LSU PETE 1010 REAL EXAM
QUESTIONS AND ANSWERS TOP
RATED STUDY MATERIAL PASS
GUARANTEE LATEST VERSION
LSU PETE 1010 REAL EXAM QUESTIONS AND |
Comprehensive Examination
LSU PETE 1010 REAL EXAM QUESTIONS AND
ANSWERS TOP RATED STUDY MATERIAL PASS
GUARANTEE LATEST VERSION 2026-2027!! actual exam
INTRODUCTION
This document is an important and very helpful preparation
resource for LSU PETE 1010 REAL EXAM QUESTIONS AND
ANSWERS TOP RATED STUDY MATERIAL PASS
GUARANTEE LATEST VERSION 2026-2027!!. Every item
uses a focused exam stem, four response options, a keyed
response, and a clear rationale, making it easier to understand
the subject, spot distractors, and build real exam confidence.
Page 1
, Question 1
A core plug has a bulk volume of 25 cm³ and a grain volume of 20 cm³. If the
pore space is fully saturated with a brine of density 1.05 g/cm³ and the plug's
dry mass is 53 g, what is the most likely interpretation of the porosity and the
dominant porosity type?
A. 20% porosity; primary intergranular porosity
B. 20% porosity; secondary moldic porosity
C. 25% porosity; primary intergranular porosity
D. 25% porosity; fracture porosity
Correct Answer: A - 20% porosity; primary intergranular
porosity
RATIONALE
Porosity = (2520)/25 = 0.20 (20%). The grain density is 53/20 = 2.65
g/cm³, consistent with quartz-dominated sandstone, which typically
exhibits primary intergranular porosity. Moldic (secondary) porosity
would imply a different diagenetic history, and fracture porosity
would yield a much lower bulk porosity with high permeability
contrast.
Question 2
During drilling, a well takes a 20-bbl kick while the bit is at 10,000 ft TVD.
The mud weight is 10.0 ppg, and the formation pressure gradient is 0.52 psi/ft.
What is the minimum kill mud weight required to balance the formation
pressure?
A. 10.0 ppg
B. 10.5 ppg
C. 11.0 ppg
D. 12.0 ppg
Correct Answer: C - 11.0 ppg
Page 2
, RATIONALE
Formation pressure = 0.52 psi/ft × 10,000 ft = 5,200 psi. Hydrostatic
pressure with 10.0 ppg mud = 0.052 × 10.0 × 10,000 = 5,200 psi, so
the well is exactly balanced; however, the kick indicates the formation
pressure is slightly higher or the mud weight is insufficient due to the
influx. To provide a safe overbalance, kill mud weight =
5,200/(0.052×10,000) = 10.0 ppg, but standard practice adds a trip
margin, yielding ~11.0 ppg. Option C is the minimum that provides a
0.5 ppg trip margin.
Question 3
A sandstone reservoir has a formation water resistivity (Rw) of 0.05 ohm-m at
reservoir temperature, and the Archie equation parameters are a=1, m=2, n=2.
If the measured true resistivity (Rt) is 20 ohm-m and porosity is 20%, what is
the water saturation (Sw)?
A. 0.25
B. 0.35
C. 0.50
D. 0.75
Correct Answer: A - 0.25
RATIONALE
Archie: Sw = [(a-Rw)/(^m-Rt)]^(1/n) = [(1×0.05)/(0.2^2×20)]^(1/2) =
[0.05/(0.04×20)]^(1/2) = [0.05/0.8]^(1/2) = (0.0625)^(1/2) = 0.25.
Thus Sw = 25%, indicating a hydrocarbon-bearing zone. Options B, C,
and D result from misapplication of exponents or incorrect porosity
values.
Page 3
, Question 4
A vertical well produces from an undersaturated oil reservoir. The current
reservoir pressure is above bubble point, but the producing bottomhole flowing
pressure (Pwf) has fallen below bubble point. Which inflow performance
relationship (IPR) model best captures the resulting productivity loss?
A. Linear PI (productivity index) model
B. Vogel's equation for solution-gas drive reservoirs
C. Fetkovich's radial flow equation for oil
D. Standing's generalized IPR for two-phase flow
Correct Answer: B - Vogel's equation for solution-gas drive
reservoirs
RATIONALE
When Pwf falls below bubble point, free gas evolves in the reservoir,
reducing oil relative permeability and causing a nonlinear IPR. Vogel's
equation is specifically designed for solution-gas drive reservoirs and
predicts the curved IPR shape. Linear PI is only valid when Pwf > Pb;
Fetkovich's equation is for gas wells; Standing's method is for gas
condensate or volatile oils, not this scenario.
Question 5
A reservoir has an initial oil-in-place of 100 MMSTB. Primary recovery by
solution gas drive is estimated at 15%, and a waterflood is expected to add
20% of the original oil-in-place. If the waterflood is implemented after primary
depletion, what is the incremental oil recovered by waterflooding?
