Reservoir Engineering II, PDB(3023)
Ch.1: Unsteady State Flow
Solution of the Diffusivity Equation Dr. Mohammed Abdalla Ayoub U n iv ers iti Tek n ologi P E T RO N A S , Ba n d a r S e r i I s k anda r, 3 1 7 50 Tr o n oh , P e r a k, Ma la ys ia | Te l:+ 6 05 3 6 8 7 0 8 6 | F a x :+ 605 3 6 5 5 6 7 0 E - ma il : ab d alla.ayoub@utp. edu. my
Learning objectives Recap on the Diffusivity Equation To apply constant terminal pressure solution (CTP)
Contents Pressure distribution as a function of time Transient (unsteady-state) flow Derivation of the Diffusivity Equation Radial Flow of Slightly Compressible Fluids Solutions to the Diffusivity Equation
Solutions to the Diffusivity Equation The diffusivity equation: ---------eq. (18)
Where, Diffusivity constant
Solutions to the Diffusivity Equation The three most common flow conditions are: ◦ Steady state ◦ P=const. at all r and t. ∂P/ ∂t=0 ◦ This is the case where natural water influx or injection of some fluid
◦ Semi-steady state ◦ The effect of the outer boundary has been felt ◦ ∂P/ ∂r=0 at r=re ◦ ∂P/ ∂t ≈ const. for all r and t
◦ Transient – early time ◦ No boundary effect (infinite acting reservoir) ◦ P = f(r, t) and ∂P/ ∂r=q(r, t)
Radial Flow of Slightly Compressible Fluids cont’d The diffusivity equation as represented by Equation 18 is essentially designed to determine the pressure as a function of time t and position r. ◦ P=f(r, t) Assumptions and limitations used in developing Equation 18: ◦ ◦ ◦ ◦ ◦
Homogeneous and isotropic porous medium Uniform thickness Single phase flow Laminar flow Rock and fluid properties independent of pressure
Radial Flow of Slightly Compressible Fluids cont’d For steady state flow: ◦ ∂P/∂t=0
Hence, eq. (18) reduces to: ---------eq. (19)
Equation 19 is Laplace’s equation for steady-state flow.
Solutions to the Diffusivity Equation To obtain a solution to the diffusivity equation (Equation 18), it is necessary to specify: ◦ An initial condition ◦ Uniform pressure Pi when production begins.
◦ Impose two boundary conditions. (outer and inner boundary) ◦ The well is producing at a constant production rate. ◦ The reservoir behaves as if it were infinite in size, i.e., re = ∞.
Solutions to the Diffusivity Equation cont’d Based on the boundary conditions imposed on, there are two generalized solutions to the
diffusivity equation: ◦ Constant-terminal-pressure solution
◦ Constant-terminal-rate solution
Constant-terminal-pressure solution The pressure is known to be constant at some particular radius and the solution is designed to provide the cumulative fluid movement across the specified radius (boundary). It is widely used in water influx calculations.
A detailed description of the solution and its practical reservoir engineering applications will be discussed in the water influx chapter.
Constant-Terminal-Rate Solution
Solution to Diffusivity Equation Constant-terminal-pressure solution is designed to provide the cumulative flow at any particular time for a reservoir in which the pressure at one boundary of the reservoir is held constant. This technique is frequently used in water influx calculations in gas and oil reservoirs. Constant-terminal-rate solution of the radial diffusivity equation solves for the pressure change throughout the radial system providing that the flow rate is held constant at one terminal end of the radial system, i.e., at the producing well. These are two commonly used forms of the constant-terminal-rate solution: The EI-function solution The dimensionless pressure pD solution 12
The Ei-function solution The Exponential integral (Ei) can be used to estimate the pressure change throughout the radial system of the reservoir. However, the following assumptions must be considered:
Infinite acting reservoir
well is producing at a constant flow rate.
uniform reservoir pressure, Pi, when production begins.
wellbore radius of rw, is centered in a cylindrical reservoir of radius re.
No flow across the outer boundary.
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The Ei-function solution 2 70.6Qo o Bo 948o ct r p (r , t ) pi Ei kh kt
This equation can be used to calculate pressures at any distance r, from the wellbore when the well is opened for production of:
-------- (20) where p (r,t) = pressure at radius r from the well after t hours t = time, hrs k = permeability, md. Qo = flow rate, STB/day
The reservoir is assumed to be infinite-acting if:
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The Ei-function solution, cont… 2 70.6Qo o Bo 948o ct r p (r , t ) pi Ei (x)kt kh
-------- (20)
When X is less than 0.02, the exponential integral Ei can be calculated using the following Equation:
when x is in the range of 0.01 < x < 3.0
• when x > 10.0, the Ei (−x) can be considered zero 15
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Example A single well is producing oil at a constant rate of 5500 STB/D. Calculate the pressure at a location 500 feet away from the well after 5 days, and 10 days of production. The drainage boundary of the
well is located 1000 feet away. Other rock and fluid properties of the reservoir are as follows:
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Solution Step 1. use the following equation to check that the well has been on production for sufficient length of time so that the accuracy is not affected. The length of time the well is on production must be greater than the time required.
