BIO 208 FLUID FLOW OPERATIONS IN BIOPROCESSING [ 3 1 0 4 ]

BIO 208 FLUID FLOW OPERATIONS IN

BIOPROCESSING [ 3 1 0 4 ]

SYLLABUS

• Fluid statics• Fluid dynamics• Bernoulli’s equation• Hagen Poiseuille's equation• Friction factor• Flow past immersed bodies• Flow thro bed of solids• Fluidization• Transportation and metering of fluids• Mixing & Agitation

References…..

• Unit Operations of Chemical Engineering, 5th edn., McCabe & Smith.

• Chemical Engineering, Vol. I, Coulson and Richardson

• Introduction to Chemical engineering, Badger and Banchero

• Unit Operations, Foust et. al.,

Why??????• Fluid mechanics is an important area of engg. science.

• The nature of flow in pipes and reactors depends on the power input to the system & physical characters of fluid

• In fermentors, fluid properties affect process energy requirements & effectiveness of mixing which can have dramatic influence on productivity & the success of equipment scale-up.

• Transport of heat and mass is often coupled with fluid flow.

• Fluids in bioprocessing often contain suspended solids, consist of more than one phase, and have non-Newtonian properties.

• All of these features complicate analysis of flow behavior and present many challenges in bioprocess design.

FLUID……• A fluid may be defined as a substance that does not

permanently resist distortion and, hence, will change its shape.

• In this text gases, liquids, and vapors are considered to have the characteristics of fluids and to obey many of the same laws.

• In the process industries, many of the materials are in fluid form and must be stored, handled, pumped and processed.

• So it is necessary that we become familiar with the principles that govern the flow of fluids and also with the equipment used.

• Typical fluids encountered include water, air, CO2, oil, slurries, and thick syrups & BIOPRODUCTS

Compressible & Incompressible

• If a fluid is inappreciably affected by changes in pressures it is said to be incompressible. Most liquids are incompressible.

• Gases are considered to be compressible fluids.

Momentum Transfer

• The study of momentum transfer, or fluid mechanics as it is often called, can be divided into two branches;

1. fluid statics, or fluids at rest, and

2. fluid dynamics or fluids in motion.

• Since in fluid dynamics momentum is being transferred, the term “momentum transfer” or “transport” is usually used.

Viscosity

• The viscosity () of a fluid measures its resistance to flow under an applied shear stress.

• Representative units for viscosity are – kg/(m.sec)– g/(cm.sec) (also known as poise designated

by P)– The centipoise (cP), one hundredth of a

poise, is also a convenient unit

• The kinematic viscosity () is the ratio of the viscosity to the density:

• = ,

• Has the units of m2/s

• Viscosity of liquids:– Viscosity of liquids in general, decreases with

increasing temperature.

• Viscosity of gases:– Viscosity of gases increases with increase in

temperature.

• @ 25oC, water = 1 cP and

• air = 1 x 10-2 cP

Fluid Statics……• In fig., a stationary column of

fluid of ht h2 and constant CSA A, where A=Ao=A1=A2, is shown.

• The pressure above the fluid is Po (which could be the press of atmos above the fluid)

• Also for a fluid at rest, the force/unit area (ie Pressure) is the same at all points with the same elevation.

Pressure in a static fluid

• We know F = mg and P = F/A• Total mass of fluid = h2 A

F=h2A g

P = F /A = h2 g

• This is the press on A2 due to the mass of fluid above it.

• To get the TOTAL PRESS P2 on A2, the Po must be added

P2 = h2 g + Po• To cal P1,

P1 = h1 g + Po• The press difference bet 2 and 1 is

P = P2-P1 = (h2 – h1) g

• Since it’s the vertical height of a fluid that determines the press of fluid, the shape of the vessel doesn’t affect the pressure

• In the above fig the press P1 at the bottom of all three vessels is the same and equal to h1 g + Po

Prob 1

• A large storage tank contains oil having a density of 917 kg/m3.The tank is 3.66 m tall and is vented (open) to the atmosphere of 1 atm abs at the top. The tank is filled with oil to a depth of 3.05 m (10 ft) and also contains 0.61 m (2.0 ft) of water in the bottom of the tank. Calculate the pressure in Pa and 3.05 m from the top of the tank and at the bottom.

Soln……

• P1 = Po + (hg)oil

= 1.288 x 105 Pa

• P2 = P1 + (hg)water

= 1.347 x 105 Pa

Head of a fluid

• Expressing the pressure in terms of head in m or ft of a particular fluid

• We know….P = hg

• h = P / g

Prob 2

• Compare the column heights of water,CCl4(=1.59 g/cc) and mercury (=13.65 g/cc) corresponding to a pressure of 50kPa

• hwater = 5.1m

• h CCl4 = 3.21m

• hHg = 0.373m