Download MECH4411 Heat Transfer Lecture Notes Wangs Part

MECH4411
Heat Transfer
L. Q. Wang
Professor
7-3 Haking Wong Building
3917-7908
[email protected]
Reference
1.
2.
Holman J.P., Heat Transfer, 10th Edition, McGraw-Hill, 2010.
Cengel Y. A., Heat Transfer: A Practical Approach, 2nd Edition,
McGraw-Hill, 2003.
1
Chapter 1
Introduction
Heat
In thermodynamics, heat is defined as a form of energy which is transferred from one
body (at a high temperature) to another body (at a lower temperature) due to the temperature
difference between the bodies.
Heat Transfer is to predict and control the rate at which heat is transferred. To find the
relation between temperature difference and the heat transfer rate forms one major purpose of
heat transfer study.
1-1
Three Ways of Heat Transfer
1.
Heat Conduction (Conduction)
The transfer of heat from one part of a substance to another part of the same substance,
or from one substance to another in physical contact with it, without macroscopic relative
motion of the medium.
T2
T1
T1
T2
The transfer of heat through solid bodies is usually by conduction alone.
2.
Heat Convection
The transfer of heat within substances in macroscopic relative motion such as within a
fluid by the mixing of one portion of the fluid with another
Fluid
T1
T2
Natural Convection
Movement due to density variation in a force field (Gravitational, Centrifugal etc.)
2
T2
T1
Air
g
Forced Convection
Movement due to external mechanical means
Tw
Water
Air
Fan
Pump
Heat transfer from a solid surface to a liquid or gas takes place partly by conduction and
partly by convection. Whenever there is an appreciable movement of the gas or liquid, the
heat transfer by conduction in the gas or liquid becomes negligibly small compared with the
heat transfer by convection. However there is always a thin boundary layer of fluid on a
surface (due to the viscosity of the fluid) and through this thin film the heat is transferred by
conduction.
3.
Radiation
3
1-2
Heat Conduction (Fourier Law)
δq
Fourier Law of Conduction
δq = −kdA ∇T
dA
δq - heat transfer rate
W(J/s), kW
dA – heat transfer area
m2
T – temperature
k, °C
∇T – temperature gradient
k/m, °C/m
k – thermal conductivity
W/m⋅k, kW/m⋅k
can be defined as the heat flow per unit area per unit time when the temperature
decreases by one degree (°C, k) in unit distance
Material with high thermal conductivity k are good conductors of heat, whereas materials
with low thermal conductivity are good thermal insulators.
k – thermodynamic state property depending on two independent, intensive state properties
for simple compressible materials (T, p etc.). But for engineering analysis,
k = f (material)
Thermal Conductivities of Some Materials
Sbstance
Pure copper
Pure aluminium
Cast iron
Lead
Rubber
Cork board
k(W/m⋅k)
386
229
52
34.6
0.15
0.043
k determined by (1) Experiments (Fourier Law)
(2) Theoretical Analysis (Microscopic analysis and statistical mechanics)
∇T determined by:
(1)
(2)
Experiments
Solving heat-conduction equation (1st Law) either analytically or numerically
4
1-3
Heat Conduction Equation
Applying 1st Law of thermodynamics to the system (element) in Fig. 1-3(a) and the
process taking in the unit time period yields
qx + q y + qz + qgen − qx + dx − q y + dy − qz + dz =
dE
dτ
By Fourier Law of Conduction and Taylor series expansion:
∂T
∂x
 ∂T ∂  ∂T  
+ k
q x + dx = − k
dx  dydz
 ∂x ∂x  ∂x  
∂T
q y = − k dxdz
∂y
q x = −kdydz
 ∂T ∂  ∂T  
+  k
q y + dy = − k
dy  dxdz
∂
∂
y
y
y
∂
 


∂T
q z = −k dxdy
∂z
 ∂T ∂  ∂T  
+ k
q z + dz = − k
dz  dxdy
 ∂z ∂z  ∂z  
qy+dy
y
qz
x
z
dy
qx
Fig. 1-3a
qz+dz
qx+dx
dz
dx
Also
qgen = q dxdy dz
qy
q gen = q dxdydz
q - energy generated per unit volume, W/m3
dE
∂T
= ρcdxdydz
dτ
∂τ
ρ - density, kg/m3
c - specific heat of material, J/kg⋅°C
(cv) (cv ≈ cp for solids)
5
dE = dkE + dPE + dU = 0 + 0 + ρdxdydz cv dT + ….. dv (dv = 0 for solids)
So that the general three-dimensional heat conduction equation is (in Cartesian Coordinates):
∂
∂T
∂
∂T
∂
∂T
∂T
(k
) + (k
) + (k
) + q = ρc
∂x ∂x
∂y ∂y
∂z ∂z
∂τ
(1-1)
For constant thermal conductivity (Uniform Conductivity) Eq. (1-1) is written
∂ 2T ∂ 2T ∂ 2T q 1 ∂T
+
+
+ =
∂x 2 ∂y 2 ∂z 2 k α ∂τ
(1-1a)
α = k/ρc is called the thermal diffusivity of the material (another thermodynamic state
property)
In cylindrical coordinates (Fig. 1-3b), Eq. (1-1a) →
∂ 2T 1 ∂T 1 ∂ 2T ∂ 2T q 1 ∂T
+
+
+ 2 + =
k α ∂τ
∂r 2 r ∂r r 2 ∂φ
∂z
(1-1b)
z
dφ
r
φ
dr
dz
y
x
Fig. 1-3b
In spherical coordinates: (Fig. 1-3c), Eq. (1-1a) →
1 ∂2
1
∂
∂T
1
∂ 2T q 1 ∂T
(
)
rT
+
(sin
)
+
θ
+ =
r ∂r 2
∂θ
r 2 sin θ ∂θ
r 2 sin 2 θ ∂φ 2 k α ∂τ
6
(1-1c)
z
φ
dφ
r
r
dr
dθ
φ
y
x
Fig. 1-3c
Special Cases
Steady-state one-dimensional ( q = 0)
d 2T
=0
dx 2
(1-2)
Steady-state 1-D in cylindrical coordinates ( q = 0) :
d 2 T 1 dT
+
=0
dr 2 r dr
(1-3)
Steady-state 1-D with heat sources:
d 2 T q
+ =0
dx 2 k
(1-4)
Two-dimensional steady-state without heat sources ( q = 0)
∂ 2T ∂ 2T
+
=0
∂x 2 ∂y 2
(1-5)
7
1-4
Heat Convection (Newton Law)
Newton Law of Cooling
q = hA(Tw - T∞)
T∞
q
Tw
q – heat transfer rate W, kW
A – heat transfer area m2
∆T – temperature difference between wall and liquid k, °C
h – convection heat-transfer coefficient W/m2⋅k, kW/m2⋅k
h – determined by temperature fields T
T – determined by (1) experiment
(2) solving governing Eqs. Analytically or numerically
(conservation of mass
conservation of momentum
conservation of energy)
h – can also be determined experimentally (Newton law of cooling)
Non-dimensional number
Theoretical analysis based on governing Eqs. (or dimensional analysis) ⇒
non-dimensional numbers which control heat transfer process (heat convection)
1.
Nusselt number
Nu =
hL
k
h – convection heat-transfer coefficient
L – characteristic length
k – thermal conductivity of fluid
Nu characterizes the strength of the convective heat transfer
8
2.
Prandtl number
Pr = ν/α
(a state property)
ν - kinematic viscosity of fluid
α - thermal diffusivity of fluid
Pr characterizes the effect of fluid properties on the heat transfer. It expresses the
relative magnitudes of diffusion of momentum and heat in the fluid
3.
Reynolds number
Re =
UL
ν
U – relative velocity of fluid (characteristic velocity)
Re characterizes the effect of macroscopic relative motion on the heat transfer. It
represents the ratio of the inertial force over the viscous force in the forcedconvection system.
10 m/s
4 m/s
4.
Grashof number
Gr =
gβ ∆T L3
ν2
g – gravitational acceleration
β - thermal (volume) expansion coefficient of fluid (a state property)
∆T – characteristic temperature difference
Gr characterizes the effect of gravitational buoyancy force on heat transfer. It
represents the ratio of the buoyancy force over the viscous force in the free convection
system.
9
Forced-Convection
Nu = f (Re, Pr, Geometry Parameters)
d
Nud = 0.023 Red 0.8 Pr n
0.4 for heating of the fluid
d
n=
0.3 for cooling of the fluid
…. ≤ Red ≤ ….
…. ≤ Pr ≤ ….
Free (Natural) – Convection
Nu = f (Gr, Pr, Geometry Parameters)
Free convection from horizontal cylinders
d
Nu d = 0.36 +
0.518(Grd Pr)1 / 4
[1 + (0.559 / Pr) 9 / 16 ] 4 / 9
10-6 < GrdPr < 109
10
Chapter 2
2-1
Steady-State Conduction: One Dimension
Introduction
In this Chapter we will use Fourier’s law of heat conduction and heat-conduction
equation to calculate heat flow in steady-state, one-dimensional systems. By steady-state we
mean that every quantities (temperature in particular) are independent of time. By onedimensional we mean that every quantities (temperature in particular) depend on space only
through one spatial coordinate.
2-2
The Plane Wall
Feature: heat transfer area is constant along heat flux direction
y
T1
k
T2

