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### 02 bsb 228 properties of fluids

• 1. Department of Basic Sciences January—May 2020 Prof. Sello Alfred Likuku Office Block 28 BSB 228 Introduction to Fluid Mechanics
• 2. BSB 228: Course Structure March 11, 2021 2  Topic 1: Introduction to Fluids and Fluid Mechanics (Basic Concepts)  Topic 2: Fluid Properties  Topic 3: Pressure and Fluid Statics  Topic 4: Bernoulli and Energy Equations  Topic 5: Momentum Analysis of Flow Systems  Topic 6: Dimensional Analysis and Modeling
• 4. 2-1 Introduction March 11, 2021 4 Any characteristic is called a Property Properties are pressure P, temperature T, volume V, and mass m. The list can include viscosity, thermal conductivity, modulus of elasticity, thermal expansion coefficient, electric resistivity, and even velocity and elevation. Properties are considered to be either intensive or extensive. Intensive properties are those that are independent of the mass of a system, such as temperature, pressure, and density. Extensive properties are those whose values depend on the size—or extent—of the system. Total mass, total volume V, and total momentum are some examples of extensive properties.
• 5. 2-1 Introduction March 11, 2021 5 To determine whether a property is intensive or extensive, divide the system into two equal parts with an imaginary partition, (Fig.). Each part will have the same value of intensive properties as the original system, but half the value of the extensive properties. Extensive properties per unit mass are called specific properties. Some examples of specific properties are specific volume (v = V/m) and specific total energy (e = E/m). That is, specifying a certain number of properties is sufficient to fix a state. The number of properties required to fix the state of a system is given by the state postulate: The state postulate is a term used in thermodynamics that defines the given number of properties to a thermodynamic system in a state of equilibrium. The state postulate allows a finite number of properties to be specified in order to fully describe a state of thermodynamic equilibrium.
• 6. 2-1 Introduction March 11, 2021 6 The state of a simple compressible system is completely specified by two independent, intensive properties. Two properties are independent if one property can be varied while the other one is held constant. Not all properties are independent, and some are defined in terms of others, as explained in the next section. Continuum: Matter is made up of atoms that are widely spaced in the gas phase. Yet it is very convenient to disregard the atomic nature of a substance and view it as a continuous, homogeneous matter with no holes, that is, a continuum. The continuum idealization allows us to treat properties as point functions and to assume that the properties vary continually in space with no jump discontinuities. This idealization is valid as long as the size of the system we deal with is large relative to the space between the molecules. This is the case in practically all problems, except some specialized ones. The continuum idealization is implicit in many statements we make, such as “the density of water in a glass is the same at any point.”
• 7. 2-2 Density & specific gravity March 11, 2021 7 Density is defined as mass per unit volume. Sometimes the density of a substance is given relative to the density of a well-known substance. Then it is called specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ρH2O = 1000 kg/m3). That is,   3 kg/m Density, V m   The reciprocal of density is the specific volume v, which is defined as volume per unit mass. That is,   /kg m 1 Volume, Specific 3    m V v less) (Dimension G Gravity, Specific O H2    S The weight of a unit volume of a substance is called specific weight and is expressed as,   on) accelerati nal gravitatio ( N/m Weight, Specific 3   g g  
• 8. 2-2 Density & specific gravity March 11, 2021 8 Property tables provide very accurate and precise information about the properties, but sometimes it is convenient to have some simple relations among the properties that are sufficiently general and accurate. Any equation that relates the pressure, temperature, and density (or specific volume) of a substance is called an equation of state. The simplest and best-known equation of state for substances in the gas phase is the ideal- gas equation of state, expressed as RT P RT Pv    or Here, P is the absolute pressure, v is the specific volume, T is the thermodynamic (absolute) temperature, ρ is the density, and R is the gas constant. The gas constant R is different for each gas and is determined from R = Ru /M, where Ru is the universal gas constant whose value is Ru = 8.314 kJ/kmol·K and M is the molar mass (also called molecular weight), in kg/mol, of the gas. The values of R and M for several substances are given in property tables (Table A–1).
• 9. 2-2 Density & specific gravity March 11, 2021 9 The thermodynamic temperature scale in the SI is the Kelvin scale, and the temperature unit on this scale is the kelvin, designated by K. 15 . 273 ) C ( ) (    T K T The ideal-gas equation of state, or simply the ideal-gas relation, and a gas that obeys this relation is called an ideal gas. For an ideal gas of volume V, mass m, and number of moles N = m/M, the ideal-gas equation of state can also be written as T NR PV mRT PV u   or For a fixed mass, where m1 = m2 = m3 = m, where; 3 3 3 3 2 2 2 2 1 1 1 1 ; ; RT V P m RT V P m RT V P m    An ideal gas at two or three different sates can be given by: C T V P T V P T V P    3 3 3 2 2 2 1 1 1
• 10. 2-2 Density & specific gravity March 11, 2021 10 Part of Table A–1
• 11. 2-2 Density & specific gravity March 11, 2021 11 It has been experimentally observed that the ideal-gas relation closely approximates the P-v- T behaviour of real gases at low densities. At low pressures and high temperatures, the density of a gas decreases and the gas behaves like an ideal gas. In the range of practical interest, many familiar gases such as air, N, O, H, He, Ar, Ne, and Kr and even heavier gases such as CO2 can be treated as ideal gases with negligible error (often less than 1 percent). Dense gases such as water vapour in steam power plants and refrigerant vapour in refrigerators, however, should not be treated as ideal gases since they usually exist at a state near saturation.
