Fig. 7.10. Annular pressure drop for turbulent flow of 3 cp, 9.5 lb/gal mud *(pipe sizes). Courtesy Hughes Tool Company.

work, however, has shown these problems to be largely due to down hole pressure variations caused by the piston-cylinder action of the pipe and borehole.

Consider Figure 7.13, which shows the drill pipe being loweredln the mud-filled borehole. The velocity profile in the annulus is characterized by the counterflow pattern shown: some mud adjacent to the pipe will flow downward due to viscous drag, while the displaced volume will flow upward. The maximum pressure surge may be calculated by assuming that the drill pipe is sealed or plugged and that its entire volume (total cross-sectional area) is displaced. While this is a somewhat extreme assumption, it is comparable to running a back pressure valve (float) such as is common practice in drill stem testing or running casing. Let us, then, analyze this situation using the hydraulic principles previously presented.

The following assumptions will be made: (1) the volume of mud flowing upward = the total volume of pipe (calculated from outside diameter) being lowered.

(2) the mud's upward velocity = velocity calculated using the total annular area (no counterflow) + 1/2 the downward pipe velocity. This is the assumption made by Ormsby13 which was found to yield sufficiently accurate results. In equation form:

_ dop2Vp , Vp _ fl , d0p2 \ Va ~ dh2 - dop2 ^ 2 Vp\2 dh2 - dop2) where va = upward flow velocity, ft/sec vp = pipe velocity, ft/sec dop = pipe outside diameter, in. dh = hole diameter, in.

Knowing va, calculations of running pressure (or pulling suction) may be made in the same manner as any other mud flow problem. Example 7.6 illustrates this procedure.

Example 7.6

(a) Given the following data, calculate the actual pressure imposed on a formation at 10,000 ft. Assume pipe is closed at lower end.

350 400 450 500 550 «00 650

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