Hydromechanical processes

Fluids: liquids (almost incompressible)

vapors (below critical temperature)

gases (above critical temperature)

Hydromechanics: hydrostatics

hydrodynamics

Continuum – a theoretical model of continuous mass distribution in spaces

real situation: discrete distribution of mass → description by statistical mechanics (difficult)

model situation: continuum → description by differential and integral calculus (“easy”)

Continuum model can be used for macroscopic systems → above certain scale given by elementary

volume of continuum δV = particle of continuum

▪ δV must be large enough to avoid effects of molecular/atomic fluctuations

▪ δV must be small enough compared to the system we want to describe

Field of a physical quantity: describes distribution of an intensive property in space and time

intensive property does not depend on the size (density, temperature)

extensive property depends on size, it is additive (mass, volume)

In steady state the quantity does not depend on time, but may depend on space, e.g. velocity of steady

state flow in pipe depends on radius

Important quantities for chemical engineering are velocity (v), pressure (p), molar concentration (c),

temperature (t), etc.

Forces acting in a fluid

�⃗�𝑉 external – act on volume or mass

�⃗�𝑆 internal – act on surfaces

external force per unit mass: 𝑓 =d�⃗�𝑉

d𝑚 … acceleration (gravitational, centrifugal)

Remark: Newton 2nd

law ∑ �⃗�𝑖 = 𝑚�⃗� is a special case of momentum balance

𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚�⃗� → ∑ �⃗�𝑖 = 𝑚d�⃗⃗�

d𝑡=

d(m�⃗⃗�)

d𝑡

Internal forces

pressure: 𝑝 = −𝜎 acts from outside the system on its boundary

Unit of σ, p and τ is N m-2

= Pa

Fluid flow: determined by velocity and pressure fields

- we will consider unidirectional flow, i.e. the vector v becomes scalar v

Volume (volumetric) flow

volume of fluid passing through a chosen cross-section per unit time t

�̇� =d𝑉

d𝑡

velocity v is volume flow per unit area

𝑣 =d�̇�

dS, d𝑆 … element of cross-sectional area perpendicular to the flow

average (mean) velocity (over the entire cross-section S)

�̇� = ∫ d�̇� =

𝑆

∫ vdS = ⟨𝑣⟩𝑆 →

𝑆

⟨𝑣⟩ = �̇�

𝑆

Character of the flow (flow pattern)

Laminar flow (at low velocities)

- filaments are smooth

- in perpendicular direction to the flow only diffusional (=molecular) transport of mass, momentum

and heat occurs

- this transport is slow and can be empirically described → viscosity

Turbulent flow (at high velocities)

- filaments are wavy → random oscillations

- flow is complex → rotation → eddies (vortices)

- in addition to diffusive transport in the perpendicular direction there is convective transport via

fluid particles much larger than molecules > δV

Transition from laminar to turbulent flow

- occurs rather suddenly

- is characterized by Reynolds number

Re = 𝑣𝑙

𝜈=𝑣𝑙𝜌

𝜂

v … velocity

l … characteristic length (e.g. diameter of a tube)

ν … kinematic viscosity

η … dynamic viscosity (𝜈 =𝜂

𝜌)

ρ … density

Re < Recritical → laminar flow

Re > Recritical → turbulent flow

For a pipe, Recritical = 2300

Structure of turbulent flow

viscosity (explained for a laminar flow)

Newton viscosity law:

𝜏𝑥(𝑦) = −𝜂d𝑣𝑥(𝑦)

d𝑦

- viscosity is a macroscopic measure of internal friction (= transport of momentum by diffusion)

- ideal fluid: η = 0, no internal friction

- Newtonian fluids: η = const.

- non-Newtonian fluids: η = η(τx)

Mass balance of a flowing fluid – continuity equation

Overall mass balance in a flow system (a pipe):

inflow + source (=0) = outflow + accumulation (source of overall mass is always zero!)

�̇�1 = �̇�2 +d𝑚𝑆

d𝑡, where 𝑚 = 𝑉𝜌 = 𝑣𝑆𝜌; v … average velocity, S … cross-section, ρ … density

continuity equation at steady state:

𝑣1𝑆1𝜌1 = 𝑣2𝑆2𝜌2

continuity equation at steady state, constant density (hold for liquids):

𝑣1𝑆1 = 𝑣2𝑆2

Energy balance of flowing fluid – Bernoulli equation

Energy E may have different forms, most inportant are:

▪ internal energy U

▪ work W

▪ heat Q

▪ kinetic energy 𝐸𝑘𝑖𝑛 = 1

2𝑚𝑣2

▪ potential energy 𝐸𝑝𝑜𝑡 = 𝑚𝑔𝑧

Total energy balance (there is no source)

