Transport phenomena
zdadach
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About This Presentation
Advanced course for undergraduate students
Slide Content
Zin Eddine Dadach
2009-2010
2009-2010
Ohm’s law
In electrical circuits, Ohm's lawstates
that the current through a conductor
between two points is directly proportional
to the potential differenceor voltage
across the two points, and inversely
proportional to the resistancebetween
them.
In electrical circuits, Ohm's lawstates
that the current through a conductor
between two points is directly proportional
to the potential differenceor voltage
across the two points, and inversely
proportional to the resistancebetween
them.
DEFINITION OF TRANSPORT
PHENOMENA
Transport phenomena are all irreversible
processesof statistical nature stemming
from the random continuous motion of
molecules, mostly observed in fluids.
They involve a net macroscopic transfer of
matter, energy or momentum in
thermodynamic systems that are not in
statistical equilibrium.
PHENOMENA
Transport phenomena are all irreversible
processesof statistical nature stemming
from the random continuous motion of
molecules, mostly observed in fluids.
They involve a net macroscopic transfer of
matter, energy or momentum in
thermodynamic systems that are not in
statistical equilibrium.
MOMENTUM, HEAT ,MASS,
KINETICS
KINETICS
Momentum, Heat & Mass
Examples include heat conduction(energy
transfer), viscosity(momentum transfer),
molecular diffusion(mass transfer),
radiationand electric chargetransfer in
semiconductors.
Apart from conduction, another heat
transfermechanism is convectionwhich is
not a transport phenomenon per se, as it
involves turbulent and bulk motion of
fluids.
Examples include heat conduction(energy
transfer), viscosity(momentum transfer),
molecular diffusion(mass transfer),
radiationand electric chargetransfer in
semiconductors.
Apart from conduction, another heat
transfermechanism is convectionwhich is
not a transport phenomenon per se, as it
involves turbulent and bulk motion of
fluids.
Analogies
An important principle in the study of
transport phenomena is analogy between
phenomena. For example, mass, energy,
and momentum can all be transported by
diffusion:
In any Dynamic System:
Rate = Driving Force/ Resistance
An important principle in the study of
transport phenomena is analogy between
phenomena. For example, mass, energy,
and momentum can all be transported by
diffusion:
In any Dynamic System:
Rate = Driving Force/ Resistance
EXAMPLES
The spreading and dissipation of odors in
air is an example of mass diffusion.
The conduction of heat in a solid material
is an example of heat diffusion.
The drag experienced by a rain drop as it
falls in the atmosphere is an example of
momentum diffusion (the rain drop
loses momentum to the surrounding air
through viscous stresses and decelerates).
The spreading and dissipation of odors in
air is an example of mass diffusion.
The conduction of heat in a solid material
is an example of heat diffusion.
The drag experienced by a rain drop as it
falls in the atmosphere is an example of
momentum diffusion (the rain drop
loses momentum to the surrounding air
through viscous stresses and decelerates).
Driving Force & Resistance in Momentum
Transfer
Driving force : Velocity gradient
Resistance : Viscosity ( μ)
* Equation of Newton:
Transfer
Driving force : Velocity gradient
Resistance : Viscosity ( μ)
* Equation of Newton:
Driving Force & Resistance in Heat
Transfer
Driving Force: Temperature Gradient
Resistance: 1/k (k=Material Conductivity)
Fourier’s Law:
Transfer
Driving Force: Temperature Gradient
Resistance: 1/k (k=Material Conductivity)
Fourier’s Law:
Example of Heat Conduction
Driving Force & Resistance in Mass
Transfer
Driving Force : Difference in Chemical
Potential. When the pressure and
temperature are the same in both sides,
the driving force is the difference in
concentration. In R.O, the pressure is
different in the sides of the cell.
Resistance: 1/ Diffusivity ( D)
Transfer
Driving Force : Difference in Chemical
Potential. When the pressure and
temperature are the same in both sides,
the driving force is the difference in
concentration. In R.O, the pressure is
different in the sides of the cell.
Resistance: 1/ Diffusivity ( D)
Fick’s law:
Transport phenomena & unit
operations
The study of transport phenomena is the
basis for most of unit operations.
In many unit operations, such as
distillation, all three transport phenomena
( i.e., fluid flow, heat transfer and mass
transfer) occur often simultaneously.
Empiricism is often required because the
exact mathematical equations cannot be
solved.
operations
The study of transport phenomena is the
basis for most of unit operations.
In many unit operations, such as
distillation, all three transport phenomena
( i.e., fluid flow, heat transfer and mass
transfer) occur often simultaneously.
Empiricism is often required because the
exact mathematical equations cannot be
solved.
EQUILIBRIUM VS RATE
PROCESSES
A system in equilibrium means that all
variables are constant with time (
Thermodynamics)
Non-equilibrium means that at least one
variable changes with time ( Transport
Phenomena).
In this case, we will have rate. They are four
different rate processes:
Rate of Momentum transfer
Rate of Heat transfer
Rate of Mass transfer
Rate of Reaction ( Not covered in this
course)
PROCESSES
A system in equilibrium means that all
variables are constant with time (
Thermodynamics)
Non-equilibrium means that at least one
variable changes with time ( Transport
Phenomena).
In this case, we will have rate. They are four
different rate processes:
Rate of Momentum transfer
Rate of Heat transfer
Rate of Mass transfer
Rate of Reaction ( Not covered in this
course)
Example of momentum
transfer
Let’s consider two tanks with two different
pressures separated by a valve.
When we open the valve, we will observe a
decrease of pressure in the tank having high
pressure and increase in pressure in the tank
with low pressuremovement of molecules
from high pressure to low pressure. Non –
Equilibrium and ∆P= Driving Force.
When the two tanks reach the same pressure.
There will be no more transfer: (Equilibrium
state: thermodynamics Study)
transfer
Let’s consider two tanks with two different
pressures separated by a valve.
When we open the valve, we will observe a
decrease of pressure in the tank having high
pressure and increase in pressure in the tank
with low pressuremovement of molecules
from high pressure to low pressure. Non –
Equilibrium and ∆P= Driving Force.
When the two tanks reach the same pressure.
There will be no more transfer: (Equilibrium
state: thermodynamics Study)
Example of molecular heat
transfer
If we put two solids having different
temperatures side by side, we will observe
a decrease of temperature of the hot solid
and increase of temperature of the cold
solid.
Molecular heat transfer by conduction
Driving force: Difference in temperature
When the two solids reach the same
temperature. No more movement of heat.
Thermal Equilibrium.
transfer
If we put two solids having different
temperatures side by side, we will observe
a decrease of temperature of the hot solid
and increase of temperature of the cold
solid.
Molecular heat transfer by conduction
Driving force: Difference in temperature
When the two solids reach the same
temperature. No more movement of heat.
Thermal Equilibrium.
Example of molecular mass
transfer
Let’s consider two tanks with two different
gases ( nitrogen & oxygen separated by a
valve.
When we open the valve, we will observe a
decrease of concentration of nitrogen in the
tank having high concentration of nitrogen and
increase in concentration of nitrogen in the
tank with oxygen movement of molecules
from high concentration to low concentration.
Non –Equilibrium and ∆C= Driving Force.
When the two tanks reach 50% of each gas,
there will be no more mass transfer:
(Equilibrium state: thermodynamics Study)
transfer
Let’s consider two tanks with two different
gases ( nitrogen & oxygen separated by a
valve.
When we open the valve, we will observe a
decrease of concentration of nitrogen in the
tank having high concentration of nitrogen and
increase in concentration of nitrogen in the
tank with oxygen movement of molecules
from high concentration to low concentration.
Non –Equilibrium and ∆C= Driving Force.
When the two tanks reach 50% of each gas,
there will be no more mass transfer:
(Equilibrium state: thermodynamics Study)
MAIN EQUATION
CASE OF HEAT TRANSFER
Driving Force: ∆T
Temperature gradient
Driving Force: ∆T
Temperature gradient
Temperature Profile
Consider a block of copper where all sides are
insulated so that heat conduction can occur only
in the x-direction ( Figure 2.2).
At time t=t
0. The initial temperature of the block
= 0
0
C
Not let’s put the block between ice (0
0
C) at the
bottom and steam ( 100
0
C) at the top. The top
temperature of the top is instantaneously ( t=t
0)
100
0
C.
Some time later ( t=t
1), we will observe a
temperature profile between 0
0
C and 100
0
C with
a parabolic shape.
At t=t
2, the profile will still be parabolic but flatter
At t=t
∞( steady state)
,the profile will become a
straight line.
Consider a block of copper where all sides are
insulated so that heat conduction can occur only
in the x-direction ( Figure 2.2).
At time t=t
0. The initial temperature of the block
= 0
0
C
Not let’s put the block between ice (0
0
C) at the
bottom and steam ( 100
0
C) at the top. The top
temperature of the top is instantaneously ( t=t
0)
100
0
C.
Some time later ( t=t
1), we will observe a
temperature profile between 0
0
C and 100
0
C with
a parabolic shape.
At t=t
2, the profile will still be parabolic but flatter
At t=t
∞( steady state)
,the profile will become a
straight line.
Observation
The linear temperature profile is an
experimental observation and , provided
enough time is allowed, this linear
temperature profile will be observed as
long as the temperature of the top and
bottom remain respectively at 100
0
C and
0
0
C.
The observation is attributed to the
French scientist Fourier and the following
equation is named after him:
(q/A)
x= -k ( ∂T/∂x)
The linear temperature profile is an
experimental observation and , provided
enough time is allowed, this linear
temperature profile will be observed as
long as the temperature of the top and
bottom remain respectively at 100
0
C and
0
0
C.
The observation is attributed to the
French scientist Fourier and the following
equation is named after him:
(q/A)
x= -k ( ∂T/∂x)
Fourier Equation
In the Fourier Equation:
1)q is the amount of heat transferred per
unit time
2)A is the area
3)x is considered the only direction of the
heat flux
4)k is the thermal conductivity
5)Since ( ∂T/∂x) and (q/A) have opposite
sign (Heat always flows from high
temperatures to low temperatures), we
need a negative sign.
In the Fourier Equation:
1)q is the amount of heat transferred per
unit time
2)A is the area
3)x is considered the only direction of the
heat flux
4)k is the thermal conductivity
5)Since ( ∂T/∂x) and (q/A) have opposite
sign (Heat always flows from high
temperatures to low temperatures), we
need a negative sign.
Class work
Calculate the steady state heat flux across
a copper block 10 cm thick if one side is
maintained at 0
0
C and the other side at
100
0
C ( See figure 2.4). We also assume
constant thermal conductivity k= 380
W.m
-1
.K
-1
Calculate the steady state heat flux across
a copper block 10 cm thick if one side is
maintained at 0
0
C and the other side at
100
0
C ( See figure 2.4). We also assume
constant thermal conductivity k= 380
W.m
-1
.K
-1
Solution
(q/A)
x= -k ( ∂T/∂x)
Integration:
(q/A)
x= -k ( ∂T/∂x)
Integration:
The integration will become:
(q/A)
x. ( x
2–x
1)= -k ( T
2-T
1)
After putting the numbers ( q/A)
2= -
3.8 x 10
5
J.m
-2
.s
-1
(q/A)
x. ( x
2–x
1)= -k ( T
2-T
1)
After putting the numbers ( q/A)
2= -
3.8 x 10
5
J.m
-2
.s
-1
Case of Mass Transfer
Let’s study the concentration profile of a
liquid A in a membrane. At time t=t
0,
the top of the membrane has a high
concentration of liquid A (C
Ain moles/
m
3
) and, in the rest of the membrane
there is no liquid A ( figure 2.2).
At time t=t
1, material had diffused
towards the bottom and the concentration
is shown in Figure 2.2 .