• 12. 2-2 Density & specific gravity March 11, 2021 12 Summary Equation: mRT PV  RT P   [ρ = density (kg/m 3 )] RT Pv  [v = specific volume (m 3 /kg)] T NR Pv u  [N = number of moles (mol) & Ru = Universal gas constant (J/mol·K)] NMRT PV  [M = Molar mass (kg/mol)] Any of these equations: the most popular is the first one, so we shall define its terms ---------------------------------------------------------------------------------------- Identification of terms and units: P = Absolute pressure of the gas = Pa (N/m 2 ) V = Volume of the gas = m 3 T = Absolute temperature of the gas [(t + 273)] = K (Kelvin) m = Mass of the gas = kg R = Gas constant = J·K –1 ·kg –1 ----------------------------------------------------------------------------------------
• 13. 2-2 Density & specific gravity March 11, 2021 13 Example: Determine the (a) density, (b) specific gravity, and (c) mass of the air in a room whose dimensions are 4 m × 5 m × 6 m at 100 kPa and 25°C. From Previous Slide it is Specified that: At low pressures and high temperatures, the density of a gas decreases and the gas behaves like an ideal gas. Then look at Table A–1; you will recognise that for air, the given pressure of 100 kPa < PCritical (3.77 MPa) and temperature of 25°C i.e., 298.15 K > Tcritical (132.5 K): Then it behaves like IDEAL gas. So ideal gas equation applies! The gas constant for air, R = 0.2870 kJ/kg·K. (a) The density: 3 kg/m 169 . 1 K) 15 . 298 ( K) kJ/kg 287 . 0 ( kPa 100      RT P  (b) The Specific gravity: (c) The mass: Note that from: PV = mRT : units for R can also be expressed from PV/mT = kPa·m3 /kg·K less) (Dimension 00169 . 0 kg/m 1000 kg/m 1.69 G 3 3 O H2      S kg 28 . 140 m ) 6 5 4 ( ) kg/m 1.69 ( Mass 3 3       V 
• 14. 2-2 Density & specific gravity March 11, 2021 14 Example: A fixed mass of gas contained in a closed system is initially at a pressure of 100 kN·m–2, a temperature of 15°C, and occupies a volume of 0.15 m3. The gas is then compressed to a volume of 0.06 m3 and a pressure of 300 kN·m–2. The gas is then expanded at constant pressure unit it reoccupies its original volume. If the characteristic constant for the gas is 189 J·kg–1·K–1, determine the (a) mass of gas in the system and the temperature at the (b) end of the compression and (c) at the end of the expansion processes. (a) The mass:    kg 10 75 . 2 K) 15 . 288 ( K) J/kg 189 ( m 15 . 0 Pa 10 100 1 3 3 1 1 1         RT V P m (b) T2 after compression: P1 = 100×103 N/m2; P2 = 300×103 N/m2: P3 = P2; m = ? T1 = (15+273.15)°C; T2 = ?; T3 = ?; R = 189 J·kg–1·K–1 V1 = 0.15 m3; V2 = 0.06 m3; V3 = V1; PV = mRT      C 17 . 73 K 32 . 346 K) J/kg 189 ( kg) 10 75 . 2 ( m 06 . 0 Pa 10 300 1 3 3 2 2 2          mR V P T (c) T3 after expansion:      C 62 . 592 K 80 . 865 K) J/kg 189 ( kg) 10 75 . 2 ( m 15 . 0 Pa 10 300 1 3 3 3 3 3          mR V P T
• 15. 2-3 Vapour pressure & cavitation March 11, 2021 15 We know that temperature and pressure are dependent properties for pure substances during phase–change processes, and there is one–to–one correspondence between temperatures and pressures. At a given pressure, the temperature at which a pure substance changes phase is called the saturation temperature Tsat. Likewise, at a given temperature, the pressure at which a pure substance changes phase is called the saturation pressure Psat. At an absolute pressure of 1 standard atmosphere (1 atm. or 101.325 kPa), for example, the saturation temperature of water is 100°C. Conversely, at a temperature of 100°C, the saturation pressure of water is 1 atm. Saturated–water temperature and Saturated–water temperature pressure tables are given in property tables Table A–4 and Table A–5, respectively.
• 16. 2-3 Vapour pressure & cavitation March 11, 2021 16
• 17. 2-3 Vapour pressure & cavitation March 11, 2021 17
• 18. 2-3 Vapour pressure & cavitation March 11, 2021 18 The vapour pressure Pv of a pure substance is defined as the pressure exerted by its vapour in phase equilibrium with its liquid at a given temperature. Pv is a property of the pure substance, and turns out to be identical to the saturation pressure Psat of the liquid (Pv = Psat). We must be careful not to confuse vapour pressure with partial pressure. Partial pressure is defined as the pressure of a gas or vapour in a mixture with other gases. For example, atmospheric air is a mixture of dry air and water vapour, and atmospheric pressure is the sum of the partial pressure of dry air and the partial pressure of water vapour. The partial pressure of water vapour constitutes a small fraction (usually under 3 percent) of the atmospheric pressure since air is mostly nitrogen and oxygen. The partial pressure of a vapour must be less than or equal to the vapor pressure if there is no liquid present.
• 19. 2-3 Vapour pressure & cavitation March 11, 2021 19 However, when both vapour and liquid are present and the system is in phase equilibrium, the partial pressure of the vapour must equal the vapour pressure, and the system is said to be saturated. The rate of evaporation from open water bodies such as lakes is controlled by the difference between the vapour pressure and the partial pressure. For example, the vapour pressure of water at 20°C is 2.34 kPa (Property Table A–4 or Table A–5). Therefore, a bucket of water at 20°C left in a room with dry air at 1 atm. will continue evaporating until one of two things happens: the water evaporates away (there is not enough water to establish phase equilibrium in the room), or the evaporation stops when the partial pressure of the water vapour in the room rises to 2.34 kPa at which point phase equilibrium is established. For phase-change processes between the liquid and vapour phases of a pure substance, the saturation pressure and the vapour pressure are equivalent since the vapour is pure.
