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• producing well W₁ with total sandface flowrate 

  LaTeX Math Inline
  body qq^{\uparrow}_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 W₀ with total sandface flowrate 

  LaTeX Math Inline
  body qq^{\downarrow}_0(t) <0

  , supporting pressure in reservoir volume

  LaTeX Math Inline
  body V_{\phi, 0}

   

...

Case #1 –  Constant flowrate production: 

LaTeX Math Inline
body

...

q^{\uparrow}_1 = \rm const >0

The bottom-hole pressure response 

\delta p_1 = - p_{u,\rm 21}(t) \cdot \delta qq^{\downarrow}_0

where

...

p_1(t) = p_i - \int_0^t p_{u,\rm 11}(t-\tau) dqdq^{\uparrow}_1(\tau) - \int_0^t p_{u,\rm 01}(t-\tau) dqdq^{\downarrow}_0(\tau) = p_i - \int_0^t p_{u,\rm 01}(t-\tau) dqdq^{\downarrow}_0(\tau)

q_0(\tau) = \delta qq^{\downarrow}_0 \cdot H(\tau)

H(\tau) = \begin{cases} 0, &  \tau <0 \\  1, &\tau \geq 0\end{cases}

d qq^{\downarrow}_0(\tau) = q_0'(\tau) d\tau = \delta qq^{\downarrow}_0 \cdot H'(\tau) \,  d\tau = \delta qq^{\downarrow}_0 \cdot \delta(\tau) \,  d\tau

\delta p_1(t) = p_1(t)-p_i = - \int_0^t p_{u,\rm 21}(t-\tau)  \delta qq^{\downarrow}_0 \cdot \delta(\tau) \,  d\tau = - p_{u,\rm 01}(t) \cdot  \delta qq^{\downarrow}_0

...

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

In this case, flowrate response 

\delta qq^{\uparrow}_1(t) = - \frac^{\uparrow} \frac{\dot p_{u,\rm 01}(t)}{\dot p_{u,\rm 11}(t)} \cdot \delta qq^{\downarrow}_0

where

...

p_1(t) = p_i - \int_0^t p_{u,\rm 11}(t-\tau) dqdq^{\uparrow}_1(\tau) - \int_0^t p_{u,\rm 01}(t-\tau) dqdq^{\downarrow}_0(\tau) = \rm const

\dot p_1(t) = - \left( \int_0^t p_{u,\rm 11}(t-\tau) dqdq^{\uparrow}_1(\tau) \right)^{\cdot} - \left( \int_0^t p_{u,\rm 01}(t-\tau) dqdq^{\downarrow}_0(\tau) \right)^{\cdot} = 0

p_{u,\rm 11}(0) \cdot \dot qq^{\uparrow}_1(t) + \int_0^t \dot p_{u,\rm 11}(t-\tau) dqdq^{\uparrow}_1(\tau)  = - p_{u,\rm 01}(0) \cdot \dot qq^{\downarrow}_0(t) -  \int_0^t \dot p_{u,\rm 01}(t-\tau) dqdq^{\downarrow}_0(\tau) 

\int_0^t \dot p_{u,\rm 11}(t-\tau) dq_dq^{\uparrow}_1(\tau)  = -  \int_0^t \dot p_{u,\rm 2101}(t-\tau) dq_2dq^{\downarrow}_0(\tau) 

\int_0^t \dot p_{u,\rm 11}(t-\tau)  \delta qq^{\uparrow}_1 \cdot \delta(\tau) \,  d\tau  = -  \int_0^t \dot p_{u,\rm 01}(t-\tau) \delta qq^{\downarrow}_0 \cdot \delta(\tau) \,  d\tau 

 \dot p_{u,\rm 11}(t)  \delta qq^{\uparrow}_1   = -  \dot p_{u,\rm 01}(t) \delta qq^{\downarrow}_0  

...

For the finite-volume drain 

\delta qq^{\uparrow}_1 / \delta qq^{\downarrow}_0 = - f_{01} = - \frac{V_{\phi, 1}}{ V_{\phi, 0}} = \rm const

...

\delta qq^{\uparrow}_1 = -\delta qq^{\downarrow}_0

which means that producer W₁ with constant BHP and finite-reservoir volume will eventually vary its rate at the same volume as injector W₀.

...

