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Consider a well-reservoir system (Fig. 1) consisting of:
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- producing well W1 draining the reservoir volume with total sandface flowrate water injecting well W2 supporting pressure in reservoir volume and BHP which includes the drainage volume , draining the reservoir volume of producer W1 and potentially other producers.
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water injecting well W0 with total sandface flowrate
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- , supporting pressure in reservoir volume
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The injection drainage volume
LaTeX Math Inline |
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body | 2} - V_{\phi, 10} >0 |
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may be related to the fact that water injection W2 is shared between includes the drainage volume and another reservoir or with another producer. ...
of producer W1 and may be equal to it
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V_{\phi, 0} = V_{\phi, 1} |
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or may be bigger The bottom-hole pressure response in producer W1 to the flowrate variation V_{\phi, 0} > V_{\phi, 1} |
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in case injector W0 supports other producers {W1 .. WN}: \delta q_2 | in injector W2: LaTeX Math Block |
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\delta p_1 = - p_{u,\rm 21}(t) \cdot \delta q_2 |
where
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V_{\phi, 0} = \sum_{k=1}^N V_{\phi, k} |
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Image Added |
Fig. 1. Location map of injector-producer pairing with 4 producers {W1, W2, W3, W4} and one injector W0. |
Case #1 – Constant flowrate production:
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body | q^{\uparrow}_1 = \rm const >0 |
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The bottom-hole pressure response
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in producer W1 to the flowrate variation LaTeX Math Inline |
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body | \delta q^{\downarrow}_0 |
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in injector W0: LaTeX Math Block |
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\delta p_1 = - p_{u,\rm 21}(t) \cdot \delta q^{\downarrow}_0 |
where
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| Consider a pressure convolution equation for the BHP in producer W1 with constant flowrate production at producer W1 LaTeX Math Inline |
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body | qq^{\uparrow}_1 = \rm const |
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| and varying injection rate at injector W20 LaTeX Math Inline |
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body | q_2q^{\downarrow}_0(t) |
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| : LaTeX Math Block |
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| p_1(t) = p_i - \int_0^t p_{u,\rm 11}(t-\tau) dqdq^{\uparrow}_1(\tau) - \int_0^t p_{u,\rm 2101}(t-\tau) dq_2dq^{\downarrow}_0(\tau) = p_i - \int_0^t p_{u,\rm 2101}(t-\tau) dq_2dq^{\downarrow}_0(\tau) |
Consider a step-change in injector's W20 flowrate LaTeX Math Inline |
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body | \delta q_2q^{\downarrow}_0 |
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| at zero time , which can be written as: LaTeX Math Inline |
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| body | dq_2writen as LaTeX Math Block |
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| q_0(\tau) = \delta | q_2 \cdot \delta q^{\downarrow}_0 \cdot H(\tau) | \, d\tauwhere . The responding pressure variation in producer W1 will be is Heaviside step function: LaTeX Math Block |
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| \delta p_1(tH(\tau) = p_1(t)-p_i = - \int_0^t p_{u,\rm 21}(t-\tau) \delta q_2 \cdot \delta(\tau) \, d\tau = - p_{u,\rm 21}(t) \cdot \delta q_2 |
which leads to LaTeX Math Block Reference |
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| .
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Case #2 – Constant BHP:
Assume that the flowrate
in producer W1 is being adjusted to compensate the bottom-hole pressure variation in response to the flowrate variation in injector W2 so that bottom-hole pressure in producer W1 stays constant at all times LaTeX Math Inline |
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body | \delta p_1(t) = \delta p_1 = \rm const |
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....
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\begin{cases} 0, & \tau <0 \\ 1, &\tau \geq 0\end{cases} |
The differential then can be written as: LaTeX Math Block |
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| d q^{\downarrow}_0(\tau) = q_0'(\tau) d\tau = \delta q^{\downarrow}_0 \cdot H'(\tau) \, d\tau = \delta q^{\downarrow}_0 \cdot \delta(\tau) \, d\tau |
The responding pressure variation |
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...
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| Case2q - \frac{\dot p_1(t)-p_i = - \int_0^t p_{u,\rm 21}(t-\tau) |
| }{\dot \delta q^{\downarrow}_0 \cdot \delta(\tau) \, d\tau = - p_{u,\rm |
| 11}q_2 |
where
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which leads to LaTeX Math Block Reference |
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| . |
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...
