Fenómenos de transporte de cantidad energía (Calor)
85
Transport Phenomena Section 2: Heat Transfer Transport Phenomena Contents of heat transfer 1. Heat conduction: Fourier law 2. Thermal conductivity of homogeneous media and heterogeneous media 3. Steady state heat conduction in slab and composite slabs 4. SS heat conduction in cylinder, sphere 5. Transient heat conduction in semi-infinite objects 6. Transient heat conduction in finite objects with symmetry conditions and asymmetry conditions 7. Transient heat conduction of objects in finite environment 8. Forced convection & heat conduction 9. Free convection 10. Equations of change 11. Heat conduction with phase change
Transcript
Transport Phenomena Section 2: Heat Transfer
Transport Phenomena
Contents of heat transfer 1. Heat conduction: Fourier law
2. Thermal conductivity of homogeneous media and heterogeneous media
3. Steady state heat conduction in slab and composite slabs
4. SS heat conduction in cylinder, sphere
5. Transient heat conduction in semi-infinite objects
6. Transient heat conduction in finite objects with symmetry conditions and asymmetry conditions
7. Transient heat conduction of objects in finite environment
8. Forced convection & heat conduction
9. Free convection
10. Equations of change
11. Heat conduction with phase change
Transport Phenomena
Heat Transfer
So far we dealt with momentum, now we turn to heat transfer (energy).
From the first principles, there is no difference between the analysis of momentum transfer and that of heat transfer, which you will see.
Transport Phenomena
Modes of heat transfer There are three modes of heat transfer
1. Conduction
2. Convection (forced & natural)
3. Radiation
Difference among the modes of heat transfer
1. Conduction: Heat is transferred due to the temperature gradient.
2. Convection: Heat is transferred by the motion of the fluid. If the fluid is forced, we talk of forced convection. On the other hand, if the motion of the fluid is induced by the temperature, we talk of natural convection.
3. Radiation: only occur at high temperature
Transport Phenomena
Heat conduction
Simplest among the three modes of heat transfer. It occurs in all three phases of matter (gas, liquid and solid). In solids, it is the only mechanism of heat transfer.
The driving force for the heat conduction is the temperature gradient, that is whenever a gradient in temperature exists there exists a conductive heat flow.
The heat flow is measured as the amount of
energy transferred through any given plane per unit area per unit time. It is called heat flux.
Transport Phenomena
Heat conduction
cross-section area A
T0
T1
Direction of heat transfer
Q
Transport Phenomena
Conductive heat flux
q k T T= −
1 0
δ
Pointwise:
q k dTdz
= −
heat flux: energy/area/time
Property of material through which the heat is conducted: Thermal conductivity
Negative sign: Heat is moving in the direction of negative temperature gradient
Temperature gradient
Fourier’s law
Transport Phenomena
q k dTdz
= −
q
dTdz
Thermal conductivity at the plane A, k
PLANE A
This is the fundamental equation in heat conduction
z
Transport Phenomena
Heat flux as a vector
Temperature is a scalar, while heat flux is a vector, of which components are
q k dTdx
q k dTdy
q k dTdzx x y y z z= − = − = −; ;
x
y
z
qx
qy
qz
T
Transport Phenomena
For isotropic materials,
kx = ky = kz = k
the heat flux vector will become:
Tkq ∇−=
where
q = qx i + qy j + qz k and
kz
jy
ix ∂
∂+∂∂+
∂∂=∇
grad operator
Transport Phenomena
Momentum vs Heat ?
q k dTdyy y= −
τ µyxxdv
dy= −
There are three components of the heat flux vector, while there are nine components of the stress tensor.
driving force
property of material
Transport Phenomena
Thermal conductivity
1. Property of material through which the heat is conducted
Liquid: no general rule; for example k of alcohols, benzene, toluene decrease with T, while glycerine, ethylene glycol increase with T
Solid: also no general rule; for example k of polyethylene, polypropylene, PTFE decrease with T, while those of insulation materials increase with T
6. Pressure dependence:
fairly independent of pressure
Temperature
k Gas Liquid
Solid
Transport Phenomena
Theory of k for monatomic gases 1. Kinetic theory of gases
a. Molecules are rigid sphere
b. Exchange of energy upon collision is translational energy
12
32
2mv TT = κ
Molar heat capacity at constant volume is
C N ddT
mv Rv T=
=
12
32
2
c. The mean free path is much shorter than the length scale of the temperature gradient
d mass = m
Boltzmann constant
Avogadro’s number
Transport Phenomena
y
v=0
p p
T(y+a)
T(y)
T(y-a)
Temperature profile
λλλλ
T
y
y + a
y - a
Transport Phenomena
d. Results of kinetics theory of gas
v TmT = 8κ
π
λπ
=1
2 2d n
Z nvT= 14
a =23λ
mean molecular thermal velocity
mean free path
wall collision frequency per unit area
the molecules reaching any plane, on the average, had their last collision at a distance a from that plane
Transport Phenomena
The heat flux qy across any plane of constant y is obtained from:
q Z mv Z mvy Ty y a
Ty y a
=
−
= − = +
12
12
2 2
According to the assumption b:
( )ayayy TTZ
23q
+−−κ=
kinetic energies of molecules that cross in the positive direction
kinetic energies of molecules that cross in the negative direction
Transport Phenomena
According to the assumption c:
T TdTdyy a y− = −
23λ
T T dTdyy a y+ = + 2
3λ
thus the heat flux qy is:
q n v dTdyy T= − 1
2κ λ
k n vd
TmT= =
12
12
3
3κ λκπ
compared with
µπ
κ= 23 3 2 2/
m Td
q k dTdyy = −
only for dilute monatomic gases at moderate pressures (usually less than 10 atm); proportional to square root of temperature and independent of pressure
Transport Phenomena
2. Chapman-Enskog theory (for gases)
k T M
k
= × −19891 10 42. /
σ Ω
µσ µ
= × −2 6693 10 52. MTΩ
cal/cm/sec
g/cm/sec
Angstrom
Angstrom
collision integral
collision integral
Transport Phenomena
Theory of k for polyatomic gases
More complex than monatomic gases due to the additional rotational and vibrational energies.
Semi-empirical formula by Eucken
k CRMp= +
54
µ
For monatomic gases, Cp = 5R/4M; thus
k RM
=
52
µ
as obtained by the kinetic theory of gases. Note: Eucken’s formula is only used as a guide.
Experimental data should be used.
heat capacity at constant pressure
viscosity
Transport Phenomena
Thermal conductivity of mixtures
Semi-empirical formula is useful:
k x k
xmix
i i
j ijj
ni
n
=
=
= ∑∑
Φ1
1
These formula equations are very similar to those for the calculation of mixture’s viscosity.
mole fraction pure component thermal conductivity
Φ iji
j
i
j
j
i
MM
MM
= +
+
−18
1 11 2 1 2 1 4
2/ / /µµ
molecular weight pure component viscosity
Transport Phenomena
Heat conduction in heterogeneous media
Of interest to engineers is the heat transfer to or from a packed column (reactor).
where φφφφ1 and φφφφ2 are monotonic decreasing function of (ks/kf)
Transport Phenomena
Zehner and Schlunder (Chem. Ing. Tech., 42, 933, 1970)
( )
( )
kk
Bkk
kk
B
kk
B Bkk
B B
Bkk
e
f
s
f
s
f
s
f
s
f
s
f
= −−
+
−
−
⋅
−
+
− + − −
−
1 11
2 1
1
1
1
1 12
1
1
1 2
1 2
2
ε
ε
/
/
ln
where
B =−
1251 10 9
./ε
ε
Modified version of Zehner-Schlunder was given by Hsu et al. (Int. J. Heat Mass Transfer, 37, 2751, 1994)
Transport Phenomena
Procedure of transport phenomena analysis
First principles
1. Draw a physical diagram as carefully as possible.
2. Identfy all possible energy transport mechanisms
3. Set a frame of coordinates and draw the direction of all transport processes identified in step 2.
4. Draw a shell in such a way that its surfaces are perpendicular to the transport direction.
5. Carry out the energy shell balance as below:
Rate ofenergy in
Rate ofenergy out
Rate of energyproduction
Accumulationof energy
−
+
=
This should give a first-order ODE in terms of heat flux
Transport Phenomena
6. Apply the Fourier law
This should give a second order ODE in terms of temperature.
7. Impose physical contraint on the boundary of the physical system. This gives rise to BCs.
Note that the number of boundary conditions must match the order of the differential equation.
8. Solve the equation for the temperature distribution.
9. Obtain the mean temperature, heat flux.
Energy production:
1. Chemical reaction
2. Electrical heat
3. Viscous dissipation (degradation of mechanical energy)
Transport Phenomena
Boundary conditions:
1. At the bounding surface, the temperature is specified. This is called boundary condition of the first kind (Dirichlet BC)
2. At the bounding surface, the heat flux equals to the flux supplied. Boundary condition of the second kind (Neumann BC)
3. At the bounding surface, the heat flux entering the domain equals the heat flux across the thin film surrounding the object. BC of the third kind (Robin BC).