• 20. 2-3 Vapour pressure & cavitation March 11, 2021 20 Note that the pressure value would be the same whether it is measured in the vapor or liquid phase (provided that it is measured at a location close to the liquid–vapor interface to avoid the hydrostatic effects). Vapour pressure increases with temperature. Thus, a substance at higher temperatures boils at higher pressures. For example, water boils at 134°C in a pressure cooker operating at 3 atm. absolute pressure, but it boils at 93°C in an ordinary pan at a 2000-m elevation, where the atmospheric pressure is 0.8 atm. The reason for our interest in vapour pressure is the possibility of the liquid pressure in liquid-flow systems dropping below the vapour pressure at some locations, and the resulting unplanned vaporization. The vapour bubbles (called cavitation bubbles since they form “cavities” in the liquid) collapse as they are swept away from the low pressure regions, generating highly destructive, extremely high-pressure waves.
• 21. March 11, 2021 21 Cavitation must be avoided (or at least minimized) in flow systems since it reduces performance, generates annoying vibrations and noise, and causes damage to equipment. The pressure spikes resulting from the large number of bubbles collapsing near a solid surface over a long period of time may cause erosion, surface pitting, fatigue failure, and the eventual destruction of the components or machinery. The presence of cavitation in a flow system can be sensed by its characteristic tumbling sound. Example: In a water distribution system, the temperature of water is observed to be as high as 30°C. Determine the minimum pressure allowed in the system to avoid cavitation. kPa 2469 . 4 C sat@30 min     P P  The vapour pressure of water at 30°C is 4.25 kPa (Property Table A–4). To avoid cavitation, the pressure anywhere in the flow should not be allowed to drop below the vapour (or saturation) pressure at the given temperature. That is, 2-3 Vapour pressure & cavitation
• 22. 2-4 Energy & specific heats March 11, 2021 22 Energy exists in many forms: i.e., thermal, mechanical, kinetic, potential, electrical, magnetic, chemical, and nuclear, and their sum constitutes the total energy E (or e on a unit mass basis) of a system: The forms of energy related to the molecular structure of a system and the degree of the molecular activity are referred to as the microscopic energy. The sum of all microscopic forms of energy is called the internal energy of a system, and is denoted by U (or u on a unit mass basis). The macroscopic energy of a system is related to motion and the influence of some external effects such as gravity, magnetism, electricity, and surface tension. (kJ/kg) m E e 
• 23. 2-4 Energy & specific heats March 11, 2021 23 The energy that a system possesses as a result of its motion relative to some reference frame is called kinetic energy. Here, v denotes the velocity of the system relative to some fixed reference frame The energy that a system possesses as a result of its elevation in a gravitational field is called potential energy. Here, g is the gravitational acceleration and z is the elevation of the centre of gravity of a system relative to some arbitrarily selected reference plane. In the analysis of systems that involve fluid flow, we normally use the combination of properties U and PV. For convenience, this combination is called enthalpy H. Here, V and v denote volume and specific volume, respectively. Do not confuse the letter v with velocity in this equation. Strictly speaking the velocity must be given by letter, v or . (kJ/kg) 2 1 : basis) mass unit (in OR (kJ) 2 1 2 2 v m KE ke mv KE    (kJ/kg) : basis) mass unit (in OR (kJ) gz m PE pe mgz PE    (kJ/kg) : basis) mass unit (in OR (kJ) Pv u h PV U H     v 
• 24. 2-4 Energy & specific heats March 11, 2021 24 Similarly, the enthalpy on unit mass basis can be given by. Exercise: Show the dimensional homogeneity of the equation above, where P/ρ is the flow energy, also called the flow work, which is the energy per unit mass needed to move the fluid and maintain flow. (kJ/kg) 2 2 Flowing gz v h pe ke h e P e          In the absence of such effects as magnetic, electric, and surface tension, a system is called a simple compressible system. The total energy of a simple compressible system consists of three parts: internal, kinetic, and potential energies. On a unit-mass basis, it is expressed as e = u + ke + pe. The fluid entering or leaving a control volume possesses an additional form of energy—the flow energy P/ρ. Then the total energy of a flowing fluid on a unit-mass basis becomes: (kJ/kg)  P u Pv u h    
• 25. March 11, 2021 25 By using the enthalpy instead of the internal energy to represent the energy of a flowing fluid, one does not need to be concerned about the flow work. The energy associated with pushing the fluid is automatically taken care of by enthalpy. In fact, this is the main reason for defining the property enthalpy. The differential and finite changes in the internal energy and enthalpy of an ideal gas can be expressed in terms of the specific heats as dT c dh dT c du p v And   Here, cv and cp are the constant-volume and constant-pressure specific heats of the ideal gas. Using specific heat values at the average temperature, the finite changes in internal energy and enthalpy can be expressed approximately as T c h T c u ave ave       , p , v And For incompressible substances, the constant-volume and constant-pressure specific heats are identical. 2-4 Energy & specific heats
• 26. March 11, 2021 26 Therefore, cp ≈ cv ≈ c for liquids, and the change in the internal energy of liquids can be expressed as: Note that ρ = constant for incompressible substances, so we can differentiate the enthalpy h : Integrating, the enthalpy change becomes. Therefore, for constant-pressure processes of liquids: And again, for constant-temperature processes of liquids: T c u ave     dP du dh P u h     gives   P T c P u h ave          T c u h ave      P h    2-4 Energy & specific heats
• 27. March 11, 2021 27 Reminder: From your level 100 Physics, the following were defined less) (Dimension Strain L L   2-5 Coefficient of compressibility  Stress: Measure of force felt by material  Strain: Measure of deformation Area ΔL L Force, F less) (Dimension L L F    Young’s Modulus, Υ: Measure of stiffness pressure) as Same : N/m or Pa (Pascal, Sress 2 A F      Strain Tensile Stress Tensile    L L A F Υ  Shear Modulus, S: a coefficient of elasticity of a substance, expressing the ratio between the force per unit area (shearing stress) that laterally deforms the substance and the shear (shearing strain) that is produced by this force.     Strain Shear Stress Shear    h x A F S
• 28. March 11, 2021 28 Reminder: From your level 100 Physics, the following were defined 2-5 Coefficient of compressibility  Bulk Modulus: Measure of the substance's resistance to uniform compression. Images of Shear Stress and bulk Modulus       V P V V V P V V A F Κ             Strain Volume Pressure in Change     h x A F S   V P V Κ    
