What is the overburden gradient at 3000 meters (9842 feet TVD-kb) if water depth is 500 meters (1640 feet) and height of the derrick floor (i.e., airgap) is 30 meters (98 feet). Assume no uplift (i.e, no erosion or tectonics) and assume a clastic section with no thick carbonates. Use the simple calculator below OR the equations presented in class.

your Answer:

Understand that OBG is the average bulk density of the section from 3000 meters to derrick floor and includes the water and airgap, and is best derived using measured formation density data for the sediment interval. The calculator below assumes a weight of 8.7 lbs/gal for sea water.

What is predicted equivalent mud density at 3000 meters at the above location if seismic-derived travel time (ITT) is 125 microseconds per foot for a thick shale at 3000 meters. Remember that slowness (interval travel time) is equal to 10^6 divided by velocity. Cable length (spread) is 8000 meters. Remember from the Eaton relationship EMW=OBG-(OBG-8.9)(ITTn/ITT)^3 where ITTn is the travel time for normal compaction and is approximately equal to 55+(140)(exp(-DBSF/6247) in EN units, if NO uplift.

your Answer:

Use Presgraf or the calculator below. Overburden is determined as noted above. Because we are evaluating an interval at a depth less than cable length, it is reasonable to make a seismic-derived pressure prediction.

What is the fracture gradient at 3000 meters. Remember that FPG is equal to k times (OBG-PPG)+PPG. Use a value of 0.33 for k for hard rocks, like low porosity carbonates and a value of 0.5 to 1 for soft shales (A good average value is 0.64.) Assume that this example is shale.

your Answer:

Fracture gradient, by definition, is the static mud density (i.e, mud pumps off) that it takes to fracture the formation and to cause loss returns, into the induced fracture. A value for FPG is measured directly with a leak-off test (LOT). For this case use Presgraf or the calculator below. If a LOT is available, calculate k using the above equation.

For the same well as above, calculate EMW using a simple 1d basin model. We know from seismic data that the top of overpressures is at 950 meters,i.e, EMW is 8.9 lbs/gallon at 950 meters. Assume that the seismic-derived average bulk density is 2.17 g/cc between 950 and 3000 meters. Remember that pore pressure (Pp) is equal to EMW times depth. And, remember that the incremental increase in pore pressure from depth (D1) to depth (D2) is equal to interval gradient (Gint) times (D2-D1), where all depths are TVD-kb. That is, EMW@D2 = (EMW@D1+(D2-D1)(Gint))/D2. Notice that if EMW and Gint are expressed in the same units, e.g. lbs/gal, the equation is dimensionless (Use a pocket calculator OR the 1-model calculator below to solve the equation.)

your Answer:

The 1d model assumes that pore pressure increases with depth at a lithostatic rate below the top of overpressures, i.e, Gint below the top of overpressures is equal to the average bulk density. Use measure density data, instead of the textbook value of 1 psi per foot (2.31 g/cc, 19.25 lbs/gal), when available. Understand that the 1d prediction is a high side case. Centroid effects or other effects can cause pore pressure in sandstones to be less.

What is the actual equivalent mud weight, if (after the well is drilled) the measured pore pressure is 7681 psia at a depth 3000 meters (9842 feet) TVD-kb. Remember that EMW equals P/D times a conversion factor where P is gauge pressure and D is explicitly defined as TVD-kb. The conversion factor is 5.867 for psig per meter and 19.25 for psig per foot.

your Answer:

This is the mud density that it would take to exactly balance formation pressure at a depth of 1500 meters. Understand that actual mud weight (MW) will be different and, in general will be a bit higher (0.5 lb/gal?) than the maximum EMW anyplace in the exposed open-hole section.

Given an onshore well that drilled a thick sand-shale section and set casing at a depth of 3032 meters. And drilled another 10 meters into the top of the gas reservoir and ran a wireline RFT tool that measured a pressure of 8000 psi at 3040 (9973 feet). Given this, What is the EMW at 3040 meters.

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What is the overburden gradient at 3040 meters (9973 feet). The derrick floor elevation is 9 meters.

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Now consider that the operator wants to drill ahead to the gas water contact, estimated from a flat spot event on seismic, at a depth of 3640 meters (11942 feet). Seismic and local geology suggests it will be a single carbonate unit with one single gas/water contact, i.e. one hydrostatic system What do you predict for a EMW at 3640 meters?. Use Presgraf or the 1d Calculator below.

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What do you predict for actual mud weight (MW) at total 3640 meters. Understand that there will not be another casing string before reaching TD.

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What do you predict for the stress ratio (k) at 3640 meters (11942 feet). The geology of the area suggests the carbonates have a porosty less than 15 percent

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What do you predict for fracture gradient at 3640 meters (11942 feet).

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Does the well make it to total depth without a major well control problem.? Yes or No or (?)

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(After Answering the Questions)