Fouling Factor: After a period of operation the heat transfer surfaces for a heat exchanger become coated with various deposits present in flow systems, or the surfaces may become corroded as a result of the interaction between the fluids and the material used for construction of the heat exchanger. This coating represents an additional resistance to the heat flow, and thus results in decreased performance. This effect is called “fouling factor”, or “fouling resistance”, Rf . The economic disadvantages of fouling can be attributed to1) Energy losses due to thermal inefficiencies.2)Additional costs associated with periodic cleaning of heat

exchangers.3) Loss of production during shutdown for cleaning. 4) Higher capital cost through oversized units.

Epstein has specified the following six categories of fouling:1) Scaling fouling : the crystallization from solution of dissolved

substance onto the heat transfer surface. 2) Particulate fouling : the accumulation of finely divided solids

suspended in the process fluid onto the heat transfer surface.3) Chemical reaction fouling : the deposit formation on the heat

transfer surface by chemical reaction.4) Corrosion fouling : the accumulation of corrosion on the heat

transfer surface. 5) Biological fouling : the attachment of microorganisms to a heat

transfer surface. 6) Solidification fouling : the crystallization of a pure liquid or one

component from the liquid phase on a sub-cooled heat transfer surface.

1- Initial cost 2- Shut-down 4- Energy loss 3- Cleaning costsDisadvantages

When a new heat exchanger is put into service, its performance decreases progressively as a result of the buildup of fouling resistance. The main factors affecting the rate of fouling on the heat transfer surface of a heat exchanger besides length of service, are: 1) The fluid velocity : an increase in the velocity decreases both

the rate of deposit and the ultimate amount of deposit on the surface.

2) The fluid temperature : increasing the fluid bulk temperature increases both the rate of buildup of fouling and its ultimate stable level.

Fouling factor must be included for both inner and outer pipe sides along with the other thermal resistances in order to determine the overall heat transfer coefficient U. Thus,

WKmhh

Rcleandirty

f /. 11 2

ooo

ofio

i

if

iiiioo AhA

R

kLrr

A

R

AhAUAU1

2ln111 ,,

ooo

ofio

i

if

ii

iioo

AhA

R

kLrr

A

R

Ah

AUAU1

2ln1

1

,,

o

ofioo

i

ifo

ii

oo

hR

krrr

r

Rr

rhr

U1ln

1

,,

oo

i

o

ofiioiif

i

i

rhr

r

Rr

krrr

Rh

U

,

,

ln11

Fouling Factors

Fouling factor

m2.K/WType of fluid

0.00009 Seawater, below 50oC

0.002 Seawater, above 50oC

0.0002 Treated boiler feed water above 50oC

0.0009 Fuel oil

0.00009 Alcohol vapors

0.00009 Steam

0.0004 Industrial air

0.0002 Refrigerating liquid

Example 3:Determine the overall heat transfer coefficient Uo and Ui based on the outer and inner surfaces, respectively, of a steel pipe (k = 54 W/m.K) with an inner diameter of di = 2.5 cm and an outer diameter of do = 3.34 cm for the following flow and fouling conditions:hi = 1800 W/m2.K , ho = 1250 W/m2.K , Rf,i = Rf,o = 0.00018 m2.K/W

Data:

steel pipe, di = 2.5 cm, do = 3.34 cm,

k = 54 W/m.K , hi =1800 W/m2.K, ho = 1250 W/m2.K, Rf,i = Rf,o = 0.00018 m2.K/W

Find:

Uo , Ui

UoAo = UiAi

Uo = Ai x ( ri/ro ) = 487.3x(2.5/3.34) = 364.75 W/m2.K

KmW

.67lnUo

./3.487

12501

00018.054100336.11

5.200018.034.3

5.2180034.3

1

2

o

ofioo

i

ifo

ii

oo

hR

krrr

r

Rr

rhr

U1ln

1

,,

Solution:

Example 4:Engine oil is to be cooled from 80 to 50oC by using a single-pass, counter-flow, concentric tube heat exchanger with cooling water available at 20oC. Water flows inside a thin tube with an inner diameter of 2.5 cm at a rate of 0.08 kg/s, and oil flows through the annulus at a rate of 0.16 kg/s. The heat transfer coefficients for the water side and oil side are 1000 W/m2.K and 80 W/m2.K, respectively. The fouling factors on the water and oil sides are the same and equal to 0.00018 m2.K/W. If the tube wall resistance is negligible, calculate the rate of heat flow, the outlet temperature of cooling water and the overall heat transfer coefficient. Take cp,w = 4180 J/kg.K and cp,oil =2090 J/kg.K.

Data: single-pass, counter-flow, concentric-tube heat exchanger. For oil, Toil,i = 80oC , Toil,o = 50oC, moil = 0.16 kg/s, ho = 80 W/m2.K , Rf,i = 0.00018 m2.K/W , cp,oil =2090 J/kg.K .