After some time ( t=t
∞
) we will observe
a straight line ( Like for the case of heat
transfer)
Let’s study the concentration profile of a
liquid A in a membrane. At time t=t
0,
the top of the membrane has a high
concentration of liquid A (C
Ain moles/
m
3
) and, in the rest of the membrane
there is no liquid A ( figure 2.2).
At time t=t
1, material had diffused
towards the bottom and the concentration
is shown in Figure 2.2 .
After some time ( t=t
∞
) we will observe
a straight line ( Like for the case of heat
transfer)
Observation
The observation is attributed to the
scientist Fick in 1855 and the following
equation is named after him:
Fick’s law:
(J
A/A)
x= -D ( ∂C
A/∂x)
The observation is attributed to the
scientist Fick in 1855 and the following
equation is named after him:
Fick’s law:
(J
A/A)
x= -D ( ∂C
A/∂x)
Fick’s law
In the Fick’s Equation:
1)J
Ais the amount of A diffused per unit time
2)A is the area of the transfer
3)x is considered the only direction of the mass
transfer
4)D is the coefficient of Diffusion or Diffusivity
5)( ∂C
A/∂x) is the concentration gradient
6)(J
A/A)
xis the molar flux.
7)Since they have opposite sign (Mass always
flows from high Concentrations to low
Concentrations), we need a negative sign.
In the Fick’s Equation:
1)J
Ais the amount of A diffused per unit time
2)A is the area of the transfer
3)x is considered the only direction of the mass
transfer
4)D is the coefficient of Diffusion or Diffusivity
5)( ∂C
A/∂x) is the concentration gradient
6)(J
A/A)
xis the molar flux.
7)Since they have opposite sign (Mass always
flows from high Concentrations to low
Concentrations), we need a negative sign.
Class Work
CO
2gas and air diffuse from two sides of
an iron tube. The concentrations of CO
2in
the two sides of the tube are respectively
0.083 kmol/m
3
and 0 kmol/m
3
. The
coefficient of diffusion of CO
2is assumed
1.56x10
-3
m
2
s
-1
.
A) Find the molar flux
B) Find the number of lbs of CO
2that pass
the tube of iron in one hour.
CO
2gas and air diffuse from two sides of
an iron tube. The concentrations of CO
2in
the two sides of the tube are respectively
0.083 kmol/m
3
and 0 kmol/m
3
. The
coefficient of diffusion of CO
2is assumed
1.56x10
-3
m
2
s
-1
.
A) Find the molar flux
B) Find the number of lbs of CO
2that pass
the tube of iron in one hour.
The case of momentum Transfer
Let’s consider a fluid maintained between two plates having
an area A. The lower plate is a stationary plate and a force
is applied to the upper plate to move with a velocity V.
Because of the internal friction in the fluid due to viscosity,
the fluid will move between the two plates at different
velocities varying from V to zero.
In momentum transfer, we need two
coordinates:
1)coordinate x is the direction of the
velocity V
x
2) coordinate y is the direction of
the change of velocity
Let’s consider a fluid maintained between two plates having
an area A. The lower plate is a stationary plate and a force
is applied to the upper plate to move with a velocity V.
Because of the internal friction in the fluid due to viscosity,
the fluid will move between the two plates at different
velocities varying from V to zero.
In momentum transfer, we need two
coordinates:
1)coordinate x is the direction of the
velocity V
x
2) coordinate y is the direction of
the change of velocity
Observation
As shown in Figure 2.2 ,at time t=0, the
upper plate has velocity V (due to force F)
and the rest of the fluid has velocity equal
to zero.
With time, some fluid between the plates
start to move . The layer of fluid which is
closest to the upper plate starts to move
first and then the others.
At time t =t
∞
, we will have a linear
velocity gradient.
The following observation is named :
Newton’s law
As shown in Figure 2.2 ,at time t=0, the
upper plate has velocity V (due to force F)
and the rest of the fluid has velocity equal
to zero.
With time, some fluid between the plates
start to move . The layer of fluid which is
closest to the upper plate starts to move
first and then the others.
At time t =t
∞
, we will have a linear
velocity gradient.
The following observation is named :
Newton’s law
Newton’s Law
F is the force applied on the upper plate
A is the area of the two plates
F/A is called Shear Stress which is called
the momentum flux ζ
Newton’s law of viscosity is :
ζ
yx= F/A= -μ(∂U
x/∂y)
F is the force applied on the upper plate
A is the area of the two plates
F/A is called Shear Stress which is called
the momentum flux ζ
Newton’s law of viscosity is :
ζ
yx= F/A= -μ(∂U
x/∂y)
Class Work
Two parallel plates are 10 cm apart. The
bottom plate is stationary and the upper
plate has a velocity of 30 cm.s
-1
If the fluid has a viscosity of 1 centipoise (
0.001 kg.m
-1
.s
-1
), calculate the
momentum flux
Two parallel plates are 10 cm apart. The
bottom plate is stationary and the upper
plate has a velocity of 30 cm.s
-1
If the fluid has a viscosity of 1 centipoise (
0.001 kg.m
-1
.s
-1
), calculate the
momentum flux
MOMENTUM, HEAT AND MASS DIFFUSIVITIES
Introduction
In the previous chapter, an analogy between the
molecular transports has been developed. The
Fourier, Fick and Newton equations are all
empirical and have the same form.
The mechanism is not the same :
1) Molecular diffusion occurs in multicomponent
mixtures with a concentration gradient as the
driving force.
2) Momentum transfer occurs perpendicular to the
direction of the flow ( the direction of the
pressure drop that causes the flow).
3) Heat transfer occurs by molecular transport (
conduction) in solids, does not involve flow or
relative motion of the molecules
In the previous chapter, an analogy between the
molecular transports has been developed. The
Fourier, Fick and Newton equations are all
empirical and have the same form.
The mechanism is not the same :
1) Molecular diffusion occurs in multicomponent
mixtures with a concentration gradient as the
driving force.
2) Momentum transfer occurs perpendicular to the
direction of the flow ( the direction of the
pressure drop that causes the flow).
3) Heat transfer occurs by molecular transport (
conduction) in solids, does not involve flow or
relative motion of the molecules
The Different Diffusivities
The respective constants in these three
equations are:
1) D: Diffusivity of Coefficient of diffusion
which unit is m
2
.s
-1
2) μ: Viscosity which unit is kg.m
-1
.s
-1
3) k: Thermal conductivity which unit is
W.m
-1
.K
-1
The respective constants in these three
equations are:
1) D: Diffusivity of Coefficient of diffusion
which unit is m
2
.s
-1
2) μ: Viscosity which unit is kg.m
-1
.s
-1
3) k: Thermal conductivity which unit is
W.m
-1
.K
-1
D: Diffusivity
I) The coefficient of diffusion is the simplest
property because other properties are not
involved. However, it is also the most difficult to
measure because:
A) there should be no velocity
B) the diffusivity is very small in magnitude ( 10
-
5
).
II) Diffusivity of binary mixtures increases with
temperature but not linearly and decreases with
pressure because pressure reduces movements
of molecules.
III) They are a large differences between
diffusivities of gases, liquids and solids ( Table
2.2 page 49).
I) The coefficient of diffusion is the simplest
property because other properties are not
involved. However, it is also the most difficult to
measure because:
A) there should be no velocity
B) the diffusivity is very small in magnitude ( 10
-
5
).
II) Diffusivity of binary mixtures increases with
temperature but not linearly and decreases with
pressure because pressure reduces movements
of molecules.
III) They are a large differences between
diffusivities of gases, liquids and solids ( Table
2.2 page 49).
Effects of temperature and pressure
of the Diffusivity
They are a number of equations available
in the literature for estimating diffusivities
in gases.
D
0known at P
0and T
0(K) and the
exponent n varies from 1.75 and 2.0 over
a range of normal temperatures and
pressures. With pressure below 5 atm,
there is no concentration dependence of
the coefficient of diffusion.
of the Diffusivity
They are a number of equations available
in the literature for estimating diffusivities
in gases.
D
0known at P
0and T
0(K) and the
exponent n varies from 1.75 and 2.0 over
a range of normal temperatures and
pressures. With pressure below 5 atm,
there is no concentration dependence of
the coefficient of diffusion.
Class Work
Predict the diffusivity of water vapor in
air at 2 atm and 75
0
C, if the coefficient of
diffusion is 0.219x10
-4
m
2
.s
-1
at 1atm and
0
0
C and assuming that the exponent n is
equal to 1.75.
Predict the diffusivity of water vapor in
air at 2 atm and 75
0
C, if the coefficient of
diffusion is 0.219x10
-4
m
2
.s
-1
at 1atm and
0
0
C and assuming that the exponent n is
equal to 1.75.
k= Thermal Conductivity
The two important properties of materials in heat
transfer are thermal conductivity and thermal
diffusivity.
We are interested in thermal conductivity. Table
2.2 gives some typical values.
The thermal conductivity of gases can be predicted
more accurately than the thermal conductivity of
liquids and solids because in gases, the energy is
directly carried by molecules while for liquids and
solids, other mechanisms are involved.
Since heat transfer is related to molecular
collisions, the thermal conductivity of liquids and
solids is much larger than gases. Water vapor at
100
0
C has k= 0.0248 W.m
-1
.K
-1
and water liquid at
100
0
C has k= 0.68 W.m
-1
.K
-1
( Table 2.2)
The two important properties of materials in heat
transfer are thermal conductivity and thermal
diffusivity.
We are interested in thermal conductivity. Table
2.2 gives some typical values.
The thermal conductivity of gases can be predicted
more accurately than the thermal conductivity of
liquids and solids because in gases, the energy is
directly carried by molecules while for liquids and
solids, other mechanisms are involved.
Since heat transfer is related to molecular
collisions, the thermal conductivity of liquids and
solids is much larger than gases. Water vapor at
100
0
C has k= 0.0248 W.m
-1
.K
-1
and water liquid at
100
0
C has k= 0.68 W.m
-1
.K
-1
( Table 2.2)
μ= Viscosity
In general, the viscosity of gases
increases with temperature at low
pressures while that of liquids usually
decreases.
For gases as rigid spheres and at low
pressures ( < 10 atm), the variation of
viscosity is related to the square root of
temperature in absolute unit ( K or R). In
reality, the power varies from 0.6 to 1.
In general, we have
Where A and B are constants.
In general, the viscosity of gases
increases with temperature at low
pressures while that of liquids usually
decreases.
For gases as rigid spheres and at low
pressures ( < 10 atm), the variation of
viscosity is related to the square root of
temperature in absolute unit ( K or R). In
reality, the power varies from 0.6 to 1.
In general, we have
Where A and B are constants.
For Liquids
The viscosity of gases is independent of pressure.
Of course the kinematic viscosity of gases
depends both on pressure and temperature.
For liquids, the theories are not very well
developed as for gases. An approximate empirical
observation for the temperature is
Where A and B are empirical constants. Because
liquids are incompressible, the viscosity does not
depend on pressure.
The fluids that follow Newton’s law are called
Newtonian fluids. The others are called non-
Newtonian fluids.
The viscosity of gases is independent of pressure.
Of course the kinematic viscosity of gases
depends both on pressure and temperature.
For liquids, the theories are not very well
developed as for gases. An approximate empirical
observation for the temperature is
Where A and B are empirical constants. Because
liquids are incompressible, the viscosity does not
depend on pressure.
The fluids that follow Newton’s law are called
Newtonian fluids. The others are called non-
Newtonian fluids.
Class Work
From Table 2.2 , estimate the constants A
and B for air at low pressures using the
viscosity at 280K and 400 K.
From Table 2.2 , estimate A and B/R for
water liquid using viscosity at 373.15K
and 273.15K.
From Table 2.2 , estimate the constants A
and B for air at low pressures using the
viscosity at 280K and 400 K.
