.8
1.4106
950
0.29001
1289.1
1423.3
1.5028
0.23321
1273.9
1403.3
1.4716
0.19245
1257.8
1382.4
1.4428
1000 0.30726
1316.1
1458.2
1.5271
0.24876
1302.8
1440.9
1.4978
0.20687
1289
1423
1.4711
1100 0.33949
1367.3
1524.4
1.5710
0.27732
1356.8
1510.8
1.5441
0.23289
1346.1
1496.9
1.5201
1200 0.36966
1416.6
1587.6
1.6103
0.30367
1408
1576.6
1.585
0.25654
1399.3
1565.4
1.5627
1400 0.42631
1513.3
1710.5
1.6802
0.35249
1507
1702.7
1.6567
0.29978
1500.7
1694.8
1.6364
1600 0.48004
1610.1
1832.2
1.7424
0.3983
1605.3
1826.4
1.7199
0.33994
1600.4
1820.5
1.7006
1800 0.53205
1708.6
1954.8
1.7991
0.44237
1704.7
1950.3
1.7773
0.37833
1700.8
1945.8
1.7586
2000 0.58295
1809.4
2079.1
1.8518
0.48532
1806.1
2075.6
1.8304
0.41561
1802.9
2072.1
1.8121
p = 4000 psia
p = 5000 psia
p = 6000 psia
650
0.02448
657.9
676.1
0.8577
0.02379
648.3
670.3
0.8485
0.02325
640.3
666.1
0.8408
700
0.02871
742.3
763.6
0.9347
0.02678
721.8
746.6
0.9156
0.02564
708.1
736.5
0.9028
750
0.06370
962.1
1009.2
1.1410
0.03373
821.8
853
1.0054
0.02981
788.7
821.8
0.9747
800
0.10520
1094.2
1172.1
1.2734
0.05937
986.9
1041.8
1.1581
0.03949
897.1
941
1.0711
850
0.12848
1156.7
1251.8
1.3355
0.08551
1092.4
1171.5
1.2593
0.05815
1018.6
1083.1
1.1819
900
0.14647
1202.5
1310.9
1.3799
0.1039
1155.9
1252.1
1.3198
0.07584
1103.5
1187.7
1.2603
950
0.16176
1240.7
1360.5
1.4157
0.11863
1203.9
1313.6
1.3643
0.0901
1163.7
1263.7
1.3153
1000 0.17538
1274.6
1404.4
1.4463
0.13128
1244
1365.5
1.4004
0.10208
1211.4
1324.7
1.3578
1100 0.19957
1335.1
1482.8
1.4983
0.15298
1312.2
1453.8
1.459
0.12211
1288.4
1424
1.4237
1200 0.22121
1390.3
1554.1
1.5426
0.17185
1372.1
1531.1
1.507
0.13911
1353.4
1507.8
1.4758
1300 0.24128
1443.0
1621.6
1.5821
0.18902
1427.8
1602.7
1.549
0.15434
1412.5
1583.8
1.5203
1400 0.26028
1494.3
1687.0
1.6182
0.20508
1481.4
1671.1
1.5868
0.16841
1468.4
1655.4
1.5598
1600 0.29620
1595.5
1814.7
1.6835
0.23505
1585.6
1803.1
1.6542
0.19438
1575.7
1791.5
1.6294
1800 0.33033
1696.8
1941.4
1.7422
0.2632
1689
1932.5
1.7142
0.21853
1681.1
1923.7
1.6907
2000 0.36335
1799.7
2068.6
1.7961
0.29023
1793.2
2061.7
1.7689
0.24155
1786.7
2054.9
1.7463