For water, Tw,i = 20oC, mw = 0.08 kg/s, hi = 1000 W/m2.K, Rf,o = 0.00018 m2.K/W, cp,w = 4180 J/kg.K.

For tube, di = do = 0.025 m.

Find: , Tc,o , U

Solution: The rate of heat flow and the outlet temperature of water can be calculated from an overall energy balance

.

Q

W TTcmQ ooilioiloilpoil 032,105080209016.0,,,

..

iwowwpw TTcmQ ,,,

..

CcmQTT owpwiwow 50418008.0032,1020,

..

,,

The temperature profiles for the hot and cold fluids are shown in the figure. It can be noted that thetemperature difference betweenthe hot and cold Fluids is constant(T = 30oC) throughout the heat exchanger. The reason is the equalcapacity rates for the hot and cold fluids.i.e. The overall heat transfer coefficient U is determined from the following equation, by neglecting the thermal resistance of the tube:

wpwoilpoil cmcm ,

.

,

.

KmW

hRRhU

oofifi

./2.7280100018.000018.010001

1

111

2

,,

Calculation of the heat transfer coefficient (h):

Laminar, hydrodynamic boundary layer development in a pipe

Laminar, thermal boundary layer development in a constant surface temperature pipe

Variation of the friction factor and the convection heat transfer coefficient in the flow direction for flow in a tube (Pr ˃ 1) for laminar flow of viscous fluids e.g. oils. (opposite gross of the boundary layers is the case for liquid metals when 1 ˃ Pr).Note that 1- for gases the two boundary layers coincide, i.e. Pr = 12- for turbulent flow the two boundary layers are about the same and independent of Pr.3- hx, fx remain constant in fully developed laminar and turbulent flow.

Hyrodynamic fully developed region:

Darcy equation for pressure drop:

Power=V.xΔP

Laminar flow:

dudu

Re mm

k

c

c kPr p

p

khd

Nu

2300Re

mm udAu m 2.

4

,4

2

.

d

mum

Re d

x

lam

hfd 05.0,

d m

Re

.

4

length entry ichydrodynamx hfd ,

Ref

C , Re

f f

164

64

factor friction Moody

tcoefficien friction Darcyf

221 dynamic

stress

)(,

m

w

f

upressure

shearwall

FanningtcoefficienfrictionSkinC

gu

dL

fp m

2

2

PrRe d

x

lam

tfd 05.0,

length entry thermalx hfd ,

Moody Chart

f

Thermal fully developed region:

For constant wall temperature: For constant heat flux:

Nusselt number for fully developed laminar flow in a circular pipeannulus with one surface insulated and the other at constant temperature . (do= outer diameter of the inner tube, Di= inner diameter of the outer tube)

contantT for Nu w 66.3 contantq for Nu w 36.4

Nui Nuo do /Di

3.66 - 0

4.06 17.46 0.05

4.11 11.56 0.10

4.23 7.37 0.25

4.43 5.74 0.50

4.86 4.86 1.00

Nusselt number and friction factor for fully developed laminar flow in pipes of different cross sections

Nu Nu

f Reh Tw = c qw = c b/a Cross Section

64 3.66 4.36 --

57 2.98 3.61 1.0

59 3.08 3.73 1.43

62 3.39 4.12 2.0

69 3.96 4.79 3.0

73 4.44 5.33 4.0

82 5.60 6.49 8.0

96 7.54 8.23 ∞53 2.47 3.11 --

Thermal entry length only: Turbulent flow:

Correlation of Hausen: Fully developed turbulent flow in a smooth pipe:Combined entry length:Correlation of Sieder and Tate: Blasius correlations:

Petukhov correlation:

3204.01

0668.066.3

PrRe Ld

PrRe Ld Nu

14.031

86.1

wdL

PrRe Nu

1670048.0

75.90044.0

Pr

constantT

w

w

000,203164.0

25.0 Re Re

f

000,20184.0

2.0 Re Re

f

264.1ln79.0 Ref

61053000 Re

10, turbhfd dx

40003000 practth Re , Re

Chilton-Colburn analogy ( j – factor):

Dittus – Boelter correlation:

3232

82Pr

PrRe Nu

Pr StfC f

nPrRe Nu 8.0023.0

cooling, for n andheating for n 3.04.0

2f32 C

Pr Stj

10

1607.0

000,10

dL

Pr

Re

Variation of local Nusselt number along a tube in turbulent flow for both uniform surface temperature and uniform surface heat flux.- values - length - similarity - equations

Flow across a single circular cylinder (tube):

Whitaker correlation: Whitaker correlated the average heat transfer coefficient hm for the flow of gases or liquids across a single cylinder (tube) by

The properties are to be evaluated at the mean film temperature Tf :

In the range of variables

25.0

4.0325.0 06.04.0

w

Pr ReRek

hdNu

30067.0

2.525.0

1040 5

Pr

Re

w

2w

f

TTT

Example 5:

A perfect insulated concentric pipe heat exchanger is used to cool unused lubricating oil at a mean temperature of 77oC for a large industrial gas turbine engine using water at an average temperature of 32oC. The flow rate of cooling water through the inner pipe (di = 25 mm, do = 30 mm, k = 16 W/m.K) is 0.2 kg/s, while the flow rate of oil through the annulus (Di = 45 mm) is 0.1 kg/s. Assuming fully developed conditions for both water and oil and constant fluid properties, calculate the convection heat transfer film coefficients and the overall heat transfer coefficient (Uo).

