PEGN513 Reservoir Simulation I
HW 1
Name: Theoza Nopranda
Due: Thursday, September 6th, 2018
Problem 1.
Poiseuille equation in a capillary tube is defined as below equation:
1 𝜕
𝜕𝑢
1 𝑑𝑝
(𝑟 ) =
=𝐶
𝑟 𝜕𝑟 𝜕𝑟
𝜇 𝑑𝑥
(1.1)
Arranging above equation to get:
𝜕
𝜕𝑢
(𝑟 ) = 𝐶𝑟
𝜕𝑟 𝜕𝑟
Integrating both hand sides in respect of 𝑟,
∫ 𝑑 (𝑟
⇒𝑟
𝜕𝑢
) = ∫ 𝐶𝑟 𝑑𝑟
𝜕𝑟
𝜕𝑢
𝑟2
= 𝐶 + 𝐶1
𝜕𝑟
2
We know that at 𝑟 = 0,
𝑟
𝑑𝑢
𝑑𝑟
= 0 and 𝐶1 = 0, so
𝜕𝑢
𝑟2
=𝐶
𝜕𝑟
2
(1.2)
Rearranging above equation, we get
𝜕𝑢
𝑟
⇒
=𝐶
𝜕𝑟
2
𝑟
⇒ 𝑑𝑢 = 𝐶 𝑑𝑟
2
Integrating both hand sides in respect of 𝑟 again,
⇒ ∫ 𝑑𝑢 = ∫ 𝐶
𝑢(𝑟) = 𝐶
𝑟
𝑑𝑟
2
𝑟2
+ 𝐶2
4
We know at 𝑟 = 𝑟𝑒 , 𝑢 = 0 and
0=𝐶
𝑟𝑒2
+ 𝐶2
4
We get
1
𝐶2 = −𝐶
𝑟𝑒2
4
𝑢(𝑟) = 𝐶
𝑟2
𝑟𝑒2
𝑟𝑒2
𝑟2
− 𝐶 = − (1 − 2 ) 𝐶
4
4
4
𝑟𝑒
(1.3)
1 𝒅𝒑
substituting 𝐶 = 𝜇 𝒅𝒙
𝒖(𝒓) = −
𝒓𝟐𝒆
𝒓𝟐 𝟏 𝒅𝒑
(𝟏 − 𝟐 )
𝟒
𝒓𝒆 𝝁 𝒅𝒙
(1.4)
Problem 2.
Average velocity 𝑢̅ is defined as the integration over the cross section area:
𝑢̅ =
∫ 𝑢(𝑟) 𝑑𝐴
𝐴
(2.1)
Area in a capillary tube can be defined as area of circle
𝐴 = 𝜋𝑟 2
⇒ 𝑑𝐴 = 2𝜋𝑟 𝑑𝑟
Substituting 𝐴, 𝑑𝐴, and 𝑢(𝑟) (from equation 1.4) to equation 2.1, we get
𝑢̅ =
1 𝑟𝑒 𝑟𝑒2
𝑟 2 𝑑𝑝
1 𝑟𝑒 𝑟𝑒2
𝑟 2 𝑑𝑝
2)
∫
−
−
𝑑(𝜋𝑟
=
∫
−
−
(1
)
(1
) 2𝜋𝑟 𝑑𝑟
4
4
𝜋𝑟𝑒2 0
𝑟𝑒2 𝑑𝑥
𝜋𝑟𝑒2 0
𝑟𝑒2 𝑑𝑥
⇒ 𝑢̅ =
2𝜋 𝑟𝑒 𝑟𝑒2
𝑟 3 𝑑𝑝
∫
−
−
(𝑟
) 𝑑𝑟
4
𝜋𝑟𝑒2 0
𝑟𝑒2 𝑑𝑥
(2.2)
Integrating the integral term in respect of 𝑟 from 0 to 𝑟𝑒
𝑟
2
𝑟𝑒2 𝑟 2 𝑟 4 𝑑𝑝 𝑒
𝑢̅ = 2 [− ( − 2 ) ]
4 2 4𝑟𝑒 𝑑𝑥 0
𝑟𝑒
⇒ 𝑢̅ =
2
𝑟𝑒2 𝑟𝑒 2 𝑟𝑒 4 𝑑𝑝
[−
( − 2) ]
4 2
𝑟𝑒2
4𝑟𝑒 𝑑𝑥
⇒ 𝑢̅ =
2
𝑟𝑒2 𝑟𝑒 2 𝑑𝑝
[−
( ) ]
4 4 𝑑𝑥
𝑟𝑒2
So, the average velocity 𝑢̅
̅=−
𝒖
𝒓𝟐𝒆 𝒅𝒑
[ ]
𝟖 𝒅𝒙
(2.3)
Problem 3.