PAGE 466

























Superheated Water (EEU)
Superheated Water (EEU) - Note: Use this table only for Nuclear Power System problems!
T
∘
ν
u
h
( F) ( f t /lbm) (Bt u /lbm) (Bt u /lbm)
3
s
(Bt u /lbm /R)
p = 500 psia (Tsat = 467.04∘ F)
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 1000 psia (Tsat = 544.65∘ F)
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 1500 psia (Tsat = 592.26∘ F)
sat. 0.019750 447.68
449.51
0.64900 0.021595
538.58
542.57
0.74341
0.023456
605.07
611.58
0.80836
32
0.015994
0.01
1.49
0.00001 0.015966
0.03
2.99
0.00005
0.015939
0.05
4.48
0.00008
50
0.015998
18.03
19.51
0.03601 0.015972
17.99
20.95
0.03593
0.015946
17.95
22.38
0.03584
100 0.016107
67.86
69.35
0.12930 0.016083
67.69
70.67
0.12899
0.016059
67.53
71.98
0.12869
150 0.016317 117.70
119.21
0.21462 0.016292
117.42
120.43
0.21416
0.016267
117.14
121.66
0.21369
200 0.016607 167.70
169.24
0.29349 0.016580
167.31
170.38
0.29289
0.016553
166.92
171.52
0.29229
250 0.016972 218.04
219.61
0.36708 0.016941
217.51
220.65
0.36634
0.016911
217.00
221.69
0.36560
300 0.017417 268.92
270.53
0.43641 0.017380
268.24
271.46
0.43551
0.017345
267.57
272.39
0.43463
350 0.017954 320.64
322.30
0.50240 0.017910
319.77
323.08
0.50132
0.017866
318.91
323.87
0.50025
400 0.018609 373.61
375.33
0.56595 0.018552
372.48
375.91
0.56463
0.018496
371.37
376.51
0.56333
450 0.019425 428.44
430.24
0.62802 0.019347
426.93
430.51
0.62635
0.019271
425.47
430.82
0.62472
500
0.42631
1513.3
1710.5
1.6802
0.020368
484.03
487.80
0.68764
0.020258
482.01
487.63
0.68550
550
0.58295
1809.4
2079.1
1.8518
0.48532
1806.1
2075.6
1.8304
0.021595
542.50
548.50
0.74731
p = 2000 psia (Tsat = 635.85∘ F)
p = 3000 psia (Tsat = 695.41∘ F)
p = 5000 psia
Sat. 0.025634 662.33
671.82
0.86224 0.034335
783.39
802.45
0.97321
0.02325
640.3
666.1
0.8408
32
0.015912
0.07
5.96
0.00010 0.015859
0.10
8.90
0.00011
0.015756
0.13
14.71
0.00002
50
0.015921
17.91
23.80
0.03574 0.015870
17.83
26.64
0.03554
0.015773
17.65
32.25
0.03505
100 0.016035
67.36
73.30
0.12838 0.015988
67.04
75.91
0.12776
0.015897
66.41
81.12
0.12652
200 0.016527 166.54
172.66
0.29170 0.016475
165.79
174.94
0.29053
0.016375
164.36
179.51
0.28824
300 0.017310 266.92
273.33
0.43376 0.017242
265.65
275.22
0.43204
0.017112
263.24
279.07
0.42874
400 0.018442 370.30
377.12
0.56205 0.018338
368.22
378.41
0.55959
0.018145
364.35
381.14
0.55492
450 0.019199 424.06
431.16
0.62314 0.019062
421.36
431.94
0.62010
0.018812
416.40
433.80
0.61445
500 0.020154 480.08
487.54
0.68346 0.019960
476.45
487.53
0.67958
0.019620
469.94
488.10
0.67254
560 0.021739 552.21
560.26
0.75692 0.021405
546.59
558.47
0.75126
0.020862
537.08
556.38
0.74154
600 0.023317 605.77
614.40
0.80898 0.022759
597.42
610.06
0.80086
0.021943
584.42
604.72
0.78803
640
0.26028
1494.3
1687.0
1.6182
0.024765
654.52
668.27
0.85476
0.023358
634.95
656.56
0.83603
680
0.29620
1595.5
1814.7
1.6835
0.028821
728.63
744.64
0.92288
0.025366
690.67
714.14
0.88745
700
0.36335
1799.7
2068.6
1.7961
0.29023
1793.2
2061.7
1.7689
0.026777
721.78
746.56
0.91564