From Table 2.2 , estimate A and B/R for
water liquid using viscosity at 373.15K
and 273.15K.
EQUATION OF CONTINUITY
Momentum transfer
In fluid mechanics, the most useful
equations are based on the principle of
mass balance or continuity equation.
The equations are first written in
differential form. The be useful
Multi-dimensional flow:
In fluid mechanics, the most useful
equations are based on the principle of
mass balance or continuity equation.
The equations are first written in
differential form. The be useful
Multi-dimensional flow:
Dρ/Dt= accumulation ( rate of density
change)
∂u/ ∂x = variation of velocity in the x
direction.
For one –dimensional flow:
dm= ρ.u.dS
ρis the density of fluid
u is the velocity in the direction of motion
dS is the cross section
dm is the change of mass flow.
change)
∂u/ ∂x = variation of velocity in the x
direction.
For one –dimensional flow:
dm= ρ.u.dS
ρis the density of fluid
u is the velocity in the direction of motion
dS is the cross section
dm is the change of mass flow.
Total flow & Mass velocity
To find the total flow, we need to
integrate the equation:
m= ρ∫u.dS
Mass velocity
G= ρ.V= m/S or m=ρ.V.S
where V is the average
velocity
To find the total flow, we need to
integrate the equation:
m= ρ∫u.dS
Mass velocity
G= ρ.V= m/S or m=ρ.V.S
where V is the average
velocity
Class work
Crude oil with SG= 0.887 flows through
two pipes. Pipe A has a diameter of 50
mm and pipe B has a diameter of 75 mm.
Each pipe C has a diameter of 38 mm.
The flow through pipe A is 6.65 m
3
/hr.
Calculate:
1) the mass flow rate in each pipe
2) the average velocity in each pipe
3) the mass velocity in each pipe
Crude oil with SG= 0.887 flows through
two pipes. Pipe A has a diameter of 50
mm and pipe B has a diameter of 75 mm.
Each pipe C has a diameter of 38 mm.
The flow through pipe A is 6.65 m
3
/hr.
Calculate:
1) the mass flow rate in each pipe
2) the average velocity in each pipe
3) the mass velocity in each pipe
Class Work
Air at 20
0
C and 2 atm enters a tube (
diameter=50 mm ) and an average
velocity of 15ms
-1
.It leaves through a
second pipe ( diameter 65 mm) at 90
0
C
and 1.6 atm.
1) Calculate the average velocity at the
outlet
Air at 20
0
C and 2 atm enters a tube (
diameter=50 mm ) and an average
velocity of 15ms
-1
.It leaves through a
second pipe ( diameter 65 mm) at 90
0
C
and 1.6 atm.
1) Calculate the average velocity at the
outlet
Heat transfer by conduction
Fourier's law is the basic law.
STEADY STATE CONDUCTION :
1)In figure 10.1a, a flat wall insulated tank
contains a refrigerant at –10
0
C while the air
is at 28
0
C.
2)The temperature falls linearly with distance
across the layer as heat flows from air to
refrigerant.
3)In figure 10.1.b, a similar tank contains
boiling water losing heat to the atmosphere
at 20
0
C.
4)As in Figure 10.1.a, there is a temperature
profile but heat flows in the reverse direction.
Fourier's law is the basic law.
STEADY STATE CONDUCTION :
1)In figure 10.1a, a flat wall insulated tank
contains a refrigerant at –10
0
C while the air
is at 28
0
C.
2)The temperature falls linearly with distance
across the layer as heat flows from air to
refrigerant.
3)In figure 10.1.b, a similar tank contains
boiling water losing heat to the atmosphere
at 20
0
C.
4)As in Figure 10.1.a, there is a temperature
profile but heat flows in the reverse direction.
Calculating heat flow
Rearranging Fourier's law:
Since T and x are the only variables, we
will have:
B is the thickness of the wall.
Rearranging Fourier's law:
Since T and x are the only variables, we
will have:
B is the thickness of the wall.
Calculating Heat Flow
Taking ( B/k) = R, we will have:
q is the rate
R is the thermal resistance
∆T is the driving force.
Taking ( B/k) = R, we will have:
q is the rate
R is the thermal resistance
∆T is the driving force.
k
*
=Average value of k
*
=Average value of k
Class work
A layer of a material of 6in ( 152 mm)
thick is used for insulation. The
temperature of the cold side is 40
0
F and
the warm side is 180
0
F. The thermal
conductivity of the material is 0.021
Btu/ft.h.
0
F at 32
0
F and 0.032 Btu/ft.h.
0
F
at 200
0
F. The area of the wall is 25 ft
2
.
1) Calculate the heat flow through the
pipe in Btu/hr.
A layer of a material of 6in ( 152 mm)
thick is used for insulation. The
temperature of the cold side is 40
0
F and
the warm side is 180
0
F. The thermal
conductivity of the material is 0.021
Btu/ft.h.
0
F at 32
0
F and 0.032 Btu/ft.h.
0
F
at 200
0
F. The area of the wall is 25 ft
2
.
1) Calculate the heat flow through the
pipe in Btu/hr.
Resistances in Series
With an insulation having different layers,
we will have different resistances to heat
conduction.
Assuming the thicknesses: B
A,B
Band B
C
The average conductivities will be : k
A
*
,k
B
*
and k
C
*
The temperature drops will be : ∆T
A; ∆T
B
and ∆T
C
Therefore: ∆T= ∆T
A+∆T
B+ ∆T
C
With an insulation having different layers,
we will have different resistances to heat
conduction.
Assuming the thicknesses: B
A,B
Band B
C
The average conductivities will be : k
A
*
,k
B
*
and k
C
*
The temperature drops will be : ∆T
A; ∆T
B
and ∆T
C
Therefore: ∆T= ∆T
A+∆T
B+ ∆T
C
Heat Flow
For every layer, we can write:
A is the cross sectional section.
For every layer, we can write:
A is the cross sectional section.
Adding the resistances, we find the
following equation:
R = R
A + R
B+ R
C
R
i= B
i/k
i
*
following equation:
R = R
A + R
B+ R
C
R
i= B
i/k
i
*
Class work
A flat furnace is composed of :
1)a Silo Cellayer of 4.5 in with k= 0.08
Btu/ft.h.
0
F
2)Common brick layer of 9 in with k= 0.8
Btu/ft.h.
0
F
The temperature of the inner face of the wall is
1400
0
F and that of the outer wall is 170
0
F.
Calculate:
a)Heat loss through the wall
b)Temperature of the interface
c)Assuming that the contact between the two
layers has a resistance of 0.50
0
F .h.ft
2
/Btu,
what is the new heat loss?
A flat furnace is composed of :
1)a Silo Cellayer of 4.5 in with k= 0.08
Btu/ft.h.
0
F
2)Common brick layer of 9 in with k= 0.8
Btu/ft.h.
0
F
The temperature of the inner face of the wall is
1400
0
F and that of the outer wall is 170
0
F.
Calculate:
a)Heat loss through the wall
b)Temperature of the interface
c)Assuming that the contact between the two
layers has a resistance of 0.50
0
F .h.ft
2
/Btu,
what is the new heat loss?
Mass Transfer
The most common cause of diffusion (
driving force) is a concentration gradient.
The driving force tends to move a
component in a direction as to equalize
the concentrations ( destroy the gradient).
In R.O. the activity gradient is the driving
force for diffusion of water from low
concentration to higher concentration of
water.
Diffusion can be at molecular transfer
through stagnant layers. Eddy diffusion
is related to mixing ( Macroscopic level)
and turbulent flow.
The most common cause of diffusion (
driving force) is a concentration gradient.
The driving force tends to move a
component in a direction as to equalize
the concentrations ( destroy the gradient).
In R.O. the activity gradient is the driving
force for diffusion of water from low
concentration to higher concentration of
water.
Diffusion can be at molecular transfer
through stagnant layers. Eddy diffusion
is related to mixing ( Macroscopic level)
and turbulent flow.
Theory of Diffusion
As we have studied, there are similarities
(analogy) between conduction of heat and
transfer of mass by diffusion. The driving
force in heat transfer is temperature
gradient and in diffusion is concentration
gradient.
Differences between mass and heat
transfer result from the fact that heat is
not a substance but energy in transit
whereas diffusion is the physical flow of
material.
As we have studied, there are similarities
(analogy) between conduction of heat and
transfer of mass by diffusion. The driving
force in heat transfer is temperature
gradient and in diffusion is concentration
gradient.
Differences between mass and heat
transfer result from the fact that heat is
not a substance but energy in transit
whereas diffusion is the physical flow of
material.
Four types of situations
1) Only one component A of the mixture is
transferred to or from the interface ( Ex:
Absorption of CO2 in amine solution)
The diffusion of component A is balanced
by an equal and opposite diffusion of a
component B ( Ex: Distillation: There is no
net volume flow in the gas
A and B diffuse in the opposite direction
with different flows ( from a surface of a
catalyst)
Different components diffuse in the same
direction with different rates ( membrane
separation).
1) Only one component A of the mixture is
transferred to or from the interface ( Ex:
Absorption of CO2 in amine solution)
The diffusion of component A is balanced
by an equal and opposite diffusion of a
component B ( Ex: Distillation: There is no
net volume flow in the gas
A and B diffuse in the opposite direction
with different flows ( from a surface of a
catalyst)
Different components diffuse in the same
direction with different rates ( membrane
separation).
Diffusion quantities
1) velocity u defined usually as
length.time
-1
2) Flux across a plane N (mol.area
-1
.time
-
1
)
3) Flux relative to a plane of zero velocity
J ( mol.area
-1
.time
-1
)
4) Concentration c and molar density ρ
M
5) concentration gradient dc/dt where b is
the length of the path perpendicular to the
area across which diffusion occurs.
1) velocity u defined usually as
length.time
-1
2) Flux across a plane N (mol.area
-1
.time
-
1
)
3) Flux relative to a plane of zero velocity
J ( mol.area
-1
.time
-1
)
4) Concentration c and molar density ρ
M
5) concentration gradient dc/dt where b is
the length of the path perpendicular to the
area across which diffusion occurs.
Velocities , Molar flow, Flux
Several velocities are needed to describe
the movements of individual substances
and the total phase. Since absolute
motion has no meaning, any velocity
should have a reference state of rest.
The total molal flow N( moles by unit time
and area)
N= ρ
M.u
0
Several velocities are needed to describe
the movements of individual substances
and the total phase. Since absolute
motion has no meaning, any velocity
should have a reference state of rest.