Next, it is important to check that the reservoir is infinite-acting after 10 days of production. This time of production must be less than the time calculated from the following equation:
The reservoir will cease to be infinite-acting after about 19.3 days of production. Hence, after 10 days of production, the reservoir is still infinite-acting. 18
Step 2: Calculate reservoir pressure after 5 days, and 10 days of production at 500 feet from well. Substituting Eq. (20) with the data provided gives:
948o ct r 2 70.6Qo o Bo p ( r , t ) pi * Ei kh kt
948 * 0.19 * 4.5 * 4 *10 6 * (500) 2 70.6 * 5500 *1.1* 4.5 p (r , t ) 19250 * Ei 7 * 345 7 * 24 * t 4.8246 p (r , t ) 19250 795.89 * Ei t At 5 days of production, the Ei (-x) is:
4.8246 4.8246 Ei E Ei (0.96) i t 5
Because x is higher than 0.1, so the value of Ei(-x) can be obtained from the table which is equal to 0.24.
p (r , t ) 19250 795.89 * 0.24 p (r , t ) 19,059 psia
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At 10 days of production, the Ei (-x) is:
4.8246 4.8246 Ei E Ei (0.48) i t 10 Because x is higher than 0.1, so the value of Ei(-x) can be obtained from the table which is equal to 0.585.
p (r , t ) 19250 795.89 * 0.585
p (r , t ) 18,784 psia
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Example An oil well is producing at a constant flow rate of 300 STB/day under unsteadystate flow conditions. The reservoir has the following rock and fluid properties:
1. Calculate pressure at radii of 0.25, 5, 10, 50, 100, 500, 1000, 1500, 2000, and 2500 feet, for 1 hour. Plot the results as: A. Pressure versus logarithm of radius B. Pressure versus radius 2 70.6Qo o Bo 948o ct r p (r , t ) pi Ei kh kt
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Solution • Step 1. From Equation (20) :
• Step 2. Perform the required calculations after one hour in the following tabulated form:
x < 0.02
0.01 < x < 3.0, use Ei chart
x >10.9, Ei = zero
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Solution, cont… • Step 3. Show results of the calculation graphically as illustrated in Figures below:
Pressure profiles as a function of time
Pressure profiles as a function of time on a semi-log scale 23
Solution, cont… • When the parameter x in the Ei-function is less than 0.01, Eq. (21) in the form of the following equation can also be used to calculate the Pwf:
-------- (21)
• For most of the transient flow calculations, engineers are primarily concerned with the behavior of the bottom-hole flowing pressure at the wellbore, i.e., r = rw
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Solution, cont… pw f
162.6Qo o Bo pi kh
kt log 2 c r o t w
3.23
-------- (22)
where k = permeability, md t = time, hr ct = total compressibility, psi−1
• Equations (21) and (22) cannot be used until the flow time t exceeds the limit imposed by: -------- (23) where t = time, hr k = permeability, md 25
Example 3 An oil well is producing at a constant flow rate of 300 STB/day under unsteadystate flow conditions. Estimate the bottom-hole flowing pressure after 10 hours
of production.The reservoir has the following rock and fluid properties:
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Solution Equation 22 can be used to calculate pwf only if the time exceeds the time limit
Since the specified time of 10 hr is greater than 0.000267 hrs, the pwf can then be estimated by Equation 22.
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The Dimensionless Pressure Drop (pD) Solution The solution describes the pressure drop as a function of time and radius for fixed values of external radius re and well bore radius rw, rock and fluid properties. It is expressed in term of dimensionless variables as:
PD = f(tD, rD, reD)
tD = dimensionless time rD = dimensionless radius reD = dimensionless external radius
This solution to the diffusivity equation at constant terminal rate under the stated initial and boundary conditions was reported by van Everdingen and Hurst. The solution is presented in terms of dimensionless variables based on Darcy’s equation in a radial form. 28
The Dimensionless Pressure Drop (pD) Solution The PD is presented in terms of dimensionless variables based on Darcy’s equation for a radial flow system.
