HEAT EXCHANGERS - EECL

```HEAT EXCHANGERS
Tutor: Nazaruddin Sinaga
email: [email protected]
Phone 024-76480655
Powerpoint Templates
Page 1
Contents
•
•
•
•
Types of Heat Exchangers
The Overall Heat Transfer Coefficient
Analysis of Heat Exchangers
The Log Mean Temperature Difference
Method
• The Effectiveness–NTU Method
Objectives
• Perform an energy balance over the heat
exchanger
• Identify key variables affecting heat transfer
and quantify the nature of any significant
effects
• Develop appropriate models to describe heat
transfer characteristics
Introduction to Heat Exchangers
What Are Heat Exchangers?
Heat exchangers are units designed to transfer heat from a hot flowing
stream to a cold flowing stream.
Why Use Heat Exchangers?
Heat exchangers and heat recovery is often used to improve process
efficiency.
Types of Heat Exchangers
• The recuperator, or through-the-wall non
storing exchanger
• The direct contact non storing exchanger
• The regenerator, accumulator, or heat storage
exchanger
Recuperators
Direct Contact
Regenerators
Heat Transfer Within a Heat Exchanger
 Heat transfer within a heat exchanger typically involves a
combination of conduction and convection
 The overall heat transfer coefficient U accounts for the overall
resistance to heat transfer from convection and conduction
1
1
1
x



UA h1 A1 h2 A2 kAmean
The schematic of a shell-and-tube heat exchanger (one-shell
pass and one-tube pass).
A plate-and-frame liquid-to-liquid heat exchanger
Thermal resistance network associated with
heat transfer in a double-pipe heat
exchanger.
The two heat transfer surface areas associated with a double-pipe heat
exchanger (for thin tubes, Di ≈ Do and thus Ai ≈ Ao).
The thermal resistance of the tube wall :
The total thermal resistance :
The rate of heat transfer between the two fluids as :
U is the overall heat transfer coefficient [ W/m2.oC]
When the wall thickness of the tube is small and the thermal
conductivity of the tube material is high, as is usually the case, the
thermal resistance of the tube is negligible (Rwall ≈ 0) and the inner and
outer surfaces of the tube are almost identical (Ai ≈ Ao ≈ As).
Equation for the overall heat transfer coefficient simplifies to:
The overall heat transfer coefficient U dominated by the smaller convection
coefficient, since the inverse of a large number is small.
 When one of the convection coefficients is much smaller than the
other (say, hi << ho), we have 1/hi >>1/ho, and thus U ≈ hi.
 Therefore, the smaller heat transfer coefficient creates a
bottleneck on the path of heat flow and seriously impedes heat
transfer.
 This situation arises frequently when one of the fluids is a gas
and the other is a liquid.
 In such cases, fins are commonly used on the gas side to enhance
the product UAs and thus the heat transfer on that side.
Type of heat exchanger
Gas-to-gas
Water-to-air in finned tubes (water in tubes)
Water-to-air in finned tubes (water in tubes)
Steam-to-air in finned tubes (steam in tubes)
Steam-to-air in finned tubes (steam in tubes)
Steam-to-heavy fuel oil
Water-to-oil
Steam-to-light fuel oil
Alcohol condensers (water cooled)
Freon condenser (water cooled)
Water-to-gasoline or kerosene
Ammonia condenser (water cooled)
Water-to-water
Feedwater heaters
Steam condenser
U, W/m2..°C
10–40
30–60†
300–850‡
30–300†
400–4000‡
50–200
100–350
200–400
250–700
300–1000
300–1000
800–1400
850–1700
1000–8500
1000–6000
*Multiply the listed values by 0.176 to convert them to Btu/h ·ft2 ·°F.
†Based on air-side surface area.
‡Based on water- or steam-side surface area.
 When the tube is finned on one side to enhance heat transfer, the total
heat transfer surface area on the finned side becomes
 For short fins of high thermal conductivity, we can use this total area in the
convection resistance relation Rconv =1/hAs since the fins in this case will
be very nearly isothermal.
 Otherwise, we should determine the effective surface area As from
Fouling Factor
• The performance of heat exchangers usually deteriorates with
time as a result of accumulation of deposits on heat transfer
surfaces.
• The layer of deposits represents additional resistance to heat
transfer and causes the rate of heat transfer in a heat
exchanger to decrease.
• The net effect of these accumulations on heat transfer is
represented by a fouling factor Rf , which is a measure of the
thermal resistance introduced by fouling.
• The most common type of fouling is the
precipitation of solid deposits in a fluid on the
heat transfer surfaces.
