Topics in Reservoir Engineering (8).pdf

(1)

Numerical Solution of

1D Single Phase Flow Equation

[ ]

Hoonyoung Jeong

Department of Energy Resources Engineering Seoul National University

[email protected]

SNU ERE Hoonyoung Jeong

(2)

Flow Equation

•

^𝜕

𝜕𝑥

𝑘 𝜇𝐵

𝜕𝑃

𝜕𝑥

+

^𝑄

𝐴𝜕𝑥

=

^𝜕

𝜕𝑡

𝜙 𝐵

• We can’t move 𝑘, 𝜇, 𝐵, 𝜙 out of 𝜕 𝜕𝑥 Τ and 𝜕 𝜕𝑡 Τ Because 𝑘, 𝜇, 𝐵, 𝜙 depends on 𝑥 and 𝑡 .

✓𝜙, 𝐵 : functions of 𝑃, 𝑃 𝑥, 𝑡

✓ Heterogeneous : k 𝑥

✓𝜇, 𝐵 : functions of 𝑃 𝑥, 𝑡

→

^𝑘

𝜇B

is dependent of 𝑥 → f 𝑥

• Here, we assume a fluid viscosity is constant

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(3)

How to Numerically Approximate Derivatives?

𝜕

𝜕𝑥

𝑘 𝜇𝐵

𝜕𝑃

𝜕𝑥 + 𝑄

𝐴𝜕𝑥 = 𝜕

𝜕𝑡

𝜙 𝐵

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(4)

Approximation of Spatial Term

𝜕

𝜕𝑥 𝑓 𝑥 𝜕𝑃

𝜕𝑥

𝑖

= 𝑓

_{𝑖+ Τ}_{1 2}

Δ𝑥

²

𝑃

_𝑖+1

− 𝑃

_𝑖

+ 𝑓

_{𝑖− Τ}_{1 2}

Δ𝑥

²

𝑃

_𝑖−1

− 𝑃

_𝑖

= 1 Δ𝑥

²

𝑘

𝜇𝐵

𝑖+1/2

𝑃

_𝑖+1

− 𝑃

_𝑖

+ 1 Δ𝑥

²

𝑘

𝜇𝐵

𝑖−1/2

𝑃

_𝑖−1

− 𝑃

_𝑖

= 1 Δ𝑥

²

1 𝜇𝐵 Τ

_𝑖+1

+ 1/ 𝜇𝐵

_𝑖

1 𝑘 Τ

_𝑖+1

+ Τ 1 𝑘

_𝑖

𝑃

_𝑖+1

− 𝑃

_𝑖

+ 1 Δ𝑥

²

1 𝜇𝐵 Τ

_𝑖−1

+ 1/ 𝜇𝐵

_𝑖

1 𝑘 Τ

_𝑖−1

+ Τ 1 𝑘

_𝑖

𝑃

_𝑖−1

− 𝑃

_𝑖

= 𝑇

_𝑖+1/2

𝑃

_𝑖+1

− 𝑃

_𝑖

+ 𝑇

_𝑖−1/2

𝑃

_𝑖−1

− 𝑃

_𝑖

This is called transmissibility 𝑇_𝑖+1/2

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SNU ERE Hoonyoung Jeong

(5)

