Systems and Units. The three systems of units are:

Transcription

1 The three systems of units are: 1. The English or the ft-lb-s System 2. The International or the m-kg-s System 3. The Laboratory or the cm-gm-s System Quantities fall into two main categories: 1. Principal Quantities Length (L) Mass (m) Time (t) Angle (θ) Temperature (T) 2. Derived Quantities Area (A) Volume (V) Velocity (ν) Frequency (Hz.) Acceleration (a) Angular Velocity (ω) Angular Acceleration (â) Volume Flow Rate (q) Mass Flow Rate ( o m ) Density (ρ) Specific Gravity (SG) Force (F) Force due to Inertia (F I ) Force due to Gravity (F G ) Force due to Viscosity (F µ ) Force due to Elasticity (F E ) Force due to Pressure (F P ) Specific Weight (γ) Energy (E) Moment of a Force (M) Work (W) Pressure (p) Stress (τ) Power (P) Dynamic Viscosity (µ) Kinematic Viscosity (ν) Systems and Units 1

2 1. Principal Quantities Quantity English System International System Length Mass Time Angle Temperature mile (mi) = 1760 yd yard (yd) = 3 ft foot (ft) = 12 in inch (in) = 2.54 cm slug (sl) = lbm pound mass (lbm) = gm year = 365 d day = 24 h hour = 60 m minute = 60 s 1 rev. = 2B radians = 360E 1 r = E 1 E = r Ordinary EF= 1.8EC+32 Absolute ER= EF ER= 1.8 K kilometer (km) = 1000 m meter (m) = 100 cm centimeter (cm) kilogram (kg) = 1000 gm gram (gm) year = 365 d day = 24 h hour = 60 m minute = 60 s 1 rev. = 2B radians = 360E 1 r = E 1 E = r Ordinary EC= (EF-32)/1.8 Absolute K= EC K= ER/1.8 2

3 2. Derived Quantities Quantity Formula English Units SI Units Lab. Units Area A = L 2 1 mi 2 = 640 acres mi 2, ft 2, in 2, acre 1 acre = 43,560 ft 2 Volume V = L 3 bbl = ft 3 bbl, ft 3, gal, qt bbl = 42 gal gal = 4 qt = ltr Velocity dl = dt m 2 m 3, ltr m 3 = 1000 ltr ltr = 1000 cc cm 2, darcy, md 1 cm 2 = 101,320,790 d 1 d = 1000 md cm 3 = cc ν ft/s m/s cm/s Frequency Hz. s -1 s -1 s -1 Acceleration dν a = dt ft/s 2 m/s 2 cm/s 2 Angle 2 rad and E 360E = 2B rad rad and E 360E = 2B rad rad and E 360E = 2B rad dθ Angular ω = Velocity dt rad/s rad/s rad/s dω Angular aˆ = Acceleration dt rad/s 2 rad/s 2 rad/s 2 Volume dv Flow Rate q = dt bbl/d, ft 3 /s m 3 /s, ltr/s cc/s Mass Flow dm m o = Rate dt slug/s, lbm/s kg/s gm/s Density m ρ = slug/ft 3, lb/ft 3, V lb/gal kg/m 3 gm/cc Specific Gravity ρ f SG = = ρ w o API D w = slug/ft 3 D w = lbm/ft 3 D w = lbm/gal D w = D w = 10 EAPI D w = 10 3 kg/m 3 D w = 1 gm/cc 3

