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HEAT EXCHANGERSHEAT EXCHANGERS

Tutor: Nazaruddin Sinaga

email: nazarsinaga@yahoo.com

Phone 024-76480655

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 ExchangersIntroduction 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

There are three broad categories:

• The recuperator, or through-the-wall non storing exchanger

• The direct contact non storing exchanger

• The regenerator, accumulator, or heat storage exchanger

RecuperatorsRecuperators

Direct ContactDirect Contact

RegeneratorsRegenerators

Heat Transfer Within a Heat ExchangerHeat Transfer Within a Heat Exchanger

The overall heat transfer coefficient U accounts for the overall resistance to heat transfer from convection and conduction

meankA

x

AhAhUA

2211

111

Heat transfer within a heat exchanger typically involves a combination of conduction and convection

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 U, W/m2..°C Gas-to-gas 10–40 Water-to-air in finned tubes (water in tubes) 30–60† Water-to-air in finned tubes (water in tubes) 300–850‡ Steam-to-air in finned tubes (steam in tubes) 30–300† Steam-to-air in finned tubes (steam in tubes) 400–4000‡ Steam-to-heavy fuel oil 50–200 Water-to-oil 100–350 Steam-to-light fuel oil 200–400 Alcohol condensers (water cooled) 250–700 Freon condenser (water cooled) 300–1000 Water-to-gasoline or kerosene 300–1000 Ammonia condenser (water cooled) 800–1400 Water-to-water 850–1700 Feedwater heaters 1000–8500 Steam condenser 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 FactorFouling 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

which is less than 4000. Therefore, the flow of oil is laminar.

The mean velocity and the Reynolds number in this case are

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.

)o

/μb

μL/D,Pr,f(Re,Nu

DIMENSIONLESS ANALYSIS TO DIMENSIONLESS ANALYSIS TO CHARACTERIZE A HEAT EXCHANGERCHARACTERIZE A HEAT EXCHANGER

..Dv

k

C p .k

Dh.

Nu a.Reb .Pr c•Further Simplification:

Can be obtained from 2 set of experiments

One set, run for constant Pr

And second set, run for constant Re q k

A(Tw T )

h

Nu D

• For laminar flow

Nu = 1.62 (Re*Pr*L/D)

•Empirical CorrelationEmpirical Correlation

14.0

3/18.0 .Pr.Re.026.0

o

bNu

Good to predict within 20%

Conditions: L/D > 10 0.6 < Pr < 16,700 Re > 20,000

• For turbulent flow

Compact Heat ExchangersCompact Heat Exchangers

• Analysis based on method NTU

• 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:

2 / 3

max

Pr

/

h

p

j St

St h Gc

G V

Results for a circular tube-continuous fin HX core

2 / 3

max

Pr

/

h

p

j St

St h Gc

G V

General ConsiderationsGeneral 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

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 .

The procedure to be followed by the selection process is:

CONCURRENT FLOWCONCURRENT FLOW

T1 T2T4 T5

T6T3

T7T8 T9

T10

P ara llel Fl ow

∆ T1∆ T2

∆ AA

1 2

Calculating U using Log Mean TemperatureCalculating U using Log Mean Temperature

ch TTT

ch dTdTTd )(h

hphh .dT.Cmdq

ccpcc dTCmdq ..

Hot Stream :

Cold Stream:

cpc

chph

h

Cm

dq

Cm

dqTd

..)(

coldhot dqdqdq dATUdq ..

cpc

hph CmCm

dATUTd.

1

.

1...)(

2

1

2

1

..

1

.

1.

)( A

Acpc

hph

T

TdA

CmCmU

T

Td

outc

inc

outh

inhch TTTT

q

AUTT

q

AU

T

T

...

ln1

2

1

2ln

12.

T

T

TTAUq

Log Mean Temperature Difference

)()( ch TTdTd

hhph T..Cmq

Calculating U using Log Mean TemperatureCalculating U using Log Mean Temperature

outc

inc

outh

inhch TTTT

q

AUTT

q

AU

T

T

...

ln1

2

1

2ln

12.

T

T

TTAUq Log Mean

Temperature Difference

outc

outh

inc

inh TTTT

q

AU

T

T

.

ln1

2

12211

2 ..ln TT

q

AUTT

q

AU

T

T

CONCURRENT FLOWCONCURRENT FLOW

1

2

12

lnT

T

TTTLn

731 TTTTT inc

inh

1062 TTTTT outc

outh

Ln

cpc

Ln

hph

TA

TTCm

TA

TTCmU

.

..

.

.. 10763

T 1 T 2T 4 T 5

T 6T 3

T 7T 8 T 9

T 10

P ara ll el Fl ow

Log Mean Temperature EvaluationLog Mean Temperature Evaluation

∆ T1

∆ T2

∆ A

A

1 2

COUNTER CURRENT FLOWCOUNTER CURRENT FLOW

1062 TTTTT inc

outh

731 TTTTT outc

inh

T 1T 2

T 4 T 5

T 3

T 7 T 8 T 9

T 10

T 6

Co un t er - C ur re n t Flow

Log Mean Temperature EvaluationLog Mean Temperature Evaluation

T1

A

1 2

T2

T3

T6

T4 T6

T7 T8

T9

T10

Wall1T

2T

T1

A

1 2

T2

T3

T6

T4 T6

T7 T8

T9

T10

Wall

q hh Ai Tlm

Tlm (T3 T1) (T6 T2)

ln(T3 T1)

(T6 T2)

q hc Ao Tlm

Tlm (T1 T7) (T2 T10)

ln(T1 T7)

(T2 T10)

The Effectiveness The Effectiveness – NTU Method– 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 Effectiveness – NTU MethodThe Effectiveness – NTU Method

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

DefinitionsDefinitions• Heat exchanger effectiveness :

max

qq

0 1

• Maximum possible heat rate :

max min , ,h i c iq C T T

min

if or

h h cC C CC

if c c hC C C

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

ExampleA 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

1. Steady operating conditions exist.

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/Dh = 40 m/0.030m = 1333.

