Introduction to Petroleum Geology
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Geology (GLGY 577-UCAL) Midterm Exam
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Question 1 |
A | Use the Ro line that is most centered to the all other Ro lines. |
B | Calibrate the Ro line with the core or well logs. |
C | Generate the most two extreme Ro lines on the Pickett Plot with the maximum Rt and minimum Rt as your reference bounds. |
D | Generate several parallel Ro lines on the Pickett Plot. |
E | Generate several Ro lines on the Pickett Plot. |
Question 2 |
A | carbon-oxygen chains. |
B | kerogen. |
C | hydrocarbons. |
D | reservoir. |
E | traps. |
Question 3 |
A | In Alberta, we commonly find oil-wet rocks in the subsurface. |
B | Computer algorithms for reservoir characterization models heavily depends on the analytical variables as opposed to empirical variables. |
C | Typically, as you move closer to mountains, frequency of stratigraphic traps increases. |
D | Typically, as you move closer to mountains, frequency of structural traps increases. |
E | In Alberta, lens traps are the most common type of hydrocarbon traps. |
Question 4 |
A | Density tools |
B | Gamma Ray tools |
C | Neutron tools |
D | Resistivity tools |
Question 5 |
A | Oil-based mud |
B | Denial-based mud |
C | Acids-based mud |
D | Synthetic mud |
Question 6 |
A | The play is an area of interest for petroleum exploration while the prospect is an area within a play that is expected to produce hydrocarbons. |
B | None of these answers are correct. |
C | The play is an area of interest for petroleum exploration and a prospect is the lowest point on the trap which can hold petroleum. |
D | The play is an area where common petroleum system characteristics exists, while the prospect is an area where the hydrocarbons can be extracted economically. |
E | The play is a large area where hydrocarbons may be found, while the prospect is an area where there is a current production in hydrocarbons. |
Question 7 |
A | Porosity and formation factor is directly related to each other. |
B | Clean granular formations have a lower formation factor than clean carbonate formations. |
C | Pore stucture is directly related to the formation factor. |
D | Formation factor is inversly related to tourtousity factor. |
Question 8 |
A | Unaltered organic matter |
B | Diagenesis |
C | Zone of gas formation |
D | Initial burial |
E | Catagenesis |
Question 9 |
A | Net sand content Hint: You need the scale for this calculation. |
B | Coarsening upward sequence |
C | Very high gamma ray kick Hint: You need the scale for this interpretation. |
D | Coarsening downward sequence |
Question 10 |
A | Exploration |
B | Reserves Growth |
C | Development |
D | Production |
E | Appraisal |
Question 11 |
A | F = 0.5 |
B | F = 2.2 |
C | F = 0.2 |
D | F = 0.9 |
E | F = 0.7 |
Question 12 |
A | To prevent well blowouts. |
B | To provide support for the Kelly Bushing. |
C | To reduce fractures near the neck. |
D | To protect aquifers from contamination. |
Question 13 |
A | Pinchout trap |
B | Salt dome trap |
C | Unconformity trap |
D | Lens trap |
E | Anticline trap |
F | Fault trap |
Question 14 |
A | Decrease friction between the pore walls and the crude oil. |
B | Increase pressure within the chamber hence driving the crude oil out of the system. |
C | Increase the conductivity between pores. |
D | It will reduce viscosity of crude oil hence allowing it to be pumped. |
E | Increase pressure within the chamber which facilitate extraction. |
Question 15 |
BHT (bottom hole temperature) = 255o F
Surface temperature = 75o F
Depth which BHT was measured: 11400 ft
A | ~ 0.015oF/ft |
B | ~ 0.016oF/ft |
C | ~ 0.029oF/ft |
D | ~ 0.018oF/ft |
E | ~ 0.028oF/ft |
Question 16 |
A | 3D seismic |
B | 4D seismic |
C | 2D seismic |
D | 4C seismic |
Question 17 |
A | increases , decrease |
B | They are not related to each other. |
C | decarses , decrease |
D | increase , increase |
Question 18 |
A | Top of the major formation |
B | Bottom of the major formation |
C | Highest point of the hydrocarbon pool |
D | Position of the Kelly Bushing |
E | Lowest point of the hydrocarbon pool |
Question 19 |
A | high , low , arbitrary |
B | high , low , at 15 API |
