Dimensional analysis

8.1
A stationary sphere in water moving at a velocity of 1.6m/s experiences a drag of 4N. Another sphere of twice the diameter is placed in a wind tunnel. Find the velocity of the air and the drag which will give dynamically similar conditions. The ratio of kinematic viscosities of air and water is 13, and the density of air 1.28 kg/m³.
[10.4m/s 0.865N]

Draw up the table of values you have for each variable:

variable	water	air
u	1.6m/s	uair
Drag	4N	Dair
n	n	13n
r	1000 kg/m3	1.28 kg/m3
d	d	2d

Kinematic viscosity is dynamic viscosity over density = n = m/r.

The Reynolds number =

Choose the three recurring (governing) variables; u, d, r.

From Buckinghams p theorem we have m-n = 5 - 3 = 2 non-dimensional groups.

As each p group is dimensionless then considering the dimensions, for the first group, p₁:

(note D is a force with dimensions MLT^-2)

M] 0 = c₁ + 1

c₁ = -1

L] 0 = a₁ + b₁ - 3c₁ + 1

-4 = a₁ + b₁

T] 0 = -a₁ - 2

a₁ = - 2

b₁ = -2

And the second group p₂ :

M] 0 = c₂

L] 0 = a₂ + b₂ - 3c₂ + 2

-2 = a₂ + b₂

T] 0 = -a₂ - 1

a₂ = -1

b₂ = -1

So the physical situation is described by this function of nondimensional numbers,

For dynamic similarity these non-dimensional numbers are the same for the both the sphere in water and in the wind tunnel i.e.

For p₁

For p₂

8.2
Explain briefly the use of the Reynolds number in the interpretation of tests on the flow of liquid in pipes.
Water flows through a 2cm diameter pipe at 1.6m/s. Calculate the Reynolds number and find also the velocity required to give the same Reynolds number when the pipe is transporting air. Obtain the ratio of pressure drops in the same length of pipe for both cases. For the water the kinematic viscosity was 1.3110^-6 m²/s and the density was 1000 kg/m³. For air those quantities were 15.110^-6 m²/s and 1.19kg/m³.
[24427, 18.4m/s, 0.157]

Draw up the table of values you have for each variable:

variable	water	air
u	1.6m/s	uair
p	pwater	pair
r	1000 kg/m3	1.19kg/m3
n	1.3110-6m2/s	15.110-6m2/s
r	1000 kg/m3	1.28 kg/m3
d	0.02m	0.02m

Kinematic viscosity is dynamic viscosity over density = n = m/r.

The Reynolds number =

Reynolds number when carrying water:

To calculate Re_air we know,

To obtain the ratio of pressure drops we must obtain an expression for the pressure drop in terms of governing variables.

Choose the three recurring (governing) variables; u, d, r.

From Buckinghams p theorem we have m-n = 5 - 3 = 2 non-dimensional groups.

As each p group is dimensionless then considering the dimensions, for the first group, p₁:

M] 0 = c₁

L] 0 = a₁ + b₁ - 3c₁ + 2

-2 = a₁ + b₁

T] 0 = -a₁ - 1

a₁ = -1

b₁ = -1

And the second group p₂ :

(note p is a pressure (force/area) with dimensions ML^-1T^-2)

M] 0 = c₂ + 1

c₂ = -1

L] 0 = a₂ + b₂ - 3c₂ - 1

-2 = a₂ + b₂

T] 0 = -a₂ - 2

a₂ = - 2

b₂ = 0

So the physical situation is described by this function of nondimensional numbers,

For dynamic similarity these non-dimensional numbers are the same for the both water and air in the pipe.

We are interested in the relationship involving the pressure i.e. p₂

8.3
Show that Reynold number, rud/m, is non-dimensional. If the discharge Q through an orifice is a function of the diameter d, the pressure difference p, the density r, and the viscosity m, show that Q = Cp^1/2d²/r^1/2 where C is some function of the non-dimensional group (dr^1/2p^1/2/m).
Draw up the table of values you have for each variable:

The dimensions of these following variables are

r ML^-3

u LT^-1

d L

m ML^-1T^-1

Re = ML^-3 LT^-1L(ML^-1T^-1)^-1 = ML^-3 LT^-1 L M^-1LT = 1

i.e. Re is dimensionless.

We are told from the question that there are 5 variables involved in the problem: d, p, r, m and Q.

Choose the three recurring (governing) variables; Q, d, r.

From Buckinghams p theorem we have m-n = 5 - 3 = 2 non-dimensional groups.