Concept of heat transfer coefficient
4. At the interface between the two domains, the temperatures and the heat fluxes of those domains are continuous. BC of the fourth kind.
Transport Phenomena
What to follow 1. Steady state heat conduction in slab
1a. Constant k
1b. Temperature-dependent k
1c. BC of 1st kind vs BC of 3rd kind
2. Steady state heat conduction in composite slab
3. Steady state heat conduction in cylinder
3a. The influence of shape
3b. Composite cylindrical walls
4. Steady state heat conduction from sphere
5. Fall of temperature in free droplet
6. Heat conduction in cooling fin (BC of the 3rd kind)
7. Heat conduction with chemical heat source.
Transport Phenomena
Steady state heat conduction in slab
Steps 1 to 4:
thin shell
constant cross section area, A
energy in energy out
z z + ∆∆∆∆z
Transport Phenomena
Steps 5 and 6: Energy balance equation
Aq Aqz z z− + =+ ∆ 0 0
steady state
rate of energy in; units: Joule/sec
rate of energy out; units: Joule/sec
no heat production
Transport Phenomena
First order ODE with respect to heat flux
− =dq
dz0
Second-order ODE with respect to T
ddz
k dTdz
= 0
q k dTdz
= −
Fourier’s law
k d Tdz
2
2 0=ddz
k T dTdz
( )
= 0
k is constant k is temperature dependent
Transport Phenomena
Step 7: Physical constraint on the boundary
Boundary conditions of the first kind.
Step 8: Temperature distribution
T z) TT T
zLL
( −−
=0
0
Temperature distribution is linear (only valid for constant k and slab geometry).
z
T(0)=T0
T(L)=TL
0
L
Transport Phenomena
Step 9: Heat flux
Apply the Fourier’s law; remember that the heat flux in general is a function of z. So to calculate the heat flux, where do we evaluate the heat flux at?
At z=0 or z=L ?
Fourier law
q kT T
LL=
−
0
q z) k dT z)dz
( (= −
T z TT T
zLL
( ) −−
=0
0
Transport Phenomena
The heat flux is proportional to the temperature difference (only valid for constant k). We see that for the case of constant “k” the temperature distribution is linear and the heat flux is proportional to the temperature difference.
Is this conclusion still valid for the case of variable thermal conductivity?
Let us look at the case when “k” takes the following functional form:
( )[ ]k k T T= + −0 01 α
Constraint on αααα: αααα(T0 - TL) < 1
T0 TL
k
αααα > 0
αααα < 0
Transport Phenomena
Step 6’: Heat balance equation
( )[ ]ddz
k dTdz
ddz
k T T dTdz
= + −
=0 01 0α
Step 7’: Boundary conditions
same as before, that is:
z = 0; T = T0
z = L; T = TL
Step 8’: Temperature distribution
( ) ( )
( ) ( )
T z T T z T
T T T T
zL
L L
( ) ( )− +
−
− +
−=
0 02
0 02
2
2
α
α
Thus the temperature distribution across the slab is not linear.
Transport Phenomena
Can you explain the shape of the
temperature distribution?
z=0
z=0 z=L
T0
TL
αααα > 0
αααα = 0
αααα < 0
z=L
T0
TL
high k; thus lower gradient
lower k; thus higher gradient
αααα > 0
Transport Phenomena
Step 9’: Heat flux
Apply the Fourier’s law
( )[ ]q k T TdTdz
= − + −0 01 α
( ) ( )q k
T T T T
L
L L
=− −
−
0
0 02
2α
Would this minus sign cause a concern to you as it could give zero flux?
Answer: Of course not, see the constraint on αααα.
k
( ) ( )
( ) ( )
T z T T z T
T T T T
zL
L L
( ) ( )− +
−
− +
−
=0 0
2
0 02
2
2
α
α
independent of z, as one would expect physically
Transport Phenomena
Note: If we define an average thermal conductivity across the slab as:
( )[ ]( )k
k T dT
dT
k T T dT
dT
k T TavgT
T
T
T
T
T
T
T L
L
L
L
L= =
+ −
= − −
∫
∫
∫
∫
( )0
0
0
0
0 0
0 0
1
12
αα
the heat flux equation is:
( )
q kT T
LavgL=
−0
which is an interesting equation.
So far so good, but are there anythings that
need to be addressed?
1. Existence of gas film surrounding the slab furnace.
2. Slab is made of many different materials.
Let’s consider these one by one.
Transport Phenomena
Heat conduction in slab: Gas film resistance
What has been done so far in steps 1 to 6 is still valid, that is the heat balance equation is:
ddz
k dTdz
= 0
Let’s study the constant k first, and then deal with temperature-dependent k.
k d Tdz
2
2 0=ddz
k T dTdz
( )
= 0
k is constant k is temperature dependent
Transport Phenomena
Step 7: Physical constraints
With the existence of the gas film, we talk of BC of the third kind.
[ ]z h T T k dTdz z
= − = −=
0 00 00
; ( )
[ ]z L h T L T k dTdzL L
z L
= − = −=
; ( )
z 0
L
T0
TL
T(0)
T(L)
unknown
( )hkL
ff= Re, Pr
Transport Phenomena
Step 8: Temperature distribution
T z TT L T
zL
( ) ( )( ) ( )
−−
=00
Note: T(0) and T(L) are yet known, but we proceed anyway.
Step 9: Heat flux
[ ]q kT T L
L=
−( ) ( )0
which must be the same as the heat fluxes through the gas film surrounding the slab, that is:
[ ] [ ] [ ]q h T T kT T L
Lh T L TL L= − =
−= −0 0 0
0( )
( ) ( )( )
this is the flux through the slab
Transport Phenomena
Rearranging the above equation as:
qT T
h
T T LLk
T L T
h
L
L
=−
= −
=−
0
0
01
01
( ) ( ) ( ) ( )
Recall the following identity
ab
cd
ef
a c eb d f
= = =+ ++ +
we get the heat flux equation
qT T
hLk h
L
L
=−
+
+
0
0
1 1
written in terms of known parameters and operating conditions.
Transport Phenomena
We can write the heat flux equation as:
( ) LTT
BiBi1kq L0
1L
10
−++
= −−
where Bi0 and BiL are nondimensionless and are defined as follows:
Bih Lk
Bih L
kLL
00= =;
which are called the Biot numbers.
Physical significance of Biot number
Biheat transfer through the film
heat transfer through the object=
1. Bi >>1: Heat transfer through the object is limiting.
2. Bi << 1: Heat transfer through the fluid film is limiting.
Transport Phenomena
Avoid the confusion
Bi and Nu
Their definitions:
Bi hLk
fluid film heat transfer coefficient lengthsolid thermal conductivity
= =( )( )
Nu hLk
fluid film heat transfer coefficient lengthfluid thermal conductivityf
= =( )( )
Bi is seen in the analysis of heat transfer to or from an object.
Nu is seen in many correlations of heat transfer coefficients; hence it involves only film properties.
More about this later.
Transport Phenomena
How about T(0) and T(L)?
Recall the following equations:
qT T
h
T T LLk
T L T
h
L
L
=−
= −
=−
0
0
01
01
( ) ( ) ( ) ( )
qT T
hLk h
L
L
=−
+
+
0
0
1 1
we get
( )T Th
hLk h
T T
L
L( )0
1
1 100
0
0= −
+
+
−
( )T L Th
hLk h
T TLL
L
L( ) = +
+
+
−
1
1 1
0
0
Transport Phenomena
Heat conduction in slab: Gas film & k(T)
Steps 1 to 6:
The heat balance equation is:
( )[ ]ddz
k dTdz
ddz
k T T dTdz
= + −
=0 01 0α
Constraint on αααα: αααα(T0 - TL) < 1
Step 7: Boundary conditions
[ ]z h T T k dTdz z
= − = −=
0 00 00
; ( )
[ ]z L h T L T k dTdzL L
z L
= − = −=
; ( )
with T(0) being the solid object temperature at z=0, and T(L) being that at z=L.
Transport Phenomena
Step 8 and 9: Temperature distribution & heat flux
We can write immediately
q k T T LLavg= −( ) ( )0
where
[ ]k k T T Lavg = − −
0 1
20α ( ) ( )
This flux must be the same as fluxes through the film at both sides of the solid object, that is
[ ] [ ] [ ]q h T T kT T L
Lh T L Tavg L L= − =
−= −0 0 0
0( )
( ) ( )( )
Transport Phenomena
Thus, we get
q T T
hL
k h
L
avg L
= −
+
+
0
0
1 1
This equation still involves unknown variables T(0) and T(L).
Like before, we solve for T(0) and T(L)
( )T( Th
hL
k h
T T
avg L
L0
1
1 10
0
0
0) = −
+
+
−
( )T(L Th
hL
k h
T TLL
avg L
L) = +
+
+
−
1
1 1
0
0
Two nonlinear algebraic equations in terms of T(0) and T(L) as kavg is a function of T(0) and T(L).
