Why is the designated field listed as the Cotton Valley at 12500'?  Is that not Bossier depth?

Sabine County

Endeavor Natural Gas completed the John O. Hall Well No. 1 to 12,500 feet 14 miles north of Hemphill in the Times Hall
Field. On a 16/64-inch choke the well potentialed 1.038 million cubic
feet of gas. Production is in the Cotton Valley.

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Just looked at the formation records in the completion report:

http://webapps.rrc.state.tx.us/CMPL/ogmappcontents/cmplcontents/pro...

Pay zone is listed at 12,500'

James Lime is listed at 6,850'

Travis Peak is listed at 7,925

Cotton Valley formation top is listed at 10,300'.

No other formations are listed.



To compare, CHK's Hezekiah well has the following formation depths listed on it's report:

BO: 11,847
HA: 12,652
Smackover: 12,776'


I'm curious if they might actually be producing from the Cotton Valley Lime (CVL) / Smackover at this location as it would generally be situated deeper than HA or BO.
The field is so new that it was not on the proration schedule last month. What I wonder is whether they are including the shale intervals in this field? Hezekiah is nearly 5 miles northwest of Hall so the depth may change several feet in that distance and the HA could be even thinner. It will be more interesting to compare Hall to the Horton or Bengal wells when they come out.
Does anyone know what the Hezekiah came on line at? Any data available on what it might be doing now?
First thing - I guess I miised the Hezekiah - what did it pot. @-? The Hall well is probably producing from a lime (whatever you call it - "Haynesville" or "CV"). I can't make DJG's link work. Is there an easy way to get to this info? I go to the GIS map, I
"Identify Wells", then to "Operator/Wellbore/PDQ" and all I see is:
"COMPLETION INFORMATION
No completion data found."
Search for completions here:

http://webapps.rrc.state.tx.us/CMPL/publicSearchAction.do?formData....

Use API # (leave out the hyphen) and leave all else blank to search for a specific well.

Hall has two reports. Hezekiah = 0
Would a vertical decline at the same rate as the horizonals?
Ken

I think a vertical well decline is much slower.

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