Computational Fluid Dynamics I

Computational Fluid Dynamics I

Exercise 4

1. The vorticity transport equation for unsteady one-dimensional flow is given:

ω_t+uω_x =νω_xx

The viscosity ν (ν >0) and the velocity u=u(x, t) are assumed to be known.

The equation should discretised for constant time and spatial steps ∆t,∆x:

x_i =i∆x, tⁿ =n∆t, ω(x_i, tⁿ) = ωⁿ_i (a) Determine with the help of Taylor series:

• ω_t for tⁿ, resp. tⁿ⁺¹ (forward, resp. backward difference)

• ω_x and ω_xx around x_i (central differences)

(b) Formulate an explicit and an implicit solution scheme for the PDE and check the consistency.

Computational Fluid Dynamics I

Exercise 4 (solution)

1. (a) Discretisation of the time derivative:

Formulate Taylor series expansion for ωⁿ⁺¹ around ωⁿ:

ω_iⁿ⁺¹ =ωⁿ_i +ω_t|ⁿ_i ∆t+ω_tt|ⁿ_i ∆t²

2! + . . . and reformulate to getforward difference:

ω_t|ⁿ_i = ωⁿ⁺¹_i −ω_iⁿ

∆t − ω_tt|ⁿ_i ∆t

2 + . . . Formulate Taylor series expansion for ωⁿ around ωⁿ⁺¹:

ωⁿ_i =ω_iⁿ⁺¹− ω_t|ⁿ⁺¹_i ∆t+ ω_tt|ⁿ⁺¹_i ∆t²

2! + . . . and reformulate to getbackward difference:

ωt|ⁿ⁺¹_i = ω_iⁿ⁺¹−ωⁿ_i

∆t + ωtt|ⁿ⁺¹_i ∆t

2 + . . . Discretisation of the spatial derivative:

Formulate Taylor series expansion for ω_i+1 and ω_i−1 around ω_i:

ω_i±1ⁿ =ω_iⁿ± ω_x|ⁿ_i ∆x+ω_xx|ⁿ_i ∆x²

2! ± ω_xxx|ⁿ_i ∆x³

3! + ω_xxxx|ⁿ_i ∆x⁴

4! + . . . Subtract ωⁿ_i−1 fromω_i+1ⁿ to get finite difference expression for ωx:

ωx|ⁿ_i = ω_i+1ⁿ −ω_i−1ⁿ

2 ∆x − ωxxx|ⁿ_i ∆x²

6 + . . . Addω_i+1ⁿ and ω_i−1ⁿ to get finite difference expression for ω_xx:

ω_xx|ⁿ_i = ωⁿ_i+1−2ωⁿ_i +ω_i−1ⁿ

∆x² − ω_xxxx|ⁿ_i ∆x²

12 + . . .

(b) • Explicit solution scheme:

ω_iⁿ⁺¹−ω_iⁿ

∆t

| {z }

forward ωt

+uⁿ_i ω_i+1ⁿ −ω_i−1ⁿ 2 ∆x

| {z }

central ωx

−ν ωⁿ_i+1−2ωⁿ_i +ω_i−1ⁿ

∆x²

| {z }

central ωxx

= 0

Explicit, since only one termω_iⁿ⁺¹is defined at the highest time level (n+1), thus the equation can be explicitly solved.

⇒ ω_iⁿ⁺¹ = fi(ωⁿ, uⁿ, ν,∆t,∆x)

truncation error τ = L(ω) − L_∆(ω)

= −ω_tt|ⁿ_i ∆t

2 − u ω_xxx|ⁿ_i ∆x²

6 + ν ω_xxxx|ⁿ_i ∆x²

12 + terms of higher order

= O(∆t,∆x²) consistent, since lim

∆t,∆x→0τ = 0

• Implicit solution scheme:

ω_iⁿ⁺¹−ω_iⁿ

∆t

| {z }

backward ωt

+uⁿ⁺¹_i ωⁿ⁺¹_i+1 −ωⁿ⁺¹_i−1 2 ∆x

| {z }

central ωx

−νω_i+1ⁿ⁺¹−2ω_iⁿ⁺¹+ω_i−1ⁿ⁺¹

∆x²

| {z }

central ωxx

= 0

Several terms ωⁿ⁺¹_i , ωⁿ⁺¹_i−1, and ω_i+1ⁿ⁺¹ are defined at the highest time level (n+ 1), therefore the equation can not be explicityly solved. The unknowns at the highest time level are implicityly coupled and build a tridiagonal system of equations:

⇒ tridiagonal system of equations aiω_i−1ⁿ⁺¹+biωⁿ⁺¹_i +ciω_i+1ⁿ⁺¹ = fi(ωⁿ, uⁿ, ν,∆t,∆x)

truncation error

τ = ωtt|ⁿ⁺¹_i ∆t

2 − u ωxxx|ⁿ⁺¹_i ∆x²

6 + ν ωxxxx|ⁿ⁺¹_i ∆x²

12 + terms of higher order

= O(∆t,∆x²) consistent, since lim

∆t,∆x→0τ = 0