Question:

Four curves designated as I, II, III, and IV are shown in the figure. \(q_f\) is the bottomhole oil flow rate, \(q_s\) is the surface oil flow rate, and \(B_o\) is the oil formation volume factor. Which ONE of the following curves depicts the minimum wellbore storage effect? \begin{center} \includegraphics[width=0.7\textwidth]{04.jpeg} \end{center}

Show Hint

In wellbore–storage diagnostics, "\(\to 1\) faster" on a \(q_f/(q_s B_o)\) vs time plot means smaller storage. Large storage gives a long early-time lag before the ratio reaches (or hovers near) 1.
Updated On: Aug 24, 2025
  • I
  • II
  • III
  • IV
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation


Step 1: Interpret the plotted quantity.
The ordinate is \( \dfrac{q_f}{q_s\,B_o} \).
If there were \emph{no} wellbore storage, then whatever rate leaves the reservoir at the sandface (\(q_f\)) appears at surface as \(q_s B_o\) (after volumetric conversion). Hence, \[ \dfrac{q_f}{q_s\,B_o} \xrightarrow[\text{no storage}]{} 1 \quad \text{(immediately).} \] Step 2: Effect of wellbore storage.
With wellbore storage, early–time surface flow is partly (or wholly) used to fill/empty the wellbore, so \(q_s B_o\) initially \emph{lags} behind \(q_f\). Therefore the ratio \(\dfrac{q_f}{q_s B_o}\) starts below 1 and gradually approaches 1 as the storage effect dies out.
\(\Rightarrow\) The weaker (smaller) the storage, the faster the curve rises to 1 and the less pronounced the early–time lag. Step 3: Select the curve with minimum storage effect.
Among I, II, III, IV, the curve that climbs to \(1.0\) \emph{most rapidly} (smallest lag) corresponds to the minimum wellbore storage. From the figure, curve IV rises the fastest and hugs \(1.0\) earliest; curve I is the slowest (largest storage). Final Answer:
\[ \boxed{\text{IV}} \]
Was this answer helpful?
0
0

Top Questions on Reservoir and channel routing

View More Questions

Questions Asked in GATE PE exam

View More Questions