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Essential fluids
- 1. KV FLUID MECHANICS {for energy conversion} Keith Vaugh BEng (AERO) MEng
- 2. KV Identify the unique vocabulary associated with fluid mechanics with a focus on energy conservation Explain the physical properties of fluids and derive the conservation laws of mass and energy for an ideal fluid (i.e. ignoring the viscous effects) Develop a comprehensive understanding of the effect of viscosity on the motion of a fluid flowing around an immersed body Determine the forces acting on an immersed body arising from the flow of fluid around it OBJECTIVES
- 3. KV BULK PROPERTIES OF FLUIDS ρ = m V ⎛ ⎝⎜ ⎞ ⎠⎟ Density (ρ) - Mass per unit volume
- 4. KV BULK PROPERTIES OF FLUIDS Pressure (P) - Force per unit area in a fluid Viscosity - Force per unit area due to internal friction ρ = m V ⎛ ⎝⎜ ⎞ ⎠⎟ Density (ρ) - Mass per unit volume
- 8. KV Fluid stream velocity (u) Stream tube streamlines MASS CONTINUITY
- 9. KV Fluid stream velocity (u) Stream tube streamlines !m = ρuA MASS CONTINUITY kg s = kg m 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ m s ⎛ ⎝⎜ ⎞ ⎠⎟ m2 ( )
- 10. KV An incompressible ideal fluid flows at a speed of 2 ms-1 through a 1.2 m2 sectional area in which a constriction of 0.25 m2 sectional area has been inserted.What is the speed of the fluid inside the constriction? Putting ρ1 = ρ2, we have u1A1 = u2A2 u2 = u1 A1 A2 = 2m s( ) 1.2 0.25 ⎛ ⎝⎜ ⎞ ⎠⎟ = 9.6m s EXAMPLE 1
- 11. KV and kinetic energy (KE) of liquid per unit mass total energy of liquid per unit mass PE = g z + p ρg ⎛ ⎝⎜ ⎞ ⎠⎟ KE = u2 2 Etotal = g z + p ρg + u2 2g ⎛ ⎝⎜ ⎞ ⎠⎟ z p ρg ENERGY OF A MOVING LIQUID
- 12. KV In many practical situations, viscous forces are much smaller that those due to gravity and pressure gradients over large regions of the flow field.We can ignore viscosity to a good approximation and derive an equation for energy conservation in a fluid known as Bernoulli’s equation. For steady flow, Bernoulli’s equation is of the form; For a stationary fluid, u = 0 ENERGY CONSERVATION IDEAL FLUID p ρ + gz + 1 2 u2 = const p ρ + gz = const
- 13. KV BERNOULLI’s EQUATION FOR STEADY FLOW (Andrews & Jelley 2007)
- 14. KV (Andrews & Jelley 2007)
- 15. KV Work done in pushing elemental mass δm a small distance δs δs = uδt δW1 = p1 A1 δs1 = p1 A1 u1 δt (Andrews & Jelley 2007)
- 16. KV Work done in pushing elemental mass δm a small distance δs δs = uδt δW1 = p1 A1 δs1 = p1 A1 u1 δt Similarly at P2 δW2 = p2 A2 u2 δt (Andrews & Jelley 2007)
- 17. KV Work done in pushing elemental mass δm a small distance δs δs = uδt δW1 = p1 A1 δs1 = p1 A1 u1 δt Similarly at P2 δW2 = p2 A2 u2 δt The net work done is δW2 +δW2 = p1 A1 u1 δt + p2 A2 u2 δt (Andrews & Jelley 2007)
- 18. KV Work done in pushing elemental mass δm a small distance δs δs = uδt δW1 = p1 A1 δs1 = p1 A1 u1 δt Similarly at P2 δW2 = p2 A2 u2 δt The net work done is δW2 +δW2 = p1 A1 u1 δt + p2 A2 u2 δt By energy conservation, this is equal to the increase in Potential Energy plus the increase in Kinetic Energy, therefore; p1 A1 u1 δt + p2 A2 u2 δt = δmg z2 − z1( )+ 1 2 δm u2 2 − u1 2 ( ) (Andrews & Jelley 2007)
- 19. KV Work done in pushing elemental mass δm a small distance δs δs = uδt δW1 = p1 A1 δs1 = p1 A1 u1 δt Similarly at P2 δW2 = p2 A2 u2 δt The net work done is δW2 +δW2 = p1 A1 u1 δt + p2 A2 u2 δt By energy conservation, this is equal to the increase in Potential Energy plus the increase in Kinetic Energy, therefore; p1 A1 u1 δt + p2 A2 u2 δt = δmg z2 − z1( )+ 1 2 δm u2 2 − u1 2 ( ) Putting δm = ρu1 A1 δt = ρu2 A2 δt and tidying p1 ρ + gz1 + 1 2 u1 2 = p2 ρ + gz2 + 1 2 u2 2 (Andrews & Jelley 2007)
- 20. KV Work done in pushing elemental mass δm a small distance δs δs = uδt δW1 = p1 A1 δs1 = p1 A1 u1 δt Similarly at P2 δW2 = p2 A2 u2 δt The net work done is δW2 +δW2 = p1 A1 u1 δt + p2 A2 u2 δt By energy conservation, this is equal to the increase in Potential Energy plus the increase in Kinetic Energy, therefore; p1 A1 u1 δt + p2 A2 u2 δt = δmg z2 − z1( )+ 1 2 δm u2 2 − u1 2 ( ) Putting δm = ρu1 A1 δt = ρu2 A2 δt and tidying p1 ρ + gz1 + 1 2 u1 2 = p2 ρ + gz2 + 1 2 u2 2 Finally, since P1 and P2 are arbitrary (Andrews & Jelley 2007)