Case #2 – Constant BHP:
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Assume that the flowrate in producer W1 is being automatically adjusted by
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to compensate the bottom-hole pressure variation
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time derivative of drawdown pressure transient response (DTR) in producer W1 to the unit-rate production in the same well
in response to the total sandface flowrate variation
LaTeX Math Inline |
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body | \delta q^{\downarrow}_0 |
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in injector W0 so that bottom-hole pressure in producer W1 stays constant at all times LaTeX Math Inline |
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body | \delta p_1(t) = \delta p_1 = \rm const |
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. In petroleum practice this happens when the formation is capable to deliver more fluid than the current lift settings in producer so that the bottom-hole pressure in producer is constantly kept at minimum value defined by the lift design..In this case, flowrate response
LaTeX Math Inline |
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body | \delta q^{\uparrow}_1 |
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in producer W1 to the flowrate variation LaTeX Math Inline |
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body | \delta q^{\downarrow}_0 |
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in injector W0 is going to be: LaTeX Math Block |
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\delta q^{\uparrow} |
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Consider a pressure convolution equation for the above 2-wells system with constant BHP:
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p_1(t) =
p_i -
\int_0^t^{\uparrow} \frac{\dot p_{u,\rm
1101}(t
-\tau) dq_1(\tau) - \int_0^t)}{\dot p_{u,\rm
2111}(t
-\tau)}
dq_2(\tau) = \rm constThe time derivative is going to be zero as the bottom-hole pressure in producer W1 stays constant at all times:
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\dot p_1(t) = - \left( \int_0^t \cdot \delta q^{\downarrow}_0
where
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| Consider a pressure convolution equation for the above 2-wells system with constant BHP: LaTeX Math Block |
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| -\tau) dq_1(\tau) \right)^{\cdot} - \left( \int_0^t p_{u,\rm 21}(t-\tau) dq_2(\tau) \right)^{\cdot} = 0 LaTeX Math Block |
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| p_{u,\rm 11}(0) \cdot q_1(t) +1(t) = p_i - \int_0^t \dot p_{u,\rm 11}(t-\tau) dqdq^{\uparrow}_1(\tau) = - p_{u,\rm 21}(0) \cdot q_2(t) - \int_0^t \dot p_{u,\rm 2101}(t-\tau) dq_2dq^{\downarrow}_0(\tau) = \rm const |
The zero-time value of DTR / CTR is zero by definition LaTeX Math Inline |
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| body | p_{u,\rm 11}(0) = 0, \, time derivative is going to be zero as the BHP in producer W1 stays constant at all times: LaTeX Math Block |
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| \dot p_1(t) = - \left( \int_0^t p_{u,\rm | 210) = 0 which leads to: LaTeX Math Block |
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anchor | Case2_PSS_p11_temp |
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alignment | left |
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| \int_0^t \dot p_{u,\rm 11}(t-t-\tau) dq^{\uparrow}_1(\tau) dq_1(\tau) =right)^{\cdot} - \left( \int_0^t \dot p_{u,\rm 2101}(t-\tau) dq_2dq^{\downarrow}_0(\tau) |
Consider a step-change in producer's W1 flowrate and injector's W2 flowrate at zero time , which can be written as LaTeX Math Inline |
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body | dq_1(\tau) = \delta q_1 \cdot \delta(\tau) \, d\tau |
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| and LaTeX Math Inline |
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body | dq_2(\tau) = \delta q_2 \cdot \delta(\tau) \, d\tau |
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| . Substituting this to LaTeX Math Block Reference |
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| leads to:
LaTeX Math Block |
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| p_{u,\rm 11}(0) \cdot \dot q^{\uparrow}_1(t) + \int_0^t \dot p_{u,\rm 11}(t-\tau) dq^{\uparrow}_1(\tau) = - p_{u,\rm 01}(0) \cdot \dot q^{\downarrow}_0(t) - \int_ | LaTeX Math Block |
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| \int_0^t \dot p_{u,\rm 1101}(t-\tau) \delta q_1 \cdot \delta(\tau) \, d\tau = - \int_0^t \dot dq^{\downarrow}_0(\tau) |
The zero-time value of DTR / CTR is zero by definition 21t-\tau) \delta q_2 \cdot \delta(\tau) \, d\tau 0) = 0, \, p_{u,\rm 01}(0) = 0 |
| which leads to: LaTeX Math Block |
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anchor | Case2_PSS_p11_temp |
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alignment | left |
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| \int_0^t \dot p_{u,\rm 11}(t-\tau) \delta q_1 dq^{\uparrow}_1(\tau) = - \int_0^t \dot p_{u,\rm 2101}(t-\tau) \delta q_2 |
which leads to LaTeX Math Block Reference |
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| .
