Geofisica

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  • Publicado : 1 de junio de 2011
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Ejercicios
1.A core sample coated with paraffin was immersed in a Russell tube. The dry sample weighed 20.0gm. The dry sample coated with paraffin weighed 20.9 gm. The paraffin coatedsample displaced 10.9 cc of liquid. Assume the density of solid paraffin is 0.9 gm/cc. What is the bulk volume of the sample?

Solution:
Weight of paraffin coating = 20.9 gm - 20.0 gm= 0.9 gm
Volume of paraffin coating = 0.9 gm / (0.9 gm/cc) = 1.0 cc
Bulk volume of sample = 10.9 cc - 1.0 cc = 9.9 cc

2.The core sample of problem I-1 was stripped of the paraffincoat, crushed to grain size, and immersed in a Russell tube. The volume of the grains was 7.7 cc. What was the porosity of the sample? Is this effective or total porosity.

Solution:Bulk Volume = 9.9 cc
Matrix Volume = 7.7 cc
f =Vb – Vm = 9.9 cc- 7.7 cc = 0.22
Vb 9.9 cc

It is total porosity.
3.Calculate the porosity of a core sample when thefollowing information is available:
Dry weight of sample = 427.3 gm
Weight of sample when saturated with water = 448.6 gm
Density of water = 1.0 gm/cm3
Weight of water saturatedsample immersed in water = 269.6 gm

Solution:
Vp = s a t. core wt. in air - dry core wt.
density of water
Vp = 4 48.6 gm - 427.3 gm
1 gm/cm3
Vp= 21.3 cm3

Vb = s a t. core wt. in air - sat. core wt. in water
density of water
Vb = 4 48.6 gm - 269.6 gm
1 gm/cm3
Vb = 179.0 cm3

=Vp=21.3 cm3= .119
Vb 179.0 cm3
= 11.9%
What is the lithology of the sample?

Vm = Vb - Vp
Vm = 179.0 cm3 - 21.3 cm3 = 157.7 cm3
m = w t . of dry sample = 4 27.3 gm =2.71 gm/(cm3)
matrix vol. 157.7 cm3
The lithology is limestone.
Is the porosity effective or total? Why?
Effective, because fluid was forced into the pore space.
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