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Case #1 – Constant flowrate production:
The bottom-hole pressure response The pressure response
in producer
W1 to the flowrate variation
in injector
W2:
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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 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 |
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Case #2 – Constant BHP:
Assume that the flowrate
in producer W1 is being adjusted to compensate the bottom-hole pressure variation in response to the flowrate variation in injector W2 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 this case, The flowrate response
in producer
W1 to the flowrate variation
in injector
W2 is going to be:
LaTeX Math Block |
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\delta q_1(t) = - \frac{\dot p_{u,\rm 21}(t)}{\dot p_{u,\rm 11}(t)} \cdot \delta q_2 |
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borderColor | wheat |
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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 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: 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 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 |
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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
LaTeX Math Inline |
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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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