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LaTeX Math Block

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

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_e}

LaTeX Math Block

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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

LaTeX Math Inline

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

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}

LaTeX Math Block

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p_1(t = 0, r_D)= 0

LaTeX Math Block

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p_1(t, r_D=1) = 1

LaTeX Math Block

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\frac{\partial p_1}{\partial r_D} \bigg|_ {r_D=r_a/r_e} = 0

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title	Derivation

Transient flow in Radial Composite Reservoir:

LaTeX Math Block

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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]

LaTeX Math Block

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

LaTeX Math Block

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

LaTeX Math Block

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\frac{\partial p_a}{\partial r} \bigg|_ {r=r_a} = 0

One can easily check that pressure from

LaTeX Math Block Reference

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honors the whole set of equations

LaTeX Math Block Reference

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–

LaTeX Math Block Reference

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and as such defines a unique solution of the above problem.

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