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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.

Related Post

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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## 8 Comments:

how if, the vessel is vertical?

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

really appreciate it!

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.

Please check the following post.

Calculate Wetted Surface Area For VERTICAL Cylindrical vessel with Elliptical Head

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?

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.

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.

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

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