Lecture 3 Spring 2000 ChE 333 1 Conduction in a Simple Fin Finned tubes are common in heat exchangers to afford greater surface area for heat transfer. How do they enhance heat transfer? Steady State Energy Balance Conductive flux at z q z z = –k d T d z z Conductive flux at z q z z+ ∆z = –k d T d z z+ ∆ z The convective flux in the x direction q x z = –hTz –T a Combining the equations, we obtain –k d T d z z – –k d T d z z 2Bw – h T z –T a 2w∆z=0 So that when we take the limit as z ---> 0 d dz k d T d z = h B Tz –T a The boundary conditions are at z = 0, T = T a and at z = L, q z L = –k d T d z L =0 L x T w z T a
Transcript
Lecture 3 Spring 2000ChE 333 1
Conduction in a Simple Fin
Finned tubes are common in heat exchangers to afford greater surface areafor heat transfer. How do they enhance heat transfer?
Steady State EnergyBalance
Conductive flux at z
qz
z
= – k dTdz z
Conductive flux at z
qz
z+ ∆z
= – k dTdz z+ ∆z
The convective flux in the x direction
qxz
= – h T z – Ta
Combining the equations, we obtain
– k dT
dz z
– – k dTdz z
2Bw – h T z – Ta 2w∆z = 0
So that when we take the limit as z ---> 0
ddz
kdTdz
= hB T z – Ta
The boundary conditions are
at z = 0, T = Ta and at z = L, qzL
= – k dTdz L
= 0
L
x
Tw
z
Ta
Lecture 3 Spring 2000ChE 333 2
Adimensionalization
If we define the following dimensions
θ = T – Ta
Tw – Ta
; ζ = zL ; N = hL2
kB
The differential equation becomes
d2θdζ2 = N2 θ
Using the boundary conditions
At ζ = 0 , θ = 1 and at ζ = 1 , dθdζ = 0
The equation becomes
θ = cosh Nζ + tanh N sinh Nζ
Rate of heat flow into fin through the base
Q = – k dT
dz z = 0
2Bw
Rate of convection from the surface
Q = 2wh T – Ta dz
0
L
Lecture 3 Spring 2000ChE 333 3
Bounds
If the fin’s temperature were the same as the base, the rate of heat transferfrom the fin surface would be
The minimum flux through an unfinned surface is
Q0 = 2wBh Tw – Ta
The maximum heat flow from the fin
Qmax = 2wLh Tw – Ta
The fractional heat flow through the fin is the efficiency
QQmax
=2wh T – Ta dz
0
L
2wLh Tw – Ta
Expressed in dimensionless terms the efficiency is
η = Q
Qmax
= θ dζ0
1
For a planar flat fin, theefficiency is given by
η = tanh N
N
Efficiency Plot
0.1
1
0.1 1 10
Modulus
Eff
icie
ncy
Lecture 3 Spring 2000ChE 333 4
Conduction with Heat Generation
Consider an insulated electrical wire of radius R1 covered with a thicknessof electrical insulation so that the outside radius is R2.
Joule heating is given by R2
Se = I 2R =I2
k e
R1
The heat flux in the copper wire is given by Fourier’s Law
q r = – k dTdr
An energy balance (steady-state) yields
2πrq r r
– 2πrq r r + ∆rL + S e 2πr dr L = 0
so that the differential equation becomes
ddr
rq r = rS e
When we invoke Fourier’s law, we obtain the conduction equation.
ddr
rdTdr
= rSe
k
When the equation is integrated in the wire region, we obtain
T = Se
4kr2 – a ln r + b
Lecture 3 Spring 2000ChE 333 5
The same can be done in the insulation to give
ddr
rdT1
d r= 0
The solution is
T1 = c ln r + d
Now there are four constants to determine four boundary conditions.
T = T1 and k dT
dr= k 1
dT1
drat r = R1
T = T1 and k dT
dr= k 1
dT1
drat r = R1
– k 1
dT1
d r= h T1 – Ta
The final condition is that the temperature T is finite at r = 0.The results are the following two equations.
T – Ta = S eR1
2
4k1 – r
R1
2– 2k
k 1
ln R1
R2
+ 2khR2
T – Ta = S eR1
2
2k 1
ln R2
r + 2khR2
Lecture 4 Spring 2000ChE 333 1
The Energy Balance
Consider a volume Ω enclosing a mass M and bounded by a surface δΩ.
At a point x, the density is ρ, the localvelocity is v, and the local Energydensity is U.
The rate of change total energy in Ω is:
ddt
ρUdVΩ
The heat flow from the body is
q • n dS∂Ω
The Work done by the body on the surroundings is
v•T•n dS∂Ω
+ ρg•vdVΩ
Since for the body
U = Q – W
An equivalent form is
ddt ρUdV
Ω+ q • n dS
∂Ω=
v•T•n dS∂Ω
+ ρg•vdVΩ
If our control volume is a differential cube, the differential equationdescribing the Energy Equation is:
∂∂tρU + ∇• ρUv + ∇•q = ∇• v•T + ρg•v
U v
ρvs.n
δΩ
Lecture 4 Spring 2000ChE 333 2
The first term is the local rate of energy changeThe second is the convective energy flowThe third is the sum of reversible work and dissipationThe last is the work done by the gravitational acceleration.
Other Conservation LawsMass
∂∂tρ + ∇• ρv = 0
Momentum
∂∂tρv + ∇• ρvv – ∇• T – ρg = 0
Mechanical EnergyThis is obtained by taking the inner product of the momentum equationand the momentum equation to yield
v• ∂
∂tρv + ∇• ρvv – ∇• T – ρg = 0
The real Energy Equation
The real Energy equation is obtained by subtracting the MechanicalEnergy Balance from the complete Energy Equation, using the massbalance and recognizing that H = U + PV.
ρ ∂
∂t H + v•∇H + ∇•q = τ••∇v – Ws + ℜα – ∆HαΣ
α =1
S
This is simplified recalling that
∂H∂T p
= Cp
ρCp
∂∂t T + v•∇T + ∇•q = τ•• ∇v – Ws + ℜ α – ∆H αΣ
α =1
S
Lecture 4 Spring 2000ChE 333 3
Applications of the Energy Equationto Steady State Conduction
The Energy Equation was
ρCp
∂∂t T + v•∇T + ∇•q = τ•• ∇v – Ws + ℜ α – ∆H αΣ
α =1
S
But for systems at steady state where there is no motion, no shaft workdone, and no chemical reaction
• time derivatives vanish• the velocity, v, is zero• the shaft work iz zero, and• the reaction rate is zero..
This means that the energy equation has a very simple form
Recall that Fourier's Law is a relation for the heat flux, q,
so that
In rectangular Cartesian coordinates, the resulting equation becomes thesteady state head conduction equation .
∇•q = 0
q = – k ∇ T
∇• k ∇ T = 0 and it follows for constantk, that ∇2 T = 0
∂ 2T
∂x2
+∂2T
∂y2
+∂2T
∂ z2
= 0
Lecture 4 Spring 2000ChE 333 4
Boundary Conditions
Types of boundary conditions in heat transfer problems
1. Constant surface temperature
On a surface S, the temperature is constant ifT(x, t) = Ts
2. Constant heat flux
a) At a surface S, the flux is continuous, finite, and constant sothat :
b) At an adiabatic surface S, the flux vanishes:
3. Convective Surface condition
At any surface, the flux leaving one body is equal to the flux leavingthe other, so that
q i = – k
∂T∂x i S
?
q i = – k
∂T∂x i S
?
= 0
– k
∂T'∂x i S
1
= – k∂T"∂x i S
2
Lecture 4 Spring 2000ChE 333 5
A Simple Steady State Conduction Problem
Consider a rectangular slab of infinite extent in the z-direction
_______ side is length L, the vetrical sides are of length W.
The differential equation for steady heat conduction in 2 dimensions is:
The boundary conditions are:
T1
T1
T1
T2
∂ 2T
∂x2+
∂ 2T
∂y2= 0
T = T1 at y = WT = T0 at y = 0
T = T0 at x = 0
T = T0 at x = L
Lecture 4 Spring 2000ChE 333 6
If the Temperature. T, and the independent variables , x and y, are madedimensionless, as
The conduction equation becomes
With the boundary conditions transformed to
The method we use to solve this partial differential equation is "the methodof separation of variables".
Assume that the solution is of the form
We obtain an equation of the form
Θ =
T – T0
T1 – T0
and η =yW
and ζ =xL
∂ 2Θ
∂ζ2+
LW
2 ∂2Θ
∂η 2= 0
Θ = 1 at η = 1Θ = 0 at ζ = 0Θ = 0 at η = 0Θ = 0 at ζ = 1
Θ = F ζ G η
∂ 2F ζ G η
∂ζ2+
LW
2∂2F ζ G η
∂η2= 0
Lecture 4 Spring 2000ChE 333 7
We group the terms that depend on each individual independent variableso that
If we divide by FG and separate variables, we obtain
G η
d2F ζ
dζ2+
LW
2
F ζd2G η
dη2= 0
For simplicity, let α 2 =
LW
1
F ζd2F ζ
dζ2= –
LW
21
G ηd2G η
dη2= constant = – λ2
Lecture 4 Spring 2000ChE 333 8
The result is that we have two ordinary differential equations to solve:
so that
The solutions to the pair are :
The entire solution is of the form
If we recognize that since at ζ = 0, θ = 0, then B = 0 and the solutionsimplifies considerably.
d2F ζ
dζ2+ λ2F ζ = 0
d2G η
dη2–
λα
2
G η = 0
Θ = F ζ G η = A sin λζ + B cos λζ
C sinhλαη + D cosh
λαη
F ζ = A sin λζ + B cos λζ
G η = C sinhλαη + D cosh
λαη
Θ = A' sinh
λαη sin λζ + B' cosh
λαη sin λζ
Lecture 4 Spring 2000ChE 333 9
The boundary condition at η = 0 gives
So that B' must be zero and the simplified solution is
There are two constants left, λ and A', and two boundary conditions.The condition at ζ = 1 leads to
And we must note thatEither A' must vanish and the solution is trivial or
This is true if and only if λ = nπ where n = 0, 1,2,3,....... That means that there are a countable infinite number of solutions. Tofind the solution we need to add all the possible solutions and determinethe coefficients (constants).
Θ = 0 at η = 0 leads to
0 = A' sinh 0 sin λζ + B' cosh 0 sin λζ
Θ = A' sinh
λαη sin λζ
0 = A' sinh
λαη sin λ
0 = sin λ
Θ = an sinh
nπα η sin nπζΣ
n= 1
∞
Lecture 4 Spring 2000ChE 333 10
The coefficients may be determined by the last boundary condition.
To determine the coefficient an, we have to recognize the orthogonalityproperty of sin functions, that is,
To determine the coefficients, we can use the orthogonality properties ofthe sine and cosine functions.
We integrate
Θ = 1 at η = 1
1 = an sinhnπα sin nπζΣ
n= 1
∞
sin(nπζ)sin(mπζ)dξ0
1
=0 for m ≠ n
π2
for m = n
ansinhnπα sin(nπζ)sin(mπζ)dζ
0
1
Σn= 1
∞
= (1)sin(mπζ)dζ0
1
Lecture 4 Spring 2000ChE 333 11
Remember that the first sine integral is non-zero if and only if n = m.Now the equation for an is
or
Finally the solution is
an
π2
sinhnπα = sin(nπζ)dζ
0
1
=– cos (nπζ)
0
1
n
an =
2π
1 – 1n
n sinhnπα
Θ =2 1 – 1
n
n π
sinhnπα η
sinhnπα
sin nπζΣn= 1
∞
ChE 333- Lecture 6 1Spring 2000
Applications of the Energy Equation
Solids with a Uniform Temperature
Suppose a metal sphere of uniform temperature, T.Heat is transferred by convection with a heat transfercoefficient, h. The temperature of the surroundings isTa
Therefore the energy balance is
ρCpVdTdt
= – hA T – Ta
(Please note the difference between this equation and 11.1.1 in the text. This is thecorrect form)
If the sphere is initially at T0, how does one describe the cooling of thesphere?
The equation is separable so that dT
T – Ta
= – hAρCpV
dt
It follows that the solution is
ln T – Ta
T0 – Ta
= – hAρCpV
t
Or more explicitly T – Ta
T0 – Ta
= e – hAρCpVt
ChE 333- Lecture 6 2Spring 2000
Adimensionalization
If I had defined the following:
θ = T – Ta
T0 – Ta
and τ =ρC pV
hA
The solution has a simple expression ... Θ = e-t/τ
Measurement of a Convective Heat Transfer Coefficient
Suppose a sphere of radius R in a stagnant gas of infinite extent. Theheat flux from the sphere through the gas is given by Fourier’s law
q r = – k gdTdr
The heat flow through a spherical shell is constant so
r2q r = – r 2k gdTdr
= C
The boundary conditions are T = Ts at r = R and T -> Ta at r -> ∞
The solution becomes
T – Ta
TR – Ta
=Rr
ChE 333- Lecture 6 3Spring 2000
We can calculate the heat flux at the surface as
q r
r = R
= – k gdTdr r = R
= k g
TR – Ta
R
We can define a heat transfer coefficient as
h ≡ q r
TR – Ta
=k g
R
The corresponding Nusselt Number for heat transfer in a stagnant gas is
This represents a lower bound for convective heattransfer, given that there is no gas flow. We expectthe heat transfer coefficient to be larger.