PAGE 467

























C OMPRESSED LIQUID WATER TABLE (EEU)
Saturated Refrigerant R-134a - Temperature (EEU)
T
(∘F)
p
νf
νg
uf
ug
hf
hg
sf
sg
(Psia) ( f t 3 /lbm) ( f t 3 /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm /R) (Btu /lbm /R)
-40
7.432
0.01130
5.7796
-0.016
89.15
0
97.10
0.00000
0.23135
-35
8.581
0.01136
5.0509
1.484
89.84
1.502
97.86
0.00355
0.23043
-30
9.869
0.01143
4.4300
2.990
90.52
3.011
98.61
0.00708
0.22956
-25
11.306
0.01150
3.8988
4.502
91.21
4.526
99.36
0.01058
0.22875
-20
12.906
0.01156
3.4426
6.019
91.89
6.047
100.12
0.01405
0.22798
-15
14.680
0.01163
3.0494
7.543
92.58
7.574
100.86
0.01749
0.22727
-10
16.642
0.01171
2.7091
9.073
93.26
9.109
101.61
0.02092
0.2266
-5
18.806
0.01178
2.4137
10.609
93.95
10.65
102.35
0.02431
0.22598
0
21.185
0.01185
2.1564
12.152
94.63
12.199
103.08
0.02769
0.22539
5
23.793
0.01193
1.9316
13.702
95.31
13.755
103.82
0.03104
0.22485
10
26.646
0.01201
1.7345
15.259
95.99
15.318
104.54
0.03438
0.22434
15
29.759
0.01209
1.5612
16.823
96.67
16.889
105.27
0.03769
0.22386
20
33.147
0.01217
1.4084
18.394
97.34
18.469
105.98
0.04098
0.22341
25
36.826
0.01225
1.2732
19.973
98.02
20.056
106.69
0.04426
0.223
30
40.813
0.01234
1.1534
21.560
98.68
21.653
107.40
0.04752
0.2226
35
45.124
0.01242
1.0470
23.154
99.35
23.258
108.09
0.05076
0.22224
40
49.776
0.01251
0.95205
24.757
100.01
24.873
108.78
0.05398
0.22189
45
54.787
0.01261
0.86727
26.369
100.67
26.497
109.46
0.05720
0.22157
50
60.175
0.01270
0.79136
27.990
101.32
28.131
110.13
0.06039
0.22127
55
65.957
0.01280
0.72323
29.619
101.97
29.775
110.79
0.06358
0.22098
60
72.152
0.01290
0.66195
31.258
102.61
31.431
111.44
0.06675
0.2207
65
78.780
0.01301
0.60671
32.908
103.24
33.097
112.09
0.06991
0.22044
70
85.858
0.01312
0.55681
34.567
103.87
34.776
112.71
0.07306
0.22019
75
93.408
0.01323
0.51165
36.237
104.49
36.466
113.33
0.07620
0.21995
80
101.45
0.01334
0.47069
37.919
105.1
38.169
113.94
0.07934
0.21972
85
110.00
0.01347
0.43348
39.612
105.7
39.886
114.53
0.08246
0.21949
90
119.08
0.01359
0.39959
41.317
106.3
41.617
115.10
0.08559
0.21926
95
128.72
0.01372
0.36869
43.036
106.88
43.363
115.66
0.08870
0.21904