q
x
o
∆x
z
A
Fig. 2-1
A.
Single-layer Wall
Under 1D steady-state condition (dimension along y, z → ∞; T1 and T2 are
independent of y, z, k uniform ect.), the heat-conduction equation for a coordinate system in
Fig. 2-1 becomes
d 2T
=0
dx 2
(qgen = 0, k = constant)
Integrating w.r.t. x yields
dT
= c1
dx
1
Integrating w.r.t. x again yields
T = c1x + c2
Applying boundary conditions
x = 0,
x = ∆x,
T = T1
T = T2
leads to
c2 = T1
T2 = c1∆x + c2
c1 =
i.e.
T2 − T1
∆x
T2 − T1
x + T1
∆x
dT T2 − T1
=
dx
∆x
∴T=
0 ≤ x ≤ ∆x
Fourier law of conduction
q = −kA ∇T = −kA
where R =
dT T1 − T2 T1 − T2
=
=
dx ∆x / kA
R
(2-1)
∆x
- thermal resistance
kA

q
T1
Heat Flow Rate =
B.
R
T2
Thermal Potential Difference
Thermal Resistance
Multilayer Wall
q12 = − k A A
T2 − T1
∆x A
q 23 = −k B A
T3 − T 2
∆x B
2
q 34 = −k C A
T 4 − T3
∆x C
Temperature
profile
A
q
q
A
Fig. 2-2a
1
B
2
C
3
4
For steady-state heat conduction
q12 = q23 = q34 = q
(1st law)
Three Eqs. for three unknowns
T2, T3, q
q=
T1 − T4
∆x A / k A A + ∆x B / k B A + ∆x C / k C A
(2-2)
q
T1
∆x A
kAA
RC
RB
RA
T2
∆x B
kB A
T3
∆x C
kC A
T4
Fig. 2-2b
3
B
A
F
E
C
G
D
(a)
1
2
3
5
4
RB
RF
q
RC
RA
T1
T2
(b)
RE
RD
T3
RG
T4
T5
Fig. 2-2c Series and parallel one-dimensional heat transfer through a composite
wall and electrical analog.
q=
2-3
∆Toverall
∑ Rth
(2-3)
R values
Thermal conductivity is an important parameter in classifying the performance of
insulation. It is, however, a common practice to use a term called the R value, which is
defined as
R=
∆T
q/ A
(2-4)
to classify the performance of insulation. The units for R by Eq. (2-4) are °C⋅m2/W. Note
that this differs from the thermal-resistance concept in that a heat flow per unit area is used.
4
It makes sense to talk about the R value rather than thermal conductivity of insulation
material, because, if we are only told about thermal conductivity of insulation material, we
would also need to know its thickness to find its resistance.
2-4
Radial Systems
Feature: Heat transfer area varies along heat flux direction.
A.
Cylinders
Consider a long cylinder of inside radius ri, outside radius ro, and length L, such as the
one shown in Fig. 2-3. We expose this cylinder to a temperature difference Ti-To and ask
what the heat flow will be.
For a cylinder with length very large compared to diameter, it may be assumed that
the heat flows only in a radial direction, so that the only space coordinate needed to specify
the system is r.
q
ro
ri
r
dr
L
q
Ti
To
Rth =
ln( ro / ri )
2πkL
Fig. 2-3 One-dimensional heat flow through a hollow cylinder and electrical analog.
Heat conduction equation (steady-state 1D in cylindrical coordinates, no heat
generation)
d 2 T 1 dT
+
=0
dr 2 r dr
which can be rewritten as
5
r2
d 2T
dT
+r
=0
2
dr
dr
(*)
Euler equation (linear ODE with variable coefficients)
Let
Then
r = et
or
t = ln r
dT dT dt 1 dT
=
=
dr
dt dr r dt
d 2 T d  1 dT
= 
dr 2 dr  r dt
1 dT 1 d  dT 

+


=− 2
r dt r dr  dt 

1 dT
1 d 2T
1  d 2 T dT 

=− 2
+
= 2  2 −
dt 
r dt r 2 dt 2
r  dt
∴ (*) becomes
d 2 T dT dT
−
+
=0
dt
dt
dt 2
i.e.
d 2T
=0
dt 2
Integrating twice w.r.t.t yields
T = c1t + c2 = c1 ln r + c2
Applying boundary conditions
r = ri
r = ro
T = Ti
T = To
gives
Ti = c1 ln ri + c2
To = c1 ln ro + c2
To − Ti
ln ro ri
T − Ti
c 2 = Ti − o
ln ri
ln(ro ri )
∴
c1 =
∴
T=
To − Ti
T − Ti
ln r + Ti − o
ln ri
ln ro ri
ln(ro ri )
6
To − Ti
r
ln + Ti
ln ro ri ri
dT To − Ti 1
=
dr ln ro ri r
=
Fourier law of conduction
q r = −kAr
T − Ti 1
dT
= −k 2πrL o
dr
ln ro ri r
(2-5)
i.e.
qr =
2πkL(Ti − To )
Ti − To
=
ln (ro ri )
ln (ro ri ) 2πkL
(2-6)
and the thermal resistance in this case is
ln(ro ri )
Rth =
2πkL
q
To
Ti
Rth =
ln( ro ri )
2πkL
The thermal-resistance concept may be used for multiple-layer cylindrical walls just as it was
used for plane walls. For the three-layer system shown in Fig. 2-4 the solution is
q=
T1 − T4
ln( r2 r1 ) 2πk A L + ln( r3 r2 ) 2πk B L + ln( r4 r3 ) 2πk C L
The thermal circuit is shown in Fig. 2-4b.
The derivation of Eq. (2-7) is left as an exercise.
7
(2-7)
q
r1
r2
T2
T1
A
r3
T3
q
r4
T4
B
T1
RA
T2
RB
T3
RC
T4
C
(a)
(b)
Fig. 2-4. One-dimensional heat flow through multiple cylindrical sections and
electrical analog.
B.
Spheres
Spherical systems may also be treated as one dimensional when the temperature is a
function of radius only. The heat flow is then
q=
4πk (Ti − To )
1 ri − 1 ro
(2-8)
(k = constant, steady-state, T = Ti at r = ri, T = To at r = ro)
The derivation of Eq. (2-8) is left as an exercise.
Example 2-1 Multilayer Cylindrical System
A tube of stainless steel (k = 19 W/m⋅°C) with 2-cm inner diameter (ID) and 4-cm
outer diameter (OD) is covered with a 3-cm layer of insulation (k =0.2 W/m⋅°C). If the inside
wall temperature of the pipe is maintained at 600°C, calculate the heat loss per meter of
length. Also calculate the tube-insulation interface temperature (T2 = 100°C)
8
Stainless steel
T1 = 600°C
r2
r1
r3
Asbestos
T2 = 100°C
T2
T1
ln( r3 r2 )
2πk a L
ln( r2 r1 )
2πk s L
Fig. Ex. 2-1
Solution:
Assume a steady-state 1D process along radial direction. The accompanying figure shows
the thermal network for this problem. The heat flow is given by
2π (T1 − T2 )
q
=
L ln( r2 r1 ) k s + ln( r3 r2 ) k i
=
2π (600 − 100)
= 680 W/m
ln(2) 19 + (ln 5 2) 0.2
This heat flow may be used to calculate the interface temperature between the outside
tube wall and the insulation. We have
Ti − T2
q
=
= 680 W/m
L ln( r3 r2 ) 2πk i
where Ti is the interface temperature, which may be obtained as
Ti = 595.8°C
The largest thermal resistance clearly results from the insulation, and thus the major
portion of the temperature drop is through that material.
9
C.
Convection Boundary Conditions
Convection heat transfer can be calculated from
qconv = hA(Tw - T∞)
An electric-resistance analogy can also be drawn for the convection process by
rewriting the equation as
q conv =
T w − T∞
1 / hA
(2-9)
where now the 1/hA term becomes the convection resistance.
D.
Fur Thickness in Small and Large Animals
Fur provides a natural thermal insulation for animals. It is interesting to see how the
thermal resistance of fur varies with its thickness. Consider different size animals that can be
treated as if they were approximately of cylindrical shape. Heat loss per unit surface area of
the animal is, by Eq. (2-6),
qsurface =
Ti − To
 ln r / r 
(2πri L) o i 
 2πkL 
where Ti is the temperature at the body surface and To is the temperature at the outer surface
of the fur. The body surface area of the animal without the fur is 2πri L ( ri is its radius
without the fur). The quantity in the denominator is the R value for the layer of fur over a
cylindrical shaped animal. This R value can be written as
 ∆r 
 ∆r 