Nu = hDk g
= 2
ChE 333- Lecture 6 4Spring 2000
The experiment
The thermal conductivity of air is 0.014 Btu/ft-°F. If the sphere was 1cm. in diameter, then h = 0.014(1/2.54(12)) = 4.2 Btu/hr-ft2-°F
Note that h = = 4.2 Btu/hr-ft2-°F = 0.00117 Btu/sec-ft2-°F
If ρ = 436 lb/ft3 and Cp = 0.12 Btu/lb.-°F, then
τ =
0.00117 6
436 0.12 1 / 30= 0.00403 sec
If the heat transfer coefficient were 5 times larger (Nu = 10) then τ =0.02 sec,
If the heat transfer coefficient were 10 times larger (Nu = 100) then τ =0.2 sec,
It should be clear that we cannot make a reasonable verification of auniform temperature until we solve for the temperature field in thesphere.
ChE 333 - Lecture 7` 1Spring 2000
Unsteady State Heat Conduction in a Bounded Solid
Consider a sphere of radius R. Initially the sphere is at a uniformtemperature T0. It is cooled by convection to an air stream at temperatureTa. What do we do to describe the conduction and the temperaturedistribution within the sphere?
You will recall that in the last lecture the uniform temperatureapproximation was quite unrealistic when it came to the time scale forcooling.
There are a number of methods that we can use to attack this problem.The first is to do a shell balance on a differential shell within the sphere.That exercise leads to
4π r 2q r r– 4π r2q r r + ∆r
= ∂∂t ρU4πr2∆r
If we divide both sides by 4πr2∆r, we obtain
r 2q r r– r2q r r + ∆r
∆r= r2 ∂
∂t ρU
Recognizing that for a solid, Cp ≅ Cv and if the density is constant, weobserve that
– 1
r2
∂∂r r 2q r = ∂
∂t ρU = ρ ∂U∂T p
∂T∂t = ρCp
∂T∂t
so that
– 1r2
∂∂r r 2q r = ρC p
∂T∂t
At this point we should introduce Fourier’s law of heat conduction
ChE 333 - Lecture 7` 2Spring 2000
q r = – k s∂T∂r
The time-dependent equation for heat conduction becomes
ρCp
∂T∂t = 1
r2
∂∂r r2k s
∂T∂r
We can define the Fourier Diffusivity or thermal diffusivity as
α = k s
ρCp
The conduction equation becomes mathematically equivalent to thediffusion equation
∂T∂t = α 1
r2
∂∂r r2 ∂T
∂r
Boundary and Initial Conditions
The problem becomes defined only when the initial conditions and theboundary conditions are specified.
For t < 0, we have T = T0 in r ε (0,R)For r = 0, the temperature field is bounded for all t > 0
or ∂T∂r
= 0 (symmetry)
For r = R, we have Newton’s Law of Cooling, that is
q r = – k s∂T∂r = h T – Ta
ChE 333 - Lecture 7` 3Spring 2000
The Dimensionless Description
My passion for a dimensionless problem exists because I’m fundamentallylazy and I don’t want to solve a problem each time I formulate it.
Let’s put the equations in dimensionless form
θ = T – Ta
T0 – Ta
; ξ = rR
XFo = αtR2 ; Bi = hR
k s
Applying these definitions to the dimensionless conduction equation andits boundary and initial conditions, we obtain
∂θ∂xFo
= 1ξ2
∂∂ξ ξ2 ∂θ
∂ξ
θ = 1 at xFo < 0 ; ∂θ∂ξ = 0 at ξ = 0
– ∂θ∂ξ = Bi θ at ξ = 1
It follows that Θ = f(ξ, xF0 , Bi). The solution procedure then uses a simpletransformation z = ξ θ to obtain the form of the problem that we havealready seen and solved in gory detail. The most common problem iswhen the Bi -> ∞ , then Θ = f(ξ, xF0)
since the boundary condition 3 becomes Θ = 0
ChE 333 - Lecture 7` 4Spring 2000
The Adimensional Problem
Let's try to solve the problem by first making the equation adimensional.Define the following:
The differential equation and the initial and boundary conditions become:
The equation and its conditions are dimensionless and parameter free.
Next Question
How do we solve the equation ?
Suppose z has the form z = Y(XFo)G(ξ)
z = ξ θ ; ξ =
rR
; XFo =α t
R2
∂z∂XFo
–∂2z
∂ξ2 = 0
z = ξ at xFo < 0 ; z = 0 at ξ = 0
–
∂z∂ξ
= Bi + 1 z at ξ = 1
∂z∂XFo
–∂2z
∂ξ2 =∂YG∂XFo
–∂2YG
∂ξ2 = GdY
dXFo
– Yd2Gdξ2 = 0
ChE 333 - Lecture 7` 5Spring 2000
The equation is separable in the form
Integrating each of the equations we obtain
The solution for y(θ,η) has the form
We can construct the exact solution using the boundary conditions
It follows that B must be 0 if the condition is true for all θ > 0Now the other boundary condition
Now this is true for all XFo > 0 if and only if sin(λ) = 0but sin(λ) = 0 only where λ = nπ where n = 0, 1, 2, .....
1Y
dYdXFo
=1G
d2Gdξ2 = – λ2
dYdXFo
= – λ2Y ;d2Gdξ2 = – λ2G
Y(XFo) = Ke–λ2XFo and G(ξ) = Asin(λξ) +Bcos(λξ)
z(0,X Fo) = Asin(0) +Bcos(0) e–λ2XFo = 0
z(1,X Fo) = Asin(λ) e–λ2XFo = 0
z(ξ, XFo) = Asin(λξ) +Bcos(λξ) e–λ2XFo
ChE 333 - Lecture 7` 6Spring 2000
This means there are a countable infinity of solutions so that
To obtain the coefficients An , we need to use the initial condition.
To determine the coefficients, we can use the orthogonality properties ofthe sine and cosine functions. (See Appendix)
z = ξ at xFo < 0
z(ξ, 0) = A n sin(nπξ) = ξΣ
n= 1
∞
z(ξ, XFo) = e –λ2XFo A n sin(nπξ)Σ
n= 1
∞
sin(nπξ)sin(mπξ)dξ
–1
1
=0 for m ≠ nπ for m = n
ChE 333 - Lecture 7` 7Spring 2000
We integrate
You might remember that the first sine integral is non-zero if and only ifn = m. Now the equation for An is
The result for the definite integrals follow from what I give above . Itfollows that
An = 4
π1 – – 1 m – 1
2
so that the solution can be described as :
z(ξ,0)sin(mπξ)dξ
0
1
= ξsin(mπξ)dξ0
1
A n sin(nπξ)sin(mπξ)dξ
0
1
Σn= 1
∞
= ξsin(mπξ)dξ0
1
A n =– ξsin(nπξ)dξ
0
1
sin2(nπξ)dξ0
1=
1nπ
xsin(x)dx0
nπ
sin2(x)dx0
nπ
z(ξ, XFo) =4π
1 – – 1n– 1
2ne –n2π2XFosin(nπξ)Σ
n= 1
∞
ChE 333 - Lecture 7` 8Spring 2000
A LITTLE EXERCISE
Prove that the following is equivalent
A Return to the Original Problem
“ How long does it take to get a fully developed temperatrurefield in the sphere?”
Recall the initial definitions
θ = T – Ta
T0 – Ta
; ξ = rR
XFo = αtR2 ; Bi = hR
k s
When θ = 0 everywhere, then the temperature profile is developed. Thesolution we saw was
z(ξ, XFo) =
4π
12n + 1
e– 2n +12π2XFosin( 2n + 1 πξ)Σ
n= 0
∞
θ(ξ,X Fo) =
z(ξ,X Fo)ξ
θ(ξ,X Fo) =
4ξπ
12n + 1
e– 2n +12π2XFosin( 2n + 1 πξ)Σ
n= 0
∞
ChE 333 - Lecture 7` 9Spring 2000
Now regardless of location in the sphere, θ is small when ξ is large. Now how large ? The first term in the series is when n = 1, thesecond is when n = 3, etc. The first term is much greater than thesubsequent one so that
so as θ→∞ y(η, θ) → 4πe–π2θsin(πη)
These solutions can be compared graphically.
Temperature Response
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.5 1 1.5
Dimensionless time
Temperature
It is apparent that by the time XFo = 0.5, θ = 0.01, that is 1 % of the originalvalue. I’ll presume this is long enough.
then XFo→∞
z(ξ,X Fo)z(ξ,0)
→ e–π2XFo
ChE 333 - Lecture 7` 10Spring 2000
You should notice though that on a semi-log plot the data are almost astraight line. What does this suggest ?
Relative Temperature Plot
0.0001
0.001
0.01
0.1
1
0 0.2 0.4 0.6 0.8 1
Time
ChE 333 - Lecture 7` 11Spring 2000
AppendixProblem --- evaluate An
The integrals in the definition of An can be proven simply by noting that eix = cos(x) + i sin(x)
and using it to prove a number of identities involving sines and cosines.
The result is a simple relation for the Fourier coefficients.The results are
cos(nx)sin(mx)dx
0
π
= 0 for all m, n
cos(nx)cos(mx)dx0
π
=
0 for m≠nπ2 for m=n > 0
π for m=n = 0
sin(nx)sin(mx)dx0
π
=
π2 for m=n > 0
0 for m≠n
xsin(mx)dx
0
π
= πm – 1 m
It shows that sin(x) and cos(x) are orthogonal functions. Some algebrawill get you the result for An.
An = 4
π1 – – 1 m – 1
2
Lecture 8ChE 333 1
Unsteady State Heat Conduction in a Bounded SolidHow Does a Solid Sphere Cool ?
We examined the cooling a sphere of radius R. Initially the sphere is at auniform temperature T0. It is cooled by convection to an air stream attemperature Ta.
How long does it take to cool to Ta?
The answer is and was simple ..... an infinitely long time. In a sensebecause it is the answer equivalent to the solution of the ArchimedeanParadox.
If I walk half the distance to a wall, how many steps will I have to take toreach the wall ?
The answer is an infinite number!However, if I get within a millimeter of the wall, for intents and purposes, Iam there.
So if the temperature is within 1 % of the final temperature, it will havereached the final temperature.What temperature am I taking about.....the center line temperature, thesurface temperature, the average temperature ???We will choose the average temperature, θ.
The first model we looked at should be valid for small Biot numbers
θ = T – Ta
T0 – Ta
= e– hAρCpV
t = e–3Bix Fo
To get this form we had to recognize that for spheres, A/V = 3/R.
If θ = 0.01, then 3 Bi xFo = ln (0.01), so that xFo = 1.535/Bi
At large Biot numbers, the suitable model was
θ sph = 0.608 e– 9.87x Fo
Lecture 8ChE 333 2
Now for this case the value of xFo it takes to reach Θ = 0.01 is
xFo = 0.415
A plot of the response is shown below
Time for a temperature drop of 99%
0.1
1
10
100
0.01 0.1 1 10 100Bi
XFo
Some Dimensional Arguments
At large Biot numbers, the dimensionless time is constant, that is, xFo = 0.415, but
xFo = αtR2
so that for two spheres one of size R1 and another R2, the ratio of thecooling times is
t1
t2= R2
R1
2
Lecture 8ChE 333 3
Heat Transfer in a Semi-Infinite region
The questions we have posed thus far and the solutions have been reallyonly applicable for “Long” times. That is, when the temperature field inthe sphere, for example, has developed to the center of the sphere. Butwhat happens at “Short” times?
Consider a large planar solid whose extent (y-direction) is very large.What is the temperature history of the slab if it is suddenly brought intocontact with a fluid at temperature Ta? The transient conduction equationis
∂T∂t = α∂ 2T
∂y2
at t = 0, T = T0
at y = 0 , T = Ta
as y → ∞ , T = T0
Let’s make the problem dimensionless.
The temperature can be expressed as
θ = T – Ta
T0 – Ta
so that the problem reposed is
∂θ∂t = α∂ 2θ
∂y2
θ = 1 at t = 0θ = 0 at y = 0θ = 1 as y → ∞
Lecture 8ChE 333 4
Solution
Let Θ = f(η(y,t)) where η = cymtn, then we can introduce that into thedifferential equation.
