1
ISAT 413

Module V:
Industrial Systems
Topic 2:
Heat Exchanger Fundamentals,
Recuperative Heat Exchangers
Heat Exchangers
:
•
UA

LMTD Design Method
•
e

乔唠䑥獩杮s䵥瑨潤
•
An Example
2
Heat Exchangers
A heat exchanger is a
device for transferring
heat from one fluid to
another. There are three
main categories:
Recuperative
, in which
the two fluids are at all
times separated by a
solid wall;
Regenerative
,
in which each fluid
transfers heat to or from
a matrix of material;
Evaporative
(direct
contact), in which the
enthalpy of vaporization
of one of the fluids is
used to provide a
cooling effect.
3
Heat Exchanger (HX) Design Methods
HX designers usually use two well

known methods
for calculating the heat transfer rate between fluid
streams
—
the
UA

LMTD
and the
effectiveness

NTU
(number of heat transfer units) methods.
Both methods can be equally employed for
designing HXs. However, the
e

乔唠浥瑨潤楳i
灲p晥牲敤景爠牡瑩湧⁰牯r汥浳⁷桥牥琠汥慳琠潮攠
exit temperature is unknown. If all inlet and outlet
temperatures are known, the UA

LMTD method
does not require an iterative procedure and is the
preferred method.
4
LMTD (Log Mean Temperature Difference)
The most commonly used type of heat exchanger is the
recuperative heat exchanger. In this type the two fluids
can flow in
counter

flow, in
parallel

flow, or in a
combination of these, and
cross

flow.
The true mean temperature difference is the Logarithmic
mean Temperature difference (
LMTD
), is defined as
2
1
2
1
t
t
ln
t
t
LMTD
5
Heat Transfer Rate
of a Heat Exchanger
The heat transferred for any recuperative heat exchanger
can be calculated as(refer to the diagram shown on the
previous slide):
LMTD
UA
t
t
c
m
t
t
c
m
Q
C
C
C
C
H
H
H
H
2
1
2
1
fluids.
two
the
g
separatin
wall
the
of
area
transfer
heat
outside
total
the
is
area.
outside
the
on
based
t
coefficien
transfer
heat
overall
the
is
fluids.
cold
and
hot
the
of
heats
specific
the
are
and
fluids.
cold
and
hot
the
of
rates
flow
mass
the
are
and
where
A
U
c
c
m
m
c
H
C
H
6
Heat Exchanger
UA

LMTD
Design Method
Where
U
is the
overall heat
transfer
coefficient (and
is assumed to be
constant
over the
whole surface
area of the heat
exchanger).
L
h
r
R
L
k
r
/
r
ln
R
L
h
r
R
UA
R
LMTD
LMTD
UA
Q
o
o
o
,
f
p
i
o
i
,
f
i
i
n
i
i
n
i
i
2
1
2
2
1
1
1
1
Heat Transfer Duty
Overall Heat Transfer
Coefficient (W/m
2
.K)
Water to condensing
R12
440830
Steam to water
9601650
Water to water
8251510
Steam to gases
252750
7
Heat Exchanger
e

乔唠
䑥獩杮s䵥瑨潤
ci
hi
min
ho
hi
h
ci
hi
min
ci
co
c
max
actual
min
t
t
C
t
t
C
t
t
C
t
t
C
Q
Q
C
UA
NTU
e
mics
Thermodyna
of
Law
Second
by the
imposed
s
limitation
the
of
because
on
based
is
min
max
C
Q
ci
hi
min
actual
t
t
C
Q
e
from
fer
heat trans
of
rate
actual
the
calculate
can
one
exchanger,
heat
a
of
right)
on the
chart
the
(from
ess
effectiven
the
Knowing
8
Row 5
Row 4
Row 3
Row 2
Row 1
t
o
t
1
T
0
T
56
T
12
T
23
T
34
T
45
t
21
t
32
t
43
t
54
0
Y
0
0
X
0
A
R
CrossCounterflow
Fluid "R" Mixed Throughout;
Fluid "A" Unmixed Throughout. (Inverted Order)
Y
0
X
0
Row 6
T
56
T
6
t
65
6

row, 6

pass
plate fin

and

tube
cross counterflow HX
9
3

row, 3

pass
plate fin

and

tube
crossflow HX
10
6Row CrossCounterflow (
C
A
/C
R
= 1.0
)
Row 1
2
3
4
5
6
0
5
10
15
20
25
30
0
1
2
3
4
5
6
7
Number of Transfer Unit (NTU)
Decrease of HX Effectiveness (%)
Relative to Many Row Counterflow
Effectiveness of a
6