Transport Phenomena
So far
Heat conduction
Thermal conductivity, k
Theory of thermal conductivity for monatomic gases
First principles of solving heat conduction problems
Heat conduction in slab object:
BC of the first kind
BC of the third kind
Constant thermal conductivity
T-dependent thermal conductivity
Now let us turn to composite objects and
objects of different geometries than slab, cylinder and sphere.
Transport Phenomena
Composite slab objects
Let deal with constant k, and N layers.
Area A
T1 T2 T3 T4
L1 L2 L3
k1 k2 k3
Transport Phenomena
Utilizing results obtained earlier for a single slab object, we can write:
( ) ( )q h T T k T TL
k T TL
k T TL
h T TNN N
NN N N= − =
−=
−= =
−= −+
+ + +0 0 1 11 2
12
2 3
2
11 1 2
or writing in the ratio form, we get ( ) ( )qT T
h
T TLk
T TLk
T TLk
T T
h
N N
N
N
N N
N
=−
=−
=−
= =−
=−
+ + +
+
0 1
0
1 2
1
1
2 3
2
2
1 1 2
1
1 1
Thus
( )qT T
hLk h
N
j
jj
N
N
=−
+
+
+
= +∑
0 2
0 1 1
1 1
The interfacial temperatures can be found by equating the above two equations.
Transport Phenomena
Heat conduction in cylinder In constrast to slab geometry, the area is
changing along the heat flow direction.
Steps 1 to 4
∆∆∆∆r
R2
R1
Transport Phenomena
Step 5:
Heat balance around the annulus shell 00)rr(q)rr(L2)r(q)rL2( =+∆+∆+π−π
A compact mathematical form:
( )[ ] ( )[ ] 0)r(qrL2)r(qrL2rrrrr=π−π
∆+==
incoming area
incoming flux
outgoing area
outgoing flux
no heat generation
means all r in the bracket are evaluated at r+∆∆∆∆r
Transport Phenomena
( )[ ] ( )[ ]2 2 0π πr q r r q rr r r r r
( ) ( )= = +
− =∆
The heat balance equation is:
f r f rr r r r r( ) ( )= = +− =∆ 0
Divide by ∆∆∆∆r:
f r f rr
r r r r r( ) ( )= = +−=∆
∆0
Now making the shell as thin as possible:
lim( ) ( ) ( )
∆
∆
∆r
r r r r rf r f rr
df rdr→
= = +−= − =
00
Define f r r q r( ) ( )= 2π
Transport Phenomena
Thus, the final heat balance equation valid at a point is:
[ ]ddr
rq r( ) = 0
Step 6:
Apply the Fourier law of heat conduction
q r k dTdr
( ) = −
into the heat balance equation to get:
ddr
r k dTdr
= 0
First-order ODE wrt q
ddr
r dTdr
= 0
ddr
r k dTdr
= 0
If k is constant If k is a function of temperature
Transport Phenomena
Step 7: Physical constraints
Let us deal with the case of constant thermal conductivity first.
r=R1; T=T1
r=R2; T=T2
Let’s start with boundary conditions of the first kind
L
Transport Phenomena
Step 8: Temperature distribution
T r TT T
rRRR
( )ln
ln
−−
=
2
1 2
2
1
2
Step 9: Heat flux
( )q r
k T T
rRR
( )ln
=−
⋅
1 2
2
1
Unlike the slab case, this heat flux is not a constant, but rather decreases as the heat moves away from the center.
always associated with cylindrical geometry
q r k dTdr
( ) = −
Fourier’s law
Transport Phenomena
This is not entirely unexpected.
r heat flux area Heat flow =
flux ×××× area small large small the same large small large the same
Although the heat flux changes with distance r, the heat flow (energy/time) must be a constant, a requirement of steady state.
Heat flow:
( ) ( )Q rL q r LkRR
T T= =
−2 2
2
1
1 2π π( )ln
constant as expected
Transport Phenomena
Compare this heat flow with that for the slab geometry.
Slab Cylinder
( )Q A k T T=
−δ 1 2 ( )Q L
RR
k T T=
−2
2
1
1 2π
ln
Irrespective of the geometry, the heat flow is proportional to
1. thermal conductivity
2. temperature difference
The only difference is the geometrical factor.
Transport Phenomena
When the annulus is very thin, that is no curvature
R2 - R1 << R1
We would expect that the solution obtained for the cylindrical geometry will reduce to that of slab.
Proof:
It is not difficult to show that: 2 2
2
1
1
2 1
π πLRR
R LR R
AreaThickness
ln
=
−=
Q.E.D.
Now let’s consider the case of non-constant thermal conductivity.
[ ]k k T T= + −1 11 α( )
Transport Phenomena
Step 6’: Heat balance equation
( )[ ]ddr
r k T T dTdr1 11 0+ −
=α
Step 7’: Same boundary constraints
r = R1; T = T1
r = R2; T = T2
Step 8’: Temperature distribution
( ) ( )
( ) ( )
T T T T
T T T T
rRRR
− +
−
− +
−=
1 12
2 1 2 12
1
2
1
2
2
α
α
ln
ln
Transport Phenomena
Step 9’: Heat flux
( ) ( )q rk
rRR
T T T T( )ln
=⋅
− −
−
1
2
1
1 2 1 22
2α
Heat flow
( ) ( )QLkRR
T T T T=
− −
−
22
1
2
1
1 2 1 22π α
ln
which is independent of r.
Transport Phenomena
Heat conduction from a sphere
r
dr
r=R; T = T0
shell
Boundary condition
Transport Phenomena
Step 6: Heat balance equation
ddr
r k dTdr
2 0
=
Step 7: Physical constraints
r R T Tr T T= =→ ∞ = ∞
;;
0
Step 8: Temperature distribution
rR
TTT)r(T
0
=−−
∞
∞
Step 9: Heat flux
( )∞−= TTRkq 0R
fluid thermal conductivity
Transport Phenomena
If the heat transfer coefficient is defined as
q = h (T0 - T∞∞∞∞)
Compare this with the solution obtained from first principles, we get:
( )h Rk
hDk
22= =
Thus, from first principles the Nusselt number of a stagnant medium is 2.
Correlation of the form:
Nu a n m= +2 Re Pr therefore, comes as no surprise.
known as the Nusselt number
Transport Phenomena
The influence of shape on heat balance
Slab Cylinder Sphere
ddr
k dTdr
= 0
ddr
r k dTdr
= 0
ddr
r k dTdr
2 0
=
1. The general form for the three shapes is:
ddr
r k dTdr
s
= 0
s is called the shape factor.
2. Those equations can be cast into the form:
dQdV
= 0
that is the rate of change of energy per unit volume is zero.
Transport Phenomena
Fall of temperature of a free droplet
Rate of mass transfer from the liquid surface is:
I = 4ππππR D (C0 - C∞∞∞∞)
If H is the latent heat of vaporization, the heat gain by the droplet is equal to the heat required to vaporize the liquid, that is:
( ) ( )4 40 0π πRk T T RD C C∞ ∞− = −
Therefore, the drop in temperature is:
( )T T HDk
C C0 0= − −∞ ∞
Example: Benzene droplet in air. The drop in temperature is 60C.
Transport Phenomena
Transient heat conduction 1. New parameter: Thermal diffusivity
2. Heat conduction in a semi-infinite slab object.
a. Penetration heat front
b. Constant surface temperature vs constant surface heat flux
3. Heat conduction in a finite object.
a. Slab geometry
b. Cylindrical geometry
c. Spherical geometry
4. Heat conduction in a finite object with asymmetry boundary conditions.
5. Heat conduction in objects within a finite environment.
6. Determination of heat flux from temperature measurements.
Transport Phenomena
Thermal diffusivity
You will feel the heat first with which system?
Chem Eng.
Engine Oil Air layer
Transport Phenomena
Answer:
The air system will transport heat quicker than the oil system.
Does this seem odd as the thermal conductivity of oil (0.14 W/m/K) is about 5 times larger than that of air (0.03 W/m/K)? Answer: No
The heat transfer in this system is transient; that is heat travels from the left surface, portion of which is conducted through and portion is absorbed by the material to increase its latent heat.
Since oil has a much higher heat capacity per unit volume, most of the heat is retained by the oil and hence it retards the movement of heat front to the right surface.
Transport Phenomena
The proper parameter characterises this transient situation is
THE THERMAL DIFFUSIVITY
It is defined as:
α
ρ= ≡k
Cability to conductability to absorbp
The larger is this parameter, the faster is the heat transport.