• Another form of fouling, which is common in the
chemical process industry, is corrosion and other
chemical fouling.
• Heat exchangers may also be fouled by the growth
of algae in warm fluids (biological fouling)
Precipitation fouling of ash particles on
superheater tubes
• The fouling factor depends on the operating temperature and
the velocity of the fluids, as well as the length of service.
• Fouling increases with increasing temperature and
decreasing velocity.
• For an unfinned shell-and-tube heat exchanger :
Rf, i and Rf, o are the fouling factors
Representative fouling factors (thermal resistance due to fouling for a unit surface area)
(Source: Tubular Exchange Manufacturers Association.)
EXAMPLE : Overall Heat Transfer Coefficient of a
Heat Exchanger
Hot oil is to be cooled in a double-tube counter-flow heat exchanger.
The copper inner tubes have a diameter of 2 cm and negligible
thickness. The inner diameter of the outer tube (the shell) is 3 cm.
Water flows through the tube at a rate of 0.5 kg/s, and the oil
through the shell at a rate of 0.8 kg/s. Taking the average
temperatures of the water and the oil to be 45°C and 80°C,
respectively, determine the overall heat transfer coefficient of this
heat exchanger.
SOLUTION
Hot oil is cooled by water in a double-tube counter-flow heat
exchanger. The overall heat transfer coefficient is to be
determined.
Assumptions
1. The thermal resistance of the inner tube is negligible since
the tube material is highly conductive and its thickness is
negligible.
2. Both the oil and water flow are fully developed.
3. Properties of the oil and water are constant.
• The properties of water at 45°C are
• The properties of oil at 80°C are
• The hydraulic diameter for a circular tube is the diameter of the tube
itself, Dh = D= 0.02 m.
• The mean velocity of water in the tube and the Reynolds number are
• Therefore, the flow of water is turbulent.
• Assuming the flow to be fully developed, the Nusselt
number can be determined from
• Now we repeat the analysis above for oil.
• The properties of oil at 80°C are :
• The hydraulic diameter for the annular space is
The mean velocity and the Reynolds number in this case are
which is less than 4000.
Therefore, the flow of oil is laminar.
Nusselt number for fully developed laminar flow in a circular annulus with
one surface insulated and the other isothermal (Kays and Perkins, Ref. 8.)
• Assuming fully developed flow, the Nusselt number on the tube
side of the annular space Nui corresponding to Di /Do =
0.02/0.03 = 0.667 can be determined from the table by
interpolation, we find :
Nu = 5.45
• and
• Then the overall heat transfer coefficient for this heat exchanger becomes
Discussion
• Note that U ≈ ho in this case, since hi >> ho.
• This confirms our earlier statement that the overall heat transfer coefficient in a
heat exchanger is dominated by the smaller heat transfer coefficient when the
difference between the two values is large.
• To improve the overall heat transfer coefficient and thus the heat transfer in
this heat exchanger, we must use some enhancement techniques on the oil side,
such as a finned surface.
DIMENSIONLESS ANALYSIS TO
CHARACTERIZE A HEAT EXCHANGER
Nu  f(Re, Pr, L/D, μ /μ )
b o
h.D
k
•Further Simplification:
v.D.
C p .

k
Nu  a.Re b .Pr c
Nu 
Can be obtained from 2 set of experiments
One set, run for constant Pr

And second set, run for constant Re
q
k

D

h
A(Tw  T )
•Empirical Correlation
• For laminar flow
Nu = 1.62 (Re*Pr*L/D)
• For turbulent flow
 b 
Nu  0.026. Re . P r . 
 o 
0.8
 Good to predict within 20%
 Conditions:
L/D > 10
0.6 < Pr < 16,700
Re > 20,000
1/ 3
0.14
Compact Heat Exchangers
• Analysis based on   NTU
method
• Convection (and friction) coefficients have been determined for selected
HX cores by Kays and London. Proprietary data have been obtained by
manufacturers of many other core configurations.
• Results for a circular tube-continuous fin HX core:
jh  St Pr 2 / 3
St  h / Gc p
G  Vmax
Results for a circular tube-continuous fin HX core
jh  St Pr 2 / 3
St  h / Gc p
G  Vmax
General Considerations
• Computational Features/Limitations of the LMTD Method:
 The LMTD method may be applied to design problems for which the fluid
flow rates and inlet temperatures, as well as a desired outlet temperature,
are prescribed.
 For a specified HX type, the required size (surface area), as well as the
other outlet temperature, are readily determined.
