Transcript Introduction - James Madison University

ISAT 413 - Module V: Industrial Systems
Topic 2:
Heat Exchanger Fundamentals,
Recuperative Heat Exchangers
 Heat Exchangers:
• UA-LMTD Design Method
• e -NTU Design Method
• An Example
1
 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.
2
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-NTU method is
preferred for rating problems where at least one
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.
3
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
t1  t 2
LMTD 
t1
ln
t 2
4
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):
 H cH t H1  t H 2   m
 C cC tC1  tC 2   UALMTD
Q  m
where m H and m C are the m ass flow rates of the hot and cold
fluids.
c H and cc are the specific heats of the hot and cold fluids.
U is the overall heat transfer coefficient based on the outsidearea.
A is the total outsideheat transfer area of the wall separating
the two fluids.
5
Heat Exchanger UA-LMTD Design Method
LMTD
Q  UALMTD  n
R
i
i 1
1

UA
n

i 1
Ri 
lnro / ri 
1
1
 R f ,i 
 R f ,o 
2ri hi L
2k p L
2ro ho L
Where U is the
overall heat
transfer
coefficient (and
is assumed to be
constant over the
whole surface
area of the heat
exchanger).
Heat Transfer Duty
Overall Heat Transfer
Coefficient (W/m2.K)
Water to condensing 440-830
R-12
Steam to water
960-1650
Water to water
825-1510
Steam to gases
25-2750
6
Heat Exchanger e-NTU Design Method
UA
NTU 
C min
Q actual
e 
Q
Q max is based on C min because
of thelimitations imposedby the
Second Law of T hermodyna
mics
max

C c t co  t ci 
C t  t 
 h hi ho
C min t hi  t ci  C min t hi  t ci 
Knowing theeffect iveness
(from thechart on t heright )
of a heat exchanger,one can
calculatetheactualrat eof
heat t rans
fer from
Q actual  eC min t hi  t ci 
7
6-row, 6-pass plate fin-and-tube
cross counterflow HX
Cross-Counterflow
Fluid "R" Mixed Throughout;
Fluid "A" Unmixed Throughout. (Inverted Order)
t21
X0
0
A
T0
Row 1
Y0
t43
Y0
T12
0
t1
t65
T23
Row 2
Row 3
T34
Row 4
T45
Row 5
Row 6
T56 T
56
T6
X0
t32
t54
to
R
8
3-row, 3-pass plate fin-and-tube
crossflow HX
9
Effectiveness of a 6-row, 6-pass plate finand-tube cross counterflow HX
6-Row Cross-Counterflow (C A /C R = 1.0 )
Decrease of HX Effectiveness (%)
Relative to Many Row Counterflow
30
Row 1
25
20
15
2
10
3
5
4
5
6
0
0
1
2
3
4
5
6
7
Number of Transfer Unit (NTU)
10
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 1,050oF (565oC). 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:
11
A hybrid central-receiver concept developed
at the NREL
12
Molten-salt-to-air HX used to
preheat combustion air
13
A HX Example: (continued)
Air flow rate: 0.503 kg/s per passage (250 lbm/s)
Air inlet temperature: 340oC (~650oF)
Air outlet temperature: 470oC (~880oF)
Salt flow rate: 0.483 kg/s per passage (240 lbm/s)
Salt inlet temperature: 565oC (~1050oF)
Salt outlet temperature: 475oC (~890oF)
Find the overall heat-transfer coefficient for this heat
exchanger. Ignore the fouling resistances.
Solution:
T heoverallheat trans
fer coefficient for thisproblemcan be calculatedfrom
1
1
t
1



, where hs and ha are theconvectionheat - transfer
UA hs A k p A ha A
coefficient on thesalt side and on theair side, respectively, k p is the
conductivity of thesteel plate,and t is its thickness.
14
Solution: (continued)
T o obtain theoverallheat - transfercoefficient (UA), we first identifyall
thermalresistances to heat trans
fer between the salt streamand air stream.
T heseresistances are shown in Slide #12,and theyconsist of convection
1
heat - transferresistancebetween the salt streamand thesteel plate(
),
hs A
conductionheat - transferresistancethrough he
t steel plate(
t
kpA
),
and theconvectionheat - transferresistancebetween the steel plateand the
1
air stream(
).
ha A
1. Calculation of air side heat trans
fer coefficient, ha :
Ta ,i  Ta ,o
340 470
T heair propertiesare calculatedat Ta 

