This article introduces a simple yet accurate way to calculate the wetted surface area of a partially filled horizontal vessel with

semi-elliptical heads. This informa-tion is often needed for the sizing of relief devices.

API Recommended Practice 521 specifies that a vessel containing liq-uid, mounted such that its lowest point is less than 7.62 m (25 ft) above ground level, must be fitted with a pressure-relief device to protect it against an external fire. The vent area needed for fire relief must always be calculated, even if this turns out not to be the limiting case.

The relieving load calculation re-quires the engineer to know the wetted surface area that would be exposed to the fire. If the vessel’s el-evation and diameter are such that the entire vessel is not within the 25-ft vertical fire zone, a partial surface area calculation is needed.

For the cylindrical portion of the drum, the wetted area can easily be determined using mensuration formulas found in sources such as the “C.R.C. Standard Mathemati-cal Tables” [1]. The wetted surface area for a partially filled hemispherical head is presented in “Machinery’s Handbook” [3].

Finding the wetted area of a partially filled ellipti-

cal head is more challenging. Experi-ence has shown that many engineer-ing firms use inaccurate curve-fitting techniques or conservative approxi-mations for this type of head. An exact mathematical relationship would be simpler and more accurate, yet the author was unable to find a published formula.

A solution was therefore developed from first principles, and is presented below. The formulas for the partially filled hemispherical head and the partially filled cylinder, found in the above references, are also included for completeness.

Elliptical headThe wetted surface area of a single el-

liptical head is given by Equation (1), below. Figure 1 shows a typical vessel with 2:1 elliptical heads, 6.62 m (21 ft 9 in.) above grade level. Since the elevation is below 7.6 m, the vessel requires fire relief under API Recom-mended Practice 521. Since the eleva-tion is above 4.8 m, we need to calcu-late the wetted area rather than the total surface area. In this case:R = 1.5 mH = 1 mF = H / (2 · R) = 1 m / (2 · 1.5 m) = 0.3333 = 0.866 for a 2:1 ellipse

From Equation (1), S = 3.64 m2

If the head is completely filled (F = 1), Equation 1 reduces to the formula given in the C.R.C. tables [2] for the surface area of half an oblate sphe-

Feature Report

56 ChemiCal engineering www.Che.Com DeCember 2007

Engineeering Practice

Accurate Wetted Areas For Partially Filled Vessels

Accurately determine wetted surface areas needed for fire relief

applications Figure 1. The elevation of this horizontal drum, partially filled with liquid, places it within the fire zone. To calcu-late the relief area, it is necessary to know the total wetted area of the drum

Richard C. DoaneS&B Engineers and Constructors, Ltd.

NomeNclature

S wetted surface area of the vessel, m2

R vessel inside radius, m

H maximum liquid depth, m

F fractional liquid level = H / (2 · R), —

eccentricity of the elliptical vessel head = 0.866 for the common case of a 2:1 ellipse, —

L tangent-to-tangent length of the cylin-drical section of the vessel, m

Equation 1

S R F FF

= ⋅ −( ) ⋅ + ⋅ −( ) + + ⋅⋅ −π

ε

ε22

20 5 1 12 0 5 1 1

4

4 0. . ln

.. .5 1 12 0 5

2 3

2( ) + + ⋅ −( )−

F

56-57 CHE 12-07.indd 56 11/28/07 1:56:36 PM

roid. For the vessel in Figure 1, the total area of each head is 9.76 m2.

Hemispherical headFor comparison, the wetted surface area of a hemispherical head is given by the concise relationship:

S R H= ⋅ ⋅π (2)

For the vessel shown in Figure 1, the wetted area with hemispherical heads is:

S = ⋅ ⋅ =π 1 5 1 4 71 2. .m m m

When the head is completely full, the formula reduces to the familiar:

S R= ⋅ ⋅ = ⋅ ⋅ ( ) =2 2 1 5 14 12 2 2π π . .m m

CylinderThe wetted surface of the cylindrical part of the vessel is found from:

S L R R HR

= ⋅ ⋅ ⋅ −

−2 1cos

(3)where all angles are measured in ra-dians.

For the cylindrical section of the vessel in Figure 1, the wetted surface area is:

S = ⋅ ⋅ ⋅ −

=

−2 6 1 5 1 5 11 5

22 1

1

2

m m m mm

m

. cos ..

.If the cylinder is completely filled

with liquid, H is equal to 2R, and Equation (3) reduces to the familiar:

S L R= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅=

2 2 6 1 5

56 6 2

π π m m

m

.

. ■Edited by Charles Butcher

References1. “C.R.C. Standard Mathematical Tables, 12th.

Edition,” p. 398, Chemical Rubber Publish-ing Co., Cleveland, Ohio, 1959.

2. Ibid., p. 401.

3. “Machinery’s Handbook,” 17th. Edition, p. 160, Industrial Press, New York, 1964.

ChemiCal engineering www.Che.Com DeCember 2007 57

AuthorRichard C. Doane is a senior process engineer with S&B Engineers and Constructors, Ltd. (7825 Park Place Bou-levard, Houston, TX 77087; Phone: 713-845-5338; Email: rcdoane@sbec.com). He has 35 years of experience in pro-cess engineering and plant operations. Doane holds B.S. and M.S. degrees in chemical engineering from Northeast-

ern 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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