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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 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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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 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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alignment | left |
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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
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