Consider a well-reservoir system (Fig. 1) consisting of:
- producing well W1 with total sandface flowrate and BHP , draining the reservoir volume
water injecting well W0 with total sandface flowrate
LaTeX Math Inline |
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body | q^{\downarrow}_0(t) <0 |
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, supporting pressure in reservoir volume
The injection drainage volume
includes the drainage volume of producer W1 and may be equal to it LaTeX Math Inline |
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body | V_{\phi, 0} = V_{\phi, 1} |
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or may be bigger LaTeX Math Inline |
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body | V_{\phi, 0} > V_{\phi, 1} |
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in case injector W0 supports other producers {W1 .. WN}: LaTeX Math Inline |
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body | V_{\phi, 0} = \sum_{k=1}^N V_{\phi, k} |
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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
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 | q^{\uparrow}_1 = \rm const |
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| and varying injection rate at injector W0 : LaTeX Math Block |
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| p_1(t) = p_i - \int_0^t p_{u,\rm 11}(t-\tau) dq^{\uparrow}_1(\tau) - \int_0^t p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\tau) = p_i - \int_0^t p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\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 writen as LaTeX Math Block |
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| q_0(\tau) = \delta q^{\downarrow}_0 \cdot H(\tau) |
where is Heaviside step function: LaTeX Math Block |
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| H(\tau) = \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 in producer W1 will be: LaTeX Math Block |
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| \delta p_1(t) = p_1(t)-p_i = - \int_0^t p_{u,\rm 21}(t-\tau) \delta q^{\downarrow}_0 \cdot \delta(\tau) \, d\tau = - p_{u,\rm 01}(t) \cdot \delta q^{\downarrow}_0 |
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 automatically adjusted by
LaTeX Math Inline |
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body | \delta q^{\uparrow}_1(t) |
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to compensate the bottom-hole pressure variation 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
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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}_1(t) = -^{\uparrow} \frac{\dot p_{u,\rm 01}(t)}{\dot p_{u,\rm 11}(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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| p_1(t) = p_i - \int_0^t p_{u,\rm 11}(t-\tau) dq^{\uparrow}_1(\tau) - \int_0^t p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\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( \int_0^t p_{u,\rm 11}(t-\tau) dq^{\uparrow}_1(\tau) \right)^{\cdot} - \left( \int_0^t p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\tau) \right)^{\cdot} = 0 |
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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_0^t \dot p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\tau) |
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, \, p_{u,\rm 01}(0) = 0 |
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| which leads to: LaTeX Math Block |
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anchor | Case2_PSS_p11_temp |
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| \int_0^t \dot p_{u,\rm 11}(t-\tau) dq^{\uparrow}_1(\tau) = - \int_0^t \dot p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\tau) |
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 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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| .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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| leads to: LaTeX Math Block |
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| \int_0^t \dot p_{u,\rm 11}(t-\tau) \delta q^{\uparrow}_1 \cdot \delta(\tau) \, d\tau = - \int_0^t \dot p_{u,\rm 01}(t-\tau) \delta q^{\downarrow}_0 \cdot \delta(\tau) \, d\tau |
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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 LaTeX Math Block Reference |
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| . |
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For the finite-volume drain
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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: LaTeX Math Block |
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anchor | Case2_PSS |
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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
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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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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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anchor | fokless1 |
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alignment | left |
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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:
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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
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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)
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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 |
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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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anchor | pre_eq |
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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) |
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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
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[UTR] [ DTR ] [ CTR ] [ Capacitance Resistance Model (CRM) ]