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...
LaTeX Math Inline |
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body | V_{\phi,1} \leq V_{\phi,2} < \infty |
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...
Consider a step-change in producer's W1 flowrate LaTeX Math Inline |
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body | \delta q^{\uparrow}_1 |
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| and injector's W0 flowrate |
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...
at zero time , which can be written as LaTeX Math Inline |
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body | dq^{\uparrow}_1(\tau) = \delta q^{\uparrow}_1 \cdot \delta(\tau) \, d\tau |
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| .Assume that a lift mechanism in producer automatically adjusts the flowrate to maintain the same flowing bottom-hole and LaTeX Math Inline |
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body | dq^{\downarrow}_0(\tau) = \delta q^{\downarrow}_0 \cdot \delta(\tau) \, d\tau |
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| .Substituting this to LaTeX Math Block Reference |
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anchor | Case2_PSS |
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alignment | left |
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\delta q_1 / \delta q_2 = f_{21} = \frac{c_{t,2} V_{\phi, 2}}{c_{t,1} V_{\phi, 1}} = \rm const |
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| For the finite-volume reservoir LaTeX Math Inline |
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body | V_{\phi,1} \leq V_{\phi,2} < \infty |
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| the DTR and CTR are both going through the PSS flow regime at late transient times:mathblock leads to: LaTeX Math Block |
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| \int_0^t \dot p_{u,\rm 11}(t | rightarrow\infty)rightarrowfrac{t{c{t,1} V_{\phi, 1}} LaTeX Math Block |
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| anchor | Case2_PSS_p21 |
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| 1 \cdot \delta(\tau) \, d\tau = - \int_0^t \dot p_{u,\rm | 21rightarrowinfty)rightarrowfrac{t}{c_{t,2} V_{\phi,2}}where | average drain-area total compressibility of formation around producer W1 |
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| average drain-area total compressibility of formation around injector W2 |
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LaTeX Math Block |
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| \dot p_{u,\rm 11}(t) \delta q^{\uparrow}_1 = - \dot p_{u,\rm 01}(t) \delta q^{\downarrow}_0 |
which leads to Substituting LaTeX Math Block Reference |
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| and . LaTeX Math Block Reference |
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| in LaTeX Math Block Reference |
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| one arrives to mathblock-ref
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For the finite-volume drain
LaTeX Math Inline |
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body | V_{\phi,1} \leq V_{\phi,0} < \infty |
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the flowrate response factor LaTeX Math Inline |
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body | \delta q^{\uparrow}_1 / \delta q^{\downarrow}_0 |
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is getting stabilised over time as:...
...
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\delta q^{\uparrow}_1 / \delta q^{\downarrow}_0 = - f_{01} = - \frac{V_{\phi, 1}}{ V_{\phi, 0}} = \rm const |
The response delay in time still exists but in usual time-scales of production analysis it becomes negligible and one can consider
LaTeX Math Block Reference |
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as constant in time. Expand |
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| For the finite-volume reservoir LaTeX Math Inline |
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body | V_{\phi,1} \leq V_{\phi,0} < \infty |
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| the DTR and CTR are both going through the PSS flow regime at late transient times:
LaTeX Math Block |
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anchor | Case2_PSS_p11 |
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alignment | left |
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| p_{u,\rm 11}(t \rightarrow \infty) \rightarrow \frac{t}{c_t V_{\phi, 1}} |
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LaTeX Math Block |
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anchor | Case2_PSS_p21 |
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alignment | left |
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| p_{u,\rm 01}(t \rightarrow \infty) \rightarrow \frac{t}{c_t V_{\phi,0}} |
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where Substituting LaTeX Math Block Reference |
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| and LaTeX Math Block Reference |
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| in LaTeX Math Block Reference |
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| one arrives to LaTeX Math Block Reference |
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| . |
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In case injector W0 supports only one producer W1, then both wells drain the same reservoir volume
LaTeX Math Inline |
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body | V_{\phi, 0} = V_{\phi, 1} |
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so that LaTeX Math Block Reference |
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leads to: LaTeX Math Block |
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\delta q^{\uparrow}_1 = -\delta q^{\downarrow}_0 |
which means that producer W1 with constant BHP and finite-reservoir volume will eventually vary its rate at the same volume as injector W0.
