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Consider a well-reservoir system (Fig. 1) consisting of:

• 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 W

  ₂

  ₀ with total sandface flowrate 

  LaTeX Math Inline
  body

  q_2

  q^{\downarrow}_0(t) <0

  , supporting pressure in reservoir volume

  LaTeX Math Inline
  body V_{\phi,

  2

  0}

   which includes the drainage volume 
   

The injection drainage volume  

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 includes the drainage volume The drainage volume difference

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or may be bigger  

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

where

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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^{\downarrow}_0

where

p_1(t

p_1(t) = p_i - \int_0^t p_{u,\rm 11}(t-\tau) dq_1(\tau) - \int_0^t p_{u,\rm 21}(t-\tau) dq_2(\tau) = p_i - \int_0^t p_{u,\rm 2111}(t-\tau) dq_2dq^{\uparrow}_1(\tau)

 - \int_0^t p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\tau) = p_i - \int_0^t p_{u,\rm 01}(t-\tau) dq^{\downarrow}_0(\tau)

\delta p_1(tq_0(\tau) = p_1(t)-p_i = - \int_0^t p_{u,\rm 21}(t-\tau)  \delta q_2 \cdot \delta\delta q^{\downarrow}_0 \cdot H(\tau)

H(\tau) = \begin{cases} 0, &  d\tau <0 =\\ - p_{u,\rm 21}(t) \cdot  \delta q_2

Case #2 – Constant BHP: 

LaTeX Math Inline
body p_1 = \rm const

...

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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

...

...

...

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\delta p_1(t) =

...

 p_1(t)-p_i = - \

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In this case, flowrate response 

\delta q_1(t) = - \frac{\dot p_{u,\rm 21}(t)}{\dot p_{u,\rm 11}(t)} \cdot \delta q_2

where

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

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 

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

where

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

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

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

\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}_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  

For the finite-volume drain 

\delta q^{\uparrow}_1 / \delta 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 

\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 - \int_0^t p_{u,\rm 11}(t-\tau) dq_1(\tau) - \int_0^t p_{u,\rm 21}(t-\tau) dq_2(\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 21}(t-\tau) dq_2(\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 21}(0) \cdot q_2(t) -  \int_0^t \dot p_{u,\rm 21}(t-\tau) dq_2(\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) 

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) 

...

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\

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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

where

Substituting 

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 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

• CRM assumptions (constant PI, constant drainage volumes with no flow boundaries and constant total compressibility) – which may or may not take place and hence may or may not make CRM applicable in a specific case

and

• CRM concept of mismatching drainage volumes between producers and injectors which is just a terminology and does not exert restrictions on well-reservoir system

See also

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[UTR] 

\delta q_1 = -\delta q_2

which means that producer W₁ with constant BHP and finite-reservoir volume will eventually vary its rate at the same volume as injector W₂.

In case injector W₂ 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_{2k} = 1

unless there is thief injection outside the drain area of all producers and in this case: 

\sum_{k=1}^N f_{2k} < 1

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\delta q_1 =\sum_k f_{k1} \delta q_k

with constant coefficients

See also

[ DTR ] [ CTR ] [ Capacitance Resistance Model (CRM) ]

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