• 29. March 11, 2021 29 The volume (or density) of a fluid changes with a change in its temperature or pressure. . But the amount of volume change is different for different fluids, and we need to define properties that relate volume changes to the changes in pressure and temperature. Two such properties are the bulk modulus of elasticity κ (Greek letter Kappa) and the coefficient of volume expansion β (Greek letter Beta). Fluid contracts when more pressure is applied on it and expands when the pressure acting on it is reduced. That is, fluids act like elastic solids with respect to pressure. Therefore, in an analogous manner to Young’s modulus of elasticity for solids, it is appropriate to define a coefficient of compressibility κ (also called the bulk modulus of compressibility or bulk modulus of elasticity) for fluids as: (Pa) T T                         P V P V 2-5 Coefficient of compressibility Solids have Young’s, Bulk, and Shear moduli. Liquids have only bulk moduli Constant) (         T P V V P    Approximating
• 30. March 11, 2021 30 It can be noted that volume and pressure are inversely proportional (volume decreases as pressure is increased and thus ∂P/∂V is a negative quantity), and the negative sign in the definition (bulk modulus of elasticity) ensures that κ is a positive quantity. Also, differentiating gives ρ = 1/V gives dρ = – dV / V 2, which can be rearranged as: V dV d     2-5 Coefficient of compressibility That is, the fractional changes in the specific volume and the density of a fluid are equal in magnitude but opposite in sign. For ideal gases: : Then ; and    P RT P RT P T              P  gas ideal  Therefore, the coefficient of compressibility of an ideal gas is equal to its absolute pressure, and the coefficient of compressibility of the gas increases with increasing pressure. Substituting κ = P into the definition of the coefficient of compressibility and rearranging gives: Constant) ( : gas Ideal     T P P  
• 31. March 11, 2021 31 Therefore, the percent increase of density of an ideal gas during isothermal compression is equal to the percent increase in pressure. For air at 1 atm. pressure, κ = P = 1 atm. and a decrease of 1% in volume (ΔV/V = –0.01) corresponds to an increase of ΔP = 0.01 atm. in pressure. But for air at 1000 atm., κ = 1000 atm. and a decrease of 1% in volume corresponds to an increase of ΔP = 10 atm. in pressure. Therefore, a small fractional change in the volume of a gas can cause a large change in pressure at very high pressures. The inverse of the coefficient of compressibility is called the isothermal compressibility α and is expressed as: 2-5 Coefficient of compressibility The isothermal compressibility of a fluid represents the fractional change in volume or density corresponding to a unit change in pressure.   Pa 1 1 1 1 T T                     P P V V    
• 32. March 11, 2021 32 The density of a fluid, in general, depends more strongly on temperature than it does on pressure. The variation of density with temperature is responsible for numerous natural phenomena such as winds, currents in oceans, rise of plumes in chimneys, the operation of hot-air balloons, heat transfer by natural convection, and even the rise of hot air. The property that provides that information is the coefficient of volume expansion (or volume expansivity) β, defined as (see Fig. in next slide): Coefficient of volume expansion 2-5 Coefficient of compressibility   K 1 1 1 P P                    T T V V    Constant) (         P T T V V    Approximating
• 33. 2-5 Coefficient of compressibility March 11, 2021 33 A large value of β for a fluid means a large change in density with temperature, and the product β×ΔT represents the fraction of volume change of a fluid that corresponds to a temperature change of ΔT at constant pressure. It can be shown easily that the volume expansion coefficient of an ideal gas (P = ρRT) at a temperature T (absolute temperature) is equivalent to the inverse of the temperature (please show as an exercise): Coefficient of volume expansion In the study of natural convection currents, the condition of the main fluid body that surrounds the finite hot or cold regions is indicated by the subscript “infinity” to serve as a reminder that this is the value at a distance where the presence of the hot or cold region is not felt.   K 1 1 gas ideal T  
• 34. March 11, 2021 34 The volume expansion coefficient can then be expressed approximately: Coefficient of volume expansion Here, ρ∞ is the density and T∞ is the temperature of the quiescent fluid away from the confined hot or cold fluid pocket. The combined effects of pressure and temperature changes on the volume change of a fluid can be determined by taking the specific volume to be a function of T and P. Differentiating V = V(T, P) and using definitions of compression and expansion coefficients α and β give: 2-5 Coefficient of compressibility                T T T T        Or  V P T dP P V dT T V dV                         T P Then the fractional change in volume (or density) due to changes in pressure and temperature can be expressed approximately as: P T V V            
• 35. March 11, 2021 35 Solution The density, ρ1 of water at 20°C and 1 atm. must be determined [Table A–2]. However, only densities for 0°C→1000 kg/m3 and 25°C→997 kg/m3 are given, so we need to interpolate for 20°C. The procedure for linear interpolation is outlined as below: Example: Consider water initially at 20°C and 1 atm. Determine the final density of water (a) if it is heated to 50°C at a constant pressure of 1 atm., and (b) if it is compressed to 100- atm. pressure at a constant temperature of 20°C. Take the isothermal compressibility of water to be α = 4.80 ×10–5 atm.–1. Given the two red points, the blue line is the linear interpolant between the points, and the value y at x may be found by linear interpolation as follows: 2-5 Coefficient of compressibility   0 0 0 1 0 1 0 1 0 1 0 0 y x x x x y y y x x y y x x y y            Let y be the required density, ρ1   3 1 kg/m 6 . 997 1000 0 20 0 25 1000 997       
• 36. 2-5 Coefficient of compressibility March 11, 2021 36 Solution—cont.:  The density of water at 20°C and 1 atm. pressure is ρ1 = 997.6 kg/m3.  The coefficient of volume expansion at the average temperature of (20°C+50°C)/2 = 35°C is β = 0.337×10–3 K–1. Determined from [Table A–3].  The isothermal compressibility of water is given to be α = 4.80×10–5 atm.–1. a) At constant pressure: T P T                      3 1 3 3 kg/m 08 . 10 K 20 50 K 10 337 . 0 kg/m 6 . 997               T   3 3 3 1 2 1 2 kg/m 52 . 987 kg/m 08 . 10 kg/m 6 . 997 then ; From                b) The change in density due to a change of pressure from 1 atm. to 100 atm. at constant temperature is:     3 1 5 3 kg/m 96 . 4 atm. 1 100 atm. 10 80 . 4 kg/m 52 . 987            P   Then density of water decreases while being heated and increases while being compressed.   3 3 1 2 kg/m 56 . 1002 kg/m 96 . 4 6 . 997         
• 37. 2-6 Viscosity March 11, 2021 37 Viscosity is a measure of a fluid's resistance to flow. It describes the internal friction of a moving fluid. A fluid with large viscosity resists motion because its molecular makeup gives it a lot of internal friction. A fluid with low viscosity flows easily because its molecular makeup results in very little friction when it is in motion. Gases also have viscosity, although it is a little harder to notice it in ordinary circumstances.