- 21. KV Work done in pushing elemental mass δm a small distance δs δs = uδt δW1 = p1 A1 δs1 = p1 A1 u1 δt Similarly at P2 δW2 = p2 A2 u2 δt The net work done is δW2 +δW2 = p1 A1 u1 δt + p2 A2 u2 δt By energy conservation, this is equal to the increase in Potential Energy plus the increase in Kinetic Energy, therefore; p1 A1 u1 δt + p2 A2 u2 δt = δmg z2 − z1( )+ 1 2 δm u2 2 − u1 2 ( ) Putting δm = ρu1 A1 δt = ρu2 A2 δt and tidying p1 ρ + gz1 + 1 2 u1 2 = p2 ρ + gz2 + 1 2 u2 2 Finally, since P1 and P2 are arbitrary p ρ + gz + 1 2 u2 = const (Andrews & Jelley 2007)
- 22. KV No losses between stations ① and ② z1 + p1 ρg + u1 2 2 = z2 + p2 ρg + u2 2 2 (1) ① ② z1 z2 p1 p2 u1 u2
- 23. KV No losses between stations ① and ② Losses between stations ① and ② z1 + p1 ρg + u1 2 2 = z2 + p2 ρg + u2 2 2 (1) z1 + p1 ρg + u1 2 2 = z2 + p2 ρg + u2 2 2 + h (2) ① ② z1 z2 p1 p2 u1 u2
- 24. KV No losses between stations ① and ② Losses between stations ① and ② Energy additions or extractions z1 + p1 ρg + u1 2 2 = z2 + p2 ρg + u2 2 2 (1) z1 + p1 ρg + u1 2 2 = z2 + p2 ρg + u2 2 2 + h (2) z1 + p1 ρg + u1 2 2 + win g = z2 + p2 ρg + u2 2 2 + wout g + h (3) ① ② z1 z2 p1 p2 u1 u2
- 25. KV (a) The atmospheric pressure on a surface is 105 Nm-2. Assuming the water is stationary, what is the pressure at a depth of 10 m ? (assume ρwater = 103 kgm-3 & g = 9.81 ms-2) (b) What is the significance of Bernoulli’s equation? (c) Assuming the pressure of stationary air is 105 Nm-2, calculate the percentage change due to a wind of 20ms-1. (assume ρair ≈ 1.2 kgm-3) EXAMPLE 2
- 26. KV A Pitot tube is a device for measuring the velocity in a fluid. Essentially, it consists of two tubes, (a) and (b). Each tube has one end open to the fluid and one end connected to a pressure gauge.Tube (a) has the open end facing the flow and tube (b) has the open end normal to the flow. In the case when the gauge (a) reads p = po+½pU2 and gauge (b) reads p = po, derive an expression for the velocity of the fluid in terms of the difference in pressure between the gauges. EXAMPLE 3
- 27. KV p ρ + gz + 1 2 u2 = const Bernoulli’s equation (Andrews & Jelley 2007)
- 28. KV Applying bernoulli’s equation p ρ + gz + 1 2 u2 = const Bernoulli’s equation (Andrews & Jelley 2007)
- 29. KV p0 ρ + 1 2 U2 = ps ρ Applying bernoulli’s equation p ρ + gz + 1 2 u2 = const Bernoulli’s equation (Andrews & Jelley 2007)
- 30. KV p0 ρ + 1 2 U2 = ps ρ Applying bernoulli’s equation p ρ + gz + 1 2 u2 = const Bernoulli’s equation Rearranging (Andrews & Jelley 2007)
- 31. KV p0 ρ + 1 2 U2 = ps ρ U = 2 ps − p0( ) ρ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 Applying bernoulli’s equation p ρ + gz + 1 2 u2 = const Bernoulli’s equation Rearranging (Andrews & Jelley 2007)
- 32. KV The velocity in a flow stream varies considerably over the cross-section.A pitot tube only measures at one particular point, therefore, in order to determine the velocity distribution over the entire cross section a number of readings are required. VELOCITY DISTRIBUTION & FLOW RATE (Bacon, D., Stephens, R. 1990)
- 33. KV The flow rate can be determine; !V = a1 u1 + a2 u2 + a3 u3 + ... = au∑ If the total cross-sectional area if the section is A; the mean velocity, U = !V A = au∑ A The velocity profile for a circular pipe is the same across any diameter Δ!V = ur × Δa where ur is the velocity at radius r !V = ur∑ × Δa and V = ur∑ × Δa πr2 The velocity at the boundary of any duct or pipe is zero.
- 34. KV In aVenturi meter an ideal fluid flows with a volume flow rate and a pressure p1 through a horizontal pipe of cross sectional area A1.A constriction of cross-sectional area A2 is inserted in the pipe and the pressure is p2 inside the constriction. Derive an expression for the volume flow rate in terms of p1, p2, A1 and A2. EXAMPLE 4
- 35. KV p1 ρ + 1 2 u1 2 = p2 ρ + 1 2 u2 2 From Bernoulli’s equation (Andrews & Jelley 2007)