ln1 +
ln1 +
ri 
ri 


Rfur = 2πri L
= ri
k
2πkL
(2-10)
where
∆r = ro − ri .
For small values of ∆r / ri , the term ln (1 + ∆r / ri ) can be approximated as ∆r / ri . Using this
approximation,
∆r
(2-11)
Rfur =
k
10
Therefore, the Rfur increases linearly with the thickness of fur ∆r for small values of
∆r relative to the radius of the animal, ri .
Eq. (2-10) is plotted in Fig. 2-5 for two values of radius of the animal, ri . It shows
that the initial linear increases in Rfur with fur thickness ∆r (as predicted by Eq. (2-11))
continues for a larger value of fur thickness when the animal is bigger (large ri ) . While for
a small animal, Rfur quickly levels off as the fur thickness increases. Thus, increasing fur
thickness beyond a certain value is not as beneficial for a small animal as that for a large
animal. Therefore, the smaller the animal, the more difficult it is to provide the insulation.
Indeed, small animals may only survive in cold climates by behavioral responses that avoid
severe stress and by having very high rates of metabolism.
Fig. 2-5 Plot of Rfur as the function of ∆r for an assumed fur thermal conductivity of
0.05W/m.K.
2-5
The Overall Heat-Transfer Coefficient
A.
Plane Wall
Consider the Plane Wall shown in Fig. 2-6 exposed to a hot fluid A on one side and a
cooler fluid B on the other side. The heat transfer is, for 1D case, expressed by
qA1 = h1A(TA – T1)
kA
q12 =
(T1 − T2 )
∆x
q2B = h2A(T2 – TB)
For steady-state heat transfer
qA1 = q12 = q2B = q
(1st law)
11
Fluid A
The heat-transfer process can be represented by the resistance network in Fig. 2-6b.
Three Eqs. for three unknowns T1, T2, q.
TA
q
Fluid B
T1
T2
h2
h1
TB
(a)
q
TA
1
h1 A
TB
T2
T1
∆x
kA
1
h2 A
(b)
Fig. 2-6 Overall heat transfer through a plane wall.
q=
T A − TB
1 h1 A + ∆x kA + 1 h2 A
(2-12)
The overall heat transfer by combined conduction and convection is often expressed
in terms of an overall heat transfer coefficient U, defined by the relation
q = UA∆Toverall
Here
(2-13)
A – some suitable area for the heat flow
∆Toverall = TA - TB
Eqs, (2-12) and (2-13) together yields
U=
1
1 h1 + ∆x k + 1 h2
12
B.
Hollow Cylinder
For a hollow cylinder exposed to a convection environment on its inner and outer
surfaces, the electric-resistance analogy would appear as in Fig. 2-7 (1D, steady state) where,
again, TA and TB are the two fluid temperatures. Note that the area for convection is not the
same for both fluids in this case, these areas depending on the inside tube diameter and wall
thickness. In this case the overall heat transfer would be expressed by
T A − TB
ln( ro ri )
1
1 hi Ai +
+
2πkL
ho Ao
(2-14)
Fluid B
q=
q
Ti
TA
Fluid A
1
1
hi Ai
2
(a)
To
ln( ro ri )
2πkL
TB
1
ho Ao
(b)
Fig. 2-7 Resistance analogy for hollow cylinder with convection boundaries.
The overall heat transfer coefficient may be based on either the inside or outside area
of the tube.
q = UiAi ∆Toverall
(a)
q = UoAo ∆Toverall
(b)
Eqs. (a) and (2-14) leads to
Ui =
1
A
ln(
r
A 1
1
o ri )
+ i
+ i
hi
2πkL
Ao ho
(2-15)
Eqs. (b) and (2-14) yields
13
Uo
1
Ao 1 Ao ln( ro ri ) 1
+
+
Ai hi
ho
2πkL
(2-16)
Eqs. (a) (b) give
U i Ao
=
≥1
U o Ai
Example 2-2. Over Heat Transfer Coefficient for a Tube
Water flows at 50°C inside a 2.5-cm-inside-diameter tube such that hi = 3500
W/m2⋅°C. The tube has a wall thickness of 0.8 mm with a thermal conductivity of 16
W/m2⋅°C. The outside of the tube loses heat by free convection with ho = 7.6 W/m2⋅°C.
Calculate the overall heat transfer coefficient and heat loss per unit length to surrounding air
at 20°C.
Solutions:
These are three resistances in series for this problem, as in Eq. (2-14). With L = 1.0 m,
di = 0.025 m and do = 0.025 + 2 × 0.008 = 0.0266 m, the resistance may be calculated as
Ri =
1
1
=
= 0.00364 o C/W
hi Ai 3500π × 0.025 × 1
ln(d o d i ) ln(0.0266 / 0.025)
=
= 0.00062 o C/W
2πkL
2π × 16 × 1
1
1
=
= 1.575 o C/W
Ro =
ho Ao 7.6π × 0.0266 × 1
Rt =
Clearly Ro is the largest. It is the controlling resistance for the total heat transfer
because the other resistance (in series) are negligible in comparison. We shall base the
overall heat transfer coefficient on the outside tube area and write
q=
∆T
= U o Ao ∆T
R
∑ th
Uo =
1
Ao ∑ Rth
=
(a)
1
π × 0.0266 × 1(0.00364 + 0.0062 + 1.575)
= 7.577 W/m 2 ⋅ o C
The heat transfer is obtained from Eq. (a), with
q = Uo Ao ∆T = 7.577π × 0.0266 × 1(50 – 20) = 19W
14
(For 1 m length)
Comment:
This example shows an important point that many practical heat transfer problems
involve multiple modes of heat transfer acting in combination; in this case, as a series of
thermal resistances. It is not unusual for one mode of heat transfer to dominate the overall
problem. In this example, the total heat transfer could have been computed very nearly by
just calculating the free convection heat loss from the outside of the tube maintained at a
temperature of 50°C. Because the inside convection and tube wall resistance are so small,
there are corresponding small temperature drops, and the outside temperature of the tube will
be very nearly that of the liquid inside, or 50°C.
2-6
Critical Thickness of Insulation
Let us consider a layer of insulation which might be installed around a circular pipe,
as shown in Fig. 2-8. The inner temperature of the insulation is fixed at Ti, and the outer
surface is exposed to a convection environment at T∞. From the thermal network the heat
transfer is
q=
2πL(Ti − T∞ )
ln( ro ri )
1
+
k
ro h
(2-17)
2πL(Ti − T∞ )  1
dq
1 


