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Note for FLUID MECHANICS AND MACHINERY - FMM by Ravichandran Rao

  • Fluid Mechanics and Machinery - FMM
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  • VELLORE INSTITUTE OF TECHNOLOGY - VIT
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MODULE-3

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FLUID DYNAMICS Forces acting on the fluids Following are the forces acting on the fluids 1. Self-Weight/ Gravity Force, Fg 2. Pressure Forces, F p 3. Viscous Force, Fv 4. Turbulent Force, Ft 5. Surface Tension Force, Fs 6. Compressibility Force, Fc Dynamics of fluid is governed by Newton’s Second law of motion, which states that the resultant force on any fluid element must be equal to the product of the mass and the acceleration of the element. ∑ F = Ma or ∑ F = Fg + Fp + Fv + Fs + Fc (1) Surface tension forces and Compressibility forces are not significant and may be neglected. Hence (1) becomes ∑ F = Fg + Fp + Fv + Ft - Reynold’s Equation of motion and used in the analysis of Turbulent flows. For laminar flows, turbulent force becomes less significant and hence (1) becomes ∑ F = Fg + Fp + Fv – Navier - Stokes Equation. If viscous forces are neglected then the (1) reduces to ∑ F = Fg + Fp = M × a – Euler’s Equation of motion. Euler equation of motion Consider a stream lime in a flowing fluid in S direction as shown in the figure. On this stream line consider a cylindrical element having a cross sectional area dA and length ds. 1

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eq.png Fluid element in stream line Forces acting on the fluid element are: Pressure forces at both ends: • Pressure force, pdA in the direction of flow • Pressure force (p+(∂ p/∂ s)ds)dA in the direction opposite to the flow direction • Weight of element ρdads acting vertically downwards Let φ be the angle between the direction of flow and the line of action of the weight of the element. The resultant force on the fluid element in the direction of s must be equal to mass of fluid element× acceleration in direction s (according to Newton’s second law of motion) pda − (pda + (∂ p/∂ s)ds)dA) − ρgd cos φ = ρdadsas (a) where as is the accelaration in direction of s now as = dv dt where v is function of s and t ∂ v ds ∂ v + ∂ s dt ∂t ∂ v ds ∂ v =v + ∂ s dt ∂t = 2

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since ds =v dt If the flow is steady, ∂v =0 ∂t hence, ∂v ∂s Substituting the valve of as in equation (a) and simplifying, as = v ∂p ∂v dsdA − ρgdsdA cos φ = ρdads × v ∂s ∂s Dividing the whole equation by ρdsdA, − ∂p ∂v − g cos φ = v ρ∂ s ∂s ∂p ∂v ⇒ + g cos φ + v = 0 ρ∂ s ∂s − But from the figure we have cos φ = dz ds Hence, 1 ∂p dz ∂v +g +v = 0 ρ ∂s ds ∂s or ∂p + gdz + vdv = 0 ρ Equation (b)is known as Euler’s equation of motion. (b) Bernoulli’s Equation of motion from Euler’s equation Statement: In a steady, incompressible fluid, the total energy remains same along a streamline throughout the reach. Bernoulli’s equation may be obtained by integrating Euler’s equation of motion i.e, equation (b) as Z dp + ρ Z Z gdz + vdv = constant If the flow is in-compressible, ρ is constant and hence, p v2 + gz + = constant ρ 2 2 p v ⇒ + + z = constant ρg 2g (c) 3

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