In a drawdown test conducted on a well, $p_D$, $t_D$ and $C_D$ denote the dimensionless pressure, the dimensionless time and the dimensionless wellbore storage coefficient, respectively. In the early time of the drawdown test, the fluid produced at the surface results purely from the unloading (expansion) of fluid already stored in the wellbore, and no fluid from the formation has reached the well yet. This period is called the pure wellbore storage period. For this period, which of the following relationships is CORRECT?

Question

A circular shaped reservoir has an external radius $r_e$. A well of radius $r_w$ lies at the centre of the reservoir. The part of the reservoir around the wellbore, up to a radius $r_s$, is damaged due to drilling and completion operations, where $r_w < r_s < r_e$. This formation damage causes an additional pressure drop, $\Delta p(r)$, that varies with the radial distance $r$ from the well. Which of the following statements is TRUE about the additional pressure drop at the radial location $r_s$ due to the formation damage?

Answer 1

Approach using the log log diagnostic unit slope line:
A standard way pressure transient analysts recognize pure wellbore storage is by plotting $\log p_D$ against $\log t_D$ for early time drawdown data. During this period the data always falls on a straight line inclined at exactly 45 degrees, commonly called the unit slope line, because $p_D$ and $t_D$ are directly proportional to one another in this regime.
Writing the unit slope line algebraically:
A straight line of slope 1 on a log log plot means $\log p_D = \log t_D + \text{constant}$, which is the same as saying $p_D = (\text{constant}) \times t_D$. Physically, this constant of proportionality for the wellbore storage period is known from the material balance of the wellbore fluid to be $1/C_D$, so the unit slope line is described by $p_D = t_D / C_D$.
Checking the physical reasonableness:
This makes sense because a well with a very small storage coefficient $C_D$ (for example a well with a packer set near the perforations) will show a larger $p_D$ for a given $t_D$, since the wellbore has almost no capacity to buffer the pressure change, causing pressure to fall quickly. A well with a large $C_D$ (a long, fluid filled wellbore with no packer) buffers the pressure change more, so $p_D$ stays smaller for the same $t_D$. This inverse dependence of $p_D$ on $C_D$ at fixed $t_D$ is exactly what the relation $p_D = t_D/C_D$ predicts.
Solving for the required variable:
Cross multiplying the unit slope relation $p_D = t_D/C_D$ by $C_D$ on both sides isolates $t_D$ on one side, giving $t_D = p_D C_D$, which is the relation asked for in the question.
Final Answer:
\[ \boxed{t_D = p_D\,C_D} \]

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The Correct Option is A

Solution and Explanation

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