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Chapter 2A: BASIC THERMAL SCIENCES:
FLUID FLOW AND THERMO
Agami Reddy (July 2016)
Basic concepts
2.1 Fluid and thermodynamic properties
- Physical properties
- Thermal properties
2.2 Determining property values
- Gibbs rule
- Ideal gas law
- Other properties
2.3 Types of flow regimes: laminar, turbulent flow- pipes and plates
2.4 Conservation of mass and momentum
2.5 First law of thermodynamics
- Applied to closed-systems
- Applied to open-systems
2.6 Second law of thermodynamics
HCB-3 Chap 2A: Fluid Flow & Thermo
1
Disciplines
A strong understanding of basic principles studied under thermal
sciences is needed:
• Fluid mechanics is the science dealing with properties of fluids,
governing laws and conditions of fluid statics and fluid motion, and
with the resistance to flow outside and inside solid surfaces.
• Thermodynamics is the science dealing with energy and its
transformations and the relationships of the various properties of a
substance as it undergoes changes in pressure and temperature.
• Heat transfer is the science and art of determining the rate at which
heat moves through substances under various externally imposed
temperature and/or boundary conditions.
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Basic Concepts
• Length- distance (m or ft)
• Area (ft2 or m2 )
• Volume (ft3 or m3)
• Velocity (m/s, ft/min, miles/h)- distance per unit time
• Acceleration (m/s2 ) velocity per unit time
Acceleration due to gravity = 9.81 m/s2 or 32.17 ft/s2
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Mass, Force, Weight and Flow Rates
– Mass of a body – quantity of matter the body contains
Unit: pound mass (lbm) and kg
– Force - push or pull that one body may exert on another
Unit: pound force (lbf) and Newton (N)
- 1 lbf = 32.174 lbm-ft/s2
- 1 Newton = Force reqd to accelerate 1 kg by 1 m/ s2
– Weight of a body – force exerted by gravity on the body
Unit: lbf or pound force and Newton
- Flow rates (2 types):
- Mass flow rate: kg/s or lbm/h
- Volume flow rate: m3/s or cfm- cubic foot per min (air)
gpm- gallons per min (water)
HCB-3 Chap 2A: Fluid Flow &
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Work and Energy
L
F
– Work: the effect created by a force when it moves a body
• Work = Force × Distance ( in the direction of the force)
Unit: ft – lbf and Newton-m or Joule (J) and kWh
1 kWh = 1000 x 60 x 60=3.6 x 10^6 Joules
• In SI units, Joule (J) is used
1 J is the work done by a force of 1 N moving by 1 m
• Other energy units: 1 kWh = 1000 x 60 x 60=3.6 x 10^6 J
– General definition of Energy: the ability to do work
This is the accepted definition even though work is just one
form of energy
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Power
– Power: time rate of doing work
or energy use per unit time
• More commonly used
Work
Power 
Time
• Unit: ft-lbf/min, horsepower (HP) and Watt (W)
- 1 W = 1 Joule per second
- 1 horse-power = 746 W or 550 ft-lb per second
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Properties
A property is any attribute or characteristic of matter that can be
observed or evaluated quantitatively.
Pressure: force per unit area
p
Unit: lbf/ft2 (psf) or lbf/in2 (psi) or Pascals (Pa)
The pressures of air and water are very important
– Absolute pressure: pressure exerted by fluid above zero pressure (vacuum)
pabs
– Gage pressure: pressure exerted by fluid above atmospheric
pg
pressure patm = 14.7 lbf/in2 (psia) at sea level (or 101 kPa)
– Vacuum pressure: pressure exerted by fluid below atmospheric
pvac
pressure patm
Atmospheric pressure at sea level = 14.7 psi = 101 kPa
HCB-3 Chap 2A: Fluid Flow &
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• Pressure: force per unit area
Pabs = Patm + Pg
Pabs = Patm - Pvac
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Area A
Pressure Exerted by Column
• Caused by weight of the liquid
• Force (F) = density (d) x volume (V)
•
•
•
•
Height H
=dxHxA
P=F/A=dxH
Can be a large number
It takes denser liquid less height to generate the same pressure
Used in monometer, mercury (high P) and water (small P) are normal
760 mm Hg = 14.7 psia (atmospheric pressure at sea level)
– Also expressed as “HEAD”
• Height of liquid usually water
– Example: What is the pressure exerted by a 300 ft vertical pipe in a
high rise building on a valve at the bottom of the pipe?
Density of the water is 62.4 lbf/ft3
P = d H = 62.4 lbf/ft3 × 300 ft = 18720 lbf/ft2 × 1 ft2 / 144 in2
= 130 lbf/ in2
HCB-3 Chap 2A: Fluid Flow &
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Measuring Air Pressure in Ducts
If H = 4” WG (typical in HVAC systems of buildings)
Pressure difference = d x H = 0.144 psi which the fan has to overcome
HCB-3 Chap 2A: Fluid Flow &
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(a) Dynamic viscosity µ is a measure of the molecular
resistance to fluid flow; for example when a solid is to be
moved through a fluid, or when a fluid is made to flow inside
a pipe or duct. This resistance directly determines the nonrecoverable energy under the above instances. For common
fluids such as water and air, the viscosity is the
proportionality constant in the expression relating shear in the
fluid to the velocity gradient (dv/dy) in the direction y
 