∂θ∂t = dθ
dη∂η∂t = dθ
dηηt = dθ
dη cnymtn– 1
∂θ∂y = dθ
dη∂η∂y = dθ
dηηy
= dθdη cmym – 1 tn
∂2θ∂y2 = dθ
dη∂2η∂y2 + d 2θ
dη 2
∂η∂y
2
∂2θ∂y2 = dθ
dη cm m – 1 ym– 2tn + d2θdη2 c2m2y 2m–2t2n
putting these derivatives into an equation we get
dθdηcnymtn –1 = α dθ
dηcmym –1tn + α dθdηcm m – 1 ym– 2tn + d 2θ
dη2c2
m2y2m –2t2n
We can divide by cymtn so that we obtain dθ
dηnt– 1 = αdθdηmy– 1 + α dθ
dηm m – 1 y –2 + d 2θdη 2cm 2y–2
Grouping we obtain a more compact form dθ
dη nt– 1– αmy–1 – α m m – 1 y–2 = d2θdη2 cm2y– 2
Note that things simplify if we pick m = 1. A bit of exercise will show thatthe appropriate choice for η is
η =y4αt
Lecture 8ChE 333 5
Solution of the Differential Equation
The equation becomes and ordinary differential equation
2η dθ
dη + d2θdη 2 = 0
The boundary conditions are
θ = 1 for η → ∞θ = 0 for η = 0
The solution we have seen is related to an error function defined as
erf (x) = 1
π e–t 2dt0
η
This solution affords us the opportunity to talk of an effective penetrationdepth, δT , that is, the distance at which the dimensionless temperature goesfrom 0 to 0.99.
The solution for Θ is Θ = erf(η), so that for the penetration depth
θ = 0.99 = erf δT
4αt
It follows that
δT
4αt= 2
This means that if δT is less than the thickness of the slab, it behaves as asemi-infinite region.
Lecture 8ChE 333 6
Heat Conduction with a Convective Boundary Condition
The boundary condition at the cooling surface can have a major effect onthe process. The problem in this instance is posed as
∂T∂t = α∂ 2T
∂y2
at t = 0, T = T0
at y = 0 , – k s∂T∂y = h T – Ta
as y → ∞ , T = T0
The problem can again be solved using combination of variables and thesame transformation as above to yield
θ = erf y
4t+ exp y + t erfc y
4t+ t
where y = hy
k s
and t = hk s
2
αt
Lecture 8ChE 333 7
Surface temperature of a Cooling Sheet
Polyethylene is extruded and coated onto an insulated substrate, moving at20 cm/sec. The molten polymer is coated at a uniform temperature T0 of400°F. Cooling is achieved by blowing air at a temperature Ta of 80°F.Earlier heat transfer studies determined that the heat transfer coefficient, h,is 0.08 cal/cm-sec-°F. The coating thickness B is 0.1 cm.
At what point downstream does the surface temperature, T(0) fall to 144°F ?
The Gurney-Lurie Chart 11.4c yields for Bi ≈ 10, the ratio of the surfacetemperature to the mid-plane temperature
However, since θ
θ10 = 0.15
, we can calculate the mid-plane temperature
from the relation for θs which is θ s = θ1
0 θθ1
0 = 0.2
This gives a midplane-temperature of θ10 = 0.2/0.15 > 1........Nonsense
What’s wrong ???We did a lot of things wrong.
Lecture 8ChE 333 8
First of all the solution we used involved only 1 term of an infinite series...
θ 1 = A 1e–β 12xFosin β 1ξ = θ1
0sin β 1ξ
We also get into trouble if we use such an equation for a “short” timesolution. Therefore avoid the charts for small xFo and large Biot numbers.
The “short time” solution we presented in the last lecture had the form.
θ = erf y
4t+ exp y + t erfc y
4t+ t
where y = hB
k s
yB and t = hB
k s
2 αtB2
Now for this case, y = 0 and we can use figure 11.3.2. We can determinethat the value of t at which Θ = 0.2.
We observe that t1/2 = 2.65 and consequently t = 7.65
Recall that ρCp = k/α = 2.5 MJ/m2-°K.
This leads to
t = hB
k s
2 αtB2 = 7.65 = h
k s
2
αt
We calculate that the time passed is t = 0.52 seconds and since d = Vt, thedistance is d = (20 cm/s) 0.52 sec = 10.4 cm.
Lecture 8ChE 333 9
An alternative method
We can use the complete Fourier expansion, not just one term.
θ =
4 sin λ n
2λn+ sin 2λn
cos λnξ e – λn2xFoΣ
n =1
∞
λntan λn = Bi
The first set of eigenvalues are
n ln1 1.4292 4.3063 7.228
If we calculate the first three terms of the Fourier expansion, we obtain
θ(1) = 0.178e–2.04x Fo + 0.155e –18.5xFo
For Θ = 0.2, by trial and error, we obtain xFo = 0.608. If we calculate thetime, we get 0.52 sec. The same as the short time solution.
This allows us a measure of “short time”. as for a slab
4 αtB2 ≤ 1 or x Fo ≤ 1
16
Lecture 9ChE 333 1
Heat Transfer in a Slab
Consider a large planar solid whose thickness (y-direction) is L. What isthe temperature history of the slab if it is suddenly brought into contactwith a fluid at temperature T? The transient conduction equation is
Let’s make the problem dimensionless.
The temperature can be expressed as
so that the problem reposed is
∂T∂t
= α∂2T
∂y2
at t = 0,T = T0
at y = 0 , T = T1 for t > 0at y = L , T = T1
θ =
T – T1
T0 – T1
; η =yL
; τ =αt
L2
∂θ∂t
= α∂2θ∂y2
θ = 1 at τ = 0θ = 0 at η = 0θ = 0 as η = 1
Lecture 9ChE 333 2
How do we solve the equation ?
Suppose z has the form Θ = Y(τ)G(η)
The equation is separable in the form
Integrating each of the equations we obtain
The solution for y(θ,η) has the form
We can construct the exact solution using the boundary conditions
It follows that B must be 0 if the condition is true for all θ > 0
∂θ∂τ –
∂2θ∂η2 =
∂YG∂τ –
∂2YG
∂η2 = GdYdτ
– Yd2Gdη2 = 0
1Y
dYdτ
=1G
d2Gdη2 = – λ2
dYdτ
= – λ2Y ;d2Gdη2 = – λ2G
Y(τ) = Ke–λ2τ and G(η) = Asin(λη) +Bcos(λη)
θ(η, τ) = Asin(λη) +Bcos(λη) e–λ2τ
θ(0, τ) = Asin(0) +Bcos(0) e –λ2τ = 0
Lecture 9ChE 333 3
Now the other boundary condition
Now this is true for all τ > 0 if and only if sin(λ) = 0but sin(λ) = 0 only where λ = nπ where n = 0, 1, 2, .....
This means there are a countable infinity of solutions so that
To obtain the coefficients An , we need to use the initial condition.
To determine the coefficients, we can use the orthogonality properties ofthe sine and cosine functions. (See Appendix)
sin(nπξ)sin(mπξ)dξ
–1
1
=0 for m ≠ nπ for m = n
θ(1, τ) = Asin(λ) e–λ2τ = 0
θ(η, τ) = e–λ2τ A n sin(nπη)Σ
n= 1
∞
θ = 1 at τ < 0
z(η, 0) = A n sin(nπη) = 1Σn= 1
∞
Lecture 9ChE 333 4
We integrate
You might remember that the first sine integral is non-zero if and only ifn = m. Now the equation for An is
The result for the definite integrals follow from what I gave above . Itfollows that
We saw earlier that the solution can be described as:
θ(η, 0)sin(mπη)dξ
0
1
= sin(mπη)dη0
1
A n sin(nπη)sin(mπη)dη
0
1
Σn= 1
∞
= sin(mπη)dη0
1
A n =– sin(nπη)dη
0
1
sin2(nπη)dη0
1= –
sin(x)dx0
nπ
sin2(x)dx0
nπ
A n =
4π
– 1n
n +12
Lecture 9ChE 333 5
We have already examined how the sum converges. For τ > 0.2, onlyone term suffices to describe the solution. We can look at many differentclasses of problems. The general problem for transient heat transfer in aslab is one posed as
The dimensionless form is:
The solution is of the form
θ(η, τ) = 4
π1
2n + 1e– 2n +1
2π 2 τsin( 2n + 1 πη)Σn = 0
∞
∂T∂t = α∂ 2T
∂y2
at t = 0, T = T0
at y = 0 , – k ∂T∂y = h T – T1 for t > 0
at y = L2 , ∂T
∂y = 0
Cn =
4 sin ζn
2ζn + sin 2 ζ n
and ζn tan ζn = Bi
θ(η, τ) = Cne–ζ n
2 τsin(ζn2η)Σ
n = 0
∞
∂θ∂t = α∂2θ
∂y2
θ = 1 at τ= 0∂θ∂η = – Bi θ at η = 0
∂θ∂η = 0 as η = 1
Lecture 9ChE 333 6
Again the approximate solution is the one-term solution
This argument is the same for any transient 1-dimensional heat transferproblems involving cylinders, planes or spheres.
Examples
Infinite Cylinder
The solution is
An approximate one-term solution is
Note that along the center line
θ(η, τ) ≈ C1e–ζ 12 τsin(ζ1
2η)
θ(η, τ) = Cne–ζ n
2 τJ 0(ζnη)Σn = 0
∞
Cn = 2
ζn
J1 ζn
2 J02 ζn + J1
2 ζn
and ζn
J1 ζn
J0 ζn
= Bi
∂θ∂ τ = 1
η∂
∂η η∂θ∂η
θ = 1 at τ= 0 in η ∋ 0, 1∂θ∂η = – Bi θ at η = 1
∂θ∂η = 0 at η = 0
θ(η, τ) = C1e–ζ 12 τJ0(ζ1η)
θ 0(0, τ) = C1e–ζ 12 τ
Lecture 9ChE 333 7
So that the simpler representation is θ(η, τ) = θ 0 J0(ζ1η)
Lecture 9ChE 333 8
Sphere
The solution is
The Approximate Solution
The center temperature is
So that the temperature can be expressed as
∂θ∂ τ = 1
η2∂
∂η η2 ∂θ∂η
θ = 1 at τ= 0 in η ∋ 0, 1∂θ∂η = – Bi θ at η = 1
∂θ∂η = 0 at η = 0
Cn =
4 sin ζ n – ζncos ζn
2 ζn + sin 2ζn
and 1 – ζ n cot ζn = Bi
θ(η, τ) = Cne–ζ n
2 τsin(ζnη)ζnη
Σn = 0
∞
θ(η, τ) = C1e–ζ 1
2 τsin(ζ1η)ζ1η
θ 0(0, τ) = C1e–ζ 12 τ
θ(η, τ) = θ 0
sin(ζ1η)ζ1η
Lecture 9ChE 333 9
Short Time Solutions
Consider a large planar solid whose extent (y-direction) is very large.What is the temperature history of the slab if it is suddenly brought intocontact with a fluid at temperature Ta? The transient conduction equationis
Let’s make the problem dimensionless.
The temperature can be expressed as
θ = T – Ta
T0 – Ta
so that the problem reposed is
∂T∂t = α∂ 2T
∂y2
at t = 0, T = T0
at y = 0 , – k ∂T∂y = h T – Ta Ta
as y → ∞ , T = T0
∂θ∂t = α∂2θ
∂y2
θ = 1 at t = 0∂θ∂y = Bi θ at y = 0
θ = 1 as y → ∞
Lecture 9ChE 333 10
Solutions
We noted earlier that the equation can be solved by a combination ofvariables supposing that Τ = Τ(η,t) and we saw that the the appropriatechoice for η is
η =y4αt
The solutions for a number of different cases are as follows:
Case 1 – Constant Surface Temperature (T = Ts)
Case 2 – Constant Surface Heat Flux (q”s = q”0)
Case 3 –Surface Convection
θ = erf y
4t+ exp y + t erfc y
4t+ t
T – Ts
Ti – Ts
= erf yαt
q"
s t =k Ts – Ti
παt
T – Ts =2 q"
0αtπ
ke–
y 2
4αt – q"0yk
erfc y4αt
Lecture 9ChE 333 11
where y = hy
k s
and t = hk s
2
αt
Lecture 9ChE 333 12
Surface temperature of a Cooling Sheet
Polyethylene is extruded and coated onto an insulated substrate, moving at20 cm/sec. The molten polymer is coated at a uniform temperature T0 of400°F. Cooling is achieved by blowing air at a temperature Ta of 80°F.Earlier heat transfer studies determined that the heat transfer coefficient, h,is 0.08 cal/cm-sec-°F. The coating thickness B is 0.1 cm.
At what point downstream does the surface temperature, T(0) fall to 144°F ?
The Gurney-Lurie Chart 11.4c yields for Bi ≈ 10, the ratio of the surfacetemperature to the mid-plane temperature
However, since θ
θ10 = 0.15
, we can calculate the mid-plane temperature
from the relation for θs which is θ s = θ1
0 θθ1
0 = 0.2
This gives a midplane-temperature of θ10 = 0.2/0.15 > 1........Nonsense
What’s wrong ???We did a lot of things wrong.
Lecture 9ChE 333 13
First of all the solution we used involved only 1 term of an infinite series...
