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?
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?
Show Hint
The additional pressure drop caused by damage is confined entirely within the damaged ring and dies out at its outer edge, \(r_s\).
Updated On: Jul 28, 2026
- \(\Delta p(r_s) = \Delta p(r_w)\)
- \(\Delta p(r_s) = 0\)
- \(\Delta p(r_s) = 2\,\Delta p(r_w)\)
- \(\Delta p(r_s) = 4\,\Delta p(r_w)\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understand where the extra pressure drop comes from:
A damaged well has lower permeability in the ring of rock between \(r_w\) and \(r_s\) compared with the rest of the reservoir. Fluid moving through this ring loses more energy than it would if the whole reservoir had the undamaged permeability. This extra energy loss shows up as an additional pressure drop \(\Delta p(r)\), defined as the difference between the actual pressure profile and the pressure profile that would exist if there were no damage at all.
Step 2: Identify the region where this additional drop exists:
Since the rock beyond \(r_s\) has the same, undamaged permeability that is used as the reference for \(\Delta p(r)\), no extra resistance is offered to flow once the fluid is outside the damaged ring. So the additional pressure drop can only build up inside the damaged ring, that is for \(r_w \le r \le r_s\). It is largest at the wellbore face \(r = r_w\), because the fluid has crossed the full thickness of the damaged ring by the time it reaches the well, and it becomes progressively smaller as \(r\) moves outward toward \(r_s\), since less of the damaged ring remains ahead of the flowing fluid.
Step 3: Apply the condition at \(r = r_s\), the outer edge of the damage:
At \(r = r_s\) the fluid has not yet entered the damaged ring, so it has not experienced any of the extra flow resistance caused by the damage. This means the additional pressure drop must be zero exactly at \(r_s\), and it accumulates only as the fluid moves further inward, reaching its full value \(\Delta p(r_w) = \dfrac{141.2\,qB\mu}{kh}\,S\) at the wellbore face, where S is the skin factor.
Step 4: Rule out the remaining options:
Option A claims \(\Delta p(r_s) = \Delta p(r_w)\), which would mean the full skin effect is already present at the outer boundary of the damaged zone, contradicting the fact that the drop builds up from zero at \(r_s\) to a maximum at \(r_w\). Options C and D claim \(\Delta p(r_s)\) is two or four times \(\Delta p(r_w)\), which is impossible since \(\Delta p(r_w)\) is the maximum value of this additional drop and \(\Delta p(r)\) decreases monotonically as \(r\) increases from \(r_w\) to \(r_s\), never exceeding the wellbore value.
Final Answer:
\[ \boxed{\Delta p(r_s) = 0} \]
A damaged well has lower permeability in the ring of rock between \(r_w\) and \(r_s\) compared with the rest of the reservoir. Fluid moving through this ring loses more energy than it would if the whole reservoir had the undamaged permeability. This extra energy loss shows up as an additional pressure drop \(\Delta p(r)\), defined as the difference between the actual pressure profile and the pressure profile that would exist if there were no damage at all.
Step 2: Identify the region where this additional drop exists:
Since the rock beyond \(r_s\) has the same, undamaged permeability that is used as the reference for \(\Delta p(r)\), no extra resistance is offered to flow once the fluid is outside the damaged ring. So the additional pressure drop can only build up inside the damaged ring, that is for \(r_w \le r \le r_s\). It is largest at the wellbore face \(r = r_w\), because the fluid has crossed the full thickness of the damaged ring by the time it reaches the well, and it becomes progressively smaller as \(r\) moves outward toward \(r_s\), since less of the damaged ring remains ahead of the flowing fluid.
Step 3: Apply the condition at \(r = r_s\), the outer edge of the damage:
At \(r = r_s\) the fluid has not yet entered the damaged ring, so it has not experienced any of the extra flow resistance caused by the damage. This means the additional pressure drop must be zero exactly at \(r_s\), and it accumulates only as the fluid moves further inward, reaching its full value \(\Delta p(r_w) = \dfrac{141.2\,qB\mu}{kh}\,S\) at the wellbore face, where S is the skin factor.
Step 4: Rule out the remaining options:
Option A claims \(\Delta p(r_s) = \Delta p(r_w)\), which would mean the full skin effect is already present at the outer boundary of the damaged zone, contradicting the fact that the drop builds up from zero at \(r_s\) to a maximum at \(r_w\). Options C and D claim \(\Delta p(r_s)\) is two or four times \(\Delta p(r_w)\), which is impossible since \(\Delta p(r_w)\) is the maximum value of this additional drop and \(\Delta p(r)\) decreases monotonically as \(r\) increases from \(r_w\) to \(r_s\), never exceeding the wellbore value.
Final Answer:
\[ \boxed{\Delta p(r_s) = 0} \]
Was this answer helpful?
0
0
Top GATE PE Petroleum Engineering Questions
- The maximum Polished Rod Load (PRL) in the operation of the sucker rod pump is observed near the
- Which ONE of the following rocks shows the highest reading in the natural gamma ray log?
Three different pressure profiles are shown in the figure. CSD is Casing Setting Depth.
Match the entries in GROUP I with the entries in GROUP II.
Match the well logging methods in GROUP I with their corresponding measured parameters in GROUP II:
- Classification of kerogen is based on the relative amount of Carbon (C), Hydrogen (H) and Oxygen (O).
Which ONE of the following options is CORRECT about Type II kerogen?
View More Questions
Top GATE PE Well Testing Questions
- 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?
- A circular reservoir with a well in the centre is producing oil at a constant rate under pseudo steady state. The rate of pressure change measured at the wellbore is 1.0 psi/day. Pressure at the external radius of the reservoir at a certain time was 3500 psi. After 500 days, the pressure (in psi) at the external radius will be __________.
- Pressure change and derivative type curves (Bourdet type curves) are used in well test analysis. Which of the following statements is/are CORRECT about these type curves?
- Which of the following statements is/are CORRECT about the skin factor?
- A drawdown test is analyzed to find the permeability and skin factor of a reservoir using a semi-log plot, where the flowing bottomhole pressure (in psi) is plotted on a linear scale against time (in hours). The flow rate is 1500 STB/day, the oil formation volume factor is 1.2 RB/STB, and the oil viscosity is 1 cP. The thickness of the producing bed is 10 ft. The slope of the straight line fitting the data in the middle time region of the semi-log plot is 60 psi/log cycle.
The permeability of the reservoir (in mD) is _______ (rounded off to one decimal place).
View More Questions
Top GATE PE Questions
- Four fair coins are tossed simultaneously. The probability that at least one tail turns up is:
- Let \( \vec{A} = 2\ell - \mathbf{j} + \mathbf{k} \) and \( \vec{B} = \ell + \mathbf{f} \), where \( \ell, \mathbf{f}, \) and \( \mathbf{k} \) are unit vectors. The projection of \( \vec{B} \) on \( \vec{A} \) is:
- The value of \( \lim_{x \to \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right) \) is:
- Which ONE of the following CANNOT be obtained from pressure transient analysis in well testing?
- The primary objective of the electrostatic grid in electrostatic heater-treater used in crude oil processing is to
View More Questions