=−
−
2 
2 
dro
 ln( ro ri )
1   kro hro 
+

k
ro h 

k

ro >
< 0
h

k

ro = = rc
= = 0
h

k

ro <
> 0
h
(2-18)
h, T
ri
Ti
ro
T∞
Ti
ln( ro ri )
2πkL
1
2πro Lh
Fig. 2-8 Critical insulation thickness
15
Example 2-3. Critical Insulation Thickness
Calculate the critical radius of insulation for asbestos (k = 0.17 W/m⋅°C) surrounding
a pipe and exposed to room air at 20°C with h = 3.0 W/m2⋅°C. Calculate the heat loss from a
200°C, 5.0-cm-diameter pipe when covered with the critical radius of insulation and without
insulation.
Solution:
From Eq. (2-18) we calculate rc as
rc =
k 0.17
=
= 0.0567 m = 5.67 cm
h 3.0
The inside radius of the insulation is 5.0/2 = 2.5 cm, so the heat transfer is calculated from Eq.
(2-17) as
q
=
L
2π ( 200 − 20)
= 105.7 W/m
 5.67 
ln

1
 2.5  +
0.17
0.0567 × 3
Without insulation the convection from the outer surface of the pipe is
q
= h(2πri)(Ti - T∞) = 3 × 2π × 0.025(200 – 20)
L
= 84.8 W/m
So, the addition of 3.17 cm (5.67 – 2.5) of insulation actually increases the heat transfer by 25
percent.
As an alternative, fiberglass having a thermal conductivity of 0.04 W/m⋅°C might be
used as the insulation material. Then, the critical radius would be
rc =
k 0.04
=
= 0.0133 m = 1.33 cm
h
3.0
Now, the value of rc is less than the outside radius of the pipe (2.5 cm), so addition of any
fiberglass insulation would cause a decrease in the heat transfer. In a practical pipe insulation
problem, the total heat loss will also be influenced by radiation as well as convection from
the outer surface of the insulation.
16
2-7
Plane Wall with Heat Sources
x=0
Consider the plane wall with uniformly distributed heat sources shown in Fig. 2-9.
The thickness of the wall in the x direction is 2L, and it is assumed that the dimensions in the
other directions are sufficiently large that heat flow may be considered as one-dimensional.
The heat generated per unit volume is q , and we assume that the thermal conductivity does
not vary with temperature. This situation might be produced in a practical situation either by
passing a current through an electrically conducting material or by bio-chemical reaction in
biological systems.
q = heat generated per
unit volume
To
Tw
L
L
Tw
x
Fig. 2-9 Sketch illustrating one-dimensional conduction problem with heat generation.
1D, steady-state heat conduction equation (for coordinates in Fig. 2-9)
d 2T q
+ =0
dx 2 k
q
(2-19)
= constant
Boundary condition
T = Tw
at
x = -L
(2-20a)
T = Tw
at
x=L
(2-20b)
Integrating (2-19) twice w.r.t. x yields
T =−
q 2
x + c1 x + c 2
2k
(2-21)
17
Applying B.C. (2-20a) and (2-20b) gives
q 2
Tw = −
L – c1L + c2
2k
Tw = −
(a)
q 2
L + c1L + c2
2k
(b)
(a) + (b) gives
2Tw = −
i.e.
q 2
L + 2c2
k
c2 = Tw +
q 2
L
2k
(c)
(a) or (b) then yields
c1 = 0
∴ T = Tw +
q 2 2
(L – x )
2k
Fourier Law:
q = − kA
dT
q
= −kA (−2 x) = Aqx
dx
2k
18
2-8
Cylinder with Heat Sources
A.
Cylinder
Consider a cylinder of radius R with uniformly distributed heat sources and constant
thermal conductivity. If the cylinder is sufficiently long that the temperature may be
considered a function of radius only. The appropriate heat conduction equation is (1D steady
state in cylindrical system).
q
dr
o r
R
L
r
d 2 T 1 dT q
+
+ =0
dr 2 r dr k
(2-22)
The boundary conditions are
T=Tw
at
r=R
(2-22a)
dT
dr
(2-22b)
q πR 2 L = −k 2πRL
r=R
we rewrite Eq. (2-22)
r
i.e.
d 2 T dT
q r
+
=−
2
dr
k
dr
d  dT
r
dr  dr
q r

=−
k

(a)
integrating (a) w.r.t. r yields
dT
qr 2
r
=−
+ c1
dr
2k
(b)
19
dT
q r c1
=−
+
dr
2k r
i.e.
(bb)
integrating (bb) w.r.t. r gives
q r 2
+ c1 ln r + c 2
4k
dT
q R c1
=−
+
2k R
dr r = R
T =−
Applying B.C. (2-22a) and (2-22b) yields
Tw = −
q R 2
+ c1 ln R + c 2
4k
qπR 2 L =
(c1)
c
2πkRL
qR − k 2πRL 1
2k
R
(c2)
Then c1 = 0
c2 = Tw +
q R 2
4k
∴ T = Tw +
q 2 2
(R – r )
4k
(2-23)
Fourier Law of conduction
q r = −k 2πrL
B.
dT
= πLqr 2
dr
Hollow Cylinder
For a hollow cylinder with uniformly distributed heat sources, the general solution is
still (1D, steady-state)
q r 2
T =−
+ c1 ln r + c 2
4k
Applying boundary conditions
T = Ti
at
r = ri (inside surface)
T = To
at
r = ro (outside surface)
leads to
20
2
q ri
+ c1 ln ri + c 2
Ti = −
4k
2
q r
To = − o + c1 ln ro + c 2
4k
c1 =
(a2)
Ti − To + q ( ri − ro ) / 4k
ln( ri ro )
2
i.e.
(a1)
2
2
q ro
c 2 = To +
− c1 ln ro
4k
(or
2
q ri
Ti +
− c1 ln ri )
4k
q
ro
∴
or
ri
r dr
L
q
r
2
( ro − r 2 ) + c1 ln
4k
ro
q 2
r
T = Ti +
( ri − r 2 ) + c1 ln
4k
ri
T = To +
(2-24)
(2-24a)
Fourier Law:
qr = -k2πrL
dT
= πLq r 2 − 2πkLc1
dr
21
2-9
Conduction-Convection Systems
The heat which is conducted through a body must often be removed (or delivered) by
some convection process. For example, the heat lost by conduction through a furnace wall
must be dissipated to the surroundings through convection. In heat exchanger applications a
finned-tube arrangement might be used to remove heat from a hot liquid. The heat transfer
from the liquid to the finned-tube is by convection. The heat is conducted through the
material and finally dissipated to the surroundings by convection. Obviously, an analysis of
combined conduction-convection systems is very important from a practical standpoint.
We shall defer part of our analysis of conduction-convection systems to Chap. 3 on
heat exchangers. For the present we wish to examine some simple extended surface
problems.
Consider 1D fin exposed to a surrounding fluid at a temperature T∞ as shown in Fig.
2-10. The temperature of the base of the fine is To. We approach the problem by making an
1st law analysis on an element of the fin of thickness dx as shown in the figure and a process
taking in the unit time period. Thus
Energy in left face = energy out right face + energy lost by convection
(for steady-state process)
dqconv = h Pdx(T - T∞)
t
A
qx
qx+dx
Z
dx
Base
L
x
Fig. 2-10. Sketch illustrating one-dimensional conduction and convection through
a rectangular fin.
22
Energy in left face = qx = -kA
dT
dx
Energy out right face = q x + dx = −kA
dT
dx
x + dx
 dT d 2T 
= −kA
+ 2 dx 
 dx dx