θ 1 = A 1e–β 12xFosin β 1ξ = θ1
0sin β 1ξ
We also get into trouble if we use such an equation for a “short” timesolution. Therefore avoid the charts for small xFo and large Biot numbers.
The “short time” solution we presented in the last lecture had the form.
θ = erf y
4t+ exp y + t erfc y
4t+ t
where y = hB
k s
yB and t = hB
k s
2 αtB2
Now for this case, y = 0 and we can use figure 11.3.2. We can determinethat the value of t at which Θ = 0.2.
We observe that t1/2 = 2.65 and consequently t = 7.65
Recall that ρCp = k/α = 2.5 MJ/m2-°K.
This leads to
t = hB
k s
2 αtB2 = 7.65 = h
k s
2
αt
We calculate that the time passed is t = 0.52 seconds and since d = Vt, thedistance is d = (20 cm/s) 0.52 sec = 10.4 cm.
Lecture 9ChE 333 14
An alternative method
We can use the complete Fourier expansion, not just one term.
θ =
4 sin λ n
2λn+ sin 2λn
cos λnξ e – λn2xFoΣ
n =1
∞
λntan λn = Bi
The first set of eigenvalues are
n λn1 1.4292 4.3063 7.228
If we calculate the first three terms of the Fourier expansion, we obtain
θ(1) = 0.178e–2.04x Fo + 0.155e –18.5xFo
For Θ = 0.2, by trial and error, we obtain xFo = 0.608. If we calculate thetime, we get 0.52 sec. The same as the short time solution.
This allows us a measure of “short time”. as for a slab
4 αtB2 ≤ 1 or x Fo ≤ 1
16
Lecture 10 3/7/00ChE 333 1
Convective Heat Transfer
Examples
1. Melt Spinning of Polymer fibers2. Heat transfer in a Condenser3. Temperature control of a Re-entry vehicle
Fiber spinning
The fiber spinning process presents a unique engineering problem,primarily due to the effects of shape variations, heat and possibly theviscoelastic behavior of the materials (polymers for example) typicallyused. This becomes evident when the design of the spinneret geometry isneeded to produce a specified fiber size and shape. Determining the properdie geometry given the desired final fiber shape is further complicated bythe heat and viscoelastic effects. In addition, since the fiber is pulled fromthe spinneret, the final dimensions of the fiber are difficult to determine.The effects of viscous heating and air cooling must be monitored to ensurethat the material does not degrade because of extreme local temperatures,often difficult to measure because of the small size. The stresses anddeformation of the material must also be predicted to avoid the fiber frombreaking. All these effects complicate the design of the fiber spinningprocess.
Lecture 10 3/7/00ChE 333 2
Definition of a Heat Transfer Coefficient
For heat transfer in conducting systems, we have seen that we can expressthe heat flux across a surface S as
q • n
s
= – k s∇T • ns
We have used a similar representation to develop detailed descriptions ofmass and transfer in a number of situations where the physics or chemistryis well-understood, e.g., permeation through a membrane, heat transfer to asphere. We showed that we could get a description of the macroscopictransfer across an interface by the use of a Heat Transfer Coefficient.
q • n
s
= h T – Tb
We do nor always have such a good model or understanding. The areother equivalent physical situations, e.g., turbulent flow in a pipe. There weuse a measure of the frictional loss in the pipe as a “momentum transfercoefficient”. The dimensionless form was the Friction Factor. Thedimensionless mass transfer coefficient is the Nusselt Number.
Nu = hLk
Lecture 10 3/7/00ChE 333 3
Methods of Analysis
1. Detailed Solution of the Conservation Laws2. Approximate Analysis3. Dimensional Analysis4. Empirical Correlation of Data
We have seen several examples of Detailed Solution and we have donesome Approximate Analysis, e.g., mass transfer to or from a flowing film,heat transfer from a solid sphere. What we did was to transform the exactproblem into a simpler more solvable one using mathematical analysis.What we do in the next few lectures is examine in greater detail the lastthree methods as tools to analyze heat transfer and to design processes.
Approximate Analysis and Film Theory
Film Theory is the simplest and oldest approach in the use of mass transfercoefficients and in their prediction. The Theory is attributed to Nernst.
Examine the neighborhood of the phase boundary. We assume that theflow field consists of two regions, a uniform region in the bulk of the fluidfar from the surface and a region in the vicinity of the boundary whereviscosity dominates (since there is no slip at the boundary).
Lecture 10 3/7/00ChE 333 4
Film Model
The film model presumes that the velocity field is linearized in some sensenear the boundary so that
τ = µ∂vx
∂y S
≈ µUδ = Fs
A
This means that the film has a thickness
δ ≈ µUτ
But recall the definition of the friction factor
f ≡ τ12ρU2
If we introduce that notion into our analysis
δ ≈ µ U
12fρU2
= 2µfρU
Not surprisingly we can relate the fractional layer thickness to theReynolds number
δL = 2µ
fρUL= 2
f Re f ≈ A Re– 1
4
Lecture 10 3/7/00ChE 333 5
Dimensionless Heat Transfer Coefficient
We defined the Heat Transfer Coefficient, h , by
q • n
s
= – k s∇T • ns
= h T – Tb
so that the Nusselt number is given as
NuL =q • n
s
k T – Tb
= hLk
Observe that
NuL = hL
k= L
δ* = Lδ
δδ* = 1
2f Re δ
δ*
so that
NuL = 1
2f Re δ
δ*
δδ* = g(Pr) ≈ Pr 1 / 3
Then we observe a relation rather like the ones we calculated in our moredetailed models.
NuL = 12
f ReSc1 / 3
Lecture 10 3/7/00ChE 333 6
The Chilton-Colburn Analogy
In the 1930s, based on the Nernst Film Theory, two duPont researchersproposed an analogy between heat transfer (and we have seen, masstransfer) and momentum transfer. They defined a dimensionless numbertermed a j-factor, jH.
jH = NuL
RePr1/ 3 = f2
For mass transfer, the relation was jD ≡ Sh L
ReSc1 / 3 = f2
Simple film theory, then, predicts that
NuL = 12
f RePr1 / 3 = 0.0792
Re–0.25RePr1 /3
or simplified
NuL = 0.04 Re0.75Pr 1 / 3
Lecture 11ChE 333 1
Exact Laminar Boundary Layer TheoryHeat Transfer from a Flat Plate
In a boundary layer, we have to describe the velocity field and thetemperature field
Conservation of Mass ∂ux
∂x +∂uy
∂y = 0
Conservation of x-component of Linear Momentum
ρux
∂ux
∂x + ρu y
∂u x
∂y = µ∂ 2ux
∂ y2
Conservation of y-component of Linear Momentum
∂P∂y = 0
Conservation of Energy
ρCpux
∂T∂x + ρCpu y
∂T∂y = k
∂2T∂ y2
The Boundary Conditions
u x = U, T = T∞ at x ≤ 0 or y → ∞u x = 0, T = T0 at y = 0 for all x
The Mass balance can be integrated to yield
uy = – ∂ux
∂x dyo
y
Lecture 11ChE 333 2
If we introduce this relation in both the temperature and velocity equationswe obtain
ux
∂ux
∂x – ∂ux
∂x dyo
y ∂ux
∂y = µρ
∂2 ux
∂ y2
ux
∂T∂x – ∂ux
∂x dyo
y ∂T∂y = k
ρCp
∂2T∂ y2
Both equations are very similar and can be expressed as
ux
∂Π∂x – ∂ux
∂x dyo
y ∂Π∂y = ν
Λ∂2Π∂ y2
where this equation represents both conservation of x-momentum andconservation of energy andwhere for the x-momentum equation
Π =
ux
U ; Λ = Λ u = 1
and where for the energy equation
Π = T – T0
T∞ – T0
; Λ = Λ T = να = Pr
Recognize that we have constructed an equation that describes both themomentum boundary layer and the energy boundary layer. Let us lookfor solutions of Π that are functions of η(x,y), that is,
We can define a combination of variables
η = y
2Uνx
1 /2
Lecture 11ChE 333 3
The equation for Π becomes an ordinary differential equation for Π = f(η).
Λ – 2Πudη
?
?dΠdy
= d 2Πdη2
where Π = 0 at η = 0
Π = 1 as η → ∞
If φ(η) is defined as
φ η = – 2Πudη
0
η
We can express the Boundary Layer equation as :
Λ φ dΠ
dy= d 2Π
dη2
This can be integrated twice to yield :
Π η, Λ =e– Λ φdη
0
η
dη0
η
e– Λ φdη0
η
dη0
∞
Now since φ is a function of Πu . and Π(η). We have to solve for Πu
numerically, but once the function is known then all is known.
We can observe similarity for the temperature and momentum boundarylayers since that Πu . = ΠΤ for Pr = 1.
Lecture 11ChE 333 4
Solution for the Momentum Boundary Layer
The Momentum Boundary Layer equation can be expressed as
Λ φ d 2φ
dη2 = d3φdη 3
since d 2φ
dη2 = – 2dΠu
dη and
d 3φdη3 = – 2d 2Πu
dη2
The boundary conditions are
at η = 0
dφdη = 0
as η → ∞ dφdη = 1
One method of solving the equation is to express the solution of theequation as a power series in η.
φ η =
γη2
2!+ –
γ2η 5
5!+ 11
γ3η 8
8!– 375
γ 4η11
11!+ ....
The equation has this form since the function and the first derivative vanishat η = 0 ( velocities vanish at the boundary ). The parameter γ must bedetermined from the free stream behavior, e.g.
as η → ∞ dφ
dη = 1
The value of γ which satisfies the free stream condition is 1.3282.
Lecture 11ChE 333 5
Approximate Analytical Solution
For the Boundary Layer equation and from the solution given above, wecan determine that the solution is very near the free stream solution whenη = 2.5.. If we define the boundary layer thickness δ where the value of
η
η = δ2
Uνx
1 / 2
= 2.5
then we can see that δ
x = Uxν
– 1 / 2
= Re x–0.5
We can do the equivalent to determine a Thermal boundary layerthickness from
– k∂T
∂y y=0
= kT0 – T∞
δT
so that
δT
x = 2Uxν
– 0.5
dΠT
dη η
From the solution above for Πυ’(0), we can determine that Π’(0) = 0.664.
Recall the solution for Π
Π η, Λ =e– Λ φdη
0
η
dη0
η
e– Λ φdη0
η
dη0
∞
Now
Π 0, Λ = e– Λ φdη
0
η
dη0
∞ –1
= e– 0.6643
ΛT2 η2dη
0
∞ – 1
Lecture 11ChE 333 6
The linear approximation for φ is within 10% of the exact solution.
We observe that ΠT(0) = 0.68ΛT1/3, then we see
δT
x = 2.94 Uxν
– 0.5
Pr– 1/ 3
The corresponding Nusselt number relation is
Nu = hx
k= 0.34 Ux
ν1 / 2
Pr1 / 3
Correlations for Heat Transfer Coefficients
Turbulent flow inside pipes
The simple film theory gives us a relation for the Chilton-Colburn Analogythat for turbulent flow
jH = f / 2
where the j-factor was defined as jH = Nu Re-1 Pr-1/3.
At high Reynolds numbers and Prandtl numbers, the Friend-Metznerrelation is useful.
Flow outside Pipes and Cylinders
Nu = 0.35 + 0.56 Re0.52
St =
NuRe Pr
=h
CpG=
f2
f2
1.2 + 11.8 f2
f2
1/ 2Pr – 1 Pr– 1 / 3
Lecture 12ChE 333 1
Design Problem
Superheater for a Polymer Solution1
Ethylene-propylene rubber (EPR) is polymerized in a solvent. Theproduct of the reaction is a 6% (by weight) solution of EPR inperchloroethylene. The polymer is recovered as "crumbs" from a drumdryer. Production capacity is limited by the capacity of the dryer. It isbelieved that concentrating the feed to the dryer will provide the sufficientincrease in capacity.
Your problem is to specify the design of a superheater that wouldheat the solution sufficiently so that upon flashing to atmospheric pressurethe solution is concentrated to at least 12%.
Task 1 Determine all the physical properties to use in this problemheat capacities, densities, chemical activity as a function of concentration
Task 2 Write a computer program or spreadsheet to calculate theprescribed pressure and temperature required at the end of the superheater
Task 3 Write a computer program or spreadsheet to calculate the sizeof the heat exchanger.
DataThe production of rubber is 300lbs./hr or 0.0379 kg/hrThe feed temperature is 35°C
Note: This is an open-ended problem with insufficientinformation given for you to solve the problem. You have to find the dataand make a reasonable set of assumptions about the fluid to be used toheat the rubber.
1 NOTE – This is an open ended, somewhat poorly defined, problem.
Lecture 12ChE 333 2
Heat Transfer Analysis in Pipe Flow
Consider the problem of flow in a long pipe of circular cross-section. Theinside diameter of the pipe is D and is maintained at a constanttemperature To. The fluid flow through the pipe at a flow rate, Q.