The total molal flow N( moles by unit time
and area)
N= ρ
M.u
0
ρ
M is the molar density of the mixture
u
0 is the volumetric average velocity
For components A & B:
N
A= c
A.u
A
N
B= c
B.u
B
Diffusivities are defined relatively to global
movement:
J
A= c
A.u
A -c
A.u
0= c
A( u
A-u
0)
J
B= c
B.u
B -c
B.u
0= c
B( u
B-u
0)
M is the molar density of the mixture
u
0 is the volumetric average velocity
For components A & B:
N
A= c
A.u
A
N
B= c
B.u
B
Diffusivities are defined relatively to global
movement:
J
A= c
A.u
A -c
A.u
0= c
A( u
A-u
0)
J
B= c
B.u
B -c
B.u
0= c
B( u
B-u
0)
Diffusion Flux: J
Diffusion of A in B
Diffusion of B in A:
For ideal gases: D
A,B= D
B,A
Diffusion of A in B
Diffusion of B in A:
For ideal gases: D
A,B= D
B,A
For Ideal Gases
Since molar density does not depend on
composition:
C
A+C
B= ρ
M= P/RT
At constant pressure and temperature:
dC
A +dC
B= dρ
M=
0
For equi-molar diffusion, we can write
N
A= J
A= ( D
A,B. ρ
M/∆x) ( y
A,1-
y
A,2)
Since molar density does not depend on
composition:
C
A+C
B= ρ
M= P/RT
At constant pressure and temperature:
dC
A +dC
B= dρ
M=
0
For equi-molar diffusion, we can write
N
A= J
A= ( D
A,B. ρ
M/∆x) ( y
A,1-
y
A,2)
For Liquids
C
A.M
A+ C
B.M
B= ρ= Constant
M
A.dC
A+ M
B.dC
B= 0
For Liquids: D
A,B= D
B,A
C
A.M
A+ C
B.M
B= ρ= Constant
M
A.dC
A+ M
B.dC
B= 0
For Liquids: D
A,B= D
B,A
General Equation
N
A= C
A.u
A
J
A= C
A.u
A-C
A.u
0= N
A-C
A.u
0
For gases:
C
A= ρ
M. y
A and u
0= N/ρ
M
For gases and sometimes for liquids, we
can apply:
N
A= y
A.N-D
A,B. (dC
A/dx)
N
A= C
A.u
A
J
A= C
A.u
A-C
A.u
0= N
A-C
A.u
0
For gases:
C
A= ρ
M. y
A and u
0= N/ρ
M
For gases and sometimes for liquids, we
can apply:
N
A= y
A.N-D
A,B. (dC
A/dx)
Diffusion of one component in a mixture of
gases
In this condition: N= N
A
Therefore N
A= y
A.N
A-D
A,B. (dC
A/dx)
N
A( 1-y
A) = -D
A,B. ρ
M(dy
A/dx)
Rearranging:
(N
A.∆x/D
A,B. ρ
M)= -∫[dy
A/(1-y
A)]= ln [(
1-y
A2)/( 1-y
A,1)]
Finally:
N
A= ( D
A,B. ρ
M/∆x). ln [( 1-y
A2)/(
1-y
A,1)]
gases
In this condition: N= N
A
Therefore N
A= y
A.N
A-D
A,B. (dC
A/dx)
N
A( 1-y
A) = -D
A,B. ρ
M(dy
A/dx)
Rearranging:
(N
A.∆x/D
A,B. ρ
M)= -∫[dy
A/(1-y
A)]= ln [(
1-y
A2)/( 1-y
A,1)]
Finally:
N
A= ( D
A,B. ρ
M/∆x). ln [( 1-y
A2)/(
1-y
A,1)]
Class Work
For a diffusion of solute A through a layer
of gas to an absorbing liquid, we have :
y
A,1= 0.20 and y
A,2= 0.10.
1) Compare the transfer rate N
A between
one-way diffusion and the equi-molar
diffusion ( hint: calculate the ratio of the
two fluxes)
2) What is the composition of A half way
through the layer for one-way diffusion
and equi-molar diffusion?
For a diffusion of solute A through a layer
of gas to an absorbing liquid, we have :
y
A,1= 0.20 and y
A,2= 0.10.
1) Compare the transfer rate N
A between
one-way diffusion and the equi-molar
diffusion ( hint: calculate the ratio of the
two fluxes)
2) What is the composition of A half way
through the layer for one-way diffusion
and equi-molar diffusion?
Incompressible
Fluids
Fluids
Introduction:
Chemical Engineers are
always concerned with the
flow of the fluids through
the pipes, tubes and
channels with non circular
cross-section
Chemical Engineers are
always concerned with the
flow of the fluids through
the pipes, tubes and
channels with non circular
cross-section
Shear Stress & Skin Friction in Pipes
Considering a steady state flow of fluid of
constant density in fully developed flow
trough a horizontal pipe.
From Figure 5.1 page 95: A small element
of the fluid is considered as disk-shaped
element of fluid with the axis of the tube,
of radius r and length dL.
Let the fluid pressures be p and (p+dp)
respectively.
Since the element has a velocity, there is
a shear force F
Sopposing the flow will
exist.
Considering a steady state flow of fluid of
constant density in fully developed flow
trough a horizontal pipe.
From Figure 5.1 page 95: A small element
of the fluid is considered as disk-shaped
element of fluid with the axis of the tube,
of radius r and length dL.
Let the fluid pressures be p and (p+dp)
respectively.
Since the element has a velocity, there is
a shear force F
Sopposing the flow will
exist.
MOMEMTUM EQUATION
Applying the momentum equation:
∑ F = p
a.S
a–p
b.S
b+ F
W–F
g
Where:
1) p
aand p
b= inlet and outlet pressures
S
aand S
b= inlet and outlet cross
sections.
F
w= net force of wall of channel on nfluid
F
g= component of force of gravity
Applying the momentum equation:
∑ F = p
a.S
a–p
b.S
b+ F
W–F
g
Where:
1) p
aand p
b= inlet and outlet pressures
S
aand S
b= inlet and outlet cross
sections.
F
w= net force of wall of channel on nfluid
F
g= component of force of gravity
Applying the momentum equation
between the two faces of the disk (Figure
5.1): ∑ F = p
a.S
a–p
b.S
b+ F
W–F
g
Since the fluid is fully developed ∑ F =
0 .
p
a.S
a= ∏r
2
p
p
b.S
b= ∏r
2
(p+dp)
F
W= 2∏.r.dL.ζ
F
g = 0 ( horizontal tube)
between the two faces of the disk (Figure
5.1): ∑ F = p
a.S
a–p
b.S
b+ F
W–F
g
Since the fluid is fully developed ∑ F =
0 .
p
a.S
a= ∏r
2
p
p
b.S
b= ∏r
2
(p+dp)
F
W= 2∏.r.dL.ζ
F
g = 0 ( horizontal tube)
Momentum Equation
∑ F= ∏r
2
p -∏r
2
(p+dp)-2∏.r.dL.ζ= 0
Dividing by ∏r
2
dL, we will obtain: ( dp/dL)
+ (2ζ/r) = 0
In steady flow, either laminar or
turbulent: the equation can be applied to
the entire cross section:
( dp/dL) + (2ζ
w/r
w) = 0 where
ζ
wis the shear stree at the wall
r
w is the radius of the tube.
∑ F= ∏r
2
p -∏r
2
(p+dp)-2∏.r.dL.ζ= 0
Dividing by ∏r
2
dL, we will obtain: ( dp/dL)
+ (2ζ/r) = 0
In steady flow, either laminar or
turbulent: the equation can be applied to
the entire cross section:
( dp/dL) + (2ζ
w/r
w) = 0 where
ζ
wis the shear stree at the wall
r
w is the radius of the tube.
Subtracting the two above equations, we
will obtain:
ζ
w/r
w = ζ/r
At r=0 ζ=0 ( Figure 5.2 page 96)
will obtain:
ζ
w/r
w = ζ/r
At r=0 ζ=0 ( Figure 5.2 page 96)
Skin friction & wall shear
Bernoulli equation:
( p
A/ρ)+ g.Z
A+ (α
A.V
A
2
/2) = ( p
B/ρ)+ g.Z
B+ (α
B
.V
B
2
/2) + h
f
Applying Bernoulli equation over a definite
length L of the complete stream in
horizontal tube, we will have :
( p
A/ρ) = ( p
A-∆p
s)/ρ+ h
fsor (∆p
s/ρ) =
h
fs
We assume :same height Z
A=Z
B ; same
velocity: V
A=V
B
p
A-p
B= ∆p= pressure drop between two
points
The only kind of friction is skin friction h
fs
Bernoulli equation:
( p
A/ρ)+ g.Z
A+ (α
A.V
A
2
/2) = ( p
B/ρ)+ g.Z
B+ (α
B
.V
B
2
/2) + h
f
Applying Bernoulli equation over a definite
length L of the complete stream in
horizontal tube, we will have :
( p
A/ρ) = ( p
A-∆p
s)/ρ+ h
fsor (∆p
s/ρ) =
h
fs
We assume :same height Z
A=Z
B ; same
velocity: V
A=V
B
p
A-p
B= ∆p= pressure drop between two
points
The only kind of friction is skin friction h
fs
Relation between skin friction and shear
stress
Combining the momentum equation and
Bernoulli equation, we will obtain:
h
fs = (2.ζ
w.L)/( ρ.r
w)= (4/ρ). Ζ
w. (L/D)
stress
Combining the momentum equation and
Bernoulli equation, we will obtain:
h
fs = (2.ζ
w.L)/( ρ.r
w)= (4/ρ). Ζ
w. (L/D)
Friction factor in turbulent flow
For turbulent flow, the fanny factor (f) is
usually used:
f= 2.ζ
w/ρ.V
2
V
2
/2 is the velocity head in Bernoulli
equation.
For turbulent flow, the fanny factor (f) is
usually used:
f= 2.ζ
w/ρ.V
2
V
2
/2 is the velocity head in Bernoulli
equation.
Relationship between f and f
hs
h
fs= (2.ζ
w.L)/( ρ.r
w)= (4/ρ). ζ
w. (L/D) =
(∆p
s/ρ)
h
fs= 4.f . ( L.V
2
/ 2D)
Where:
f= ( ∆p
s. D)/( 2L.ρ.V
2
) or : (∆p
s/ L)=
2f..ρ.V
2
/D
hs
h
fs= (2.ζ
w.L)/( ρ.r
w)= (4/ρ). ζ
w. (L/D) =
(∆p
s/ρ)
h
fs= 4.f . ( L.V
2
/ 2D)
Where:
f= ( ∆p
s. D)/( 2L.ρ.V
2
) or : (∆p
s/ L)=
2f..ρ.V
2
/D
Friction Factor Chart
For design purposes, we need a friction
factor chart. The chart is a log-log chart
relating the friction factor f to the
Reynolds number.
For turbulent flow, it is more convenient
to use the following equations than the
chart.
50,000 < Re< 10
6
f= 0.046 Re
-0.2
3000 < Re< 3.10
6
f= 0.0014 + (
0.125/Re
0.32
)
R
e= D.V.ρ/ μ( Dimensionless)
(∆p
s/ L)= 2f.ρ.V
2
/D
For design purposes, we need a friction
factor chart. The chart is a log-log chart
relating the friction factor f to the
Reynolds number.
For turbulent flow, it is more convenient
to use the following equations than the
chart.
50,000 < Re< 10
6
f= 0.046 Re
-0.2
3000 < Re< 3.10
6
f= 0.0014 + (
0.125/Re
0.32
)
R
e= D.V.ρ/ μ( Dimensionless)
(∆p
s/ L)= 2f.ρ.V
2
/D
Class work ( SI UNITS)
Water flows at 50
0
F (μ= 1.307. 10
-3
N.s
/ m
2
) through a long horizontal plastic
pipe having an inside diameter of 3in.
1) If the velocity is 8ft/s, calculate the
pressure drop in lbf/in
2
per 100 ft of
length.
A) Reynolds number,
B) f from equation for turbulent flow or
from chart,
C) Apply equation.
2) If the pressure drop must be limited to 2
lbf/in
2
per 100 ft of pipe, what is the
maximum allowable velocity?
Water flows at 50
0
F (μ= 1.307. 10
-3
N.s
/ m
2
) through a long horizontal plastic
pipe having an inside diameter of 3in.
1) If the velocity is 8ft/s, calculate the
pressure drop in lbf/in
2
per 100 ft of
length.
A) Reynolds number,
B) f from equation for turbulent flow or
from chart,
C) Apply equation.
2) If the pressure drop must be limited to 2
lbf/in
2
per 100 ft of pipe, what is the
maximum allowable velocity?
Form frictions losses
Bernoulli equation:
( p
A/ρ)+ g.Z
A+ (α
A.V
A
2
/2) = ( p
B/ρ)+ g.Z
B+
(α
B.V
B
2
/2) + h
f
Form friction losses or singularities should
be incorporated in the formula for friction
losses as:
h
f= ( 4.f.(L/D) + K
c+ K
e+ K
f) . ( V
2
/2)
K
c.V
2
/2 is the contraction loss at the
entrance of the pipe
K
e.V
2
/2is the expansion loss at the exit of
the pipe.
