Presentation on theme: "BIO 208 FLUID FLOW OPERATIONS IN BIOPROCESSING [ ]"— Presentation transcript:
1 BIO 208 FLUID FLOW OPERATIONS IN BIOPROCESSING [ 3 1 0 4 ]
2 SYLLABUS Fluid statics Fluid dynamics Bernoulli’s equation Hagen Poiseuille's equationFriction factorFlow past immersed bodiesFlow thro bed of solidsFluidizationTransportation and metering of fluidsMixing & Agitation
3 References…..Unit Operations of Chemical Engineering, 5th edn., McCabe & Smith.Chemical Engineering, Vol. I, Coulson and RichardsonIntroduction to Chemical engineering, Badger and BancheroUnit Operations, Foust et. al.,
4 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 fluidIn 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.
5 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
6 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.
7 Momentum TransferThe study of momentum transfer, or fluid mechanics as it is often called, can be divided into two branches;fluid statics, or fluids at rest, andfluid dynamics or fluids in motion.Since in fluid dynamics momentum is being transferred, the term “momentum transfer” or “transport” is usually used.
8 ViscosityThe viscosity (m) of a fluid measures its resistance to flow under an applied shear stress.Representative units for viscosity arekg/(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
9 The kinematic viscosity (n) is the ratio of the viscosity to the density: n = m/r,Has the units of m2/s
10 Viscosity of liquids: Viscosity of gases: @ 25oC, mwater = 1 cP and Viscosity of liquids in general, decreases with increasing temperature.Viscosity of gases:Viscosity of gases increases with increase in temperature.@ 25oC, mwater = 1 cP andmair = 1 x 10-2 cP
11 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
12 We know F = mg and P = F/ATotal mass of fluid = h2 A rF=h2A r gP = F /A = h2 r gThis 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 addedP2 = h2 r g + PoTo cal P1,P1 = h1r g + PoThe press difference bet 2 and 1 isDP = P2-P1 = (h2 – h1)r g
13 Since it’s the vertical height of a fluid that determines the press of fluid, the shape of the vessel doesn’t affect the pressureIn the above fig the press P1 at the bottom of all three vessels is the same and equal to h1r g + Po
14 Prob 1A 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.
15 Soln…… P1 = Po + (hrg)oil P2 = P1 + (hrg)water = 1.347 x 105 Pa
16 Head of a fluidExpressing the pressure in terms of head in m or ft of a particular fluidWe know….P = hrgh = P / rg
17 Prob 2Compare the column heights of water,CCl4(r=1.59 g/cc) and mercury (r=13.65 g/cc) corresponding to a pressure of 50kPahwater = 5.1mh CCl4 = 3.21mhHg = 0.373m