dv
dy
Oil is an example of a highly viscous fluid whereas air has
low viscosity. Under certain circumstances such as fluid
flow far from from any solid surfaces, the effect of
viscosity can neglected. The fluid is then said to be an
ideal fluid or a inviscid fluid .
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Temperature
Temperature: A measure of the thermal activity in a body
• Thermal activity depends on the velocity of the molecules and
other particles of which a matter is composed.
• Thermometer is used to measure temperature
– rely on the fact that most liquids expand and contract when
their temperature is raised or lowered
• Temperature scale: Fahrenheit(˚F) and
Celsius (˚C), Rankine (˚R) and Kelvin (K)
HCB-3 Chap 2A: Fluid Flow &
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Temperature Scales
– Fahrenheit
• 0°F as the stabilized
temperature when equal
amount of ice, water, and
salt are mixed
– Celsius
• 0°C as melting point of ice
(water) and 100°C as
boiling point of water
• ˚ F = 1.8 ˚ C + 32
– Kelvin
• 0 K as absolute zero
• K = ˚ C + 273.15
– Rankine
• ˚ R = ˚ F + 459.67
HCB-3 Chap 2A: Fluid Flow &
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Density and Specific Volume
–
Density – mass/volume (used for solids and liquids)
3
3
“d”
Unit: lbm/ft and kg/m
Density of water and air
air =1.2 kg/m3, water = 1000 kg/m3 
Ratio= 833
– Specific volume – volume/mass (used for liquids and gases)
3
3
n=1/d
Unit: ft / lbm and m /kg
Changes slightly with temperature, why? Because of volume
change
– Specific gravity – weight of the substance/weight of an equal
volume of water at 39 °F
HCB-3 Chap 2A: Fluid Flow &
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Thermal Properties
Specific Heat
– Amount of heat that is required to change the temperature of
1 lbm (or 1 kg) of the substance by 1 °F ( or 1 °C ).
– Units of Btu/lbm-°F) or
– Without phase change !! (J/kg-°C)
– Property of material which changes slightly with temperature
Specific heat for water is 1.0 Btu/lbm-°F at 60 °F
or 4.186 kJ/(kg-K)
air is 0.24 Btu/lbm-°F at 70 °F
or 1.00 kJ/(kg-K)
HCB-3 Chap 2A: Fluid Flow &
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Thermal Energy
• Internal energy (U): microscopic energy possessed by a system
caused by the motion/vibration of the molecules and/or
intermolecular forces- the motion/vibration increases with temperature
• Internal energy is thus often measured by the body’s temperature
(this is not true when the body is a liquid or a solid (such as ice)
which is changing phase!)this leads to sensible and latent heat discussed later
• Total energy of a substance:
E = U + KE + PE+…
(Important: a body does
not contain heat; it contains
thermal energy)
HCB-3 Chap 2A: Fluid Flow &
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Latent Heat
– How to calculate latent heat at a given temperature
• Inter-molecular changes although temperature does not
change
• Use latent heat equation:
Q  m  h fg
where Q is the stored energy (kJ or Btu)
m is the mass flow rate (kg or lbm)
hfg is the latent heat of vaporization ( kJ/kg or Btu/lbm)
(determined from steam tables- discussed later)
HCB-3 Chap 2A: Fluid Flow &
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Enthalpy (h)
– A property of a body that measures its heat content
– Enthalpy includes: (i) Internal energy U and
(ii) pV or energy due to flow work
– Enthalpy is a combined property which is widely used in thermal
analysis
– When T, p or V changes, H changes
– Enthalpy H = U + V.p
in kJ or Btu (V is total volume)
specific enthalpy h = u + p.v in kJ or Btu/lb (v is specific volume)