 If the LMTD method is used in performance calculations for which both
outlet temperatures must be determined from knowledge of the inlet
temperatures, the solution procedure is iterative.
 For both design and performance calculations, the effectiveness-NTU
method may be used without iteration.
LMTD Method
Q = U As Tlm
The procedure to be followed by the selection process is:
1. Select the type of heat exchanger suitable for the application.
2. Determine any unknown inlet or outlet temperature and the heat transfer rate
using an energy balance.
3. Calculate the log mean temperature difference Tlm and the correction factor F, if
necessary.
4. Obtain (select or calculate) the value of the overall heat transfer coefficient U.
5. Calculate the heat transfer surface area As .
CONCURRENT FLOW
1
2
∆T1
∆T2
∆A
A
T10
T1
T4
T5
T2
T6
T3
T9
T8
T7
P ara llel Fl ow
Calculating U using Log Mean Temperature
Hot Stream :
 h.Cph.dTh
dqh  m
Cold Stream:
 c .C pc .dTc
dqc  m
dq  dqhot  dqcold
 dq  U .T .dA

T2
T1
d (T )  dTh  dTc
d (T )  d (Th  Tc )
T  Th  Tc
 dqh
dqc
d (T )  

 m .C h m .C c
c
p
 h p
 1
1
d (T )  U .T .dA.

 m .C h m .C c
c
p
 h p
 1
d (T )
1

 U .

h
c

T
m
.
C
m
.
C
c
p
 h p
 A2
.
dA
 A1

 T2 
U . A.
  
Th  Tc    U . A Thin  Thout  Tcin  Tcout
ln
q
q
 T1 

 h.Cph.Th
qm
Log Mean Temperature
Difference




 

T  T
2
1
q U . A


 T

2


ln 

 T 
1





Calculating U using Log Mean Temperature
 T2 
U . A.
U . A in
out
in
out




ln



T


T


T

T

T

T
h
c
h
h
c
c


T
q
q
1 


 
 T2
ln
 T1

U .A
  
Thin  Tcin  Thout  Tcout
q

 T2
ln
 T1

U .A
U .A
  
T1  T2  
T2  T1 
q
q


 
T  T
2
1
q  U .A


 T

2

ln 

 T 
1


Log Mean
Temperature
Difference

Log Mean Temperature Evaluation
CONCURRENT FLOW
1
T1  Thin  Tcin  T3  T7
2
T2  Thout  Tcout  T6  T10
∆ T1
∆ T2
TLn 
∆A
A
U
T 10
T1
T4
T5
T2
T6
T3
T9
T8
T7
P ara ll el Fl ow
T2  T1
 T2 

ln
 T1 
 h .C ph .T3  T6 
m
A.TLn

 c .C pc .T7  T10 
m
A.TLn
Log Mean Temperature Evaluation
COUNTER CURRENT FLOW
1
T1
2
T3
T4
T6
T1
T6
Wall
T7
T2
T8
T9
T10
A
T 10
T1
T4
T2
T5
T3
T6
T7
T8
Co un t er - C u r re n t F low
T9
T2
T1  Thin  Tcout  T3  T7
T2  Thout  Tcin  T6  T10
q  hh Ai Tlm
(T  T1 )  (T6  T2 )
Tlm  3
(T  T1)
ln 3
(T6  T2 )

1
2
T3
T4
T1
T6
T6
Wall
T2
T7
T8
T9
T10

q  hc Ao Tlm
Tlm 
(T1  T7 )  (T2  T10 )
(T1  T7 )
ln
(T2  T10 )
A
The Effectiveness
– NTU Method
The Effectiveness – NTU Method
 In an attempt to eliminate the iterations from the solution of such
problems, Kays and London came up with a method in 1955
called the effectiveness–NTU method, which greatly simplified
heat exchanger analysis.
 This method is based on a dimensionless parameter called the
heat transfer effectiveness, defined as
The actual heat transfer rate in a heat exchanger can be determined
from an energy balance on the hot or cold fluids and can be
expressed as
Definitions
• Heat exchanger effectiveness : 
 
q
qmax
0   1
• Maximum possible heat rate :
qmax  Cmin Th,i  Tc,i 
Cmin

Ch if Ch  Cc
 or

Cc if Cc  Ch
 Will the fluid characterized by Cmin or Cmax experience the largest
possible temperature change in transit through the HX?
 Why is Cmin and not Cmax used in the definition of qmax?