• 38. 2-6 Viscosity March 11, 2021 38 To obtain a relation for viscosity, consider a fluid layer between two very large parallel plates (or equivalently, two parallel plates immersed in a large body of a fluid) separated by a distance l (Fig.). Now a constant parallel force F is applied to the upper plate while the lower plate is held fixed. A F   After the initial transients, it is observed that the upper plate moves continuously under the influence of this force at a constant velocity V. The fluid in contact with the upper plate (where A is the contact area between the plate and the fluid) sticks to the plate surface and moves with it at the same velocity, and the shear stress τ acting on this fluid layer is:
• 39. 2-6 Viscosity March 11, 2021 39 The fluid layer deforms continuously under the influence of shear stress. The fluid in contact with the lower plate assumes the velocity of that plate, which is zero (again because of the no-slip condition). In steady laminar flow, the fluid velocity between the plates varies linearly between 0 and V, and thus the velocity profile and the velocity gradient (where y is the vertical distance from the lower plate) are, respectively: l V dy du V l y y u   and ) ( During a differential time interval dt, the sides of fluid particles along a vertical line MN (figure on previous slide) rotate through a differential angle dβ while the upper plate moves a differential distance da = V dt. The angular displacement or deformation (or shear strain) can be expressed as: dt dy du l Vdt l da d       tan
• 40. 2-6 Viscosity March 11, 2021 40 After re-arranging the expression in previous slide, we get: dy du dt d      or Thus, the rate of deformation of a fluid element is equivalent to the velocity gradient du/dy. Further, for most fluids the rate of deformation (and thus the velocity gradient) is directly proportional to the shear stress dy du dt d   Fluids for which the rate of deformation is proportional to the shear stress are called Newtonian fluids (after Sir Isaac Newton). Most common fluids such as water, air, gasoline, and oils are Newtonian fluids. Blood and liquid plastics are examples of non- Newtonian fluids. In one-dimensional shear flow of Newtonian fluids, shear stress can be expressed by the linear relationship   2 N/m dy du    Shear stress:
• 41. 2-6 Viscosity March 11, 2021 41 In the expression: The constant of proportionality μ is called the coefficient of viscosity or the dynamic (or absolute) viscosity of the fluid, whose unit is kg/m·s, or equivalently, N·s/m2 (or Pa·s where Pa is the pressure unit pascal). dy du dt d   A common viscosity unit is poise, which is equivalent to 0.1 Pa·s (or centipoise, which is one-hundredth of a poise). The viscosity of water at 20°C is 1 centipoise, and thus the unit centipoise serves as a useful reference. A plot of shear stress versus the rate of deformation (velocity gradient) for a Newtonian fluid is a straight line whose slope is the viscosity of the fluid, as shown in Fig.
• 42. 2-6 Viscosity March 11, 2021 42 Note that viscosity is independent of the rate of deformation. The shear force acting on a Newtonian fluid layer (or, by Newton’s third law, the force acting on the plate) is (Fig. in slide 41) :   N dy du A A F     Shear force:   N l V A F    For non-Newtonian fluids, the relationship between shear stress τ and rate of deformation du/dy is not linear, and referred to as the apparent viscosity of the fluid.
• 43. 2-6 Viscosity March 11, 2021 43 Fluids for which the apparent viscosity increases with the rate of deformation (such as solutions with suspended starch or sand) are referred to as dilatant or shear thickening fluids, and those that exhibit the opposite behavior (the fluid becoming less viscous as it is sheared harder, such as some paints, polymer solutions, and fluids with suspended particles) are referred to as pseudoplastic or shear thinning fluids. Some materials such as toothpaste can resist a finite shear stress and thus behave as a solid, but deform continuously when the shear stress exceeds the yield stress and behave as a fluid. Such materials are referred to as Bingham plastic.
• 44. 2-6 Viscosity March 11, 2021 44 The ratio of dynamic viscosity to density is used more often in fluid mechanics and its for convenience called kinematic viscosity ν and is expressed as ν = μ/ρ. The two most common units of ν are m2/s and stoke (1 stoke = 1 cm2/s = 0.0001 m2/s). In general, the viscosity of a fluid depends on both temperature and pressure, although the dependence on pressure is rather weak. For liquids, both the dynamic and kinematic viscosities are practically independent of pressure, and any small variation with pressure is usually disregarded, except at extremely high pressures. For gases, this is also the case for dynamic viscosity (at low to moderate pressures), but not for kinematic viscosity since the density of a gas is proportional to its pressure (Fig.).
• 45. 2-6 Viscosity March 11, 2021 45 The viscosity of a fluid is directly related to the pumping power needed to transport a fluid in a pipe or to move a body through a fluid. Viscosity is caused by the cohesive forces between the molecules in liquids and by the molecular collisions in gases, and it varies greatly with temperature. For liquids, the molecules possess more energy at higher temperatures, and they can oppose the large cohesive intermolecular forces more strongly. As a result, the energized liquid molecules can move more freely. For gases, the intermolecular forces are negligible, and the gas molecules at high temperatures move randomly at higher velocities. This results in more molecular collisions per unit volume per unit time and therefore in greater resistance to flow (Fig.).