Instead of sensible or latent heat equations,
enthalpy equation is widely used since one
does not have to worry about state of fluid:
Q  m  (h final  hinitial )
V
P- pressure
V- volume
T- temperature
HCB-3 Chap 2A: Fluid Flow &
Thermo
T P
18
Entropy
Specific entropy s is another important property which cannot be directly
measured (such as internal energy or enthalpy).
- Defined from the second law of thermodynamics.
Entropy is a measure of the energy that is not available for work during a
thermodynamic process due to the fact that natural processes tend not to be
reversible. For example, thermal energy always flows spontaneously as heat
from regions of higher temperature to regions of lower temperature.
Such processes reduces the state of order of the initial system by
homogenization, and therefore entropy is an expression of the degree of
disorder or chaos at the miscopscopic level within the system.
Units of specific entropy are kJ/(kg.K) or Btu/(lbm.oF)
HCB-3 Chap 2A: Fluid Flow &
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Changes of State (Phase)
– Substances can exist in three states: solid, liquid, and gas (vapor)
– Two factors that affect state: temperature and pressure
– Process of state change
- Temperature change or phase change
- Pressure dependent
– Molecular Theory of Liquids and Gases suggested to explain
observed phenomena
– Concept of “saturated state”
subcooled, saturated and superheated
Saturated steam tables used to determine state
HCB-3 Chap 2A: Fluid Flow &
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Change of State
Molecular theory or Kinetic theory
• Temperature is a measure
of average molecule
speed
• Some molecules have
faster speed and escape
• Resisting pressure plays a
role
• During boiling process,
average speed reaches a
level at which the link
between molecules break
HCB-3 Chap 2A: Fluid Flow &
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Gibbs Phase Rule
For pure substance
An useful rule which specifies the number of independent intensive
properties (or degrees of freedom F) needed to completely specify the
thermodynamic state of a fluid (liquid or gaseous).
It is expressed as F = 2+ N –P where
P is the number of phases and N the number of components.
The thermodynamic state of a single-substance system—e.g., air in a
building, steam in a boiler, or refrigerant in an air conditioner—is
defined by specifying F = 2+1-1=2 i.e., two independent, intensive
thermodynamic coordinates or properties.
For moist air which is a mixture of dry air and water vapor, F = 2+2-1 =
3, i.e., three independent properties.
HCB-3 Chap 2A: Fluid Flow &
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Types of Gases
• Real gases
• Ideal gases
– Semi-perfect:
pv=f(T), u=f(T), Specific Heat c=f(T)
– Perfect gases:
Specific heat c=cte, follows ideal gas law
HCB-3 Chap 2A: Fluid Flow &
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Ideal Gas Law
– A gas is a perfect ideal gas when the following relation holds:
PV=mRT
where P – pressure, lbf/ft2
V – volume, ft3
m – mass, lbm
R – gas constant, (for air: 55.16 ft-lbf/lbm-R)
T – absolute temperature, °R
= 287 (J/kg.K)
– A more powerful relation (independent of substance)
P V = (m/MW) R*T
where (m/MW) is the atomic or molecular weight in moles, lbmol or mol
R*is called universal gas constant= R x MW
=1545 ft-lbf /( lbmol-°R )= 8.314 J/(mol. K)
MW is the molecular weight of the substance
HCB-3 Chap 2A: Fluid Flow &
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Ideal Gas Law contd…
- Different forms:
• Same gas: m1 = m2 and R1 = R2 then
Boyle’s Law
P1V1 P2 V2