For a parallel-flow heat exchanger can be rearranged as
Effectiveness relations of the heat exchangers typically involve the
dimensionless group UAs /Cmin.
This quantity is called the number of transfer units NTU and is
expressed as
In heat exchanger analysis, it is also convenient to define another
dimensionless quantity called the capacity ratio c as
Example
A counter-flow double-pipe heat exchanger is to heat water from 20°C
to 80°C at a rate of 1.2 kg/s. The heating is to be accomplished by
geothermal water available at 160°C at a mass flow rate of 2 kg/s. The
inner tube is thin-walled and has a diameter of 1.5 cm. If the overall
heat transfer coefficient of the heat exchanger is 640 W/m2 .°C,
determine the length of the heat exchanger required to achieve the
desired heating.
Assumptions
2. The heat exchanger is well insulated so that heat loss to the
surroundings is negligible and thus heat transfer from the hot fluid is
equal to the heat transfer to the cold fluid.
3. Changes in the kinetic and potential energies of fluid streams are
negligible.
4. There is no fouling.
5. Fluid properties are constant.
Problem 11.28: Use of twin-tube (brazed) heat exchanger to heat air
by extracting energy from a hot water supply.
KNOWN: Counterflow heat exchanger formed by two brazed tubes with prescribed hot and
cold fluid inlet temperatures and flow rates.
FIND: Outlet temperature of the air.
SCHEMATIC:
Problem: Twin-Tube Heat Exchanger (cont.)
ASSUMPTIONS: (1) Negligible loss/gain from tubes to surroundings, (2) Negligible
changes in kinetic and potential energy, (3) Flow in tubes is fully developed since L/D h = 40
m/0.030m = 1333.
PROPERTIES: Table A-6, Water ( Th = 335 K): ch = cp,h = 4186 J/kgK,  = 453  10-6
Ns/m2, k = 0.656 W/mK, Pr = 2.88; Table A-4, Air (300 K): cc = cp,c = 1007 J/kgK,  =
184.6  10-7 Ns/m2, k = 0.0263 W/mK, Pr = 0.707; Table A-1, Nickel ( T = (23 + 85)C/2 =
327 K): k = 88 W/mK.
ANALYSIS: Using the NTU -  method, from Eq. 11.30a,

1  exp   NTU 1  Cr  
1  Cr exp   NTU 1  Cr  
NTU  UA / Cmin
Cr  Cmin / Cmax .
(1,2,3)
and the outlet temperature is determined from the expression
  Cc  Tc,o  Tc,i  / Cmin  Th,i  Tc,i  .
(4)
From Eq. 11.1, the overall heat transfer coefficient is
1
1
1
1



UA o hA h Kt L o hA c
(5)
Since circumferential conduction may be significant in the tube walls, o needs to be evaluated for each of the tubes.
Problem: Twin-Tube Heat Exchanger (cont.)
The convection coefficients are obtained as follows:
4mh
4  0.04 kg / s
Water-side:
ReD 

 11, 243.
 D   0.010m  453 106 N  s / m2
The flow is turbulent, and since fully developed, the Dittus-Boelter correlation may be used,
0.3  0.023 11, 243
Nu h  h h D / k  0.023Re0.8


D Pr
0.8
 2.880.3  54.99
h h  54.99  0.656 W / m  K / 0.01m  3, 607 W / m2  K.
ReD 
Air-side:
4mc
4  0.120 kg / s

 275,890.
 D   0.030m 184.6 107 N  s / m2
The flow is turbulent and, since fully developed,
0.4  0.023 275,890
Nu c  h c D / K  0.023Re0.8


D Pr
0.8
 0.707 0.4  450.9
h c  450.9  0.0263 W / m  K / 0.030m  395.3 W / m 2  K.
Water-side temperature effectiveness:
A h   Dh L    0.010m  40m  1.257 m 2
o,h  f ,h  tanh  mLh  / mLh

m  3607 W / m2  K / 88 W / m  K  0.002m
m   hh P / kA 
1/ 2

1/ 2
 143.2 m 1
  h h / kt 
1/ 2
Problem: Twin-Tube Heat Exchanger (cont.)
With Lh = 0.5 Dh, o,h = tanh(143.2 m-1  0.5   0.010m)/143.2 m-1  0.5   0.010 m =
0.435.