We shall see that this parameter will evolve naturally during the analysis.
divide by A∆∆∆∆x and take the limit when ∆∆∆∆x →→→→ 0
q r k dTdr
( ) = −
Fourier’s law
thermal diffusivity evolves naturally
Transport Phenomena
Initial and boundary conditions
t T Tx T Tx T T
s
= == =→ ∞ =
00
0
0
;;
;
Solution method: Combination of variables: The essence of this
method is that the temperature behaves with respect to x and t in a special way (MATCHING OF CONDITIONS).
condition of the object before the surface temperature is changed
This condition is maintained for t > 0+
far away from the surface, the temperature is not yet disturbed by the change in surface temperature
Transport Phenomena
Temperature distribution
T x t TT T
erf xt
s
s
( , ) −−
=
0 4α
where erf is the error function.
Error function is defined as:
( )erf z e dxxz
= −∫2 2
0π
It has the following properties:
1. erf(0) = 0; erf(∞∞∞∞) = 1
2. ( )ddz
erf z ez= 2 2
π
x erf(x) 0 0.01 0.1 0.5 1.0 1.5 2.0
0 0.01128 0.11246 0.52049 0.84270 0.96610 0.99532
special combination of x and t
Transport Phenomena
The heat penetration front can be defined as the front at which the temperature is such that:
T t TT T
erft
s
s
( , ).
δ δα
−−
=
=
0 40 995
From the table of error function, erf(2)=0.995; thus
δ α= ⋅4 t
T(x,t)
x
at a given time
heat penetration front
T0
Ts
Transport Phenomena
δ αρ
= ⋅ = ⋅4 4t kC
tp
The penetration front is proportional to
1 the square root of time,
2. the square root of k
3. the inverse of the square root of ρρρρCp
Front travels fast Front travels slow 1. high conductivity 2. low volumetric heat capacity
1. low conductivity 2. high volumetric heat capacity
The square root dependence of the heat front with respect to time is the characteristics of conduction (diffusion) type problems.
Transport Phenomena
Heat flux:
Obtained from the Fourier law
( )q x t
k T Tt
xt
s( , ) exp=−
−
0
2
4πα α
What we can observe from this are:
1. Heat flux decays rapidly with x for a given time
2. Heat flux increases to a maximum and then decays with respect to time.
Of interest to engineers is the heat flux entering the object at x=0:
( )
q tk T T
ts( , )0 0=−πα
This heat flux is infinite at t=0! Is this acceptable physically?
Transport Phenomena
Answer:
The infinite flux at t=0 is due to the sudden exposure of the surface to a new temperature, that is
infinite temperature gradient at t=0.
If we are interested in the amount of heat has entered the object up to time t, we find:
( )Q t A q t dt
Ak T Tt
ts( ) ( , )= =−
∫ 02
0
0
πα
or
( )Q t
A T Tk C ts
p( ) =−
⋅ ⋅2 0
πρ
The amount of heat transfer is proportional to the square root of
1. The thermal conductivity
2. The volumetric heat capacity
Transport Phenomena
If there is a film heat transfer resistance at x=0, the solution for the temperature distribution is (Levenspiel, 1984):
T(x t TT T
erf xt
erf xt
h tk
hxk
h tk
s
s
, )
exp
−−
=
+
− +
+
0
2
2
4
14
α
αα α
When heat transfer coefficient h is infinite (that is no film resistance), the second term in the RHS is zero, as erf(∞∞∞∞)=1.
Transport Phenomena
Case 2: Constant heat flux at x=0
The heat balance equation is still the same as before as
the equation describes the pointwise heat balance within the domain (i.e. object) and it is not affected by the way how we impose the system.
How we impose the system is through the boundary condition.
The boundary condition for this case is:
x kTx
q s= − =0;∂∂
Transport Phenomena
The solution for the heat flux is:
q x t q erf xts( , ) = −
1
4α
and for the temperature is:
α−
−
α
−πα+=
t4xerf1
kxq
t4xexpt4
kqT)t,x(T
s
2s
0
The quantity of interest is the rise in temperature at the x=0 surface:
πα+= t4
kqT)t,0(T s
0
Due to the constant supply of heat flux, the x=0 temperature increases with time and follows the square root dependence.
Transport Phenomena
Heat conduction in a finite slab object
The slab object is symmetrical, so it is only necessary to consider half of the object.
The origin is chosen as the center of the object as shown.
time
x x=0
T1
x=L
shell center
Transport Phenomena
The heat balance will be exactly the same as before. Why.
Because the heat balance is a pointwise equation. It is not affected by the boundary condition as well as the size of the domain.
The heat balance equation is:
α ∂∂
∂∂
2
2
Tx
Tt
=
assuming a constant thermal diffusivity
The initial and boundary conditions are:
t T T
xTx
x L T T
= =
= =
= =
0
0 0
0
1
;
;
;
∂∂
symmetry at center
no film resistance. If there is the new BC is
[ ]x L kTx
h T Tx= − = −=;∂∂ 0 1
Transport Phenomena
Solution method:
Either by
1. Separation of variables method
2. Laplace transform
Temperature distribution:
The Laplace solution is:
( )T
Ts
T Ts
xs
Ls
= +−
0 1 0
cosh
cosh
α
α
The inverse is found by the method of residues:
( )T T(x tT T
xL
tL
n
nn n
n
1
1 0
22
1
2−−
=
−
=
∞
∑, ) sincos exp
λλ
λ λα
where λλλλn is called the eigenvalue and is defined as
λ πn n= −
12
Transport Phenomena
Heat flux into the particle: ( )∑
∞
==
αλ−−−=
∂∂−=
1n2
2n
01
Lx Ltexp
LTTk2
xTkq
Amount of heat transfer:
The amount of energy up to time t passing through the surface at x=L
( )Q t A q t dt AL C T T
tLt
p
n
nn
( ) ( )exp
= = − −− −
∫ ∑=
∞
01 0
22
21
21
ρλ α
λ
This is basically the amount received by half of the object up to time t
When time is sufficient large, this amount is:
Q(∞∞∞∞) = ALρρρρCp (T1 - T0) This is physically expected as this is the
sensible heat required to bring the object from T0 to T1.
Transport Phenomena
The solution of the amount of heat transfer passing through the x=L surface valid for very short time is:
( )Q A
k T Ts
=−1 03 2α /
or the inverse is:
( )Q
A T Tk C tp=
−⋅ ⋅
2 1 0
πρ
This solution for short time is exactly the same as the solution we obtained earlier for a semi-infinite object.
WHY?
independent of the size of the object. Why?
Transport Phenomena
Answer:
Because for short time the heat does not penetrate far into the object, and hence the center is not yet felt by the heat front.
Thus, the object behaves just like a semi-infinite object.
We have learnt that the solutions for the semi-inifinite object are simpler than the finite object. They are recommended for the description of the system under study during the initial stage of transfer.
Transport Phenomena
Heat conduction in a finite cylindrical or spherical object
The heat balance equation is:
α ∂∂
∂∂
∂∂
1x x
x Tx
Tts
s
=
with
1. s = 0 for slab
2. s = 1 for cylinder
3. s = 2 for sphere
The same set of initial and boundary conditions:
t T T
xTx
x L T T
= =
= =
= =
0
0 0
0
1
;
;
;
∂∂
symmetry at center
no film resistance. If there is the new BC is
[ ]x L kTx
h T Tx= − = −=;∂∂ 0 1
Transport Phenomena
The temperature distribution is: T T(x t
T Ta K x t
Ln n nn
1
1 0
22
1
−−
= −
=
∞
∑, )( ) exp λ α
1. The parameter an is called coefficient.
2. The function Kn(x) is called eigenfunction.
3. The parameter λλλλn is called the eigenvalue.
shape an Kn(x) λλλλn
Slab ( )2
sin λλ
n
n cos λn
xL
n −
12
π
Cylinder ( )
2
1λ λn nJ J xLn0 λ
J 0 0( )• =
Sphere ( )−2cos λ
λn
n sin λn
xL
xL
nπ
Transport Phenomena
The heat flux solution is:
( )q
k T TL
tLn
n
= −−
−
=
∞
∑2 1 0 2
21
exp λα
The form is exactly the same for all shape. The only difference is the eigenvalue (characteristic of the system, i.e. shape).
The amount of heat received by the object is given by:
( )[ ] ( )Q t s V C T T
tL
object p
n
nn
( ) ( )exp
= + ⋅ ⋅ ⋅ −− −
=
∞
∑2 11
1 0
22
21
ρλ α
λ
where Vobject is the volume of the object.
For sufficiently large time, the total amount of
heat absorbed by the object
Q(∞∞∞∞) = Vobject ρρρρCp (T1 - T0)
which is expected physically.
