ANDREW GRANT ASSIGNMENT # 1 DUE DATE: 30
TH
SEPTEMBER 2014
ID# 65188 RESR2003
1. Assuming a sandstone grain density of 2.65 g/cm3, calculate the porosity of a 3 inch long sandstone core sample of 1.5 in. width and breadth, respectively, if the grain weighs 250.0 g?
(4 marks)
Solution:
Sandstone grain density: 2.65 g/cm3 Sandstone core sample: 3 inches in length 1.5 inches in width 1.5 inches in breath Grain weights:
250.0g
Bulk volume = L x B x W = 3 x 1.5 x 1.5 = 6.75 in Converting inches to ft: 6.75 in
1
ft 3
1
3
12 in
Therefore V b = 0.003906325 ft Grain volume of sand pack: 3
3
Vg = 250.0 g/ 2.65 g/cm = 94.34 cm 3
3
Converting cm to ft : 3
-5
3
1 cm = 3.53 x 10 ft So
-5
Vg = 94.34 x 3.53 x 10 = 0.003331522 ft
3
= .003906 ft
= Vb – Vg
Vb Therefore: 0.003906325 – 0.003331522
= 14.71 %
0.003906325
2.
Calculate the bulk volume of a preserved (paraffin-coated) core sample immersed in water, given the following data: weight of dry sample in air: 20 g, weight of dry sample coated with paraffin: 20.9 g (density of paraffin- 0.9g/cm3), weight of coated sample immersed in water: 10 g (density of water: 1 g/cm3).
Determine the rock’s porosity,
assuming a sand grain density of 2.67 g/cm3. Solution
WT dry = dry weight in air = 20 g WTsat = weight of saturated sample in air = 20 g WTcoated =
weight of dry sample coated with paraffin = 20.9 g
WTsub = weight of coated sample immersed in wa ter = 10g Weight of paraffin = WTcoated – WTsat = 20.9 – 20 = 0.9 g Density of paraffin = 0.9 gm/cc Volume of paraffin: = m/v v = m/ = WTparafin
= 0.9
Density of paraffin
= 1.0 cc
0.9
Weight of water displaced: WTcoated – WTsub = 20.9 – 10 = 10.9 g Volume of water displaced = 10.9 / 1.0 = 10.9 cc Volume of water minus displaced-volume of paraffin = 10.9 – 1.0 = 9.9 g/cc Therefore bulk volume = 9.9 g/cc Porosity = Vb – Vg/ Vb Grain volume of sand pack: = m /v V = m/ 3
Therefore Vg = 20/ 2.67 = 7.49 cm
So = 9.9 -7.49 / 9.9 = 0.243 or 24.3 %
3.
A cylindrical core, 2.54 cm long and 2.54 cm in diameter has a porosity of 22%. It is saturated with oil and water, where the oil content is 1.5 cm3. a) What is the pore volume of the core? b) What are the oil and water saturations of the core ?
Solution: 3a. Pore
volume = bulk volume – grain volume 2
Grain volume: /4 x d h Vb =
2
3
(2.54) (2.54)
5
1 ft
= 3.64 x 10 ft 3
4
3.53 x 10-5 cm
Now porosity: = Vb – Vg/ Vb Since porosity is given we can find Vg Therefore: 5
0.22 = 3.64 x 10 – Vg 5
3.64 x 10
Thus Vg = 283920 ft Vb – Vg = Vp 5
5
3
3.64 x 10 – 2.84 x 10 = 80,000ft or 2.824 cm 3
-5
3
NB: 1 cm = 3.53 x 10 ft 3b
Sw = water cc/ pore volume 3
3
So = volume oil/ pore volume = 1.5 cm / 28.24 cm = 0.053 Since So + Sg + Sw = 1.0 So + Sw = 1.0 0.053 + Sw = 1.0 Sw = 1.0 – 0.053 = 0.947
4. What is the next preferred gas (or gases) if helium is not available for the Boyle’s Law Porosimeter? Why? Solution:
Nitrogen (N2) Nitrogen is a relatively inert gas, and will not contribute to any absorption effects on the pore surfaces. Adsorption will cause erroneously low values of grain volume an d subsequent overestimation of porosity. 5. Calculate the weight of 1 m3 sandstone of 14% porosity, assuming a sand grain density of 2.65 g/cm3. Solution:
= m /v
3
2.65 = m/ 1 m x 1.0 x 10 -6
-6
-6
3
NB: 1 cm 3 = 1.0 x 10 m -6
m = x v = 2.65 x 1 m 3 x 1.0 x 10 g = 2.65 x 10 g
6. List at least 2 sources of error in porosity measurements from cores. What are the problems that result from these errors? Solution:
I. II.
Bias core sampling Incomplete core recovery
There is a temptation to pre-select the best pa rts of the core for analysis. This leads to sample alterations either during coring in the subsurface or while being transported to the lab. Incomplete recovery can lead to a misrepresentation of the entire reservoir.