• 46. 2-6 Viscosity March 11, 2021 46 In the kinetic theory of gases the viscosity of gases is predicted to be proportional to the square root of temperature, T, i.e.: Measuring viscosities at two different temperatures is sufficient to determine the constants. For air, the values of these constants are a = 1.458×10–6 kg/(m·s·K1/2) and b = 110.4 K at atmospheric conditions. The viscosity of gases is independent of pressure at low to moderate pressures (from a few percent of 1 atm. to several atm.). But viscosity increases at high pressures due to the increase in density. The viscosity of gases is expressed as a function of temperature by the Sutherland correlation (from The U.S. Standard Atmosphere; where T is absolute temperature and a & b are experimentally determined constants) as: T gases   T b T a   1  For gases:
• 47. 2-6 Viscosity March 11, 2021 47 For liquids, the viscosity is approximated as (where again T is absolute temperature and a, b, and c are experimentally determined constants : For water, using the values a = 2.414 ×10–5 N·s/m2, b = 247.8 K, and c = 140 K results in less than 2.5 percent error in viscosity in the temperature range of 0°C to 370°C. ) /( 10 c T b a    For liquids Can you Please give us the formula for obtaining the AREA of a CYLINDER? Okay, Class, give us the formula for obtaining the AREA of a CYLINDER?
• 48. 2-6 Viscosity March 11, 2021 48 Consider a fluid layer of thickness l within a small gap between two concentric cylinders, such as the thin layer of oil in a journal bearing. The gap between the cylinders can be modelled as two parallel flat plates separated by a fluid. From torque, Τ = FR (force times the moment arm, which is the radius R of the inner cylinder in this case), the tangential velocity, V = ωR, and taking the wetted surface area of the inner cylinder to be A = 2πRL by disregarding the shear stress acting on the two ends of the inner cylinder: l L n R l L R FR Τ  3 2 3 4 2         Here, where L is the length of the cylinder and ṅ is the number of revolutions per unit time. The angular distance travelled during one rotation is 2ω rad, and thus the angular velocity in rad/min and the rpm is ω = 2π ṅ
• 49. 2-6 Viscosity March 11, 2021 49 The two concentric cylinders, and the equation given, can therefore be used as a viscometer, a device that measures viscosity. The viscosities of some fluids at room temperature *and their plots) are presented in next slide. Note that the viscosities of different fluids differ by several orders of magnitude. Also note that it is more difficult to move an object in a higher-viscosity fluid such as engine oil than it is in a lower-viscosity fluid such as water. Liquids, in general, are much more viscous than gases.
• 50. 2-6 Viscosity March 11, 2021 50 The variation of dynamic (absolute) viscosity of common fluids with temperature at 1 atm. (1 Ns/m2 = 1 kg/ms)
• 51. 2-6 Viscosity March 11, 2021 51 Example: The viscosity of a fluid is to be measured by a viscometer constructed of two 0.4- m long concentric cylinders The outer diameter of the inner cylinder is 0.12 m, and the gap between the two cylinders is 0.0015 m. The inner cylinder is rotated at 5 revolutions per second, and the torque is measured to be 1.8 N·m. Determine the viscosity of the fluid. L n R Τl l L n R l L R FR Τ   3 2 3 2 3 4 4 2                     2 1 3 2 3 2 s/m N 158 . 0 m 4 . 0 s 5 m 06 . 0 4 m 0015 . 0 m N 8 . 1 4          L n R Τl  Note: Since viscosity is a strong function of temperature, the temperature of the fluid should have also been measured during this experiment, and reported with this calculation.
• 52. 2-6 Viscosity March 11, 2021 52 Consider a round flat disc of radius R rotating at angular velocity ω rad/s on top of a flat surface and separated from it by an oil film of thickness t. Assuming a linear velocity gradient, show that the TORQUE, T in terms of Diameter, D is given by: t D Τ 2 3 4     Assume the velocity gradient is linear, in which case is given as: du/dy = u/t = ωr/t at any raduis r
• 53. 2-7 Surface tension and capillary effect March 11, 2021 53 Liquid droplets behave like small balloons filled with the liquid on a solid surface, and the surface of the liquid acts like a stretched elastic membrane under tension. The pulling force that causes this tension acts parallel to the surface and is due to the attractive forces between the molecules of the liquid. The magnitude of this force per unit length is called surface tension, σs (or coefficient of surface tension) and is usually expressed in the unit N/m. This effect is also called surface energy [per unit area] and is expressed in the equivalent unit of N·m/m2 or J/m2. In this case, σs represents the stretching work that needs to be done to increase the surface area of the liquid by a unit amount. Some consequences of surface tension: (a) drops of water beading up on a leaf, (b) a water strider sitting on top of the surface of water,
• 54. 2-7 Surface tension and capillary effect March 11, 2021 54 Consider a liquid film (such as the film of a soap bubble) suspended on a U-shaped wire frame with a movable side (Fig.). Normally, the liquid film tends to pull the movable wire inward in order to minimize its surface area. A force F needs to be applied on the movable wire in the opposite direction to balance this pulling effect. The thin film in the device has two surfaces (the top and bottom surfaces) exposed to air, and thus the length along which the tension acts in this case is 2b. Then a force balance on the movable wire gives F = 2bσs, and thus the surface tension can be expressed as: b F s 2  
• 55. 2-7 Surface tension and capillary effect March 11, 2021 55 In the U-shaped wire, the force F remains constant as the movable wire is pulled to stretch the film and increase its surface area. When the movable wire is pulled a distance Δx, the surface area increases by ΔA = 2bΔx, and the work done W during this stretching process with constant force is: A x b x F W s s           2 2 Distance Force This result can also be interpreted as the surface energy of the film is increased by an amount σsΔA during this stretching process, which is consistent with the alternative interpretation of σs as surface energy. This is similar to a rubber band having more potential (elastic) energy after it is stretched further. In the case of liquid film, the work is used to move liquid molecules from the interior parts to the surface against the attraction forces of other molecules. Therefore, surface tension also can be defined as the work done per unit increase in the surface area of the liquid. The surface tension varies greatly from substance to substance, and with temperature for a given substance. At 20°C, for example, the surface tension is 0.073 N/m for water and 0.440 N/m for mercury surrounded by atmospheric air.
