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 01}(t-\tau) dq_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) dq_1(\tau) \right)^{\cdot} - \left( \int_0^t p_{u,\rm 01}(t-\tau) dq_0(\tau) \right)^{\cdot} = 0 |
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| p_{u,\rm 11}(0) \cdot \dot q_1(t) + \int_0^t \dot p_{u,\rm 11}(t-\tau) dq_1(\tau) = - p_{u,\rm 01}(0) \cdot \dot q_0(t) - \int_0^t \dot p_{u,\rm 01}(t-\tau) dq_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) 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 W0 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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.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_0(\tau) = \delta q_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 01}(t-\tau) \delta q_0 \cdot \delta(\tau) \, d\tau |
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| \dot p_{u,\rm 11}(t) \delta q_1 = - \dot p_{u,\rm 01}(t) \delta q_0 |
which leads to | LaTeX Math Block Reference |
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