K
f. V
2
/2 is the friction loss in a globe valve.
Bernoulli equation:
( p
A/ρ)+ g.Z
A+ (α
A.V
A
2
/2) = ( p
B/ρ)+ g.Z
B+
(α
B.V
B
2
/2) + h
f
Form friction losses or singularities should
be incorporated in the formula for friction
losses as:
h
f= ( 4.f.(L/D) + K
c+ K
e+ K
f) . ( V
2
/2)
K
c.V
2
/2 is the contraction loss at the
entrance of the pipe
K
e.V
2
/2is the expansion loss at the exit of
the pipe.
K
f. V
2
/2 is the friction loss in a globe valve.
Class Work ( SI UNITS)
Crude oil having a specific gravity of 0.93
and a viscosity of 4 cP is draining by
gravity from the bottom of a tank. The
depth of liquid from the draw off
connection is 6 ft. the line from the draw
off is 3 in diameter schedule 40 pipe. Its
length is 45 m and it contains one elbow
and two globe valves. The oil discharges
into atmosphere 9 m below the draw off
connection of the tank.
1) What flow rate can be expected
through the line.
Using Bernouilli equation assume:
Crude oil having a specific gravity of 0.93
and a viscosity of 4 cP is draining by
gravity from the bottom of a tank. The
depth of liquid from the draw off
connection is 6 ft. the line from the draw
off is 3 in diameter schedule 40 pipe. Its
length is 45 m and it contains one elbow
and two globe valves. The oil discharges
into atmosphere 9 m below the draw off
connection of the tank.
1) What flow rate can be expected
through the line.
Using Bernouilli equation assume:
Mach Number
The Mach Number ( Ma) is defined as the
ratio of u, speed of the fluid, to a, speed
of the sound in the fluid under the
conditions of flow.
Ma = u/a
Sonic Fluid: When the Mach number is
equal to unity. Speed of the fluid is equal
speed of the sound in the same at the
pressure and temperature of the fluid.
The Mach Number ( Ma) is defined as the
ratio of u, speed of the fluid, to a, speed
of the sound in the fluid under the
conditions of flow.
Ma = u/a
Sonic Fluid: When the Mach number is
equal to unity. Speed of the fluid is equal
speed of the sound in the same at the
pressure and temperature of the fluid.
Subsonic Flow: Mach number is less
than unity
Supersonic Flow:Mach number is higher
than unity.
The most important problem in
compressible fluids flow lie in the high
velocity range where Mach numbers are
higher to unity and flow are supersonic.
than unity
Supersonic Flow:Mach number is higher
than unity.
The most important problem in
compressible fluids flow lie in the high
velocity range where Mach numbers are
higher to unity and flow are supersonic.
Assumptions made
The flow is steady.
The flow is one-dimensional.
Velocity gradients within cross section are
neglected.
Friction only in walls
Shaft work is zero
Gravitational effects are negligeable
Fluid is an ideal gas.
The flow is steady.
The flow is one-dimensional.
Velocity gradients within cross section are
neglected.
Friction only in walls
Shaft work is zero
Gravitational effects are negligeable
Fluid is an ideal gas.
Basic relations
The continuity equation
Steady flow total-energy balance
The mechanical energy balance with wall
friction
The equation of the velocity of sound
The equation of state of ideal gas
The continuity equation
Steady flow total-energy balance
The mechanical energy balance with wall
friction
The equation of the velocity of sound
The equation of state of ideal gas
Continuity equation
dm= ρ.u.dS can be rewritten in the
logarithmic form:
ln ρ+ ln u + ln S =
constant
Differentiating the equation:
(dρ/ρ) + ( du/u) +
(dS/S) = 0
dm= ρ.u.dS can be rewritten in the
logarithmic form:
ln ρ+ ln u + ln S =
constant
Differentiating the equation:
(dρ/ρ) + ( du/u) +
(dS/S) = 0
Total-Energy Balance
A fluid in steady flow through a system,
entering at station a with velocity u
aand
enthalpy H
aand leaving at velocity u
band
enthalpy H
b. Assuming only heat added,
we will have:
(Q/m)= (H
b-H
a) + [( u
2
b
–u
2
a) /2]
A fluid in steady flow through a system,
entering at station a with velocity u
aand
enthalpy H
aand leaving at velocity u
band
enthalpy H
b. Assuming only heat added,
we will have:
(Q/m)= (H
b-H
a) + [( u
2
b
–u
2
a) /2]
Mechanical Energy balance
After using our assumptions, we will have:
(dp/ρ) + d(u
2
/2) +
(u
2
.f.dL)/2.r
H) = 0
After using our assumptions, we will have:
(dp/ρ) + d(u
2
/2) +
(u
2
.f.dL)/2.r
H) = 0
Velocity of Sound
The velocity of sound through a
continuous material is also called the
acoustical velocity.
The acoustical velocity is the velocity of a
very small compression wave moving
adiabatically and frictionlessly through the
medium.
Thermodynamically, the motion of a
sound wave is isentropic.
The velocity of sound through a
continuous material is also called the
acoustical velocity.
The acoustical velocity is the velocity of a
very small compression wave moving
adiabatically and frictionlessly through the
medium.
Thermodynamically, the motion of a
sound wave is isentropic.
Ideal Gas Equation
p= (R/M).ρ.T
For a pure substance or no
change in composition, we will
have:
(dp/p)= (dρ/ρ) +
(dT/T)
p= (R/M).ρ.T
For a pure substance or no
change in composition, we will
have:
(dp/p)= (dρ/ρ) +
(dT/T)
Enthalpy of a gas
Assuming c
pindependent of temperature,
the enthalpy of the gas at temperature T
is:
H= H
0+ c
p( T-T
0)
Where T
0is the reference
temperature. The differential
form is:
dH= c
p.dT
Assuming c
pindependent of temperature,
the enthalpy of the gas at temperature T
is:
H= H
0+ c
p( T-T
0)
Where T
0is the reference
temperature. The differential
form is:
dH= c
p.dT
Acoustic velocity in ideal gas
γ= c
p/c
v
γ= c
p/c
v
Mach Number in Ideal Gas
Asterisk conditions: The conditions were
u=a ( acoustic velocity) or Ma=1 are
called asterisk conditions. They are
denoted p
*
, T
*
, ρ
*
and H
*
.
Asterisk conditions: The conditions were
u=a ( acoustic velocity) or Ma=1 are
called asterisk conditions. They are
denoted p
*
, T
*
, ρ
*
and H
*
.
Stagnation Temperature
The stagnation temperature of a high
speed fluid is defined as the temperature
of the fluid brought adiabatically to a rest
position ( v=0) without shaft work.
Using the total energy equation without
heat exchange Q, we will obtain:
H
s-H = (u
2
/2)= where H
sis the
stagnation enthalpy where v
S=0
The stagnation temperature of a high
speed fluid is defined as the temperature
of the fluid brought adiabatically to a rest
position ( v=0) without shaft work.
Using the total energy equation without
heat exchange Q, we will obtain:
H
s-H = (u
2
/2)= where H
sis the
stagnation enthalpy where v
S=0
Using the enthalpy equation: H= H
0+ c
p.
( T-T
0), we will get: T
s= T + (u
2
/2c
p)
and H
s= H + (u
2
/2)
0+ c
p.
( T-T
0), we will get: T
s= T + (u
2
/2c
p)
and H
s= H + (u
2
/2)
Changes of gas properties
Equations to be applied for subsonic and
supersonic flows
Equations to be applied for subsonic and
supersonic flows
Velocity in Nozzle
In absence of friction, we can rewrite the
mechanical energy balance:
(dp/ρ)= -d( u
2
/2)
replacing ρby the previous equation, we
will obtain:
In absence of friction, we can rewrite the
mechanical energy balance:
(dp/ρ)= -d( u
2
/2)
replacing ρby the previous equation, we
will obtain:
Mach Number and pressure for isentropic flow
Critical Pressure Ratio r
c
c
Mass Velocity
Class work
Air enters a convergent-divergent nozzle at
555.6K and 20 atm. The throat area is one-
half that of the discharge of divergent section.
Assume isentropic flow.
1) Assume Mach number at the throat 0.8,
what are the values of the following quantities
at the throat: p, T, linear velocity, density and
mass velocity.
2) What are the values of p
*,
T
*
, u
*
,G
*
3) Assuming supersonic flow, what is the
Mach number at the discharge .
γ= 1.4; M=29 , R= 82.056 . 10
-3
atm.m
3
/kgmol.K
Air enters a convergent-divergent nozzle at
555.6K and 20 atm. The throat area is one-
half that of the discharge of divergent section.
Assume isentropic flow.
1) Assume Mach number at the throat 0.8,
what are the values of the following quantities
at the throat: p, T, linear velocity, density and
mass velocity.
2) What are the values of p
*,
T
*
, u
*
,G
*
3) Assuming supersonic flow, what is the
Mach number at the discharge .
γ= 1.4; M=29 , R= 82.056 . 10
-3
atm.m
3
/kgmol.K
INTRODUCTION
Heat transfer from a warmer fluid to a cooler
fluid, usually through a solid wall, is common
in chemical engineering.
Latent heat: heat transfer causing phase
change such as condensers and evaporators.
Sensible heat: rise in temperature in solid,
liquid or gas phases without change of phase.
Equipments:
we have :
a)Double pipe heat exchanger ( Figure 11.3
page 317)
b)Shell and tube heat exchanger ( Figure 11.1
page 316)
Heat transfer from a warmer fluid to a cooler
fluid, usually through a solid wall, is common
in chemical engineering.
Latent heat: heat transfer causing phase
change such as condensers and evaporators.
Sensible heat: rise in temperature in solid,
liquid or gas phases without change of phase.
Equipments:
we have :
a)Double pipe heat exchanger ( Figure 11.3
page 317)
b)Shell and tube heat exchanger ( Figure 11.1
page 316)
Counter-Current & Parallel
Flow
Parallel Flow:
When the two fluids enter at the same end
of the heat exchanger ( Figure 11.4 b
page 318)
Counter-current Flow:
When the two fluids enter at different ends
of the heat exchanger ( Figure 11.4.a
page 318)
Flow
Parallel Flow:
When the two fluids enter at the same end
of the heat exchanger ( Figure 11.4 b
page 318)
Counter-current Flow:
When the two fluids enter at different ends
of the heat exchanger ( Figure 11.4.a
page 318)
Energy Balances ( steady state)
Rates of heat transfer are based on energy
balances.
In heat exchangers, we can neglect any shaft
work, mechanical, potential and kinetic
energies compared to the heat exchange.
Therefore, the total energy conservation for
each stream can be written as:
m ( H
b-H
a) = q
m= mass flow rate of the considered stream
H
b,H
a= enthalpies by unit mass at the exit
and entrance of the considered stream.
q= rate of heat transfer from or to the
considered stream
Rates of heat transfer are based on energy
balances.
In heat exchangers, we can neglect any shaft
work, mechanical, potential and kinetic
energies compared to the heat exchange.