PAGE 468

















S AT U R AT E D R - 13 4 A - T E M P E R AT U R E TA B L E ( E E U )
∘
( F)
p
νf
νg
uf
ug
hf
hg
sf
sg
(Psia) ( f t 3 /lbm) ( f t 3 /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm /R) (Btu /lbm /R)
100
138.93
0.01386
0.34045
44.768
107.45
45.124
116.20
0.09182
0.21881
105
149.73
0.01400
0.31460
46.514
108.01
46.902
116.73
0.09493
0.21858
110
161.16
0.01415
0.29090
48.276
108.56
48.698
117.23
0.09804
0.21834
115
173.23
0.01430
0.26913
50.054
109.08
50.512
117.71
0.10116
0.21809
120
185.96
0.01446
0.24909
51.849
109.6
52.346
118.17
0.10428
0.21782
130
213.53
0.01482
0.21356
55.495
110.57
56.08
119.00
0.11054
0.21724
140
244.06
0.01521
0.18315
59.226
111.44
59.913
119.71
0.11684
0.21655
150
277.79
0.01567
0.15692
63.059
112.2
63.864
120.27
0.12321
0.21572
160
314.94
0.01619
0.13410
67.014
112.81
67.958
120.63
0.12970
0.21469
170
355.80
0.01681
0.11405
71.126
113.22
72.233
120.73
0.13634
0.21335
180
400.66
0.01759
0.09618
75.448
113.35
76.752
120.48
0.14323
0.21158
190
449.90
0.01860
0.07990
80.082
113.03
81.631
119.68
0.15055
0.20911
200
504.00
0.02009
0.06441
85.267
111.92
87.14
117.93
0.15867
0.20533
210
563.76
0.02309
0.04722
91.986
108.48
94.395
113.41
0.16922
0.19761
T

PAGE 469

















Saturated Refrigerant R-134a - Temperature (EEU)
Saturated Refrigerant R-134a - Pressure (EEU)
p
(Psia)
νf
νg
uf
ug
hf
hg
sf
sg
(∘F) ( f t 3 /lbm) ( f t 3 /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm /R) (Btu /lbm /R)
T
5
-53.09 0.01113
8.3785
-3.918
87.36
-3.907
95.11
-0.00945
0.23408
10
-29.52 0.01144
4.3753
3.135
90.59
3.156
98.68
0.00742
0.22948
15
-14.15 0.01165
2.9880
7.803
92.70
7.835
100.99
0.01808
0.22715
20
-2.43
0.01182
2.2772
11.401
94.30
11.445
102.73
0.02605
0.22567
25
7.17
0.01196
1.8429
14.377
95.61
14.432
104.13
0.03249
0.22462
30
15.37
0.01209
1.5492
16.939
96.72
17.006
105.32
0.03793
0.22383
35
22.57
0.01221
1.3369
19.205
97.69
19.284
106.35
0.04267
0.22319
40
29.01
0.01232
1.1760
21.246
98.55
21.337
107.26
0.04688
0.22268
45
34.86
0.01242
1.0497
23.110
99.33
23.214
108.07
0.05067
0.22225
50
40.23
0.01252
0.94791
24.832
100.04
24.948
108.81
0.05413
0.22188
55
45.20
0.01261
0.86400
26.435
100.69
26.564
109.49
0.05733
0.22156
60
49.84
0.01270
0.79361
27.939
101.30
28.08
110.11
0.06029
0.22127
65
54.20
0.01279
0.73370
29.357
101.86
29.51
110.69
0.06307
0.22102
70
58.30
0.01287
0.68205
30.700
102.39
30.867
111.22
0.06567
0.2208
75
62.19
0.01295
0.63706
31.979
102.88
32.159
111.73
0.06813
0.22059
80
65.89
0.01303
0.59750
33.201
103.35
33.394
112.20
0.07047
0.2204
85
69.41
0.01310
0.56244
34.371
103.79
34.577
112.64
0.07269
0.22022
90
72.78
0.01318
0.53113
35.495
104.21
35.715
113.06
0.07481
0.22006
95
76.02
0.01325
0.50301
36.578
104.61
36.811
113.46
0.07684
0.21991
100
79.12
0.01332
0.47760
37.623
104.99
37.869
113.83
0.07879
0.21976
110
85.00
0.01347
0.43347
39.612
105.70
39.886
114.53
0.08246
0.21949
120
90.49
0.01360
0.39644
41.485
106.35
41.787
115.16
0.08589
0.21924
130
95.64
0.01374
0.36491
43.258
106.95
43.589
115.73
0.08911
0.21901
140
100.51 0.01387
0.33771
44.945
107.51
45.304
116.26
0.09214
0.21879
150
105.12 0.01400
0.31401
46.556
108.02
46.945
116.74
0.09501
0.21857
160
109.50 0.01413
0.29316
48.101
108.50
48.519
117.18
0.09774
0.21836
170
113.69 0.01426
0.27466
49.586
108.95
50.035
117.59
0.10034
0.21815
180
117.69 0.01439
0.25813
51.018
109.36
51.497
117.96
0.10284
0.21795

