Fundamental of Fluid Mechanics PDF

**GET205** **Fundamental of Fluid Mechanics** **Definition:** Fluid mechanics is the study of the behaviors of fluids (Liquid and gas) under the influence of forces. The study is classified under two broad topics; fluid statics and fluid dynamics. Fluid statics is the branch of fluid mechanics tha...

**GET205** **Fundamental of Fluid Mechanics** **Definition:** Fluid mechanics is the study of the behaviors of fluids (Liquid and gas) under the influence of forces. The study is classified under two broad topics; fluid statics and fluid dynamics. Fluid statics is the branch of fluid mechanics that studies incompressible fluid at rest. It encompasses the study of the conditions under which fluids are at rest in stable equilibrium. That is, the study of forces which kept the fluid statics in static equilibrium. Fluid dynamics deals with the motion of fluids and the forces which kept them in motion. **Fluid Statics** Fluid statics is the study of the forces which keep a body of fluids in static equilibrium. Static equilibrium implies that the resultant of all the forces does not result in translation or rotation of the body. Consequently, the resultant moment must be zero. Linear and angular acceleration are therefore zero and shear stresses are also zero. Since there is no rotation, the body of fluid is in equilibrium under the influence of normal forces. In the absence of mechanical, electrical, thermal, magnetic and gravitational forces; the normal forces are hydrostatic forces which are directly proportional to the weight of the fluid above the body. The most important characteristics of a static fluid is the pressure it exerts on bodies floating on or immersed in it. Such knowledge is essential for the design, safety and performance of such devices as aircrafts and submarines and the suits of deep-sea divers. ***Pressure head*** The pressure on a given plane in a fluid is defined as the fluid force acting perpendicularly to the plane, that is \ [\$\$P\\mspace{6mu} = \\mspace{6mu}\\frac{F}{A}\$\$]{.math.display}\ Very often, it is convenient to express the pressure at a point in a fluid in terms of the height h, of fluid above the point. The height h is usually referred to as head of liquid above the point. The relationship between the pressure at a point and the height of fluid above that point can be deduced as follows: Consider the column of fluid of arbitrary cross-sectional area [*Δ*]{.math.inline}A and height h standing perpendicularly on area A in the fluid as shown below: **Figure 1:** Pressure head The force on[ *Δ*]{.math.inline}A is solely that due to the weight of the element. Column, which is *ρ(hA)g*. therefore, the pressure due to this weight is \ [\$\$P\\mspace{6mu} = \\mspace{6mu}\\frac{\\rho\\left( \\text{hΔA} \\right)g}{\\text{ΔA}}\$\$]{.math.display}\ [*P* = *ρhg*]{.math.inline} ***Terminal Velocity*** In a free fall, the object will attain a constant velocity at which the net gravitational accelerating force, V~g~ is equal to resistance upward drag force, F~r~. This constant velocity is called terminal velocity (V~t~). If the particle density is greater than the fluid density, the particle motion will be downward. However, if particle density is smaller than the fluid density, the particle will rise. F~r~ = F~g~ Where V = V~t~ \ [\$\$M\_{p}g\\left\\lbrack \\frac{\\rho\_{p} - \\rho\_{f}}{\\rho\_{p}} \\right\\rbrack = 1/2CA\_{p}\\rho\_{f}{V\_{t}}\^{2}\\ \$\$]{.math.display}\ [*V*~*t*~ = ]{.math.inline} [\$\\left\\lbrack \\frac{{2W(\\rho}\_{p} - \\rho\_{f})}{\\rho\_{p}\\rho\_{f}A\_{p}C} \\right\\rbrack\^{2}\$]{.math.inline} Where g = acceleration due to gravity, *M~p~* = mass of particle, *ρ~p~* = density of particle, *ρ~f~* = density of fluid, *A~p~* = projected area of particle, w = weight of particle, C = drag co-efficient. **Fluid Dynamics** Fluid Dynamics can be subdivided into hydrodynamics and gas dynamics: (1). Hydrodynamics is the study of fluid flows where there are no density changes. It deals with fluid motion, forces on bodies immersed in fluid, and the motion of a body relative to the motion of fluid. The flow of liquids falls in this category mainly but it also includes the flow of gases