A new family of practical fin profiles for surface tension drained condensation has been
described by Kedzierski and Webb (1990). More precise models for surface tension con-
trolled condensation have been developed by Honda and Nozu (1987b), Honda et al.
(1987) and Adamek and Webb (1990).
To account for the condensate flooding, Owen et al. (1983) and Webb et al. (1985)
suggested that the heat transfer coefficients for the unflooded and flooded parts be
computed separately. They recommended that eq 90 should be modified as follows:
B Ar
A B
hη = 1 -
+ ηf hf f + hb
hh
(98)
π
A π
A
where 1 (B/π) and B/π represent the fractions of the circumference unflooded and flood-
ed, respectively, and hb is the heat transfer coefficient for the flooded region. Webb et al.
(1985) found that for steam condensing on a 19-mm-diam. tube having 203 fins/m, heat
transfer through the flooded region was only 1.6 % of the total. Thus the second term in eq
98 can be neglected for most practical purposes.
In the most recent work, Rose (1994) used some simplifying assumptions, together
with dimensional analysis, to develop an equation for calculating the ratio hfined tube/
neous paper, Briggs and Rose (1994) modified the equation given by Rose (1994) to
include the effect of fin efficiency.
Example 6
Saturated refrigerant R-12 at 32C condenses on a horizontal integral-fin tube of outer
diameter do = 19.1 mm and a root diameter dr = 15.88 mm. The tube has 748 fins/m. The
temperature at the root of the fin is maintained at 22C. The fins are trapezoidal in shape,
0.38 mm thick at the base and 0.23 mm thick at the tip. Assuming the fin efficiency ηf = 1,
calculate the condensation heat transfer coefficient, and the enhancement ratio using 1)
the Beatty and Katz (1948) gravity drained model, and 2) the Webb et al. (1985) surface
tension drained model. For the surface tension drained model, assume that the conden-
sate film shape is described by the Adamek profile parameters, ξ = 0.857, sm = 1.5936
temperature of 27C (300 K) are
ρl
1305.8 kg/m3
=
ρv
40 kg/m3
=
νl
19.5 108 m2/s
=
kl
=
0.072 W/m K
σ
=
0.0158 N/m
hfg
=
133.79 kJ/kg.
Solution
The fin surface per meter of tube length is
π 2
(
)
do - dr2 n + πdowtn = 0.1458 m2/m .
Af = 2
4
The unfinned surface area of the tube per meter is
Ar = πdr - nπ dr wb = 0.0357 m2/m .
The total surface of the finned tube per meter is
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