1. 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 to
1) 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.

2. 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 Shut-down Energy loss Cleaning costs
Disadvantages

3. 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,

5. Fouling Factors Fouling factor m2.K/W Type of fluid 0.00009
Seawater, below 50oC
0.002
Seawater, above 50oC
0.0002
Treated boiler feed water above 50oC
0.0009
Fuel oil
Alcohol vapors
Steam
0.0004
Industrial air
Refrigerating liquid

6. 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 = 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 = m2.K/W
Find:
Uo , Ui

7. UoAo = UiAi Uo = Ai x ( ri/ro ) = 487.3x(2.5/3.34) = 364.75 W/m2.K
Solution:

8. 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 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.

9. 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 = 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 = m2.K/W, cp,w = 4180 J/kg.K.
For tube, di = do = 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

10. The temperature profiles for the hot and cold fluids are shown in the figure. It can be noted that the
temperature difference between
the hot and cold Fluids is constant
(T = 30oC) throughout the heat
exchanger. The reason is the equal
capacity 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:

11. Laminar, hydrodynamic boundary layer development in a pipe
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

12. 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 = 1 2- 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.

13. Hyrodynamic fully developed region:
Darcy equation for pressure drop:
Power=V.xΔP
Laminar flow:

14. Moody Chart

15. Nusselt number for fully developed laminar flow in a circular pipe
Thermal fully developed region:
For constant wall temperature: For constant heat flux:
Nusselt number for fully developed laminar flow in a circular pipe
annulus with one surface insulated and the other at constant temperature .
(do= outer diameter of the inner tube, Di= inner diameter of the outer tube)
Nui
Nuo
do /Di
3.66 -
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
1.00

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

17. Thermal entry length only: Turbulent flow: Fully developed turbulent
Correlation of Hausen:
Fully developed turbulent
flow in a smooth pipe:
Combined entry length:
Correlation of Sieder and Tate: Blasius correlations:
Petukhov correlation:

18. Chilton-Colburn analogy ( j – factor): Dittus – Boelter correlation:

19. 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

20. 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

21. 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.

22. 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:

23. 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

24. The overall heat transfer coefficient is then

25. 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 = m , L = 1 m.

26. 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)

27. 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:

28. The heat loss per unit length of tube is

29. 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.