at low speeds. Simple hydrodynamics theory deals with fluid without viscosity. In a non-viscous fluid, a deeply submerged body experiences no resistance. Although, the fluid is disturbed by the passage of the body, it returns to its original state of rest once the body has passed. A subdivision of hydrodynamics is called hydraulics which is the study of the flow of liquids in pipes and open channels. (2). Gas dynamics is the study of flows where density changes occur such as high-speed gas flows through conduits or over solid surfaces. This is a study of the kinetic theory of gases, often leading to the study of gas diffusion, statistical mechanics, chemical thermodynamics and non-equilibrium thermodynamics. Gas dynamics is synonymous with aerodynamics when the gas field is air and the subject of study is flight. **Type of Fluid** Basically, there are two types of fluids namely ideal and real or non-ideal fluids. 1. **Ideal fluid** is a fluid that is incompressible and no internal resistance to flow (zero viscosity). In addition, ideal fluid particles undergo no rotation about their center of mass (irrotational). Since it does not exhibit viscous properties and cannot sustain frictional and shear stresses when in motion. The forces sustaining its motion are purely pressure forces. In practice, ideal fluids are imaginary in nature and does not exist. 2. Real fluid possesses viscous properties, sustains frictional and shear stresses and dissipates mechanical energy into heat. The flow of many real fluids can be analysed by assuming that they are ideal especially if their viscosities are low. Examples of real fluids are viscous (Newtonian, non-Newtonian), and compressible fluids. Viscosity may be thought of as a liquid's internal resistance to flow. A liquid can be envisaged as having a series of layers and when it flows over a surface, the uppermost layer flows faster and drags the next layer along at a slightly lower velocity and so on through the layers until the one next to the surface is stationary. The force that moves the liquid is known as the shearing force or shear stress and the velocity gradient is known as the shear rate of strain. If shear stress is plotted against shear rate, as in most liquids and gases shows a linear relationship as shown in A and these are termed Newtonian fluids. Examples of this include water, most oils, gases and simple solutions of sugars and salts. Where the relationship is non-linear as shown in B and C, the fluids are termed non-Newtonian. For all liquids, viscosity decreases with an increase in temperature but for most gases, it increases with temperature. Many liquids are non-Newtonian, including emulsions and suspensions, concentrated solutions etc. These liquids often display Newtonian properties at low concentrations but as the concentration of the solution is increased, the viscosity increases rapidly and there is a transition to non-Newtonian properties. B, C, D and E are called time independent non-Newtonian fluids. \ **Figure 2:** Changes in in Viscosity of Newtonian Fluid and different Types of non-Newtonian Fluid A = Newtonian fluid B = Pseudoplastic fluid C = Dilatant fluid D = Bingham fluid E = Casson fluid When force or shear stress is applied to fluid, the fluid flows and nature of the flow is determined. In any system in which fluids flow, there exists a boundary film or surface film of fluid next to the surface over which the fluid flows. The thickness of the boundary film is influenced by a number of factors, including the velocity, viscosity, density and temperature of the fluid. Fluids which have a low flow rate or high viscosity may be thought of as a series of layers which move over one another without mixing, this produces movement of the fluid, which is termed streamline or laminar flow. In a pipe, the velocity of the fluid is highest at the center and zero at the pipe wall. Above a certain flow rate, which is determined by the nature of fluid and the pipe, the layers of liquid mix together and turbulent flow is established in the bulk of the fluid, although the flow remains streamline in the boundary film. Higher flow rate produces more turbulent flow and hence thinner boundary films. Fluid