The matrix method of dimensional analysis

Example 11.2a of (page 347-349) of the prescribed text book (Chadwick and Morfett) is used to explain how the method works. The hydraulic head loss hf, in a simple pipeline is assumed to depend on the following quantities: The density (ρ) and viscosity (µ) of the fluid The diameter D, length L, and roughness ks of the pipe A typical flow velocity – the mean velocity V.

Develop the appropriate dimensionless groups to describe the flow.

A dimensionless group Π incorporating all the variables would take the form,

Π = ρ k 1 D k 2V k 3h f k 4 µ k 5 Lk 6 k s k 7

where k1, k2, ……, k7 are the unknown indices.

(1)

The variables considered important; ρ, D and V, are used as governing variables and are placed to allow them to locate on the first three columns of the matrix of dimensional equations.

Substituting the dimensions of the variables, noting that equation,

Π is dimensionless we obtain the dimensional

(2)

[M

0 0

L T 0 = ML−3

] [

]

k1

[L]k 2 [LT −1 ]k 3 [L]k 4 [ML−1T −1 ]k 5 [L]k 6 [L]k 7

Three equations relating to the dimensions are obtained: M: 0 = k1+k5 L: 0=-3k1+k2+k3+k4-k5+k6+k7 T: 0=-k3-k5 The dimensional equations (3) may be presented in matrix form as: (3)

M L T

ρ

1 −3 0

D 0 1 0

V 0 1 −1

hf 0 1 0

µ

L

1 0 −1 1 −1 0

k1 k 2 ks k 3 0 0 k4 = 0 1 k 5 0 0 k6 k 7

(4)

The matrices of equation 4 are partitioned at the locations indicated below separating the governing and the other variables

1

K

M L T

ρ D V hf µ L

1 0 0 0 1 0 −3 1 1 1 −1 1 0 0 −1 0 −1 0

A B

k1 k 2 ks k3 0 0 k4 = 0 1 k5 0 0 k6 k 7

K

M A= L T

ρ

1 −3 0

D 0 1 0

V 0 1 − 1

M B= L T

hf 0 1 0

L ks 0 0 −1 1 1 −1 0 0 1

µ

k1 KU = k 2 k3

Therefore AKU...