Wetted Surface Area of 2:1 SE Vertical Tanks

[TransferPhenomena]

Hi,

I've seen some previous threads on this topic, but I took D. Moss book and implemented equations for surface area 2:1 SE (Appendix E, Table E-1).

It appears that the surface area of the head in h=0 reduces to D^2*pi()/4, which is the cross section of the cylidnical part. At h=0 the surface area should be zero?

Have anyone implemented surface areas of vertical vessel?

Cheers,
Luke

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[mk3223]

For vertical vessel, just turn the vessel 90-deg of the horizontal vessel per App.E.
Also, may search on-line for the tech support, such as link

[link ]

http://webwormcpt.blogspot.com/2009/12/calculate-wetted-surface-area-for.html

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[TransferPhenomena]

Hi,

What I was looking for is a wetted area of partially filled tank head 2:1 elipsoidal, which in case of horizontal tanks is differerent than in case of vertical by obvious reasons.
I've found the articles from Kolev on horizontal ones, but couldnt find any on vertical.
Eventually I've evalulated elipsoid equation with 2:1 axis ratio and integrated symbolically.

For anyone interested for SE 2:1, the wetted area is:
4*pi()*y/6 *(3*y*sqrt(1 + (3*y^2)/b^2) + sqrt(3)*b*sinh^(-1)((sqrt(3)*y)/b))

where y- is a liquid height and b is a total height.
The height is measured from the cylindrical part- so it's top head. For bottom one needs replacing y for b-y.
The equation simplifies to 1.084D^2 when y=b.

Case closed.