C | low , high , at 15 API |
D | low , high , at 75 API |
E | high , low , at 75 API |
F | low , high , arbitrary |
Question 20 |
A | Rw is obtained from the upper boundary (green line above) and the Sw is obtained from the upper limit of the lower boundary (red line above). |
B | You cannot take these measurements from a Pickett Plot. You must use a well log to determine such properties. |
C | Rw is obtained from the upper limit of the lower boundary (red line above) and So maximum is obtained from the upper boundary (green line above). |
D | Rw is the intersect point between the lower boundary (green line above) and read off the x-values straight down (first brown dash line down). The So maximum is the x-value of the upper most point and read off the x-axis (second brown dash line). |
E | So maximum is the intersect point between the lower boundary (green line above) and read off the x-values straight down (first brown dash line down). The Rw is the x-value of the upper most point and read off the x-axis (second brown dash line). |
Question 21 |
A | is directly proportional to the resistivity. |
B | is most commonly found to be equals to 1.0. |
C | is highly dependent on the formation factor. |
D | varies between 0.35 and 4.5. |
Question 22 |
A | It measures the true density of the material. |
B | Typically, silicon has a higher neutron reading than carbon. |
C | It measures the slow-down length of neutrons when they interacts with atoms. |
D | The thermal phase of the tool uses energy at around 10 eV |
E | A typical tool uses uranium as the energy source. |
Question 23 |
A | green , blue , orange |
B | blue , green , orange |
C | orange , blue , green |
D | blue , orange , green |
E | orange , green , blue |
F | green , orange , blue |
Question 24 |
A | Point bars |
B | Tidal flats |
C | Deep marine |
D | Beach marine |
Question 25 |
A | the beds are extremly thick. |
B | there are multiple sleeply dippinng beds. |
C | the beds has no dip, hence almost horizontal. |
D | the seismic signals have to travel through fluids (eg. oceanic seismic). |
Question 26 |
A | The topography of the area may be distorted. |
B | The calculated net sand volume may be incorrect. |
C | The obvious features of the well logs may be ignored. |
D | The key well logs features may not be able to correlate further away from the reference wells. |
E | The overall assessment of the geology may be completely wrong. |
Question 27 |
A | 200 Hz |
B | 2 kHz |
C | 20 MHz |
D | 2 MHz |
E | 100 kHz |
Question 28 |
A | It is easier to frack on soft sediments. |
B | It is a relatively very new technology. |
C | It may result in earthquakes. |
D | It will not affect aquifers. |
Question 29 |
A | Fluvial point bar deposits |
B | Progradating delta deposits |
C | Progradating marine shelf |
D | Deep marine sediments |
Question 30 |
A | 15% |
B | 12% |
C | 10% |
D | 5% |
E | 6% |
Question 31 |
net sand = 15.5 m
shale = 5.3 m
gross thickness = 26.3 m
A | Net:Gross = 59 and Sand:Shale = 29 |
B | Net:Gross = 0.59 and Sand:Shale = 0.34 |
C | Net:Gross = 59 and Sand:Shale = 34 |
D | Net:Gross = 0.59 and Sand:Shale = 2.92 |
E | Net:Gross = 0.79 and Sand:Shale = 0.34 |
Question 32 |
A | ~ 2.5 m |
B | ~ 80 cm |
C | ~ 50 cm |
D | ~ 1.5 m |
E | ~ 3 m |
Question 33 |
A | Crosswell seismic data acquisition |
B | 4D seismic data acquisition |
C | 2D seismic data acquisition |
D | 3D seismic data acquisition |
Question 34 |
A | Lignin and Carbohydrates |
B | Carbohydrates and Proteins |
C | Proteins and Lignin |
D | Carbohydrates, Proteins and Lipids |
E | Lignin, Carbohydrates, Proteins and Lipids |
F | Proteins and Lipids |
Question 35 |
A | 2 - 3 m |
B | 1 - 5 m |
C | 0.5 - 1.5 m |
D | 0.8 - 1.5 m |
E | 0.5 - 2m |
Question 36 |
A | transitional metals and carbons. |
B | carbon-oxygen bonds. |
C | carbon-hydrogen bonds. |
D | kerogen. |
E | inorganic compounds. |
Question 37 |
A | 0.09 |
B | 1.00 |
C | 0.65 |
D | 1.08 |
E | 0.08 |
Question 38 |
A | Caliper log |
B | Sonic log |
C | Gamma ray log |
D | Dipmeter log |
E | Neuron log |
Question 39 |
A | Residuum |
B | Heavy gas oil |
C | Gasoline |
D | Lubricating oil |
E | Diesel fuel |
F | Kerosene |
Question 40 |
A | The m and porosity are directly related to each other. |