In case injector W0 supports many producers {W1 .. WN} then all injection shares towards producers are going to sum up to a unit value:
LaTeX Math Block |
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\sum_{k=1}^N f_{0k} = 1 \quad \Leftrightarrow \quad \sum_{k=1}^N V_{\phi,k} = V_{\phi,0} |
with constant coefficients LaTeX Math Inline |
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body | f_{0k} \geq 0, \ {k=\{i..N \} } |
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, unless there is a thief injection outside the drainage area of all producers and in this case:
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\sum_{k=1}^N f_{0k} < 1 \quad \Leftrightarrow \quad \sum_{k=1}^N V_{\phi,k} < V_{\phi,0} |
If pressure around producer W1 is supported by several injectors
then production response in producer W1 is going to be: LaTeX Math Block |
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\delta q^{\uparrow}_1 =-\sum_i f_{i1} \delta q^{\downarrow}_i
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with constant coefficients LaTeX Math Inline |
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body | f_{i1} \geq 0, \ {i=\{0..N_{\rm inj} \} } |
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.
The equations LaTeX Math Block Reference |
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, LaTeX Math Block Reference |
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and LaTeX Math Block Reference |
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make one of the key assumptions in Capacitance Resistance Model (CRM).
It is important to note that CRM assumption that injector W0 may drain bigger volume than producer W1 LaTeX Math Inline |
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body | V_{\phi, 0}> V_{\phi, 1} |
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is a misnomer in most practical cases.
When wells (producers and injectors) are placed into the same connected reservoir volume they drain the same total volume all together and all UTRs will have the same LTR asymptotic:
LaTeX Math Block |
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p_{u,\rm ik}(t \rightarrow \infty ) \rightarrow \frac{t}{\rm RS}, \quad \forall i \in N_{\rm inj}, k \in N. |
where LaTeX Math Inline |
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body | \rm RS = \int_V c_t \, \phi \, dV |
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is total reservoir storage connecting all the wells.
Moreover, if each well is placed in different reservoir volumes which are only connected through wellbores then again they will all drain the same volume which is the sum of all connected volumes through the wellbores and all UTRs will again trend to the same LTR asymptotic.
In order to relate true UTRs (from numerical grid simulations or from deconvolution) to the CRM injection share constants one needs to implement a certain workflow:
- Start with true UTRs
LaTeX Math Inline |
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body | \displaystyle p_{u, ik}(t) |
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with the same LTR asymptotic LaTeX Math Inline |
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body | \displaystyle p_{u, ik}(t) \rightarrow \frac{t}{RS} |
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. - Select injector W0
- Select producer W1
- Perform a convolution tests to account for the impact from {W2 .. WN} production and from {W-1 .. W-M} on to CTR_01 :
LaTeX Math Inline |
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body | \displaystyle p^*_{u, 01}(t) = p_{u, 01}(t) + \sum_{i \neq 0 \in {\rm inj}} p_{u, i1}(t) \cdot q^{\downarrow}_i(t) |
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- Perform two convolution tests to account for the impact from {W2 .. WN} production on to DTR_11 and CTR_01 :
- Test #1 – DTR_11
- Calculate interfering DTR_11:
LaTeX Math Inline |
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body | \displaystyle p^*_{u, 11}(t) = p_{u, 11}(t) + \sum_{k \neq 1 \in {\rm prod}} p_{u, k1}(t) \cdot q^{\uparrow}_k(t) |
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, meaning that all injectors W0 are shut-down and all producers were working with their historical rates , except producer W1 which is working with unit-rate
- Test #2 – CTR_01
- Calculate interfering CTR_01:
LaTeX Math Inline |
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body | \displaystyle p^*_{u, 01}(t) = p_{u, 01}(t) + \sum_{i \neq 0 \in {\rm inj}} p_{u, i1}(t) \cdot q^{\downarrow}_i(t) |
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, meaning that all producers are shut-down and all injectors are working with their historical rates , except injector W0 which is working with unit-rate
- Calculate injection share constant:
LaTeX Math Inline |
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body | \displaystyle f_{01} = \frac{\dot p^*_{01}(t)}{\dot p^*_{11}(t)} \Bigg|_{t \rightarrow \infty} |
