...

• 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 q_2q^{\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

 q^{\downarrow}_0 \cdot

 H(\tau)

H(\delta p_1(t) = p_1(t)-p_i = - \int_0^t p_{u,\rm 21}(t-\tau)  \delta q_0 \cdot \delta(\tau) \,  d\tau = - p_{u,\rm 01}(t) \cdot  \delta q_0

Case #2 – Constant BHP: 

LaTeX Math Inline
body p_1 = \rm const

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 01}(t)}{\dot p_{u,\rm 11}(t)} \cdot \delta q_0

where

...

...

...

...

...

time derivative of drawdown pressure transient response (DTR) in producer W₁ to the unit-rate production in the same well

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

d q^{\downarrow}_0(\tau) = q_0'(\tau) d\tau = \delta q^{\downarrow}_0 \cdot H'(\tau) \,  d\tau = \delta q^{\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 q^{\downarrow}_0 \cdot \delta(\tau) \,  d\tau = - p_{u,\rm 01}(t) \cdot  \delta q^{\downarrow}_0

Case #2 – Constant BHP: 

LaTeX Math Inline
body p_1 = \rm const

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

In this case, flowrate response 

Consider a pressure convolution equation for the above 2-wells system with constant BHP:

 p_{u,\rm 

01}(t

)}{\dot p_{u,\rm 

11}(t

)} 

The time derivative is going to be zero as the BHP in producer W₁ stays constant at all times:

\cdot \delta q^{\downarrow}_0

\delta q^{\uparrow}_1(t) = -^{\uparrow} \frac{\dot

where

...

...

...

p_{u,\rm 11}(0) \cdot q_1(t) +1(t) = p_i - \int_0^t \dot p_{u,\rm 11}(t-\tau) dqdq^{\uparrow}_1(\tau)  = - p_{u,\rm 01}(0) \cdot q_0(t) -  \int_0^t \dot p_{u,\rm 01}(t-\tau) dqdq^{\downarrow}_0(\tau) = \rm const

\dot p_1(t) = - \left( \int_0^t p_{u,\rm

 11}(

\int_0^t \dot p_{u,\rm 11}(t-t-\tau) dq^{\uparrow}_1(\tau) dq_1(\tauright)  = ^{\cdot} - \left( \int_0^t \dot p_{u,\rm 2101}(t-\tau) dq_2dq^{\downarrow}_0(\tau) 

\right)^{\cdot} = 0

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

\int_0^t \dot p_{u,\rm 1101}(t-\tau)  \delta q_1 \cdot \delta(\tau) \,  d\tau  = -  \int_0^t \dot dq^{\downarrow}_0(\tau) 

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

...

...

...

...

...

\delta q_1 / \delta q_0 = - f_{01} = - \frac{V_{\phi, 1}}{ V_{\phi, 0}} = \rm const

...

...

\int_0^t \dot 

p_{u,\rm 11}(t-\tau) 

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

-\tau) \

delta q^{\

downarrow}

where

_0 \cdot \delta(\tau) \,  d\tau 

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

which leads to Substituting 

...

...

For the finite-volume drain 

...

...

...

...

...

\delta qq^{\uparrow}_1 =/ -\delta q_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.

In case injector W₀ supports many producers {W₁ .. W_N} then all injection shares towards producers are going to sum up to a unit value:

\sum_{k=1}^N f_{0k} = 1

with constant coefficients

\sum_{k=1}^N f_{0k} < 1

...

...

\delta q_1 =-\sum_i f_{i1} \delta q_i

with constant coefficients

...

...

q^{\downarrow}_0 = - f_{01} = - \frac{V_{\phi, 1}}{ V_{\phi, 0}} = \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 01}(t \rightarrow \infty) \rightarrow \frac{t}{c_t V_{\phi,0}}

...

In case injector W₀ supports only one producer W₁, then both wells drain the same reservoir volume fk1 make one of the key assumptions in Capacitance Resistance Model (CRM).It is important to note that CRM assumption that injector W₀ may drain bigger volume than producer W₁  

p_{u,\rm ik}(t \rightarrow \infty ) \rightarrow \frac{t}{\rm RS}, \quad \forall i \in N_{\rm inj}, k \in N.

where 

...

\delta q^{\uparrow}_1 = -\delta q^{\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.

In case injector W₀ supports many producers {W₁ .. W_N} then all injection shares towards producers are going to sum up to a unit value:

\sum_{k=1}^N f_{0k} = 1 	\quad \Leftrightarrow \quad \sum_{k=1}^N V_{\phi,k} = V_{\phi,0}

with constant coefficients

\sum_{k=1}^N f_{0k} < 1	\quad \Leftrightarrow \quad \sum_{k=1}^N V_{\phi,k} < V_{\phi,0}

If pressure around producer W₁ is supported by several injectors 

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

with constant coefficients

The equations 

It is important to note that CRM assumption that injector W₀ may drain bigger volume than producer W₁  

When wells (producers and injectors) are placed into the same connected reservoir volume they drain the same total volume 

p_{u,\rm ik}(t \rightarrow \infty ) \rightarrow \frac{t}{\rm RS}, \quad \forall i \in N_{\rm inj}, k \in N.

where 

Moreover, if each well is placed in different reservoir volumes which are only connected through wellbores then again they will all drain the same volume which is the sum of all connected volumes through the wellbores and all UTRs will again trend to the same LTR asymptotic.

In order to relate true UTRs (from numerical grid simulations or from deconvolution) to the CRM injection share constants 

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 a convolution tests to account for the impact from {W₂ .. W_N} production and from {W_-1 .. W_-M} on to CTR_01

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

         : 

         LaTeX Math Inline
         body \displaystyle p^*_{u, 01}(t) = p_{u, 01}(t) + \sum_{i \neq 0 \in {\rm inj}} p_{u, i1}(t) \cdot q^{\downarrow}_i(t)

      2. Perform two convolution tests to account for the impact from {W₂ .. W_N} production on to DTR_11

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

          and CTR_01

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

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

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

             , meaning that all injectors W₀ are shut-down and all producers were working with their historical rates 

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

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

             LaTeX Math Inline
             body \displaystyle p^*_{u, 01}(t) = p_{u, 01}(t) + \sum_{i \neq 0 \in {\rm inj}} p_{u, i1}(t) \cdot q^{\downarrow}_i(t)

             , 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
      3. 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)
3. 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) 

...

In order to relate true UTRs (from numerical grid simulations or from deconvolution) to the CRM injection share constants 

...

...

...

1. Select producer W₁
   1. Perform two convolution tests to account for the impact from {W₂ .. W_N} production on to DTR_11

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

       and CTR_01

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

      :
      • Test #1 – DTR_11
        
        • Calculate historically-averaged rate for each producer: 

          LaTeX Math Inline
          body \displaystyle q^*_k = \frac{1}{N_k} \sum_{m=1}^{N_k} q_k(t_m)

        • Calculate interfering DTR_11: 

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

          , 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 – CTR_01
        • Calculate historically-averaged rate for each producer: 

          LaTeX Math Inline
          body \displaystyle q^*_k = \frac{1}{N_k} \sum_{m=1}^{N_k} q_k(t_m)

        • Calculate interfering CTR_01: 

          LaTeX Math Inline
          body \displaystyle p^*_{u, 01}(t) = p_{u, 01}(t) + \sum_{k} p_{u, k1}(t) \cdot q^*_k

          , meaning that injector W₀ is working with unit-rate and all producers are working with constant rates 

          LaTeX Math Inline
          body q^*_k

   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)

...

Again it is important to note a difference between

...