�̇�1 + �̇� + �̇� = �̇�2 +𝑑𝐸𝑆𝑑𝑡

Work:

i) flow work (volume work) – net work done by pressure forces as they push the fluid through intput

and output, i.e 𝑝1𝑉1 − 𝑝2𝑉2

ii) external mechanical work – done by stirring, pumping etc., flow of mechanical work is called

power P

Remark: in closed systems where kinetic and potential energy can be neglected →

�̇� + �̇� =𝑑𝑈

𝑑𝑡 or 𝑄 +𝑊 = ∆𝑈

Assume an adiabatic system (�̇� = 0), then at steady state:

�̇�1 + �̇�𝑘𝑖𝑛,1 + �̇�𝑝𝑜𝑡,1 + 𝑝1�̇�1 + 𝑃 = �̇�2 + �̇�𝑘𝑖𝑛,2 + �̇�𝑝𝑜𝑡,2 + 𝑝2�̇�2

Assume constant density, �̇�1 = �̇�2 = �̇�, at steady state also �̇�1 = �̇�2 = �̇�.

�̇�1 +1

2�̇�𝑣1

2 + �̇�𝑔𝑧1 + 𝑝1�̇�

𝜌+ 𝑃 = �̇�2 +

1

2�̇�𝑣2

2 + �̇�𝑔𝑧2 + 𝑝2�̇�

𝜌

𝑣12

2+ 𝑔𝑧1 +

𝑝1𝜌+𝑃

�̇�= 𝑣22

2+ 𝑔𝑧2 +

𝑝2𝜌+�̇�2 − �̇�1�̇�

𝒗𝟏𝟐

𝟐+ 𝒈𝒛𝟏 +

𝒑𝟏𝝆+𝑷

�̇�= 𝒗𝟐𝟐

𝟐+ 𝒈𝒛𝟐 +

𝒑𝟐𝝆+ 𝒆𝒅𝒊𝒔

′

This is Bernoulli equation for a flow of incompressible fluid in a pipe with a pump.

The energy loss 𝑒𝑑𝑖𝑠′ is due to friction in the fluid in the pipe and in the pump (dis ≡ dissipation ≡ loss of

mechanical energy)

𝑒𝑑𝑖𝑠′ = 𝑒𝑑𝑖𝑠 + 𝑒𝑑𝑖𝑠,𝑝𝑢𝑚𝑝

Thus the Bernoulli equation can be written as:

𝑣12

2+ 𝑔𝑧1 +

𝑝1𝜌+ (

𝑃

�̇�− 𝑒𝑑𝑖𝑠,𝑝𝑢𝑚𝑝) =

𝑣22

2+ 𝑔𝑧2 +

𝑝2𝜌+ 𝑒𝑑𝑖𝑠

𝑣12

2+ 𝑔𝑧1 +

𝑝1𝜌+ 𝑒𝑝𝑢𝑚𝑝 =

𝑣22

2+ 𝑔𝑧2 +

𝑝2𝜌+ 𝑒𝑑𝑖𝑠 ; 𝑒𝑝𝑢𝑚𝑝 =

𝑁

�̇�

where

P … input power to the pump

N … output power from the pump to the liquid

epump … specific output power

Efficiency of the pump: 𝜂𝑝𝑢𝑚𝑝 =𝑁

𝑃

Versions of the Bernoulli equation

1) energy form

𝒗𝟏𝟐

𝟐+ 𝒈𝒛𝟏 +

𝒑𝟏𝝆+ 𝒆𝒑𝒖𝒎𝒑 =

𝒗𝟐𝟐

𝟐+ 𝒈𝒛𝟐 +

𝒑𝟐𝝆+ 𝒆𝒅𝒊𝒔

2) height form

𝑣12

2𝑔+ 𝑧1 +

𝑝1𝜌𝑔+𝑒𝑝𝑢𝑚𝑝𝑔

= 𝑣22

2𝑔+ 𝑧2 +

𝑝2𝑔𝜌+𝑒𝑑𝑖𝑠𝑔

𝒗𝟏𝟐

𝟐𝒈+ 𝒛𝟏 +

𝒑𝟏𝝆𝒈

+𝑯𝒑𝒖𝒎𝒑 = 𝒗𝟐𝟐

𝟐𝒈+ 𝒛𝟐 +

𝒑𝟐𝒈𝝆

+ 𝒉𝒅𝒊𝒔

3) pressure form

𝜌𝑣12

2+ 𝜌𝑔𝑧1 + 𝑝1 + 𝜌𝑒𝑝𝑢𝑚𝑝 =

𝜌𝑣22

2+ 𝜌𝑔𝑧2 + 𝑝2 + 𝜌𝑒𝑑𝑖𝑠

𝝆𝒗𝟏𝟐

𝟐+ 𝝆𝒈𝒛𝟏 + 𝒑𝟏 + ∆𝒑𝒑𝒖𝒎𝒑 =

𝝆𝒗𝟐𝟐

𝟐+ 𝝆𝒈𝒛𝟐 + 𝒑𝟐 + ∆𝒑𝒅𝒊𝒔

Remarks:

i) For ideal fluids → no friction → edis = 0

ii) If there is no pump → epump = 0

iii) If the fluid does not flow → v = 0 → edis = 0 and epump = 0

For v = 0 Bernoulli equation reduces to equation of hydrostatics:

𝑔𝑧1 +𝑝1𝜌= 𝑔𝑧2 +

𝑝2𝜌= 𝑐𝑜𝑛𝑠𝑡 = 𝑔𝑧 +

𝑝

𝜌

hydrostatic pressure: 𝑝 − 𝑝1 = 𝜌𝑔(𝑧 − 𝑧1) = 𝜌𝑔ℎ

Application of Bernoulli equation to a fluid with internal friction (= real fluid)

We need to be able to express the energy loss edis in terms of characteristics of the fluid.

a) for a straight pipe

𝑒𝑑𝑖𝑠 = 𝜆𝑙

𝑑

𝑣2

2 (empirical formula)

λ … friction factor (or coefficient)

l … length of the pipe

d … diameter of the pipe

𝑣 = �̇� 𝑆⁄ … average velocity

λ depends on the flow

▪ laminar flow: 𝜆 =𝑐𝑜𝑛𝑠𝑡

Re, Re ≤ 2300, const = 64 (for circular cross-section)

- where Re = 𝑣𝑑𝜌

𝜂

- for non-circular cross-sections: 𝑑 → 𝑑𝑒𝑞𝑢𝑖𝑣 =4𝑆

𝑠

S … area of flow cross-section

s … wetted circumference

▪ turbulent flow (Re >2300)

- λ = λ(Re,ε/d)

ε/d … relative roughness of the pipe material

ε … roughness (can be found in tables)

1

√𝜆= −2log (

2.51

Re√𝜆+ 0.28

𝜀

𝑑) Moody

𝜆 =0.25

{−log [(6.81Re

)0.9

+𝜀𝑑⁄

3.7]}

2

b) for fittings (= resistances in the pipeline, such as valves, turns, bends, etc.)

𝑒𝑑𝑖𝑠 = 𝜁𝑣2

2, 𝜁 … loss coefficient (found in tables)

or 𝑒𝑑𝑖𝑠 = 𝜆𝑙𝑒𝑞𝑢𝑖𝑣𝑑

𝑣2

2, 𝑙𝑒𝑞𝑢𝑖𝑣 … length of a tube that has equivalent resistance

For pipelines with several fittings we have:

𝑒𝑑𝑖𝑠 = (𝜆𝑙

𝑑+∑𝜁𝑗

𝑗

)𝑣2

2 𝑜𝑟 𝑒𝑑𝑖𝑠 = (𝜆

𝑙 + ∑ 𝑙𝑒𝑞𝑢𝑖𝑣,𝑗𝑗

𝑑)𝑣2

2

Energy loss during discharge of fluid from container

For ideal fluid:

𝑣2

2+ 𝑔𝑧 +

𝑝

𝜌= �̃�02

2+ 𝑔𝑧0 +

𝑝0𝜌

𝑣𝑆 = �̃�0𝑆0, �̃�0… velocity of ideal fluid

On expressing v0:

�̃�0 = √

2

1 − (𝑆0𝑆 )

2 (𝑝 − 𝑝0𝜌

+ 𝑔ℎ)

Correction for a real fluid:

𝑣0 = µ�̃�0 , µ … discharge coefficient (empirical)

For a pipeline with no pumps we can solve three kinds of problems:

a) calculation of p1 or p2 → simple use of Bernoulli equation

b) calculation of v1 or v2 → since λ depends on v, this must be solved by iterations of a formula

obtained from Bernoulli equation or by Kármán number + graph

c) calculation of d1 or d2 → again λ depends on Re and therefore on d → iterations from Bernoulli

equation or graph

Example:

- given: S1 = S2, l, p1, p2, z1, z2

- calculate: v1 = v2 = v

𝑣12

2+ 𝑔𝑧1 +

𝑝1𝜌= 𝑣22

2+ 𝑔𝑧2 +

𝑝2𝜌+ 𝑒𝑑𝑖𝑠

𝑒𝑑𝑖𝑠 = (𝜆𝑙

𝑑+∑𝜁𝑗

𝑗

)𝑣2

2

By expressing v from Bernoulli equation we get an iterative formula:

𝑣 = √(𝑔(𝑧1−𝑧2)+

𝑝1−𝑝2𝜌

1

2(𝜆

𝑙

𝑑+∑𝜁𝑗)

)

estimate v0 → calculate Re → λ → using the formula C v1; repeat until |𝒗𝒏+𝟏 − 𝒗𝒏| < 𝜀

Transport of fluids by using pumps

Pumps

- displacement pumps – e.g. piston pumps, rotary pumps

- centrifugal pumps – most common

Power output N can be calculated from Bernoulli equation:

𝑣12

2+ 𝑔𝑧1 +

𝑝1𝜌+ 𝑒𝑝𝑢𝑚𝑝 =

𝑣22

2+ 𝑔𝑧2 +

𝑝2𝜌+ 𝑒𝑑𝑖𝑠

by expressing 𝑒𝑝𝑢𝑚𝑝, and then 𝑒𝑝𝑢𝑚𝑝 = 𝑁

�̇�=

𝑁

𝜌�̇�.

Power input P then is 𝑃 = 𝑁

𝜂𝑝𝑢𝑚𝑝, where the efficiency 𝜂𝑝𝑢𝑚𝑝 < 1.

Suction + discharge height, cavitation

Problem is that in point 2 there is a low pressure which can be so low that the liquid starts to evaporate

→ cavitation

Maximum suction height to avoid evaporation can be calculated from Bernoulli equation applied to the

suction branch:

𝑣12

2+ 𝑔𝑧1 +

𝑝1𝜌= 𝑣22

2+ 𝑔𝑧2 +

𝑝2𝜌+ 𝑒𝑑𝑖𝑠

we set p2=p0 … vapor pressure of the liquid (can be found in tables, depends on temperature)

ℎ𝑚𝑎𝑥 = 𝑧2 − 𝑧1 =𝑝1 − 𝑝

0

𝜌𝑔−𝑣12 − 𝑣2

2

2𝑔−𝑒𝑑𝑖𝑠𝑔

Example: p0 for water at t = 20 C is small (≈ 0) if p1 = 10

5 Pa then ℎ𝑚𝑎𝑥 =

105−0

1000∙9.81=̇ 10 m

Maximum discharge height is given by the construction of the pump

Characteristics of the pipe and the pump

characteristic of the pipe e−:

- energy (per unit mass) necessary to provide so that the fluid flows through pipe at

volumetric flow V, i.e. e− = e−(�̇�) or H− = H−(�̇�)

- e− can be calculated from Bernoulli equation (simply e− = e𝑝𝑢𝑚𝑝)

- typically: e− = 𝐾1 + 𝐾2�̇�2 (quadratic form)

characteristic of the pump e+:

- energy (per unit mass), which is provided by pump to the fluid at a given �̇�, i.e. e+ =

e+(�̇�) or H+ = H+(�̇�) - this characteristic must be measured in the pump (cannot be calculated from Bernoulli eq)

At steady state e− = e+ or H− = H+ → working point of the pump,

Pumps can be combined

- in series 𝐻𝑜𝑣𝑒𝑟𝑎𝑙𝑙+ = ∑ 𝐻𝑗

+𝑗

- in parallel 𝑉𝑜𝑣𝑒𝑟𝑎𝑙𝑙= ∑ 𝑉𝑗𝑗 , 𝐻1

+(�̇�1) = 𝐻2+(�̇�2) = 𝐻𝑁

+(�̇�𝑁) = 𝐻𝑜𝑣𝑒𝑟𝑎𝑙𝑙+ (�̇�1 + �̇�2 +⋯�̇�𝑁)

Fluid flow through a bed of solids

Application:

- filtration

- heterogeneous reactors

- adsorption and chromatography

Aim is to calculate pressure drop. Bed of solid particles is characterized by:

- random arrangement of particles

- irregular channels within the bed

Properties:

cross-sectional area: S

volume of fluid: Vf

volume of particles: VS

volume of the bed: VB=hS=Vf+VS

surface of particles: AS

void fraction (porosity): 𝜀 =𝑉𝑓

𝑉𝑆 (= volume fraction of fluid)

volume fraction of particles: 𝑉𝑆

𝑉𝐵= 1 − 𝜀

superficial velocity: 𝑣 =�̇�

𝑆

interstitial velocity: 𝑣𝜀 =�̇�

𝜀𝑆=𝑣

𝜀

specific surface: 𝑎 =𝐴𝑆

𝑉𝐵= (1 − 𝜀)

𝐴𝑆

𝑉𝑆, for spheres 𝑎 = (1 − 𝜀)

6

𝑑𝑝

for non-spherical particles 𝑎 = (1 − 𝜀)6

𝜓𝑑𝑝

ψ … sphericity, a measure of non-spherical shape (ψ < 1)

dp … characteristic size (= diameter for sphere)

Flow in real bed is difficult to describe, therefore we use a model.