• 56. 2-7 Surface tension and capillary effect March 11, 2021 56 Mercury droplets form spherical balls that can be rolled like a solid ball on a surface without wetting the surface. The surface tension of a liquid, in general, decreases with temperature and becomes zero at the critical point (and thus there is no distinct liquid–vapour interface at temperatures above the critical point). The effect of pressure on surface tension is usually negligible. The surface tension of a substance can be changed considerably by impurities. Therefore, certain chemicals, called surfactants, can be added to a liquid to decrease its surface tension. For example, soaps and detergents lower the surface tension of water and enable it to penetrate through the small openings between fibres for more effective washing.
• 57. 2-7 Surface tension and capillary effect March 11, 2021 57 We speak of surface tension for liquids only at liquid–liquid or liquid–gas interfaces. Therefore, it is important to specify the adjacent liquid or gas when specifying surface tension. Also, surface tension determines the size of the liquid droplets that form. A droplet that keeps growing by the addition of more mass will break down when the surface tension can no longer hold it together. This is like a balloon that will burst while being inflated when the pressure inside rises above the strength of the balloon material. A curved interface indicates a pressure difference (or “pressure jump”) across the interface with pressure being higher on the concave side.
• 58. 2-7 Surface tension and capillary effect March 11, 2021 58 The excess pressure ΔP inside a droplet or bubble above the atmospheric pressure, for example, can be determined by considering the free-body diagram of half a droplet or air bubble (Fig.). Noting that surface tension acts along the circumference and the pressure acts on the area, horizontal force balances for the droplet and the bubble (where Pi and Po are the pressures inside and outside the droplet or bubble, respectively) give:     R P P P P πR σ πR s i s  2 Δ 2 : Droplet 0 droplet droplet 2           R P P P P πR σ πR s i s  4 Δ 2 2 : Bubble 0 bubble bubble 2       Free-body diagram of: (a) half a droplet or air bubble, and (b) half a soap bubble
• 59. 2-7 Surface tension and capillary effect March 11, 2021 59 When the droplet or bubble is in the atmosphere, Po is simply atmospheric pressure. The factor 2 in the force balance for the bubble is due to the bubble consisting of a film with two surfaces (inner and outer surfaces) and thus two circumferences in the cross section. The excess pressure in a droplet (or bubble) also can be determined by considering a differential increase in the radius of the droplet due to the addition of a differential amount of mass and interpreting the surface tension as the increase in the surface energy per unit area. Then the increase in the surface energy of the droplet during this differential expansion process becomes:   dR Rσ πR d σ dA σ W s s s   8 4 2 surface    The expansion work done during this differential process is determined by multiplying the force by distance to obtain:   PdR R dR PA FdR W        2 expansion 4 Distance Force   Note that Area of Sphere, A = 4πr2
• 60. 2-7 Surface tension and capillary effect March 11, 2021 60 Equating the two expressions (in previous slide) That is: PdR R W dR Rσ W s    2 expansion surface 4 and 8     We obtain: R P s  2 droplet   This equation is the same relation obtained before and given in slide 58. Note that the excess pressure in a droplet or bubble is inversely proportional to the radius.
• 61. 2-7 Surface tension and capillary effect March 11, 2021 61 Capillary effect Another consequence of surface tension is the capillary effect. Capillary effect: The rise or fall of a liquid in a small-diameter tube inserted into the liquid. Capillaries: Such narrow tubes or confined flow channels. The capillary effect is partially responsible for the rise of water to the top of tall trees. Meniscus: The curved free surface of a liquid in a capillary tube. The strength of the capillary effect is quantified by the contact (or wetting) angle ϕ, defined as the angle that the tangent to the liquid surface makes with the solid surface at the point of contact.
• 62. 2-7 Surface tension and capillary effect March 11, 2021 62 Capillary effect The surface tension force acts along this tangent line toward the solid surface. A liquid is said to wet the surface when ϕ < 90° and not to wet the surface when ϕ > 90°. In atmospheric air (and most other organic liquids) the contact angle with glass, ϕ ≈ 0°. The magnitude of the capillary rise in a circular tube can be determined from a force balance on the cylindrical liquid column of height h in the tube (Fig.). The bottom of the liquid column is at the same level as the free surface of the reservoir, and thus P = Patm. This balances the atmospheric pressure acting at the top surface, and thus these two effects cancel each other.
• 63. 2-7 Surface tension and capillary effect March 11, 2021 63 Capillary effect The weight of the liquid column is approximately: This relation is also valid for non-wetting liquids (such as mercury in glass) and gives the capillary drop. In this case ϕ < 90° and thus cos ϕ > 0, which makes h negative. Therefore, a negative value of capillary rise corresponds to a capillary drop (Fig.). Capillary rise is inversely proportional to the radius of the tube and density of the liquid.   h R g Vg mg 2 W       Equating the vertical component of the surface tension force to the weight gives        cos 2 W 2 surface s R h R g F      constant R cos 2 : rise capillary The      gR h s
• 64. 2-7 Surface tension and capillary effect March 11, 2021 64 Example: A 0.6-mm-diameter glass tube is inserted into water at 20°C, ρ = 1000 kg/m3, in a cup. Determine the capillary rise of water in the tube (Fig.). The surface tension of water at 20°C is 0.073 N/m (Table in slide 52) and the contact angle of water with glass is 0° (from preceding text, slide 58).       m 05 . 0 0 cos m 10 3 . 0 m/s 81 . 9 kg/m 000 1 /m m/s kg 073 . 0 2 cos 2 : rise capillary The 3 2 3 2            gR s
• 65. Topic 2: Tutorial Questions March 11, 2021 65 1. What is the difference between intensive and extensive properties? 2. What is specific gravity? How is it related to density? 3. Under what conditions is the ideal-gas assumption suitable for real gases? 4. What is the difference between R and Ru? How are these two related? 5. What is vapour pressure? How is it related to saturation pressure? 6. Does water boil at higher temperatures at higher pressures? Explain. 7. If the pressure of a substance is increased during a boiling process, will the temperature also increase or will it remain constant? Why? 8. What is cavitation? What causes it? In a piping system, the water temperature remains under 40°C. Determine the minimum pressure allowed in the system to avoid cavitation. 9. The analysis of a propeller that operates in water at 20°C shows that the pressure at the tips of the propeller drops to 2 kPa at high speeds. Determine if there is a danger of cavitation for this propeller. 10. What is the difference between the macroscopic and microscopic forms of energy? 11. What is total energy? Identify the different forms of energy that constitute the total energy. 12. List the forms of energy that contribute to the internal energy of a system.