row, 6

pass
plate fin

and

tube
cross counterflow HX
11
A HX Example:
A schematic representation of a hybrid central receiver is
shown in the following slide (Slide #12). In this system,
molten nitrate salt is heated in a central receiver to
temperature as high as 1,050
o
F (565
o
C). The molten salt is
then passed through a heat exchanger, where it is used to
preheat combustion air for a combined

cycle power plant.
For more information about this cycle, refer to Bharathan
et. Al. (1995) and Bohn et al. (1995). The heat exchanger
used for this purpose is shown in Slide #13. The plates of
the heat exchanger are made of steel and are
2 mm
thick.
The overall flow is counter

flow arrangement where the air
and molten salt both flow duct

shape passages (unmixed).
The shell side, where the air flow takes place, is baffled to
provide cross flow between the lateral baffles. The
baseline design conditions are:
12
A hybrid central

receiver
concept developed
at the
NREL
13
Molten

salt

to

air HX
used to
preheat combustion air
14
A HX Example:
(continued)
Air flow rate: 0.503 kg/s per passage (250 lbm/s)
Air inlet temperature: 340
o
C (~650
o
F)
Air outlet temperature: 470
o
C (~880
o
F)
Salt flow rate: 0.483 kg/s per passage (240 lbm/s)
Salt inlet temperature: 565
o
C (~1050
o
F)
Salt outlet temperature: 475
o
C (~890
o
F)
Find the overall heat

transfer coefficient for this heat
exchanger. Ignore the fouling resistances.
Solution:
t
k
h
h
A
h
A
k
t
A
h
UA
p
a
s
a
p
s
.
thickness
its
is
and
plate,
steel
the
of
ty
conductivi
the
is
ly,
respective
side,
air
on the
and
side
salt
on the
t
coefficien
transfer

heat
convection
the
are
and
where
,
1
1
1
from
calculated
be
can
problem
for this
t
coefficien
fer
heat trans
overall
The
15
Solution:
(continued)
.
from
re
temperatu
at this
Pr
,
,
,
properties
obtain
may
we
where
405
2
470
340
2
at
calculated
are
properties
air
The
:
t,
coefficien
fer
heat trans
of
n
Calculatio
1.
).
1
(
stream
air
the
and
plate
steel
e
between th
resistance
transfer

heat
convection
the
and
),
(
plate
steel
he
through t
resistance
transfer

heat
conduction
),
1
(
plate
steel
the
and
stream
salt
e
between th
resistance
transfer

heat
convection
of
consist
they
and
#12,
Slide
in
shown
are
s
resistance
These
stream.
air
and
stream
salt
e
between th
fer
heat trans
to
s
resistance
thermal
all
identify
first
we
),
(
t
coefficien
transfer

heat
overall
obtain the
To
EES
,
C
T
T
T
h
air side
A
h
A
k
t
A
h
UA
o
o
,
a
i
,
a
a
a
a
p
s
16
Solution:
(continued)
.
.
.
,
.
Pr
Re
.
NU
.
UA
m
,
m
.
.
.
P
A
D
D
,
s
m
.
kg
.
m
.
m
.
s
kg
A
D
m
D
A
m
UD
Re
Nu
h
.
.
.
.
s
b
.
.
D
wet
a
H
H
H
H
H
a
H
30
53
1
69
0
335
15
027
0
027
0
:
ducts
and
tubes
long
smooth,
through
flow
turbulent
developed
fully
a
For
rate
flow
mass
and
006
0
003
0
2
2
003
0
2
4
4
as
calculated
be
can
and
diameter
hydraulic
the
is
where
,
335
15
10
28
3
006
0
006
0
0.503
is
number
Reynolds
The
.
number
Nusselt
for
expression
e
appropriat
an
choose
then
and
Reynolds
the
calculate
first
we
,
obtain
To
14
0
33
0
8
0
14
0
33
0
8
0
5
2
17
Solution:
(continued)
are
re
temperatu
average
at this
properties
salt
molten
The
520
2
475
565
:
for the
procedure
similar
a
follow
We
:
t,
coefficien
fer
heat trans
of
n
Calculatio
2.
448
006
0
0504
0
30
53
or
is
t
coefficien
transfer