B | All or other things being equal, as m increases, the prore geometry become complicated. |
C | All or other things being equal, increase in m will result in decrease in a. |
D | Value of m is independent of the surface area of the pores. |
E | All or other things being equal, increase in m will result in increase in formation factor, F. |
Question 41 |
A | ~ 80 cm |
B | ~ 40 cm |
C | ~ 10 cm |
D | ~ 30 cm |
Question 42 |
A | Low net pay regions |
B | Valleys |
C | Hills or mountains |
D | High concentration of hydrocarbons |
E | High net pay regions |
Question 43 |
A | Resistivity of the formation water is same above and below the contact. |
B | Oil and gas acts as conduits for electrical signals. |
C | Formation must be clean and free of clay. |
D | Formation is completely saturacted with fulids. |
E | Proposity must be the same both above and below the oil-water contect. |
Question 44 |
A | Rmc |
B | Ro |
C | Rw |
D | Rt |
E | Rm |
Question 45 |
A | This part of the core is most likely harder and stronger relative to other areas of the core. |
B | This part of the core is most likely saturated with hydrocarbons. |
C | This part of the core is most likely saturated with organic materials. |
D | This part of the core is most likely have a very high permeability. |
E | This part of the core is most likely have a very low permeability. |
F | This part of the core is most likely concentrated with bitumen. |
Question 46 |
A | Directionally focused tools. |
B | High concentration of formation fluids. |
C | Low viscosity mud. |
D | Low mud invasion zone. |
Question 47 |
A | Geostatistical models such as kriging |
B | Core samples |
C | Gravity readings |
D | Wireline well logs |
E | Seismic sections |
Question 48 |
A | Epi-thermal phase of the scatter |
B | Electric energy |
C | Thermal phase of the scatter |
D | Uranium source |
Question 49 |
A | Natural gas produced from reservoirs after the peak production of the play. In other words, low yield gas produced after production has been in place for a period of time. |
B | Natural gas produced from reservoirs that are under very high pressures making it tight and easier to extract. |
C | Natural gas produced from reservoir rocks with such low permeability that massive hydraulic fracturing is necessary to produce the well at economic rates. |
D | Natural gas produced from reservoir rocks that are interbeded with alternating permeable and non-permeable layers. |
Question 50 |
A | Section 4 |
B | Section 5 |
C | Section 10 |
D | Section 3 |
E | Section 9 |
Question 51 |
A | Increase with sand interbedding. |
B | Decrease with depth. |
C | Decrease with increase in clay interbedding. |
D | Becomes almost zero with hydrocarbon saturation. |
E | Decrease with increase in hydrocarbon saturation. |
Question 52 |
A | 120.0 Ωm |
B | 0.102 Ωm |
C | 102.0 Ωm |
D | 9.80 Ωm |
Question 53 |
A | Porosity and permeability of the core. |
B | Total length of the recovered core. |
C | Well diversion trajectory angles. |
D | Length of the drill string. |
Question 54 |
A | ~ 3.50 |
B | ~ 0.64 |
C | ~ 4.00 |
D | None of the answers are correct. |
E | ~ 0.20 |
Net sand interval = 3550 m - 1000 m = 2550 m (read off the GR curve)
Hence, net:gross = 2550 m/4000 m = 0.6375
Question 55 |
A | Very low formation pressures. |
B | Very high formation pressures. |
C | Very high porosity in the formation. |
D | Very high permeability in the formation. |
Question 56 |
A | ~ 1.2 to 1.25 m |
B | ~30 to 50 cm |
C | ~ 1.35 to 1.75 m |
D | ~ 1.1 to 2 m |
E | ~ 26 to 30 cm |
Question 57 |
A | ...the formation is saturated with fresh water. |
B | ...the formation is composed of highly complex pore structures. |
C | ...the formation is low in clay minerals. |
D | ...the formation is concentrated with clay minerals in saline waters. |
Question 58 |
A | ~ 2.81 |
B | ~ 3.50 |
C | ~ 1.76 |
D | ~ 0.71 |
E | ~ 0.20 |
Net shale interval = 5000 m - 3550 m = 1450 m (read off the GR curve)
Hence, sand:shale = 2550 m/1450 m = 1.7586...
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Credits: Based on the excellent class notes provided by, Dr. Stephen M. Hubbard during Winter 2015
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