Channel model

We assume that void fraction 𝜺 and specific surface 𝒂 are the same for the channels and the real porous

bed. Flow in channel can be described by Bernoulli equation.

𝑣𝑐ℎ2

2+ 𝑔𝑧1 +

𝑝1𝜌= 𝑣𝑐ℎ2

2+ 𝑔𝑧2 +

𝑝2𝜌+ 𝑒𝑑𝑖𝑠

𝑝1 − 𝑝2 = ∆𝑝 = 𝜌𝑔(𝑧2 − 𝑧1)⏟ ≈0

+ 𝜌𝑒𝑑𝑖𝑠

where 𝑒𝑑𝑖𝑠 = 𝜆𝑙𝑐ℎ

𝑑𝑒𝑞𝑢𝑖𝑣

𝑣𝑐ℎ2

2.

It can be shown that: 𝑣𝑐ℎ =ℎ𝑣

𝜀, 𝑑𝑒𝑞𝑢𝑖𝑣 =

4𝑆𝑐ℎ

𝑠𝑐ℎ=4𝜀

𝑎

On substitution:

𝑒𝑑𝑖𝑠 = 𝜆3

2

𝑘ℎ(1 − 𝜀)

𝜀ψ𝑑𝑝

𝑘2𝑣2

2𝜀2=3

2𝜆𝑘3

(1 − 𝜀)𝑣2

2ψ𝜀3𝑑𝑝ℎ =

3

2𝜆𝑆(1 − 𝜀)𝑣2

2ψ𝜀3𝑑𝑝ℎ

∆𝑝 = 𝜌𝑒𝑑𝑖𝑠 =3

2𝜆𝑆𝜌

(1 − 𝜀)𝑣2

2ψ𝜀3𝑑𝑝ℎ

𝜆𝑆 … friction factor in porous bed

𝜆𝑆 = 𝐴

Re⏟𝑙𝑎𝑚𝑖𝑛𝑎𝑟𝑓𝑙𝑜𝑤

+ 𝐵⏟𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡𝑓𝑙𝑜𝑤

… 𝐸𝑟𝑔𝑢𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝐴 = 133,𝐵 = 2.34

Filtration

- separation of solid particles from a suspension on a porous membrane or medium

pressure difference across cake: ∆𝑝𝐾 = 𝑝1 − 𝑝2

pressure difference across medium: ∆𝑝𝑀 = 𝑝2 − 𝑝3

overall pressure difference: ∆𝑝𝐹 = ∆𝑝𝐾 + ∆𝑝𝑀

Filtration rate 𝑣𝐹 =�̇�𝐹

𝑆𝐹=

1

𝑆𝐹

d𝑉𝐹

d𝜏=

d𝑞𝐹

d𝜏

VF … volume of filtrate at given time

qF … volume of filtrate per unit area

Pressure difference is typically made by pumps or in centrifuge or by vacuum at the filtrate side

Stages of filtration:

- filtration

- washing of the cake

- removal of the cake

filtration

- continuous (e.g. drum filter)

- discontinuous (e.g. filter press)

Mass balance of a filter

- total balance 𝑚𝑆 = 𝑚𝐹 +𝑚𝐾

- balance on the solid 𝑚𝑆𝑤𝑆 = 𝑚𝐾𝑤𝐾

thus:

𝑚𝐹 = 𝑚𝑆 (1 −𝑤𝑆

𝑤𝐾) or 𝑉𝐹 = 𝑉𝑆 (1 −

𝑋𝑉𝑆

𝑥𝑉𝐾)

Kinetics of filtration

general rate equation: 𝒗𝑭 =∆𝒑𝑭

𝑹𝑭

∆𝒑𝑭 … driving force = pressure difference

𝑹𝑭 … resistance of cake + medium

also: 𝒗𝑭 =∆𝒑𝑲

𝑹𝑲=∆𝒑𝑴

𝑹𝑴=∆𝒑𝑲+∆𝒑𝑴

𝑹𝑲+𝑹𝑴 (the rate is the same in cake and medium)