• 66. Topic 2: Tutorial Questions March 11, 2021 66 13. How are heat, internal energy, and thermal energy related to each other? 14. What is flow energy? Do fluids at rest possess any flow energy? 15. How do the energies of a flowing fluid and a fluid at rest compare? Name the specific forms of energy associated with each case. 16. Using average specific heats, explain how internal energy changes of ideal gases and incompressible substances can be determined. 17. Using average specific heats, explain how enthalpy changes of ideal gases and incompressible substances can be determined. 18. What does the coefficient of compressibility of a fluid represent? How does it differ from isothermal compressibility? 19. What does the coefficient of volume expansion of a fluid represent? How does it differ from the coefficient of compressibility? 20. Can the coefficient of compressibility of a fluid be negative? How about the coefficient of volume expansion? 21. What is viscosity? What is the cause of it in liquids and in gases? Do liquids or gases have higher dynamic viscosities?
• 67. Topic 2: Tutorial Questions March 11, 2021 67 22. What is a Newtonian fluid? Is water a Newtonian fluid? 23. Consider two identical small glass balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first? Why? 24. How does the dynamic viscosity of (a) liquids and (b) gases vary with temperature? 25. How does the kinematic viscosity of (a) liquids and (b) gases vary with temperature? 26. What is surface tension? What is it caused by? Why is the surface tension also called surface energy? 27. Consider a soap bubble. Is the pressure inside the bubble higher or lower than the pressure outside? 28. What is the capillary effect? What is it caused by? How is it affected by the contact angle? 29. A small-diameter tube is inserted into a liquid whose contact angle is 110°. Will the level of liquid in the tube rise or drop? Explain. 30. Is the capillary rise greater in small- or large-diameter tubes? Problems 1–30 are concept questions, and you are encouraged to answer them all
• 68. Topic 2: Tutorial Questions March 11, 2021 68 31. A 3-kg plastic tank that has a volume of 0.2 m3 is filled with liquid water. Assuming the density of water is 1000 kg/m3, determine the weight of the combined system. 32. The density of atmospheric air varies with elevation, decreasing with increasing altitude. (a) Using the data given in the table, obtain a relation for the variation of density with elevation, and calculate the density at an elevation of 7000 m. (b) Calculate the mass of the atmosphere using the correlation you obtained. Assume the earth to be a perfect sphere with a radius of 6377 km, and take the thickness of the atmosphere to be 25 km. 33. Water at 1 atm. pressure is compressed to 800 atm. pressure isothermally. Determine the increase in the density of water. Take the isothermal compressibility of water to be 4.80×10–5 atm.–1. 34. Water at 15°C and 1 atm. pressure is heated to 100°C at constant pressure. Using coefficient of volume expansion data (Table A-3), determine the change in the density of water. [–38.7 kg/m3]. 35. The density of seawater at a free surface where the pressure is 98 kPa is approximately 1030 kg/m3. Taking the bulk modulus of elasticity of seawater to be 2.34×109 N/m2 and expressing variation of pressure with depth z as dP = ρgdz determine the density and pressure at a depth of 2500 m. Disregard the effect of temperature..
• 69. Topic 2: Tutorial Questions March 11, 2021 69 36. A 0.5 m × 0.3 m × 0.2 m block weighing 150 N is to be moved at a constant velocity of 0.8 m/s on an inclined surface with a friction coefficient of 0.27 (Fig.). (a) Determine the force F that needs to be applied in the horizontal direction. (b) If a 0.4×10–3 m-thick oil film with a dynamic viscosity of 0.012 Pa·s is applied between the block and inclined surface, determine the percent reduction in the required force. 37. A thin 20-cm×20-cm flat plate is pulled at 1 m/s horizontally through a 3.6-mm-thick oil layer sandwiched between two plates, one stationary and the other moving at a constant velocity of 0.3 m/s, as shown in Fig. The dynamic viscosity of oil is 0.027 Pa·s. Assuming the velocity in each oil layer to vary linearly, (a) plot the velocity profile and find the location where the oil velocity is zero and (b) determine the force that needs to be applied on the plate to maintain this motion..
• 70. Topic 2: Tutorial Questions March 11, 2021 70 38. The clutch system shown in Fig. is used to transmit torque through a 3-mm-thick oil film with μ = 0.38 N·s/m2 between two identical 0.30 m-diameter disks. When the driving shaft rotates at a speed of 1450 rpm, the driven shaft is observed to rotate at 1398 rpm. Assuming a linear velocity profile for the oil film, determine the transmitted torque. 39. The viscosity of a fluid is to be measured by a viscometer constructed of two 75-cm-long concentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the two cylinders is 0.12 cm. The inner cylinder is rotated at 200 rpm, and the torque is measured to be 0.8 N·m. Determine the viscosity of the fluid.
• 71. Topic 2: Tutorial Questions March 11, 2021 71 40. Nutrients dissolved in water are carried to upper parts of plants by tiny tubes partly because of the capillary effect. Determine how high the water solution will rise in a tree in a 0.005-mm- diameter tube as a result of the capillary effect. Treat the solution as water at 20°C with a contact angle of 15°.
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