Ram preventers

Dear Colleagues,

I would like to gain a better understanding of the meaning and implications of the closing ratio of ram preventers.  My understanding is lacking and I am supposed to be teaching this.

IWCF questions used to ask a candidate to calculate a new minimum bottle pressure required to close a ram on RWP.  For example, a 15,000 Ram with a closing ratio of 7:1 would require a minimum hydraulic pressure of 2,143 psi (15,000 / 7).  The question would then go on to ask something like how many gallons of useable fluid is now available from a bottle.  The simplified Gas Law would be used to calculate V2 Nitrogen, and oil volume at minimum pressure would be inferred from the standard bottle volume of 10 gallons. 

API Standard 53 defines closing ratio as that between the surface area of the closing piston and cross-sectional area of the ram shaft exposed to well-bore pressure.  It makes sense until you consider friction is not mentioned.  To my mind, if you are closing a ram on RWP (highly unlikely, but possible) you might get the piston moving at a lower hydraulic pressure but as the ram block approached the pipe and the flow area past the block decreased, the friction needing to be overcome would rise because the top seal is increasingly being forced onto the seal face.  Assumes well not already closed in above.

My questions:

• Is Standard 53 correct?
• Is friction accounted for by the manufacturer in the closing ratio?
• Why would any surface ram, apart from shears, need more than standard manifold pressure to be closed successfully?
• Is closing ratio only relevant in the extreme case of closing on RWP concurrent with loss of primary and secondary charging pumps?

Thanks in advance,
Chris

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