- 36. KV p1 ρ + 1 2 u1 2 = p2 ρ + 1 2 u2 2 From Bernoulli’s equation Also by mass continuity u2 = u1 A1 A2 (Andrews & Jelley 2007)
- 37. KV p1 ρ + 1 2 u1 2 = p2 ρ + 1 2 u2 2 From Bernoulli’s equation Also by mass continuity u2 = u1 A1 A2 Eliminating u2 we obtain the volume flow rate as !V = A1 u1 = A1 A2 A1 2 − A2 2 ( ) − 1 2⎛ ⎝⎜ ⎞ ⎠⎟ 2 p1 − p2( ) ρ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 (Andrews & Jelley 2007)
- 38. KV p1 ρg + u1 2 2g = p2 ρg + u2 2 2g Applying Bernoulli’s equation to stations ① and ② ORIFICE PLATE (Bacon, D., Stephens, R. 1990)
- 39. KV p1 ρg + u1 2 2g = p2 ρg + u2 2 2g Applying Bernoulli’s equation to stations ① and ② Mass continuity A1 u1 = A2 u2 ⇒ u2 = u1 A1 A2 ORIFICE PLATE (Bacon, D., Stephens, R. 1990)
- 40. KV p1 ρg + u1 2 2g = p2 ρg + u2 2 2g Applying Bernoulli’s equation to stations ① and ② Mass continuity A1 u1 = A2 u2 ⇒ u2 = u1 A1 A2 ∴ !V = A1 u1 = A1 A2 A1 2 − A2 2 ( ) − 1 2⎛ ⎝⎜ ⎞ ⎠⎟ 2 p1 − p2( ) ρ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 = k h ORIFICE PLATE (Bacon, D., Stephens, R. 1990)
- 41. KV In an engine test 0.04 kg/s of air flows through a 50 mm diameter pipe, into which is fitted a 40 mm diameter Orifice plate.The density of air is 1.2 kg/m3 and the coefficient of discharge for the orifice is 0.63.The pressure drop across the orifice plate is measured by a U-tube manometer. Calculate the manometer reading. EXAMPLE 5
- 42. KV Mass flow rate - !m = ρ !V ⇒∴ !V=0.033m3 s (Bacon, D., Stephens, R. 1990)
- 43. KV Mass flow rate - !m = ρ !V ⇒∴ !V=0.033m3 s !V = k h !V = π × 0.052 4 ⎛ ⎝⎜ ⎞ ⎠⎟ π × 0.042 4 ⎛ ⎝⎜ ⎞ ⎠⎟ π × 0.042 4 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 − π × 0.042 4 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 1 2 2 × 9.81× h !V = 0.007244 h (Bacon, D., Stephens, R. 1990)
- 44. KV Mass flow rate - The actual discharge is !m = ρ !V ⇒∴ !V=0.033m3 s !V = k h !V = π × 0.052 4 ⎛ ⎝⎜ ⎞ ⎠⎟ π × 0.042 4 ⎛ ⎝⎜ ⎞ ⎠⎟ π × 0.042 4 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 − π × 0.042 4 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 1 2 2 × 9.81× h !V = 0.007244 h 0.63× 0.007244 h = 0.0333 ∴h = 53.24m of air (Bacon, D., Stephens, R. 1990)
- 45. KV Mass flow rate - The actual discharge is !m = ρ !V ⇒∴ !V=0.033m3 s !V = k h !V = π × 0.052 4 ⎛ ⎝⎜ ⎞ ⎠⎟ π × 0.042 4 ⎛ ⎝⎜ ⎞ ⎠⎟ π × 0.042 4 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 − π × 0.042 4 ⎛ ⎝⎜ ⎞ ⎠⎟ 2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 1 2 2 × 9.81× h !V = 0.007244 h 0.63× 0.007244 h = 0.0333 ∴h = 53.24m of air Equating pressures at Level XX in the U-tube 1.2g × 53.34 =103 g × x ∴ x = 64 ×103 m (Bacon, D., Stephens, R. 1990)
- 46. KV The motion of a viscous fluid is more complicated than that of an inviscid fluid. DYNAMICS OF A VISCOUS FLUID (Andrews & Jelley 2007)
- 47. KV The motion of a viscous fluid is more complicated than that of an inviscid fluid. DYNAMICS OF A VISCOUS FLUID u y( )= U y d (0 ≤ y ≤ d) (Andrews & Jelley 2007)
- 48. KV The motion of a viscous fluid is more complicated than that of an inviscid fluid. DYNAMICS OF A VISCOUS FLUID u y( )= U y d (0 ≤ y ≤ d) The viscous shear force per unit area in the fluid is proportional to the velocity gradient (Andrews & Jelley 2007)
- 49. KV The motion of a viscous fluid is more complicated than that of an inviscid fluid. DYNAMICS OF A VISCOUS FLUID u y( )= U y d (0 ≤ y ≤ d) The viscous shear force per unit area in the fluid is proportional to the velocity gradient F A = −µ du dy = µ u d μ - coefficient of dynamic viscosity (Andrews & Jelley 2007)
- 50. KV Laminar flow in a pipe Turbulent flow in a pipe Typical flow regimes (Andrews & Jelley 2007)
- 51. KV Laminar flow in a pipe Turbulent flow in a pipe Typical flow regimes Reynolds Number, Re = ρUL µ = UL v (Andrews & Jelley 2007)
- 52. KV Flow around a cylinder for an inviscid fluid Flow around a cylinder for a viscous fluid (Andrews & Jelley 2007)
- 53. KV Lift and Drag (Andrews & Jelley 2007)
- 54. KV Lift and Drag Lift = 1 2 CL ρU2 A (Andrews & Jelley 2007)
- 55. KV Lift and Drag Lift = 1 2 CL ρU2 A Drag = 1 2 CD ρU2 A (Andrews & Jelley 2007)
