## A GENERAL THEORY OF THE HYDRAULIC

TRANSPORT OF SOLIDS IN
FULL SUSPENSION

By A C Bonapace

**VI. APPLICATIONS**

Let one apply to the below development the conditions x > 0 andμ ≠ 0, sometimes separately, sometimes superimposed (cf. Ch V at b and
at c).

From the energetic point of view scalar superimposition of effects is
to be considered as physically legitimate.

Thus one deals with symbols of the form Ω_{μx} ...etc. Further in the μ, x plane the particle diameter has a direct role in
defining the volumetric concentration x as per Eqs. (5.5), (5.11) and (5.23).

1^{o}) In a first worked example one calculates in the generating plane the hydraulic gradient required for the conveyance of a cylindrical body
inside a smooth wall pipe.

Putting for the size of the cylindrical body the condition → i.e. for δ10 → –η being η a small quantity, it results from the
calculations that a finite hydraulic gradient is required, of an order of
magnitude, which can be verified experimentally.

Denoting by Ilx the hydraulic gradient one gets from Eq. (5.21) and
(3.2*)

(6.1)

For x → 1 - η one writes:

(6.2)

Developing in a power series expansion, one writes for (1 – x^{2} )^{3} while neglecting terms of power greater than one:

(6.3)

With the denominator of Eq. (6.1) expressed by (6.3) and putting because of the arbitrariness of η and of μ_{10}, (both small quantities):

(6.4)

one obtains for a cylinder (of length Λ = Δ°_{10} )

(6.5)

which is the finite value of the hydraulic gradient insuring transport in absence of inertial impacts upon the walls.

From the technical literature dealing with capsule transport it is of
interest to mention the work of Ellis [4]. In Fig. 2 of the referred paper
the hydraulic gradient required to support a capsule of size = 0,95 is equal to = 0,083, this for a capsule of relative density = 1 + μ = 1,033.^{#}

2°) In this second worked example one departs from conditions Ω_{10} and Δ_{10} defined in the generating plane and calculate for a certain particle of size .

This will be applied to the transport of sand in an air stream.

The data of the problem are given as below:

Air temperature 30°C, kinematics viscosity of the air 16 x .

For the velocity scale (cf. Eq. 2.2):

For the distance scale (cf. Eq. 2.3)

For a sand/air density ratio = 1 + μ = 2285

For a water/air density ratio

one gets from Eq. (5.2):

f_{μo} = 0,1214_{μ}(1+μ)^{1−3α}

Select for α the value 0,730 i.e.

1 – = 1 – 3 x 0,730 = – 1,19

Hence from Eq. (5.2).

f_{μo} = 0,1214 x 2284 x 2285^{−1,19} = 0,02791.

Let one consider an unspecified small particle and a smooth wall
pipe of diameter Δ°_{10} as below:

= 3,370.

Further on account of Eq. (3.12).

Then from Eqs. (4.3) and (4.4)

Expressing the Reynolds number of the stream RE_{μo}, this is given by:

At the above Reynolds number the value f_{μo} = 0,02791 is in agreement with the value selected.

Hence one can write as per Eq. (2.6)

and

Let assume that the very small particle has been increased to a reasonable size, e.g. to 1 mm i.e. with = 3,37 the stream carries a line of cylindrical particles of diameter ratio

as per Eq. (5.11).

Further taking into account the factor β = 1,11 as per Eq. (5.23) one gets from Eq. (5.20) applied to spherical particle, i.e. for βx^{1/2} =
1,11x0,1237 = 0,137

Analogously for the hydraulic gradient

Passing to dimensional symbols by the scale factors introduced above one gets for the dimensional velocity and pipe diameter

and

Further expressing the hydraulic gradient in meters of water column per meter of pipe, i.e. dividing by 862,3 one gets

Finally for x^{1/2} = 0,1237 one gets x = 0,0153 as the actual volumetric concentration of sand particles transported.

3°) As a third worked example one calculates the suspension velocity Ω_{μo} for a buoyant particle of excess density μ = -0,50 i.e. for 1 +μ = 1 – 0,50 = 0,50

Let one consider a flow in a smooth pipe

For a stream Reynolds number

one gets a friction factor

Let one introduce the convention that to a buoyant particle a negative friction factor f_{μo} applies. Hence writing in Eq. (5.2) f_{μo} =
-0,0240 one gets:

with = - 0,113 substituted in Eqs. (4.3) and (4.4) one gets with (3.12):

(I)

Further

(II)

Hence from (I) and (II):

= 42,32

= 708,8

and from Eqs. (4.3) and (4.4) one gets in the generating plane:

and

with values of Ω_{10} and of Δ°_{10} satisfying Eq. (3.12) as it should be.

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# Applying to = 0.083 the values of the experiments η = 1 – 0,95 = 0,050 and μ = 0,033 one gets for the calculated value .

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