Energy lost by convection = h Pdx(T - T∞)
Here
∴
A – cross-sectional area of the fin, m2
P – the perimeter of the fin, m
d 2T
− m 2 (T − T∞ ) = 0
2
dx
where m2 =
(2-25a)
hP
≥0
kA
Let θ = T - T∞ (T∞ = constant)
Then Eq. (2-25a) becomes
d 2θ
− m 2θ = 0
dx 2
(2-25b)
This is a second-order, linear, homogenous ODE with constant coefficient. The two roots of
the auxiliary equation
r2 – m2 = 0
are
r1 = -m
(m =
hP
- constant)
kA
r2 = +m
The general solution for Eq. (2-25b) is thus,
θ = c1 e-mx + c2 emx
(2-26)
c1 and c2 are to be determined by two boundary conditions.
One boundary condition is
θ = θo = To - T∞ at x = 0
(*)
23
The other boundary condition depends on the physical situation. Several cases may be
considered:
Case 1: The fin is very long, and the temperature at the end of the fin is essentially that of the
surrounding fluid.
To
T(x)
T
x →∞
= T∞
T∞
∞
x
0
θ=0
at x = ∞
(a)
Case 2: The fin is of finite length and loses heat by convection from its end
−k
To
dT
= hL (TL − T∞ )
dx x = L
T∞
L
x
−k
dT
dx
= h L (T x = L − T∞ )
(b)
x=L
24
Case 3: The end of the fin is insulated so that dT/dx = 0 at x = L.
T(x)
Ts
dT
dx
=0
x =L
T∞
L
x
dT
= 0 at x = L
dx
(c)
Case 1:
Applying (*) and (a) in (2-26)
∴
∴
c1 + c2 = θo
(a1)
c1 e-m∞ + c2 em∞ = 0
(a2)
c2 = 0
c1 = θo
T − T∞
θ
=
= e − mx
θ o To − T∞
(2-27)
All of the heat lost by the fin for steady-state process, must be conducted into the base at x =
0. The heat loss (Fourier Law of heat conduction):
q = −kA
dT
dx
= −kA
x =0
= kAθ o m e − mx
x =0
dθ
dx
x =0
= kAθ o
(2-28)
hP
= hP kA θ o
kA
Case 2:
Applying (*) and (b) in (2-26)
25
c1 + c2 = θo
(b1)
− k ( −c1 me − mx + c 2 me + mx )
x=L
= h L (T x = L − T∞ )
i.e.
c1 e − mL − c 2 e mL =
∴
hL
( c1 e − mL + c 2 e mL )
km
(b2)
c1 = …..
c2 = …..
(left as an exercise)
T − T∞
θ
=
θ o To − T∞
cosh m( L − x) + (hL / mk ) sinh m( L − x)
=
cosh mL + (hL / mk ) sinh mL
(2-29)
Fourier Law of heat conduction
q = − kA
dT
dx
= −kA
x =0
dθ
dx
x =0
sinh mL + (hL / mk ) cosh mL
= hP kA (To − T∞ )
cosh mL + (hL / mk ) sinh mL
(2-30)
Case 3:
Using (*) and (c) in (2-26)
∴
c1 + c2 = θo
(c1)
0 = m(-c1 e-mL + c2 emL)
(c2)
θo
c1 =
1 + e − 2 mL
c2 = θ o −
θ=
i.e.
θo
1+ e
θo
1+ e
− 2 mL
− 2 mL
=
e − mx +
θo
1 + e 2 mL
θo
1+ e
2 mL
e mx
e − mx
e mx
θ
=
+
θ o 1 + e − 2 mL 1 + e 2 mL
=
(2-31a)
cosh[m( L − x)]
cosh mL
(2-32b)
26
Fourier Law of heat conduction
q = − kA
dT
dx
= − kA
x =0
dθ
dx
x =0
1