T1
T2
• Same gas, same temperature, then
P1V1  P2 V2
• Same gas, same volume, then
P1 P2

T1 T2
• Same gas, same pressure, then
V1 V2

T1 T2
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Example 2.1
Determine the mass of air in a room of dimensions
(10 m x 10 m x 2.5 m) at 100 kPa and 20o C?
SOLUTION
The ideal gas law applies to this problem. First we find the density from Eq. (2.11b),
p
101,325 Pa


=1.20 kg/m3
RT
287 J/(kg· K)  (20+273) K
The mass can then be deduced as the product of the density and room volume V:
mair  V  1.20 kg/m3  (10 m  10 m  2.5 m)  300 kg (660 lb m )
HCB-3 Chap 2A: Fluid Flow &
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Other properties of ideal gases can be found by assuming the
perfect gas law to hold. For example, the specific heat at
constant volume cυ and at constant pressure cp can be used to
find changes in the specific internal energy u and the specific
enthalpy h. If the specific heats are constant,
u  uo  cv T  To 
and
(2.13)
h  ho  c p T  To 
(2.14)
where the subscripted variables indicate a specified but arbitrary
reference state o.
The final property of an ideal gas that is needed in secondlaw calculations is the entropy. The specific entropy of an ideal
gas (relative to the reference state To, vo, so) can be found from
s  so  cv ln
T
v
 R ln
To
vo
(2.15)
The variation of the above properties with temperature
The variation
of the
above
properties
with temperature
is relatively
small
for the
temperature
differences is
relatively
and can in
beHVAC
approximated
by constant
mostlysmall
encountered
calculations,
and
theyaverage
can be approximated
by constant
average
values for HVAC
calculations
values.
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Table 2.2
Properties of Common Gases
Gas
Molecular
Weight
cp
Btu/(lbm •
°F)
cv
kJ/(kg •
C)
Btu/(lbm •
°F)
R
kJ/(kg •
°C)
ft•lb/(lbm •
°R)
J/(kg •
K)
Air
28.97
0.240
1.005
0.1715
0.718
53.35
287.1
Hydrogen (H2)
2.016
3.42
14.32
2.43
10.17
767.0
4127
Helium (He)
4.003
1.25
5.234
0.75
3.14
386.3
2078
Methane
(CH4)
16.04
0.532
2.227
0.403
1.687
96.4
518.7
Water vapor
(H2O)
18.02
0.446
1.867
0.336
1.407
85.6
460.6
Acetylene
(C2H2)
26.04
0.409
1.712
0.333
1.394
59.4
319.6
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Flow Regimes
Laminar and turbulent
The Reynolds number (Re) is a dimensionless number
which provides an indication of the type of flow regime. It is
defined as the ratio of inertia to viscous forces, and is given by:
 vD
Re 

where  is the density of the fluid,  the dynamic
viscosity, v the fluid average velocity and D a
characteristic length of the solid.
HCB-3 Chap 2A: Fluid Flow &
Thermo
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TABLE 2.3 Range of Reynold numbers for two common
geometries
If viscous force is
strong, laminar flow
Critical Re numberdepends on
geometry
Flow inside
pipes
Laminar
< 2,100
Flow over a flat <
plate
300,000
Turbulent Characterisc length
> 4,000 Pipe diamater (round
pipe) or hydraulic
diameter (for noncircular)
Distance from leading
edge of plate
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Conservation of Mass and Momentum
The conservation of mass principle simply states that mass
of a substance can be neither created nor destroyed in the
processes analyzed. Consider a simple flow system, when a fluid
stream flows into and out of a control volumethe conservation of
mass principle states:
m(t )   min  m(t   t )   mout
or

dm
 min  mout
dt

m
where m 
t
Momentum is also conserved. The linear momentum of a
rigid body is defined as the mass multiplied by its velocity of the
object. The momentum remains constant; it can neither be created
nor destroyed, but only changed through the action of forces as
described by Newton's laws of motion. Angular momentum
considerations arise while studying the design of impellers of
pumps and fans.
HCB-3 Chap 2A: Fluid Flow &
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Continuity equation
Assuming steady state flow:
Mass balance in terms of volumetric flow rate:


1 V 1   2 V 2
In terms of velocity,
1v1 A1   2 v 2 A2
where the areas are represented by A.
HCB-3 Chap 2A: Fluid Flow &
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Continuity Equation
Example 2.2
Air flows in a duct of 0.03 m2 (46.5 in2) cross-sectional area with a
velocity of 15.24 m/s (3000 ft/min). This high velocity results in a
distrurbing noise. If the velocity is to be reduced to 6.1 m/s (1200
ft/min), what should be the cross-section of the duct be increased to?
Given: v1 = 15.24 m/s, v2 = 10.1 m/s, A1= 0.03 m2
Find: A2
Assumption: Density remains constant,
SOLUTION
A2  A1.(v1/v 2 )  0.03(15.24 /10.1)  0.045 m 2 (70.2 in 2 )
Note: Density assumed constant
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Forms of Mechanical Energy
1) Potential energy: energy possessed by a system due to its
elevation
PE = force x distance = weight x height = (m.g). H
• Example- A crane lifts a block of concrete weighing 1 Ton to
a height of 30 m. How much energy has been expended?
1 Ton = 1000 kg
PE = 1000 kg x 9.81 m/s2 x 30 m = 294300 J = 294.3 kJ
• The above is accomplished in 10 seconds. What is the power
of the crane?
Power = Energy/time = 294.3/10 = 29.43 kW
HCB-3 Chap 2A: Fluid Flow &
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Forms of Mechanical Energy
2) Kinetic energy: energy possessed by a system caused
by the velocity of the molecules.
KE = mass x half of velocity squared = m V2/2
• A truck of mass 1000 kg travels at 30 m/s. What is its
kinetic energy?
KE= 0.5 x 1000 kg x (30) 2 (m/s) 2 = 450,000 J = 450 kJ
• What is the corresponding power?
Power = energy / time

Energy  K .E.  450 kJ

 450 kW

Time
1s
 Time 
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Conversion of Potential to Kinetic Energy
Consider a small spring near a
mountain cabin. If 120 kg/min
flows down a height of 15 m:
(a) What is the velocity of water
at the bottom of the hill
1
mgh  mv 2  v  (2 gh)1/ 2 
2
= (2 x9.81x15)1/ 2  17.15 m/s
(b)How much power can be ideally Power  mgh  120 kg x 9.81 m x15 m=294.3 W
delivered?
60 s
s2
Energy  Power x time  294.3 W x 30 days x 24 hr x 3600 s
(c) How much energy can be
ideally delivered in a month?
=7.6 x108  760 MJ =
HCB-3 Chap 2A: Fluid Flow &
Thermo
760
kWh
 212
3.6
month
36
Stored Energy and Energy Transfer
Body
Forms of
Stored
Energy
Internal energy
Pressure energy
Chemical energy
Potential energy
Kinetic energy
Other forms
Heat (Q)
Another body
Work (W)
Forms of
Energy in
Transfer
Three forms of energy transfer:
Conduction, Convection, and Radiation
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Laws of Thermodynamics
In simplest terms, the Laws of Thermodynamics dictate the specifics for the
movement of heat and work.
Basically, the First Law is a statement of the conservation of energy –
the Second Law is a statement about the quality of energy or direction of
that conservation –
and the Third Law is a statement about reaching Absolute Zero (0 K).
However, since their conception, these laws have become some of the most
important laws of all science - and are often associated with concepts far
beyond what is directly stated in the wording.
-Heat is the lowest form of energy
-Work (from which electricity is produced) is a higher form
-One unit of thermal energy at a high temperature is more VALUABLE
than the same amount of energy at a lower temperature
HCB-3 Chap 2A: Fluid Flow &
Thermo
38
First Law of Thermodynamics
Energy Balance or energy conservation law
– Different ways to express it
A closed system
is one
where the fluid
does not cross
the system
boundaries
– Closed System: Change in the internal energy U is the difference in
the heat Q added to the system minus work done by the system
– If no work involved: the change in total energy in a system equals the
energy added to the system minus the energy removed from the system
dU = Qin – Qout
dU: change in internal or stored energy in the system
Qin
Qin: heat added (entering ) to the system
Qout: heat removed (leaving) from the system
dU
Qout
Sign: +ve if energy added to system
–ve if energy removed from system
HCB-3 Chap 2A: Fluid Flow &
Thermo
39
Closed system- for refrigerant inside piping
Consider Fig. 2.7 which depicts the primary components of an
air-conditioning system. The refrigerant is contained inside the
piping and so this is a closed system. Further this is a cyclic


