Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

No Downloads

Total views

1,270

On SlideShare

0

From Embeds

0

Number of Embeds

3

Shares

0

Downloads

0

Comments

0

Likes

2

No embeds

No notes for slide

- 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

No public clipboards found for this slide

Be the first to comment