PAGE 470

















S AT U R AT E D R - 13 4 A - P R E S S U R E TA B L E ( E E U )
p
(Psia)
νf
νg
uf
ug
hf
hg
sf
sg
( F) ( f t 3 /lbm) ( f t 3 /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm) (Btu /lbm /R) (Btu /lbm /R)
190
121.53 0.01452
0.24327
52.402
109.75
52.912
118.30
0.10524
0.21774
200
125.22 0.01464
0.22983
53.743
110.11
54.285
118.62
0.10754
0.21753
220
132.21 0.01490
0.20645
56.310
110.77
56.917
119.17
0.11192
0.2171
240
138.73 0.01516
0.18677
58.746
111.34
59.419
119.63
0.11603
0.21665
260
144.85 0.01543
0.16996
61.071
111.83
61.813
120.00
0.11992
0.21617
280
150.62 0.01570
0.15541
63.301
112.25
64.115
120.30
0.12362
0.21567
300
156.09 0.01598
0.14266
65.452
112.60
66.339
120.52
0.12715
0.21512
350
168.64 0.01672
0.11664
70.554
113.18
71.638
120.74
0.13542
0.21356
400
179.86 0.01757
0.09642
75.385
113.35
76.686
120.48
0.14314
0.21161
450
190.02 0.01860
0.07987
80.092
113.03
81.641
119.68
0.15056
0.20911
500
199.29 0.01995
0.06551
84.871
112.04
86.718
118.10
0.15805
0.20566
T
∘