The goal is to describe the average temperature as a function of distance inthe pipe.
MODEL Energy balance
ρCpuz
∂T∂z
= k1r
∂∂r
r∂T∂r
+
∂2T
∂z2
Momentum balance
0 = – ∂p∂z
+ u 1r
∂∂r
r∂uz
∂r
Initial and Boundary conditions
at z = 0 T = Ti for all r
at r = 0 ∂T∂r
= ∂uz
∂r = 0
at r = R T = TR ; uz = 0
Lecture 12ChE 333 3
Adimensionalization and Scaling
The convective heat transfer equation
ρCpuz
∂T∂z
= k1r
∂∂r
r∂T∂r
+
∂2T
∂z2
can be scaled using a set of reference parameters
θ = T – TR
T1 – TR
; ζ = zL ; η = r
R ; v = u z
vThe equation is
ρCp v v T1 – TR
L∂θ∂ζ =
k T1 – TR
R21η
∂∂η η∂θ
∂η + R2
L2
∂2θ∂ζ2
which after some multiplication becomes
v∂θ∂ζ =
kL
ρCp v R2
1η
∂∂η η∂θ
∂η + R2
L2
∂2θ∂ζ2
It follows that if R/L is amall then only the first term in the Laplacian isimportant and the equation can be written ignoring axial heat conduction.
Lecture 12ChE 333 4
The dimensional form of the equation is:
ρCpuz∂T∂z
= k 1r
∂∂r
r∂T
∂r
along with the boundary and initial conditionsT = T1 at z = 0
T = TR at r = R
∂T∂r
= 0 at r = 0
We could solve for the temperature profile in detail, but it might be better if weseek a solution for the average temperature by integrating over the crossection atposition z.
∫ ρ0
R
Cpuz∂T∂z
r dr =∫ k ∂∂r
r∂T
∂rdr
0
R
If the velocity field is independent of axial position, we can write
ρCp∂∂z
uzTr dr = k0R∫ r
∂T∂r
0
R
= kR∂T∂r R
– 0
Examine an average temperature <T>, the mixing cup temperature, themean temprature of the fluid that leaves cross section at z = z
T =
uzT 2πr dr0
R
u z 2πr dr0
R=
u zT 2πr dr0
R
Q
Lecture 12ChE 333 5
This last equation can be re-written as
∫ uzTr dr0
R
= Q2π
T
The integrated energy equation is :
ρCpddz
Q2π
T
= kR∂T∂r R
The material balance teaches us that
ρQ = w = constant
So that we can write:
wCpd Tdz
= k2πR∂T∂r R
Recall that we can define a heat transfer coefficient by an expression suchas :
−k∂T∂r R
=h T – TR[ ]
and the equation for the mixing cup temperature is :
wCpd Tdz
= πDh TR – T[ ]
Lecture 12ChE 333 6
This can be prepared for integration2:
d TR – T( )TR – T
= –πDhwCp
dz
The relation is integrated readily if h is not a function of z3
T – TR
T1 – T R
= exp –πDhzwCp
= exp –4St
zD
The definition of the Stanton Number is :
St =h
ρCpU=
NuRePr
=NuPe
where Pe = Re Pr
T2 – TR
T1 – TR
= exp –4StLD
= exp –
πDLhwCp
2 From here on, ,we drop the brackets on T<>T> for convenience and the experinced playersbenefit!!!
3 We can still use the same relation for St = h/ρCpU where
h= 1L
hdz0
L
Lecture 12ChE 333 7
Other Ways of Defining and using Heat Transfer Coefficients
QH = hA(∆T)
Questions
What is QH and what is ∆T
We know that an energy balance contains:
QH = +wCp(T1–T2)We can rewrite as
∆T = +wCp T1 – T2( )
πDLh
After integrating we find that
–πDLhwCp
= lnT2 – TR
T1 – TR
It follows that I can write
QH = hA(∆T)
if the temperature difference is ∆T
∆T =T2 – T1
lnT2 – TR
T1 – TR
=T1 – T2
lnT1 – TR
T2 – TR
=T1 – TR( ) – T2 – TR( )
lnT1 – TR
T2 – TR
= ∆T ln
Lecture 13 3/18/00ChE 333 1
Parallel Plate Heat Exchanger
Parallel Plate Heat Exchangers are use in a number of thermal
processing applications. The characteristics are that the fluids flow in the
narrow gap, between two parallel sheets. The flow is usually laminar.
For our example, assume a fluid flows confined between two parallel
planes. both held at a fixed temperature, TH. The fluid enters at T1 into the
heated section at a mean velocity U. The flow is laminar so that the
velocity profile is given by
u y( ) =32
U 1–yH
2
The volumetric flow rate per unit width is given by
qw = 2UH
Reasonable assumptions include, Steady State, no viscous dissipation,constant thermal properties, etc.
Lecture 13 3/18/00ChE 333 2
Fluid in laminar flow
x
y = H
y = – H
Uniform wall temperature T H
x
Uniform inlet temperature T 1
x=0Temperature profile T(x,y)
y
Figure 12.5.1 Laminar flow between heated parallel planes
The convective energy balance is given by
u y( )∂T∂x
= α∂2T∂y2 +
∂2T∂x2
As we saw in the case of a tubular channel, radial conduction is muchlarger than axial conduction so the equation can be simplified to
u(y)∂T∂x
= α∂2TDy2
Boundary Conditions
T = T1 at x = 0 for all y Initial temperature profile
∂T∂y
= 0 at y = 0 for all x Symmetry at the channel axis
T = TH at y ≠H for all x ≥ 0 Wall temperature
Lecture 13 3/18/00ChE 333 3
Non-Dimensional Form of the Equation
We can make the equation dimensionless with these definitions
Θ = T – TH
T1 – TH
; y* = yH ; x*= αx
UH2
So that the equation and its boundary conditions become
32
1 – y* 2 ∂Θ∂x*
= ∂2Θ∂y*2
Θ = 1 at x* = 0 for all y*
∂Θ∂y*
= 0 at y* = 0 for all x*
Θ = 0 at y* = ±1 for all x* ≥ 0
The solution of the equation can be expressed as a series solution
Θ = A nexp – 2
3λn
2x* anmy*nΣn= 0
∞
Σm= 0
∞
The heating rate per unit width of channel is
QH = 2 U H [ρ Cp (T1 – Tcm)]
where Tcm is the mixing cup average temperature.
Lecture 13 3/18/00ChE 333 4
The mixing cup average obtained from the temperature profile is
Θcm ≡
32
U 1 –yH
2
Θ x, y dy0
H
32 U 1 –
yH
2
dy0
H
It can, therefore, be written as
Θcm = Gm exp–2λm
2 x*3
m=0
∞∑
The first three coefficients and eigenvalues are:
m G m λ m2
0 0.910 2.83
1 0.0533 32.1
2 0.0153 93.4
Everything we want to know about the temperature profile is in thesolution given above.
There is a simple one-term approximation since the second eigenvalues isso much larger than the first.
Θcm = 0.91 exp –1.89x*( )
Lecture 13 3/18/00ChE 333 5
The Local Heat Transfer Coefficient
The local heat transfer coefficient is defined in terms of the heat transferredto the fluid in a differential distance along the exchanger.
dQH = - UH ρCp dTcm = hln(Tcm - TH) dx
Recalling the non-dimensionalization, we can express the local heattransfer coefficient, hln, as
hln = – UHΘcm
k
UH2
dΘcm
dx *
From the solution for the dimensionless mixing cup temperature, we obtain
4h lnHk
= Nu ln =
83
λ m2 Gm exp –2λm
2 x */3( )m=0
∞∑
Gm exp –2λm2 x*/3( )
m=0
∞∑
Here the Nusselt number is defined with a length scale 4 H, that is the“hydraulic diameter”.
DH = 4 (cross-sectional are)/(wetted perimeter)1
The one-term appoximation to Θn provides the limiting value for theNusselt number for large x*
Nu ln =83
λ02 = 7.55
This is generally valid if x* > 0.1
1
Note that DH =
4 2HW2W + 4H = 4H
1+ 2HW
Lecture 13 3/18/00ChE 333 6
We will find shortly that, though the local Nusselt is useful, we will have
recourse to an average value over the length of the heat exchanger
NuL ≡1
xL* Nu ln x *( )0
xL*
∫ dx*
Though it would appear that we would have to return to the detailedsolution for Θcm, the energy balance yields us a simpler form to evaluate.
NuL ≡4
xL* ln
1
Θcm xL*( )
Short Time Solutions
The equations describing the entrance region of the Parallel Plate HeatExchanger are identical save for the the midplane condition
u(y)∂T∂x
= α∂2TDy2
Boundary Conditions
T = T1 at x = 0 for all y Initial temperature profile
T → T1 as y → ∞ Free stream condition
T = TH at y ≠H for all x ≥ 0 Wall temperature
Lecture 13 3/18/00ChE 333 7
Approximate Velocity Profile
In the entrance region, the thermal boundary layer is thin compared to thewidth of the channel so that the velocity profile can be approximated by alinear profile and the problem is then posed as
βy∂ T
∂ x = α∂2 T∂y2
Boundary Conditions
T = T1 at x = 0 for all y Initial temperature profile
T → T1 as y → ∞ Free stream condition
T = TH at y ≠H for all x ≥ 0 Wall temperature
Here from an expansion of the velocity profile near the wall, β = 3U/H.
The problem is identical in statement to the falling film problem discussedin Mass Transfer and the solution is the same. We can use a combinationof variables
η = y
β9αx
1/ 3
Thsolution for Θ is
Θ =exp – ξ3 dξ
0
η
exp – ξ3 dξ0
∞ =exp – ξ3 dξ
0
η
Γ 43
Lecture 13 3/18/00ChE 333 8
Nusselt Number Relations
The heat flux at the boundary is
qy x = – kdT
dy y = 0
= k TH – T1
β9αx*
1 /3
The heat transferred per unit width is
QH
W = – k dTdy y= 0
dx0
L
=3k TH – T1
2Γ 43
β9α
1 / 3
so that the local heat transfer coefficient can be expressed as
h x =qy
TH – T1
= k
β9αx
1 /3
Γ 43
The local Nusselt Number relation is
Nu(x) = 4h(x)Hk
=4 UH2
3αx
1 /3
Γ 43
= 3.12x*
1/ 3
Expressed as an average Nusselt Number the relation becomes
NuL = 4hL H
k= 2.95 Re Pr H
4L
1 / 3
Lecture 14 3/18/00ChE 333 1
Correlations for Heat Exchange
Forced Convection Heat Transfer in Laminar Flow in a Tube
There are two measures of the state of a system in Heat Transfer...Reynoldsnumber and the Graetz number
Graetz Number Reynolds number(local)
Gz = Dx Re Pr
Re =
ρUDµ
(average) Gz = D
L Re Pr
Re =ρUD
µ
Developing Boundary layer with Fully Established VelocityProfileUniform Wall Temperature
Local Heat Transfer Coefficient Nu = 1.077 Gz 1/ 3 for Gz > 100
Average Heat Transfer Coefficient Nu = 1.61 Gz 1/ 3 for Gz > 1000
Uniform Heat Flux
Local Heat Transfer Coefficient Nu = 1.302 Gz 1/ 3 for Gz > 104
Average Heat Transfer Coefficient Nu = 1.953 Gz 1/ 3 for Gz > 104
Lecture 14 3/18/00ChE 333 2
Simultaneously Developing Temperature and Velocity Profiles
Nu = 1.86 Gz 1/ 3 η b
η w
0.14
for Gz η b
η w
0.14
≥ 2
Fully Developed Laminar Forced Convection Heat TransferConstant Wall Temperature
Nu = 3.66
Constant Wall Heat Flux
Nu = 4.36
Overall Heat Transfer
Nu = 3.662 + 1.61 3Gz
1 /3
Forced Convection Heat Transfer in Turbulent Pipe FlowHeat Transfer in Entry Length
ChE 333 - Lecture 15 1Heat and Mass Transfer 3/18/00
Some Observations on Natural or Free ConvectionSometimes a single physical process in nature can explain a variety of events. Free convection isone such process. It functions because heated fluids, due to their lower density, rise and cooledfluids fall. A heated fluid will rise to the top of a column, radiate heat away and then fall to be re-heated, rise and so on. Gasses, like our atmosphere, are fluids, too. A packet of fluid can becometrapped in this cycle. When it does, it becomes part of a convection cell.
Convection cells can form at all scales. They can be millimeters across or larger than Earth. They allwork the same way. The convection that you are most likely to have observed is in cumulonimbusclouds or "thunderheads." These towering vertical clouds can be seen to evolve over a few minutes.The tops of the clouds have a sort of cauliflower appearance as warm moist air rises through thecenter of the cloud. The moisture in the cloud condenses as it cools. The air gives up some of itsheat to the cold high altitude air and begins to fall.