 405o C ,
2
2
where we may obtain properties, , , P r at thistemperatu
re from EES .
15
Solution: (continued)
T o obt ain ha , we first calculat et heReynoldsand t henchoosean appropriate
expressionfor Nusselt number Nu. T heReynoldsnumber is
m
kg

DH
0.503
 0.006m
UDH
m DH
A
s
Re 



 15,335,
kg
.
m


A
0.006m 2  3.28 105
s
where DH is t hehydraulicdiamet erand can be calculat edas
DH 
4 Aa 42  0.003

 0.006m , and mass flow rat em  UA.
Pwet 22  0.003
For a fully developedt urbulentflow t hroughsmoot h,long t ubes and duct s :
NU DH  0.027 Re
0. 8
Pr
 0.02715,335
0.33   b
0.8

 
 s 
0.14
0.690.33 10.14  53.30.
16
Solution: (continued)
0.14
 
T he term b  representthe ratioof air viscosities at bulk and
 s 
surface temperatu
res, respectively, this termis close to unity and
thereforeis ignored here.T he Nusselt number in termof theheat - transfer
coefficient is NuDH 
ha DH
or
k
W
NuDH k
m .K  448 W .
ha 

DH
0.006m
m 2 .K
2. Calculation of salt side heat trans
fer coefficient, hs :
53.30 0.0504
We followa similar procedurefor thesalt side :
565 475
Ts 
 520o C .
2
T hemoltensalt propertiesat thisaverage temperatu
re are
17
Solution: (continued)
kg.m
W
kg
, k  0.543
, and   1756 3 .
s
m .K
m
T o obt ain hs , we first calculat et heReynoldsand t henchoosean appropriate
μ  1.2510- 3
expressionfor Nusselt number Nu. T heReynoldsnumber is

m
kg

DH
0.483
 0.006m
 DH
UDH
m
A
s
Re 



 773,
kg.m


A
0.003m 2  1.25 103
s
4 Aa
40.003
  UA.
where DH 

 0.006m , and mass flow rat em
Pwet 21  0.003
T heflow on t hesalt side is laminar,and t he Nusselt number for a laminarflow
inside a duct wit h verylarge aspect rat iois NuDH  8, as given by Shah and
London[1978].Wit h t hecorrelat ion we can calculat e
W
NuDH k 8  0.543 m .K
W
hs 

 724 2 .
DH
0.006m
m .K
18
Solution: (continued)
T he thicknessof thesteel plateis t  0.002m , and its thermalconductivity
W
at about 450o C is k p  22
. Now, by substituting for all thesein the
m .K
first equation we have we can obtain theoverallheat trans
fer coefficient
for thisheat exchanger.Note that theheat - flow area A is thesame for
convectionand conductionterms;therefore, it can be canceledfrom both
sides of theequation.
T herefore
1
1
0.002 1



U 724
22
448
W
or U  270 2 .
m .K
19
 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.
20
Waste-Heat
water-Tube
Boiler
Shell Boiler
using Waste
Gas
21
Furnace
Gas Air
Pre-Heater
22
Two-Pass Shell-and-Tube
Heat Exchanger
23
Gas-to-Gas Heat Recovery with a
Plate-Fin Heat Exchanger
A
A
24
Liquid-to-Liquid
Plate-Fin Heat Exchanger
25
Basic Equations
T hebasic equationsfor any recuperatve
i heat excahnger(HX)
are given as
Q  m H c H t H 1  t H 2   m C cC t C1  t C 2   UAo LMTDK
t1  t 2
where LMTD 
ln Δt1 / t 2 
and K is thecorrect ionfactordepends on t hetypeof flow.
Also
1
1
1