PAGE 471

















Saturated Refrigerant R-134a - Pressure (EEU)
Superheated R-134a (EEU)
T
∘
ν
u
h
( F) ( f t /lbm) (Bt u /lbm) (Bt u /lbm)
3
s
(Btu/lbm/R)
p = 10.0 psia (Tsat = − 29.52∘ F)
sat.
4.3753
90.59
98.68
0.22948
-20
4.4856
92.13
100.43
0.2335
0
4.7135
95.41
104.14
20
4.938
98.77
40
5.16
60
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Btu/lbm/R)
p = 15.0 psia (Tsat = − 14.15∘ F)
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Btu/lbm/R)
p = 20.0 psia (Tsat = − 2.43∘ F)
2.988
92.7
100.99
0.22715
2.2772
94.3
102.73
0.22567
0.24174
3.1001
95.08
103.68
0.2331
2.2922
94.72
103.2
0.22671
107.91
0.24976
3.2551
98.48
107.52
0.24127
2.413
98.19
107.12
0.23504
1u02.2
111.75
0.25761
3.4074
101.95
111.41
0.24922
2.5306
101.7
111.07
0.24311
5.3802
105.72
115.67
0.26531
3.5577
105.5
115.38
0.257
2.6461
105.28
115.07
0.25097
80
5.5989
109.32
119.68
0.27288
3.7064
109.13
119.42
0.26463
2.76
108.93
119.15
0.25866
100
5.8165
113.01
123.78
0.28033
3.854
112.84
123.54
0.27212
2.8726
112.66
123.29
0.26621
120
6.0331
116.79
127.96
0.28767
4.0006
116.63
127.74
0.2795
2.9842
116.47
127.52
0.27363
140
6.249
120.66
132.22
0.2949
4.1464
120.51
132.02
0.28677
3.095
120.37
131.82
0.28093
160
6.4642
124.61
136.57
0.30203
4.2915
124.48
136.39
0.29393
3.2051
124.35
136.21
0.28812
180
6.6789
128.65
141.01
0.30908
4.4361
128.53
140.84
0.301
3.3146
128.41
140.67
0.29521
200
6.893
132.77
145.53
0.31604
4.5802
132.66
145.37
0.30798
3.4237
132.55
145.22
0.30221
220
7.1068
136.98
150.13
0.32292
4.7239
136.88
149.99
0.31487
3.5324
136.78
149.85
0.30912
p = 30.0 psia (Tsat = 15.37∘ F)
sat.
1.5492
96.72
105.32
0.22383
20
1.5691
97.56
106.27
0.22581
40
1.6528
101.17
110.35
60
1.7338
104.82
80
1.813
100
p = 40.0 psia (Tsat = 29.01∘ F)
1.176
98.55
107.26
0.22268
0.23414
1.2126
100.61
109.58
0.22738
114.45
0.24219
1.2768
104.34
113.79
108.53
118.59
0.25002
1.3389
108.11
1.8908
112.3
122.8
0.25767
1.3995
120
1.9675
116.15
127.07
0.26517
140
2.0434
120.08
131.42
160
2.1185
124.08
180
2.1931
200
p = 50.0 psia (Tsat = 40.23∘ F)
0.9479
100.04
108.81
0.22188
0.23565
1.0019
103.84
113.11
0.23031
118.02
0.24363
1.054
107.68
117.43
0.23847
111.93
122.29
0.2514
1.1043
111.55
121.77
0.24637
1.4588
115.82
126.62
0.259
1.1534
115.48
126.16
0.25406
0.27254
1.5173
119.78
131.01
0.26644
1.2015
119.47
130.59
0.26159
135.84
0.27979
1.575
123.81
135.47
0.27375
1.2488
123.53
135.09
0.26896
128.16
140.34
0.28693
1.6321
127.91
140
0.28095
1.2955
127.66
139.65
0.27621
2.2671
132.32
144.91
0.29398
1.6887
132.1
144.6
0.28803
1.3416
131.87
144.28
0.28333
220
2.3408
136.57
149.56
0.30092
1.7449
136.36
149.27
0.29501
1.3873
136.15
148.98
0.29036
240
2.4141
140.89
154.29
0.30778
1.8007
140.7
154.03
0.3019
1.4326
140.5
153.76
0.29728
260
2.4871
145.3
159.1
0.31456
1.8562
145.12
158.86
0.30871
1.4776
144.93
158.6
0.30411
280
2.5598
149.78
163.99
0.32126
1.9114
149.61
163.76
0.31543
1.5223
149.44
163.53
0.31086