As the air falls along the exterior of the cloud, it returns to warmer low altitudes where it can becaught up in the rising column of air in the center of the cloud. This fountain-like cell can formalongside other cells, and a packet can move between cells. Hail forms when water droplets, carriedby the strong updrafts, freeze, fall through the cloud and are caught in the updraft again. Anadditional layer of water freezes around the ice ball each time it makes a trip up through the cloud.Eventually, the hail becomes too heavy to be carried up anymore, so it falls to the ground. Largehailstones, when cut apart, show multiple layers, indicating the number of vertical trips the stonemade while it was caught in the convection cell.
Convection also occurs on the Sun. A high resolution white light image of the Sun shows a patternthat looks something like rice grains. Very large convection cells cause this granulation. The brightcenter of each cell is the top of a rising column of hot gas. The dark edges of each grain are thecooled gas beginning its descent to be re-heated. These granules are the size of Earth and larger.They constantly evolve and change.
Thunderheads and granulation are large-scale examples of convection. Fortunately, there areexamples of convection that fit into a classroom. An excellent example can be seen in hot JapaneseMiso (soybean paste) soup.
The interior of the broth is hot. The surface of the soup is exposed to cool air. Hot packets of fluidrise out of the interior of the soup to the surface where they give off heat. Now cooled, they falldown into the bowl to be re-heated. Left alone, the soup will dissipate its heat in this way (andthrough conduction with the sides of the bowl) and reach room temperature.
The soybean paste granules and other ingredients will highlight the convection cells vividly. Asstudents gaze into their soup, they will see the rising and descending columns of fluid. The cellswill evolve and change their positions. Dark bottomed bowls show the effect best. If the soup isstirred up, students can observe the cells reform. Of course, the demonstration material can beconsumed at the conclusion of the demonstration.
Free onvection acts as described in the examples above where gravity's effects are present (so thatwarm, low density fluids can rise and cool, high density fluids can fall). What happens in theweightlessness of space where up (rise) and down (fall) have no meaning?
ChE 333 - Lecture 15 2Heat and Mass Transfer 3/18/00
How do We Describe Free convection?
Free convection is driven by density differences. In order to describe theprocess, we can express the change in density as a function of temperatuerin terms of a Taylor’s Series
ρ T = ρ T +
∂ρ∂T
T
T – T + ....
The coefficient of volumetric expansion is
β = 1ρ
∂ρ∂T
T
so that the Taylor’s series can be expressed more simply.
ρ T = ρ T 1 + β T – T + ....
Role of buoyancy on the flow field
If there is a gravitational field, then there is a buoyancy force that acts onthe fluid to impart motion to the fluid. The induced motion would beinfluenced by the viscosty or “viscous drag”. Such motion we term“natural convection” or “free convection”.
In a column of fluid of height L, with a cross section A (normal to x-axis),the body force acting across the ends of the column is BAL. If the bodyforce were unopposed by drag or viscous forces the momentum of the fluidwould be of order ρ(∆u)2. The buoyancy driven velocity ub would be nolarger than
ub = OgL ∆ρ
ρ
1 /2
ChE 333 - Lecture 15 3Heat and Mass Transfer 3/18/00
One can express the velocity in terms of a Reynolds’ Number
Reb =ubLν = gL3
ν2
∆ρρ
1 /2
The natural number to describe natural convection is the Grashof Number,a merasure of the buoyancy forces to inertial forces.
Gr =
gL3
ν2
∆ρρ
but recognizing that ρ behaves ∆ρρ = – β∆T
leads to a simpler expression.
Gr = gL3
ν2 β∆T
Buoyancy-induced Flow in a Confined Space
Suppose two vertical plates located 2b apart, One palte is heated andmaintained at T1 The second plate is set at T2. Suppose as well that all thephysical properties are independent of temperature. The temperature fielddoes not change appreciably. Now the energy equation looks like L:
vy
∂T∂y + vz
∂T∂z = α
∂ 2T
∂y2 +∂2T
∂z2
The velocity field is one dimensional so that
v ≈ 0,vz(y) = εvz0,vz
0(y)
ChE 333 - Lecture 15 4Heat and Mass Transfer 3/18/00
Then the temperature equation simplifies to
d2Tdy 2 = 0 where T = T1 at y = – b
T = T2 at y = + b
and the solution can be expressed as
T = Tm – 1
2∆T
yb
∆T = T1 – T2 ; Tm = 12
T1 + T2
The Navier-Stokes equation describing the flow field in the z-direction is
µ
d 2vz
d y2 =dpdz
+ ρg
For this mode, we make the Boussinesq approximation that is allproperties are T-independednt save density.
µ
d 2vz
d y2 =dpdz
+ ρg – ρβg T – T
Now if we examine the perturbation expansion of vz applied to the N-Seqaution, we observe that since
vz = vz0 + εvz
1
The leading term of the perturbation expansion is the solution to
0 = = dp
dz+ ρg
That is, in the zeroth approximation, there is no flow in the z-direction as a result ofthe temperature field... the pressure field is given by the “hydrostatic law”.... thepressure field is balanced by the gravitational force.However, in the first approximation, the velocity field is governesd ny the balance ofviscous forces and buoyancy forces.
µ
d 2vz1
d y2 = – ρβg T – T
ChE 333 - Lecture 15 5Heat and Mass Transfer 3/18/00
The conservation equations are actually coupled, but in this firstapproximation, we may treat them as uncoupled
µ
d 2vz1
d y2 = – ρβg Tm – T – 12
∆Tyb
The boundary conditions are vz = 0 at y = - b and at y = +b
The solution is simple.
vz
1 =ρβgb 2∆T
12µyb
3
– A yb
2
– yb
+ A
where
A = 6Tm – T
∆T
However, there should be no net flowhat is :
vzdy
–b
b
= 0
This demands that A = 0, so
vz
1 =ρβgb 2∆T
12µyb
3
–yb
We can make the velocity dimensionless
vz
1 ρbµ = 1
12Gr y
b
3
– yb
Gr is the Grashof number and is a measure of the buoyancy forces to theviscous forces.
ChE 333 - Lecture 16 1Heat Transfer 3/18/00
Buoyancy-induced Flow:Natural Convection in a Unconfined Space
If we examine the flow induced by heat transfer from a single vertical flatplat, we observe that the flow resembles that of a boundary layer. Theappropriate description begins with the conservation laws.
Mass
∂vy
∂y +∂vz
∂z = 0
Momentum
ρ vy
∂vz
∂y + vz
∂vz
∂z = µ∂ 2vz
∂y2 +∂2vz
∂z 2 –∂p∂z – ρg
Energy
vy∂T∂y + vz
∂T∂z = α
∂ 2T
∂y2 +∂2T
∂z2
The velocity field is such that the induced flow is viewed as a perturbationof the steady flow.
v ≈ εvy
1,vz0(y) + εvy
1
Now if we examine the perturbation expansion of vz applied to the Navier-Stokes equation, the leading term of the perturbation expansion is thesolution to
0 = = dp
dz+ ρg
That is, in the zeroth approximation, there is no flow in the z-direction as a result ofthe temperature field... the pressure field is given by the “hydrostatic law”.... thepressure field is balanced by the gravitational force.
ChE 333 - Lecture 16 2Heat Transfer 3/18/00
Velocity Field
The velocity field is governed by.
ρ vy
∂vz
∂y + vz
∂vz
∂z = µ∂ 2vz
∂y2 +∂2vz
∂z 2 –∂p∂z – ρg – ρβg T – T
The conservation equations are clearly coupled. The solution mirrors theresults obtained in Boundary Layer theory analysis.
The boundary conditions areat y = 0 vz = vy = 0 T = T0at y = ∞ vz = 0 T = T1at z = -∞ vz = vy = 0 T = T1
The solution is not simple, but involves combination of variables solutionsas in Boundary Layer theory. The problem is best solved by putting theequations in dimensionless form.
Y = Hαµ
B
1 / 4
; Vy = Bα3
µH
1/ 4
; Vz = BHαµ
1 / 2
B = ρg∆T
ChE 333 - Lecture 16 3Heat Transfer 3/18/00
Dimensionless Equations
Without belaboring the procedure in detail, the results in dimensionlessform are.The differential equations
∂ϕ y
∂η +∂ϕ z
∂ζ = 0
Pr– 1 ϕy
∂ϕ z
∂η + ϕz
∂ϕ z
∂ζ =∂2ϕ z
∂η2 + ε2∂ 2ϕz
∂ζ2 + θ
ϕy
∂θ∂η + ϕz
∂θ∂ζ =
∂2θ∂η2 + ε2
∂2 θ∂ζ 2
For sufficiently small ε, the equations are simpler. It simply means thatchanges in the ζ are less important than chasnges in the η direction.
We can use the solution or at least the form of the solution of theequations, as we note the definition of the heat flux per unit length.
q' = – k∂T
∂y η =0
dz
0
H
= – k∆THY
∂θ∂η η =0
dζ0
1
It is apparent that the definite integral is a function only of Pr since
θ = θ η, ζ, Pr
It follows that since the definite intergral is a constant C = C(Pr), that wecan express the flux q’ as
q' = Ck∆T Gr Pr 1 /4
where the Grashof number here is
Gr =
ρ2βgH3∆Tµ 2
It is easy to recall the definition of a heat transfer coefficient.
q'H = h∆T
and we obtain for our model
Nu = C Gr Pr 1 / 4 = CRa1 /4
ChE 333 - Lecture 16 5Heat Transfer 3/18/00
An Empirical Correlation
Chu and Churchill developed a more useful quasi-empirical model
Nu1/ 2 = Nu1 / 2 +
RaL 300RaL 300
1 + 0.5Pr0.5 Pr9 /16
16/9
1 / 6
Geometry L Nu0Vertical Surface L 0.68Vertical cylinder L 0.68Horizontal cylinder πD 0.36πSphere πD/2 πInclined disk 9D/11 0.56
ChE 333 - Lecture 16 6Heat Transfer 3/18/00
An Example
Heat Loss from a Horizontal Pipe
An uninsulated pipe, 5 cm in outside diam., runs horizontally through alaboratory maintained at 30°C.. Air enter the pipe at 80°C at a rate of 300kg/hr. The pipe is 20 meters.. The down stream pressure in the pipe is 105
Pa gauge. What is the heat loss from the pipe and what is the exittemperatrure of the air ?
Assume the principal resistance to heat transfer is on the outside of thepipe. We assume as well that he external surface is 10°C lower than theinside of the pipe. (Can we get a better estimate ? How?
From the Chu-Churchill equation, we obtain Nu = 10, so that
h = kD Nu = 5.8 watts
m 2 – °C
In order to calculate the heat loss we need to know the ∆T
∆T lm = Th – Tc
lm
We can make a first approximation that ∆T lm = Th – Tc
lm≈ Th – Tc
max
This allows us to get a quick estimate of the heat loss and the exittemperature from the Heat Exchanger Design Equation.
wairCp Tinh – Tout
h = hA∆T lm
ChE 333 - Lecture 16 7Heat Transfer 3/18/00
Tin
h – Touth = hA∆T lm
wairCp
=5.18 3.14 40
300 / 3600 1005= 8.3°C
That means that ∆T ~ 36°C so (Tin - Tout) = 7.5 °C
ChE 333 - Lecture 17 1Heat Transfer 3/18/00
Boiling Bubbles, Condensing Drops and Films:Heat Transfer with a Phase Transition
It was clear in the last examination that many of you had littleunderstanding of what happens with heat transfer during a phase transition,that is, in boiling, or in condensation. We saw that heat transfer duringcondensation happens at a single temperature for a single componentsystem. The same is true of boiling. This makes analysis of condensersand evaporators, and reboilers a bit easier than liquid liquid heatexchangers. However, it isn’t all that simple because there are lot ofpheniomenolgy in these processes.
Boiling
Lte’s examine Boiling first. Imagine a horizontal heated flat plate of largeextent on which is a liquid at its boiling point. The temperature of the plateis controlled.
What does the liquid look likeat very low heating rates ?at moderate heating rates ?at high heating rates ?
We measure the heat flux to the fluid as a function of the temperaturedifference between the plate and the fluid. The heat flux from the plate attemperature Ts to the saturated fluid at Tsat is described by
q • n = h Tb – Ts = h ∆T e
The term ∆Te is the excess temperature, that is, the temperature differencebetween the surface and the boiling liquid.
The Boiling process is characterized by the formation of vapor bubbleswhich grow and detach from the surface. The character of the boiling weobserve depends largely on magnitude of the excess temperature.
ChE 333 - Lecture 17 2Heat Transfer 3/18/00
Modes of Boiling
We observe three principal modes of boiling :
• Free ConvectionBoiling• Nucleate Boiling• Film Boiling
There are of course, transitions between each regime. This is clear in thefollowing Boiling Curve
Free Convection Boiling occurs when ∆Te ≤ ∆Te,A ~ 5°CDuring Free Convection Boiling, bubbles are formed onisolated spots on the surface and are swept away by thebuoyancy forces ( free convection determines the motion).Depending on whether the flow is laminar or tubulent, the heattransfer coefficient varies as ∆Te
1/4 or ∆Te1/3 and therefore the
heat flux q•n is proportional to ∆Te5/4 or ∆Te
4/3. The magnitude
is much larger than free convection in the absence of a phasechange.