The resistance of the cake 𝑹𝑲 can be determined by using the equation for pressure drop in porous bed:

∆𝑝𝐾 =3

2𝜆𝑆𝜌𝐹

(1 − 𝜀)𝑣𝐹2

2ψ𝜀3𝑑𝑝ℎ𝐾

we assume laminar flow: 𝜆𝑆 =𝐴

Re=

𝐴𝜂𝐹

𝑑𝑝𝑣𝐹𝜌𝐹 and thus

∆𝑝𝐾 = 𝑐𝑜𝑛𝑠𝑡𝜂𝐹

𝑑𝑝𝑣𝐹𝜌𝐹

3

2𝜆𝑆𝜌𝐹

(1 − 𝜀)𝑣𝐹2

2ψ𝜀3𝑑𝑝ℎ𝐾

𝑣𝐹 =∆𝑝𝐾𝛼ℎ𝐾𝜂𝐹

=∆𝑝𝐾

𝛽𝑉𝐹𝑆𝐹𝜂𝐹⏟

𝑅𝐾

where 𝛼 and 𝛽 are constants.

We introduce 𝑞𝐹 =𝑉𝐹

𝑆𝐹 and by recalling that 𝑣𝐹 =

d𝑞𝐹

d𝜏 we can write final rate equation for filtration:

𝑣𝐹 =d𝑞𝐹d𝜏

= ∆𝑝𝐹

𝑅𝐾 + 𝑅𝑀=

∆𝑝𝐹𝛽𝜂𝐹𝑞𝐹 + 𝑅𝑀

=

∆𝑝𝐹𝛽𝜂𝐹

𝑞𝐹 +𝑅𝑀𝛽𝜂𝐹

=𝐾𝐹

𝑞𝐹 + 𝑞𝑀

KF can be recalculated (since RM qM does not change!):

𝐾𝐹,1 =∆𝑝𝐹,1𝛽𝜂𝐹,1

; 𝐾𝐹,2 =∆𝑝𝐹,2𝛽𝜂𝐹,2

→ 𝐾𝐹,2𝐾𝐹,1

=∆𝑝𝐹,2∆𝑝𝐹,1

𝜂𝐹,1𝜂𝐹,2

Applications of the rate equation for filtration

- discontinuous filtration

a) 𝑣𝐹 is constant (occurs at the beginning of filtration)

d𝑞𝐹d𝜏

=𝐾𝐹

𝑞𝐹 + 𝑞𝑀= 𝑐𝑜𝑛𝑠𝑡

∫ d𝑞𝐹 =𝐾𝐹

𝑞𝐹 + 𝑞𝑀∫ d𝜏𝜏𝐹

0

𝑞𝐹

0

𝑞𝐹2 + 𝑞𝐹𝑞𝑀 − 𝐾𝐹𝜏𝐹 = 0

𝐾𝐹 is determined for ∆𝑝𝐹 at 𝜏 = 𝜏𝐹

b) ∆𝑝𝐹 is constant – common case

d𝑞𝐹d𝜏

=𝐾𝐹

𝑞𝐹 + 𝑞𝑀 where 𝐾𝐹 =

∆𝑝𝐹𝛽𝜂𝐹

= 𝑐𝑜𝑛𝑠𝑡

∫ (𝑞𝐹 + 𝑞𝑀)d𝑞𝐹 = 𝐾𝐹∫ d𝜏𝜏𝐹

0

𝑞𝐹

0

1

2𝑞𝐹2 + 𝑞𝐹𝑞𝑀 − 𝐾𝐹𝜏𝐹 = 0

Remark: lower bound in the integral could be final state of i)

Washing of the cake

- the cake has constant thickness and thus at constant pressure difference the rate of

washing is also constant

𝑣𝑃 =1

𝑆𝑃

d𝑉𝑃d𝜏

=d𝑞𝑃d𝜏

= 𝑐𝑜𝑛𝑠𝑡.

thus the rate equation

d𝑞𝑃d𝜏

= ∆𝑝𝑃

𝛽𝜂𝑃𝑞𝐹 + 𝑅𝑀=

𝐾𝑃𝑞𝐹 + 𝑞𝑀

= 𝑐𝑜𝑛𝑠𝑡.

here 𝐾𝑃 =∆𝑝𝑃

𝛽𝜂𝑃→

𝐾𝑃

𝐾𝐹=∆𝑝𝑃

∆𝑝𝐹

𝜂𝐹

𝜂𝑃

since 𝑅𝑀 = 𝑀𝜂𝑃 then 𝑞𝑀 =𝑀

𝛽= 𝑐𝑜𝑛𝑠𝑡.