A. 5 MMSTB
B. 15 MMSTB
C. 20 MMSTB
D. 35 MMSTB
Correct Answer: C - 20 MMSTB
Page 4
QUESTIONS AND ANSWERS TOP
RATED STUDY MATERIAL PASS
GUARANTEE LATEST VERSION
LSU PETE 1010 REAL EXAM QUESTIONS AND |
Comprehensive Examination
LSU PETE 1010 REAL EXAM QUESTIONS AND
ANSWERS TOP RATED STUDY MATERIAL PASS
GUARANTEE LATEST VERSION 2026-2027!! actual exam
INTRODUCTION
This document is an important and very helpful preparation
resource for LSU PETE 1010 REAL EXAM QUESTIONS AND
ANSWERS TOP RATED STUDY MATERIAL PASS
GUARANTEE LATEST VERSION 2026-2027!!. Every item
uses a focused exam stem, four response options, a keyed
response, and a clear rationale, making it easier to understand
the subject, spot distractors, and build real exam confidence.
Page 1
, Question 1
A core plug has a bulk volume of 25 cm³ and a grain volume of 20 cm³. If the
pore space is fully saturated with a brine of density 1.05 g/cm³ and the plug's
dry mass is 53 g, what is the most likely interpretation of the porosity and the
dominant porosity type?
A. 20% porosity; primary intergranular porosity
B. 20% porosity; secondary moldic porosity
C. 25% porosity; primary intergranular porosity
D. 25% porosity; fracture porosity
Correct Answer: A - 20% porosity; primary intergranular
porosity
RATIONALE
Porosity = (2520)/25 = 0.20 (20%). The grain density is 53/20 = 2.65
g/cm³, consistent with quartz-dominated sandstone, which typically
exhibits primary intergranular porosity. Moldic (secondary) porosity
would imply a different diagenetic history, and fracture porosity
would yield a much lower bulk porosity with high permeability
contrast.
Question 2
During drilling, a well takes a 20-bbl kick while the bit is at 10,000 ft TVD.
The mud weight is 10.0 ppg, and the formation pressure gradient is 0.52 psi/ft.
What is the minimum kill mud weight required to balance the formation
pressure?
A. 10.0 ppg
B. 10.5 ppg
C. 11.0 ppg
D. 12.0 ppg
Correct Answer: C - 11.0 ppg
Page 2
, RATIONALE
Formation pressure = 0.52 psi/ft × 10,000 ft = 5,200 psi. Hydrostatic
pressure with 10.0 ppg mud = 0.052 × 10.0 × 10,000 = 5,200 psi, so
the well is exactly balanced; however, the kick indicates the formation
pressure is slightly higher or the mud weight is insufficient due to the
influx. To provide a safe overbalance, kill mud weight =
5,200/(0.052×10,000) = 10.0 ppg, but standard practice adds a trip
margin, yielding ~11.0 ppg. Option C is the minimum that provides a
0.5 ppg trip margin.
Question 3
A sandstone reservoir has a formation water resistivity (Rw) of 0.05 ohm-m at
reservoir temperature, and the Archie equation parameters are a=1, m=2, n=2.
If the measured true resistivity (Rt) is 20 ohm-m and porosity is 20%, what is
the water saturation (Sw)?
A. 0.25
B. 0.35
C. 0.50
D. 0.75
Correct Answer: A - 0.25
RATIONALE
Archie: Sw = [(a-Rw)/(^m-Rt)]^(1/n) = [(1×0.05)/(0.2^2×20)]^(1/2) =
[0.05/(0.04×20)]^(1/2) = [0.05/0.8]^(1/2) = (0.0625)^(1/2) = 0.25.
Thus Sw = 25%, indicating a hydrocarbon-bearing zone. Options B, C,
and D result from misapplication of exponents or incorrect porosity
values.
Page 3
, Question 4
A vertical well produces from an undersaturated oil reservoir. The current
reservoir pressure is above bubble point, but the producing bottomhole flowing
pressure (Pwf) has fallen below bubble point. Which inflow performance
relationship (IPR) model best captures the resulting productivity loss?
A. Linear PI (productivity index) model
B. Vogel's equation for solution-gas drive reservoirs
C. Fetkovich's radial flow equation for oil
D. Standing's generalized IPR for two-phase flow
Correct Answer: B - Vogel's equation for solution-gas drive
reservoirs
RATIONALE
When Pwf falls below bubble point, free gas evolves in the reservoir,
reducing oil relative permeability and causing a nonlinear IPR. Vogel's
equation is specifically designed for solution-gas drive reservoirs and
predicts the curved IPR shape. Linear PI is only valid when Pwf > Pb;
Fetkovich's equation is for gas wells; Standing's method is for gas
condensate or volatile oils, not this scenario.
Question 5
A reservoir has an initial oil-in-place of 100 MMSTB. Primary recovery by
solution gas drive is estimated at 15%, and a waterflood is expected to add
20% of the original oil-in-place. If the waterflood is implemented after primary
depletion, what is the incremental oil recovered by waterflooding?
A. 5 MMSTB
B. 15 MMSTB
C. 20 MMSTB
D. 35 MMSTB
Correct Answer: C - 20 MMSTB
Page 4
