Reading Materials: Chapter 6 Fluid Flow LECTURES 16-18 CHEM ENG 1007

What is a Fluid Material that continually deforms under a shear stress Divided into two groups Liquid Gases CHEM ENG 1007

GASES Characterized as loosely-associated molecules which are normally not close together and which travel through space for long distances before colliding with each other. The velocity of their travel depends on the temperature of the gas. Characteristic of gases: Readily compressible Expand quickly and fill a container To convert from volume to mass/mole use PV=nRT CHEM ENG 1007

LIQUIDS Density = mass/volume Characterized by molecules which are very close together and which are in collision with each other very frequently as they move around each other. The velocity of that motion and the rate of that collision depend on the temperature of the liquid. Characteristic of liquids: Slightly compressible Takes shape of container sides and bottom To convert from volume to mass/mole use Density = mass/volume CHEM ENG 1007

Fluid-Like Systems Solids present in fluids Slurries fine particles suspended in liquid Solids in a fluidized bed particles moving with the fluid in a tall reactor. CHEM ENG 1007

Variables associated with fluids Others Thermal conductivity Electrical conductivity Boiling point Freezing point Heat capacity Enthalpy Density Flow rate Pressure Viscosity Surface tension CHEM ENG 1007

Flow Rate Rate at which a material is transported through a process line. Mass flow rate: Molar flow rate: Volume flow rate: where: Chapter 7 CHEM ENG 1007

Pressure The pressure of the fluid is defined as the total force (exerted on the boundary by the fluid molecules) divided by the surface area of the boundary it is acting upon. {F} {A} {P} SI N m2 Nm-2 (Pa) cgs dyne cm2 dynecm-2 American lbf in2 lbfin-2 (psi) Chapter 7 CHEM ENG 1007

Puzzle Pressure = force/area A woman’s high heels sink into the soft ground, but the larger shoes of the much bigger man do not. Pressure = force/area Chapter 7 CHEM ENG 1007

Pressure We express pressure in two ways: absolute and gauge pressures Gauge Pressure: fluid pressure that is measured relative to the atmospheric pressure. Absolute Pressure: the total magnitude of force exerted per unit area In process calculations: Pabs = Pgauge + Patmosphere Pabs = 0 in complete vacuum Letter “a” or “g” is added to designate absolute or gauge, thus psia or psig Chapter 7 CHEM ENG 1007

Standard Atmospheric Pressure At sea level, 0oC and 45o latitude: Patm = 1 atm = 101,325 Pa = 14.7 psi = 1.01325 bars = 760 mmHg = 10.333 m H2O = 33. 9 ft H2O Chapter 7 CHEM ENG 1007

Standard Atmospheric Pressure How much does the atmosphere heigh? Answer: the same as 76 cm of mercury. How? Torricelli filled a tube with mercury and inverted it into an open container of mercury. Air pressure acting on the mercury in the dish can support a column of mercury 76 cm in height. Chapter 7 CHEM ENG 1007

Example 7.1 A man pumps his automobile tire until the tire gauge reads 34.0 psi. If the atmosphere in his community is 14.2 psia, what is the absolute pressure of the air in the tire? Solution: Pg = 34.0 psig Patm = 14.2 psia Pabs = 34.0 + 14.2 = 48.2 psia Chapter 7 CHEM ENG 1007

Hydrostatic Pressure It is the pressure (P) of the fluid at the base of the column. That is the force (F) exerted on the base divided by the base area (A). F thus equals the force on the top surface plus the weight of the fluid in the column. P = P0+ gh h = height of a hypothetical column of the fluid A pressure may also be expressed as a head of a particular fluid (Ph) Ph = P0 + h Chapter 7 CHEM ENG 1007

Example 7.2 For the tank depicted in Fig. 7.2, if the NaOH solution is 8 ft high, what is the pressure at the bottom of the tank? Assume that the density of the NaOH solution is the same as that of water. Perform the calculation in metric units. Solution: P1 = 0 Pa (gauge), so P2 = 23,896 Pa (gauge) or P2 = 23,896 + 101,325 = 125,221 Pa (absolute) Chapter 7 CHEM ENG 1007

Quick Quiz 1 What is the pressure 30.0 m below the surface of a lake? Assume the atmospheric pressure (the pressure at the surface) is 10.4 m H2O? Express your answer in atm. Solution: Ph = P0 + h = 10.4 + 30.0 = 40.4 m H2O Chapter 7 CHEM ENG 1007