7 a) Describe, with the aid of a sketch, the principle of the Boyle’s Law Gas Expansion Porosimeter.
(5 marks)
Solution:
7.A)
Boyle’s Law Method: A Boyle’s Law porosimeter consists of two sample chambers as shown above. The first step is to calibrate the volumes of the sample chambers by injecting inert gas such as helium or nitrogen and recording the pressure differences when the valve between the two chambers is open and equalization occurs. The next step is to place the core sample in one chamber at some pressure, p1, which is isolated from the second chamber at p2. When the valve is opened pressure equilibrium occurs at some final pressure, pf. The pore space of the sample is penetrated by the gas; therefore the gas volume difference between the two tests is a measure of the grain volume.
The total moles of gas is constant, thus
nt = n1 + n2
Substituting the ideal gas equation, Pf Vf /RT = P1V1/RT + P2V2/RT Isothermal conditions prevail, Pf Vf = P1V1 + P2V2 Substituting for the volumes, P f (V1 + V2- Vg) = P1 (V1-Vg) + P2V2 Rearranging results in an expression for grain volume Vg = V1 (Pf – P1) + V2 (P f – P2)/ Pf – P1 Where V1 and V2 are the calibrated chamber volumes. 7. B) i.
Vg = 75 (700-0) + 100 (700 -900) 700
Vg = 46.43 cc
ii.
Now Vb - Vg = Vp 72 cm 3 – 46.43 cc = 57.57 cc
Vp = 25.57 cc
iii.
= Vp/ Vb = 25.57 / 72 = 35.51 %
8. The Dean-Stark Extraction Method, was used to determine the saturations of core samples taken from the Herrerra sands from the Moruga West Field, at approximately 4000 ft with areal extent of 950 acres. The following information was determined: Mass of saturated sample = 59 gms Bulk volume determined by nondestructive means = 42 cc Oil density = 0.80 gm/cc Volume of water collected when solvent was heated = 3.3 ml Dry weight obtained, after core sample was
removed and dried = 48 gms Density of water used to re-saturate sample = 1.00 gm/cc Weight of sample after saturated with fresh water = 62 gms Solution:
Mass of saturated sample: 59 g Bulk volume
: 42 cc
Oil density
: 0.80 gm/cc
Volume of water collected when solvent was heated: 3.3 ml Dry weight obtained
: 48 g
(After core sample was removed and dried) Density of water used to re-saturate sample
: 1.0g/cc
Weight of sample after saturated with fresh water
: 62g
(a) Pore volume Vp : Vb -Vg Vp = 62 -48 / 1.00 = 14 cc
(b) = 14/ 42= 33.33% (c) Water saturation Sw = Vw/ Vp = 3.3 / 14 = 23.57% (d) Oil saturation So = (wt of wet core, gm-wt of dr y cor e, gm- wt of wat er , gm) (pore volu me, cc) (density of oi l, gm/cc)
So = 59- 48 -3.3 x 1.00
=
68.75%
14 x 0.80 (e) Sg = 1 – Sw - So Sg = 1- 0.2357 -0.6875 = 0.0 768 or 7.68%
9. An oil well, is producing at a stabilized rate of 570 stb/day at a stabilized bottom- hole flowing pressure of 1750 psi. The well drains and area of approx. 50 acres with a uniform thickness of 30 ft. With the following additional data is available, determine the average permeability:
Solution: 10a
K avg = (10) (120) ln (100/1.25) (10) ln (100/1.25) + (120) ln (20/1.25) Where Ka = 10 mD, ra = 20 ft, rw = 1.25ft, re = 100ft, ke = 120 mD K avg = 13.97 mD 10. In a reservoir with two layered beds, calculate the following, assuming that there is no cross flow between the layers: a. the average k of the entire system. individual layer contribution to volumetric flow. (4 marks) Solution:
Kavg = ∑ ∑ = K1 x h1 + K2 x h2/ ht Where ht = h1 + h2 ∑ / ∑ = (100)(5) + (50) + (10)/ 5+10 .Kavg = 66.67 mD 10b -3
Q1 = 1.127 x 10
100 * 5 * 150
* 125 = 25.16 bbl/day
(200)(2.1) Where A = h x W -3
Q2 = 1.127 x 10
50 * 10 * 150
* 125 = 25.16 bbl/day
200 * 2.1 qt = q1 +q2 = 25.1 + 25.1 = 50.2 bbl/day
(3 marks) b. The
