Chapter 4M. Darwish (darwish@aub.edu.lb)

American University of BeirutMECH 663

The Finite Volume Method

The General Scalar Equation

�

∂ ρφ( )∂t

transient term

+ ∇. ρuφ( )convective term

= ∇ ⋅ Γ∇φ( )diffusion term

+ Qφ φ( )source term

Time derivative

Source term

Control VolumeCV

Advection

Diffusion

The scalar equation is a balance equation written in differential form

Balance Form

�

∂ ρφ( )∂t

dVV∫ + ρuφ( ) ⋅ dS

∂V∫ = Γ∇φ( ) ⋅ dS

∂V∫ + Q φ( )dV

V∫

Time derivative

Source term

Control VolumeCV

Advection

Diffusion

∂V

V

�

∂ ρφ( )∂t

dVVP

∫ + J ⋅ dSf∫

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

f = nb VP( )∑ = Qφ φ( )dVP

VP

∫

To recover the balance form we integrate over some CV

�

∂ ρφ( )∂t

dVV∫ + ρuφ( ) ⋅ dS

f∫

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

f = nb V( )∑ − Γ∇φ( ) ⋅ dS

f∫

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

f = nb V( )∑ = Q φ( )dV

V∫

�

J = JC + JD

= ρuφ( ) − Γ∇φ( )P

Sf1f1

Sf2

f2

f3 f4

Sf4

Sf3

discrete CV

Flux Integration

�

ρuφ( ) ⋅ dSf∫

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

f = nb VP( )∑ − Γ∇φ( ) ⋅ dS

f∫

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

f = nb VP( )∑

P

Nf

P

Nf

P

Nf

1 point gauss integration

2 point gauss integration

3 point gauss integration

�

J ⋅ dSf∫ = J ⋅S( )ip

ip( f )∑

�

= Jip ⋅ wipS fip( f )∑

�

ρuφ( ) f ⋅S f

�

ρuφ( )1 f( ) ⋅ w1 f( )S f + ρuφ( )2 f( ) ⋅ w2 f( )S f

Volume Integration

P

1 point gauss integration

4 point gauss integration

9 point gauss integration

P P

�

Q φ( )dVPVP

∫

�

∂ ρφ( )∂t

dVVP

∫

�

= QipVip( )ip(P )∑

�

QPVP

�

Q1 P( )V1 P( ) + Q2 P( )V2 P( ) + Q3 P( )V3 P( ) + Q4 P( )V4 P( )

Semi-Discretized Equation

�

wip ρuφ( )ip ⋅S f( )ip~ ip( f )∑

f = nb VP( )∑ − wip Γ∇φ( )ip ⋅S f( )

ip~ ip( f )∑

f ~ faces VP( )∑ = wipQipΩP

ip~ ip VP( )∑

PN1

N4

N3

N2

J1

J2

J3

J4

�

ρuφ −Γ∇φ( )b ⋅Sb

BC: Flux Specified

bnt

P

Sb

e

dPb

qb,specified

�

−Γ∇φ( )b ⋅nbSpecified Flux

= qb,specified

�

ub = 0

�

= −Γ∇φ( )b ⋅nbSb

�

= qb,specified Sb

Wall

BC: Value Specified

b

n

tP

Sb

Φb,specified

Inlet

�

φb = φb,specified

�

ρu( )b known

�

Jb ⋅Sb = ρuφ( )b ⋅Sb

No Diffusion

�

= ρbub ⋅Sb( )φb,specified

�

= Fbφb,specified

Spatial Variation

�

φ(x) = φP + x − xP( ) ⋅ ∇φ( )P + 12 x − xP( )2 : ∇∇φ( )P

+ 13! x − xP( )3 :: ∇∇∇φ( )P + …

+ 1n! x − xP( )n:::

n(∇∇∇

n φ)P + …

�

φ(x) = φP + x − xP( ) ⋅ ∇φ( )P + O Δx 2( )ΦP

Φ(x)

∇ΦPxP

x

∇∇ΦP

Mean Value Theorem

�

φ P = 1ΩP

φ( )dΩΩP

∫

= 1ΩP

φP + x − xP( ) ⋅ ∇φ( )P + O Δx 2( )[ ]dΩΩP

∫

= φP1ΩP

dΩΩP

∫ + 1ΩP

x − xP( )dΩΩP

∫ ⋅ ∇φ( )P + 1ΩP

O Δx 2( )dΩΩP

∫

= φP + O Δx 2( )

ΦP

Problem 3.4Consider one-dimensional diffusion in the calculation domain shown in the figure below with ∆x =1 for all control volumes. Assume that the source term Q=50x, and that Γ is constant with a value Γ=1. Let the boundary values of f be Φ0=100 and Φ4=500a Write the discrete equations for each of the cells 1, 2 and 3.Solve the discrete equation set using Gauss-Seidel iteration and report the resulting cell-centroid valuesCompute the diffusion flux at each of the cell faces f0, f12, f23 and f4 using the same discretization approximations made in obtaining the discrete equationsShow that the conservation principle is satisfied on each discrete cell and on the whole domainFind the exact solution to this problem. Find the percentage error in the computed solution at each cell centroid

21 3

(δx) (δx)

(∆x)

(δx)/2 (δx)/2

(∆x)(∆x)

Φ0=1000 4

Φ4=500

Problem 3.5Consider the computational domain in the figure below. Using the conservation of mass principle compute the mass flux at faces f12, f23 and f4.

Write the steady state transport equation for a scalar, Φ, being advected through the domain with no diffusion and no source terma. Write the discrete equations for cells 1, 2 and 3 (Use the upwind profile),b. If we assume that (dΦ/dx)4 = 0 compute the value of Φ4.

c. If (∆x=1) and (∆y=0.5) compute the values of Φ1, Φ2, Φ3 and Φ4.

21 3

(δx) (δx)

(∆x)

(δx)/2 (δx)/2

(∆x)

(∆x)

0 4(∆y)

(3∆y)

Problem 3.6Consider the computational domain shown below. Write the diffusion equation for this domain for Γ=1 and Q=0Write discrete equations for cells 1,2 and 3Can you find a unique solution for Φ1, Φ2 and Φ3. Explain ?

21 3

(δx) (δx)

(∆x)

(δx)/2 (δx)/2

(∆x)(∆x)

0 4