1.
Types of materials
Gravel
Medium-grained sand
Poorly cemented
sandstone
Well cemented
sandstone
Clay
Shale
Limestone
Unfractured granite
Fractured granite
Volume of sample
(cm3)
500
600
650
Volume of pore space
(cm3)
210
270
163
Porosity (%)
800
40
5%
825
435
950
500
700
404
57
123
5
35
48.97%
13.10%
12.95%
1%
5%
42%
45%
27.08%
Solution:
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 =
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
𝑥𝑥 100
𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣
210
𝑥𝑥 100 = 42%
500
57
𝑥𝑥 100 = 13.10%
435
163
𝑥𝑥 100 = 27.08%
650
5
𝑥𝑥 100 = 1%
500
270
𝑥𝑥 100 = 45%
600
40
𝑥𝑥 100 = 5%
800
404
𝑥𝑥 100 = 48.97%
825
123
𝑥𝑥 100 = 12.95%
950
35
𝑥𝑥 100 = 5%
700
2.
Wtdry = 20.0 gm
Wtimm = 15.0 gm
Ρwater = 1.0 gm/cc
Wt(displaced) = Wtdry - Wtimm
Wt(displaced) = 20.0 gm – 15.0 gm
Wt(displaced) = 5.0 gm
Vgrain =
𝑊𝑊𝑡𝑡(𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑)
𝛲𝛲𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
5.0 𝑔𝑔𝑔𝑔
Vgrain = 1.0 𝑔𝑔𝑔𝑔/𝑐𝑐𝑐𝑐 = 5.0 cc
3.
V1 = 100 cc
P2 = 60.0 psi
V2 = 100 cc
Pf = 39.0 psi
P1 = 15.0 psi
Vgrain =
𝑉𝑉1 (𝑃𝑃𝑃𝑃 − 𝑃𝑃1) + 𝑉𝑉2 (𝑃𝑃𝑃𝑃−𝑃𝑃2)
(𝑃𝑃𝑃𝑃−𝑃𝑃1)
Vgrain =
100 𝑐𝑐𝑐𝑐 (39.0 𝑝𝑝𝑝𝑝𝑝𝑝−15.0 𝑝𝑝𝑝𝑝𝑝𝑝) + 100𝑐𝑐𝑐𝑐 (39.0𝑝𝑝𝑝𝑝𝑝𝑝−60.0𝑝𝑝𝑝𝑝𝑝𝑝)
(39.0 𝑝𝑝𝑝𝑝𝑝𝑝−15.0 𝑝𝑝𝑝𝑝𝑝𝑝)
Vgrain = 12.5 cc
4.
Wtdry = 20.0 gm
Wtsat = 22.5 gm
Ρwater = 1.0 gm/cc
Wt(water) = Wtsat - Wtdry
Vpore =
Wt(water) = 22.5 gm – 20.0 gm
𝑊𝑊𝑡𝑡(𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤)
𝛲𝛲𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
2.5 𝑔𝑔𝑔𝑔
Vpore = 1.0 𝑔𝑔𝑔𝑔/𝑐𝑐𝑐𝑐
Wt(water) = 2.5 gm
Vpore = 2.5 cc
5.
P = 0.83 atm
PV = nRT
𝑑𝑑𝑑𝑑
( 𝑑𝑑𝑑𝑑 )t = −
1
𝑛𝑛𝑛𝑛𝑛𝑛
𝑃𝑃2
Kt = − 𝑉𝑉 (−
Kt =
𝑛𝑛𝑛𝑛𝑛𝑛
𝑛𝑛𝑛𝑛𝑛𝑛
𝑃𝑃2( 𝑃𝑃 )
𝑛𝑛𝑛𝑛𝑛𝑛
)
𝑃𝑃2
1 𝑑𝑑𝑑𝑑
T = 311 K
Kt = − 𝑉𝑉 (𝑑𝑑𝑑𝑑)
1
Kt = 𝑃𝑃
1
Kt = 0.83 𝑎𝑎𝑎𝑎𝑎𝑎 = 1.2048 atm-1
Kt = 1.2048 atm-1 x
1 𝑎𝑎𝑎𝑎𝑎𝑎
= 1.1891 Pa-1
101325 𝑃𝑃𝑃𝑃
6.
Kt = 2.21 x 10-6 atm-1
T = 293K
𝑑𝑑𝑑𝑑
= −.0008
𝑉𝑉
1 𝑑𝑑𝑑𝑑
Kt = − 𝑉𝑉 (𝑑𝑑𝑑𝑑)
𝑑𝑑𝑑𝑑
Kt = − 𝑉𝑉𝑑𝑑𝑑𝑑
𝑑𝑑𝑉𝑉
dP = - 𝑉𝑉 𝐾𝐾𝐾𝐾
(−0.0008)
dP = − 2.21𝑥𝑥10−6 𝑎𝑎𝑎𝑎𝑎𝑎−1
dP = 361.9910 atm