- 56. KV Variation of the lift and drag coefficients CL and CD with angle of attack, (Andrews & Jelley 2007)
- 57. KV Bulk properties of fluids Streamlines & stream-tubes Mass continuity Energy conservation in an ideal fluid Bernoulli’s equation for steady flow ! Applied example’s (pitot tube,Venturi meter & questions Dynamics of a viscous fluid ! Flow regimes ! Laminar and turbulent flow ! Vortices ! Lift & Drag
- 58. KV Andrews, J., Jelley, N. 2007 Energy science: principles, technologies and impacts, Oxford University Press Bacon, D., Stephens, R. 1990 MechanicalTechnology, second edition, Butterworth Heinemann Boyle, G. 2004 Renewable Energy: Power for a sustainable future, second edition, Oxford University Press Çengel,Y.,Turner, R., Cimbala, J. 2008 Fundamentals of thermal fluid sciences, Third edition, McGraw Hill Turns, S. 2006 Thermal fluid sciences:An integrated approach, Cambridge University Press Twidell, J. and Weir,T. 2006 Renewable energy resources, second edition, Oxon: Taylor and Francis Young, D., Munson, B., Okiishi,T., Huebsch,W., 2011 Introduction to Fluid Mechanics, Fifth edition, John Wiley & Sons, Inc. Illustrations taken from Energy science: principles, technologies and impacts & Mechanical Technology
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- There is essential text for this module. The module content, as in the theory that will be addressed is in a fluidic state and therefore being development on week by week basis. No one text book covers the essential elements.\n
- There is essential text for this module. The module content, as in the theory that will be addressed is in a fluidic state and therefore being development on week by week basis. No one text book covers the essential elements.\n
- There is essential text for this module. The module content, as in the theory that will be addressed is in a fluidic state and therefore being development on week by week basis. No one text book covers the essential elements.\n
- There is essential text for this module. The module content, as in the theory that will be addressed is in a fluidic state and therefore being development on week by week basis. No one text book covers the essential elements.\n
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- There is essential text for this module. The module content, as in the theory that will be addressed is in a fluidic state and therefore being development on week by week basis. No one text book covers the essential elements.\n
- There is essential text for this module. The module content, as in the theory that will be addressed is in a fluidic state and therefore being development on week by week basis. No one text book covers the essential elements.\n
- There is essential text for this module. The module content, as in the theory that will be addressed is in a fluidic state and therefore being development on week by week basis. No one text book covers the essential elements.\n
- There is essential text for this module. The module content, as in the theory that will be addressed is in a fluidic state and therefore being development on week by week basis. No one text book covers the essential elements.\n
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- To appreciate energy conversion such as hydro, wave, tidal and wind power a detailed knowledge of fluid mechanics is essential.\n\nDuring the course of this lecture, a brief summary of the basic physical properties of fluids is provided and the conservation laws of mass and energy for an ideal (or inviscid) fluid are derived.\n\nThe application of the conservation laws to situations of practical interest are also explored to illustrate how useful information about the flow can be derived. \n\nFinally, the effect of viscosity on the motion of a fluid around an immersed body (such as a turbine blade) and how the flow determines the forces acting on the body of interest. \n
