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Thursday, April 9, 2009

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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).






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 :


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

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

Ref : "Accurate Wetted Areas for Partially Filled Vessels", by Richard C. Doane, "Chemical Engineering", December 2007


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    Labels: ,

    posted by Webworm, 3:11 PM

    8 Comments:

    Anonymous Anonymous said...

    how if, the vessel is vertical?

    August 24, 2009 at 11:16 PM  
    Anonymous Anonymous said...

    hey it really helped me. thanks a lot mr. ankur
    really appreciate it!

    August 26, 2009 at 12:31 PM  
    Anonymous Anonymous said...

    The equation for surface area of partially filled heads is not correct. I did not see the referenced article, so I don't know if the mistake started there.
    The error can easily be visualized - an incremental height at the bottom of the head will have less surface area than the same incremental height at the middle of the head, but the given equation indicates the opposite.

    August 27, 2009 at 9:00 AM  
    Blogger Webworm said...

    Please check the following post.

    Calculate Wetted Surface Area For VERTICAL Cylindrical vessel with Elliptical Head

    December 27, 2009 at 8:44 AM  
    Anonymous Anonymous said...

    Can this topic be presented for other types of vessel heads? Or can you provide a reference for the eccentricity factor for other types of vessel heads?

    March 10, 2010 at 11:22 AM  
    Anonymous Webworm said...

    Thanks for your inputs.

    Will consider to include calculation for other type of vessel head.

    Any pressure vessel design book will contain the eccentricity factor.

    Good luck.

    March 10, 2010 at 12:12 PM  
    Anonymous Anonymous said...

    The equation for the wetted surface area of a cylindrical vessel is not correct. This is obvious if H=2R is input and the result is 360*L*R. The correct result is obtained if the presented equation is divided by 180 and multiplied by PI.

    October 9, 2010 at 1:12 PM  
    Anonymous JoeWong said...

    Arc COS return as numeric instead of degree. Therefore, do not required to divide by 180 and multiply by PI.

    October 10, 2010 at 3:50 AM  

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