Transport Phenomena
Heat conduction in slab with asymmetry condition
The heat balance equation is:
α ∂∂
∂∂
2
2T
xTt
=
The initial and boundary conditions are: t T Tx T Tx L T T
s
= == == =
00
0
0
;;;
T0
Ts
x=0 x=L
time
Transport Phenomena
Solution method:
1. Laplace transform or
2. Separation of variables
Temperature distribution
T T x tT T
xL
nn x
Ln t
L
s
s
n
−−
= +
−
=
∞
∑
( , )
sin exp
0
2 22
1
2 1π
π π α
linear steady state profile
transient term, will decay to zero as time is getting large
Transport Phenomena
Heat flux at x=0 and x=L
( ) ( )q
k T TL
k T TL
n tL
s s
n0
0 0 2 22
1
2=
−+
−−
=
∞
∑ exp π α
( ) ( )q
k T TL
k T TL
n n tLL
s s
n
=−
+−
⋅ −
=
∞
∑0 0 2 22
1
2cos( ) expπ π α
At steady state, the heat flux entering the surface x=0 must be the same as that leaving the surface x=L.
During the trasient operation, the heat flux entering the x=0 surface is greater than that leaving at x=L. The difference is absorbed by the object as sensible heat.
steady state heat flux
Transport Phenomena
Amount of heat entering and leaving the object
( )
( )Q
Ak T TL
t
T T AL Cn
n tL
s
s p
n
00
02 2
2 22
136 1
=−
+
−− −
=
∞
∑ρ
ππ αexp
and
( )
( ) ( )
QAk T T
Lt
T T AL Cn
nn t
L
Ls
s pn
=−
+
− − −
=
∞
∑
0
0 2 22 2
21
21ρ
ππ
π αcosexp
When time is sufficiently large
( )Q
Ak T TL
t Ls0
02
3=
−+
α
( )Q
Ak T TL
t LL
s=−
−
0
2
6α
Transport Phenomena
This method is known as the time lag method.
QL
time 0 L2/6αααα
This is called the lag time, which is the time required for the heat to penetrate the medium
Transport Phenomena
Heat conduction in a finite environment
Question: Cooling or heating time of a collection of solid objects in a finite environment.
time t <0 time t > 0+
Transport Phenomena
Two sub systems to write down heat balance equations:
1. Heat conduction in a sphere
2. Heat transfer from the environment into the collection of sphere.
Heat balance of a sphere:
α ∂∂
∂∂
∂∂
12
2
x xx T
xTt
=
Heat balance in the reservoir:
RxP
Pfpfff x
TkmR3
dtdTCV
=∂∂
ρ
−=ρ
The initial and boundary conditions are:
t T T T T
x Tx
x L T T
f
f
= = =
= =
= =
0
0 0
0 1; ;
;
;
∂∂
Transport Phenomena
Solution method:
Laplace transform.
Solution for the fluid temperature T TT T
BB
BB B
tR
f
nn
n
−−
=+
++ +
−
=
∞
∑0
1 02 2
22
116
19 1( )
expλ
λα
where
BV Cm Cf f pf
p p
=ρ
The eigenvalues are determined from the following transcendental equation:
λ λ λcot( ) = +
13
2B
Solution for the object average temperature T TT T
BB
BB B
tRn
nn
−−
=+
−+ +
−
=
∞
∑0
1 0
22 2
22
116 1
9 1( )exp
λλ α
ratio of the heat capacities of the two phases
Transport Phenomena
Parameter determination T TT T
BB
BB B
tR
f
nn
n
−−
=+
++ +
−
=
∞
∑0
1 02 2
22
116
19 1( )
expλ
λα
1. Plot the LHS versus ττττ = ααααt/R2
2. From experimental data, obtain the LHS and then use the above plot to obtain ττττ corresponding to experimental t.
3. Plot ττττ versus t, the slope is αααα/R2.
experimentally measured
function of only heat capacities
contain dynamic parameters: thermal conductivity
Transport Phenomena
Heat convection
So far, we deal with heat conduction whereby the driving force is the temperature gradient
Now we consider the second mode of heat transfer
CONVECTION
1. As the name suggests, the heat transfer in this mode is induced by some form of fluid movement around the object.
2. There are two types of heat convection:
a. Forced Convection
b. Free Convection
Transport Phenomena
Flow pattern: by external force
Flow pattern: by the bouyancy effects of heated fluid
1. Velocity profile, then
2. Temperature profile
1. Velocity and temp. profiles are connected
Nu = f(Re, Pr) Nu = f(Gr, Pr)
Nu hDk
u D Ck
Gr gD
f
p
f
= = = =; Re ; Pr ;ρµ
µ ρµ
3
2
∆ρ
object object
Forced convection by forced stream of air
Free convection by heated air which rises
Transport Phenomena
Forced convection
Two broad classes of heat convection: 1. Internal forced convection
2. External forced convection
velocity
temperature
Transport Phenomena
Internal forced convection with constant surface temperature
The thin shell and the direction of
energy transport
r
∆∆∆∆r
In by convection and conduction
out by convection & conduction
In by conduction
Out by conduction
z z+∆∆∆∆z
Transport Phenomena
The velocity profile distribution for Newtonian fluids is:
v r vrRz ( ) max= −
12
Step 5: Energy balance equation
( )
( )EnergyIn
r z q r z
r r q r z
C v r r r T T
r r
z z
p z z
=
+
+
⋅ −
2
2
2 0
π
π
ρ π
∆
∆
∆
( , )
( , )
( )( ) ( )
sensible heat carried by the fluid
heat conduction in at r-surface
Transport Phenomena
and
( )
( )EnergyOut
r z q r z
r r q r z
C v r r r T T
r r r
z z z
p z z z
=
+
+
⋅ −
+
+
+
2
2
2 0
π
π
ρ π
∆
∆
∆
∆
∆
∆
( , )
( , )
( )( ) ( )
Heat balance equation of the finite shell is:
( ) ( ) ( ) ( )
00z
TT)r(vCr
zrqrq
rrqrq
zzzzp
zzzzzrrrrr
=+∆−
ρ
−∆−
−∆−
−
∆+
∆+∆+
no heat production in the tube
steady state
Transport Phenomena
Now taking ∆∆∆∆r and ∆∆∆∆z to zero, we obtain the following heat balance equation valid at any point inside the tube
( )z
qrqrr
1zT)r(vC z
rzp ∂∂−
∂∂−=
∂∂ρ
Step 6: Apply the Fourier law
q k Tr
q k Tzr z= − = −∂
∂∂∂
;
we get
∂∂+
∂∂
∂∂=
∂∂ρ 2
2
zp zT
rTr
rr1k
zT)r(vC
Usually
2
2
zp zT
zT)r(vC
∂∂>>
∂∂ρ
Thus, the heat balance equation is:
∂∂
∂∂=
∂∂
−ρ
rTr
rr1k
zT
Rr1vC
2
maxp
Transport Phenomena
Step 7: Physical constraints
1. At the entrance, the fluid temperature is equal to the inlet temperature.
2. At the center of the tube, we have the symmetry condition.
3. At the tube surface, the temperature is equal to the surface temperature, Tw.
z=0; T=T0
r=R; T = Tw
r=0; ∂∂∂∂T/∂∂∂∂r=0
Transport Phenomena
Step 8: Temperature distribution
( )T r z TT T
A K r kzR C v
w
wn n
nn
p
( , )exp
max
−−
= −
=
∞
∑0 1
22λρ
Step 9: Quantities of interest
Mixing cup temperature is defined:
( ) ( )Enthalpy R v C T Tp m= −π ρ20
How is enthalpy calculated?
coefficient
eigenfunction
eigenvalue
average velocity
mixing cup temperature
Transport Phenomena
[ ]Enthalpy C rv r T r z T drp z
R
= −∫20
0πρ ( ) ( , )
Mixing cup temperature:
T z TT T
E kzR C v
m w
wn
nn
p
( )exp
max
−−
= −
=
∞
∑0 1
22λρ
r
dr
vz(r)
Transport Phenomena
Heat flux at tube surface
Apply the Fourier law at the tube surface
( )
q z k T r zr
k T TR
A Y kzR C v
Rr R
wn n
nn
p
( ) ( , )
( ) exp'
max
= −
= −−
−
=
=
∞
∑
∂∂
λρ
0
1
221
This is the heat flux at the tube surface which is a function of downstream distance.
qR(z)
z
Transport Phenomena
The necessary constants in solutions are:
n λλλλn An En Y’n(1) 1 2 3 4 5
2.70436 6.67903 10.6734 14.6723 18.5149
+1.4764 -0.8061 +0.5888 -0.4764 +0.3591
0.81905 0.09753 0.03250 0.01547 0.00703
-0.10443 +1.3429 -1.5723 +1.7479 -1.6766
We see that using the first principles, we can obtain
1. the temperature distribution
2. the cup-mixing temperature
3. the heat flux at tube surface
without resort to any correlations of heat transfer.
Transport Phenomena
Approximate mixing cup temperature
Note that all solutions are in the form of infinite series.
To obtain the mixing cup temperature at large distance, we see that all the terms in the series are very small compared to the first term
Thus
lim( )
exparg
maxz l e
m w
w p
T z TT T
E kzR C v→
−−
= −
01 1
22λρ
Solving for the length required to achieve a cup-mixing temperature of Tm:
kLR C v E
T L TT Tp
m w
w2
12
1 0
1 1ρ λmax
ln( )
= −−
−
Transport Phenomena
Thus, if we want (Tm-Tw)/(T0-Tw) = 0.1, the length of the tube required is:
L
R v Ck
p= 0 292
. maxρ
or written in terms of flow rate F:
L
F Ck
p= 0185.ρ
An amazing simple result.