ChE 333 - Lecture 17 3Heat Transfer 3/18/00
Nucleate Boiling
Nucleate Boiling occurs when ∆Te,A ≥ ∆Te ≤ ∆Te,C where ∆Te,C ~ 50°C.Vapor nucleates on the surface and the bubble rise towards the freesurface. Most of the heat exchange is directly from the surface to the fluidwith extensive mixing of the fluid. As ∆Te rises, the bubble densityincreases until a maximum concentration of bubbles occur as aconsequence of bubble coalescence. This corresponds to a maximum inthe heat flux . This Critical Heat Flux is in excess of 1 MW/m2.
q • n
critical~ π
24ρ V∆H fg
σg ρL – ρ V
ρ V2
1 / 4
ρL + ρ V
ρL
1 / 2
The Nucleate Boiling regime is characterized by extremely large heattransfer coefficients, h, often larger than 10 kW/m2-°K . As ∆Te continuesto rise beyond the critical ∆Te, the flux will decrease as the coaleschenceincreases as the bubble collapse dominates.
Transition Boiling occurs when ∆Te,C ≥ ∆Te ≤ ∆Te,D where ∆Te,D ~ 150°C
ChE 333 - Lecture 17 4Heat Transfer 3/18/00
Film Boiling
Film Boiling occurs when ∆Te ≥ ∆Te,D. Heat Transfer in Film Boiling isdue to both radiation and convection.
h4 /3 = h conv4 / 3 + h radh rad
1 /3
where the radiative heat transfer coefficient is
h rad = 5.67x10 –8ε
Ts4 – Tsat
4
Te – Tsat
and the convective heat transfer coefficient is described by
h conv = 0.62k v
3ρv ρL – ρv g ∆Hv + 0.4Cp,v∆Te
µvD∆Te
1 /4
Lecture 18 1ChE 333 – Heat Transfer
Condensation and the Nusselt's Film Theory
Condensation is a rather complicated process. It was Wilhelm Nusselt's ideato reduce the complexity of the real process to a rather simple model,namely that the only resistance for the removal of the heat released duringcondensation occurs in the condensate film. The following gives anexplanation of the Nusselt theory at the example of condensation on avertical wall.
Condensation occurs if a vapor is cooled below its (pressure dependent)saturation temperature. The heat of evaporation which is released duringcondensation must be removed by heat transfer, e.g. at a cooled wall. Figure1 shows how saturated vapor at temperature Ts is condensing on a verticalwall whose temperature Tw is constant and lower than the saturationtemperature.
A condensate film developswhich flows downwards underthe influence of gravity. Ascondensation occurs over thewhole surface the thickness ofthe film increases.
Lecture 18 2ChE 333 – Heat Transfer
For laminar film flow heat can be transferred from the film surface to thewall only by heat conduction through the film (Figure 2).
From the definition of the local heat transfer coefficient hloc ,
it follows that
The problem is reduced to the calculation of the film thickness profile . If is known integration of equations (1) and (3) over the whole surface yields thetotal heat flow and the mean heat transfer coefficient.
qz = h loc Ts – Tw
h loc = kδ z
The local heat flux at position zthrough the film due toconduction is
qz = kδ z
Ts – Tw
where k is the thermalconductivity of the condensate(which is assumed to beconstant) and is the filmthickness at position z.
Lecture 18 3ChE 333 – Heat Transfer
The first step in determining is the calculation of the velocity profile w(y)in the condensate film (see Figure 3).
and the solution is
Using the velocity profile equation (5) we can now calculate the condensatemass flow rate by integrating from y=0 to y= δ:
The result is:
w y = –
ρ – ρv g
µδy – y 2
2
m =
ρ ρ – ρv gB
µδ3
3
w(y) can either be determined byapplying the Navier-Stokes equation ordirectly from a force balance for a fluidelement in the film (force exerted by theshear stress equals force of gravityminus buoyancy
µd 2w
d y2 = – ρ – ρv g
dwdy y= δ
= 0 ; w 0 = 0
Lecture 18 4ChE 333 – Heat Transfer
By differentiating equation (6) we can also determine the change in the massflow rate with the film thickness:
The change of the condensate mass flow rate results from the condensationof vapor and requires the heat flow
to be removed ( = enthalpy of evaporation). Using equations (1) and (7) thedifferential equation for the film thickness as a function of the coordinate zis:
Integration of equation (8) with the boundary condition , δ(0) = 0, yields
The film thickness increases with the fourth root of the coordinate z.
By substituting δ, according to equation (9) into equation (3) the local heat
transfer coefficient follows:
dmdδ =
ρ ρ – ρv gB
µδ2
dQ = ∆Hv dm = qBdz
δ2 dδ
dz= k µ
ρ ρ – ρv gTs – Tw
δ =
k µ Ts – Tw
∆Hvρ ρ – ρv g
1 / 4
h loc = k
δ=
∆Hvρ ρ – ρv gk3
4µ Ts – Tw z
1 / 4
Lecture 18 5ChE 333 – Heat Transfer
Finally, the mean heat transfer coefficient for a wall of height L can becalculated by integrating the local heat transfer coefficient, hloc,from z = 0 to z = L:
As we can see from this equation, the heat transfer coefficients are large forsmall temperature differences ts-tw and heights L. In both cases thecondensate film is thin and hence the heat transfer resistance is low.Equation (11) can also be used for film condensation at the inner or outerwalls of vertical tubes if the tube diameter is large compared to the filmthickness. All fluid properties in equation (11) with the exception of thevapor density are best evaluated at the mean temperature
ρv is evaluated at the saturation temperature Ts.
Nusselt derived a similar equation for film condensation on horizontal tubesusing a numerical integration. The mean heat transfer coefficient for a singlehorizontal tube of diameter D is
h m = 1
L h locdz0
L
= 0.943∆Hvρ ρ – ρv gk3
4µ Ts – Tw L
1 / 4
Tm = 34
Tw + 34
Ts
h m = 0.728
∆Hvρ ρ – ρv gk3
µ Ts – Tw D
1 / 4
Lecture 18 6ChE 333 – Heat Transfer
In spite of the simplifications heat transfer coefficients from the Nusselttheory are surprising accurate. Measured heat transfer coefficients are up to+25% higher than the values calculated from above equations. The mainreason for the deviation is the formation of waves on the film surface whichisn't considered in the Nusselt theory. These waves lead to an improvementin the heat transfer. Whitaker recommends for the rippling fallingcondensate film a value 20% larger so that we might use
The flow in the film could become turbulent at Re > 1800, in which case inthe coefficient might be represented as
so that the heat transfer coefficient is
h m = 1.137
∆Hvρ ρ – ρv gk3
µ Ts – Tw L
1 / 4
Re = 4
3gν2
4νk Ts – Tw L
∆Hvρg
h m = 0.0076
ρ ρ – ρv g
µ 2
1 / 3
Re– 0.4
Lecture 19ChE 333 1
Heat Exchangers - Introduction
Concentric Pipe Heat Exchange
Energy Balance on Cold Stream (differential)
dQC = wCp CdTC = CCdTC
Energy Balance on Hot Stream (differential)
dQH = wCp HdTH = CHdTH
Overall Energy Balance (differential)
For an adiabatic heat exchanger, the energy lost to the surroundings iszero so what is lost by one stream is gathered by the other.
dQC + dQH = 0
Tc1 Tc2
Th1
Th1
Lecture 19ChE 333 2
Heat Exchange Equation
It follows that the heat exchange from the hot to the cold is expressed interms of the temperature difference between the two streams.
dQH = U TH – TC dA
The proportionality constant is the “Overall” heat transfer coefficient( discussion later)
Solution of the Energy Balances
The Energy Balance on the two streams provides a delation for thedifferential temperature change.
dTH =
dQH
CH
and dTC =dQC
CC
However, we should recall that we have an adiabatic heat exchanger sothat
d ∆T = –
dQH
CH
1 +CH
CC
Overall Energy balances on each stream
Hot Fluid QH = CH TH1 – TH2
Cold fluid QC = CC TC2 – TC1
Overall Energy balance on the Exchanger QC + QH = 0
Lecture 19ChE 333 3
The equation for ∆T can be modified using the overall energy balances toyield
d ∆T =
dQ H
CH
∆T2 – ∆T1
TH1 – TH2
The denominator is the energy lost by the hot stream, so
d ∆T =
dQ H
QH
∆T2 – ∆T1
Application of the relation for energy transfer between the two streamsyields
d ∆T = –
UdA∆TQH
∆T2 – ∆T1
Integration of the relation is the basis of a design equation for a heatexchanger.
ln ∆T2
∆T1
=UAQ H
∆T2 – ∆T1
Rearrangement of the equation leads to
The Design Equation for a Heat Exchanger
QH = UA
∆T2 – ∆T1
ln ∆T2
∆T1
= UA∆T lm
Lecture 19ChE 333 4
Design of a Parallel Tube Heat Exchanger
The Exchanger
The Design Equation for a Heat Exchanger
QH = UA
∆T2 – ∆T1
ln ∆T2
∆T1
= UA∆T lm
Glycerin-water solution with a Pr = 50 (at 70 °C) flows through a set ofparallel tubes that are plumbed between common headers. We must heatthis liquid from 20 °C to 60°C with a uniform wall temperature of 100 °C.The flow rate, F, is 0.002 m3/sec (31.6 gal/sec.).
• How many parallel tubes are required ?• How do we select L and D for these tubes ?
DataThe heat capacity, Cp, is 4.2 kJ/kg-°KThe density, ρ, is 1100 kg/m3
The liquid has a kinematic viscosity, ν = 10−3 cm2/sec.
Tc1 Tc2
Th1
Th1
Lecture 19ChE 333 5
Step 1Calculate the heat load
Qc = ρFCp Tout – Tout
Qc = 1100
kgm3 0.002 m3
sec 4.2 kJkg–°K
1°K1°C
60 – 20 °C
Qc = 369.6 kJsec = 369.6 kWatts
Step 2Calculate the heat transfer coefficient
If the flow is laminar, likely since glycerin is quite viscous,and the Re < 2000the Nusselt number relation for laminar flow can be expressed as
Nu = 3.66 3 + 1.61 3Gz
1 /3
The Graetz number is Gz = Re Pr D
L
If the flow is turbulent (Re > 2000), the Nusselt numberr is given by Nu = 0.023 Re 0.8Pr0.4
We do not know the flow per tube and therefore we do not know the Re.However we don’t need to know that. In Lecture 27 we observed for HeatTransfer in a Tube that
T – TR
T1 – T R
= exp –πDhzwCp
= exp –4St
zD
Lecture 19ChE 333 6
The definition of the Stanton Number is :
St =h
ρCpU=
NuRePr
=NuPe
Given a Re and Pr, we can calculate the Nu and the Stanton Number, thelatter prviding us with the temperature at length L from the previousequation. Let’s examine several configurations at L/D = 50, 100, 200.The Excel table below can be used to specify a design chart.
We can observe that the flow rate per tube is given by
Fnt = Fn t
so that the Reynolds’ number is Re = 4F
πDυn t
Lecture 19ChE 333 9
As a consequence we can observe that the total length of tubing is notdependent on D alone but on othere considerations that might set acondition for Re, e.g. a pressure drop limitation. Wv find that for this basecase, we find
Maximum Cooling Capacity of an Exchanger of Fixed Area
Water is available for use as a coolant for an oil stream in a double-pipeheat exchanger.
The flow rate of the water is 500 lbm/hr.The heat exchanger has an area of 15 ft2.The oil heat capacity, Cpo, is 0.5 BTU/lb-°FThe overall heat transfer coefficient, U, is 50 BTU/hr-ft2-°F
The initial temperature of the water, Tw0, is 100°FThe maximum temperature of the water is 210°FThe initial temperature of the oil, Tw0, is 250°FThe minimum temperature of the oil, Tw0, is 140°F
Estimate the maximum flow rate of oil that may be cooled assuming afixed flow rate of water at 500 lbm/hr
There are two possible modes of operationCo-current flowCounter-current flow
Using the temperatures, we obtain T0max = 221 °Fand from the heat balance for oil, we obtain the oil flow rate as 3800lbm/hr.
I thought that countercurrent flow was supposed to be more efficient.What is the problem ?
Lecture 20ChE 333 1
Design of a Parallel Tube Heat Exchanger
The Exchanger
100°F
Water: 70°F5 ft/s
Benzene180 °F7500 lb/h
The Design Equation for a Heat Exchanger
QH = UA
∆T2 – ∆T1
ln ∆T2
∆T1
= UA∆T lm
Problem Find the Required Length of a Heat Exchanger with Specified Flows:Turbulent Flow in Both Streams
The design constraints are given in the schematic above. We show this asa countercurrent configuration, but we will examine the cocurrent case aswell. The benzene flow is specified as a mass flow rate (in pound massunits), and the water flow is given as a linear velocity. Heat transfercoefficients are not provided; we will have to calculate them based on ourearlier discussions and the correlations presented in earlier lectures.The inside tube is specified as "Schedule 40––1-14 inch steel."