- Density - Mass per unit volume of a fluid. For the purposes of this module, Density is assumed to be constant i.e. that the flow of the fluid is incompressible (small variations in pressure arising from fluid motion in comparison to atmospheric pressure).\n\nPressure - Force per unit area, and acts in the normal direction to the surface of a body in a fluid. Its units are the pascal or N-m^2\n\nViscosity - Force per unit area due to internal friction in a fluid arising from the relative motion between neighbouring elements in a fluid. Viscous forces act in the tangential direction to the surface of a body immersed in a flow. (consider a deck of cards.\n
- A useful concept to visualize a velocity field is to imagine a set of streamlines para;;e; to the direction of motion at all points in the fluid. Any element of mass in the fluid flows along a notational stream-tube bounded by neighbouring streamlines. In practice, streamlines can be visualised by injecting small particles into the fluid e.g. smoke can be used in wind tunnels to visualise fluid flow around objects.\n
- Also known as conservation of mass.\n\nConsider flow along a stream tube in a steady velocity field. The direction of flow is parallel to the boundries and at any point within the stream tube, the speed of the fluid (u) and the cross-sectional area (A). Therefore the fluid is confined to the stream tube and the mass flow per second is constant. Therefore: \n\nDensity * Velocity * Cross-sectional Area = Constant\n\nTherefore the speed of fluid is inversely proportional to the cross-sectional area of the stream tube.\n
- Also known as conservation of mass.\n\nConsider flow along a stream tube in a steady velocity field. The direction of flow is parallel to the boundries and at any point within the stream tube, the speed of the fluid (u) and the cross-sectional area (A). Therefore the fluid is confined to the stream tube and the mass flow per second is constant. Therefore: \n\nDensity * Velocity * Cross-sectional Area = Constant\n\nTherefore the speed of fluid is inversely proportional to the cross-sectional area of the stream tube.\n
- Also known as conservation of mass.\n\nConsider flow along a stream tube in a steady velocity field. The direction of flow is parallel to the boundries and at any point within the stream tube, the speed of the fluid (u) and the cross-sectional area (A). Therefore the fluid is confined to the stream tube and the mass flow per second is constant. Therefore: \n\nDensity * Velocity * Cross-sectional Area = Constant\n\nTherefore the speed of fluid is inversely proportional to the cross-sectional area of the stream tube.\n
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- Consider a liquid at a pressure p, moving with a velocity (u) at a height Z above datum level\n\nIf a tube were inserted in the top of the pipe, the liquid would rise up the tube a distance of p/ρg and this is equivalent to an additional height of liquid relative to datum level.\n\nEtotal equation represents the specific energy of the liquid. Each of the quantities in the brackets have the units of length and are termed heads, i.e Z is referred to as the potential head of the liquid, p/ρg as the pressure head and V2/2g as the velocity head. \n\n
- For a stationary fluid, u = 0 everywhere in the fluid therefore this equation reduces to \nwhich is the equation for hydrostatic pressure. It shows that the fluid at a given depth is all the same at the same pressure. \n
- Consider the steady flow of an ideal fluid in the control volume shown. \nThe height (z), cross sectional area (A), speed (u) and pressure are denoted.\nThe increase in gravitational potential energy of a mass δm of fluid between z1 and z2 is δmg(z2-z1)\nIn a short time interval (δt) the mass of fluid entering the control volume at P1 is δm=ρu1A1δt and the mass exiting at P2 is δm=ρu2A2δt\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- In order for the fluid to enter the control volume it has to do work to overcome the pressure p1 exerted by the fluid. The work done in pushing the elemental mass δm a small distance δs = uδt \nat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity * Change in time. Remember F = P*A\n\nSimilarly, the work done in pushing the elemental mass out of the control volume at P2 is δW2=-p2A2u2δt.\n\nThe net work done is δW1+δW2\n