 1
= −kA θ o m
−
− 2 mL 
2 mL
1+ e

1+ e
(2-33)
= hP kA θ o tanh mL
 e mL − e − mL

 mL
= tanh(mL) 
− mL
e +e

Note:
In the above development it has been assumed that the temperature gradients occur
only in the x-direction. This assumption will be satisfied if the fin is sufficiently thin. For
most fins of practical interest the error introduced by this assumption is less than 1 percent.
The overall accuracy of practical fin calculations will normally be limited by uncertainness in
values of h. The convection coefficient is seldom uniform over the entire surface, as has been
assumed above. If severe nonuniform behavior is encountered, numerical techniques must be
used to solve the problem.
2-10
Fins
A.
Different Types of Finned Surface
In 2-9 we derived relations for the heat transfer from a rod or fin of uniform cross-sectional
area from a flat wall. In practical applications, fins may have varying cross-sectional areas
and may be attached to circular surface. Fig. 2-11 and Fig. 2-12 show some fins used in
practical engineering.
Fig. 2-11. Different types of finned surfaces. (a) Straight fin of rectangular profile on plane
wall, (b) straight fin of rectangular profile on circular tube, (c) cylindrical tube with radial fin
of rectangular profile, (d) cylindrical-spine or circular-rod fin.
27
Fig. 2-12. Some fin arrangements used in electronic cooling applications.
Wakefield Engineering Inc., Wakefield, Mass.)
(Courtesy
As the cross-sectional area varies, solution of the basic differential equation and the
mathematical techniques become more tedious. Please refer to:
1.
Schneider P.J. Conduction Heat Transfer, Addison-Wesley, Reading, Mass., 1955.
2.
Kern D.Q. and Kraus A.D. Extended Surface Heat Transfer, 2nd Ed., McGraw-Hill,
New York, 1972.
28
B.
Fins and Bio-Heat Transfer
Warm blooded animals appear to have adapted to new or changing environments by
varying the size and shape of their bodies and extremities. These physical changes facilitate
heat conservation or dissipation as dictated by ambient conditions. Larger animals need to
develop means of dealing with great amounts of heat that they produce. The African elephant
(Fig. 2-14), the largest land mammal, has accordingly developed the largest thermoregulatory
organ known in any animal, the pinna or external ear. The pinna is considered to be the main
external organ responsible for the temperature regulation of the body.
Fig. 2-14 An African elephant with a large external ear or pinna that is important for
thermoregulation.
The combined surface area of both sides of both ears of an African elephant is about
20% of its total surface area. The high surface to volume ratio (see discussion following Eq.
(2-35)), large surface area, and extensive vascular network of subcutaneous vessels in the
medial side of the ear make it behave like a fin and pay a role in temperature regulation.
Temperature distribution measured in a pinna is shown in Fig. 2-15. The temperature changes
from where the ear attaches to the head (base of the fin) to the outer edges show the
characteristics of a fin. Movement of the pinna (flapping) also increases the heat loss due to
the increased air flow.
Fig. 2-15 Temperature distribution in the right pinna at two different ambient temperatures
of 18 0 C (left) and 32.1 0 C . The change in pattern indicates that a change in blood flow
occurs at higher temperatures.
29
A 4000 kg elephant needs to maintain a heat loss of 4.65 kw or more while moving
and feeding. This large amount of heat cannot be dissipated by surface evaporation since
elephants do not have sweat glands. Thus, the pinna plays a important role in the heat
dissipation and by some estimates, up to 100% of an African elephant’s heat loss can be met
by movement of its pinna and by vasodilatation. Use of pinna for heat loss by convection and
radiation is not unique to elephants. New Zealand white rabbits are known to do the same.
30
Chapter 3
Heat Exchangers
In this Chapter, we will discuss the methods of predicting heat-exchanger
performance, along with methods which may be used to estimate the heat-exchanger size and
type necessary to accomplish a particular task. We limit our discussion to heat exchangers
where the primary modes of heat transfer are conduction and convection.
3-1
Some Types of Heat Exchangers
Heat Exchangers are devices where two moving fluid streams exchange heat without
mixing (TA ≠ TB).
Double-Pipe Heat Exchangers
(c)
(d)
Fig. 3-1. Double-pipe heat exchange: (a) schematic; (b) thermal-resistance
network for overall heat transfer; (c) miniature coiled tube-in-tube
exchanger; (d) detail of inlet-outlet fluid connections for
miniature tube-in-tube exchanger.
1
Either hot or cold fluid occupying the annular space and the other fluid occupying the
inside of the inner pipe.
Either counterflow or parallel flow may be used in this type of exchanger.
Shell-and-Tube Heat Exchangers
Fig. 3-2.
Photos of commercial heat exchanges. (a) shell-and-tube heat
exchanger with one tube pass; (b) head arrangement for
exchanger with two tube passes; (c) miniature shell and tube
exchanger with one shell pass and one tube pass; (d) internal
construction of miniature exchanger.
2
One fluid flows on the inside of the tubes, while the other fluid is forced through the
shell and over the outside of the tubes.
One or more tube passes may be used depending on the head arrangement at the ends
of the exchanger (one tube pass in Fig.3-2a; two tube passes in Fig.3-2b).
Cross-flow Heat Exchanger
Commonly used in air or gas heating and cooling (Figs. 3-3 and 3-4).
Fig. 3-3. Cross-flow heat exchanger: one fluid mixed and one unmixed.
Fig. 3-4. Cross-flow heat exchanger: both fluids unmixed.
3
3-2
Overall Heat Transfer Coefficient
The overall heat transfer coefficient U is one important parameter to characterize the
performance of heat exchangers.We have already discussed the overall heat-transfer
coefficient in Sec. 2-5 with the heat transfer through the plane wall of Fig. 3-5.
T
TA
Fluid A
q
q
T1
TA
k
T1
1
h1 A
T2
h1 A
A
∆x
h2
TB
T2
∆x
kA
1
h2 A
TB
Fluid B
Fig. 3-5
Definition of U
q= UA ∆Toverall
(∆Toverall = TA – TB)
as
q=
T A − TB
1
∆x
1
+
+
h1 A kA h2 A
1
(How to increase U?)
1 ∆x 1
+
+
h1
k
h2
To calculate U, we need to know h1, k, h2 and ∆x.
∴
U=
********************************************************************
q1 = h1A (TA – T1)
q 2 = kA
T1 − T2
∆x
q3 = h2A(T2 – TB)
(Newton law of cooling)
(Fourier law of heat conduction: one-D, constant k)
(Newton law of cooling)
Steady heat transfer process:
q1 = q2 = q3 = q
4
Solving for q by eliminating T1 and T2 to obtain
q=
T A − TB
1
∆x
1
+
+
h1 A kA h2 A
For double-pipe heat exchangers (Fig. 3-1),
q = Ui Ai(TA – TB)
q = Uo Ao(TA - TB)
Heat transfer area between Fluid A and Fluid B: Ai or Ao

q=
∴
Ui =
TA − TB
1
ln(ro / ri )
1
+
+
hi Ai
2πkL
ho Ao
1
1 Ai ln(ro / ri ) Ai 1
+
+
hi
2πkL
Ao ho
1
Uo =
Ao 1 Ao ln(ro / ri ) 1
+
+
Ai hi
2πkL
ho
In most practical problems the conduction resistance is small compared with the
convection resistance. Then, if one value of h is much lower than the other values, it will
tend to dominate the equation for U. (We usually want large U, ?).
5
Table 3-1
Approximate Values of Overall Heat-transfer Coefficients.
Heat Exchangers
Brick exterior wall, plaster interior, uninsulated
Frame exterior wall, plaster interior: Uninsulated
With rock-wool insulation
Plate-glass window
Double plate-glass window
Steam condenser
Feedwater heater
Freon-12 condenser with water coolant
Water-to-water heat exchanger
Finned-tube heat exchanger, water in tubes, air
Across tubes
Water-to-oil heat exchanger
Steam to light fuel oil
Steam to heavy fuel oil
Steam to kerosone or gasoline
Finned-tube heat exchanger, steam in tubes,
Air over tubes
Ammonia condenser, water in tubes
Alcohol condenser, water in tubes
Gas-to-gas heat exchanger
6
U
W/m 2 K
2.55
1.42
0.4
6.2
2.3
1100-5600
1100-8500
280-850
850-1700
25-55
110-350
170-340
56-170
280-1140
28-280
850-1400
255-680
10-40
3-3
The Log Mean Temperature Difference
For heat exchangers, we propose to relate q (heat transfer) to temperature difference
by introducing the notation of overall heat-transfer coefficient U.
q = U A ∆Tm
(3-1)
where: U = overall heat-transfer coefficient
A = surface area for heat transfer consistent with definition of U
∆Tm = suitable mean temperature difference across heat exchanger (between hot and
cold fluids)
This plays the same role as Fourier law of heat conduction for heat conduction and Newton
law of cooling for heat convection.
In last section, we assume that temperature of two fluids is constant to relate U to the
other parameters which we already know. However, in real heat exchangers, temperature of
two fluids changes from inlet to outlet. Why? ∆Tm?
 Parallel flow 
For Double-pipe heat exchanger 

 counterflow 
∆Tm =
(Th 2 − Tc 2 ) − (Th1 − Tc1 )
ln[(Th 2 − Tc 2 ) (Th1 − Tc1 )]
Log mean temperature difference (LMTD)
(3-2)
Under assumptions:
(1)
(2)
(3)
(4)
steady flow and heat transfer
fluid specific heats do not vary with T and P
heat exchanger is well-insulated
convective heat-transfer coefficients h are constant throughout the heat exchanger
Fig. 3-6. Temperature profiles for (a) parallel flow and (b) counterflow in double-pipe heat
exchanger
7
Proof:
Parallel flow
δq = −m h c h dTh = m c c c dTc
(3-3)
(Assumptions: heat exchanger is well insulated, steady process, neglect the 2nd term in dh)
(1st law of thermodynamics for two fluids)
Also
δq = U(Th – Tc)dA
(3-4)
By (3-3),
dTh =
dTc =
Thus
− δq
m h c h
δq
m c c c
 1
1
dTh - dTc = d(Th – Tc) = -δq 
+
 m h c h m c c c



(3-5)
Eqs. (3-4) and (3-5) yield
 1
d (Th − Tc )
1 
dA
= −U 
+
Th − Tc
 m h ch m c cc 
(3-6)
Integration with respect to A from inlet 1 to outlet 2:
ln
 1
T h 2 − Tc 2
1
= −UA
+
Th1 − Tc1
 m h c h m c c c



