Feature Report
Engineeering
Practice
Accurate Wetted Areas
For Partially Filled Vessels
N
Accurately determine
wetted surface areas
needed for fire relief
applications
Richard C. Doane
S&B Engineers
and Constructors, Ltd.
T
N
&YQPTFEMJRVJEMFWFMN
Nomenclature
S wetted surface area of the vessel, m2
R vessel inside radius, m
Figure 1.
The elevation of
this horizontal
drum, partially
filled with liquid,
places it within the
fire zone. To calculate the relief area,
it is necessary
to know the total
wetted area of the
drum
N
GUJO
his article introduces a simple
H maximum liquid depth, m
yet accurate way to calculate the
F fractional liquid level = H / (2 · R),
wetted surface area of a partially
—
filled horizontal vessel with
eccentricity of the elliptical vessel
semi-elliptical heads. This informahead = 0.866 for the common case
tion is often needed for the sizing of
of a 2:1 ellipse, —
relief devices.
Ltangent-to-tangent length of the cylinAPI Recommended Practice 521
drical section of the vessel, m
specifies that a vessel containing liquid, mounted such that its lowest
point is less than 7.62 m (25 ft) above
ground level, must be fitted with a cal head is more challenging. Experi- liptical head is given by Equation (1),
pressure-relief device to protect it ence has shown that many engineer- below. Figure 1 shows a typical vessel
against an external fire. The vent area ing firms use inaccurate curve-fitting with 2:1 elliptical heads, 6.62 m (21 ft
needed for fire relief must always be techniques or conservative approxi- 9 in.) above grade level. Since the
calculated, even if this turns out not mations for this type of head. An exact elevation is below 7.6 m, the vessel
to be the limiting case.
mathematical relationship would be requires fire relief under API RecomThe relieving load calculation re- simpler and more accurate, yet the mended Practice 521. Since the elevaquires the engineer to know the author was unable to find a published tion is above 4.8 m, we need to calcuwetted surface area that would be formula.
late the wetted area rather than the
exposed to the fire. If the vessel’s elA solution was therefore developed total surface area. In this case:
evation and diameter are such that from first principles, and is presented R = 1.5 m
the entire vessel is not within the 25- below. The formulas for the partially H = 1 m
ft vertical fire zone, a partial surface filled hemispherical head and the F = H / (2 · R) = 1 m / (2 · 1.5 m) = 0.3333
area calculation is needed.
partially filled cylinder, found in the
= 0.866 for a 2:1 ellipse
For the cylindrical portion of the above references, are also included for From Equation (1), S = 3.64 m2
drum, the wetted area can easily completeness.
If the head is completely filled (F =
1), Equation 1 reduces to the formula
be determined using mensuration
formulas found in sources such as Elliptical head
given in the C.R.C. tables [2] for the
the “C.R.C. Standard Mathemati- The wetted surface area of a single el- surface area of half an oblate sphecal Tables” [1]. The wetted
surface area for a partially
Equation 1
filled hemispherical head is


2 
presented in “Machinery’s
 4ε ⋅ ( F − 0.5) + 1 + 12 ⋅ ( F − 0.5) 
2
π ⋅ R2 
1
S=
⋅ ln 
( F − 0.5) ⋅ 1 + 12 ⋅ ( F − 0.5) + 1 +

Handbook” [3].
2 
4ε
2− 3


Finding the wetted area



of a partially filled ellipti56
Chemical Engineering www.che.com December 2007
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CPD-5
8/20/07
10:34 AM
Page 1
The best way to heat and cool
the most corrosive materials.
roid. For the vessel in Figure 1, the
total area of each head is 9.76 m2.
Hemispherical head
For comparison, the wetted surface
area of a hemispherical head is given
by the concise relationship:
(2)
S = π ⋅ R⋅ H For the vessel shown in Figure 1,
the wetted area with hemispherical
heads is:
S = π ⋅ 1.5m ⋅ 1m = 4.71m 2
When the head is completely full,
the formula reduces to the familiar:
S = 2 ⋅ π ⋅ R2 = 2 ⋅ π ⋅ (1.5m ) = 14.1m 2
2
Cylinder
The wetted surface of the cylindrical
part of the vessel is found from:
R− H
S = 2 ⋅ L ⋅ R ⋅ cos−1 

 R 
(3)
where all angles are measured in radians.
For the cylindrical section of the
vessel in Figure 1, the wetted surface
area is:
 1.5m − 1m 
S = 2 ⋅ 6m ⋅ 1.5m ⋅ cos−1 

 1.5m 
= 22.1m 2
If the cylinder is completely filled
with liquid, H is equal to 2R, and
Equation (3) reduces to the familiar:
S = 2 ⋅ π ⋅ L ⋅ R = 2 ⋅ π ⋅ 6m ⋅ 1.5m
2
= 56.6m
■
Edited by Charles Butcher
References
1. “C.R.C. Standard Mathematical Tables, 12th.
Edition,” p. 398, Chemical Rubber Publishing Co., Cleveland, Ohio, 1959.
2. Ibid., p. 401.
3. “Machinery’s Handbook,” 17th. Edition,
p. 160, Industrial Press, New York, 1964.
Author
Richard C. Doane is a senior
process engineer with S&B
Engineers and Constructors,
Ltd. (7825 Park Place Boulevard, Houston, TX 77087;
Phone: 713-845-5338; Email:
rcdoane@sbec.com). He has
35 years of experience in process engineering and plant
operations. Doane holds B.S.
and M.S. degrees in chemical
engineering from Northeastern University and an M.S. degree in accounting
from the University of Houston, Clear Lake. He
is a professional engineer in the state of Texas.
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