Therefore, the total energy conservation for
each stream can be written as:
m ( H
b-H
a) = q
m= mass flow rate of the considered stream
H
b,H
a= enthalpies by unit mass at the exit
and entrance of the considered stream.
q= rate of heat transfer from or to the
considered stream
Energy balances
For the hot stream:
m
h( H
hb–H
ha) = q
h
For the cold stream:
m
c( H
cb–H
ca) = q
c
Since the hot stream gives heat :
q
his negative and q
cis positive q
h= -q
c
For the hot stream:
m
h( H
hb–H
ha) = q
h
For the cold stream:
m
c( H
cb–H
ca) = q
c
Since the hot stream gives heat :
q
his negative and q
cis positive q
h= -q
c
Energy balances
m
h( H
ha–H
hb)= m
c( H
cb–
H
ca) = q
m
h( H
ha–H
hb)= m
c( H
cb–
H
ca) = q
Energy balances for sensible heat
m
h.c
ph( T
ha–T
hb)= m
c.c
pc( T
cb
–T
ca) = q
m
h.c
ph( T
ha–T
hb)= m
c.c
pc( T
cb
–T
ca) = q
In a Total Condenser
Condensation of a saturated vapor to saturated
liquid and increase of temperature of the
other fluid:
m
h. λ= m
c.c
pc( T
cb–T
ca)
= q
if condensate leaves condenser at temperature
lower than T
hb, we will have:
m
h.[λ+ c
ph(T
h-T
hb)] = m
c.c
pc(T
cb
–T
ca)= q
Condensation of a saturated vapor to saturated
liquid and increase of temperature of the
other fluid:
m
h. λ= m
c.c
pc( T
cb–T
ca)
= q
if condensate leaves condenser at temperature
lower than T
hb, we will have:
m
h.[λ+ c
ph(T
h-T
hb)] = m
c.c
pc(T
cb
–T
ca)= q
DEFINITIONS
Heat flux: rate of heat transfer per unit
area. Heat fluxes can be based on the
internal or external area. The heat flux will
not be the same.
Average temperature of fluid stream:
If the fluid is heated, the maximum
temperature is located at the wall of the
tube and decreases towards the center of
the stream.
Heat flux: rate of heat transfer per unit
area. Heat fluxes can be based on the
internal or external area. The heat flux will
not be the same.
Average temperature of fluid stream:
If the fluid is heated, the maximum
temperature is located at the wall of the
tube and decreases towards the center of
the stream.
DEFINITIONS
Overall Heat-Transfer coefficient:
In a heat exchanger, the driving force is
taken as ( T
h-T
c), where T
his the average
temperature of the hot fluid and T
cis the
average temperature of the cold fluid.
The difference (T
h-T
c) is the overall local
temperature difference ∆T. Using Figure
11.4, ∆T changes considerably from point to
point along the tube.
Since the flux depends on ∆T, the flux also
varies along the tube.
The local flux is therefore related to the local
∆T.
Overall Heat-Transfer coefficient:
In a heat exchanger, the driving force is
taken as ( T
h-T
c), where T
his the average
temperature of the hot fluid and T
cis the
average temperature of the cold fluid.
The difference (T
h-T
c) is the overall local
temperature difference ∆T. Using Figure
11.4, ∆T changes considerably from point to
point along the tube.
Since the flux depends on ∆T, the flux also
varies along the tube.
The local flux is therefore related to the local
∆T.
Local overall heat-transfer coefficient
(dq/dA)= U.∆T= U.( T
h-T
c)
Or:
U= (dq/dA)/∆T
dq/dA= local flux
∆T= local difference of
temperature
(dq/dA)= U.∆T= U.( T
h-T
c)
Or:
U= (dq/dA)/∆T
dq/dA= local flux
∆T= local difference of
temperature
DEFINITION OF U
For a tubular heat exchanger, it is
necessary to specify the area ( inside or
outside).
For the outside area A
0U
0
For the inside area A
iU
i
Since ∆T and dq are independent:
(U
0/U
i) = ( D
i/D
0)
For a tubular heat exchanger, it is
necessary to specify the area ( inside or
outside).
For the outside area A
0U
0
For the inside area A
iU
i
Since ∆T and dq are independent:
(U
0/U
i) = ( D
i/D
0)
AFTER INTEGRATION
ln ( ∆T
1/ ∆T
2)= [U(∆T
2-∆T
1].A
T/q
T
q
T= [U(∆T
2-∆T
1)].A
T/ ln ( ∆T
1/
∆T
2)= U.A
T.∆T
L
∆T
L = ∆T
2-∆T
1/ ln ( ∆T
1/ ∆T
2)
ln ( ∆T
1/ ∆T
2)= [U(∆T
2-∆T
1].A
T/q
T
q
T= [U(∆T
2-∆T
1)].A
T/ ln ( ∆T
1/
∆T
2)= U.A
T.∆T
L
∆T
L = ∆T
2-∆T
1/ ln ( ∆T
1/ ∆T
2)
Individual heat transfer coefficient
The film coefficient for the warm and
cold fluids:
h
h= (dq/dA)/(T
h-T
w); h
c= (dq/dA)/( T
w-
T
c).
Conduction through the wall:
dq/dA= -k( dT/dy)
w
Eliminating dq/dA
h=-k (dT/dy)
w/ (T-T
w)
The film coefficient for the warm and
cold fluids:
h
h= (dq/dA)/(T
h-T
w); h
c= (dq/dA)/( T
w-
T
c).
Conduction through the wall:
dq/dA= -k( dT/dy)
w
Eliminating dq/dA
h=-k (dT/dy)
w/ (T-T
w)
dq/dA
0= (T
h-T
c)/ [(D
0/h
iD
i)+ x
w.D
0/k
mD
L+
1/h
0]
D
L= (D
0-D
i)/ln(D
0/D
i)]
0= (T
h-T
c)/ [(D
0/h
iD
i)+ x
w.D
0/k
mD
L+
1/h
0]
D
L= (D
0-D
i)/ln(D
0/D
i)]
CLASS WORK
Methyl alcohol flowing in the inner pipe of
a double pipe exchanger is cooled with
water flowing in the jacket. The inner pipe
is 1in ( 25mm) Schedule 40 steel pipe .
The thermal conductivity could be taken
from table 11.1 page 331. thermal
conductivity of steel is 45 W/m.
0
C.
1) What is the overall coefficient based on
the outside area of the inner pipe?
1in Schedule 40 : Do= 1.315 in ; Di=
1.049 in; thickness w= 0.133 in
Methyl alcohol flowing in the inner pipe of
a double pipe exchanger is cooled with
water flowing in the jacket. The inner pipe
is 1in ( 25mm) Schedule 40 steel pipe .
The thermal conductivity could be taken
from table 11.1 page 331. thermal
conductivity of steel is 45 W/m.
0
C.
1) What is the overall coefficient based on
the outside area of the inner pipe?
1in Schedule 40 : Do= 1.315 in ; Di=
1.049 in; thickness w= 0.133 in
Class work
Aniline is to be cooled from 200 to 150F in
a double-pipe heat exchanger having an
area of 70ft
2
. Toluene is used for cooling
at 8600lb/h at temperature of 100F. The
heat exchanger consists of 1.25 in and 2
in schedule 40 pipe. The aniline flow is
10000lb/hr.
1) WHAT IS THE OUTLET TEMPERATURE
OF TOLUENE, LMTD AND U FOR
COUNTERCURRENT AND PARALLEL
FLOWS.
Aniline is to be cooled from 200 to 150F in
a double-pipe heat exchanger having an
area of 70ft
2
. Toluene is used for cooling
at 8600lb/h at temperature of 100F. The
heat exchanger consists of 1.25 in and 2
in schedule 40 pipe. The aniline flow is
10000lb/hr.
1) WHAT IS THE OUTLET TEMPERATURE
OF TOLUENE, LMTD AND U FOR
COUNTERCURRENT AND PARALLEL
FLOWS.
CLASSIFICATIONS
I) Convection can be classified as
natural or forced depending on how the
fluid motion is initiated.
a) In forced convection; the fluid is forced
to move over a surface or in a pipe by
external means such as a pump or a fan.
b) In natural convection, any fluid motion is
caused by natural means such as
buoyancy effect, which manifests itself as
the rise of warmer fluid and the fall of
cooler fluid.
I) Convection can be classified as
natural or forced depending on how the
fluid motion is initiated.
a) In forced convection; the fluid is forced
to move over a surface or in a pipe by
external means such as a pump or a fan.
b) In natural convection, any fluid motion is
caused by natural means such as
buoyancy effect, which manifests itself as
the rise of warmer fluid and the fall of
cooler fluid.
II) Convection can also be classified as
external or internal:
a)External: Fluid forced to flow over a
surface
b)Internal: Fluid forced to flow inside a pipe
external or internal:
a)External: Fluid forced to flow over a
surface
b)Internal: Fluid forced to flow inside a pipe
Heat transfer by convection
Heat transfer by convection is similar to
heat transfer by conduction because both
mechanism require a presence of material
medium.
However, conduction requires a solid
surface but convection requires a fluid in
motion.
Heat transfer by convection is similar to
heat transfer by conduction because both
mechanism require a presence of material
medium.
However, conduction requires a solid
surface but convection requires a fluid in
motion.
Newton’s law of cooling
Despite the complexity of convection, the rate
of heat transfer is observed to be proportional
to the temperature difference and is
expressed by the Newton’s law of cooling:
Q
conv= h.A
S(T
S-T
∞)
h= convection heat transfer coefficient (
W/m
2
.
0
C)
A
S= Surface area of heat transfer m
2
T
S= Temperature of the surface
0
C
T
∞= Temperature of the fluid far from the
surface
Despite the complexity of convection, the rate
of heat transfer is observed to be proportional
to the temperature difference and is
expressed by the Newton’s law of cooling:
Q
conv= h.A
S(T
S-T
∞)
h= convection heat transfer coefficient (
W/m
2
.
0
C)
A
S= Surface area of heat transfer m
2
T
S= Temperature of the surface
0
C
T
∞= Temperature of the fluid far from the
surface
No Slip Condition?
Observations show that fluid in motion comes to
a complete stop at the surface and zero velocity
is assumed.
Therefore, the fluid in direct contact with the
surface sticks to the surface due to viscous
effects and there is no slip. This is known as
No-Slip Condition.
An implication of no-slip condition is that heat
transfer from the surface to the fluid is pure
conduction and Fourier's law is applied.
q
conv=q
cond= -k ( dT/dy)
y=0
Observations show that fluid in motion comes to
a complete stop at the surface and zero velocity
is assumed.
Therefore, the fluid in direct contact with the
surface sticks to the surface due to viscous
effects and there is no slip. This is known as
No-Slip Condition.
An implication of no-slip condition is that heat
transfer from the surface to the fluid is pure
conduction and Fourier's law is applied.
q
conv=q
cond= -k ( dT/dy)
y=0
H=convection heat ransfer coefficient
Equating the two equations, we will
obtain:
h= -k [( dT/dy)
y=0/ ( T
S-T
∞)]
Equating the two equations, we will
obtain:
h= -k [( dT/dy)
y=0/ ( T
S-T
∞)]
Nusselt Number?
A dimensionless number that measures
the effects of heat convection on heat
conduction:
Nu= h.L
C/ k
L
c= is a characteristic length.
1) High Nusselt number means convection
more important than conduction.
2) Low Nusselt numbers means conduction
more important that convection.
A dimensionless number that measures
the effects of heat convection on heat
conduction:
Nu= h.L
C/ k
L
c= is a characteristic length.
1) High Nusselt number means convection
more important than conduction.
2) Low Nusselt numbers means conduction
more important that convection.
Nusselt Number
q
conv= h.∆T and q
cond= k. ∆T/L
(q
conv/ q
cond)= h.∆T.L/k.∆T=
hL/k= Nu
q
conv= h.∆T and q
cond= k. ∆T/L
(q
conv/ q
cond)= h.∆T.L/k.∆T=
hL/k= Nu
Viscous & inviscid regions of flow
Convection heat transfer is related to
fluid mechanics and fluid flows need
to be included.
Two layers of a fluid in motion create
friction or an internal resistance to
flow called viscosity.
There is no fluid with zero viscosity.
Neglecting the viscous effects, we call
the flow inviscid.
Convection heat transfer is related to
fluid mechanics and fluid flows need
to be included.
Two layers of a fluid in motion create
friction or an internal resistance to
flow called viscosity.
There is no fluid with zero viscosity.