heat
the
of
in term
number
Nusselt
The
here.
ignored
is
therefore
and
unity
to
close
is
term
this
ly,
respective
res,
temperatu
surface
and
bulk
at
ities
air viscos
of
ratio
the
represent
term
The
2
0.14
s
b
.
C
T
salt side
h
salt side
.
K
.
m
W
m
.
K
.
m
W
.
.
D
k
Nu
h
k
D
h
Nu
o
s
s
H
D
a
H
a
D
H
H
18
Solution:
(continued)
.
K
.
m
W
m
.
K
.
m
W
.
D
k
Nu
h
Nu
.
UA
m
,
m
.
.
.
P
A
D
s
m
.
kg
.
m
.
m
.
s
kg
A
D
m
D
A
m
UD
Re
Nu
h
m
kg
K
.
m
W
.
k
s
kg.m
.
μ
H
D
s
D
wet
a
H
H
H
H
s

H
H
2
3
2
3
3
724
006
0
543
0
8
calculate
can
n we
correlatio
With the
[1978].
London
and
Shah
by
given
as
,
8
is
ratio
aspect
large
very
duct with
a
inside
flow
laminar
a
for
number
Nusselt
the
and
laminar,
is
side
salt
on the
flow
The
rate
flow
mass
and
006
0
003
0
1
2
003
0
4
4
where
,
773
10
25
1
003
0
006
0
0.483
is
number
Reynolds
The
.
number
Nusselt
for
expression
e
appropriat
an
choose
then
and
Reynolds
the
calculate
first
we
,
obtain
To
.
1756
and
,
543
0
,
10
25
1
19
Solution:
(continued)
.
K
.
m
W
U
.
U
A
K
.
m
W
k
m
.
t
p
2
o
270
or
448
1
22
002
0
724
1
1
Therefore
equation.
the
of
sides
both
from
canceled
be
can
it
,
therefore
terms;
conduction
and
convection
for
same
the
is
area
flow

heat
that the
Note
exchanger.
heat
for this
t
coefficien
fer
heat trans
overall
obtain the
can
we
have
e
equation w
first
in the
these
all
for
ng
substituti
by
Now,
.
22
is
C
450
about
at
ty
conductivi
thermal
its
and
,
002
0
is
plate
steel
the
of
thickness
The
20
Recuperative Heat Exchangers
:
•
Definition of Recuperative HX
•
Types of Recuperative HX
•
Design Factors
•
Examples
A
Recuperative
Heat Exchanger (HX)
is one in
which the two fluids are separated at all times
by a solid barrier.
21
Waste

Heat
water

Tube
Boiler
Shell Boiler
using Waste
Gas
22
Furnace
Gas Air
Pre

Heater
23
Two

Pass Shell

and

Tube
Heat Exchanger
24
Gas

to

Gas Heat Recovery with a
Plate

Fin Heat Exchanger
A
A
25
Liquid

to

Liquid
Plate

Fin Heat Exchanger
26
Basic Equations
factors
fouling
surface
outside
and
inside
are
and
where
1
1
1
Also
flow.
of
type
on the
depends
factor
correction
the
is
and
where
as
given
are
(HX)
excahnger
heat
ve
recuperati
any
for
equations
basic
The
2
1
2
1
2
1
2
1
o
i
o
i
o
o
w
i
i
o
o
C
C
C
C
H
H
H
H
F
F
F
F
A
h
R
A
h
UA
K
t
/
Δt
ln
t
t
LMTD
K
LMTD
UA
t
t
c
m
t
t
c
m
Q
27
Heat Exchanger
Configurations
2
1
2
1
t
/
Δt
ln
t
t
LMTD
H
C
c
m
c
m
flow

Counter
H
C
c
m
c
m
flow

Counter
flow

Parallel
28
Extended Surfaces:
Fins, fpi (fins per inch)
.
efficiency
fin
the
is
and
fins,
the
of
area
surface
total
the
is
surface;
base
unfinned
of
area
the
is
fluid;
the
and
surface
base
e
between th
difference
re
temperatu
the
is
where
f
fin
base
f
f
base
A
A
T
T
A
A
h
Q
29
Example 5.4 (Eastop & Croft) Fin Surface
30
Example 5.4
A flat surface as shown in the previous slide has a base
temperature of 90
o
C when the air mean bulk
temperature is 20
o
C. Air is blown across the surface
and the mean heat transfer coefficient is 30 W/m
2

