Feature Report Engineeering Practice
Accurate Wetted Areas For Partially Filled Vessels
Accurately determine wetted surface areas needed for fire relief applications
Richard C. Doane
NOMENCLATURE
S&B Engineers and Constructors, Ltd.
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 cylindrical section of the vessel, m S
T
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
his article introduces a simple yet accurate way to calculate the wetted surface area of a partially filled horizontal vessel with semi-elliptical heads. This information is often needed for the sizing of relief devices. API Recomme Recommended nded Practice 521 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 g rade level. le vel. 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 el A 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 25below. 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 = = be determined using mensuration 1), Equation 1 reduces to the formula 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 elsurface area of half an oblate sphecal Tables” [ 1]. The wetted surface area for a partially Equation 1 filled hemispherical head is 2 2 ε ⋅ ( F − 0.5) + 1 + 12 ⋅ ( F − 0.5) 4 presented in “Machinery’s 2 π ⋅ R 1 = − ⋅ + ⋅ − + + ⋅ S F 0 . 5 1 1 2 F 0 . 5 1 ln l n ( ) ( ) Handbook” [ 3]. 2 4ε 2− 3 Finding the wetted area of a partially filled ellipti56
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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 m 2.
Hemispherical head For comparison, the wetted surface area of a hemispherical head is given by the concise relationship: S
= π ⋅ R ⋅ H
(2)
For the vessel shown in Figure 1, the wetted area with hemispherical heads is: S
= π ⋅ 1.5m ⋅ 1m = 4.71m2
When the head is completely full, the formula reduces to the familiar: S
2
= 2 ⋅ π ⋅ R2 = 2 ⋅ π ⋅ (1.5m) = 14.1m2
Cylinder The wetted surface of the cylindrical part of the vessel is found from: S
= 2 ⋅ L ⋅ R ⋅ cos−1 R − H R
(3) where all angles are measured in radians. For the cylindrical section of the vessel in Figure 1, the wetted surface area is: S
If the cylinder is completely filled with liquid, H is equal to 2 R, and Equation (3) reduces to the familiar:
= 2 ⋅ π ⋅ L ⋅ R = 2 ⋅ π ⋅ 6m ⋅ 1.5m
= 56.6m 2
■
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.
Circle 26 on p. 74 or go to adlinks.che.com/6902-26
= 2 ⋅ 6m ⋅ 1.5m ⋅ cos−1 1.5m − 1m 1.5m
= 22.1m 2
S
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“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:
[email protected]). 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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