PROPERTIES: Table A-6, Water ( hT = 335 K): ch = cp,h = 4186 J/kgK, = 453 10-6

Ns/m2, k = 0.656 W/mK, Pr = 2.88; Table A-4, Air (300 K): cc = cp,c = 1007 J/kgK, = 184.6 10-7 Ns/m2, k = 0.0263 W/mK, Pr = 0.707; Table A-1, Nickel ( T = (23 + 85)C/2 = 327 K): k = 88 W/mK.

ANALYSIS: Using the NTU - method, from Eq. 11.30a,

r min max

rmin C C / C .

r r

1 exp NTU 1 CNTU UA / C

1 C exp NTU 1 C

(1,2,3)

and the outlet temperature is determined from the expression c c,o c,i min h,i c,iC T T / C T T . (4)

From Eq. 11.1, the overall heat transfer coefficient is

o t oh c

1 1 1 1

UA hA K L hA

(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:

Water-side: hD 6 2

4m 4 0.04 kg / sRe 11,243.

D 0.010m 453 10 N s / m

The flow is turbulent, and since fully developed, the Dittus-Boelter correlation may be used,

h0.8 0.30.8 0.3

h DNu h D / k 0.023Re Pr 0.023 11,243 2.88 54.99

2hh 54.99 0.656 W / m K / 0.01m 3,607 W / m K.

Air-side: cD 7 2

4m 4 0.120 kg / sRe 275,890.

D 0.030m 184.6 10 N s / m

The flow is turbulent and, since fully developed,

c0.8 0.40.8 0.4

c DNu h D / K 0.023Re Pr 0.023 275,890 0.707 450.9

2ch 450.9 0.0263 W / m K / 0.030m 395.3 W / m K.

Water-side temperature effectiveness: 2h hA D L 0.010m 40m 1.257 m

1/ 2 1/ 2o,h f ,h h h h htanh mL / mL m h P / kA h / kt

1/ 22 1m 3607 W / m K / 88 W / m K 0.002m 143.2m

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

1/ 22 1o,c f ,c c ctanh mL / mL m 395.3 W / m K / 88 W / m K 0.002m 47.39 m

With Lc = 0.5Dc, 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

2 2 2 2

1 1 1 1

UA 100 W / m K 40m0.435 3607 W / m K 1.257 m 0.438 395.3 W / m K 3.770 m

14 4 3UA 5.070 10 2.50 10 1.533 10 W / K 437 W / K.

With

h h h maxr min max

c c c min

C m c 0.040kg / s 4186 J / kg K 167.4 W / K C C C / C 0.722C m c 0.120kg / s 1007 J / kg K 120.8 W / K C

min

437 W / KUANTU 3.62

C 120.8 W / K

Problem: Twin-Tube Heat Exchanger (cont.)

1 exp 3.62 1 0.7220.862

1 0.722 exp 3.62 1 0.722

Hence, from Eq. (4), with Cmin = Cc,

c c,oc,o

c

C T 23 C0.862 T 76.4 C

C 85 23 C

<

COMMENTS: (1) Using the overall energy balance, the water outlet temperature is h,o h,i c h c,o c,iT T C / C T T 85 C 0.722 76.4 23 C 46.4 C.

(2) To initially evaluate the properties, we assumed that hT 335 K and cT 300 K. From

the calculated values of Th,o and Tc,o, more appropriate estimates of hT and cT 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.

and from Eq. (1) the effectiveness is

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:

L

T = 40 Ch,o o

T = 15 Cc,i o

T = 100 Ch,i o

.m h = 0.5 kg/s

.m c = 0.5 kg/s D = 0.075 mi

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 hT 70 C : cp,h = 2606 J/kgK; Table A-6,

Water cT 35 C : cp,c = 4178 J/kgK.

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

h h,i h,oq C T T 1303 W / K 100 40 C 78,180 W

max min h,i c,iq C T T 1303 W / K 100 15 C 110,755 W

the effectiveness is = q/qmax = 0.706. From Eq. 11.30b,

r r

1 1 0.294NTU ln 2.66 ln 1.71

C 1 C 1 0.559

Hence, with A = DL and NTU = UA/Cmin,

min2i

C NTU 1303 W / K 1.71L 9.46m

D U 0.075m 1000 W / m K

(b) Since c, h, h,i, h,o c,im m T T and T are unchanged, Cr, and NTU are unchanged. Hence,

with U = 2000 W/m2K, 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 = Tc,i + q/Cc = 15C + 78,180 W/2089

W/K = 52.4C. The mean temperature cT 33.7 C is close to that used to evaluate the

specific heat of water.

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