Flow in an ideal fracture is defined as below equation
2
𝜕 2 𝑢 1 𝑑𝑝
=
=𝐶
𝜕𝑦 2 𝜇 𝑑𝑥
(3.1)
Integrating both sides twice, we get
𝑢(𝑦) = 𝐶
𝑦2
+ 𝐶1 𝑦 + 𝐶2
2
We know when 𝑦 =
𝑤𝑓
2
and −
𝑤𝑓
2
, 𝑢 = 0, by substituting those two known-values, we get two equations:
𝑤𝑓2
𝑤𝑓
𝑤𝑓
𝑢( ) = 𝐶
+ 𝐶1
+ 𝐶2 = 0
2
8
2
𝑢 (−
(3.2)
𝑤𝑓2
𝑤𝑓
𝑤𝑓
)=𝐶
− 𝐶1
+ 𝐶2 = 0
2
8
2
(3.3)
Based on equation 3.2 and 3.3, we find that 𝐶1 = 0 and 𝐶2 = −𝐶
𝑤𝑓2
8
, so
𝑤𝑓2
𝑦2
1 𝑤𝑓2
𝑢(𝑦) = 𝐶
−𝐶
= − ( − 𝑦2) 𝐶
2
8
2 4
Substituting 𝐶 =
𝑑𝑝
,
𝑑𝑥
(3.4)
the flow inside two parallel plates can be calculated using below equation:
𝟏 𝒘𝟐𝒇
𝟏 𝒅𝒑
𝒖(𝒚) = − ( − 𝒚𝟐 )
𝟐 𝟒
𝝁 𝒅𝒙
(3.5)
Problem 4.
Average velocity 𝑢̅ is defined as:
𝑢̅ =
∫ 𝑢(𝑦) 𝑑𝐴
𝐴
(4.1)
Substituting the expression of 𝑢(𝑦) from equation 3.5 and 𝐴 as the area of rectangle with dimension of
𝑤𝑓 × 𝐿 to equation 4.1, we get
𝑤𝑓
1
2
1 𝑤𝑓2
1 𝑑𝑝
𝑢̅ =
∫ 𝑤 − ( − 𝑦2)
𝑑(𝑦𝐿)
𝑤𝑓 𝐿 − 𝑓 2 4
𝜇 𝑑𝑥
2
𝑤𝑓
𝑤𝑓
1
2
1 𝑤𝑓2
1 𝑑𝑝
1
1 𝑤𝑓2 𝑦 𝑦 3 1 𝑑𝑝 2
⇒ 𝑢̅ =
∫ 𝑤 − ( − 𝑦2)
𝑑𝑦 = [− (
− )
]
𝑤𝑓 − 𝑓 2 4
𝜇 𝑑𝑥
𝑤𝑓
2 4
3 𝜇 𝑑𝑥 −𝑤𝑓
2
Integrating the integral term in respect of 𝑦 from −
𝑢̅ =
(4.2)
2
𝑤𝑓
2
to
𝑤𝑓
2
𝑤𝑓3 𝑤𝑓3 1 𝑑𝑝
1
1 𝑤𝑓3 𝑤𝑓3 1 𝑑𝑝 1
[− ( − )
+ (−
+ )
]
𝑤𝑓
2 8
24 𝜇 𝑑𝑥 2
8
24 𝜇 𝑑𝑥
3
𝑤𝑓3 1 𝑑𝑝
1
⟹ 𝑢̅ =
[− ( )
]
𝑤𝑓
12 𝜇 𝑑𝑥
So, the average velocity 𝑢̅ between two parallel plates
̅ = −(
𝒖
𝒘𝟐𝒇 𝟏 𝒅𝒑
)
𝟏𝟐 𝝁 𝒅𝒙
(4.3)
Problem 5.
General Darcy equation is defined as:
𝑢̅ = −
𝑘 𝑑𝑝
𝜇 𝑑𝑥
(5.1)
By comparing equation 5.1 and 4.3, we can conclude that
𝑘𝑓 =
𝑤𝑓2
12
(5.2)
However, in porous media, fracture volume depends on the fracture porosity. With the assumption of
tortuosity of fracture (𝜏) is equal to 1, the effective fracture permeability inside porous media can be
written as:
𝑘𝑓,𝑒𝑓𝑓 =
𝜙𝑓 𝑤𝑓2 𝜙𝑓 𝑤𝑓2
=
𝜏𝑓 12
12
(5.3)
4