PAGE 472






















S U P E R H E AT E D R - 13 4 A TA B L E S ( E E U )
T
∘
ν
u
h
( F) ( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 60.0 psia (Tsat = 49.84∘ F)
ν
u
h
( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 70.0 psia (Tsat = 58.30∘ F)
sat.
0.7946
101.31
110.13
0.22132
0.6829
102.40
111.25
0.22084
60
0.8179
103.31
112.39
0.22572
0.6857
102.74
111.62
0.22157
80
0.8636
107.24
116.82
0.23408
0.7271
106.77
116.18
100
0.9072
111.17
121.24
0.24212
0.7662
110.77
120
0.9495
115.14
125.69
0.24992
0.8037
140
0.9908
119.17
130.17
0.25753
160
1.0312
123.26
134.71
180
1.0709
127.42
200
1.1101
220
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 80.0 psia (Tsat = 65.89∘ F)
0.5982
103.36
112.22
0.22045
0.23018
0.6243
106.27
115.51
0.22663
120.69
0.23838
0.6601
110.35
120.12
0.23501
114.79
125.20
0.24630
0.6941
114.43
124.70
0.24305
0.8401
118.86
129.74
0.25399
0.7270
118.53
129.29
0.25084
0.26497
0.8756
122.98
134.32
0.26151
0.7589
122.69
133.92
0.25843
139.31
0.27227
0.9105
127.16
138.95
0.26886
0.7900
126.89
138.59
0.26585
131.64
143.97
0.27945
0.9447
131.40
143.64
0.27608
0.8206
131.17
143.31
0.27312
1.1489
135.94
148.69
0.28651
0.9785
135.72
148.40
0.28318
0.8507
135.50
148.09
0.28026
240
1.1872
140.31
153.49
0.29346
1.0118
140.11
153.22
0.29017
0.8803
139.91
152.94
0.28728
260
1.2252
144.76
158.36
0.30032
1.0449
144.57
158.10
0.29706
0.9096
144.38
157.85
0.29420
280
1.2629
149.28
163.30
0.30709
1.0776
149.10
163.06
0.30386
0.9386
148.93
162.82
0.30102
300
1.3004
153.49
168.31
0.31378
1.1101
153.71
168.09
0.31057
0.9674
153.55
167.87
0.30775
320
1.3377
158.36
173.40
0.32039
1.1424
158.40
173.20
0.31720
0.9959
158.25
172.99
0.31440
p = 90.0 psia (Tsat = 72.78∘ F)
p = 100 psia (Tsat = 79.12∘ F)
sat.
0.53113
104.21
113.06
0.22006
0.4776
104.99
113.83
0.21976
80
0.54388
105.74
114.8
0.2233
0.47906
105.18
114.05
0.22016
100
0.57729
109.91
119.52
0.23189
0.51076
109.45
118.9
120
0.60874
114.04
124.18
0.24008
0.54022
113.66
140
0.63885
118.19
128.83
0.24797
0.56821
160
0.66796
122.38
133.51
0.25563
180
0.69629
126.62
138.22
200
0.72399
130.92
220
0.75119
240
p = 120 psia (Tsat = 90.49∘ F)
0.39644
106.35
115.16
0.21924
0.229
0.41013
108.48
117.59
0.22362
123.65
0.23733
0.43692
112.84
122.54
0.23232
117.86
128.37
0.24534
0.4619
117.15
127.41
0.24058
0.59513
122.08
133.09
0.25309
0.48563
121.46
132.25
0.24851
0.26311
0.62122
126.35
137.85
0.26063
0.50844
125.79
137.09
0.25619
142.97
0.27043
0.64667
130.67
142.64
0.26801
0.53054
130.17
141.95
0.26368
135.27
147.78
0.27762
0.67158
135.05
147.47
0.27523
0.55206
134.59
146.85
0.271
0.77796
139.69
152.65
0.28468
0.69605
139.49
152.37
0.28233
0.57312
139.07
151.8
0.27817
260
0.80437
144.19
157.58
0.29162
0.72016
143.99
157.32
0.28931
0.59379
143.61
156.79
0.28521
280
0.83048
148.75
162.58
0.29847
0.74396
148.57
162.34
0.29618
0.61413
148.21
161.85
0.29214
300
0.85633
153.38
167.64
0.30522
0.76749
153.21
167.42
0.30296
0.6342
152.88
166.96
0.29896
320
0.88195
158.08
172.77
0.31189
0.79079
157.93
172.56
0.30964
0.65402
157.62
172.14
0.30569







