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Given: (^) Equations Describing pipe flow

Find: (^) Non-dimensionalized equation; Dimensionless groups

Solution:

Nondimensionalizing the velocity, pressure, spatial measures, and time:

L

V t t L

r r L

x x p

p p V

u u = = = Δ

*= *= * * *

Hence

• • • • t * V

L u = Vu p =Δ pp x = Lx r = Dr t =

Substituting into the governing equation

⎟

⎟ ⎠

⎞ ⎜

⎜ ⎝

⎛

∂

∂

∂

∂

∂

∂ =− Δ ∂

∂

∂

∂

1

1 *

1 1 *

2

2

(^2) r

u

r r

u

D

V x

p

L

p t

u

L

V V t

u ν ρ

The final dimensionless equation is

⎟

⎟ ⎠

⎞ ⎜

⎜ ⎝

⎛

∂

∂

∂

∂ ⎟ ⎠

⎞ ⎜ ⎝

⎛ ⎟ ⎠

⎞ ⎜ ⎝

⎛

∂

Δ ∂ =− ∂

∂

1

2

2

(^2) r

u

r r

u

D

L

x DV

p

V

p

t

u ν

ρ

The dimensionless groups are

D

L

V DV

p ν

ρ 2

Δ

Given: (^) Equation for unsteady, 2D compressible, inviscid flow

Find: (^) Dimensionless groups

Solution:

Denoting nondimensional quantities by an asterisk

0

0

0 0 0

L Lc

tc t c

c c c

v v c

u u L

y y L

x x

Note that the stream function indicates volume flow rate/unit depth!

Hence

0

0 0 0 ψ^ Lc^ ψ

c

Lt x = Lx y = Ly u = c u v = c v c = c c t = =

Substituting into the governing equation

3 2 0 2

2 2 2

3 0 2

2 2 2

3 0

3 2 2 0 2

3 2 0 = ∂ ∂

x y

u v L

c

y

v c L

c

x

u c L

c

t

u v

L

c

L t

c ψ ψ ψ ψ

The final dimensionless equation is

2

2

2 2 2 2

2 2 2

2 2

2

2

∂ ∂

x y

u v y

v c x

u c t

u v

t

No dimensionless group is needed for this equation!

Given: (^) That drag depends on speed, air density and frontal area

Find: (^) How drag force depend on speed

Solution:

Apply the Buckingham Π procedure

c F V ρ A n = 4 parameters

d Select primary dimensions M , L , t

e

2 2 3

L L

M

t

L

t

ML

F V ρ A

r = 3 primary dimensions

f V ρ A m = r = 3 repeat parameters

g Then n – m = 1 dimensionless groups will result. Setting up a dimensional equation,

( ) 000 2

2 3

1

M L t t

ML L L

M

t

L

V AF

c

a b

a b c

⎟^ = ⎠

⎞ ⎜ ⎝

⎛ ⎟ ⎠

⎞ ⎜ ⎝

⎛

Π = ρ

Summing exponents,

: 2 0 2

: 3 2 1 0 1

: 1 0 1

− − = = −

− + + = =−

• = =−

t a a

L a b c c

M b b

Hence

V A

F (^1) ρ 2 Π =

h Check using F , L , t as primary dimensions

[ ] 1 2 2

2

4

(^1 ) Π = =

L t

L

L

Ft

F

The relation between drag force F and speed V must then be

2 2 F ∝ρ V A ∝ V

The drag is proportional to the square of the speed.

Π 1 = =[ ] 1 FL

FL

[ ] 1

2 Π 2 = = F

t

L

L

Ft

Π 3 = = [ ] 1 F

L

L

F

Note: Any combination of Π1, Π 2 and Π 3 is a Π group, e.g., (^3)

2

1

e

T

, so Π1, Π 2 and Π 3 are not unique!