By integration: 𝑞𝑃(𝑞𝐹 + 𝑞𝑀) − 𝐾𝑃𝜏𝑃 = 0

Remark: for a filter press resistance is doubled: 𝑅𝑀 = 2𝑅𝑀 and 𝛽 = 2𝛽

area is halved 𝑆𝑃 = 1

2𝑆𝐹 → 𝑞𝑃

′ =𝑉𝑃

𝑆𝐹=𝑞𝑃

2

Therefore it holds

𝑞𝑃′ (𝑞𝐹 + 𝑞𝑀) − 𝐾𝑃𝜏𝑃 = 0

Continuous filtration

Basic equation at constant pressure difference holds 1

2𝑞𝐹2 + 𝑞𝐹𝑞𝑀 − 𝐾𝐹𝜏𝐹 = 0

but because the drum is rotating with frequency n, the time 𝜏𝐹 is

𝜏𝐹 = 𝜑

2𝜋𝑛

Therefore 1

2𝑞𝐹2 + 𝑞𝐹𝑞𝑀 − 𝐾𝐹

𝜑

2𝜋𝑛= 0

Sedimentation (settling)

- separation of particles disperse in a medium by an external force

External forces: mostly gravitational, can be also centrifugal, electric or magnetic

Types of dispersions

- suspension: solids in liquid medium

solids in gaseous medium

- mist: liquid in gaseous medium

- emulsion: liquid in liquid medium

Equipment

- gravitational settlers: vertical flow

horizontal flow

- rotational settlers: centrifuge

cyclone

- electrostatic or magnetic separators

Gravitational settling

forces action a particles

- external: Fg … gravitational

- internal: 𝐹𝐴 = 𝐹𝐴𝑟⏟𝒃𝒖𝒐𝒚𝒂𝒏𝒕

(𝐴𝑟𝑐ℎ𝑖𝑚𝑒𝑑𝑒𝑠)

+ 𝐹𝑅⏟𝒅𝒓𝒂𝒈

(𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒)

Newton second law (= momentum balance)

−𝐹𝑔 + 𝐹𝐴𝑟 + 𝐹𝑅 = 𝑚𝑝d𝑣𝑝d𝜏

𝑣𝑝 … velocity of the particle

At steady state d𝑣𝑝

d𝜏= 0 → 𝑣𝑝 = 𝑣𝑢 … terminal (settling velocity)

- balance of forces

𝐹𝑔 − 𝐹𝐴𝑟 = 𝐹𝑅

𝜌𝑝𝑔𝑉𝑝 − 𝜌𝑓𝑔𝑉𝑝 = 𝜁𝑢𝜌𝑓𝑣𝑢2

2𝑆𝑝

for a sphere 𝑉𝑝 = 𝜋𝑑𝑝3

6, 𝑆𝑝 = 𝜋

𝑑𝑝2

4 projected area in the direction of the fall

(𝜌𝑝 − 𝜌𝑓)𝑔𝜋𝑑𝑝3

6= 𝜁𝑢𝜌𝑓

𝑣𝑢2

2𝜋𝑑𝑝2

4

thus

𝒗𝒖𝟐 =

𝟒

𝟑

𝒈𝒅𝒑

𝜻𝒖

𝝆𝒑 − 𝝆𝒇

𝝆𝒇 (Eq. A)

𝜁𝑢 … drag coefficient

𝜌𝑝, 𝜌𝑓 … densities of particles and fluid

It is convenient to rewrite Eq. A in a dimensionless form upon introducing:

- Reynolds number: Re = 𝑣𝑢𝑑𝑝𝜌𝑝

𝜂=

𝑣𝑢𝑑𝑝

𝜈

- Archimedes number: Ar = 𝑔𝑑𝑝

3(𝜌𝑝−𝜌𝑓)𝜌𝑓

𝜂2=𝑔𝑑𝑝

3(𝜌𝑝−𝜌𝑓)

𝜈2𝜌𝑓

On multiplying Eq. A by 𝑑𝑝2

𝜈2 we obtain

𝑑𝑝2

𝜈2𝑣𝑢2 =

4

3𝜁𝑢

𝑔𝑑𝑝3

𝜈2𝜌𝑝 − 𝜌𝑓

𝜌𝑓

Re2 =4

3𝜁𝑢 Ar → 𝐀𝐫 =

𝟑

𝟒𝜻𝒖Re𝟐

However, the drag coefficient 𝜁𝑢 depends on Re (similarly as the friction factor 𝜆 depends on Re in fluid

flow in pipes) and hence this relation must be given → obtained empirically (graph or ”correlations”

between 𝜁𝑢 and Re) .

