Static Head

[Mash18]

Hi All, im currently Studying for API 510 & trying to calculate metal loss as per ASME VIII Div.1 UG-27(c)(1), However I am struggling to understand the Pressure conversion part. (refer to attachment)

Does this mean that Static Head = Given Value x ID + 2 X Corrosion Allowance??
If as stated, the ID is relaative to static head value, why was 0.433 used instead of the ID? Or is that just part of the example

 

[SnTMan]

Mash18, Density of water = 62.4 lb / ft^3. Per cubic inch = 62.4 / 12^3 in^3 / ft^3 = 0.036 lb / in^3

Pressure of column 1 ft high = 0.036 lb / in^3 * 12 in / ft = 0.433 lb / in^2 / ft

Diameter of vessel, ft = (48 + 2 * 0.125) / 12 = 4.0208 ft.

You can take it from there.

Regards,

Mike

The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[JStephen]

Note that the 2*CA should have been divided by 12 also, as SnTMan has above.
The 0.433 is for water at "normal" conditions, and may vary depending on what is in the vessel.