- Consider stations ① and ② of the inclined pipe illustrated. If there are no losses between these two sections, and no energy changes resulting from heat transfer of work, then the total energy will remain constant, therefore stations ① and ② can be set equal to one another, equ (1)\n\nIf however losses do occur between the two stations, then equation equ (2)\n\nWhen energy is added to the fluid by a device such as a pump, or when energy is extracted by a turbine, these need also be accounted for equ (3). w is the specific work in J/kg\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- Consider stations ① and ② of the inclined pipe illustrated. If there are no losses between these two sections, and no energy changes resulting from heat transfer of work, then the total energy will remain constant, therefore stations ① and ② can be set equal to one another, equ (1)\n\nIf however losses do occur between the two stations, then equation equ (2)\n\nWhen energy is added to the fluid by a device such as a pump, or when energy is extracted by a turbine, these need also be accounted for equ (3). w is the specific work in J/kg\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- Consider stations ① and ② of the inclined pipe illustrated. If there are no losses between these two sections, and no energy changes resulting from heat transfer of work, then the total energy will remain constant, therefore stations ① and ② can be set equal to one another, equ (1)\n\nIf however losses do occur between the two stations, then equation equ (2)\n\nWhen energy is added to the fluid by a device such as a pump, or when energy is extracted by a turbine, these need also be accounted for equ (3). w is the specific work in J/kg\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- Consider stations ① and ② of the inclined pipe illustrated. If there are no losses between these two sections, and no energy changes resulting from heat transfer of work, then the total energy will remain constant, therefore stations ① and ② can be set equal to one another, equ (1)\n\nIf however losses do occur between the two stations, then equation equ (2)\n\nWhen energy is added to the fluid by a device such as a pump, or when energy is extracted by a turbine, these need also be accounted for equ (3). w is the specific work in J/kg\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
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- Consider the fluid flowing along the stream-line AB. In case (a), the fluid slows down as it approaches the stagnation point B. Putting u = U, p = po at A and u = 0, p = ps at B and applying bernoulli’s equation we get\n\nIn example (and the previous question), ps is measured by tube (a) and p0 measured by tube (b). \nNOTE - ps is larger than p0 by an amount ps - p0 = ρU2. The quantities ρU2, p0 and ps = p0 + ρU2 are called the dynamic pressure, static pressure and total pressure respectively. \n
- Consider the fluid flowing along the stream-line AB. In case (a), the fluid slows down as it approaches the stagnation point B. Putting u = U, p = po at A and u = 0, p = ps at B and applying bernoulli’s equation we get\n\nIn example (and the previous question), ps is measured by tube (a) and p0 measured by tube (b). \nNOTE - ps is larger than p0 by an amount ps - p0 = ρU2. The quantities ρU2, p0 and ps = p0 + ρU2 are called the dynamic pressure, static pressure and total pressure respectively. \n
- Consider the fluid flowing along the stream-line AB. In case (a), the fluid slows down as it approaches the stagnation point B. Putting u = U, p = po at A and u = 0, p = ps at B and applying bernoulli’s equation we get\n\nIn example (and the previous question), ps is measured by tube (a) and p0 measured by tube (b). \nNOTE - ps is larger than p0 by an amount ps - p0 = ρU2. The quantities ρU2, p0 and ps = p0 + ρU2 are called the dynamic pressure, static pressure and total pressure respectively. \n
