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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 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) 

Again it is important to note a difference between

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