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Wetted Surface Area In Fire Psv Sizing


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#1 Propacket

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Posted 26 October 2009 - 12:39 AM

Hi,
I am sizing a PSV for fire case for which I need to calculate wetted surface area of vessel. I am confused what does it mean here? I have found equation for horizontal vessels but i can’t find any equation for vertical vessels. Can anybody help me? Thanks

#2 ankur2061

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Posted 26 October 2009 - 09:33 AM

Hi,
I am sizing a PSV for fire case for which I need to calculate wetted surface area of vessel. I am confused what does it mean here? I have found equation for horizontal vessels but i can't find any equation for vertical vessels. Can anybody help me? Thanks


P. Engr,

Here is the equation for a vertical vessel with 2:1 ellipsoidal head
At = Aev + Acv = (1.2*Pi*D2/4) + (Pi*D*H) Where

At = total wetted area of vertical vessel Aev = wetted area of bottom ellipsoidal head

Acv = wetted area of cylindrical portion

D = diameter of vessel

H = Depth (Height) of liquid in cylindrical portion Use consistent units. I am using this equation for our fire case relief valve sizing for vertical vessels.

This formula can be traced from: "Chapter 7, Page 452, Applied Process Design for Chemical & Petrochemical Plants" by E. Ludwig

Hope this helps.

Regards,
Ankur




#3 Propacket

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Posted 26 October 2009 - 10:46 PM

Ankur,
Thanks for taking interest. The equation you have provided seems very simple. Take a look to the equation I was using for horizontal vessels.
A=Awet, cyl=2LR x COS-1[(R-H)/R]
Awet, head= (2.178/2п) (2R) 2 COS-1(R-H/R)

Where,

Awet, cyl = wetted surface area of the cylindrical portion of vessel, m2
Awet, head = wetted surface area of one elliptical head, m2
L = tangent-to-tangent length of the cylindrical section of the vessel, m
R = vessel inside radius, m (also equal to D/2 where D is the vessel inside diameter)
H = maximum liquid depth, m

It’s a complex equation. Have you ever used this equation for horizontal vessels?

#4 ankur2061

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Posted 26 October 2009 - 11:15 PM

Ankur,
Thanks for taking interest. The equation you have provided seems very simple. Take a look to the equation I was using for horizontal vessels.
A=Awet, cyl=2LR x COS-1[(R-H)/R]
Awet, head= (2.178/2п) (2R) 2 COS-1(R-H/R)

Where,

Awet, cyl = wetted surface area of the cylindrical portion of vessel, m2
Awet, head = wetted surface area of one elliptical head, m2
L = tangent-to-tangent length of the cylindrical section of the vessel, m
R = vessel inside radius, m (also equal to D/2 where D is the vessel inside diameter)
H = maximum liquid depth, m

It's a complex equation. Have you ever used this equation for horizontal vessels?


Hi,

I thought your question was for vertical vessels. For horizontal vessels with 2:1 ellipsoidal post I have prepared an excel sheet which is posted at the following link:

http://webwormcpt.bl...e-area-for.html

This should solve your query.

Regards,
Ankur.

#5 Propacket

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Posted 26 October 2009 - 11:55 PM

Hi,

I thought your question was for vertical vessels. For horizontal vessels with 2:1 ellipsoidal post I have prepared an excel sheet which is posted at the following link:

http://webwormcpt.bl...e-area-for.html

This should solve your query.

Regards,
Ankur.
[/quote]
My question was definitely for vertical vessels but i was asking for simpler equation for horizontal vessels like your equation for vertical vessels.

#6 aaejaz

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Posted 16 November 2009 - 04:13 PM

Ankur,

I tried to find the reference you made in your response earlier to page 452 of the book by Ludwig. I was unable to find those equations or anything in reference to that topic in that book. I am also working on calculating surface area for hemispherical heads and spherical tanks. Do you or anyone else know if the following is correct:

FOR Vertical Vessel: A(wet,hemispherical head) = 0.5*PI*D^2

FOR Horizontal Vessel: A(wet,hemispherical head) = 0.5*PI*H*D

FOR Spherical Vessel: A(wet) = PI*D*H

Thanks!

Abby

#7 ankur2061

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Posted 18 November 2009 - 03:16 AM

Ankur,

I tried to find the reference you made in your response earlier to page 452 of the book by Ludwig. I was unable to find those equations or anything in reference to that topic in that book. I am also working on calculating surface area for hemispherical heads and spherical tanks. Do you or anyone else know if the following is correct:

FOR Vertical Vessel: A(wet,hemispherical head) = 0.5*PI*D^2

FOR Horizontal Vessel: A(wet,hemispherical head) = 0.5*PI*H*D

FOR Spherical Vessel: A(wet) = PI*D*H

Thanks!

Abby


Abby,

The wetted area & vessel section volume questions have been debated so many times on the forum that I now feel that some sort of attachment giving information on this matter has become an absolute necessity. The attachment is a relevant page from the "Pressure Vessel Design Manual" by D.R. Moss for volumes & surface areas of vessel sections.

Hopefully this should go a long way in resolving the frequent questions on volumes & surface areas.

Regards,
Ankur

Attached Files



#8 aaejaz

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Posted 18 November 2009 - 08:51 AM

Thanks Ankur,

That was definitely helpful although I am working on surface areas of partially filled vessels (wetted area). Your other postings for a horizontal vessel with 2:1 Ellipsoidal Head was a great help. I have been searching for days and this task has been more difficult than I anticipated. I found an equation from an article for calculating the surface area for a hemispherical head of a partially filled vessel (A=PI*R*H), but I have unable to verify that from any other source.

Thanks.

Abby

#9 chatpal_2005

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Posted 24 May 2012 - 11:37 PM

I am verfying a PSV design....In which vessel(T/T) height is 3.048 m and diameter of vessel id 1.118 m. By using the above eqns I am getting answer for wetted surface area for 2:1 ellipsoidal head 13.41 m2, but according to vendor spec its 15.71 m2. Couldnot find the answer why is it so?
Please help me up. thanks in advace.

Saad

#10 breizh

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Posted 25 May 2012 - 12:10 AM

A=2*t*D^2+pi*D*J ft2 t=0.931 2;1 ellipsoidal head J = 10 ft D 3.67 ft A 140.3755 ft2 or
my calculation : 13.04 m2

Breizh

Edited by breizh, 25 May 2012 - 12:19 AM.





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