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- producing well W1 with total sandface flowrate and BHP , draining the reservoir volume
- water injecting well W20 with total sandface flowrate , supporting pressure in reservoir volume which includes the drainage volume of producer W1 and potentially other producers.
The drainage volume difference
LaTeX Math Inline |
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body | \delta V_{\phi} = V_{\phi, 20} - V_{\phi, 1} >0 |
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may be related to the fact that water injection
W2 is shared between
and another reservoir or with another producer.
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The bottom-hole pressure response
in producer
W1 to the flowrate variation
in injector
W20:
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\delta p_1 = - p_{u,\rm 21}(t) \cdot \delta q_2 |
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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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| p_1(t) = p_i - \int_0^t p_{u,\rm 11}(t-\tau) dq_1(\tau) - \int_0^t p_{u,\rm 2101}(t-\tau) dq_20(\tau) = p_i - \int_0^t p_{u,\rm 2101}(t-\tau) dq_20(\tau) |
Consider a step-change in injector's W20 flowrate at zero time , which can be written as: LaTeX Math Inline |
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body | dq_20 (\tau) = \delta q_2 0 \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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| \delta p_1(t) = p_1(t)-p_i = - \int_0^t p_{u,\rm 21}(t-\tau) \delta q_20 \cdot \delta(\tau) \, d\tau = - p_{u,\rm 2101}(t) \cdot \delta q_20 |
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
to compensate the
bottom-hole pressure variation
in response to the
total sandface flowrate variation
in injector
W20 so that
bottom-hole pressure in producer
W1 stays constant at all times
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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
W20 is going to be:
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\delta q_1(t) = - \frac{\dot p_{u,\rm 2101}(t)}{\dot p_{u,\rm 11}(t)} \cdot \delta q_20 |
where
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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) dq_1(\tau) - \int_0^t p_{u,\rm 2101}(t-\tau) dq_20(\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) dq_1(\tau) \right)^{\cdot} - \left( \int_0^t p_{u,\rm 2101}(t-\tau) dq_20(\tau) \right)^{\cdot} = 0 |
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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 2101}(0) \cdot q_20(t) - \int_0^t \dot p_{u,\rm 2101}(t-\tau) dq_20(\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 2101}(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 W20 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 .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_20(\tau) = \delta q_2 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 q_1 \cdot \delta(\tau) \, d\tau = - \int_0^t \dot p_{u,\rm 2101}(t-\tau) \delta q_20 \cdot \delta(\tau) \, d\tau |
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| \dot p_{u,\rm 11}(t) \delta q_1 = - \dot p_{u,\rm 2101}(t) \delta q_20 |
which leads to LaTeX Math Block Reference |
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| . |
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For the finite-volume drain
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body | V_{\phi,1} \leq V_{\phi,20} < \infty |
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the flowrate response factor
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body | \delta q_1 / \delta q_20 |
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is getting stabilised over time as:
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anchor | Case2_PSS |
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alignment | left |
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\delta q_1 / \delta q_20 = - f_{21} = - \frac{V_{\phi, 1}}{ V_{\phi, 20}} = \rm const |
The response delay in time still exists but in usual time-scales of production analysis it becomes negligible and one can consider
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as constant in time.
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| For the finite-volume reservoir LaTeX Math Inline |
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body | V_{\phi,1} \leq V_{\phi,20} < \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 V_{\phi, 1}} |
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anchor | Case2_PSS_p21 |
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alignment | left |
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| p_{u,\rm 2101}(t \rightarrow \infty) \rightarrow \frac{t}{c_t 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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| . |
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In case injector W20 supports only one producer W1 , then both wells drain the same volume and
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body | V_{\phi, 20} = V_{\phi, 1} |
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so that
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leads to:
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\delta q_1 = -\delta q_20 |
which means that producer W1 with constant BHP and finite-reservoir volume will eventually vary its rate at the same volume as injector W20.
In case injector W20 supports many producers {W1 .. WN } then all injection shares towards producers are going to sum up to a unit value:
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