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The bottom-hole pressure response
in producer
W1 to the flowrate variation
in injector
W2:
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| anchor | DD4HWCase1 |
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| alignment | left |
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\delta p_1 = - p_{u,\rm 21}(t) \cdot \delta q_2 |
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| borderColor | wheat |
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| Consider a pressure convolution equation for the BHP in producer W1 with constant flowrate production at producer W1 and varying injection rate at injector W2 :| LaTeX Math Block |
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| anchor | Case2_PSS_p11 |
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| alignment | left |
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| p_1(t) = p_i - \int_0^t p_{u,\rm 11}(t-\tau) dq_1(\tau) - \int_0^t p_{u,\rm 21}(t-\tau) dq_2(\tau) = p_i - \int_0^t p_{u,\rm 21}(t-\tau) dq_2(\tau) |
Consider a step-change in injector's W2 flowrate at zero time , which can be written as: | LaTeX Math Inline |
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| body | dq_2(\tau) = \delta q_2 \cdot \delta(\tau) \, d\tau |
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.The responding pressure variation in producer W1 will be:| LaTeX Math Block |
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| anchor | Case2_PSS_p11 |
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| alignment | left |
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| \delta p_1(t) = 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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Case #2 – Constant BHP:
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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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| anchor | Case2_PSS_p11 |
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| alignment | left |
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| p_1(t) = p_i - \int_0^t p_{u,\rm 11}(t-\tau) dq_1(\tau) - \int_0^t p_{u,\rm 21}(t-\tau) dq_2(\tau) = \rm const |
The time derivative is going to be zero as the bottom-hole pressure in producer W1 stays constant at all times: | LaTeX Math Block |
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| anchor | Case2_PSS_p11 |
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| alignment | left |
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| \dot p_1(t) = - \left( \int_0^t p_{u,\rm 11}(t-\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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| anchor | Case2_PSS_p11 |
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| alignment | left |
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| p_{u,\rm 11}(0) \cdot q_1(t) + \int_0^t \dot p_{u,\rm 11}(t-\tau) dq_1(\tau) = - p_{u,\rm 21}(0) \cdot q_2(t) - \int_0^t \dot p_{u,\rm 21}(t-\tau) dq_2(\tau) |
The zero-time value of of DTR / CTR is zero by definition | LaTeX Math Inline |
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| body | p_{u,\rm 11}(0) = 0, \, p_{u,\rm 21}(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) dq_1(\tau) = - \int_0^t \dot p_{u,\rm 21}(t-\tau) dq_2(\tau) |
Consider a step-change in producer's W1 flowrate and injector's W2 flowrate at zero time , which can be written as: 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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. Substituing Substituting this to | LaTeX Math Block Reference |
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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 \cdot \delta(\tau) \, d\tau = - \int_0^t \dot p_{u,\rm 21}(t-\tau) \delta q_2 \cdot \delta(\tau) \, d\tau |
| LaTeX Math Block |
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| anchor | Case2_PSS_p11_temp |
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| alignment | left |
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| \dot p_{u,\rm 11}(t) \delta q_1 = - \dot p_{u,\rm 21}(t) \delta q_2 |
which leads to | LaTeX Math Block Reference |
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For the finite-volume drain
| LaTeX Math Inline |
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| body | V_{\phi,1} \leq V_{\phi,2} < \infty |
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the flowrate response factor
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| body | \delta q_1 / \delta q_2 |
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is getting stabilised over time as:
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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:
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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,1} V_{\phi, 1}} |
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| anchor | Case2_PSS_p21 |
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| alignment | left |
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| p_{u,\rm 21}(t \rightarrow \infty) \rightarrow \frac{t}{c_{t,2} V_{\phi,2}} |
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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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Capacitance Resistance Model (CRM)
[ DTR ] [ CTR ]