Fluid Pressure Measurement A hollow tube closed at one end and bent into a C configuration. The open end of the tube is exposed to the fluid whose pressure is to be measured. As the pressure increases, the tube tends to straighten, causing a pointer attached to the tube to rotate. Figure 3.4-3 (p. 57) of Felder and Rousseau Bourdon gauge. It is used to measured fluid pressure from nearly perfect vacuums to about 7000 atm. Chapter 7 CHEM ENG 1007

Fluid Pressure Measurement Gauge pressure Pressure difference Absolute pressure Figure 3.4-4 (p. 58) Manometers. It is used for more accurate measurements of pressure (below about 3 atm) Chapter 7 CHEM ENG 1007

Figure 3.4-5 (p. 58) Differential Manometer variables. Fluid Pressure Measurement If  is gas then  may be neglected Figure 3.4-5 (p. 58) Differential Manometer variables. Chapter 7 CHEM ENG 1007

Illustration 1 A manometer reading gives 100 mmHg, calculate the absolute pressure. Solution: Measured gauge pressure: Pabs = 13,328 + 101,325 = 114,653 Pa = 115 kPa Chapter 7 CHEM ENG 1007

How does pressure relate to flow? In the absence of other forces, fluids tend to flow from regions of high pressure to regions where the pressure is lower. Therefore, pressure differences provide a driving force for fluid flow. Example: when a tire is punctured, air flows out of the high pressure tire to the atmosphere, which is a low pressure. Chapter 7 CHEM ENG 1007

Viscosity (m) Characterises its resistance to flow A measure of “stickiness” of a fluid A “frictional force” Low viscosity fluid High viscosity fluid Less energy required for mixing More energy required for mixing Chapter 7 CHEM ENG 1007

Viscosity (m) Measurement - “Shear stress/shear rate” Units SI: kg/(m.s) = N.s/m2 = Pa.s cgs: cp (centipoise) 1 cP = 10-2 poise = 10-3 Pa.s = 1 mPa.s Chapter 7 CHEM ENG 1007

Viscosity of various fluids Temperature (oC) Viscosity Pa.s Water 15.6 1.1x10-3 Gasoline 0.3x10-3 SAE 30 oil 383x10-3 Air 15 1.8x10-5 Methane 20 1.1x10-5 Chapter 7 CHEM ENG 1007

Viscosity of various fluids Substance (25°C) Viscosity (Pa.s) Water 1 x 10-3 Mercury 1.5 x 10-3 Air 1.8 x 10-5 Castor oil 0.99 Chapter 7 CHEM ENG 1007

Kinematic viscosity The property viscosity may also be combined with the fluid’s density to give the property kinematic viscosity Chapter 7 CHEM ENG 1007

Types of viscous fluids Chapter 7 CHEM ENG 1007

Types of viscous fluids Newtonian fluid e.g. water, air, other gases. 2-4 Non-Newtonian fluids Bingham-plastic. e.g. toothpaste, margarine, soap Pseudo-plastic (shear thinning) e.g. mayonnaise, polymer melts, paints. Dilatant (shear thickening) e.g. wet beach sand, starch in water Chapter 7 CHEM ENG 1007

Newtonian fluids Question: How would you determine the viscosity from a plot of shear stress against shear rate? With Newtonian fluids, shear stress increases proportionately with the shear rate Shear rate Shear stress Shear rate Viscosity Examples Water Glycerine Alcohol Air Chapter 7 CHEM ENG 1007

Other types of Non-Newtonian fluids Dilatant Pseudoplastic Shear rate Viscosity Viscosity Shear rate Viscosity changes with power input Chapter 7 CHEM ENG 1007

Other types of Non-Newtonian fluids Thixotropic Rheopetic Viscosity Viscosity Time Time Viscosity changes with time at constant shear rate Chapter 7 CHEM ENG 1007

Non-Newtonian fluids can have more than one property Example: Damp gypsum Pseudoplastic, Thixotropic and Viscoelastic Cream Dilatant, Rheopectic and not Viscoelastic Viscoelastic: Fluid returns to original viscosity after power input ceases Chapter 7 CHEM ENG 1007

Ideal / Inviscid Fluid Hypothetical fluid which is incompressible and has zero viscosity. Chapter 7 CHEM ENG 1007

