# HW1 PEGN513 Reservoir Simulation I - Theoza Nopranda

```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 π
ππ’
1 ππ
(π ) =
=πΆ
π ππ ππ
π ππ₯
(1.1)
Arranging above equation to get:
π
ππ’
(π ) = πΆπ
ππ ππ
Integrating both hand sides in respect of π,
∫ π (π
⇒π
ππ’
) = ∫ πΆπ ππ
ππ
ππ’
π2
= πΆ + πΆ1
ππ
2
We know that at π = 0,
π
ππ’
ππ
= 0 and πΆ1 = 0, so
ππ’
π2
=πΆ
ππ
2
(1.2)
Rearranging above equation, we get
ππ’
π
⇒
=πΆ
ππ
2
π
⇒ ππ’ = πΆ ππ
2
Integrating both hand sides in respect of π again,
⇒ ∫ ππ’ = ∫ πΆ
π’(π) = πΆ
π
ππ
2
π2
+ πΆ2
4
We know at π = ππ , π’ = 0 and
0=πΆ
ππ2
+ πΆ2
4
We get
1
πΆ2 = −πΆ
ππ2
4
π’(π) = πΆ
π2
ππ2
ππ2
π2
− πΆ = − (1 − 2 ) πΆ
4
4
4
ππ
(1.3)
1 ππ
substituting πΆ = π ππ
π(π) = −
πππ
ππ π ππ
(π − π )
π
ππ π ππ
(1.4)
Problem 2.
Average velocity π’Μ is defined as the integration over the cross section area:
π’Μ =
∫ π’(π) ππ΄
π΄
(2.1)
Area in a capillary tube can be defined as area of circle
π΄ = ππ 2
⇒ ππ΄ = 2ππ ππ
Substituting π΄, ππ΄, and π’(π) (from equation 1.4) to equation 2.1, we get
π’Μ =
1 ππ ππ2
π 2 ππ
1 ππ ππ2
π 2 ππ
2)
∫
−
−
π(ππ
=
∫
−
−
(1
)
(1
) 2ππ ππ
4
4
πππ2 0
ππ2 ππ₯
πππ2 0
ππ2 ππ₯
⇒ π’Μ =
2π ππ ππ2
π 3 ππ
∫
−
−
(π
) ππ
4
πππ2 0
ππ2 ππ₯
(2.2)
Integrating the integral term in respect of π from 0 to ππ
π
2
ππ2 π 2 π 4 ππ π
π’Μ = 2 [− ( − 2 ) ]
4 2 4ππ ππ₯ 0
ππ
⇒ π’Μ =
2
ππ2 ππ 2 ππ 4 ππ
[−
( − 2) ]
4 2
ππ2
4ππ ππ₯
⇒ π’Μ =
2
ππ2 ππ 2 ππ
[−
( ) ]
4 4 ππ₯
ππ2
So, the average velocity π’Μ
Μ=−
π
πππ ππ
[ ]
π ππ
(2.3)
Problem 3.
Flow in an ideal fracture is defined as below equation
2
π 2 π’ 1 ππ
=
=πΆ
ππ¦ 2 π ππ₯
(3.1)
Integrating both sides twice, we get
π’(π¦) = πΆ
π¦2
+ πΆ1 π¦ + πΆ2
2
We know when π¦ =
π€π
2
and −
π€π
2
, π’ = 0, by substituting those two known-values, we get two equations:
π€π2
π€π
π€π
π’( ) = πΆ
+ πΆ1
+ πΆ2 = 0
2
8
2
π’ (−
(3.2)
π€π2
π€π
π€π
)=πΆ
− πΆ1
+ πΆ2 = 0
2
8
2
(3.3)
Based on equation 3.2 and 3.3, we find that πΆ1 = 0 and πΆ2 = −πΆ
π€π2
8
, so
π€π2
π¦2
1 π€π2
π’(π¦) = πΆ
−πΆ
= − ( − π¦2) πΆ
2
8
2 4
Substituting πΆ =
ππ
,
ππ₯
(3.4)
the flow inside two parallel plates can be calculated using below equation:
π πππ
π ππ
π(π) = − ( − ππ )
π π
π ππ
(3.5)
Problem 4.
Average velocity π’Μ is defined as:
π’Μ =
∫ π’(π¦) ππ΄
π΄
(4.1)
Substituting the expression of π’(π¦) from equation 3.5 and π΄ as the area of rectangle with dimension of
π€π &times; πΏ to equation 4.1, we get
π€π
1
2
1 π€π2
1 ππ
π’Μ =
∫ π€ − ( − π¦2)
π(π¦πΏ)
π€π πΏ − π 2 4
π ππ₯
2
π€π
π€π
1
2
1 π€π2
1 ππ
1
1 π€π2 π¦ π¦ 3 1 ππ 2
⇒ π’Μ =
∫ π€ − ( − π¦2)
ππ¦ = [− (
− )
]
π€π − π 2 4
π ππ₯
π€π
2 4
3 π ππ₯ −π€π
2
Integrating the integral term in respect of π¦ from −
π’Μ =
(4.2)
2
π€π
2
to
π€π
2
π€π3 π€π3 1 ππ
1
1 π€π3 π€π3 1 ππ 1
[− ( − )
+ (−
+ )
]
π€π
2 8
24 π ππ₯ 2
8
24 π ππ₯
3
π€π3 1 ππ
1
βΉ π’Μ =
[− ( )
]
π€π
12 π ππ₯
So, the average velocity π’Μ between two parallel plates
Μ = −(
π
πππ π ππ
)
ππ π ππ
(4.3)
Problem 5.
General Darcy equation is defined as:
π’Μ = −
π ππ
π ππ₯
(5.1)
By comparing equation 5.1 and 4.3, we can conclude that
ππ =
π€π2
12
(5.2)
However, in porous media, fracture volume depends on the fracture porosity. With the assumption of
tortuosity of fracture (π) is equal to 1, the effective fracture permeability inside porous media can be
written as:
ππ,πππ =
ππ π€π2 ππ π€π2
=
ππ 12
12
(5.3)
4
```