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- 1. Mathematical Models for FLUID MECHANICS P M V Subbarao Associate Professor Mechanical Engineering Department IIT Delhi Convert Ideas into A Precise Blue Print before feeling the same....
- 2. A path line is the trace of the path followed by a selected fluid particle
- 3. Few things to know about streamlines • At all points the direction of the streamline is the direction of the fluid velocity: this is how they are defined. • Close to the wall the velocity is parallel to the wall so the streamline is also parallel to the wall. • It is also important to recognize that the position of streamlines can change with time - this is the case in unsteady flow. • In steady flow, the position of streamlines does not change • Because the fluid is moving in the same direction as the streamlines, fluid can not cross a streamline. • Streamlines can not cross each other. • If they were to cross this would indicate two different velocities at the same point. • This is not physically possible. • The above point implies that any particles of fluid starting on one streamline will stay on that same streamline throughout the fluid.
- 4. A useful technique in fluid flow analysis is to consider only a part of the total fluid in isolation from the rest. This can be done by imagining a tubular surface formed by streamlines along which the fluid flows. This tubular surface is known as a streamtube. A Streamtube A two dimensional version of the streamtube The "walls" of a streamtube are made of streamlines. As we have seen above, fluid cannot flow across a streamline, so fluid cannot cross a streamtube wall. The streamtube can often be viewed as a solid walled pipe. A streamtube is not a pipe - it differs in unsteady flow as the walls will move with time. And it differs because the "wall" is moving with the fluid
- 5. Fluid Kinematics • The acceleration of a fluid particle is the rate of change of its velocity. • In the Lagrangian approach the velocity of a fluid particle is a function of time only since we have described its motion in terms of its position vector.
- 6. In the Eulerian approach the velocity is a function of both space and time; consequently, V ˆ j ˆ u ( x, y, z, t )i v( x, y, z, t ) ˆ w( x, y, z, t )k Velocityco mponents x,y,z are f(t) since we must follow the total derivative approach in evaluating du/dt.
- 7. Similarly for ay & az, In vector notation this can be written concisely
- 8. x Conservation laws can be applied to an infinitesimal element or cube, or may be integrated over a large control volume.
- 9. Basic Control-Volume Approach
- 10. Control Volume • In fluid mechanics we are usually interested in a region of space, i.e, control volume and not particular systems. • Therefore, we need to transform GDE’s from a system to a control volume. • This is accomplished through the use of Reynolds Transport Theorem. • Actually derived in thermodynamics for CV forms of continuity and 1st and 2nd laws.
- 11. Flowing Fluid Through A CV • A typical control volume for flow in an funnel-shaped pipe is bounded by the pipe wall and the broken lines. • At time t0, all the fluid (control mass) is inside the control volume.
- 12. The fluid that was in the control volume at time t0 will be seen at time t0 + t as: .
- 13. The control volume at time t0 + t . The control mass at time t0 + t . The differences between the fluid (control mass) and the control volume at time t0 + t .
- 14. • Consider a system and a control volume (C.V.) as follows: • the system occupies region I and C.V. (region II) at time t0. • Fluid particles of region – I are trying to enter C.V. (II) at time t0. III II I • the same system occupies regions (II+III) at t0 + t • Fluid particles of I will enter CV-II in a time t. •Few more fluid particles which belong to CV – II at t0 will occupy III at time t0 + t.
- 15. The control volume may move as time passes. III has left CV at time t0+ t III II I is trying to enter CV at time t0 II At time t0+ t I VCV At time t0
- 16. Reynolds' Transport Theorem • Consider a fluid scalar property b which is the amount of this property per unit mass of fluid. • For example, b might be a thermodynamic property, such as the internal energy unit mass, or the electric charge per unit mass of fluid. • The laws of physics are expressed as applying to a fixed mass of material. • But most of the real devices are control volumes. • The total amount of the property b inside the material volume M , designated by B, may be found by integrating the property per unit volume, M ,over the material volume :
- 17. Conservation of B • total rate of change of any extensive property B of a system(C.M.) occupying a control volume C.V. at time t is equal to the sum of • a) the temporal rate of change of B within the C.V. • b) the net flux of B through the control surface C.S. that surrounds the C.V. • The change of property B of system (C.M.) during Dt is BCM Bt t Bt 0 0 BCM BII t0 t BIII t0 t BI t0 BII t0 add and subtract B t t 0
- 18. BCM BII t0 t BIII t0 t BI t0 BII t0 BI t0 t BI t0 t BCM BI BII t0 t BIII t0 t BI BII t0 BI t0 t BCM BCV t0 t BCV t0 BIII t0 t BI t0 t The above mentioned change has occurred over a time t, therefore Time averaged change in BCM is BCM BCV t0 t BCV t0 BIII t0 t BI t0 t t t t t
- 19. For and infinitesimal time duration BCM BCV t0 t BCV t0 BIII t0 t BI t0 t lim lim lim lim t o t t o t t o t t o t • The rate of change of property B of the system. dBCM dBCV BIII BI dt dt
- 20. Conservation of Mass • Let b=1, the B = mass of the system, m. dmCM dmCV mout min dt dt The rate of change of mass in a control mass should be zero. dmCV mout min 0 dt
- 21. Conservation of Momentum • Let b=V, the B = momentum of the system, mV. d mV d mV CM CV mV out mV in dt dt The rate of change of momentum for a control mass should be equal to resultant external force. d mV CV mV out mV in F dt
- 22. Conservation of Energy • Let b=e, the B = Energy of the system, mV. d me d me CM CV me out me in dt dt The rate of change of energy of a control mass should be equal to difference of work and heat transfers. d me CV me out me in Q W dt
- 23. Applications of Momentum Analysis M out M in Vn A Vout Vn A Vin F out in This is a vector equation and will have three components in x, y and z Directions. X – component of momentum equation: UA U out UA U in Fx out in
- 24. X – component of momentum equation: UA U out UA U in Fx out in Y – component of momentum equation: VA Vout VA Vin Fy out in Z – component of momentum equation: WA Wout WA Win Fz out in For a fluid, which is static or moving with uniform velocity, the Resultant forces in all directions should be individually equal to zero.