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as LLS over equation: LaTeX Math Inline |
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body | \displaystyle \dot p^*_{01}(t) = f_{01} \cdot \dot p^*_{11}(t) |
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- Repeat the same for other producers (starting from point 2a onwards)
- Repeat the same for other injectors (starting from point 2 onwards)
Show If |
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| Consider a pressure convolution equation for the well W1 with constant BHP in a multi-well system : LaTeX Math Block |
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| p_1(t) = p_i - \sum_{k \in {\rm prod}} \int_0^t p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) - \sum_{i \in {\rm inj}} \int_0^t p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) = \rm const |
The time derivative is going to be zero as the BHP in producer W1 stays constant at all times: LaTeX Math Block |
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| \dot p_1(t) = - \left( \sum_{k \in {\rm prod}} \int_0^t p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) \right)^\cdot -
\left( \sum_{i \in {\rm inj}} \int_0^t p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) \right)^\cdot = 0 |
LaTeX Math Block |
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| \sum_{k \in {\rm prod}} p_{u,\rm k1}(0) \dot q^{\uparrow}_k(t) +
\sum_{k \in {\rm prod}} \int_0^t \dot p_{u,\rm kk}(t-\tau) dq^{\uparrow}_k(\tau) =
- \sum_{i \in {\rm inj}} p_{u,\rm i1}(0) \dot q^{\downarrow}_i(t)
- \sum_{i \in {\rm inj}} \int_0^t \dot p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) |
The zero-time value of DTR / CTR is zero by definition LaTeX Math Inline |
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body | p_{u,\rm kj}(0) = 0, \ \forall k,j \in \mathbb{Z} |
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| which leads to: LaTeX Math Block |
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| \sum_{k \in {\rm prod}} \int_0^t \dot p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) =
- \sum_{i \in {\rm inj}} \int_0^t \dot p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) |
Let's separate producer W1 and injector W0 terms: LaTeX Math Block |
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| \int_0^t \dot p_{u,\rm 11}(t-\tau) dq^{\uparrow}_1(\tau) + \sum_{k \neq 1 \in {\rm prod}} \int_0^t \dot p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) =
- \int_0^t \dot p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\tau) - \sum_{i \neq 0 \in {\rm inj}} \int_0^t \dot p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) |
Consider a step-change in injector's W0 flowrate LaTeX Math Inline |
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body | \delta q^{\downarrow}_0 |
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| at zero time , which can be written as LaTeX Math Inline |
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body | dq^{\uparrow}_1(\tau) = \delta q^{\uparrow}_1 \cdot \delta(\tau) \, d\tau |
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| , leading to a step-change in production rate in producer W1 LaTeX Math Inline |
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body | dq^{\uparrow}_1(\tau) = \delta q^{\uparrow}_1 \cdot \delta(\tau) \, d\tau |
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| .Substituting this to LaTeX Math Block Reference |
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| leads to: LaTeX Math Block |
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| \dot p_{u,\rm 11}(t) \delta q^{\uparrow}_1 + \sum_{k \neq 1 \in {\rm prod}} \int_0^t \dot p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) =
- \dot p_{u,\rm 01}(t) \delta q^{\downarrow}_0 - \sum_{i \neq 0 \in {\rm inj}} \int_0^t \dot p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) |
|
|
|
Again it is important to note a difference between
- CRM assumptions (constant PI, constant drainage volumes with no flow boundaries and constant total compressibility) – which may or may not take place and hence may or may not make CRM applicable in a specific case
and
- CRM concept of mismatching drainage volumes between producers and injectors which is just a terminology and does not exert restrictions on well-reservoir system
See also
...
[UTR]
LaTeX Math Block |
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\delta q_1 =\sum_k f_{k1} \delta q_k
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with constant coefficients LaTeX Math Inline |
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body | f_{1k}, \ {k=\{1..N_{\rm inj} \} } |
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, which makes one of the key assumptions in Capacitance Resistance Model (CRM).
See also
[ DTR ] [ CTR ] [ Capacitance Resistance Model (CRM) ]
...