Page History

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• producing well W₁ with total sandface flowrate 

  LaTeX Math Inline
  body q_1(t)>0

   and BHP 

  LaTeX Math Inline
  body p_1(t)>0

  , draining the reservoir volume

  LaTeX Math Inline
  body V_{\phi, 1}

   
  
  
• water injecting well W₂₀ with total sandface flowrate 

  LaTeX Math Inline
  body q_20(t) <0

  , supporting pressure in reservoir volume

  LaTeX Math Inline
  body V_{\phi, 20}

   which includes the drainage volume 

  LaTeX Math Inline
  body V_{\phi, 1}

   of producer W₁ and potentially other producers. 

The drainage volume difference

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The bottom-hole pressure response 

\delta p_1 = - p_{u,\rm 21}(t) \cdot \delta q_2

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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)

\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

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Assume that the flowrate in producer W₁ is being automatically adjusted by

In this case, flowrate response 

\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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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

\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

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) 

\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) 

\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 

 \dot p_{u,\rm 11}(t)  \delta q_1   = -  \dot p_{u,\rm 2101}(t) \delta q_20  

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For the finite-volume drain 

\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

p_{u,\rm 11}(t \rightarrow \infty) \rightarrow \frac{t}{c_t V_{\phi, 1}}

p_{u,\rm 2101}(t \rightarrow \infty) \rightarrow \frac{t}{c_t V_{\phi,2}}

In case injector W₂₀ supports only one producer W₁ , then both wells drain the same volume and 

\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 .. W_N } then all injection shares towards producers are going to sum up to a unit value:

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