\delta qq^{\uparrow}_1 =-\sum_i f_{i1} \delta qq^{\downarrow}_i

with constant coefficients

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1. Start with true UTRs 

   LaTeX Math Inline
   body \displaystyle p_{u, ik}(t)

   with the same LTR asymptotic

   LaTeX Math Inline
   body \displaystyle p_{u, ik}(t) \rightarrow \frac{t}{RS}

   .
2. Select injector W₀ 
   1. Select producer W₁
      1. Perform two a convolution tests to account for the impact from {W₂ .. W_N} production and from {W_-1 .. W_-M} on to DTRCTR_11 01

         LaTeX Math Inline
         body p_{u, 11}(t)

          and CTR_01

         LaTeX Math Inline
         body p_{u, 01}(t)

         :Test #1 – DTR_11Calculate interfering DTR_11: 

         LaTeX Math Inline
         body \displaystyle p^*_{u,

         11

         01}(t) = p_{u,

         11

         01}(t) + \sum_{

         k

         i \neq

         1

         0 \in {\rm inj}} p_{u,

         k1

         i1}(t) \cdot q^{\downarrow}_

         k

         i(t)

      2. Perform two convolution tests to account for the impact from {W₂ .. W_N} production on to DTR_11 , meaning that injector W₀ is shut-down and all producers are working with constant rates 

         LaTeX Math Inline
         body

         q^*_k, except producer W₁ which is working with unit-rate Test #2 –

         p_{u, 11}(t)

          and CTR_01 Calculate interfering CTR_01: 

         LaTeX Math Inline
         body

         \displaystyle p^*

         p_{u, 01}(t)

         :
         • Test #1 – DTR_11
           
           • Calculate interfering DTR_11: 

             LaTeX Math Inline
             body \displaystyle p^*_{u, 11}(t) = = p_{u, 0111}(t) + \sum_{k \neq 1 \in {\rm prod}} p_{u, k1}(t) \cdot qq^{\uparrow}_k(t)

             , meaning that injector all injectors W₀ is working with unit-rate  are shut-down and all producers are were working with constant their historical rates 

             LaTeX Math Inline
             body q^*{\uparrow}_k(t)

             , except producer W₁ which is working with unit-rate
         Calculate injection share constant
         •  Test #2 – CTR_01
           
           • Calculate interfering CTR_01: 

             LaTeX Math Inline
             body \displaystyle

         f_{01} = \frac{\dot

         • • p^*_{u, 01}(t)

         }{\dot p^*

         • • = p_{

         11

         • • u, 01}(t)

         }

         • • + \

         Bigg|

         • • sum_{

         t

         • • i \

         rightarrow \infty} as LLS over equation:  LaTeX Math Inlinebody\displaystyle \dot p^*_{01

         • • neq 0 \in {\rm inj}} p_{u, i1}(t)

         = f_{01}

         • • \cdot

         \dot p^*_{11}

         • • q^{\downarrow}_i(t)

   2. Repeat the same for other producers (starting from point 2a onwards)
Repeat the same for other injectors (starting from point 2 onwards)
3. 1. 1. • • , meaning that all producers are shut-down and all injectors are working with their historical rates 

             LaTeX Math Inline
             body q^{\downarrow}_i(t)

             , except injector W₀ which is working with unit-rate
      2. Calculate injection share constant: 

         LaTeX Math Inline
         body \displaystyle f_{01} = \frac{\dot p^*_{01}(t)}{\dot p^*_{11}(t)} \Bigg|_{t \rightarrow \infty}

          as LLS over equation: 

         LaTeX Math Inline
         body \displaystyle \dot p^*_{01}(t) = f_{01} \cdot \dot p^*_{11}(t)

   2. Repeat the same for other producers (starting from point 2a onwards)
4. Repeat the same for other injectors (starting from point 2 onwards)

p_1(t) = p_i - \sum_{k \in {\rm prod}} \int_0^t p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) - \sum_{i \in {\rm inj}} \int_0^t p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) = \rm const

\dot p_1(t) = - \left( \sum_{k \in {\rm prod}} \int_0^t p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) \right)^\cdot - 
\left( \sum_{i \in {\rm inj}} \int_0^t p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) \right)^\cdot = 0

\sum_{k \in {\rm prod}} p_{u,\rm k1}(0) \dot q^{\uparrow}_k(t) + 
\sum_{k \in {\rm prod}} \int_0^t \dot p_{u,\rm kk}(t-\tau) dq^{\uparrow}_k(\tau) = 
- \sum_{i \in {\rm inj}} p_{u,\rm i1}(0) \dot q^{\downarrow}_i(t) 
-  \sum_{i \in {\rm inj}} \int_0^t \dot p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) 

\sum_{k \in {\rm prod}} \int_0^t \dot p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) = 
-  \sum_{i \in {\rm inj}} \int_0^t \dot p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) 

 \int_0^t \dot p_{u,\rm 11}(t-\tau) dq^{\uparrow}_1(\tau) + \sum_{k \neq 1 \in {\rm prod}} \int_0^t \dot p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) = 
-  \int_0^t \dot p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\tau)  - \sum_{i \neq 0 \in {\rm inj}} \int_0^t \dot p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) 

\dot p_{u,\rm 11}(t) \delta q^{\uparrow}_1 + \sum_{k \neq 1 \in {\rm prod}} \int_0^t \dot p_{u,\rm k1}(t-\tau) dq^{\uparrow}_k(\tau) = 
-  \dot p_{u,\rm 01}(t) \delta q^{\downarrow}_0  - \sum_{i \neq 0 \in {\rm inj}} \int_0^t \dot p_{u,\rm i1}(t-\tau) dq^{\downarrow}_i(\tau) 

Again it is important to note a difference between

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