Chapter 2 Examples & Solutions: Example 2-1. Steady-State Production Rate Calculation and Rate Improvement (Stimulation) [PDF]

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Chapter 2 Examples & Solutions 2.1 Introduction NGEN 4396 2.2 Steady-State Well Performance Natural Gas Production 2.3 Transient Flow of Undersaturated Oil Prepared by Dr. Zhaoqi Fan 2.4 Pseudosteady-State Flow 2.5 Wells Draining Irregular Patterns 2.6 Inflow Performance Relationship 2.7 Effects of Water Production, Relative Permeability 2.8 Summary of Single-Phase Oil Inflow erformance Relationships Example 2-1. Steady-State Production Rate Calculation and Rate Improvement (Stimulation) Assume that a well in the reservoir described in Appendix A has a drainage area equal to 640 acres (re = 2980 ft) and is producing at steady state with an outer boundary (constant) pressure equal to 5651 psi. Calculate the steady-state production rate if the flowing bottomhole pressure is equal to 4500 psi. Use a skin effect equal to +10. Describe two mechanisms to increase the flow rate by 50%. Show calculations. Solution: From Equation (2-9),

To increase the production rate by 50%, one possibility is to increase the drawdown, pe – pwf, by 50%. Therefore, leading to pwf = 3925 psi. A second possibility is to increase JD by reducing the skin effect. In this case,

leading to s2 = 3.6. Example 2-2. Effect of Drainage Area on Well Performance Demonstrate the effect of drainage area on oilwell production rate by calculating the ratios of production rates from 80-, 160-, and 640-acre drainage areas to that obtained from a 40-acre drainage area. The well radius is 0.328 ft. Solution: Assuming that the skin effect is zero (this would result in the most pronounced difference in the production rate), the ratios of the production rates (or productivity indices) can be given by an New Section 1 Page 1

expression of the form

The drainage radius for a given drainage area is calculated by assuming that the well is in the center of a circular drainage area. Thus,

The results are shown in Table 2-1. These ratios indicate that the drainage area assigned to a well has a small impact on the production rate. For tight reservoirs, cumulative production differences are particularly immune to the drainage area because transient behavior is evident for much of the time. Table 2-1. Production Rate Increases (over a 40-acre spacing)

Example 2-3. Prediction of Production Rate in an Infinite-Acting Oil Well Using the well and reservoir variables in Appendix A, develop a production rate profile for 1 year assuming that no boundary effects emerge. Do this in increments of 2 months and use a flowing bottomhole pressure equal to 3500 psi. Solution From Equation (2-25) and substitution of the appropriate variables in Appendix A, the well production rate is given by

For t = 2 months, for Equation (2-27) the production rate q = 627 STB/d.

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Figure 2-2. Rate decline for an infinite-acting oil reservoir (Example 2-3). Example 2-4. Production from a No-Flow Boundary Reservoir What would be the average reservoir pressure if the outer boundary pressure is 6000 psi, the flowing bottomhole pressure is 3000 psi, the drainage area is 640 acres, and the well radius is 0.328 ft? What would be the ratio of the flow rates before (q1) and after (q2) the average reservoir pressure drops by 1000 psi? Assume that s = 0. Solution A ratio of Equations (2-29) and (2-33) results in

The drainage area A = 640 and therefore re = 2980 ft. Substituting the given variables in Equation (2-38) results in

The flow-rate ratio after the 1000 psi average pressure decline would be

Example 2-5. Impact of Irregular Well Positioning on Production Rate Assume that two wells in the reservoir described in Appendix A each drain 640 acres. Furthermore, assume that 𝑝̅ = 5651 psi (same as pi) and that s = 0. The flowing bottomhole pressure in both is 3500 psi. However, well A is placed at the center of a square, whereas well B is at the center of the upper right quadrant of a square drainage shape. Calculate the production rates from the two wells at the onset of pseudosteady state. (This calculation is valid only at very early time. At late time, either drainage shapes will change, if they are artificially New Section 1 Page 3

induced, or the average reservoir pressure will not decline uniformly within the drainage areas because of different production rates and resulting different rates of depletion.) Solution Well A The shape factor, CA, from Figure 2-3 is equal to 30.9. Therefore, from Equation (2-44),

