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Fire relief estimation for two phase gas liquid in horizontal pressure vessel, wetted surface area is required to estimate total heat input into the horizontal pressure vessel. A horizontal pressure vessel with radius R and length L (tan-tan) with liquid height of H, the wetted surface area can be the total wetted surface area of cylindrical section and elliptical head (2 heads).
Fire relief estimation for two phase gas liquid in horizontal pressure vessel, wetted surface area is required to estimate total heat input into the horizontal pressure vessel. A horizontal pressure vessel with radius R and length L (tan-tan) with liquid height of H, the wetted surface area can be the total wetted surface area of cylindrical section and elliptical head (2 heads).
Wetted Surface Area (Cylindrical section)
Wetted Surface Area for Cylindrical section can be calculated with following equation :
Wetted Surface Area (Elliptical head)
Wetted Surface Area for Elliptical head (one head) can be calculated with following equation :
Example
L = 6m
R = D / 2 = 1.5m
F = H / 2R = 1 / (2 x 1.5) = 0.333
A = F - 0.5 = 0.333 - 0.5 = -0.167
B = SQRT[1+12A^2] = SQRT[1+12x(-0.167)^2] = 1.1547
Awet,Cyl = 2LRxAcos[(R-H)/R]
Awet,Cyl = 2x6x1.5xAcos[(1.5-1)/1.5]= 22.16 m2
Awet,Head = (PIxR^2/2)x[AxB+1+1/(4e)xLn[((4exA)+B)/(2-3^0.5)]=3.64 m2
Total wetted surface area, S = Scyl + 2 x Shd = 29.43 m2
Ref : "Accurate Wetted Areas for Partially Filled Vessels", by Richard C. Doane, "Chemical Engineering", December 2007
Above equations have been programmed by Ankur, a experience Chemical Engineer, share with readers of Chemical and Process Technology. You may download here.
Thanks to Ankur
Download
*If you have any useful program and would like to share within our community, please send to me.
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Wetted Surface Area for Cylindrical section can be calculated with following equation :
Wetted Surface Area (Elliptical head)
Wetted Surface Area for Elliptical head (one head) can be calculated with following equation :
where
R = Vessel inside radius (m)
H = Liquid height from bottom (m)
L = Vessel tangent-to-tangent length (m)
e (epsilon) = Eccentricity (0.866 for 2:1 Ellipsoidal head)
F = Fraction Liquid Level, = H / 2R
R = Vessel inside radius (m)
H = Liquid height from bottom (m)
L = Vessel tangent-to-tangent length (m)
e (epsilon) = Eccentricity (0.866 for 2:1 Ellipsoidal head)
F = Fraction Liquid Level, = H / 2R
Example
An ellipsoidal heads horizontal vessel with internal diameter (D) of 3m and tangent-tangent length is 6m. Determine wetted surface area when maximum liquid level is at 1m above vessel bottom.
L = 6m
R = D / 2 = 1.5m
F = H / 2R = 1 / (2 x 1.5) = 0.333
A = F - 0.5 = 0.333 - 0.5 = -0.167
B = SQRT[1+12A^2] = SQRT[1+12x(-0.167)^2] = 1.1547
Awet,Cyl = 2LRxAcos[(R-H)/R]
Awet,Cyl = 2x6x1.5xAcos[(1.5-1)/1.5]= 22.16 m2
Awet,Head = (PIxR^2/2)x[AxB+1+1/(4e)xLn[((4exA)+B)/(2-3^0.5)]=3.64 m2
Total wetted surface area, S = Scyl + 2 x Shd = 29.43 m2
**********************************
Above equations have been programmed by Ankur, a experience Chemical Engineer, share with readers of Chemical and Process Technology. You may download here.
Thanks to Ankur
Download
*If you have any useful program and would like to share within our community, please send to me.
Related Post
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