Neglecting the viscous effects, we call
the flow inviscid.
Internal & external Flow
Internal flow: Fluid forced to flow
in confined channel. Flow
completely bounded by solid
surfaces. Viscous effects
important.
External flow: fluid forced to flow
over a surface. In external flow,
viscous effects are limited
Internal flow: Fluid forced to flow
in confined channel. Flow
completely bounded by solid
surfaces. Viscous effects
important.
External flow: fluid forced to flow
over a surface. In external flow,
viscous effects are limited
Compressible/incompressible fluids
Compressible fluids (Gases): important
density change with pressure. A pressure
change of 0.01 atm can cause density of air
changes by 1%. Of atmospheric air.
Incompressible fluids ( Liquids): almost
no density change with pressure. For
example, a pressure of 210 atm causes the
density change of water at 1 atm by just 1%.
Incompressible gases approximation
depends on the Mach Number
Ma= u/a
Compressible fluids (Gases): important
density change with pressure. A pressure
change of 0.01 atm can cause density of air
changes by 1%. Of atmospheric air.
Incompressible fluids ( Liquids): almost
no density change with pressure. For
example, a pressure of 210 atm causes the
density change of water at 1 atm by just 1%.
Incompressible gases approximation
depends on the Mach Number
Ma= u/a
Condition for Incompressible gases
If density changes by a
maximum of 5% which is usually
the case at Ma < 0.3
Incompressibility effects of air
can be neglected at speeds lower
than 100 m/s.
If density changes by a
maximum of 5% which is usually
the case at Ma < 0.3
Incompressibility effects of air
can be neglected at speeds lower
than 100 m/s.
Laminar & Turbulent Flows
Laminar flow is smooth and
parabolic because of low
velocities ( high viscosity
liquids).
Turbulent is chaotic because of
high velocities ( low viscosity
gases).
Laminar flow is smooth and
parabolic because of low
velocities ( high viscosity
liquids).
Turbulent is chaotic because of
high velocities ( low viscosity
gases).
Steady & unsteady state flows
Steady means no change at a
point with time dx/dt=0
Unsteady means change at a
point with time dx/dt ≠ 0
Steady means no change at a
point with time dx/dt=0
Unsteady means change at a
point with time dx/dt ≠ 0
Velocity boundary layer
Velocity boundary layer develops when a fluid
flows over a surface as a result of the fluid
layer adjacent to the surface assuming zero
velocity.
Considering the movement of a flow over a
flat plate.
Fluid flows with velocity V
Because of No-slip condition, velocity at the
surface is zero.
The velocity boundary layer is defined as a
thickness δfrom the surface, where the local
velocity u = 0.99V.
Velocity boundary layer develops when a fluid
flows over a surface as a result of the fluid
layer adjacent to the surface assuming zero
velocity.
Considering the movement of a flow over a
flat plate.
Fluid flows with velocity V
Because of No-slip condition, velocity at the
surface is zero.
The velocity boundary layer is defined as a
thickness δfrom the surface, where the local
velocity u = 0.99V.
Surface shear stress
Newton’s law:
ζ
S= -μ(du/dy)
y=o
Newton’s law:
ζ
S= -μ(du/dy)
y=o
Thermal Boundary Layer
In the same way as Velocity Boundary
layer, Thermal boundary layer develops
when a fluid at a specified temperature T
∞
flows over a surface at different
temperature T
S
Thermal boundary layer δ
Tis defined as
the position in the fluid from the surface
where T-T
S= 0.99 ( T-T
∞)
If T
S=0 T=0.99. T
∞
In the same way as Velocity Boundary
layer, Thermal boundary layer develops
when a fluid at a specified temperature T
∞
flows over a surface at different
temperature T
S
Thermal boundary layer δ
Tis defined as
the position in the fluid from the surface
where T-T
S= 0.99 ( T-T
∞)
If T
S=0 T=0.99. T
∞
Prandtl Number
The relative thickness of the velocity and
the thermal boundary layers is best
described by the dimensionless parameter
Prandtl number defined as :
Pr= ( molecular diffusivity of momentum)/
molecular diffusivity of heat)
Pr= ν/α= μc
p/k
The relative thickness of the velocity and
the thermal boundary layers is best
described by the dimensionless parameter
Prandtl number defined as :
Pr= ( molecular diffusivity of momentum)/
molecular diffusivity of heat)
Pr= ν/α= μc
p/k
Reynolds Number
By definition: N
Re= Inertia forces/ Viscous
forces
Inertia forces ρV
2
L
2
Viscous forces: VL/νwith ν= μ/ρ
N
re= ρVL/μ
Laminar flows at low Reynolds numbers
Turbulent flows at high Reynolds
numbers.
By definition: N
Re= Inertia forces/ Viscous
forces
Inertia forces ρV
2
L
2
Viscous forces: VL/νwith ν= μ/ρ
N
re= ρVL/μ
Laminar flows at low Reynolds numbers
Turbulent flows at high Reynolds
numbers.
Continuity equation=Mass balance
X-axis flow direction and y-axis the
normal direction:
∂u/∂x + ∂v/∂y =0 if the flow is parallel
then v=0
One-direction flow variation
∂u/∂x =0 u= f (y)
X-axis flow direction and y-axis the
normal direction:
∂u/∂x + ∂v/∂y =0 if the flow is parallel
then v=0
One-direction flow variation
∂u/∂x =0 u= f (y)
Momentum Equation
We will study one –direction ( x-direction)
ρ( u. ∂u/dx +v. ∂u/∂y) = μ∂
2
u/ ∂y
2
-∂P/∂x
We will study one –direction ( x-direction)
ρ( u. ∂u/dx +v. ∂u/∂y) = μ∂
2
u/ ∂y
2
-∂P/∂x
Energy Equation
Steady state accumulation equal zero
One direction
0= k∂
2
T/ ∂y
2
+μ( ∂u/∂y)
2
kd
2
T/dy
2
= -μ( V/L)
2
Steady state accumulation equal zero
One direction
0= k∂
2
T/ ∂y
2
+μ( ∂u/∂y)
2
kd
2
T/dy
2
= -μ( V/L)
2
INTRODUCTION
Heat transfer is governed by three distinct
mechanisms: convection, conduction, and
radiation.
Unlike convection or conduction, heat transfer
through radiation does not occur through a
particular medium.
To understand this phenomenon one must enter
into the atomic or quantum realm. All atoms, at
finite temperatures, are continuously in motion.
Consequently, it may be understood that the
mechanism of radiation is derived from the
energetic vibrations and oscillations of these
atomic particles, namely electrons.
Heat transfer is governed by three distinct
mechanisms: convection, conduction, and
radiation.
Unlike convection or conduction, heat transfer
through radiation does not occur through a
particular medium.
To understand this phenomenon one must enter
into the atomic or quantum realm. All atoms, at
finite temperatures, are continuously in motion.
Consequently, it may be understood that the
mechanism of radiation is derived from the
energetic vibrations and oscillations of these
atomic particles, namely electrons.
METAL WORK BY HEAT
Wavelength of Thermal
Radiation
Radiation
THEORY OF RADIATION
James Clark Maxwell established in 1864 the theory of
radiation.
Thermal radiation is a direct result of the movements of
atoms and molecules in a material. Since these atoms and
molecules are composed of charged particles (protons and
electrons), their movements result in the emission of
electromagnetic radiation, which carries energy away
from the surface.
Since the amount of emitted radiation increases with
increasing temperature, a net transfer of energy from
higher temperatures to lower temperatures results.
James Clark Maxwell established in 1864 the theory of
radiation.
Thermal radiation is a direct result of the movements of
atoms and molecules in a material. Since these atoms and
molecules are composed of charged particles (protons and
electrons), their movements result in the emission of
electromagnetic radiation, which carries energy away
from the surface.
Since the amount of emitted radiation increases with
increasing temperature, a net transfer of energy from
higher temperatures to lower temperatures results.
Electromagnetic radiation
Electromagnetic radiation or waves
transport energy and all electromagnetic
waves travel at the speed of light in a
vacuum, which is c
0= 2.9979. 10
8
m/s.
Electromagnetic waves are characterized
by their frequency νor wavelength λ.
Electromagnetic radiation or waves
transport energy and all electromagnetic
waves travel at the speed of light in a
vacuum, which is c
0= 2.9979. 10
8
m/s.
Electromagnetic waves are characterized
by their frequency νor wavelength λ.
C= SPEED OF PROPAGATION
OF HEAT
The speed of propagation of a wave in a medium
is defined by : λ=c/ν
the speed of propagation in a medium is related
to the speed of light in a vacuum by: c=c
0/n
where n is the index of refraction of that
medium.
The refractive index is essentially:
a)unity for air and most gases.
b)1.5 for glass
c)1.33 for water
OF HEAT
The speed of propagation of a wave in a medium
is defined by : λ=c/ν
the speed of propagation in a medium is related
to the speed of light in a vacuum by: c=c
0/n
where n is the index of refraction of that
medium.
The refractive index is essentially:
a)unity for air and most gases.
b)1.5 for glass
c)1.33 for water
ν=frequency of an electromagnetic
wave
The commonly used unit of wavelength is
micrometer (μm).
Unlike wavelength (λ) and speed of
propagation (c), the frequency of an
electromagnetic wave ( ν) depends only
on the source and independent on the
medium through which the wave travels.
the frequency ( oscillations per second) of
an electromagnetic wave can range from
less than a million (10
6
) Hz to a septilion
(10
24
) Hz or higher.
wave
The commonly used unit of wavelength is
micrometer (μm).
Unlike wavelength (λ) and speed of
propagation (c), the frequency of an
electromagnetic wave ( ν) depends only
on the source and independent on the
medium through which the wave travels.
the frequency ( oscillations per second) of
an electromagnetic wave can range from
less than a million (10
6
) Hz to a septilion
(10
24
) Hz or higher.
Max Plank: Energy of Photons
In 1900, Max Planck proposed that
Electromagnetic radiation is considered as
the propagation of a collection of discrete
packets of energy called photons or
quanta.
Each photon of frequency νis considered
to have an energy of :
e= h.ν= h.c/λ
h is the Plank’s constant = 6.626069 10
-34
J.s
In 1900, Max Planck proposed that
Electromagnetic radiation is considered as
the propagation of a collection of discrete
packets of energy called photons or
quanta.
Each photon of frequency νis considered
to have an energy of :
e= h.ν= h.c/λ
h is the Plank’s constant = 6.626069 10
-34
J.s
Thermal Radiation
Thermal radiation is continuously emitted by
all matter whose temperature is above
absolute zero.
Thermal radiation is also defined as the
portion of the electromagnetic spectrum that
extends from about 0.1 to 100 μm.
Thermal radiation includes all visible and
Infrared (IR) radiation as well as a portion of
ultraviolet (UV) radiation.
Light is the visible portion of electromagnetic
waves from 0.4 to 0.76 μm. Solar radiation
falls in the range 0.3-3 μm
Thermal radiation is continuously emitted by
all matter whose temperature is above
absolute zero.
Thermal radiation is also defined as the
portion of the electromagnetic spectrum that
extends from about 0.1 to 100 μm.
Thermal radiation includes all visible and
Infrared (IR) radiation as well as a portion of
ultraviolet (UV) radiation.
Light is the visible portion of electromagnetic
waves from 0.4 to 0.76 μm. Solar radiation
falls in the range 0.3-3 μm
BLACKBODY RADIATION
A blackbody is an idealized body to serve
as standard against which the radiative
properties of real surfaces may be
compared.
A blackbody is defined as perfect emitter
and absorber of radiation.
At a specified temperature and
wavelength, no surface can emit more
energy than a blackbody.
A blackbody absorbs all incident radiation.
a blackbody emits radiation uniformly in
all directions.
A blackbody is an idealized body to serve
as standard against which the radiative
properties of real surfaces may be
compared.