(3-7)
(U, m h , ch, m c , cc are constant w.r.t.A;
d( )
= ln( ) + c) )
∫
()
Integrate (3-3) w.r.t. Th or Tc from 1 to 2
m h c h =
q
Th1 − Th 2
m c c c =
q
Tc 2 − Tc1
8
( m h , ch , m c , cc are constant w.r.t. Th or Tc)
Thus (3-7) gives
ln
Th 2 − Tc 2
UA
=−
(Th1 − Th 2 + Tc 2 − Tc1 )
Th1 − Tc1
q
(3-8)
= UA[(Th 2 − Tc 2 ) − (Th1 − Tc1 )] / q
i.e.
q = UA(LMTD)
LMTD =
(Th 2 − Tc 2 ) − (Th1 − Tc1 )
ln[(Th 2 − Tc 2 ) /(Th1 − Tc 2 )]
Comparing Eq. (3-8) with Eq. (3-1) (definition of U)
q = UA∆Tm
we have
∆Tm = LMTD =
(Th 2 − Tc 2 ) − (Th1 − Tc1 )
ln[(Th 2 − Tc 2 ) /(Th1 − Tc1 )]
Counterflow:
Still (3-9), left as an exercise. (under the same notation in Fig. 3-6)
Other heat Exchangers
∆Tm = F(LMTD)counterflow
F – correction factor
(LMTD)counterflow – LMTD for counterflow double
– pipe heat exchangers
(LMTD)counterflow – f(t1, t2, T1, T2)
t – tube side fluid
T – shell side fluid
1 – inlet
2 – outlet
F = f1(P, R)
9
(3-9)
P=
t 2 − t1
T1 − t1
T1 − T2
t 2 − t1
f1 depends on type of heat exchangers, see Figs. 3-7 and 3-8 for example.
R=
Fig. 3-7. Correction-factor plot for exchanger with one shell pass and two, four,
or any multiple of tube passes.
Fig. 3-8. Correction-factor plot for single-pass cross flow exchanger: one fluid mixed, the
other unmixed.
10
Example 3-1.
Water at the rate of 68 kg/min is heated from 35 to 75°C by an oil having a specific
heat of 1.9 kJ/kg⋅°C. The fluids are used in a counterflow double-pipe heat exchanger, and
the oil enters the exchanger at 110° and leaves at 75°C. The overall heat-transfer coefficient
is 320 W/m2⋅°C. Calculate the heat-exchanger area.
Solution.
The total heat transfer is determined from the energy absorbed by the water.
q = m w c w ∆Tw = (68)( 4180)(75 − 35) = 11.37 MJ/ min
= 189.5 kW
(a)
Since all the fluid temperatures are known, the LMTD can be calculate by using the
temperature scheme in Fig. 3-6:
∆Tm =
(110 − 75) − (75 − 35)
= 37.44°C
ln[(110 − 75) /(75 − 35)]
(b)
Then since q = UA∆Tm,
1.895 ×105
=
A = 15.82 m 2
(320)(37.44)
Example 3-2.
Instead of the double-pipe heat exchanger of Example 3-1, it is desired to use a shelland-tube exchanger with the water making one shell pass and the oil making two tube passes.
Calculate the area required for this exchanger, assuming that the overall heat-transfer
coefficient remains at 320 W/m2⋅°C.
Solution
To solve this problem, we determine a correction factor from Fig. 3-7 to be used with
the LMTD calculated on the basis of a counterflow exchanger. The parameters according to
the nomenclature of Fig. 3-7 are
T1 = 35°C
T2 = 75°C
t1 = 110°C
P=
t 2 − t1 75 − 110
=
= 0.467
T1 − t1 35 − 110
R=
T1 − T2 35 − 75
=
= 1.143
t 2 − t1 75 − 110
11
t2 = 75°C
so the correction factor is
F = 0.81
And the heat transfer is
q = UAF(LMTD)
so that A
=
1.895 ×105
= 19.53 m 2
(320)(0.81)(37.44)
12
3-4
Effectiveness – NTU Method
LMTD approach to heat exchanger analysis is useful when Tc1, Tc2, Th1, Th2 are
known or are easily determined. When the inlet or exit temperatures are to be evaluated for a
given heat exchanger, effectiveness – NTU method is more easily performed.
1.
Capacity ratio C, NTU and effectiveness ε
When m c cc ≤ m h ch
Cmin = m c cc
Cmax = m h ch
C=
NTU=
ε=
Cmin m c cc
=
≤ 1 (Capacity ratio)
Cmax m h ch
UA
UA
(Number of transfer units)
=
Cmin m c cc
Tc 2 − Tc1 m c cc (Tc 2 − Tc1 )
q
=
=
≤ 1 (Effectiveness)
Th1 − Tc1 m c cc (Th1 − Tc1 ) maximum possible q
When m h ch ≤ m c cc
Cmin = m h ch
Cmax = m c cc
C=
NTU=
ε=
Cmin m h ch
=
≤ 1 (Capacity ratio)
Cmax m c cc
UA
UA
(Number of transfer units)
=
Cmin m h ch
Th1 − Th 2 m h ch (Th1 − Th 2 )
q
=
=
≤ 1 (Effectiveness)
Th1 − Tc1 m h ch (Th1 − Tc1 ) maximum possible q
13
2.
ε-C-NTU relation for Double-pipe Heat Exchanger
Parallel Flow
T
Th1
dTh
δq
Th2
dTc
Tc2
dA
Tc1
2
1
δq = - m h ch dTh = m c cc dTc
dTh =
dTc =
− δq
m h c h
δq
m c c c
 1
1
+
dTh − dTc = −δq


 m h c h m c c c
d (Th − Tc )



 δq = U (Th − Tc )dA
 1
1
+
∴ d (Th − Tc ) = −U (Th − Tc )dA
 m h c h m c cc
i.e.
 1
d (Th − Tc )
1
= −UdA
+
Th − Tc
 m h c h m c cc



Integrate with respect to A from 1 to 2
ln
Th 2 − Tc 2
 1
1
= −UA
+
Th1 − Tc1
 m h c h m c cc



14



A
m c cc ≤ m h c h ,
Assuming
ln
Th 2 − Tc 2 − UA 
m c
1 + c c
=
Th1 − Tc1 m c cc  m h c h
Th 2 − Tc 2
Th1 − Tc1
or







− UA

= Exp
(1 + c)

 m c cc

 

 c min
 m h c h (Th1 − Th 2 ) = m c cc (Tc 2 − Tc1 )
Th 2 = Th1 + C (Tc1 − Tc 2 )
T − Tc 2 Th1 + C (Tc1 − Tc 2 ) − Tc 2
∴ h2
=
Th1 − Tc1
Th1 − Tc1
=
(Th1 − Tc1 ) + C (Tc1 − Tc 2 ) + (Tc1 − Tc 2 )
Th1 − Tc1
= 1−
∴ε =
Tc 2 − Tc1
(1 + C ) = 1 − ε (1 + C )
Th1 − Tc1
1 + Exp[(− UA C min )(1 + C )]
1+ C
Or
ε=
1 − Exp[− NTU (1 + C )]
1+ C
(a)
Assuming
m c cc ≥ m h c h
ln
 1
Th 2 − Tc 2
1
= −UA
+
Th1 − Tc1
 m h c h mc cc
=



− UA
(1 + C )
m h c h
or
Th 2 − Tc 2
= Exp[− NTU (1 + c)]
Th1 − Tc1
 m c cc (Tc 2 − Tc1 ) = m h c h (Th1 − Th 2 )
15
Tc2 = Tc1 + C(Th1 – Th2)
∴
Th 2 − Tc 2 Th 2 − Tc1 − C (Th1 − Th 2 )
=
Th1 − Tc1
Th1 − Tc1
=
(Th1 − Tc1 ) − (Th1 − Th 2 ) − C (Th1 − Th 2 )
Th1 − Tc1
= 1−
∴ε =
Th1 − Th 2
(1 + c) = 1 − ε (1 + c)
Th1 − Tc1
1 − Exp[− NTU (1 + c)]
1+ C
(b)
same as (a)
Counterflow
A similar analysis leads to
∴ε =
3.
1 − Exp[− NTU (1 − C )]
1 − C Exp[− NTU (1 − C )]
ε-C-NTU relation for the other heat exchangers
Table 3-2
ε = f(NTU, C)
Table 3-3
NTU = f1(ε, C)
16
Table 3-2. Heat-exchanger Effectiveness Relations.
N = NTU =
UA
C min
C=
Flow geometry
Double pipe:
Parallel flow
Relation
1 − exp[− N (1 + C )]
1+ C
1 − exp[− N (1 − C )]
ε=
1 − C exp[− N (1 − C )]
N
ε=
N +1
ε=
Counterflow
Counterflow, C = 1
Cross flow:
Both fluids unmixed
Both fluids mixed
Cmax mixed, Cmin unmixed
Cmax unmixed, Cmin mixed
Shell and tube:
One shell pass, 2, 4, 6, tube passes
C min
C max
 exp(− NCn) − 1

