1. Grain density= 2.65g/cm3

1 inch= 2.54cm

Length= 3 in = 7.62cm Width, breadth= 1.5 in = 3.8cm

Area =L*B*H = 7.62 * 3.8 * 3.8 = 110.61 cm

Density()= mass/Grain Volume (Vg)

Therefore:

= 2.50/2.65 = 94.34 cm3

2.

Weight of dry sample = 20g

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: 1g/cm3)

Mass of Coated Paraffin Sample = Coated Paraffin Sample- Weight of dry

Sample = 20.9 – 20 = 0.9 g

Coated Paraffin Volume = Mass/Density = 0.9/0.9

= 1cm3

Volume of Coated Sample in water = 10/1 = 10 cm3 Bulk Volume =10 -1

= 9 cm3

3.

Cylindrical Core: Length = 2.54cm Diameter = 2.54 cm

Porosity = 22%

a) Pore Volume:

b) Oil and water saturations of the core:

4. If helium is not available for the Boyle’s Law porosimeter the next

preferred gas or gases that would be used are nitrogen and air because they are inert and would not react with the apparatus’ material or the

reservoir fluids that may be found in the core sample.

5. Weight of sandstone (Vb) = 1m3 = 1000 000cc

Φ = 14%

Sand grain density = 2.56g/ cc

Φ =Vp/Vb

Vp= .14 x 1000 000= 140 000cc

Grain volume = Vb - Vp

= 1000 000 – 140 000

= 860 000cc

Grain density = Grain mass/ Grain Volume

Grain mass = Grain volume x Grain density

= 860 000 x 2.65

= 2279kg

6. When the core sample contains clay minerals it may swell in the

lab due to its contact with water as a result the porosity

measurements that will be taken will be inaccurate.

Another source of error is while the core samples are being moved

fine particles may be lost from it as a result the porosity value

measured for the core would be higher than the actual real value of

the reservoir. This would cause a wrong value in calculating the

potential reserve.

7. a)

This method relies on the ideal gas law, or rather Boyle’s law. The rock is sealed in a container of known volume V1 at atmospheric pressure P1

This container is attached by a valve to another container of known volume, V2, containing gas at a known pressure, P2. When the valve that connects the two volumes is opened slowly so that the system remains

isothermal, the gas pressure in the two volumes equalizes to P3. The value of the equilibrium pressure can be used to calculate the volume of grains in the rock.

b) V1= 75cm3 P1= 0kPa

V2= 100cm3 P2=900kPA

Pf= 700kPa

i)

ii)

iii)

o

8. Dean Stark Method:

Mass of saturated sample = 59 gms

Bulk volume determined by nondestructive means, Vb= 42 cc

Oil density = 0.80 gm/cc

Volume of water collected when solvent was heated, Vw = 3.3 ml

Dry weight obtained, after core sample was removed and dried, Vg = 48

gms

Density of water used to re-saturate sample = 1.00 gm/cc

Weight of sample after saturated with fresh water, Vb = 62 gms

a.

b.

c.

d.

e.

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:

re = 100 ft ra = 20 ft rw = 1.25 ft ke = 120mD ka = 10mD

10.

a.

k1 = 100 mD k2 = 50 mD h1 = 5 ft h2 = 10 ft

L = 200 ft w = 150 ft = 2.1 cp

b.