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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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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 qq^{\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_2q^{\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 01}(t-\tau) dqdq^{\downarrow}_0(\tau) = p_i - \int_0^t p_{u,\rm 01}(t-\tau) dqdq^{\downarrow}_0(\tau) |
Consider a step-change in injector's W0 flowrate LaTeX Math Inline |
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body | \delta qq^{\downarrow}_0 |
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| at zero time , which can be written as: LaTeX Math Inline |
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| body | dqwriten as LaTeX Math Block |
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| q_0(\tau) = \delta | q \deltawhere \, d\tau.The responding pressure variation in producer W1 will be is Heaviside step function: LaTeX Math Block |
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| H(\delta p_1(t) = p_1(t)-p_i = - \int_0^t p_{u,\rm 21}(t-\tau) \delta q_0 \cdot \delta(\tau) \, d\tau = - p_{u,\rm 01}(t) \cdot \delta q_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
to compensate the bottom-hole pressure variation in response to the total sandface flowrate variation 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
in producer W1 to the flowrate variation in injector W0 is going to be: LaTeX Math Block |
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\delta q_1(t) = - \frac{\dot p_{u,\rm 01}(t)}{\dot p_{u,\rm 11}(t)} \cdot \delta q_0 |
where
...
...
time since injector's W0 rate has changed by
....
...
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time derivative of drawdown pressure transient response (DTR) in producer W1 to the unit-rate production in the same well
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
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}_1(t) = -^{\uparrow} \frac{\dot |
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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
1101}(t
-\tau) dq_1(\tau) - \int_0^t )}{\dot p_{u,\rm
0111}(t
-\tau)}
dq_0(\tau) = \rm constThe 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 \cdot \delta q^{\downarrow}_0
where
| time since injector's W0 rate has changed by LaTeX Math Inline |
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body | \delta q^{\downarrow}_0 |
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| . |
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| Consider a pressure convolution equation for the above 2-wells system with constant BHP: -\tau) dq_0(\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 01}(0) \cdot q_0(t) - \int_0^t \dot p_{u,\rm 01}(t-\tau) dqdq^{\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 | 010) = 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(\tauright) = ^{\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 W0 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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| .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_0(\tau) = \delta q_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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| 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 | 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 01t-\tau) \delta q_0 \cdot \delta(\tau) \, d\tau 0) = 0, \, p_{u,\rm 01}(0) = 0 |
| which leads to: LaTeX Math Block |
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anchor | 1Case2_PSS_p11_temp |
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alignment | left |
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| \int_0^t \dot p_{u,\rm 11}(t-\tau) \delta qdq^{\uparrow}_1(\tau) = - \int_0^t \dot p_{u,\rm 01}(t-\tau) \delta qdq^{\downarrow}_0(\tau) |
which leads to LaTeX Math Block Reference |
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| .
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|
...
Consider a step-change in producer's W1 flowrate |
|
...
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and injector's W0 flowrate |
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...
...
LaTeX Math Block |
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anchor | Case2_PSS |
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alignment | left |
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\delta q_1 / \delta q_0 = - f_{01} = - \frac{V_{\phi, 1}}{ V_{\phi, 0}} = \rm const |
...
LaTeX Math Block Reference |
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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 |
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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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| \rightarrow \infty) \rightarrow \frac{t}{c_t V_{\phi, 1}} LaTeX Math Block |
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anchor | Case2_PSS_p21 |
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alignment | left |
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| \delta q^{\uparrow}_1 \cdot \delta(\tau) \, d\tau = - \int_0^t \dot p_{u,\rm 01}(t | \rightarrow \inftyrightarrowfrac{t{c_t V_{\phi,2}}where | average drain-area total compressibility of formation within which is jointly drained by producer W1 and injector W0 |
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_0 \cdot \delta(\tau) \, d\tau |
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 LaTeX Math Block Reference |
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| .
|
...
...
For the finite-volume drain
...
...
...
...
LaTeX Math Block Reference |
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|
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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 | 1Case2_PSS |
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alignment | left |
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|
\delta qq^{\uparrow}_1 =/ -\delta q_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 |
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:
LaTeX Math Block |
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anchor | fokless1 |
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alignment | left |
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\sum_{k=1}^N f_{0k} < 1 |
...
...
LaTeX Math Block |
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\delta q_1 =-\sum_i f_{i1} \delta q_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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|
.
...
LaTeX Math Block Reference |
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|
...
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}} |
|
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}} |
|
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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| . |
|
In case injector W0 supports only one producer W1, then both wells drain the same reservoir volume
fk1 | 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 > 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 so that LaTeX Math Block Reference |
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|
leads to: 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.
...
\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:
LaTeX Math Block |
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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
|
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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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) |
|
|
|
...
In order to relate the UTR from numerical grid simulations or from deconvolution to the CRM injection share constants one needs to implement a certain workflow.
...
LaTeX Math Inline |
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body | \displaystyle p_{u, ik}(t) |
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|
...
LaTeX Math Inline |
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body | \displaystyle p_{u, ik}(t) \rightarrow \frac{t}{RS} |
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|
...
- Test #1 – DTR_11
- Calculate historically-averaged rate for each producer:
LaTeX Math Inline |
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body | \displaystyle q^*_k = \frac{1}{N_k} \sum_{m=1}^{N_k} q_k(t_m) |
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|
- Calculate DTR_11:
LaTeX Math Inline |
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body | \displaystyle p^*_{u, 11}(t) = p_{u, 11}(t) + \sum_{k \neq 1} p_{u, k1}(t) \cdot q^*_k |
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|
(meaning that injector W0 is shut-down and all producers are working with constant rates , except producer W1 which is working with unit-rate)
- Test #2 – CTR_01
- Calculate historically-averaged rate for each producer:
LaTeX Math Inline |
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body | \displaystyle q^*_k = \frac{1}{N_k} \sum_{m=1}^{N_k} q_k(t_m) |
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|
- Calculate CTR_01:
LaTeX Math Inline |
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body | \displaystyle p^*_{u, 01}(t) = p_{u, 01}(t) + \sum_{k} p_{u, k1}(t) \cdot q^*_k |
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|
(meaning that injector W0 is working with unit-rate and all producers are working with constant rates )
...
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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|
...
LaTeX Math Inline |
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body | \displaystyle \dot p^*_{01}(t) = f_{01} \cdot \dot p^*_{11}(t) |
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|
...
Again it is important to note a difference between
...