flow is characterized by a dimensionless group named the Reynolds number (Re). This is calculated using: \ [\$\$\\text{Re}\\mspace{6mu} = \\mspace{6mu}\\frac{\\text{DV}\\rho}{µ}\$\$]{.math.display}\ Where D (m) = the diameter of the pipe, V (ms^-1^) = average velocity, ρ (kgm^-3^) = fluid density and µ (Nsm^2^) = fluid viscosity. A Reynolds number of less than or equal to 2100 describes streamline or laminar flow and a Reynolds number greater than or equal to 4000 describes turbulent flow. For Reynolds numbers between 2100 and 4000, transitional flow is present, which can be either laminar or turbulent at different times. These different flow characteristics have important implication for heat transfer and mixing operations; turbulent flow produces thinner boundary layers, which in turn permit higher rates of heat transfer. The Reynolds number can also be used to determines the power requirement to pumps and mixers used for blending and mixing operations. In turbulent flow, particles of fluid move in all directions and solids are retained in suspension more readily. This reduces the formation of deposits on heat exchangers and prevent solids from settling out in pipework. Streamline flow produces a larger range of residence times for individual particles flowing in a tube, this is especially important when calculating the residence time required for heat treatment of liquid foods, as it is necessary to ensure that all parts of the food receive the required amount of heat. Turbulent flow causes higher friction losses than streamline flow does and therefore requires higher energy inputs from pumps. **Example:** Two fluids, brake oil and engine oil are flowing along pipes of the same diameter (5 cm) at 20^o^C and at the same flow velocity of 3 ms^-1^. Determine whether the flow is streamline or turbulent in each fluid. Given that for brake oil; viscosity = 2.10 x 10^-3^ Nsm^-2^, density = 1030 kgm^-3^. For engine oil; viscosity = 118 x 10^-3^ Nsm^-2^, density = 900 kgm^-3^. For brake oil; [\$\\text{Re}\\mspace{6mu} = \\mspace{6mu}\\frac{\\text{DV}\\rho}{µ}\$]{.math.inline} \ [\$\$\\text{Re}\\mspace{6mu} = \\mspace{6mu}\\frac{0.05\\ x\\ 3\\ x\\ 1030}{0.00210}\$\$]{.math.display}\ \ [Re = 73, 571]{.math.display}\ Thus, the flow is turbulent because Re is greater than 4,000 For engine oil; [\$\\text{Re}\\mspace{6mu} = \\mspace{6mu}\\frac{\\text{DV}\\rho}{µ}\$]{.math.inline} \ [\$\$\\text{Re}\\mspace{6mu} = \\mspace{6mu}\\frac{0.05\\ x\\ 3\\ x\\ 900}{0.118}\$\$]{.math.display}\ \ [Re = 1144]{.math.display}\ Thus, the nature of the flow is streamline because Re is less than 2100. **Compressibility of Liquids** Being a rigid body with the particles closely packed together, the solid can be compressed or stretched and it will return to its original shape as long as the yield stress or the elastic limit is not reached. The nature of the particles of fluids does not permit handling them in the same way as the solid. The volume of a given mass of fluid can be changed by increasing or decreasing the pressure on it. In particular, if the fluid is gas, its volume can also be altered by changing its temperature. The new state of the perfect gas can be predicted using the perfect gas law. For many practical purposes, the liquid can be regarded as incompressible. When subjected to large pressure, the compressibility of the liquid cannot be ignored. Suppose a liquid of volume V is subjected to sudden increase in pressure, *dP* with a corresponding decrease in volume *-dV*. The compressibility of the liquid is gives as [\$K\\mspace{6mu} = \\mspace{6mu} - \\frac{\\text{dP}}{dV/V}\$]{.math.inline}........(1) Where K is the bulk modulus of elasticity. The negative sign is to take care of the fact that *dV* is negative since K cannot be negative. Equation 1 can be expressed in terms of density as follow: by differentiation \ [*M* = ρV]{.math.display}\ We have [dM = ρdV]{.math.inline} *+ Vdρ* Since M is constant, *dm = 0* and \ [\$\$\\mspace{6mu} - \\frac{\\text{dV}}{V} = \\ \\frac{d\\rho}{\\rho}\$\$]{.math.display}\ Substituting *dρ/ρ* for *-dV/V* in equation (1) yields [\$K\\mspace{6mu} = \\mspace{6mu}\\frac{\\text{dP}}{d\\rho/\\rho}\$]{.math.inline}......