- Consider the fluid flowing along the stream-line AB. In case (a), the fluid slows down as it approaches the stagnation point B. Putting u = U, p = po at A and u = 0, p = ps at B and applying bernoulli’s equation we get\n\nIn example (and the previous question), ps is measured by tube (a) and p0 measured by tube (b). \nNOTE - ps is larger than p0 by an amount ps - p0 = ρU2. The quantities ρU2, p0 and ps = p0 + ρU2 are called the dynamic pressure, static pressure and total pressure respectively. \n
- The velocity profile is obtained by traversing the pitot tube along the line AA. The cross-section is divided into convenient areas, a1, a2, a3, etc... and the velocity at the centre of each area is determined\n
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- A venturi meter is a device which is used to measure the rate of flow through a pipe.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- A venturi meter is a device which is used to measure the rate of flow through a pipe.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- A venturi meter is a device which is used to measure the rate of flow through a pipe.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- A venturi meter is a device which is used to measure the rate of flow through a pipe.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- A venturi meter is a device which is used to measure the rate of flow through a pipe.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- A venturi meter is a device which is used to measure the rate of flow through a pipe.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- A venturi meter is a device which is used to measure the rate of flow through a pipe.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
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- An Orifice plate is another means of determining the fluid flow in a pipe and works on the same principle of the Venturi meter.\n\nThe position of the Vena contract can be difficult establish and therefore the area A2. The constant k is determined experimentally when it will incorporate the coefficient of discharge.\n\nIf h is small so that ρ is approximately constant then the this equation can be used for compressible fluid flow.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- An Orifice plate is another means of determining the fluid flow in a pipe and works on the same principle of the Venturi meter.\n\nThe position of the Vena contract can be difficult establish and therefore the area A2. The constant k is determined experimentally when it will incorporate the coefficient of discharge.\n\nIf h is small so that ρ is approximately constant then the this equation can be used for compressible fluid flow.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
- An Orifice plate is another means of determining the fluid flow in a pipe and works on the same principle of the Venturi meter.\n\nThe position of the Vena contract can be difficult establish and therefore the area A2. The constant k is determined experimentally when it will incorporate the coefficient of discharge.\n\nIf h is small so that ρ is approximately constant then the this equation can be used for compressible fluid flow.\n\nNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation\n
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- Let us consider a simple case of laminar flow between two parallel plates separated by a small distance d. \n\nThe upper plate moves at a constant velocity U while the lower plate remains at rest. At the plate fluid interface in both cases there is no velocity due to the strong forces of attraction. Therefore the velocity profile in the fluid is given by\n\n\n
- Let us consider a simple case of laminar flow between two parallel plates separated by a small distance d. \n\nThe upper plate moves at a constant velocity U while the lower plate remains at rest. At the plate fluid interface in both cases there is no velocity due to the strong forces of attraction. Therefore the velocity profile in the fluid is given by\n\n\n