process, and 1 law simplifies to Q  W  0. Physically this
implies that, if the loses/gains of the refrigerant piping are
neglected , heat picked up in the evaporator (Qin,evap) plus the
work input at the compressor have to be rejected at the
condenser (Qout,cond). Thus for a closed cyclic process:
st
•
•
•
Qout,cond -Qin,evap = Wcomp
The power consumed by the two fans are external to the system
and should not be included in the energy balance of the
refrigerant.
HCB-3 Chap 2A: Fluid Flow &
Thermo
40
Example of Closed System
• A business equipment room has 1000 watts of lighting and some small motors
with a total output of 10 HP. All of the energy in the lighting and from the
motors is converted into heat. What is the increase in enthalpy of the room air
from these sources?
• Analysis: 1) system: room air
2) energy added to the system: from lighting and motor
3) added energy is in form of heat
4) effect of added energy is to increase the air temp.,
i.e. increase in enthalpy
• Solution:
Ech = Ein – Eout
Ein = Ein-light + Ein-motor = 1000 W + 10 HP = … = 28,860 Btu/hr
Eout = 0 Btu/hr
Ech = 28,860 Btu/hr – 0 Btu/hr = 28,860 Btu/hr
Δh = Ech = 28,860 Btu/hr
HCB-3 Chap 2A: Fluid Flow &
Thermo
41
First Law of Thermodynamics contd.
– For a OPEN system:
On a total-mass (extensive) basis, the open-system first law is
v in2
v 2out
min ( gzin 
 hin )  Q  mout ( gzout 
 hout )  W
2
2
where
min, mout = inlet and outlet mass flow amounts, kg ( lbm)
zin, zout = inlet and outlet system port elevations, m (ft)
vin,vout = inlet and outlet fluid average velocities, m/s (ft/s)
hin, hout = inlet and outlet specific enthalpies, kJ/kg (Btu/lbm)
HCB-3 Chap 2A: Fluid Flow &
Thermo
42
First Law of Thermodynamics contd.
For a OPEN system where KE, PE and other energy sources are negligible
- when W=0:
Q = m (hout – hin)
boiler
- when Q=0:
W = m (hin– hout) turbine
where
hin: enthalpy of fluid entering the system
hout: enthalpy of fluid leaving the system
Sign convention for Q:
+ve when added to the system
–ve when removed from the system
Sign convention for W is opposite
HCB-3 Chap 2A: Fluid Flow &
Thermo
43
Sensible Heat
Sensible heat of a body is the energy associated with its
temperature.
Example- heating water
For liquids and solids, the change in stored energy when its mass
undergoes a temperature change is given by:
Qs  m  c  T  m  c  (t2  t1 )
where Qs is the stored energy, (kJ or Btu)
 is the mass (kg or lb)
m
c is the specific heat, (kJ/kg-C or Btu/lbm-°F)
The same equation also applies when a fluid flow is involved. In
that case, Q is the rate of heat transfer (Btu/hr) and m is the mass
flow rate (kg/s)
HCB-3 Chap 2A: Fluid Flow &
Thermo
44
Sensible Heat Example
A hot water re-heater heats duct air from 55 °F to 70 °F before it
enters a room when the valve is 100 % open. The air flow rate is
500 cfm. What is the heating capability of the reheater?
Solution: Use sensible heat equation:
55ºF
From AHU
Qs  m  c  T  m  c  (t2  t1 )
- Check to see whether we have all the
necessary variables.
– mass flow rate unknown
- Use density (d) to obtain mass flow rate from
volumetric flow rate and then use equation
Qs  m  c  T  V  d  c  (t2  t1 )
ft 3
lbm
Btu
min
 500
 0.075 3  0.24
 (70  55)0 F  60
 8100 Btu / hr
min
ft
lbm  F
hr
HCB-3 Chap 2A: Fluid Flow &