K.
The fins are made of an aluminum alloy; the fin
thickness is 1.6 mm, the fin height is 19 mm, and the
fin pitch is 13.5 mm. Calculate the heat loss per m
2
of
primary surface with and without the fins assuming
that the same mean heat transfer coefficient applies in
each case. Neglect the heat loss from the fin tips and
take a fin efficiency of 71%.
31
.
200
of
increase
an
shows
2100
of
surface
unfinned
for
value
with the
this
Comparing
6219
70
2.930
0.71
0.881
30
by
given
is
surface
finned
from
loss
heat
the
Therefore,
930
2
1
0.0016
74
1
0.019
74
2
881
0
1
0.0016

0.0135
74
:
are
areas
relevant
The
74
0135
0
1
as
calculated
be
can
length
m
1
a
on
fins
of
numbr
the
figure,
the
to
Referring
2100
20
90
1
1
30
by
given
is
fins
no
with
surface
primary
of
area
unit
per
,
loss,
heat
The
2
2
%
W
W
Q
m
.
A
m
.
A
.
.
/
W
T
T
hA
Q
Q
fin
,
loss
fin
base
a
s
loss
loss
Example 5.4
(continued)
32
e

NTU Method
(Effectiveness
—
Number of Thermal Units Method)
min
o
max
min
min
,
C
max
,
H
min
H
C
c
m
UA
NTU
,
C
C
R
NTU
R
.
t
t
c
m
Q
Q
where
,
units,
transfer
of
number
the
and
fluids,
two
the
of
capacities
thermal
the
of
ratio
of
function
the
as
expressed
be
can
It
or
Transfer
heat
possible
Maximum
transfer
heat
Actual
as
defined
is
,
ess,
effectiven
exchanger
Heat
e
e
33
e

NTU
(
Effectiveness against NTU)
for shell

and

tube heat exchangers
(with 2 shell passes and
4, 8, 12 tube passes)
R
NTU
R
NTU
Re
e
1
1
1
1
e
34
Characteristics of
e

乔唠䍨慲C
For given mass flow rates and specific heats of two
fluids the value of
e
摥灥湤猠潮⁴桥乔慮搠桥湣攠潮
瑨攠灲潤畣琠⡕(
o
). Thus for a given value of U the NTU
is proportional to A
o
. It can then be seen from the
e

乔唠
捨慲琠瑨慴t楮捲敡獩湧eA
o
increases
e
慮搠桥湣攠瑨攠t慶楮朠
楮略i.
The capital cost of the heat exchanger increases as the
area increases and
e

乔唠捨慲琠獨潷猠瑨慴t慴a桩杨h
values of
e
污牧攠楮捲敡獥e楮i慲敡e灲潤畣攠潮汹l愠獭慬氠
楮捲敡獥e楮i
e
.
The NTU and hence the effectiveness,
e
捡渠扥b
楮捲敡獥e
景f晩硥搠癡汵攠潦o慲敡e批楮捲敡獩湧e瑨攠
癡汵攠潦o瑨攠潶敲慬氠桥慴h瑲慮獦tr捯敦c楣楥湴Ⱐ售
35
Increasing HX
e
睩瑨w䙩F敤A
o
(1)
o
i
o
o
w
i
i
o
F
F
A
h
R
A
h
UA
1
1
1
The NTU, and hence
e
捡渠扥湣c敡獥搠景f硥搠癡汵攠潦瑨攠
area by increasing the value of the overall heat transfer
coefficient, U, which can be increased by increasing the heat
transfer coefficient for one or both of the individual fluids.
The h
eat transfer coefficient can be increased by reducing the
tube diameter, and/or increasing the mass flow rate per tube.
8
1
8
0
8
1
8
0
4
0
8
0
2
4
0
8
0
(constant)
4
023
0
4
4
where
)
(recall
023
0
tube
a
in
fer
heat trans
turbulent
typical
a
For
.
i
.
t
.
i
.
t
.
.
i
t
i
i
t
i
i
i
.
.
d
/
m
d
/
m
Pr
/
k
.
h
d
m
/
d
d
m
A
d
Au
ud
Re
k
/
hd
Nu
Pr
Re
.
Nu
36
Increasing HX
e
睩瑨w䙩F敤eA
o
(2)
Since , for a constant total mass flow rate the
number of tubes per pass must be reduced
correspondingly if the mass flow rate per tube is
increased.
Also, the heat transfer area is given by ,
where n is the number of tubes per pass, and p is the
number of tube passes. Therefore, to maintain the same
total heat transfer area for a reduced tube diameter in a
given type of heat exchanger, it is necessary to increase
the length of the tubes per pass, L, and/or the number of
tubes per pass (which will reduce the heat transfer rate.)
The design process is therefore
an iterative process
in
order to arrive at the optimum arrangement of tube
diameter, tube length, and number of tubes.
t
m
n
m
L
d
np
A
o
o
37
o
i
o
o
w
i
i
o
F
F
A
h
R
A
h
UA
1
1
1
Overall HX Design Considerations
Altering the inside diameter of a tube to increase the
heat transfer coefficient for flow through the tube will
alter the heat transfer on the shell side.
A full economic analysis also requires consideration of
the pumping power for both fluids. Pressure losses in
fluid flow due to friction, turbulence, and fittings such as
valves, bends etc. are proportional to the square of the
flow velocity. The higher the fluid velocity and the more
turbulent the flow the higher is the heat transfer
coefficient but the greater the pumping power.
38
Example 5.5
(a) A shell