 Rw 
 Fi  Fo
UAo hi Ai
ho Ao
where Fi and Fo are inside and outside surface fouling factors
26
Heat Exchanger
Configurations
Counter- flow
m c C  m c H
t1  t 2
LMTD 
lnΔt1 / t 2 
Counter- flow
m c C  m c H
Parallel - flow
27
Extended Surfaces: Fins, fpi (fins per inch)


Q  h Abase   f A f T
where T is the temperatu
re
differencebetween the base
surface and thefluid;
Abase is thearea of unfinned
base surface;
A fin is the totalsurface area
of thefins, and
 f is thefin efficiency.
28
Example 5.4 (Eastop & Croft) Fin Surface
29
Example 5.4
A flat surface as shown in the previous slide has a base
temperature of 90oC when the air mean bulk
temperature is 20oC. Air is blown across the surface
and the mean heat transfer coefficient is 30 W/m2-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 m2 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%.
30
Example 5.4 (continued)
The heat loss,Q loss , per unitarea of primary surface with no fins
is givenby Q
 hA T  T  30 11 90  20  2100W
loss
s
a

Referring to the figure, the numbr of fins on a 1 m lengthcan be
calculatedas 1 / 0.0135 74.
The relevantareas are :
Abase  74 0.0135- 0.0016 1  0.881m 2
A fin  2  74 0.019 1  74 0.0016 1  2.930m 2
Therefore, the heat lossfrom finned surface is givenby
Q
 30 0.881 0.71 2.930 70  6219W
loss, fin
Comparingthis with thevaluefor unfinnedsurface of
2100 W showsan increaseof 200%.
31
e-NTU Method
(Effectiveness — Number of Thermal Units Method)
Heat exchangereffectiveness, e , is definedas
Actualheat transfer
e
Maximumpossibleheat T ransfer
Q C or Q H

.
m c min t H ,max  tC ,min 
It can be expressed as the functionof ratio of the
thermalcapacitiesof the two fluids,R and the number
of transfer units, NTU, where
Cmin
UAo
R
, NTU 
m c min
Cmax
32
e-NTU (Effectiveness against NTU)
for shell-and-tube heat exchangers
1  e  NTU 1 R 
e
1  Re NTU 1 R 
(with 2 shell passes and
4, 8, 12 tube passes)
33
Characteristics of e-NTU Chart
 For given mass flow rates and specific heats of two
fluids the value of e depends on the NTU and hence on
the product (UAo). Thus for a given value of U the NTU
is proportional to Ao. It can then be seen from the e-NTU
chart that increasing Ao increases e and hence the saving
in fuel.
 The capital cost of the heat exchanger increases as the
area increases and e-NTU chart shows that at high
values of e large increase in area produce only a small
increase in e.
 The NTU and hence the effectiveness, e can be
increased for a fixed value of area by increasing the
value of the overall heat transfer coefficient, U.
34
Increasing HX e with Fixed Ao (1)
1
1
1

 Rw 
 Fi  Fo
UAo hi Ai
ho Ao
The NTU, and hence e can be increased for a fixed value of the
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.
For a t ypicalt urbulent heat t rans
fer in a t ube
Nu  0.023Re 0.8 Pr 0.4 (recall Nu  hd i / k )
ud i Au d i
m t d i
4m t
where Re 



2

A
 d i / 4 d i 




0.8
1.8
0.8
1. 8


 h  0.023k 4 /  Pr m t / d i  (const antm
) t / di
The heat transfer coefficient can be increased by reducing the
tube diameter, and/or increasing the mass flow rate per tube.
0. 8
0.4
35
Increasing HX e with Fixed Ao (2)
  nm
 t , for a constant total mass flow rate the
 Sincem
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 byAo  npdo L ,
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.
36
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.
1
1
1

 Rw 
 Fi  Fo
UAo hi Ai
ho Ao
37
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 150oC, and a flow of water of 2.4 kg/s
entering at 40oC, 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?
38
Example 5.5 (continued)
39
”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}
40
{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]
41
{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]
42