View
220
Download
1
Category
Preview:
Citation preview
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.
Recommended