The length of the tube is proportional to:
1. flow rate
2. volumetric heat capacity
and inversely proportional to
1. fluid thermal conductivity
Transport Phenomena
Heat transfer coefficient & Nu number
Recall the heat flux at the surface and the cup mixing temperature:
( )
q z k T r zr
k T TR
A Y kzR C v
Rr R
wn n
nn
p
( ) ( , )
( ) exp'
max
= −
= −−
−
=
=
∞
∑
∂∂
λρ
0
1
221
and
T z TT T
E kzR C v
m w
wn
nn
p
( )exp
max
−−
= −
=
∞
∑0 1
22λρ
There is one quantity that engineers like to use is the heat transfer coefficient. It is defined as follows:
( )q z h z T TR m w( ) ( )= −
Transport Phenomena
Thus the heat transfer coefficient can be evaluated as:
h z kD
A Y kzR C v
E kzR C v
n nn
np
nn
np
( )' ( ) exp
exp
max
max
=
− −
−
=
∞
=
∞
∑
∑
2 11
22
1
22
λρ
λρ
Making use of the definition of the Nusselt number, we get:
Nu zA Y kz
R C v
E kzR C v
n nn
np
nn
np
( )' ( ) exp
exp
max
max
=
− −
−
=
∞
=
∞
∑
∑
2 11
22
1
22
λρ
λρ
which must be computed numerically.
Transport Phenomena
This Nusselt number is a function of distance, and from this we can get the asymptotic Nusselt number
76.3E
)1('YA2)(Nu1
11 =−=∞
If we keep two terms, we will obtain how Nusselt number would decay along the axis z.
( )Nu z Nu kzR C vp
( ) . exp .max
= ∞ + −
1 0 583 37 3 2ρ
3.76
full solution
two-term solution Nu
z
Transport Phenomena
Before we solve the next problem, we recall the solution for the cup-mixing temperature:
T z TT T
E kzR C v
m w
w p
( )exp
max
−−
≈ −
01 1
22λρ
The LHS is dimensionless, E1 is dimensionless, λλλλ1
2 is also dimensionless; so the group
kz
R C vp2ρ max
is also dimensionless. Rearranging this group as follows:
kzR C v
zR v R
kC
zR
p p2
1
ρµρ µmax max
Re Pr
=
⋅
⋅
=
⋅
⋅
This shows that the dimensionless groups Re and Pr are generally appeared in forced convection problems.
Transport Phenomena
Heat conduction & convection in tube with constant wall heat flux
In the previous example, we dealt with conduction and convection in tube with constant wall temperature.
Now we deal with constant wall heat flux.
These wall conditions (either constant temperature or heat flux) only affect the boundary conditions. So the heat balance equation obtained earlier will still be applicable here:
∂∂
∂∂=
∂∂
−ρ
rTr
rr1k
zT
Rr1vC
2
maxp
Transport Phenomena
Step 7: Physical constraints
@ z = 0; T = T0
@ r = 0; ∂∂∂∂T/∂∂∂∂r = 0
@ r = R; - k∂∂∂∂T/∂∂∂∂r = q1
Remember, the constant heat flux q1 is an algebraic quantity. It is negative, if the fluid is heated, and is positive when the fluid is cooled.
Transport Phenomena
Step 8: Temperature distribution
By defining
( )θ ζρ
=−
= =T Tq R k
xrR
kzC v Rp
0
12/
; ;max
the heat balance equation and the boundary conditions will become:
( )1 12− =
xx x
xx
∂θ∂ζ
∂∂
∂θ∂
@ ζζζζ = 0; θθθθ = 0
@ x = 0; ∂∂∂∂θθθθ/∂∂∂∂x = 0
@ x = 1; ∂∂∂∂θθθθ/∂∂∂∂x = -1
Exact solution to this problem is possible, but we are interest in the solution at distance far away from the entrance.
Entrance solution is not so much of interest to engineers.
Transport Phenomena
Because of the constant heat flux, the temperature is expected to rise linearly, i.e.
θ ζ ζ ψ∞ = ⋅ +( , ) ( )x C x0
that is, the temperature at any point across the tube section increases linearly at the same rate.
This long distance solution must satisfies the center and wall boundary conditions, as expected. But it will not satisfy the entrance condition
@ ζζζζ = 0; θθθθ = 0
To make up for this, the solution must, however, satisfy the overall heat balance equation.
Transport Phenomena
To find the overall heat balance equation, we start with
( )1 12− =
xx x
xx
∂θ∂ζ
∂∂
∂θ∂
( )dd
x x dx xx
xxx xζ
θ ∂θ∂
∂θ∂
1 2
0
1
1 0
− =
−
∫
= =
( )dd
x x dxζ
θ1 12
0
1
− = −∫
( )x x dx1 2
0
1
− = −∫ θ ζ
multiply by xdx and integrate wrt x from 0 to 1
apply the BCs in x
integrate wrt ζζζζ from 0 to ζζζζ
Transport Phenomena
This is the overall heat balance equation. It simply states that the heat input through the wall is equal to the sensible heat gained by the fluid.
Written in dimensional terms, that equation is:
( ) ( )v rR
T T rdr Rz qmax 1 2 22
00
1
1−
− = − ⋅∫ π π
Transport Phenomena
Summary
Thus, the long distance solution
θ ζ ζ ψ∞ = ⋅ +( , ) ( )x C x0
must satisfy the differential heat balance equation
( )1 12− =
xx x
xx
∂θ∂ζ
∂∂
∂θ∂
the two boundary conditions in x
@ x = 0; ∂∂∂∂θθθθ/∂∂∂∂x = 0
@ x = 1; ∂∂∂∂θθθθ/∂∂∂∂x = -1
and the overall heat balance equation
( )x x dx1 2
0
1
− = −∫ θ ζ
Transport Phenomena
The solution finally is:
θ ζ ζ∞ = − ⋅ + − +( , )x x x44
724
42
Step 9: Desired quantities Surface temperature:
θ ζ ζ∞ = − ⋅ −( , )1 4 1124
Center temperature:
θ ζ ζ∞ = − ⋅ +( , )0 4 724
Cup-mixing temperature:
θ ζθ
ζ∞
∞
=−
−= − ⋅
∫
∫( )
( )
( )
x x dx
x x dx
1
14
2
0
1
2
0
1
Transport Phenomena
Heat transfer coefficient: It is defined as:
( )h T T qR− = 1
Rearrange:
( ) ( )[ ]h T T T T
q Rk
qq R
k
R− − −
=
0 0
1
1
1
that is:
( )h kR
θ θ∞ − =1
Hence
Nu hDk
= = =4811
4 36.
cup-mixing temperature
surface temperature
Transport Phenomena
Length required
To determine the length required to heat the fluid to some desired temperature, we use the solution:
θ ζ ζ∞ = − ⋅( ) 4
Rewrite the equation in dimensional quantities, we get:
( )
( )LT T R C v
qp
=−
−0
14
ρ max
Written this in terms of the volumetric flow rate, F, we have:
( )
( )LT T C F
R qp
=−
−016 0
1
.ρ
Transport Phenomena
Free convection
What we have dealt with so far:
1. Various problems of heat conduction, steady state as well as un-steady state.
2. Combined heat conduction and heat convection.
Now we will deal with a combined heat conduction and free convection problem.
The problem is two parallel plates with a fluid confined between them. The fluid region closer to the hot plate will rise, while the fluid region close to the cold plate will descend.
The movement of fluid due to the temperature difference is determined by the momentum balance equation:
µ ρd v ydy
dpdz
gz2
2( ) = +
In this equation, we will assume the viscosity is a constant, and the density to follow Taylor expansion:
( )ρ ρ ∂ρ∂
( ( )T) TT
T TT
≈ + −0 00
where T0 is yet an unspecified temperature.
Def. of the coefficient of volume expansion:
1
0ρ
∂ρ∂
βT T
= −
the above Taylor series will become:
( )ρ ρ β ρ( ( ) ( )T) T T T T≈ − ⋅ ⋅ −0 0 0
Transport Phenomena
Put this in the momentum eqn., we get:
( )
µ ρ
β ρ
d v ydy
dpdz
g T
g T T T
z2
2 0
0 0
( ) ( )
( )
= +
− ⋅ ⋅ −
If the pressure gradient is solely due to the weight of the fluid, then:
dpdz
g T= − ρ( )0
The momentum equation is:
( )µ β ρd v ydy
g T T Tz2
2 0 0( ) ( )= − ⋅ ⋅ −
The physical meaning: The viscous forces (LHS) are balanced by the buoyancy forces.
