Heat Exchanger Fundamentals and ... - CISAT Sharepoint

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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
R-12
440-830
Steam to water
960-1650
Water to water
825-1510
Steam to gases
25-2750
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
Cross-Counterflow
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

6-Row Cross-Counterflow (
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]