Data: Fully developed conditions.

K CT oh 3502737777

K CT oc 3052733232

skg m , skg m hc /1.0/2.0..

m D , KmW k , m d , m d ipoi 45.0./163.025.0

Find: hi , ho , Uo

Solution: From tables of properties,For unused engine oil at For water at

For water flow through the inner pipe,

Accordingly, the flow is turbulent and Nusselt number may be computed from Dittus – Boelter correlation:

K T h 350W/m.K k , N.s/m 103.56 , kJ/kg.K c 22-

p 138.0118.2 K T c 305

W/m.K k ,N.s/m 107 , kJ/kg.K c 26-p 62.069178.4

2.5Pr

7.1324510769

2.0446

.

0.025 d

mdu Re

i

cm

4.08.0023.0 PrRe Nu

For the flow of oil through the annulus, the hydraulic diameter is

The Reynolds number is

The annular flow is therefore laminar. Nusselt number may be obtained from the table for annular flow. With do /Di = 0.667, linear interpolation provides

27.882.57.13245023.0 4.08.0 Nu

KW/m dk

Nuh 2

ii 1.2189

025.062.027.88

m dDD oih 015.0030.0045.0

oioi

oihm

dDm

dDmdD Du

Re

.

22

.

44

69.471056.3030.0045.0

1.042

Re

45.5kDh

Nu hoo

The overall heat transfer coefficient is then

KW/m h 2o 14.50

015.0138.0

45.5

op

ioo

ii

oo

hkrrr

rhr

U1ln

1

14.50

116

025.0030.0ln015.0025.01.2189

030.01

oU

KW/m U 2o 4.48

019944.0000171.0000548.01

Example 6:

Water at a mean temperature of Tm = 80oC and a mean velocity of um = 0.15 m/s flows inside a 2.5 cm inner diameter, thin –walled copper tube. Atmospheric air at T∞ = 20o C and a velocity u∞=10 m/s flows across the tube. Neglecting the tube wall resistance, calculate the inner and outer heat transfer coefficients, the overall heat transfer coefficient and the rate of heat loss per unit length ( one meter) of the tube. Data: A single tube, cross-flow heat exchanger for water Tm = 80oC, um = 0.15 m/s, for air T∞ = 20o C, u∞ =10 m/s

for tube d = di = do = 0.025 m , L = 1 m.

Find:

Solution: The physical properties of water at Tm = 80oC are The Reynolds number for water flow is

We use the Dittus–Boelter correlation to determine hi for water flow: , with n = 0.3 (cooling of water)

.

Q , U , h , h oi

22.26680/10364.0 6 Pr , W/m.K .k , sm 2

1030210364.0025.015.0

6

du

Re m

nPrRe Nu 8.0023.0

4.4722.210302023.0 3.08.0 Nu

KW/m dk

Nuh 2

ii 1267

025.0668.0

4.47

CTT owatermw 80,

CTTT owf 50280202

The physical properties of air at Tf = 50oC are

The Reynolds number for the air flow becomes

The Nusselt number for cross-flow of air across a single circular tube is determined from the Whitaker correlation:

Csmkg

Csmkg

W/m.K.s

o

w

o

80at ./ 10075.2

,20at ./ 10983.1

,703.0Pr , 02810k ,/m 1022.18

5

5

26

721,131022.18025.010

6

du

Re

25.0

4.0325.0 06.04.0

w

Pr ReRek

hdNu

77.69

10075.210983.1

7030721,1306.0721134.025.0

5

54.0325.0

. ,Nu

The heat loss per unit length of tube is

KW/m dk

Nuh 2

oo 4.78

025.00281.0

77.69

KmW hh

Uoi

./8.734.78112671

111

1 2

TUAQ .

mW

TTdL UQ airwater

/8.34720801025.08.73

.

Free Convection on the Outer Surface of a Tube:

The free convection heat transfer coefficient (ho) at the outer surface of a tube can be calculated from the following simplified expression:

where,ho = outer heat transfer coefficient [W/m2.K],

do = outer diameter of the tube,

Tw = wall temperature,

T∞ = ambient temperature.

25.0

32.1

o

wo d

TTh