The dimensionless force balance Ar =3

4𝜁𝑢Re2 in particular region reads:

a) Stokes region:

Ar = 𝟏𝟖Re; Re<0.2, Ar<3.6

b) Allen region

Ar = 𝟏𝟖Re(1+0.125Re𝟎.𝟕𝟐)

0.2 < Re < 103; 3.6 < Ar < 3.43×105

c) Newton region

Ar = 𝟎. 𝟑𝟑Re𝟐

103 < Re < 1.5 × 105; 3.43×105 < Ar < 7.4×109

Discussion:

There are two tasks:

- dp is given and we want to find vu

If dp is given then Re can be expressed from a) or c) and. Re cannot be expressed from b)

and its value must be obtained by iteration

- vu is given and we want to find dp

Iteration is required in all three cases (that is, dp is estimated → Re → Ar → new estimate

...)

We can avoid iteration by introducing Lyaschenko number:

𝐿𝑦 = 𝑣𝑢3𝜌𝑓2

𝑔𝜂(𝜌𝑓 − 𝜌𝑝)=

𝑣𝑢3𝜌𝑓

𝑔𝜈(𝜌𝑓 − 𝜌𝑝)

This number contains only vu and not dp (unlike Re), then:

a) Stokes: Ar = 76.4 Ly1

2

c) Newton: Ar = 0.036 Ly1

2

In Allen region we can use a less accurate correlation

b) Allen: : Ar = 138.6 Ly0.875

Remark: Terminal velocity of a nonspherical particle can be found by using:

1) dynamical shape factor 𝜑𝐴𝑟 according to 𝑣𝑢 = 𝜑𝐴𝑟𝑣𝑢,0

𝑣𝑢,0 … terminal velocity of an equivalent spherical particles having the same volume as

nonspherical particle

𝜑𝐴𝑟 is an empirical function of a factor 𝐹 =𝑙𝑝𝑙𝑑⁄ and Ar

𝑙𝑝 … maximum length of nonspherical particle

𝑙𝑝 … diameter of equivalent spherical particle

Ar is evaluated using dp

2) sphericity 𝜓 = 𝐴0

𝐴⁄

𝐴0 … surface area of equivalent spherical particle

𝐴 … surface area of nonspherical particle

𝑣𝑢 is found from Ly which is given graphically as function of Ar and

Design of a settler

Calculation of settling area

First step is mass balance (notation S, F, K as in filtration)

- total mass balance

�̇�𝑆 = �̇�𝐹 + �̇�𝐾

- balance on solids

�̇�𝑆𝑤𝑆 = �̇�𝐾𝑤𝐾

�̇�𝐹 = �̇�𝑆(1 −𝑤𝑆𝑤𝐾)

Then

�̇�𝐹 = �̇�𝐹𝜌𝑓

Remark: Balance on volume can be also used

Settling area:

- settler with vertical flow

condition: the fluid average velocity 𝑣𝑓 =�̇�𝐹

𝐴 must be equal to the

terminal velocity

𝑣𝑓 =�̇�𝐹

𝐴= 𝑣𝑢 → 𝐴 =

�̇�𝐹

𝑣𝑢 , where 𝐴 =

𝜋𝐷2

4 is the area for settling

- settler with horizontal flow

-

condition: settling time 𝜏𝑢 =ℎ

𝑣𝑢 must be equal to the mean residence time for the fluid

�̅� =𝑉

�̇�=̇ℎ𝑙𝑏

�̇�𝐹, thus 𝐴 =

�̇�𝐹

𝑣𝑢

Mixing

- homogenization of mixtures

- typically mixing is achieved by using a stirrer (= impeller)

- to enhance mixing, baffles are used

Calculation of power input

Euler number: Eu = 𝑃

𝑛3𝑑5𝜌

Reynolds number: Re = 𝑛𝑑2𝜌

𝜂

Geometrical simplex: Γ, e.g number of blades

n … frequency of stirring

d … diameter of the impeller

Dimensional analysis

Dimensional equation are transformed to dimensionless equations → new dimensionless parameters in

these equation are for example:

- Reynolds number

- Archimedes number

- Friction factor

- Drag coefficient

and many more.

Advantage is, that:

- description in terms of dimensionless quantities is more universal

- solutions of the equations, which may be hard to obtain are replaced by empirical

relations such as:

𝜆 = 𝜆(Re, 𝜀 𝑑⁄ )

𝜁𝑢 = 𝜁𝑢(Re)

Ar = Ar(Ly)

Eu = Eu(Re,Γ1,Γ2, … )