Transport Phenomena
Summary:
The governing equations are:
k d T
dy
2
2 0=
@ y = -b; T = T2
@ y = +b; T = T1
and
( )µ β ρd v ydy
g T T Tz2
2 0 0( ) ( )= − ⋅ ⋅ −
@ y = -b; vz =0
@ y = +b; vz =0
Transport Phenomena
Step 8: Temperature & velocity distribution
The temperature distribution is:
T( y T T ybm) = − ⋅
∆2
where
T T T T T Tm = + = −1 22 12
; ∆
The velocity distribution:
v y T g b T
yb
A yb
yb
A
z ( ) ( )=
×
−
−
+
ρ β02
3 2
12∆
where
( )AT T
Tm=−6 0
∆
Transport Phenomena
Now we require that the net volume flow is zero, that is:
v y dyzb
b
( )−∫ = 0
Solving this equation, we get:
A = 0
that is the reference T0 used in the Taylor series expansion for density is equal to the mean temperature, Tm.
The velocity distribution now is:
v y T g b T yb
ybz ( ) ( )=
⋅
−
ρ β02 3
12∆
Transport Phenomena
Equation of change You have seen the analysis of a number
of simple heat transfer problems by using the first principles on shell elements.
NOW
You will learn how to generalize the shel energy balance to obtain the equation of energy, which describe energy transport in homogeneous fluid or solid.
The advantage of using this equation of energy is that we do not need to set up shell balance every time we solve a problem
Transport Phenomena
Equations of energy
Rate of accummulationof ernal and kineticenergy
Rate of ernaland kinetic energyin by convection
Rate of ernaland kinetic energyout by convection
Net rate of heataddition byconduction
Net rate of workdone by systemon surroundings
intint int
=
−
+
−
(x,y,z)
x
y
z
Transport Phenomena
This is the first law of thermodynamics.
Kinetic energy: is energy associated with
the fluid motion, i.e. ρρρρv2 on a per unit volume basis.
Internal energy: is energy associated with
the random translational and internal motions of the molecules plus the energy of interaction between the molecules.
The internal energy depends on the local temperature and density of the fluid.
Transport Phenomena
Now back to the energy equation:
Rate of accummulationof ernal and kineticenergy
Rate of ernaland kinetic energyin by convection
Rate of ernaland kinetic energyout by convection
Net rate of heataddition byconduction
Net rate of workdone by systemon surroundings
intint int
=
−
+
−
Let do this term-by-term.
Term 1: Accummulation
( )Rate of accummulationof ernal and kineticenergy
x y zt
U vint^
= +
∆ ∆ ∆ ∂∂
ρ ρ12
2
where U^
is the internal energy per unit mass.
Transport Phenomena
Term 2 & 3: Convection
( )
( )
( )
Rate of ernaland kinetic energyin by convection
Rate of ernaland kinetic energyout by convection
y z v U v v U v
x z v U v v U v
x y v U v v U
xx
xx x
yy
yy y
zz
z
int int
^ ^
^ ^
^ ^
−
=
+
− +
+
+
− +
+
+
− +
+
+
∆ ∆
∆ ∆
∆ ∆
∆
∆
ρ ρ ρ ρ
ρ ρ ρ ρ
ρ ρ ρ
12
12
12
12
12
2 2
2 2
2 12
2ρvz z
+∆
Term 4: Conduction
( )[ ]
( )[ ] ( )[ ]
Net rate of heatby conduction
y z q q
x z q q x y q q
x x x x x
y y y y y z z z z z
= − +
− + −
+
+ +
∆ ∆
∆ ∆ ∆ ∆
∆
∆ ∆
where qx, qy, qz are components of the
heat flux vector q
Transport Phenomena
Term 5: Work done on the surrounding.
Recall the following basic formula
( ) ( )Work Forcedis ce in the directionof the force
= ×
tan
( )The rate ofdoing work
Forcevelocity in the directionof the force
= ×
Consists of two parts:
a. Work against the volume forces, e.g. gravity.
b. Work against surface forces, e.g. pressure & viscous forces
Transport Phenomena
The work against the volume forces (gravity)
The rate of doing work against the three components of the gravity force:
( )( )Rate of work againstgravity force
x y z v g v g v gx x y y z z
= − + +ρ ∆ ∆ ∆
The work against the pressure
The rate of doing work against the static pressure:
( ) ( ) ( )[ ]
( ) ( ) ( ) ( ) ( ) ( )[ ]
Rate of work againstpressure force
y z pv pv
x z pv pv x y pv pv
x x x x x
y y y y y z z z z z
= − +
−
+ −
+
+ +
∆ ∆
∆ ∆ ∆ ∆
∆
∆ ∆
the minus sign because work is done against gravity
Transport Phenomena
The work against the viscous forces
The rate of doing work against the viscous forces is:
( ) ( ) ( )[ ]
( ) ( ) ( )
( ) ( ) ( )[ ]
Rate of work againstviscous force
y z v v v v v v
x z v v v v v v
x y v v v v v v
xx x xy y xz z x xx x xy y xz z x x
yx x yy y yz z y yx x yy y yz z y y
zx x zy y zz z z zx x zy y zz z z z
=
+ + − + + +
+ + − + +
+
+ + − + +
+
+
+
∆ ∆
∆ ∆
∆ ∆
∆
∆
∆
τ τ τ τ τ τ
τ τ τ τ τ τ
τ τ τ τ τ τ
Putting all those terms into the shell energy balance equation, then dividing by ∆∆∆∆x∆∆∆∆y∆∆∆∆z, we get the following energy balance equation per unit volume:
Transport Phenomena
( )( ) ( ) ( )
∂∂
ρ ρ
∂∂
ρ ρ
∂∂
ρ ρ
∂∂
ρ ρ
∂∂
∂∂
∂∂
ρ
∂∂
∂∂
∂∂
tU v
xv U v
yv U v
zv U v
qx
qy
qz
v g v g v g
xp v
yp v
zp v
x
y
z
x y z
x x y y z z
x y z
^
^
^
^
+
=
−
+
+
+
+
+
− + +
+ + +
− + +
12
12
12
12
2
2
2
2
( )( )
( )
−
+ + +
+ + +
+ +
∂∂
τ τ τ
∂∂
τ τ τ
∂∂
τ τ τ
xv v v
yv v v
zv v v
x x x x y y x z z
yx x yy y yz z
z x x z y y z z z
convection
conduction
gravity
pressure
viscous
Transport Phenomena
Such a massive energy equation can be written in a compact vector-tensor:
( )( )
( )[ ]( )
∂∂
ρ
ρ
ρ
τ
tU v
v U v
q
v g
pv
v
^
^
+
=
− ∇ • +
− ∇ •
+ •
− ∇ •
− ∇ • •
12
12
2
2
The equation obtained so far is fairly general.
rate of gain of energy per unit volume
rate of energy input per unit volume by convection
rate of energy input per unit volume by conduction
rate of work done on fluid per unit volume by gravitational forces
rate of work done on fluid per unit volume by pressure forces
rate of work done on fluid per unit volume by viscous forces
Transport Phenomena
BUT we could make use of the continuity equation to obtain another form:
Combine the accummulation term & the convection term, we get:
( )
( ) ( ) ( ) [ ]( )ρ ∂
∂∂ρ∂
ρ
ρ τ
tU v v U v U v
tv
q v g pv v
^ ^ ^+
+ • ∇ +
+ +
+ ∇ •
=
− ∇ • + • − ∇ • − ∇ • •
12
12
12
2 2 2
Recall the continuity equation:
( )∂ρ∂
ρt
v+ ∇ •
= 0
the above equation will become:
( ) ( ) ( ) [ ]( )ρ ρ τDDt
U v q v g pv v^+
= − ∇ • + • − ∇ • − ∇ • •1
22
in which we have used the definition of the substantial derivative.
DDt t
vx
vy
vzx y z= + + +∂
∂∂∂
∂∂
∂∂
Transport Phenomena
Now we recall the mechanical energy equation:
( ) ( ) ( )[ ]( ) ( )
ρ ρ
τ τ
DDt
v p v pv v g
v v
12
2
= ∇ • − ∇ • + •
− ∇ • • + ∇:
Substracting this mechanical energy equation from the energy equation:
( )( )
( )
ρ
τ
D UDt
q
p v
v
^
:
=
− ∇ •
− ∇ •
− ∇
This is called thermal energy equation.
rate of gain of internal energy per unit volume
rate of internal energy input by conduction
reversible rate of internal energy increase per unit volume by compression
irreversible rate of internal energy increase by viscous
Transport Phenomena
Equations written in terms of internal energy are not directly too useful.