Fluid Flow Principles Flow patterns vary with: velocity geometry of surface, and fluid properties such as viscosity, density. Classic experiment by Osborne Reynolds (1883) observed two types of fluid flow: Laminar flow – low flow rates Turbulent flow – higher flow rates Chapter 7 CHEM ENG 1007

Reynolds’ experiment (i) Low flow rates: fluid moves in parallel layers. (ii) High flow rates: cross currents (eddies) develop Chapter 7 CHEM ENG 1007

Reynolds Number Reynolds’ key variables Arrange into single “dimensionless group”: Chapter 7 CHEM ENG 1007

Reynolds Number For pipe flow: Re < 2,100 - laminar flow 2,100 < Re < 10,000 - transition region Re > 10,000 - turbulent flow Pipe-flow systems with the same Re are said to be dynamically similar. Chapter 7 CHEM ENG 1007

Fluid flow through a pipe Illustration 2 Fluid flow through a pipe Water at 25oC flows through a pipe of internal diameter 0.1 m at a velocity 0.2 m/s. Is the flow laminar flow or turbulent? What is the effect of reducing the velocity by a factor of 10? Chapter 7 CHEM ENG 1007

Solution Chapter 7 CHEM ENG 1007

Velocity Profiles: Laminar Flow Velocity profile is parabolic with the maximum velocity occurring in the centre of the pipe (r = 0) (i) Laminar Flow Chapter 7 CHEM ENG 1007

Velocity Profiles: Laminar Flow (i) Velocities of a fluid in laminar flow through a circular pipe Chapter 7 CHEM ENG 1007

Velocity Profiles: Laminar Flow Volumetric flow rate: Mean velocity: Chapter 7 CHEM ENG 1007

Velocity Profiles: Turbulent Flow Difficult to mathematically model due to its complex and rapidly changing flow patterns. Experimental measurements show for time-averaged velocity and mean velocity (ii) Turbulent Flow Chapter 7 CHEM ENG 1007

Velocity Profiles: Plug Flow Common assumption for highly turbulent flow is that velocity does not vary over cross-section: (iii) Plug Flow Chapter 7 CHEM ENG 1007

Mass Conservation in Fluid Flow Consider steady-state, one-dimensional fluid through pipe Chapter 7 CHEM ENG 1007

Mass Conservation in Fluid Flow Is v2 in (a) less than v2 in (b)? No, they are both the same. Chapter 7 CHEM ENG 1007

Fluid Friction Frictional force cause pressure drop during fluid flow through a constant diameter horizontal pipe. Consider flow situation in pipe below. Chapter 7 CHEM ENG 1007

Fluid Friction Momentum balance: f = friction factor; L = length of pipe; D = diameter of pipe; Chapter 7 CHEM ENG 1007

Fluid Friction Chapter 7 CHEM ENG 1007

Example 7.7 What value of the friction per mass of fluid (ef) is necessary to cause a decrease in pressure equal to 10 psi (answer in Btu/lbm)? Chapter 7 CHEM ENG 1007

Example 7.7 What value of the friction per mass of fluid (ef) is necessary to cause a decrease in pressure equal to b) 68,900 Pa (answer in J/kg)? Chapter 7 CHEM ENG 1007

Laminar Flow f determined analytically from velocity profile. Experimental data confirmed for Re < 2,100 (see Figure 2.10-3) Chapter 7 CHEM ENG 1007

Figure 2.10-3: Friction Factor Chart Chapter 7 CHEM ENG 1007

Turbulent Flow f cannot be determined analytically. Experimental curves have been devised. f also depends on surface roughness factor, e Pipe e (mm) Concrete 0.3-3.0 Cast iron 0.26 Galvanized iron 0.15 Plastic 0.0 (smooth) Chapter 7 CHEM ENG 1007

Turbulent Flow For smooth pipe & 2,100 < Re < 105 Blassius equation may be used: Chapter 7 CHEM ENG 1007

Illustration 3 Frictional losses in a pipe Pipeline 5 km long & 30 cm internal diameter Conveys water at 25oC at a rate of 180 kg/s. Roughness factor e/D = 0.001 Estimate the pressure drop across the pipe due to friction. Chapter 7 CHEM ENG 1007

Solution For water Chapter 7 CHEM ENG 1007

Solution From Figure 2.10-3, f = 0.005 Hence, Chapter 7 CHEM ENG 1007

Chapter 7 CHEM ENG 1007