Air-side temperature effectiveness:
Ac = DcL = (0.030m)40m = 3.770 m2

o,c  f ,c  tanh  mLc  / mLc m  395.3 W / m 2  K / 88 W / m  K  0.002m

1/ 2
 47.39 m 1
With Lc = 0.5Dc, o,c = tanh(47.39 m-1  0.5   0.030m)/47.39 m-1  0.5   0.030m =
0.438.
Hence, from Eq. (5) the UA product is
1
UA
1

2
0.435  3607 W / m  K  1.257 m
2

1
100 W / m  K  40m 
UA  5.070 104  2.50 104  1.533 103 


With
1

1
2
0.438  395.3 W / m  K  3.770 m
2
W / K  437 W / K.

Ch  mh ch  0.040 kg / s  4186 J / kg  K  167.4 W / K  C max
Cr  Cmin / Cmax  0.722
Cc  mccc  0.120 kg / s 1007 J / kg  K  120.8 W / K  C min
NTU 
437 W / K
UA

 3.62
Cmin
120.8 W / K
Problem: Twin-Tube Heat Exchanger (cont.)
and from Eq. (1) the effectiveness is
 
1  exp  3.62 1  0.722  
1  0.722 exp  3.62 1  0.722  
 0.862
Hence, from Eq. (4), with Cmin = Cc,
0.862 

Cc Tc,o  23C
Cc  85  23 C

Tc,o  76.4C
<
COMMENTS: (1) Using the overall energy balance, the water outlet temperature is


Th,o  Th,i   Cc / Ch  Tc,o  Tc,i  85C  0.722  76.4  23 C  46.4C.
(2) To initially evaluate the properties, we assumed that Th  335 K and Tc  300 K. From
the calculated values of Th,o and Tc,o, more appropriate estimates of Th and Tc are 338 K and
322 K, respectively. We conclude that proper thermophysical properties were used for water
but that the estimates could be improved for air.
Problem: Heat Transfer Enhancement
Problem 11.65: Use of fluted spheres and solid spheres to enhance the performance
of a concentric tube, water/glycol heat exchanger.
KNOWN: Flow rates and inlet temperatures of water and glycol in counterflow heat
exchanger. Desired glycol outlet temperature. Heat exchanger diameter and overall heat
transfer coefficient without and with spherical inserts.
FIND: (a) Required length without spheres, (b) Required length with spheres, (c)
Explanation for reduction in fouling and pump power associated with using spheres.
SCHEMATIC:
Th,i = 100oC
Th,o = 40oC
.
mh = 0.5 kg/s
L
Tc,i = 15oC
.
mc = 0.5 kg/s
Di = 0.075 m
Problem: Heat Transfer Enhancement (cont.)
ASSUMPTIONS: (1) Negligible kinetic energy, potential energy and flow work changes,
(2) Negligible heat loss to surroundings, (3) Constant properties, (4) Negligible tube wall
thickness.
PROPERTIES: Table A-5, Ethylene glycol  Th  70C  : cp,h = 2606 J/kgK; Table A-6,
Water  Tc  35C  : cp,c = 4178 J/kgK.
ANALYSIS: (a) With Ch = Cmin = 1303 W/K and Cc = Cmax = 2089 W/K, Cr = 0.624. With
actual and maximum possible heat rates of
q  Ch Th,i  Th,o  1303 W / K 100  40  C  78,180 W




q max  Cmin Th,i  Tc,i  1303 W / K 100  15  C  110, 755 W
the effectiveness is  = q/qmax = 0.706. From Eq. 11.30b,
  1 
1
 0.294 
NTU 
ln 
  2.66 ln 
  1.71
Cr  1   Cr  1 
 0.559 
Hence, with A = DL and NTU = UA/Cmin,
L
1303 W / K 1.71
Cmin NTU
 9.46m

2
 Di U
  0.075m 1000 W / m  K
(b) Since mc, mh, Th,i, Th,o and Tc,i are unchanged, Cr,  and NTU are unchanged. Hence,
with U = 2000 W/m2K,
L  4.73m
<
Problem: Heat Transfer Enhancement (cont.)
(c) Because the spheres induce mixing of the flows, the potential for contaminant build-up on
the surfaces, and hence fouling, is reduced. Although the obstruction to flow imposed by the
spheres acts to increase the pressure drop, the reduction in the heat exchanger length reduces
the pressure drop. The second effect may exceed that of the first, thereby reducing pump
power requirements.
COMMENTS: The water outlet temperature is Tc,o = T c,i + q/Cc = 15C + 78,180 W/2089
W/K = 52.4C. The mean temperature  Tc  33.7C  is close to that used to evaluate the
specific heat of water.
The End
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