A blackbody is defined as perfect emitter
and absorber of radiation.
At a specified temperature and
wavelength, no surface can emit more
energy than a blackbody.
A blackbody absorbs all incident radiation.
a blackbody emits radiation uniformly in
all directions.
RADIATION EMITTED BY A
BLACKBODY
In 1879, Joseph Stefan determined
experimentally that the radiation emitted
by a blackbody per unit time and unit
surface can be described by the blackbody
emisive power E
b(T) :
E
b(T)= σ.T
4
The Stefan-Boltzmann Constant σ= 5.67 10
8
W/m
2
.K
4
Equation theoretically verified in 1884 by Ludwig Boltzmann.
BLACKBODY
In 1879, Joseph Stefan determined
experimentally that the radiation emitted
by a blackbody per unit time and unit
surface can be described by the blackbody
emisive power E
b(T) :
E
b(T)= σ.T
4
The Stefan-Boltzmann Constant σ= 5.67 10
8
W/m
2
.K
4
Equation theoretically verified in 1884 by Ludwig Boltzmann.
Definition of Large capacity
body
Another type of body that closely
resembles to a blackbody is a large
capacity with a small opening ( Figure
12.8, page 667).
Radiation coming through the small
opening undergoes multiple reflections
and thus it has several chances to be
absorbed by the interior surfaces of the
cavity before any part of it can escape.
body
Another type of body that closely
resembles to a blackbody is a large
capacity with a small opening ( Figure
12.8, page 667).
Radiation coming through the small
opening undergoes multiple reflections
and thus it has several chances to be
absorbed by the interior surfaces of the
cavity before any part of it can escape.
Spectral blackbody emissive
power
It is defined as the amount of radiation
energy emitted by a blackbody at a
thermodynamic temperature T per unit
time and unit area and per unit
wavelength about the wavelength λ.
Plank’s Law:
power
It is defined as the amount of radiation
energy emitted by a blackbody at a
thermodynamic temperature T per unit
time and unit area and per unit
wavelength about the wavelength λ.
Plank’s Law:
Plank’s Law
E
b,λ(λ,T) is in ( W/m
2
.μm)
C
1= 2.π.h.c
0
2
= 3.74177 .10
8
W.μm
4
/m
2
C
2= hc
0/k = 1.43878 .10
4
μm.K
K ( Boltzmann’s Constant)= 1.38065. 10
-23
J/K
T is the absolute temperature of the surface
λ=wavelength of the radiation emitted
E
b,λ(λ,T) is in ( W/m
2
.μm)
C
1= 2.π.h.c
0
2
= 3.74177 .10
8
W.μm
4
/m
2
C
2= hc
0/k = 1.43878 .10
4
μm.K
K ( Boltzmann’s Constant)= 1.38065. 10
-23
J/K
T is the absolute temperature of the surface
λ=wavelength of the radiation emitted
Variation of spectral Blackbody
Emissive power with
wavelength
Emissive power with
wavelength
From the figure
The emitted radiation is a continuous
function of wavelength
At a specific temperature, it increases with
wavelength, reaches a peak and then
decreases with increasing wavelength.
At any wavelength, the amount of emitted
radiation increases with temperature.
Radiation emitted by sun , blackbody, at
5780 K reaches its peak in the visible
region.
The emitted radiation is a continuous
function of wavelength
At a specific temperature, it increases with
wavelength, reaches a peak and then
decreases with increasing wavelength.
At any wavelength, the amount of emitted
radiation increases with temperature.
Radiation emitted by sun , blackbody, at
5780 K reaches its peak in the visible
region.
Maximum Power
As the temperature increases, the peak of
the curves shifts to the shorter
wavelength.
The wavelength at which the peak occurs
for a specified temperature is given by
Wien’s displacement law:
(λ.T)
max.power= 2897.8 μm.K
The peak for solar radiation occurs at 2897.8/
5780 = 0.50 μm.
The peak of radiation emitted by a surface at
room temperature ( 298K) is 2897.8/298 = 9.7
μm
As the temperature increases, the peak of
the curves shifts to the shorter
wavelength.
The wavelength at which the peak occurs
for a specified temperature is given by
Wien’s displacement law:
(λ.T)
max.power= 2897.8 μm.K
The peak for solar radiation occurs at 2897.8/
5780 = 0.50 μm.
The peak of radiation emitted by a surface at
room temperature ( 298K) is 2897.8/298 = 9.7
μm
Total blackbody emissive power
Integration is needed of the spectral
blackbody emissive power E
bover the
entire wavelength spectrum:
Integration is needed of the spectral
blackbody emissive power E
bover the
entire wavelength spectrum:
f
λ= Blackbody radiation
function
f
λis useful because it is difficult to
perform the integration knowing that each
time we need the value E
b,0-λ
f
λis dimensionless
λ= Blackbody radiation
function
f
λis useful because it is difficult to
perform the integration knowing that each
time we need the value E
b,0-λ
f
λis dimensionless
F
λ1-λ2 = f
λ2(T)-f
λ1(T)
λ1-λ2 = f
λ2(T)-f
λ1(T)
Class Work
Example 12-1 and 12.2
Example 12-1 and 12.2
Fick’s law
Proposed in 1855, the law states that the
rate of diffusion of a chemical species at a
location in a gas mixture ( or liquid, or
solid) is proportional to the concentration
gradient of that species at that location.
N
A= -D
AB.dC
A/dx
Proposed in 1855, the law states that the
rate of diffusion of a chemical species at a
location in a gas mixture ( or liquid, or
solid) is proportional to the concentration
gradient of that species at that location.
N
A= -D
AB.dC
A/dx
Mass Basis
Partial density of the species (i):
ρ
i=m
i/V
Total density:
ρ= m/V= Σ(mi)/V= Σρ
i
Mass Fraction of (i)= ω
i= m
i/m=
(m
i/V)/(m/V)= ρ
i/ ρ
ω
i = ρ
i/ ρ
Σω
i=1
Partial density of the species (i):
ρ
i=m
i/V
Total density:
ρ= m/V= Σ(mi)/V= Σρ
i
Mass Fraction of (i)= ω
i= m
i/m=
(m
i/V)/(m/V)= ρ
i/ ρ
ω
i = ρ
i/ ρ
Σω
i=1
Molar Basis
Partial molar concentration:
C
i= N
i/V
Total molar concentration:
C=N/V
Molar Fraction: y
i= N
i/N= (N
i/V)/(N/V)
y
i= C
i/C
Σy
i=1
Partial molar concentration:
C
i= N
i/V
Total molar concentration:
C=N/V
Molar Fraction: y
i= N
i/N= (N
i/V)/(N/V)
y
i= C
i/C
Σy
i=1
Relation between m and N
m= N.M
ρ = m/V= N.M/V
C=N/V ρ= C.M & ρ
i=C
i.M
i
m= N.M
ρ = m/V= N.M/V
C=N/V ρ= C.M & ρ
i=C
i.M
i
Molecular weight of the mixture
M= m/N= ΣN
i.M
i/N= Σ(N
i/N).M
i
M= Σy
i.M
i
M= m/N= ΣN
i.M
i/N= Σ(N
i/N).M
i
M= Σy
i.M
i
Relation between ω
iand y
i
ω
i=ρ
i/ρ= C
i.M
i/C.M = (C
i/C). ( M
i/M)
ω
i= y
i.(M
i/M)
iand y
i
ω
i=ρ
i/ρ= C
i.M
i/C.M = (C
i/C). ( M
i/M)
ω
i= y
i.(M
i/M)
Ideal gas mixture
y
i= N
i/N= P
i/P
y
i= N
i/N= P
i/P
Boundary Conditions
1) At the surface of water, the partial
pressure of water is the vapor pressure of
water at the surface conditions.
2) using Henry’s law for ideal gas
mixtures, the composition of the gas in
the liquid side is defined by:
x
i= P
i/H
Using Raoult’s law:
P
i=y
i.P= x
i.P
sat
1) At the surface of water, the partial
pressure of water is the vapor pressure of
water at the surface conditions.
2) using Henry’s law for ideal gas
mixtures, the composition of the gas in
the liquid side is defined by:
x
i= P
i/H
Using Raoult’s law:
P
i=y
i.P= x
i.P
sat
Boundary Conditions
The concentration of a gas in
a solid is given by the
solubility (S) :
C
i= S.P
i
The concentration of a gas in
a solid is given by the
solubility (S) :
C
i= S.P
i
Stationary binary medium
Mass Basis:
j
i= -ρ.D
ij.(dω
i/dx) kg/s.m
2
j
i= -D
ij.(dρ
i/dx) at constant ρ
Molar basis:
J
i= -C.D
ij.(dy
i/dx) kmol/s.m
2
J
i= -D
ij.(dC
i/dx) at constant C
Mass Basis:
j
i= -ρ.D
ij.(dω
i/dx) kg/s.m
2
j
i= -D
ij.(dρ
i/dx) at constant ρ
Molar basis:
J
i= -C.D
ij.(dy
i/dx) kmol/s.m
2
J
i= -D
ij.(dC
i/dx) at constant C
Diffusivity D
Because of the complex nature of mass
diffusion, the diffusivity is determined
experimentally:
For water and air, Marrero and Mason
(1972) proposed:280K< T< 450 K
D
H2O,Air= 1.87.10
-10
. T
2.072
/P (m/s)
In general: D
A,B,1/D
AB,2= ( P
2/P
1). (T
1/T
2)
1.5
Because of the complex nature of mass
diffusion, the diffusivity is determined
experimentally:
For water and air, Marrero and Mason
(1972) proposed:280K< T< 450 K
D
H2O,Air= 1.87.10
-10
. T
2.072
/P (m/s)
In general: D
A,B,1/D
AB,2= ( P
2/P
1). (T
1/T
2)
1.5
Steady state diffusion through a
wall, cylinder and sphere
Knowing the boundary conditions:
I) Wall
m=ρ.D
ij.A. (ω
i1-ω
i2)/L=D
ij.A.(ρ
i1-ρ
i2) (kg/s)
m= (ω
i1-ω
i2)/(L/ρ.D
ij.A) or (ω
i1-ω
i2)/R
wall
With R
wall= (L/ρ.D
ij.A)
II) Cylinder:
m=2πL.ρ.D
ij. (ω
i1-ω
i2)/ln(r
2/r
1)
m =2πL.D
ij.(ρ
i1-ρ
i2)/ln(r
2/r
1) (kg/s)
wall, cylinder and sphere
Knowing the boundary conditions:
I) Wall
m=ρ.D
ij.A. (ω
i1-ω
i2)/L=D
ij.A.(ρ
i1-ρ
i2) (kg/s)
m= (ω
i1-ω
i2)/(L/ρ.D
ij.A) or (ω
i1-ω
i2)/R
wall
With R
wall= (L/ρ.D
ij.A)
II) Cylinder:
m=2πL.ρ.D
ij. (ω
i1-ω
i2)/ln(r
2/r
1)
m =2πL.D
ij.(ρ
i1-ρ
i2)/ln(r
2/r
1) (kg/s)
III) Sphere:
m=2πr
1r
2.ρ.D
ij. (ω
i1-ω
i2)/(r
2/r
1)
m =2πr
1r
2.D
ij.(ρ
i1-ρ
i2)/(r
2/r
1) (kg/s)
SIMILAR EQUATIONS CAN BE FOUND FOR
MOLAR FLOWS REPLACING ρBY C AND ωBY y
m=2πr
1r
2.ρ.D
ij. (ω
i1-ω
i2)/(r
2/r
1)
m =2πr
1r
2.D
ij.(ρ
i1-ρ
i2)/(r
2/r
1) (kg/s)
SIMILAR EQUATIONS CAN BE FOUND FOR
MOLAR FLOWS REPLACING ρBY C AND ωBY y
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