Cn


− 0.22
where n = N
ε = 1 − exp 

C
1
1
+
− 
ε =
1 − exp(− N ) 1 − exp(− NC ) N 
ε = (1/C){1 – exp[-C(1 – e-N)]}
ε = 1 – exp{-(1/C)[1 – exp(-NC)]}

ε = 21 + C + (1 + C 2 )1 / 2

1 + exp[− N (1 + C 2 )1 / 2 ] 
×

1 − exp[− N (1 + C 2 )1 / 2 ] 
Multiple shell passes, 2n, 4n, 6n
tube passes (εp = effectiveness of
each shell pass, n = number of shell
passes)
Special case for C = 1
All exchangers with C = 0
ε=
ε=
[(1 − ε p C ) /(1 − ε p )]n − 1
[(1 − ε p C ) /(1 − ε p )]n − C
nε p
1 + (n − 1)ε p
ε = 1 – e-N
17
−1
−1
Table 3-3. NTU Relations for Heat Exchangers
C = Cmin/Cmax
ε = effectiveness
Flow geometry
Double pipe:
Parallel flow
N = NTU = UA/Cmin
Relation
− ln[1 − (1 + C )ε ]
1+ C
1
 ε −1 
N=
ln

C − 1  Cε − 1 
N=
Counterflow
Counterflow, C = 1
N=
Cross flow:
Cmax mixed, Cmin unmixed
ε
1− ε
 1

N = − ln 1 + ln(1 − Cε )
 C

−1
N=
ln[1 + C ln(1 − ε )]
C
Cmax unmixed, Cmin mixed
Shell and tube:
One shell pass, 2, 4, 6, tube passes
N = −(1 + C 2 ) −1 / 2
All exchangers, C = 0
 2 / ε − 1 − C − (1 + C 2 )1 / 2 
× ln 
2 1/ 2 
 2 / ε − 1 − C + (1 + C ) 
N = -ln(1 - ε)
Evaporators and Condensers
In a boiling or condensation process, the fluid temperature stays essentially constant, or the
fluid acts as if it had infinite specific heat. In these cases Cmin/Cmax → 0 and all the heatexchanger effectiveness relations approach a single simple equation.
ε = 1 – e-NTU
N = -ln(1 - ε)
Example 3-3.
Hot oil at 100°C is used to heat air in a shell-and-tube heat exchanger. The oil makes
six tube passes and the air makes one shell pass; 2.0 kg/s of air are to be heated from 20 to
80°C. The specific heat of the oil is 2100 J/kg K, and its flow rate is 3.0 kg/s. Calculate the
area required for the heat exchanger for U = 200 W/m2 K.
Solution
The basic energy balance is
18
m o co ∆To = m a c pa ∆Ta
or
(3.0)(2100)(100 – Toe) = (2.0)(1009)(80 – 20)
Toe = 80.78°C
m h c h = (3.0)(2100) = 6300 W/K
m c cc = (2.0)(1009) = 2018 W/K
We have
so the air is the minimum fluid and
C=
C min 2018
=
= 0.3203
C max 6300
The effectiveness is
ε=
∆Tc
80 − 20
=
= 0.75
∆Tmax 100 − 20
Now, we may use the analytical relation from Table 3-3 to obtain NTU,
2 −1 / 2
N = −(1 + 0.3203 )
 2 / 0.75 − 1 − 0.3203 − (1 + 0.3203 2 )1 / 2 
ln 
2 1/ 2 
 2 / 0.75 − 1 − 0.3203 + (1 + 0.3203 ) 
= 1.99
Now, with U = 200 we calculate the area as
A = NTU
C min (1.99)(2018)
=
= 20.09 m 2
U
200
Example 3-4.
A counterflow double-pipe heat exchanger is used to heat 1.25 kg/s of water from 35
to 80°C by cooling an oil [cp = 2.0 kJ/kg K] from 150 to 85°C. The overall heat-transfer
coefficient is 850 W/m2 K. A similar arrangement is to be built at another plant location, but
it is desired to compare the performance of the single counterflow heat exchanger with two
smaller counterflow heat exchangers connected in series on the water side and in parallel on
the oil side, as shown in the sketch. The oil flow is split equally between the two exchangers,
and it may be assumed that the overall heat-transfer coefficient for the smaller exchangers is
the same as for the large exchanger. If the smaller exchangers cost 20 percent more per unit
heat-transfer area, which would be the most economical arrangement – the one large
exchanger or two equal-sized small exchangers?
19
Exchanger 1
= 150°C
Tw3
= 80°C
Exchanger 2
Tw1
= 35°C
Toi
Tw2
Toe1
Toe
Toe2
= 85°C
Fig. Ex. 3-4
Solution
We calculate the surface area required for both alternatives and then compare costs.
For the one large exchanger
q = m c cc ∆Tc = m h c h ∆Th
= (1.25)(4180)(80 − 35) = m h c h (150 − 85)
= 2.31 × 10 5 W
m c cc = 5225 W/K, m h c h = 3617 W/K
so that the oil is the minimum fluid:
εh =
∆Th
150 − 85
=
= 0.565
150 − 35 150 − 35
C min 3617
=
= 0.692
C max 5225
From Table 3-3, NTUlarge = 1.09, so that
20
A = NTU large
C min (1.09)(3617)
=
= 4.638 m 2
U
850
We now wish to calculate the surface-area requirement for the two small exchangers shown
in the sketch. We have
3617
= 1809 W/K
2
m c cc = 5225 W/K
m h c h =
C min 1809
=
= 0.347
C max 5225
The number of transfer units is the same for each heat exchanger because UA and Cmin are
the same for each exchanger. This requires that the effectiveness be the same for each
exchanger. Thus,
ε1 =
ε1 =
Toi − Toe ,1
Toi − Tw,1
150 − Toe ,1
150 − 35
= ε2 =
Toi − Toe , 2
= ε2 =
Toi − Tw, 2
(a)
150 − Toe , 2
150 − Tw, 2
where the nomenclature for the temperatures is indicated in the sketch. Because the oil flow
is the same in each exchanger and the average exit oil temperature must be 85°C, we may
write
Toe,1 + Toe , 2
2
= 85
(b)
An energy balance on the second heat exchanger gives
(5225)(Tw3 – Tw2) = (1809)(Toi – Toe,2)
(5225)(80 – Tw2) = (1809)(150 – Toe,2)
(c)
We now have the three equations (a), (b), and (c) which may be solved for the three
unknowns Toe,1, Toe,2, and Tw2. The solutions are
Toe,1 = 76.98°C
Toe,2 = 93.02°C
Tw2 = 60.26°C
The effectiveness can then be calculated as
ε1 = ε 2 =
150 − 76.98
= 0.635
150 − 35
From Table 3-3, we obtain NTUsmall = 1.16, so that
21
A = NTU small
C min (1.16)(1809)
=
= 2.47 m 2
U
850
We thus find that 2.47 m2 of area is required for each of the small exchangers, of a total of
4.94 m2. This is greater than the 4.638 m2 required in the one larger exchanger; in addition,
the cost per unit area is greater so that the most economical choice would be the single larger
exchanger. It may be noted, however, that the pumping costs for the oil would probably be
less with the two smaller exchangers, so that this could precipitate a decision in favor of the
smaller exchangers if pumping costs represented a sizable economic factor.
22