One prefers equations written in terms of temperature and heat capacities
From thermodynamics:
d U U
Vd V U
TdT
p T pT
d V C dT
T V
Vv
^^
^
^^
^ ^
^
^
=
+
= − +
+
∂
∂
∂∂
∂∂
The thermal energy equation will become:
( ) ( ) ( )ρ ∂∂
τC DTDt
q T pT
v vvV
^
^:= − ∇ • −
∇ • − ∇
If we express q in terms of temperature gradient and stress tensors in terms of velocity gradient, the above equation will be in terms of temperature & velocity
Transport Phenomena
Special cases:
For a Newtonian fluid with constant k:
( ) ( )ρ ∂∂
µΦC DTDt
k T T pT
vvV
v
^
^= ∇ −
∇ • +2
where ΦΦΦΦv is the dissipation function, and it takes the form for rectangular coordinates:
Φ vx y z
y x z y x z
x y z
vx
vy
vz
vx
vy
vy
vz
vz
vx
vx
vy
vz
=
+
+
+
+
+ +
+ +
−
+ +
2
23
2 2 2
2 2 2
2
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
Transport Phenomena
Special case: Ideal gas
For ideal gas, the change of pressure versus temperature at constant volume
∂∂pT
pTV
=^
the thermal energy equation is:
( ) ( )ρ µΦC DTDt
k T p vv v
^= ∇ − ∇ • +2
Special case: Constant pressure fluid
From thermodynamics:
d U pd V C dTp^ ^ ^
= − +
and the thermal energy equation is:
( )ρC DTDt
k Tp^
= ∇ 2
neglecting viscous dissipation.
Transport Phenomena
Special case: Constant density fluid
We have for this case:
( )ρC DTDt
k Tp^
= ∇ 2
Special case: Solids
( )ρ ∂∂
C Tt
k Tp^
= ∇ 2
Transport Phenomena
Heat transfer coefficient What we have learnt?
1. Modes of heat transfer
2. Thermal conductivity & thermal diffusivity
3. Shell heat balance procedure
4. Simple examples
5. Temperature dist’n ⇒⇒⇒⇒ heat flux
6. Steady state vs transient
7. Shape and size of objects
8. Similarity between heat and momentum transfer.
9. Heat transfer coefficient (HTC)of simple problems can be obtained from first principles.
Now consider HTC a bit further.
Transport Phenomena
Definition of heat transfer coefficient
General definition:
Q=hA(Tw - T0)
stagnant film
hot surface, Tw
bulk fluid, T0
hot object
Transport Phenomena
The heat transfer coefficient, unlike thermal conductivity, does vary with the situation.
Two situations where HTC is defined
1. Flow in conduit
2. Flow around submerged objects
HTC:
1. Local heat transfer coefficient: vary along the direction of flow
2. Overall heat transfer coefficient: a combination of HTCs of processes in series
Transport Phenomena
Flow in conduits
There are three definitions:
( )( ) ( )( )Q h DL T T h DL T Tb b= − = −1 0 1 1 2 0 2 2π π, , , ,
( ) ( ) ( )Q h DL
T T T Tb b=− + −
10 1 1 0 2 2
2π , , , ,
( ) ( ) ( )( ) ( )[ ]Q h DLT T T T
T T T Tb b
b b
=− − −
− −
ln
, , , ,
, , , ,ln /π 0 1 1 0 2 2
0 1 1 0 2 2
T0,1 T0,2
Tb,1 Tb,2
Transport Phenomena
First definition:
( )( ) ( )( )Q h DL T T h DL T Tb b= − = −1 0 1 1 2 0 2 2π π, , , ,
This definition is based on information at one point. In general, h1 is different from h2.
Second definition:
( ) ( ) ( )Q h DL
T T T Tb b=− + −
10 1 1 0 2 2
2π , , , ,
The second definition is based on the average of the driving force.
The third definition:
( ) ( ) ( )( ) ( )[ ]Q h DLT T T T
T T T Tb b
b b
=− − −
− −
ln
, , , ,
, , , ,ln /π 0 1 1 0 2 2
0 1 1 0 2 2
The third definition is based on the log mean driving force.
Transport Phenomena
If the temperature of the fluid changes appreciably, the local heat transfer coefficient is preferable:
dQ = hloc (ππππDdz) (T0 - Tb )
The heat transfer coefficient obtained in the analysis of flow in a tube is local, i.e.
h = h(z)
Transport Phenomena
Flow around an object
Definition of the mean HTC
Q = hm (4ππππR2) (T0 - T∞∞∞∞ )
Definition of the local HTC
dQ = hloc (dA) (T0 - T∞∞∞∞ )
T0 T∞∞∞∞
Transport Phenomena
Warning about the use of HTC
To use the heat transfer coefficient, make sure that you know exactly:
1. The def. of temperature driving force
2. The def. of heat transfer area
Heat transfer coefficient is a function of:
1. the fluid properties (k, Cp, ρρρρ, µµµµ)
2. the system’s geometry
3. the flow velocity
4. the driving force
5. the surface temperature variation
h = h(k, Cp, ρρρρ, µµµµ, v, T0, Tb, D, L)
Transport Phenomena
Order of magnitude of HTC
HTC varies widely depending on the situation.
Situation h (kcal/m2/hr/K) Free convection Gases Liquids Boiling water
3 - 20 100 - 600 1000 - 20,000
Forced convection Gases Viscous liquids Water
10 - 100 50 - 500 500 - 10,000
Condensing vapors 1,000 - 100,000
Transport Phenomena
Correlations for HTC in tube
As we have seen, HTC is a function of
h = h(k, Cp, ρρρρ, µµµµ, v, T0, Tb, D, L)
There are many parameters in RHS. We need a means to reduce the number of parameters.
NON-DIMENSIONALIZATION
Transport Phenomena
For a tube of radius R and length L, the total heat flow into the fluid is:
Q kT r z
rRd dz
L
r R
=
∫∫
=
∂ θ∂
θπ ( , , )
0
2
0
Using the definition of heat transfer coefficient
( )( )Q h DL T Tb= −1 0 1 1π , ,
we get
( )( )hDL T T
k T r zr
Rd dzb
L
r R1
0 1 1 0
2
0
1=−
∫∫
=π∂ θ
∂θ
π
, ,
( , , )
For a constant physical properties system, the thermal energy balance equation is:
ρ µΦC DTDt
k Tp v= ∇ +2
In principles, by solving this equation and substitute the result into the HTC equation we will obtain h1.
Transport Phenomena
NONDIMENSIONLIZE:
By defining
t tVD
r rR
z zL
TT T
T Tb
* * * * ,
, ,
; ; ;= = = =−−
0 1
1 0 1
( )Nuh D
kDV C
kBr
Vk T T
p
b1
12
1 0 1
= = = =−
; Re ; Pr ;, ,
ρµ
µ µ
the HTC and thermal energy eqns become
( )NuL D
T r zr
d dzr
10
2
0
1
1
12
= −
∫∫
=π
∂ θ∂
θπ
/( , , )* * *
**
*
DTDt
T Brv
*
** * *
Re Pr Re Pr= ∇ +1 2 Φ
THUS, we see that
Nu Nu Br L D1 1= (Re, Pr, , / )
If we neglect the viscous dissipation
Nu Nu L D1 1= (Re, Pr, / )
Transport Phenomena
The functional dependence
Nu Nu L D1 1= (Re, Pr, / )
was obtained assuming constant physical properties.
For large temperature difference, the following functional dependence is expected to allow for temperature variation of the viscosity:
Nu Nu L D b1 1 0= (Re, Pr, / , / )µ µ
Transport Phenomena
Thus, we see that the process of nondimensionalization has reduced
h = h(k, Cp, ρρρρ, µµµµ, v, T0, Tb, D, L) to
Nu Nu L D1 1= (Re, Pr, / )
A significant reduction in the number of parameter dependence.
Similarly, we have:
Nu Nu L Da a= (Re, Pr, / )
Nu Nu L Dln ln (Re, Pr, / )=
Nu Nu z Dloc loc= (Re, Pr, / )
Transport Phenomena
Some correlations in the literature Highly turbulent flow:
Nu bln
. /.
. Re Pr=
0 026 0 8 1 3
0
0 14µµ
All physical properties are evaluated at (Tb,1 + Tb,2)/2, except µµµµ0 which is evaluated at (T0,1 + T0,2)/2.
Laminar flow
Nu DL
bln
/ // .
. Re Pr=
186 1 3 1 3
1 3
0
0 14µµ
Theoretical analysis by Leveque (1928) gives:
Nu DLln
/ //
. Re Pr=
162 1 3 1 3
1 3
which is remarkable result.
Transport Phenomena
Correlations for HTC around a submerged object & packed bed
The commonly used correlation for a sphere in an infinite environment:
h Dk
Dv Ck
m
f
f
f
p
f
= +
∞2 0 0 60
1 2 1 3
. ./ /
ρµ
µ
where all physical properties are evaluated at the film temperature.
The correlation for a packed bed is:
h Dk
Dv Ck
m
f
f
f
p
f
= +
∞2 0 11
0 6 1 3
. .. /
ρµ
µ
where all physical properties are evaluated at the film temperature.
The coefficient of 1.1 (instead of 0.6) and exponent 0.6 (instead of 0.5) are due to the density of packing in the fixed bed.