Lecture 20ChE 333 2
Pipe "schedules" are simply agreed-upon standards for pipe constructionthat specify the wall thickness of the pipe. Perry’s Handbook specifies thefollowing dimensions for
the inside pipe :Schedule 40 1 1/4” pipeDo = 1.66 in. = 0.138 ft. Sc = πD2/4 = 0.0104 ft2
Di = 1.38 in = 0.115 ft. (cross-sectional area for flow)the outside pipe :Schedule 40 2” pipeDi = 2.07 in = 0.115 ft.
To calculate the heat exchanger area, we must find Ao = πDL. We knowthe diameter; what is the length ?
The Design Equation isQh = Uo Ao ∆Tln
The overall heat transfer coefficient, Uo, is given by
Uo = 1ro
1roh o
+ ln ro / r i
k+ 1
r ih i
– 1
We can write it as:
Uo
Ao
L = 1ho Ao / L
+ln ro / r i
2πk+ 1
h i Ai / L
–1
= ΣR–1
To evaluate the parameters of the problem, we need the physical andthermal properties and conditions for flow in the system
Tb = 140˚F ρb = 52.3 lbm/ft3 Cp = 0.45 BTU/lb-°F
k b = 0.085 Btu / h · ft · °F
Pr =Cpbµk b
= 5
µ b = 0.39 cP = 0.391000
147.88 = 8.1 x 10– 6lb f · s / ft 2
= 2.6 X 104lbm/ft·s
Lecture 20ChE 333 3
Internal Film ResistanceThe Nusselt number on the inside of the inner pipe is given by the Dittus-Boelter equation.
Nu = h iD i
k b
= 0.023 Re0.8Pr0.3 = 337
so that the film heat transfer coefficienthi = 249 Btu/h·ft2·˚F
The heat transfer area per unit length is A i
L = π(0.115) = 0.361 ft2/ftso that the inner film resistance is :
h i
A i
L
– 1
= 249 (0.361)–1
= 0.011 h · ft · °F / Btu
The other tube dimensions areDoi = 0.138 ft and Dio = 0.172 ft
Calculation of the Water Flow RateThe hydraulic diameter is
Deq = 4
π D2i,o – Do,i
2 / 4
π D i,o + D o,i
= D i,o – Do,i = 0.034 ft
Given the water velocity of 5 ft/s, we can solve for the water flow rateWwater = 9300 lbm/h
The Overall Energy balance (wCp ∆T)benz = (wCp∆T)water
Solving for the outlet water temperature:7500 (0.45) (100 – 180) = 9300 (1) (70 – Tout)
gives the exit temperature as:Tout = 99˚F
Lecture 20ChE 333 4
External Film ResistanceThe physical properties of the water must be estimated in order todetermine the film heat transfer coefficient in the annular shell. Theaverage water temperature Tb is calculated as 84.7 °F
µ = 0.8 cp k = 0.34 BTU/h-ft-°F ρ = 62.4 lb/ft3
so that the Reynolds number can be calculated.
Re ≡ρVDeq
µ =
62.4 lb m/ft3
32.2 lb m · f/lb f · s2 (5 ft / s) 0.034 ft
1.67 x 10–5 lb f · s / ft 2 = 2 x 10 4
From the Dittus-Boelter equation, the Nusselt number is given as:Nu = 0.023 Re0.8Pr0.4 = 127
so that the external film coefficient, ho , isho = 1270 Btu/h·ft2·˚F
The external area/length is A 0
L = π(0.138) = 0.434 ft 2 / ftso that the external film resistance is
h 0
A 0
L
– 1
= 1270 (0.434)–1
= 0.00181·hft·°F / Btu
Conduction Resistance
The last term in the equation for the overall heat transfer coefficient is ln r0 / r i
Given the heat load, we can calculate the length of tubing so that
L = Qh
3640= 2.7 x 105
3640= 74 ft
The case we considered was countercurrent flow, but we noted in anearlier example that in co-current flow we could be more fluid. Now is thepipe longer or shorter ?
Lecture 20ChE 333 6
A Co-current Flow Heat Exchanger
The Design Equation for a Heat Exchanger
QH = UA
∆T2 – ∆T1
ln ∆T2
∆T1
= UA∆T lm
The heat loads are identical, the Overall Resistances to heat transfer (UA)-1
are no different since the film coefficients do not change, but the ∆Tlm aredifferent.
There are two observations to be made. First that the tube length requiredfor co-current flow is more than twice as long. Secondly that the approachtemperature for co-current flow becomes diminishingly small.
Lecture 20ChE 333 7
Questions
Question 1.To have a single concentric pipe heat exchanger 73 ft. long may beimpractical. Why ?
Question 2.What are the alternatives and can you make a rapid evaluation of the theirrequirements ?
Question 3.What if we use more tubes, do I need more area ? How do I estimate thenumber of tubes and the required area for a single pass heat exchanger.
Question 4.If we use more tubes, should we specify the tubes to be smaller. Why?How do we estimate the effect ?
Lecture 20ChE 333 8
Question 1.To have a single concentric pipe heat exchanger 73 ft. long may beimpractical. Why ?
Where do I put a 73 ft. piece of pipe ? Can I fold it up? Can I cut it intoshoter pieces and have them in parallel. ?
Question 2.What are the alternatives and can you make a rapid evaluation of the theirrequirements ?
One alternative is to cut the pipe into 12 equal length, place them in aheader and put a shell around the bundle of tubes.
Question 3.If we use more tubes, do I need more area ? How do I estimate the numberof tubes and the required area for a single pass heat exchanger.
If we use N identical tubes, Renew = Reold/N since
Re =
ρUDµ =
ρQ4πDµ
From the Dittus-Boelter equation we have
Nu = 0.023 Re0.8Pr0.4 = 127
The internal film heat tarnsfer coefficient hi ~ QIf the new flow rate is Q/N then hi ~ (1/ Nnew)0.8
So that for 12 identical tubes hinew = 0.137 hiold
The overall resistance is now
ΣR = 0.011/(0.137) + 0.00181 + 0.00116 = 0.0833
The required length is Lnew/Lold = Rnew/ Rold/ = 0.0833/0.014 = 5.95so that Lnew is 73(5.95) = 434.2 ft.
Lecture 20ChE 333 9
Question 4.If we use more tubes, should we specify the tubes to be smaller. Why?How do we estimate the effect ?
When we introduced the use of multiple tubes, we decreased the Re tosignificantly reduce the internal film resistance. We can then use similararguments in decreasing the tube diameter, but we have the followingconsequences
1. we reduce the area/length for heat transfer.2. we increase the Reynolds number and the heat transfer coefficient3. we increase the pressure drop4. we make it harder to clean
How do we do evaluate the trade-offs ?
Lecture 21ChE 333 1
Effectiveness Concept for Heat Exchangers
The Design Equation for a Heat Exchanger
QH = UA
∆T2 – ∆T1
ln ∆T2
∆T1
= UA∆T lm
A typical problem in the analysis of a heat exchanger is the Performancecalculation. That is, we are asked , given inlet conditions to evaluate howthe exchanger performs, that is what are the outlet temperatures. With theequation given above, the solution may be reached only by trial-and-error.
EffectivenessAn alternate approach lies in the notion of exchanger effectiveness, E.
E = actual heat transfermaximum possible heat transfer
Overall Energy BalanceThe actual heat transfer is given by the energy balance
(wCp ∆T)hot = (wCp∆T)cold
The maximum possible temperature rise is the difference between thetemperatures of the two entering streams (Thin - Tcin). Which fluidundergoes the maximum temperature rise ? Of course, it is the one withthe least heat capacitance
(wCp)min (∆T)max = (wCp)max (∆T)min
It follows then thatQmax = (wCp)min (Thin - Tcin)max
Lecture 21ChE 333 2
Definitions of Effectiveness
For the Double-Pipe Heat Exchanger there are four possible cases:
Co-Current Counter-CurrentHot Fluid minimum Case 1 Case 3Cold Fluid minimum Case 2 Case 4
Case 1- Co-Current flow, hot fluid minimal
E =
wCp hTh1 – Th2
wCp hTh1 – Tc1
=Th1 – Th2
Th1 – Tc1
Case 2- Co-Current flow, cold fluid minimal
E =
wCp cTc2 – Tc1
wCp cTh1 – Tc1
=Tc2 – Tc1
Th1 – Tc1
Case 3- Counter-Current flow, hot fluid minimal
E =
wCp hTh1 – Th2
wCp hTh1 – Tc2
=Th1 – Th2
Th1 – Tc2
Case 4- Counter-Current flow, cold fluid minimal
E =
wCp cTc1 – Tc2
wCp cTh1 – Tc2
=Tc1 – Tc2
Th1 – Tc2
Lecture 21ChE 333 3
Number of Transfer Units
Recall the definition of the ratio of thermal capacitances
R =
WCp c
WCp h
= Cc
Ch
= Th1 – Th2
Tc2 – Tc1
Also we can reexamine the Design Equation and rewrite it in the followingform:
lnTh2 – Tc2
Th1 – Tc1
= – UACc
1 + R
or
Th2 – Tc2
Th1 – Tc1
= e– UAC c
1 + R
We need to express this temperature ratio in terms of the effectiveness, E.
A good deal of algebra leads for Case 1 to
E = 1 + e– UA
C c1 + R
1 + R
For case 2 the equation is the relation is very similar and indeed would bethe same if R were replaced by Rmin = Cmin/Cmax.
E = 1 – e– UA
Cmin1 + R min
1 + Rmin
Lecture 21ChE 333 4
For case 3 and case 4, the equation can be expressed as a single relation.
E = 1 – e– UA
C min1 – Rmin
1 + Rmine– UACmin
1 – Rmin
We can define a dimensionless group as the Number of Transfer Units(NTU)
NTU = UACmin
This whole concept can be extended to all kinds of exchangerconfigurations, e.g.,.shell and tube with n tube passes and one shell pass; across-flow exchanger.
The Device is a 1 Shell Pass and a 2 Tube Pass Exchanger
The Design Equation for a Heat Exchanger
QH = UAF
∆T2 – ∆T1
ln ∆T2
∆T1
= UAF∆T lm
F is the correction to the ∆T for a non-ideal flow path. To determine thisfor this exchanger , noting that it is at the same time a co-current and acounter-current exchange, we have to solve some enrgy balances.
Overall Energy Balance
(wCp ∆T)hot = (wCp∆T)cold
This leads to a ratio of thermal capacitances
R =
WCp C
WCp H
= CC
CH
= TH1 – TH2
TC2 – TC1
Lecture 22ChE 333 2
or it can be written as
R =WCp tube
WCp shell
= C tube
Cshell
= ∆Tshell
∆Ttube
Shell Balance on a first section of tube (cold stream)
CC TC
'
z
– TC'
z+ ∆z
= U TC' – TH
dA2
Shell Balance on a second section of tube (cold stream)
CC Tz"
z
– TC"
z+ ∆z
= U TH – TC" dA
2
Overall Shell Balance on a second section of tube (hot stream)
CH TH
z
– THz+ ∆z
= U TH – TC" dA
2+ U TH – TC
' dA2
The corresponding differential equations.
CC
UdTC
'
dA= TH – TC
'
2
CC
UdTC
"
dA= – TH – TC
"
2
CH
dTH
dA= – U
TH – TC"
2– U
TH – TC'
2
Lecture 22ChE 333 3
If we normalize the distance term to
dn ≡ UdACC
In this representation the equations are easier to formulate:
dTC'
dn= TH – TC
'
2
dTC"
dn= – TH – TC
"
2
1R
dTH
dn=
TC" – TH
2+
TC' – TH
2
Energy balance from z to L
CH TH – THz = C C TC" – TC
'
This becomes 1
R TH – THz = TC" – TC
'
Lecture 22ChE 333 4
Eliminate all the TC variables, from the equations and the overall energybalance, we obtain
dTC
"
dn= dTC
'
dn+ 1
RdTH
dn= – TH
2+
TC' + 1
R TH – THz
2
and
–TC
' – TH
2+ 1
RdTH
dn= – TH
2+
TC' + 1
R TH – THz
2
The final form of the temperature equation is 1
Rd 2TH
dn2 + dTH
dn– 1
4RTH – THz = 0
Boundary ConditionsTH = TH1 at n = 0TH = TH2 at n = nL where nL = UA/CC
If we set a dimensionless TH, we obtain
θ = TH – TH2
TH1 – TH2and the equation
1R
d 2θdn2 + dθ
dn– 1
4Rθ = 0
Boundary Conditions
θ = θ 1 at n = 0θ = θ 2 at n = nL
Lecture 22ChE 333 5
The solution requires algebraic gymnastics, but it produces