PAGE 473

























Superheated R-134a (EEU)
T
∘
ν
u
h
( F) ( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 140 psia (Tsat = 100.50∘ F)
ν
u
h
( f t 3 /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 160 psia (Tsat = 58.30∘ F)
3
ν
u
h
( f t /lbm) (Bt u /lbm) (Bt u /lbm)
s
(Bt u /lbm /R)
p = 180 psia (Tsat = 117.69∘ F)
sat.
0.33771
107.51
116.26
0.21879
0.29316
108.5
117.18
0.21836
0.25813
109.36
117.96
0.21795
120
0.36243
111.96
121.35
0.22773
0.30578
111.01
120.06
0.22337
0.26083
109.94
118.63
0.2191
140
0.38551
116.41
126.4
0.23628
0.32774
115.62
125.32
0.2323
0.28231
114.77
124.17
0.2285
160
0.40711
120.81
131.36
0.24443
0.3479
120.13
130.43
0.24069
0.30154
119.42
129.46
0.23718
180
0.42766
125.22
136.3
0.25227
0.36686
124.62
135.49
0.24871
0.31936
124
134.64
0.2454
200
0.44743
129.65
141.24
0.25988
0.38494
129.12
140.52
0.25645
0.33619
128.57
139.77
0.2533
220
0.46657
134.12
146.21
0.2673
0.40234
133.64
145.55
0.26397
0.35228
133.15
144.88
0.26094
240
0.48522
138.64
151.21
0.27455
0.41921
138.2
150.62
0.27131
0.36779
137.76
150.01
0.26837
260
0.50345
143.21
156.26
0.28166
0.43564
142.81
155.71
0.27849
0.38284
142.4
155.16
0.27562
280
0.52134
147.85
161.35
0.28864
0.45171
147.48
160.85
0.28554
0.39751
147.1
160.34
0.28273
300
0.53895
152.54
166.5
0.29551
0.46748
152.2
166.04
0.29246
0.41186
151.85
165.57
0.2897
320
0.5563
157.3
171.71
0.30228
0.48299
156.98
171.28
0.29927
0.42594
156.66
170.85
0.29656
340
0.57345
162.13
176.98
0.30896
0.49828
161.83
176.58
0.30598
0.4398
161.53
176.18
0.30331
360
0.59041
167.02
182.32
0.31555
0.51338
166.74
181.94
0.3126
0.45347
166.46
181.56
0.30996
p = 200 psia (Tsat = 125.22∘ F)
sat.
0.22983
110.11
118.62
0.21753
140
0.24541
113.85
122.93
0.22481
160
0.26412
118.66
128.44
180
0.28115
123.35
200
0.29704
220
p = 300 psia (Tsat = 156.09∘ F)
0.14266
112.6
120.52
0.21512
0.23384
0.14656
113.82
121.95
0.21745
133.76
0.24229
0.16355
119.52
128.6
128
138.99
0.25035
0.17776
124.78
0.31212
132.64
144.19
0.25812
0.19044
240
0.32658
137.3
149.38
0.26565
260
0.34054
141.99
154.59
280
0.3541
146.72
300
0.36733
320
p = 400 psia (Tsat = 179.86∘ F)
0.09642
113.35
120.48
0.21161
0.22802
0.09658
113.41
120.56
0.21173
134.65
0.23733
0.1144
120.52
128.99
0.22471
129.85
140.42
0.24594
0.12746
126.44
135.88
0.235
0.20211
134.83
146.05
0.2541
0.13853
131.95
142.2
0.24418
0.27298
0.21306
139.77
151.59
0.26192
0.14844
137.26
148.25
0.2527
159.82
0.28015
0.22347
144.7
157.11
0.26947
0.15756
142.48
154.14
0.26077
151.5
165.09
0.28718
0.23346
149.65
162.61
0.27681
0.16611
147.65
159.94
0.26851
0.38029
156.33
170.4
0.29408
0.2431
154.63
168.12
0.28398
0.17423
152.8
165.7
0.27599
340
0.393
161.22
175.77
0.30087
0.25246
159.64
173.66
0.29098
0.18201
157.97
171.44
0.28326
360
0.40552
166.17
181.18
0.30756
0.26159
164.7
179.22
0.29786
0.18951
163.15
177.18
0.29035







PAG E 4 74

























Superheated R-134a (EEU)