(2) K varies with temperature and at any particular temperature, the higher the compression pressure, the higher the resistance to compression and the higher the value of K. **Fluid Properties** a. ***Density and Specific Weight*** The density of a fluid at any given temperature and pressure can be defined in two ways in terms of mass and weight. The mass density is defined as mass per unit volume while specific weight is defined as weight per unit volume. Using the symbols *ρ* and ω for density and specific weight respectively, the definitions are: \ [\$\$\\rho\\mspace{6mu} = \\mspace{6mu}\\frac{M}{V}\$\$]{.math.display}\ \ [\$\$\\omega\\mspace{6mu} = \\mspace{6mu}\\frac{\\text{Mg}}{V}\$\$]{.math.display}\ We can deduce from the two equation that \ [*ω*  = *ρg*]{.math.display}\ On the SI system, the units of *ω* are N/m^3^ and those of *ρ* are kg/m^3^. By deduction from the units of *ω*, we can also see that the unit of *ρ* are N/s^2^/m^4^. The reciprocal of specific weight is called specific volume, that is, it is the volume of a unit weight of the fluid. The densities of fluids are often expressed as ratio of their actual density to that of water at a standard temperature (usually, 4^o^C). Such ratio is referred to as specific gravity, S.g: \ [\$\$\\text{S.g}\\mspace{6mu} = \\mspace{6mu}\\frac{\\rho\\text{\\ liquid}}{\\rho\\text{\\ water}}\$\$]{.math.display}\ b. ***Properties of perfect gas*** i. **Gas constant** The perfect gas is one which obeys the equation of state \ [pv  = *mRT*]{.math.display}\ Where p = pressure of the gas, v = volume, m = mass, R = gas constant or universal constant, T = absolute temperature. The density of the gas is obtained from [\$p\\mspace{6mu} = \\mspace{6mu}\\frac{m}{v}\\text{RT}\$]{.math.inline} *= ρRT* The gas constant R is different for different gasses. ii. **Specific heat** **Figure.** Volume -- Pressure -- Temperature relationship for a perfect gas. At constant volume, addition of heat energy to a system, results in increase in internal energy, U, of the system. If just enough heat is added to raise the temperature of the system by one degree, we can write [\$\\text{Cv}\\mspace{6mu} = \\left( \\mspace{6mu}\\frac{\\delta U}{\\delta T} \\right)\$]{.math.inline}~v~[ *ΔT*]{.math.inline} Since [ΔT ]{.math.inline}= 1, the equation reduces to [\$\\text{Cv}\\mspace{6mu} = \\left( \\mspace{6mu}\\frac{\\delta U}{\\delta T} \\right)\$]{.math.inline}~v~[.....(1)]{.math.inline} At constant pressure, the addition of heat results in increase in internal energy as well as increase in work of expansion represented by the product pv. The sum of these two quantities is called enthalpy, h. Thus \ [\$\$\\text{h\\;} = u + pv\\ldots\\ldots(2)\\text{\\ \\ \\ \\ \\ \\ \\ }\$\$]{.math.display}\ by using the equation of state, \ [\$\$\\text{h\\;} = u + RT\\ldots\\ldots.(3)\$\$]{.math.display}\ Consequently, following the argument for Cv, the expression for Cp is [\$\\text{Cp}\\mspace{6mu} = \\left( \\mspace{6mu}\\frac{\\text{δh}}{\\delta T} \\right)\$]{.math.inline}~p~[.........(4) ]{.math.inline} Differentiation of equation (3) yields [\$\\frac{\\delta h}{\\delta T} = \\ \\frac{\\text{δu}}{\\delta T}\$]{.math.inline} + R Substituting for the differential terms from equation (1) and (4) yields \ [Cp  = *Cv* + *R*.........(5)]{.math.display}\ In this equation, the three quantities must be in the same units, either thermal unit (J/kg. K) or mechanical unit (m.N/kg). The specific heat ratio or Adiabatic index (1.4) \ [\$\$\\gamma\\mspace{6mu} = \\mspace{6mu}\\frac{\\text{Cp}}{\\text{Cv}}\\ldots\\ldots\\ldots\\ldots(6)\$\$]{.math.display}\ **Example:** Calculate the values of Cv and Cp for a perfect gas whose universal constant is 519.5, given that Cp/Cv = 1.4. Solution. From equation 5 and 6 \ [Cp  = *Cv* + *R*.....(1)]{.math.display}\ \ [Cp  = *γ*Cv......(2)]{.math.display}\ Therefore, *\ *[*Cv* + *R*]{.math.inline} *= γCv* *Divide equation (1) through by Cv* [\$\\mspace{6mu}\\frac{\\text{Cp}}{\\text{Cv}}\$]{.math.inline} = 1 + [\$\\frac{R}{\\text{Cv}}\$]{.math.inline} 1.4= 1 + [\$\\frac{519,5}{\\text{Cv}}\$]{.math.inline} [\$\\frac{519,5}{\\text{Cv}}\$]{.math.inline} = 1.4 - 1 Cv = 1298.75 J/kg. K From equation (1) Cp = 1298.75 + 519.5 Cp = 1818.25 J/kg. K