Thermo
45
Sensible Heat and Latent Heat
– Why the need to distinguish between them?
• During a process with phase change, temperature is not the
only variable that determines heat transfer rate.
– What is sensible heat?
• Energy (heat) that is added or removed during a process
where the temperature of a substance changes but there is NO
change in state (phase) of the substance.
– What is latent heat change?
• Energy (heat) absorbed or released during a process of state
(phase) change.
Both these can be treated together by using the enthalpy
equations
HCB-3 Chap 2A: Fluid Flow &
Thermo
46
Example of Using Superheat Tables
Boiler example:
Q  m  (h final  hinitial )
A boiler heats up 130 lb/min of water from 80 F to 400 F at a pressure of 60
psia. What is the needed heat input?
From the saturated table, the enthalpy of saturated liquid at 80 F
h
 48.02 Btu/lb
initial
From the superheat table, corresponding to the sub- table of P=60 psia and
temperature of 400 F, h
=1,233.5 Btu/lb
final
Finally, heat input
Q= (130 lb/min) (1233.5-48.02) Btu/lb x 60 min/h
= 9.25 x 10^6 Btu/h
HCB-3 Chap 2A: Fluid Flow &
Thermo
47
All forms of energy are not equal!
Thermal energy (or heat) is a more “disordered” form of energy-
HCB-3 Chap 2A: Fluid Flow &
Thermo
From Cengal and Boles
48
Entropy and Second Law
2nd Law of Thermo places limit on energy conversion and direction of flow:
– Various statements:
• Heat will not flow spontaneously from cold object to hot object.
• Any system which is free of external influences becomes more
disordered with time. This disorder can be expressed in terms of the
quantity called entropy.
• All work can be converted in heat but all heat cannot be converted in to
work (alternative statement: you cannot create a heat engine which
extracts heat and converts it all to useful work).
– Understanding:
• Work is needed to move heat from low temp. to high temp.
• Not all the thermal energy can be fully used.
• Concept of entropy: a measure of the irreversibility of the process (due to
friction, heat transfer across a temperature difference,…)
a property (just like temperature, pressure, enthalpy)
Proper understanding important for energy resources sustainability
HCB-3 Chap 2A: Fluid Flow &
Thermo
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Outcomes
•
•
•
•
•
•
•
•
•
•
•
Competence in basic physical properties: mass, volume, pressure, temperature,
density, viscosity
Competence in basic thermal properties: specific heat, heat of vaporization, internal
energy, enthalpy, entropy
Understand phase changes and kinetic theory.
Familiarity with Gibbs phase rule and its usefulness
Familiarity with real and ideal gases, and Ideal Gas Law
Familiarity with different flow regimes, turbulent and laminar flows through pipes
and flat surfaces
Understand how to apply conservation of mass and momentum principles
Familiarity with different forms of stored energy
Understand the difference between stored energy and energy transfer
Understand the application of the first law of thermodynamics to closed and open
systems
Familiarity with the second law of thermodynamic and its usefulness which limits
energy conservation and the direction of flow
HCB-3 Chap 2A: Fluid Flow &
Thermo
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