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LaTeX Math Block
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\frac{d Q^{\downarrow}_{AQ}}{dt} = q^{\downarrow}_{AQ}(t)= B \cdot \int_0^t W_{eD}(t - \tau) \dot p d\tau




LaTeX Math Block
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q^{\downarrow}_{AQ}(t)= C_a \cdot \frac{\partial p_a(t,r)}{\partial r} \bigg|_{r=r_edQ^{\downarrow}_{AQ}}{dt}



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pW_a{eD}(t, r)= p(0) + \int_0^t p_1 \left(0^{t} \frac{(t-\tau) \cdot \chi}{r_e^2}, \frac{r}{r_e} \right) \dot p(\tau) d\tau
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\dot p(\tau) = \frac{d p}{d \tau}partial p_1}{\partial r_D} \bigg|_{r_D = 1} dt_D 







LaTeX Math Block
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\frac{\partial p_1}{\partial t_D} =  \frac{\partial^2 p_1}{\partial r_D^2} + \frac{1}{r_D}\cdot \frac{\partial p_1}{\partial r_D}



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p_1(t_D = 0, r_D)= 0



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p_1(t_D, r_D=1) = 1




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\frac{\partial p_1}{\partial r_D} 
\bigg|_{(t_D, r_D=r_a/r_e)} = 0


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Expand
titleDerivation

Transient flow in Radial Composite Reservoir:


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\frac{\partial p_a}{\partial t} = \chi \cdot \left[ \frac{\partial^2 p_a}{\partial r^2} + \frac{1}{r}\cdot \frac{\partial p_a}{\partial r} \right]



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p_a(t = 0, r)= p(0)



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p_a(t, r=r_e) = p(t)



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\frac{\partial p_a}{\partial r} 
\bigg|_{(t, r=r_a)} = 0



Consider a pressure convolution:


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p_a(t, r) = p(0) + \int_0^t p_1 \left(\frac{(t-\tau) \cdot \chi}{r_e^2}, \frac{r}{r_e} \right) \dot p(\tau) d\tau



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\dot p(\tau) = \frac{d p}{d \tau}



One can easily check that

pressure from

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honors the whole set of equations
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and as such defines a unique solution of the above problem.

The the water flowrate at interface with oil reservoir will be:

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q^{\downarrow}_{AQ}(t)= C_a \cdot \frac{\partial p_a(t,r)}{\partial r} \bigg|_{r=r_e}

and cumulative flux:

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\frac{d Q^{\downarrow}_{AQ}}{dt} = q^{\downarrow}_{AQ}(t)



See Also

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Petroleum Industry / Upstream / Subsurface E&P Disciplines / Field Study & Modelling / Aquifer Drive / Aquifer Drive Models

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