Rearrange Dimensionless pressure PD
Dimensionless external radius reD
In Dimensionless format the previous equation can be written as:
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The Dimensionless Pressure Drop (pD) Solution In transient flow analysis, the dimensionless pressure PD =
f(tD, rD, reD)
If the reservoir is fixed in size. i.e. reD is a particular value, then the dimensionless pressure drop PD at the wellbore is only a function of dimensionless time, PD = f(tD) The dimensionless pressure variable PD is defined as:
pe pwf PD Qo Bo o 0.00708kh
-------- (25)
-------- (24)
The dimensionless time variable
Based on total drainage area, tDA
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The Dimensionless Pressure Drop (pD) Solution ----- (25a)
----- (25b)
where A = total drainage area = π re2 re = drainage radius, ft rw = wellbore radius, ft
reD = dimensionless external radius reD = dimensionless external radius
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The Dimensionless Pressure Drop (pD) Solution The dimensions units can be obtained by substituting the appropriate dimensions such as M for mass, L for length and T for the time for each of the quantities in the original Darcy low: pe pwf PD Qo Bo o 0.00708kh
force ML M p 2 2 2 area T L T L PD
L3 Q T
M / T L L / T L / L M / LT 2
3
3
3
0.00708 * L2 * L
After cancelation, the PD is dimensionless
L3 Bo 3 L
M LT
A L2 hL K L2 32
Dimensionless Diffusivity Equation 0.000264kt tD ct rw2
0.000264 * L2 * T tD M / LT LT 2 / M L2
After cancelation, the tD is dimensionless Introduce pD, tD, and rD into the diffusivity equation:
-------- (26)
Dimensionless Diffusivity Equation
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Assumptions: Perfectly radial reservoir system The producing well is in the center and producing at a constant production rate of Q Uniform pressure pi throughout the reservoir before production No flow across the external radius re
Van Everdingen and Hurst (1949) proposed an analytical solution to the above equation For an Infinite-acting reservoir as:
-------- (27)
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Infinite-Acting Reservoir For an infinite-acting reservoir, i.e., reD = ∞, the dimensionless pressure drop function pD is strictly a function of the dimensionless time tD, or: Chatas and Lee tabulated the pD values for the infinite-acting reservoir as shown here. For tD < 0.01
-------- (28)
For 100 < tD < 0.25 r2eD -------- (29)
For 0.02 < tD < 1000
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Computational procedure The computational procedure of using the pD-function in determining the bottom-hole flowing pressure: 1
Calculate the dimensionless time tD
2
Calculate the dimensionless radius reD
3
Using the calculated values of tD and reD, determine the corresponding pressure function pD from the appropriate table or equation
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Solve for the pressure at the desired radius, i.e., rw
It should be pointed out that, for an infinite acting reservoir with tD > 100, the pD-function is related to the Ei-function by the following relation:
-------- (27) -------- (27a) 36
Example 4 A well is producing at a constant flow rate of 300 STB/day under unsteady-state flow condition. Assuming an infinite acting reservoir, calculate the bottom-hole flowing pressure after one hour of production by using the dimensionless pressure approach. Assume the reservoir boundary is 1000 ft and the reservoir has the following rock and fluid properties
Calculate the dimensionless time tD
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Solution, cont… Since 100 < tD < 0.25 r2eD, use Equation 29 to calculate the dimensionless pressure drop function
Calculate the bottom-hole pressure after 1 hour by applying Equation 27:
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Example 5 An oil well located at X field is supposed to produce at a constant production rate from a drainage
boundary of 304.8 meters away from the wellbore. After 10 days of production, the bottom hole flowing pressure was found to be 8134 psia at a wellbore radius of 0.5 ft. Estimate the well production rate in bbl/day. other rock and fluid properties of the reservoir are as follows:
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Solution Step 1: Calculate the dimensionless time tD using Eq. 25.
tD
0.000263kt 0.000263 * 7 * 24 *10 518,147 2 6 2 c t rw 0.19 * 4.5 * 4 *10 * (0.5)
Step 2: Calculate the dimensionless radius reD .
reD
re 304.8 * 3.2808 2000 rw 0 .5
Step 3: Calculate the dimensionless pressure PD using Eq. 29.
PD 0.5ln t D 0.80907 6.9835 40
Solution, cont… Step 4: Calculate the oil production rate using Eq. 27.
Pwf
Qo Bo o Pi PD 0.00708kh
Q 5500 STB / day
Step 5: convert the oil production rate from STB/day to bbl/day
Q 5500 STB / day *1.1bbl / STB 6050bbl / day
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Finite-Radial Reservoir The arrival of the pressure disturbance at the well drainage boundary marks the end of the transient flow period and the beginning of the semi (pseudo)steady state. During this flow state, the reservoir boundaries and the shape of the drainage area influence the wellbore pressure response as well as the behavior of the pressure distribution throughout the reservoir.
There is a short period of time that separates the transient state from the semisteady state that is called late-transient state. Due to its complexity and short duration, the late transient flow is not used in practical well test analysis.
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Finite-Radial Reservoir
where
-------- (30) For 25 < tD and 0.25 r2eD < tD
-------- (31)
when r2eD >> 1
-------- (32)
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Example 6 A well is producing at a constant flow rate of 300 STB/day under unsteady-state flow condition. The reservoir has the following rock and fluid properties
Assuming an infinite acting reservoir, i.e., reD = ∞, calculate the bottom-hole flowing pressure after one hour of production by using the dimensionless pressure approach. Solution Step 1. Calculate the dimensionless time tD from Equation:
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Solution, cont… Step 2. Since tD > 100, use Equation:
Step 3. Calculate the bottom-hole pressure after 1 hour by applying
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Important Note! The main difference between the two formulations is that the pDfunction can be only used to calculate the pressure at radius r when the flow rate Q is constant and known. In that case, the pD-function application is essentially restricted to the wellbore
radius because the rate is usually known. On the other hand, the Ei-function approach can be used to calculate the pressure at any radius in the reservoir by using the well flow rate Q. 46
Questions
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