4 2. Derived Quantities (cont.) Quantity Formula English Units SI Units Lab. Units Force F = m a lbf = slug ft/s 2 lbf = 444,822 dyne lbf = N Force due to Inertia Force due to Gravity=Wt. Force due to Viscosity F N = kg m/s 2 1 N = 10 5 dynes dyne = gm cm/s 2 F I = m a lbf N dyne F G = m g µ lbf g = ft/s 2 N g = m/s 2 dyne g = cm/s 2 dν = µ A lbf N dyne dy Force due to Elasticity Force due to Pressure F E = E A lbf N dyne F P = p A lbf N dyne Specific Wt. ( = F G /V lbf/ft 3 N/m 3 dyne/cc Energy E = F L lbf-ft J = N-m erg = dyne cm Moment of a Force M = F L lbf-ft J erg = dyne cm Work W = F L lbf-ft J erg = dyne cm Pressure p = F N /A lbf/ft 2, lbf/in 2 = psi p sc = psi 4 Pa = N/m 2 bar = 10 5 Pa p sc = bar p sc = Pa dyne/cm 2 p sc = d/cm 2 p sc = 1 atm p sc = 76 cm Hg Stress J = F T /A lbf/ft 2, lbf/in 2 = psi bar, Pa dyne/cm 2 Power P = E/t lbf-ft/s, hp 1 hp = 550 lbf-ft/s F Dynamic F Viscosity µ lbf-s/ft 2 = 47,880 µ dν / dya cp Kinematic Viscosity W = J/s N s/m 2 = 10 poise N s/m 2 = 1000 cp < = µ/d ft 2 /s = 929 stoke m 2 /s = 10 4 stoke erg/s dyne s/cm 2 = poise 1 poise = 100 cp cm 2 /s = stoke

5 EPS-441: Petroleum Development Geology Units and Conversion Semester: Homework #: Name: SS#: Problem #1: Do the following unit conversions: From 38 o API 12 o API 56 o API 40 o API 28 o API 31 o API SG = 0.76 SG = 1.10 SG = 0.74 SG = 1.08 SG = lb/ft lb/ft lb/ft lb/ft lb/gal 8.33 lb/gal 8.33 lb/gal 10.4 lb/gal 0.82 gm/cc 1.02 gm/cc 0.87 gm/cc 0.91 gm/cc To lb/ft 3 SG SG gm/cc lb/ft 3 lb/gal gm/cc o API o API lb/ft 3 lb/gal o API SG o API gm/cc gm/cc lb/ft 3 gm/cc SG lb/ft 3 Problem #2: Given a rectangular solid with dimensions 1000 ft x 400 ft x 40 ft. Calculate its volume in ft 3, bbl, acre-ft?. 5

6 Problem #3: Calculate pressure gradients of the following liquids: SG = 1.00 D w = 66.3 lb/ft 3 SG = 1.15 D w = 9.5 lb/gal SG = 0.85 EAPI = 42 EAPI = 32 EAPI = 45 D o = 58.1 lb/ft 3 Problem #4: A well drilled to 3000 ft penetrates a formation containing 28 EAPI oil. If reservoir pressure is 1300 psia, what is the shut-in surface pressure?. If reservoir pressure is 1000 psia, how many ft of oil will be standing in the wellbore?. Problem #5: In an area where ambient temperature is 78 EF, two wells, A and B, were drilled. The depth of well A is 7250 ft and the depth of well B is 8000 ft. The bottom hole temperature (BHT) in well B is 180 EF. What is the BHT in well A?. Problem #6: The areal extent of a reservoir as determined with seismic data is 1500 acres. From logs the following reservoir properties were determined: Zone N h, ft S w a) Calculate the pore volume in the reservoir. Give your answers in acre-ft, ft 3, and bbl. b) Calculate the volume of oil in the reservoir. Give your answers in acre-ft, ft 3, and bbl. c) Calculate the STB of oil if B o = 1.34 RB/STB. 6

7 Problem #7: The average reservoir properties of a 300 acre reservoir are: N = 18% and S w = 36%. Estimated volume of oil originally in place = 410 MM STB. Formation thickness = 2000 ft. Determine the oil formation volume factor. Problem #8: Consider the sketch below. Given EAPI of oil = 35. water SG = If well A penetrates the oil zone, find the shut-in surface pressure at well A?. 7