As each p group is dimensionless then considering the dimensions, for the first group, p₁:

M] 0 = c₁ + 1

c₁ = -1

L] 0 = 3a₁ + b₁ - 3c₁ - 1

-2 = 3a₁ + b₁

T] 0 = -a₁ - 1

a₁ = -1

b₁ = 1

And the second group p₂ :

(note p is a pressure (force/area) with dimensions ML^-1T^-2)

M] 0 = c₂ + 1

c₂ = -1

L] 0 = 3a₂ + b₂ - 3c₂ - 1

-2 = 3a₂ + b₂

T] 0 = -a₂ - 2

a₂ = - 2

b₂ = 4

So the physical situation is described by this function of non-dimensional numbers,

The question wants us to show :

Take the reciprocal of square root of p₂: ,

Convert p₁ by multiplying by this number

then we can say

8.4
A cylinder 0.16m in diameter is to be mounted in a stream of water in order to estimate the force on a tall chimney of 1m diameter which is subject to wind of 33m/s. Calculate (A) the speed of the stream necessary to give dynamic similarity between the model and chimney, (b) the ratio of forces.

Chimney: r = 1.12kg/m³ m = 1610^-6 kg/ms

Model: r = 1000kg/m³ m = 810^-4 kg/ms

[11.55m/s, 0.057]

Draw up the table of values you have for each variable:

variable	water	air
u	uwater	33m/s
F	Fwater	Fair
r	1000 kg/m3	1.12kg/m3
m	810-4 kg/ms	1610-6kg/ms
d	0.16m	1m

Kinematic viscosity is dynamic viscosity over density = n = m/r.

The Reynolds number =

For dynamic similarity:

To obtain the ratio of forces we must obtain an expression for the force in terms of governing variables.

Choose the three recurring (governing) variables; u, d, r, F, m.

From Buckinghams p theorem we have m-n = 5 - 3 = 2 non-dimensional groups.

As each p group is dimensionless then considering the dimensions, for the first group, p₁:

M] 0 = c₁ + 1

c₁ = -1

L] 0 = a₁ + b₁ - 3c₁ - 1

-2 = a₁ + b₁

T] 0 = -a₁ - 1

a₁ = -1

b₁ = -1

i.e. the (inverse of) Reynolds number

And the second group p₂ :

M] 0 = c₂ + 1

c₂ = -1

L] 0 = a₂ + b₂ - 3c₂ - 1

-3 = a₂ + b₂

T] 0 = -a₂ - 2

a₂ = - 2

b₂ = -1

So the physical situation is described by this function of nondimensional numbers,

For dynamic similarity these non-dimensional numbers are the same for the both water and air in the pipe.

To find the ratio of forces for the different fluids use p₂

8.5
If the resistance to motion, R, of a sphere through a fluid is a function of the density r and viscosity m of the fluid, and the radius r and velocity u of the sphere, show that R is given by

Hence show that if at very low velocities the resistance R is proportional to the velocity u, then R = kmru where k is a dimensionless constant.
A fine granular material of specific gravity 2.5 is in uniform suspension in still water of depth 3.3m. Regarding the particles as spheres of diameter 0.002cm find how long it will take for the water to clear. Take k=6p and m=0.0013 kg/ms.
[218mins 39.3sec]

Choose the three recurring (governing) variables; u, r, r, R, m.

From Buckinghams p theorem we have m-n = 5 - 3 = 2 non-dimensional groups.

As each p group is dimensionless then considering the dimensions, for the first group, p₁:

M] 0 = c₁ + 1

c₁ = -1

L] 0 = a₁ + b₁ - 3c₁ - 1

-2 = a₁ + b₁

T] 0 = -a₁ - 1

a₁ = -1

b₁ = -1

i.e. the (inverse of) Reynolds number

And the second group p₂ :

M] 0 = c₂ + 1

c₂ = -1

L] 0 = a₂ + b₂ - 3c₂ - 1

-3 = a₂ + b₂

T] 0 = -a₂ - 2

a₂ = - 2

b₂ = -1

So the physical situation is described by this function of nondimensional numbers,

or

he question asks us to show or

Multiply the LHS by the square of the RHS: (i.e. p_2(1/p₁²) )

So

The question tells us that R is proportional to u so the function f must be a constant, k

The water will clear when the particle moving from the water surface reaches the bottom.

At terminal velocity there is no acceleration - the force R = mg - upthrust.

From the question:

s = 2.5 so r = 2500kg/m³ m = 0.0013 kg/ms k = 6p

r = 0.00001m depth = 3.3m