Well B Since it is located at the center of the upper right quadrant, its shape factor (from Figure 2-3) is equal to 4.5. All other variables in Equation (2-44) remain the same. The flow rate calculated is then equal to 574 STB/d, representing a 10% reduction. Example 2-6. Determining Average Reservoir Pressure within Adjoining Drainage Areas The following data were obtained on a three-well fault block. A map with well locations is shown in Figure 2-4. (Use properties as for the reservoir described in Appendix A.)

Figure 2-4. Three-well fault block for Example 2-6. (a problem by H. Dykstra, class notes, 1976) Each well produced for 200 days since the previous shut-in. At the end of the 200 days, the following rates and skin effects were obtained from each well:

If the bottomhole pressure is 2000 psi for each well, calculate the average reservoir pressure within each drainage area. Solution

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The drainage volumes formed by the production rate of each well are related by

Since h varies, the volumes can be replaced by the product hiAi, where i refers to each well. Finally, a third equation is needed:

Equations (2-46), (2-47), and (2-48) result in AA = 129 acres (5.6 × 106 ft2) AB = 243 acres (1.06 × 107 ft2) AC = 108 acres (4.7 × 106 ft2) The next step is to sketch these areas on the fault block map. From Figure 2-3, well A is shape no. 12 (CA = 10.8), well B is shape no. 7 (CA = 30.9), and well C is shape no. 10 (CA = 3.3). From Equation (2-44) and for well A,

The average pressures in the drainage areas for wells B and C are calculated to be 2838 and 2643 psi, respectively. Example 2-7. Transient IPR Using the well and reservoir data in Appendix A, construct transient IPR curves for 1, 6, and 24 months. Assume zero skin. Solution Equation (2-24) with substituted variables takes the form

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Figure 2-5. Transient IPR curves for Example 2-7. Example 2-8. Steady-State IPR: Influence of the Skin Effect Assume that the initial reservoir pressure of the well described in Appendix A is also the constant pressure of the outer boundary, pe (steady state). Draw IPR curves for skin effects equal to 0, 5, 10, and 50, respectively. Use a drainage radius of 2980 ft (A = 640 acres). Solution Equation (2-9) describes a straight-line relationship between q and pwf for any skin effect. For example, after substitution of variables for a skin equal to 5, Equation (2-9) becomes Similarly, the multipliers of the rate for the 0, 10, and 50 skin effects are 3.66, 7.67, and 23.7, respectively.

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Figure 2-6. Steady-state and impact of skin effect for Example 2-8. Example 2-9. Pseudosteady-State IPR: Influence of Average Reservoir Pressure This calculation is the most useful and the one most commonly done for the forecast of well performance. Each IPR curve reflects a “snapshot” of well performance at a given reservoir pressure. This is a time-dependent calculation, done in discrete intervals. Combination with volumetric material balances (treated in Chapter 10) will allow the forecast of rate and cumulative production versus time. For this exercise, calculate the IPR curves for zero skin effect but for average reservoir pressures in increments of 500 psi from the “initial” 5651 to 3500 psi. Use all other variables from Appendix A. Drainage radius is 2980 ft. Solution Equation (2-44) is the generalized pseudosteady-state equation for any drainage shape and well position. For a circular drainage shape, Equation (2-34) is sufficient. Substituting the variables from Appendix A into Equation (2-34) results (for 𝑝̅

5651 𝑝𝑠𝑖) in

For all average reservoir pressures, the slope in Equation (2-51) will remain the same. The intercept will simply be the average pressure. Therefore, as shown in Figure 2-7, the pseudosteady-state IPR curves are depicted as parallel straight lines, each reflecting an average reservoir pressure.

the slope of the IPR is the reciprocal of the productivity index, J.

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Figure 2-7. Pseudosteady-state IPR curves for a range of average reservoir pressures.

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