8 EPS-441: Petroleum Development Geology Units and Conversion Semester: Homework #: Name: SS#: Problem #1: A reservoir has an areal extent of 500 acres, an average thickness of 90 ft, and an average porosity of 20%. a) What is the reservoir volume available for hydrocarbons?. Answer in acre-ft, bbl, and ft 3. b) If the average water saturation is 35%, what is the reservoir volume available for oil?. Answer in acre-ft, bbl, and ft 3. c) Same as (b) except reservoir fluid is gas. d) If B o = 1.34 RB/STB. What is the oil volume from (b) in surface barrels?. in surface cubic feet?. e) If B g = 310 SCF/CF. What is the gas volume from (c) in surface cubic feet?. in surface barrels?. Problem #2: A well is being drilled to a depth of 10,000 ft. The wellbore diameter is 9 inches. The drill string has an inside diameter of 4.5 inches and an outer diameter of 5 inches. What is the volume of mud in the hole when it is at total depth (TD) and there is a complete drill string in the hole?. Ignore collars and bit volume and assume the mud is incompressible. Answer in cubic feet and barrels. 8

9 Problem #3: Data for the diagram shown below are: a) oil gravity = 38 EAPI b) water specific gravity = 1.05 c) average gas gradient = d) only oil flows in oil zone e) reservoir B is in water communication with the surface f) pressure at oil zone datum in reservoir A = 3820 psig g) temperature gradient = 1.5 EF/100 ft. h) mean surface temperature = 61 EF i) ground level is 3000 ft above sea level j) elevations given are sub-sea but answers should be in subsurface a) Calculate pressure and temperature at each of the 7 zones indicated in the diagram. b) What is the bubble point pressure for the oil in reservoir A?. c) Did the oil accumulated before or after faulting in both reservoirs?. 9

10 Problem #4: Seismic data indicates an areal extent of a reservoir as one square mile. Other reservoir properties are: Zone N, % h, ft S w, % a) Calculate hydrocarbon pore volume in the reservoir. Give your answers in acre-ft, ft 3, and bbl. b) Calculate the volume of oil in the reservoir. Give your answers in acre-ft, ft 3, bbl, and STB if B o = 1.18 RB/STB. c) How many SCF of gas in the reservoir if B g = 263 SCF/CF. Problem #5: A well drilled in an area had the following tests run: DST, temperature survey and a suite of logs. A surface gas and oil samples were taken and recombined at simulated reservoir conditions and a PVT analysis was done. From these measurements, the following reservoir rock and fluid properties were determined: EAPI of oil = 29 EAPI B o = 1.38 RB/STB SG g = 0.78 reservoir depth = 4520 ft initial reservoir pressure = 2952 psia bubble point pressure = 1831 psia reservoir temperature = 151 EF surface temperature = 70EF reservoir acreage = 640 acres Zone h, ft N (%) S w (%) a) calculate the geothermal gradient in this area?. b) calculate the original oil in place (OOIP)?. 10

11 Problem #6: Consider the sketch below. If BHT at well A is 200 EF and BHT at well B is 232 EF. a) What is the BHT at well C?. b) What is the ambient temperature in this area?. 11

12 EPS-441: Petroleum Development Geology Units and Conversion Semester: Homework #: Name: SS#: Problem #1: A well was drilled offshore into an unconsolidated formation. The used sand control methods were not successful so the well was shut-in. The average annual surface temperature in that area is 84 EF and the temperature gradient is 1.6 EF/100 ft. The shut-in surface pressure is 850 psig and the shut-in bottom hole pressure is 2800 psig. It is known that the wellbore (diameter = 5.5 inches) is filled with gas (SG = 0.78, z = 0.85), oil (API = 32E), water (SG = 1.08), and water saturated sand (D = 35 gm/cc). Knowing that there is a 2500 ft of gas, a 2000 ft of sand-free water, and a 500 ft of water-saturated sand, calculate the following: a) Total depth of the well. b) B g in SCF/CF of the gas using average pressure and temperature. Problem #2: Define the following using equations only - include units: a) Oil formation volume factor (B o ) b) Porosity (N) c) Temperature gradient (G T ) d) Gas saturation (S g ) e) Oil gravity (EAPI) f) Fluid pressure gradient (() 12

13 Problem #3: Using the figure below and the following data: D w D o D g p atm = 1.02 gm/cc = 0.80 gm/cc = 0.10 gm/cc = 14.7 psia What does gauge A read?. 13