and

tube heat exchanger is used to recover energy
from engine oil and consists of two shell passes for water and
four tube passes for the engine oil as shown diagrammatically in
the following figure. The effectiveness can be calculated based
on Eastop Equation (3.33). For a flow of oil of 2.3 kg/s entering
at a temperature of 150
o
C, and a flow of water of 2.4 kg/s
entering at 40
o
C, use the data given to calculate:
(i) the total number of tubes required;
(ii) the length of the tubes;
(iii) the exit temperatures of the water and oil;
(iv) the fuel cost saving per year if water heating is currently
provided by a gas boiler of efficiency 0.8.
(b) What would be the effectiveness and fuel saving per year
with eight tube passes?
39
Example 5.5
(continued)
40
”Use EES for
Eastop Example 5.5
"
{hot oil}
m_dot_H=2.3
{oil mass flow rate, kg/s}
t_H1=150
{hot oil inlet temperature, C}
cp_H=2.19
{mean spefici heat of oil, kJ/kg

K}
rho_H=840
{mean oil density, kg/m^3}
C_H=m_dot_H*cp_H
{hot fluid capacity, kW/K}
{cold water}
m_dot_C=2.4
{water mass flow rate, kg/s}
t_C2=40
{cold water inlet temperature, C}
cp_C=4.19
{mean specific heat of water, kJ/kg

K}
C_C=m_dot_C*cp_C
{cold fluid capacity, kW/K}
{data}
eta_boiler=0.8
{gas boiler efficiency}
v_H=0.8
{oil velocity in the tube, m/s}
eta_Hx=0.7
{require HX effectiveness}
n_pass=4
{four pass heat exchanger}
d_i=0.005
{tube inside diameter, m}
d_o=0.007
{tube outside diameter, m}
U=0.400
{overall heat transfer coefficient, kW/m^2

K}
t_hours=4000
{annual usage, h}
cost=1.2
{cost of water heating, p/kWh}
41
{a(i): calculate the total number of tubes required}
V_dot_H=m_dot_H/rho_H
A_cross=V_dot_h/v_H
A_1=PI*d_i^2/4
n_tube=A_cross/A_1*n_pass
n_tube=697 [tubes]
{a(ii): calculate the length of the tubes}
R=min(C_H,C_C)/max(C_H,C_C)
eta_HX=(1

exp(

NTU*(1

R)))/(1

R*exp(

NTU*(1

R)))
NTU=U*A_o/min(C_H,C_C)
A_o=PI*d_o*n_tube*L_tube
L_tube=1.27 [m]
{a(iii): calculate the exit temperature of oil and water}
Eta_HX=C_H*(t_H1

t_H2)/(min(C_h,C_C)*(t_H1

t_C2))
C_H*(t_H1

t_H2)=C_C*(t_C1

t_C2)
t_C1=78.6 [C];
t_H2=73.0 [C]
42
{a(iv): calculate the total heat transfer and fuel cost saving per year}
Q_dot=C_C*(t_C1

t_C2)
Fuel_saving=Q_dot*t_hours*cost/(eta_boiler*100)
Fuel_saving=23271 [British Pounds]
{b(v): calculate eta2_hx if double Ao}
NTU2=2*NTU
eta2_HX=(1

exp(

NTU2*(1

R)))/(1

R*exp(

NTU2*(1

R)))
eta2_HX=0.881
{b(vi): calulate t_H2, t_C2 and fuel_saving}
eta2_HX=C_H*(t_H1

t2_H2)/(min(C_h,C_C)*(t_H1

t_C2))
C_H*(t_H1

t2_H2)=C_C*(t2_C1

t_C2)
Q2_dot=C_C*(t2_C1

t_C2)
Fuel2_saving=Q2_dot*t_hours*cost/(eta_boiler*100)
Fuel2_saving=29279 [British Pounds]
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