Heat Exchangers:

Design Considerations

Common Operational Problems

1. Fouling:

As discussed by Mueller, fouling can be caused by:

(1) Precipitation of dissolved substances,

(2) Deposit of particulate matter,

(3) Solidification of material through chemical reaction,

(4) Corrosion of the surface,

(5) Attachment and growth of biological organisms, and

(6) Solidification by freezing.

The term "baffle cut" is used to specify the dimensions of a segmental

baffle. The baffle cut is the height of the segment removed to form the

baffle, expressed as a percentage of the baffle disc diameter. Baffle

cuts from 15 to 45 per cent are used. Generally, a baffle cut of 20 to

25 per cent will be the optimum, giving good heat-transfer rates,

without excessive drop. There will be some leakage of fluid round the

baffle as a clearance must be allowed for assembly. The clearance

needed will depend on the shell diameter; typical values, and

tolerances, are given in Table 12.5.

Baffle

The minimum thickness to be used for baffles and

support plates are given in the standards. The baffle

spacing used range from 0.2 to 1.0 shell diameters. A

close baffle spacing will give higher heat transfer

coefficients but at the expense of higher pressure drop.

The optimum spacing will usually be between 0.3 to 0.5

times the shell diameter.

Design an exchanger to subcool condensate from a methanol condenser from 95°C to

40°C. Flow rate of methanol 100,000 kg/h. Brackish water will be used as the

coolant, with a temperature rise from 25° to 40°C.

Example 12.1

Only the thermal design will be considered.

This example illustrates Kern’s method.

Coolant is corrosive, so assign to tube side.

Solution

Use one shell pass and two tube passes:

From Figure 12.19:

From Figure 12.1

Choose 20mm o.d., 16mm i.d., 4.88-m long tubes (¾ in ×16 ft), cupro-nickel.

Allowing for tube-sheet thickness, take

Use a split-ring floating-head type.

From Figure 12.10, bundle diametrical clearance = 68 mm, shell diameter,

Ds = 826 +68 =894 mm.

Note:

• Nearest standard pipe sizes are 863.6 or 914.4 mm.

• Shell size could be read from standard tube count tables.

Tube-Side Coefficient

The coefficient can also be calculated using equation 12.15; this is done to

illustrate use of this method.

The coefficient can also be calculated using equation 12.15; this is done to

illustrate use of this method.

From Figure 12.23, jh = 3.9× 10–3

0.2 x Ds

1.25 do

Shell-Side Coefficient

Shell-Side Coefficient

Shell-Side Coefficient

To calculate the shell-side Reynolds number, given by:

Choose 25% baffle cut, from Figure 12.29

This shows that the correction for a low-viscosity fluid is not significant.

Thermal conductivity of cupro-nickel alloys = 50W/m°C.

Take the fouling coefficients from Table 12.2; methanol (light organic) 5000Wm2°C–1,

brackish water (sea water), take as highest value, 3000Wm2°C–1

Overall Coefficient

For heat exchange across a typical heat exchanger tube, the relationship between

the overall coefficient and the individual coefficients, is given by:

Overall Coefficient

well above assumed value of 600 Wm2°C–1.

Pressure Drop

Tube-Side

From Figure 12.24, for Re = 14,925

jf = 4.3 × 10–3

Neglecting the viscosity correction term

low; could consider increasing the number of tube passes.

Shell-Side

could be reduced by increasing the baffle pitch. Doubling the pitch halves the shellside

velocity, which reduces the pressure drop by a factor of approximately (1/2)2

Design a shell and tube exchanger for the following duty: 20,000 kg/h of kerosene (42° API)

leaves the base of a kerosene side-stripping column at 2008C and is to be cooled to 90 °C by

exchange with 70,000 kg/h light crude oil (34° API) coming from storage at 40 °C. The

kerosene enters the exchanger at a pressure of 5 bar and the crude oil at 6.5 bar. A pressure

drop of 0.8 bar is permissible on both streams. Allowance should be made for fouling by

including a fouling factor of 0.0003 (W/m2 °C)-1 on the crude stream and 0.0002 (W/m2 °C) -1

on the kerosene stream.

Example

The tube velocity needs to be reduced. This will reduce the heat transfer

coefficient, so the number of tubes must be increased to compensate. There

will be a pressure drop across the inlet and outlet nozzles. Allow 0.1 bar for

this, a typical figure (about 15% of the total), which leaves 0.7 bar across the

tubes. Pressure drop is roughly proportional to the square of the velocity, and

t is proportional to the number of tubes per-pass. So the pressure drop

calculated for 240 tubes can be used to estimate the number of tubes required.

Modified Design

Tubes needed = 240/(0.6/1.4)0.5 = 365

Say, 360 with four passes.

Retain four passes, as the heat transfer coefficient will be too low with two

passes.

Second trial design: 360 tubes 19.05mm o.d., 14.83mm i.d., 5m long, triangular

pitch 23.81 mm.

Tube side recalculated

This result is well within specification.

Keep the same baffle cut and spacing.

This result is too high; the specification allowed only 0.8 overall, including the loss over

the nozzles. Check the overall coefficient to see if there is room to modify the shell-side

design.

So, to check the overall coefficient to see if there is room to modify the shell-side

design.

FIND: (a) Evaporator area, (b) Water flow rate.

SCHEMATIC:

Design of a two-pass, shell-and-tube heat exchanger to supply vapor for the

turbine of an ocean thermal energy conversion system based on a standard

(Rankine) power cycle. The power cycle is to generate 2 MWe at an efficiency

of 3%. Ocean water enters the tubes of the exchanger at 300K, and its desired

outlet temperature is 292K. The working fluid of the power cycle is evaporated

in the tubes of the exchanger at its phase change temperature of 290K, and the

overall heat transfer coefficient is known.

Example:

ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and

potential energy changes, (3) Constant properties.

PROPERTIES: Table A-6, Water ( mT = 296 K): cp = 4181 J/kgK.

ANALYSIS: (a) The efficiency is

W 2MW

0.03.q q

Hence the required heat transfer rate is

2MW

q 66.7MW.0.03

Also

m,CF300 290 292 290 C

T 5 C300 290

n292 290

and, with P = 0 and R = , from Fig. 11.10 it follows that F = 1. Hence

7

2m,CF

q 6.67 10 WA

U F T 1200 W / m K 1 5 C

2A 11,100m .

b) The water flow rate through the evaporator is

7

hp,h h,i h,o

q 6.67 10 Wm

4181 J / kg K 300 292c T T

hm 1994 kg / s.

COMMENTS: (1) The required heat exchanger size is enormous due to the small

temperature differences involved,

(2) The concept was considered during the energy crisis of the mid 1970s but has not since

been implemented.