- 25. X – component of momentum equation: UA U out UA U in Fx out in Y – component of momentum equation: VA Vout VA Vin Fy out in Z – component of momentum equation: WA Wout WA Win Fz out in For a fluid, which is static or moving with uniform velocity, the Resultant forces in all directions should be individually equal to zero.
- 26. X – component of momentum equation: Max Fx FB, x FS , x Y – component of momentum equation: May Fy FB, y FS , y Z – component of momentum equation: Maz Fz FB, z FS , z For a fluid, which is static or moving with uniform velocity, the Resultant forces in all directions should be individually equal to zero.
- 27. Vector equation for momentum: Ma F FB FS Vector momentum equation per unit volume: a f fB fS f Body force per unit volume:B Gravitational force: f B oi 0 ˆ ˆ j ˆ gk
- 28. Electrostatic Precipitators Electric body force: Lorentz force density The total electrical force acting on a group of free charges (charged ash particles) . Supporting an applied volumetric charge density. fe fE Jf B Where = Volumetric charge density f E = Local electric field B = Local Magnetic flux density field Jf = Current density
- 29. Electric Body Force • This is also called electrical force density. • This represents the body force density on a ponderable medium. • The Coulomb force on the ions becomes an electrical body force on gaseous medium. • This ion-drag effect on the fluid is called as electrohydrodynamic body force.
- 30. 0 Ideal Fluids….
- 31. Pressure Variation in Flowing Fluids • For fluids in motion, the pressure variation is no longer hydrostatic and is determined from and is determined from application of Newton’s 2nd Law to a fluid element.
- 32. Various Forces in A Flow field • For fluids in motion, various forces are important: • Inertia Force per unit volume : finertia a • Body Force: f body ˆ gk • Hydrostatic Surface Force: f surface, pressure p • Viscous Surface Force: f surface ,viscous . • Relative magnitudes of Inertial Forces and Viscous Surface Force are very important in design of basic fluid devices.
- 33. Comparison of Magnitudes of Inertia Force and Viscous Force • Internal vs. External Flows • Internal flows = completely wall bounded; • Both viscous and Inertial Forces are important. • External flows = unbounded; i.e., at some distance from body or wall flow is uniform. • External Flow exhibits flow-field regions such that both inviscid and viscous analysis can be used depending on the body shape.
- 34. Ideal or Inviscid Flows Euler’s Momentum Equation X – Momentum Equation:
- 35. Euler’s Equation for One Dimensional Flow Define an exclusive direction along the axis of the pipe and corresponding unit ˆ direction vector el Along a path of zero acceleration the pressure variation is hydrostatic
- 36. Pressure Variation Due to Acceleration V V p z V t l l For steady flow along l – direction (stream line) V P z V l l l Integration of above equation yields
- 37. Momentum Transfer in A Pump • Shaft power Disc Power Fluid Power. 2 TN P Td Td m vdp or m pdv 60 • Flow Machines & Non Flow Machines. • Compressible fluids & Incompressible Fluids. • Rotary Machines & Reciprocating Machines.
- 38. Pump • Rotate a cylinder containing fluid at constant speed. • Supply continuously fluid from bottom. • See What happens? Flow in •Any More Ideas?
- 39. Momentum Principle P M V Subbarao Associate Professor Mechanical Engineering Department IIT Delhi A primary basis for the design of flow devices ..
- 40. Momentum Equation
- 41. Applications of of the Momentum Equation Initial Setup and Signs • 1. Jet deflected by a plate or a vane • 2. Flow through a nozzle • 3. Forces on bends • 4. Problems involving non-uniform velocity distribution • 5. Motion of a rocket • 6. Force on rectangular sluice gate • 7. Water hammer
- 42. Navier-Stokes Equations Differential form of momentum equation X-component: 2 2 2 u u u u (p z) u u u u v w t x y z x x2 y2 z2 Y-component: 2 2 2 v v v v (p z) v v v u v w t x y z y x2 y2 z2
- 43. z-component: 2 2 2 w w w w (p z) w w w u v w t x y z z x2 y2 z2
- 44. Applications of Momentum Equation
- 45. Generation of Motive Power Through Newton’s Second Law
- 46. Jet Deflected by a Plate or Blade Consider a jet of gas/steam/water turned through an angle CV and CS are for jet so that Fx and Fy are blade reactions forces on fluid. 2 2 2 u u u u (p z) u u u u v w t x y z x x2 y2 z2
- 47. Steady 2 Dimensional Flow X-component: 2 2 u u (p z) u u u v x y x x2 y2 Y-component: 2 2 v v (p z) v v u v x y y x2 y2 Continuity equation: u v u v 0 0 x y x y
- 48. Steady 2 Dimensional Invisicid Flow X-component: u u p u v x y x Y-component: v v p u v x y y Continuity equation: u v 0 x y Inlet conditions : u = U & v = 0
- 49. Pure Impulse Blade Pressure remains constant along the entire jet. u u X-component: u v o x y Y-component: v v u v 0 x y Continuity equation: u v u v 0 x y x y

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