Approximation of Temporal Term

𝜕

𝜕𝑡 𝜙 𝐵 𝑖

= 1 𝐵

𝜕𝜙

𝜕𝑡 + 𝜙 𝜕 1 𝐵Τ

𝜕𝑡 𝑖

= 1 𝐵

𝜕𝜙

𝜕𝑃

𝜕𝑡 + 𝜙 𝜕 1 𝐵Τ

𝜕𝑃

𝜕𝑡 𝑖

= 𝜙𝐶_𝑟 𝐵

𝜕𝑃

𝜕𝑡 + 𝜙𝐶_𝑓 𝐵

𝜕𝑃

𝜕𝑡 𝑖

= 𝜙(𝐶_𝑟+𝐶_𝑓) 𝐵

𝜕𝑃

𝜕𝑡 𝑖

= 𝜙𝐶_𝑡 𝐵

𝜕𝑃

𝜕𝑡 𝑖

= 𝜙_𝑖𝐶_𝑡,𝑖 𝐵_𝑖

𝑃_𝑖^𝑡+Δ𝑡 − 𝑃_𝑖^𝑡

Δ𝑡 = 𝜙_𝑖𝐶_𝑡,𝑖

𝐵_𝑖Δ𝑡 𝑃_𝑖^𝑡+Δ𝑡 − 𝑃_𝑖^𝑡 = 𝐶_𝑝,𝑖 𝑃_𝑖^𝑡+Δ𝑡 − 𝑃_𝑖^𝑡

𝐶_𝑟 = 1 𝜙

𝑑𝜙 𝑑𝑃 𝐶_𝑓 = 1

𝜌 𝑑𝜌

𝑑𝑃 = 1

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐵Τ

𝑑 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐵Τ

𝑑𝑃 = 𝐵𝑑 Τ1 𝐵 𝑑𝑃

Chain rule

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(6)

Explicit and Implicit Methods

• 𝑇

_{𝑖+ Τ}_{1 2}

𝑃

_𝑖+1

− 𝑃

_𝑖

+ 𝑇

_{𝑖− Τ}_{1 2}

𝑃

_𝑖−1

− 𝑃

_𝑖

+

^𝑄^𝑖

𝐴Δ𝑥_𝑖

= 𝐶

_𝑝,𝑖

𝑃

_𝑖^𝑡+Δ𝑡

− 𝑃

_𝑖^𝑡

• 𝑃

₁^𝑡

, ⋯ , 𝑃

_𝑁^𝑡

are given

we know pressure at current time(t) at all grid blocks.

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(7)

Explicit Method

• Explicit : 𝑃

_𝑖+1^𝑡

, 𝑃

_𝑖^𝑡

, P

_i−1^t

, Only 𝑃

_𝑖^𝑡+Δ𝑡

is unknown

• 𝑇

_{𝑖+ Τ}_{1 2}

𝑃

_𝑖+1^𝑡

− 𝑃

_𝑖^𝑡

+ 𝑇

_{𝑖− Τ}_{1 2}

𝑃

_𝑖−1^𝑡

− 𝑃

_𝑖^𝑡

+

^𝑄^𝑖

𝐴Δ𝑥_𝑖

= 𝐶

_𝑝,𝑖

𝑃

_𝑖^𝑡+Δ𝑡

− 𝑃

_𝑖^𝑡

• Very Easy to solve!

✓ John Von Neumann stability analysis for homogeneous rock

✓Δ𝑡 ≤

¹

2

𝜙𝜇𝐶_𝑡

𝑘

Δ𝑥

²

✓ Conditionally Stable

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(8)

Implicit Method

• Implicit : 𝑃

_𝑖+1^𝑡+Δ𝑡

, 𝑃

_𝑖^𝑡+Δ𝑡

, P

_i−1^t+Δ𝑡

✓ Should solve inversion of a matrix

✓ but theoretically stable,

✓ but unstable for too large Δ𝑡, Δ𝑥,

✓ but much more stable than explicit

• 𝑇

_{𝑖+ Τ}_{1 2}

𝑃

_𝑖+1^𝑡+Δ𝑡

− 𝑃

_𝑖^𝑡+Δ𝑡

+ 𝑇

_{𝑖− Τ}_{1 2}

𝑃

_𝑖−1^𝑡+Δ𝑡

− 𝑃

_𝑖^𝑡+Δ𝑡

+

^𝑄^𝑖

𝐴Δ𝑥_𝑖

= 𝐶

_𝑝,𝑖

𝑃

_𝑖^𝑡+Δ𝑡

− 𝑃

_𝑖^𝑡

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(9)

