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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 qq^{\downarrow}_0 |
where
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Panel |
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borderColor | wheat |
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bgColor | mintcream |
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borderWidth | 7 |
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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 writen as LaTeX Math Block |
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| q_0(\tau) = \delta qq^{\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 qq^{\downarrow}_0(\tau) = q_0'(\tau) d\tau = \delta qq^{\downarrow}_0 \cdot H'(\tau) \, d\tau = \delta qq^{\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 qq^{\downarrow}_0 \cdot \delta(\tau) \, d\tau = - p_{u,\rm 01}(t) \cdot \delta qq^{\downarrow}_0 |
which leads to LaTeX Math Block Reference |
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| . |
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Assume that the flowrate in producer W1 is being automatically adjusted by
LaTeX Math Inline |
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body | \delta qq^{\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_2q^{\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 qq^{\uparrow}_1 |
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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 is going to be:
LaTeX Math Block |
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\delta qq^{\uparrow}_1(t) = -^{\uparrow} \frac{\dot p_{u,\rm 01}(t)}{\dot p_{u,\rm 11}(t)} \cdot \delta qq^{\downarrow}_0 |
where
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Panel |
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borderColor | wheat |
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bgColor | mintcream |
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borderWidth | 7 |
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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) dqdq^{\uparrow}_1(\tau) - \int_0^t p_{u,\rm 01}(t-\tau) dqdq^{\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) dqdq^{\uparrow}_1(\tau) \right)^{\cdot} - \left( \int_0^t p_{u,\rm 01}(t-\tau) dqdq^{\downarrow}_0(\tau) \right)^{\cdot} = 0 |
LaTeX Math Block |
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| p_{u,\rm 11}(0) \cdot \dot qq^{\uparrow}_1(t) + \int_0^t \dot p_{u,\rm 11}(t-\tau) dqdq^{\uparrow}_1(\tau) = - p_{u,\rm 01}(0) \cdot \dot qq^{\downarrow}_0(t) - \int_0^t \dot p_{u,\rm 01}(t-\tau) dqdq^{\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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alignment | left |
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| \int_0^t \dot p_{u,\rm 11}(t-\tau) dqdq^{\uparrow}_1(\tau) = - \int_0^t \dot p_{u,\rm 2101}(t-\tau) dq_2dq^{\downarrow}_0(\tau) |
Consider a step-change in producer's W1 flowrate LaTeX Math Inline |
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body | \delta qq^{\uparrow}_1 |
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| and 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 | dqdq^{\uparrow}_1(\tau) = \delta qq^{\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 | dqdq^{\downarrow}_0(\tau) = \delta qq^{\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 qq^{\uparrow}_1 \cdot \delta(\tau) \, d\tau = - \int_0^t \dot p_{u,\rm 01}(t-\tau) \delta qq^{\downarrow}_0 \cdot \delta(\tau) \, d\tau |
LaTeX Math Block |
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| \dot p_{u,\rm 11}(t) \delta qq^{\uparrow}_1 = - \dot p_{u,\rm 01}(t) \delta qq^{\downarrow}_0 |
which leads to LaTeX Math Block Reference |
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| . |
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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 qq^{\uparrow}_1 / \delta qq^{\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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alignment | left |
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\delta qq^{\uparrow}_1 / \delta qq^{\downarrow}_0 = - f_{01} = - \frac{V_{\phi, 1}}{ V_{\phi, 0}} = \rm const |
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LaTeX Math Block |
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\delta qq^{\uparrow}_1 = -\delta qq^{\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.
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LaTeX Math Block |
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\delta qq^{\uparrow}_1 =-\sum_i f_{i1} \delta q_i
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- 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 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 qq^{\uparrow}_k(t) |
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, meaning that injector that all injectors W0 is are shut-down and all producers are were working with constant 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_{ki \neq 0 \in {\rm inj}} p_{u, k1i1}(t) \cdot q_kq^{\downarrow}_i(t) |
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, meaning that injector W0 is working with unit-rate and all producers are working with constant all producers are shut-down and all injectors are working with their historical rates LaTeX Math Inline |
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body | q^*_k{\downarrow}_i(t) |
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, 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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