Formulation of Implicit Method

Implicit

𝑇_{𝑖+ Τ}_{1 2} 𝑃_𝑖+1^𝑡+Δ𝑡 − 𝑃_𝑖^𝑡+Δ𝑡 + 𝑇_{𝑖− Τ}_{1 2} 𝑃_𝑖−1^𝑡+Δ𝑡 − 𝑃_𝑖^𝑡+Δ𝑡 + 𝑄_𝑖 𝐴Δ𝑥_𝑖

= 𝐶_𝑝,𝑖 𝑃_𝑖^𝑡+Δ𝑡 − 𝑃_𝑖^𝑡

𝑎_𝑖𝑃_𝑖−1^𝑡+Δ𝑡 + 𝑏_𝑖𝑃_𝑖^𝑡+Δ𝑡 + 𝑐_𝑖𝑃_𝑖+1^𝑡+Δ𝑡 = 𝑑_𝑖 , 𝑖 = 1,2, … , 𝑁

𝑇_𝑖−1/2𝑃_𝑖−1^𝑡+Δ𝑡 + −𝑇_{𝑖+ Τ}_{1 2} − 𝑇_𝑖−1/2 − 𝐶_𝑝,𝑖 𝑃_𝑖^𝑡+Δ𝑡 + 𝑇_𝑖+1/2𝑃_𝑖+1^𝑡+Δ𝑡

= −𝐶_𝑝,𝑖𝑃_𝑖^𝑡 − 𝑄_𝑖Τ𝐴Δ𝑥 𝑇_𝑖+1/2 = 1

Δ𝑥²

1 𝜇𝐵Τ _𝑖+1 + 1/ 𝜇𝐵 _𝑖

1 𝑘Τ _𝑖+1 + 1/𝑘_𝑖 , 𝑄_𝑖 ቊ 𝑖𝑛𝑗𝑒𝑐𝑡𝑖𝑜𝑛 ∶ + 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 ∶ − 𝐶_𝑝,𝑖 = 𝜙_𝑖𝐶_𝑡,𝑖

𝐵_𝑖Δt , 𝐶_𝑡,𝑖 = 𝐶_𝑟 + 𝐶_𝑓

(# of grid blocks)

𝑎_𝑖 𝑏_𝑖 𝑐_𝑖

𝑑_𝑖

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(10)

Simultaneous Linear Equations

• Assume we know 𝑃

_𝑖^𝑡

, calculate 𝑃

_𝑖^𝑡+Δ𝑡

𝑖 = 1,2,3,4,5

𝑖 = 1, 𝑎₁𝑃₀ + 𝑏₁𝑃₁ + 𝑐₁𝑃₂ = 𝑑₁ 𝑖 = 2, 𝑎₂𝑃₁ + 𝑏₂𝑃₂ + 𝑐₂𝑃₃ = 𝑑₂ 𝑖 = 3, 𝑎₃𝑃₂ + 𝑏₃𝑃₃ + 𝑐₃𝑃₄ = 𝑑₃ 𝑖 = 4, 𝑎₄𝑃₃ + 𝑏₄𝑃₄ + 𝑐₄𝑃₅ = 𝑑₄ 𝑖 = 5, 𝑎₅𝑃₄ + 𝑏₅𝑃₅ + 𝑐₅𝑃₆ = 𝑑₅

𝑏₁ 𝑐₁ 0 0 0 𝑎₂ 𝑏₂ 𝑐₂ 0 0 0 𝑎₃ 𝑏₃ 𝑐₃ 0 0 0 𝑎₄ 𝑏₄ 𝑐₄ 0 0 0 𝑎₅ 𝑏₅

𝑃₁ 𝑃₂ 𝑃₃ 𝑃₄ 𝑃₅

= 𝑑₁ 𝑑₂ 𝑑₃ 𝑑₄ 𝑑₅

𝐴 𝑥 𝑑

𝑥 = 𝐴⁻¹𝑑

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