- Let us consider a simple case of laminar flow between two parallel plates separated by a small distance d. \n\nThe upper plate moves at a constant velocity U while the lower plate remains at rest. At the plate fluid interface in both cases there is no velocity due to the strong forces of attraction. Therefore the velocity profile in the fluid is given by\n\n\n
- Let us consider a simple case of laminar flow between two parallel plates separated by a small distance d. \n\nThe upper plate moves at a constant velocity U while the lower plate remains at rest. At the plate fluid interface in both cases there is no velocity due to the strong forces of attraction. Therefore the velocity profile in the fluid is given by\n\n\n
- A viscous fluid can exhibit two different kinds of flow regimes, Laminar and Turbulent\n\nIn laminar flow, the fluid slides along distinct stream-tubes and tends to be quite stable, Turbulent flow is disorderly and unstable\n\nThe flow that exists in an y given situation depends on the ratio of the inertial force to the viscous force. The magnitude of this ratio is given by Reynolds number, where U is the velocity, L is the length and v = μ/ρ is the kinematic viscosity. Reynolds observed that flows at small Re where laminar while flow at high Re contained regions of turbulence\n
- For the inviscid flow the velocity fields in the upstream and downstream regions are symmetrical, therefore the corresponding pressure distribution is symmetrical. It follows that the net force exerted by the fluid on the cylinder is zero. This is an example of d’Alembert’s paradox\n\nFor a body immersed in a viscious fluid, the component of velocity tangential to the surface of the fluid is zero at all points. At large Re numbers (Re>>1), the viscous force is negligible in the bulk of the fluid but us very significant in a viscous boundary layer close to the surface of the body. Rotational components of flow know as vorticity are generated within the boundary layer. As a certain point, the seperation point, the boundary layer becomes detached from the surface and vorticity are discharged into the body of the fluid. The vorticity are transported downstream side of the cylinder in the wake. Therefore the pressure distributions on the upstream and downstream sides of the cylinder are not symmetrical in the case of a viscous fluid. As a result, the cylinder experiences a net force in the direction of motion known as drag. In the case of a spinning cylinder, a force called life arises at right angles to the direction of flow. \n
- CL and CD are dimensionless constants know as the coefficient of Lift and the coefficient of Drag respectively.\n\nLift and drag can be changed by altering the shape of a wing i.e. ailerons. For small angles of attack, the pressure distribution ont he upper surface of an aerofoil is significantly lower than that on the lower surface which results is in a net lift force. \n
- CL and CD are dimensionless constants know as the coefficient of Lift and the coefficient of Drag respectively.\n\nLift and drag can be changed by altering the shape of a wing i.e. ailerons. For small angles of attack, the pressure distribution ont he upper surface of an aerofoil is significantly lower than that on the lower surface which results is in a net lift force. \n
- CL and CD are dimensionless constants know as the coefficient of Lift and the coefficient of Drag respectively.\n\nLift and drag can be changed by altering the shape of a wing i.e. ailerons. For small angles of attack, the pressure distribution on the upper surface of an aerofoil is significantly lower than that on the lower surface which results is in a net lift force. \n
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- The significance of Euler’s turbine equation is that the details of the flow inside the turbine are irrelevant. All that matters is the total change in the angular momentum of the fluid between the inlet and the outlet. The maximum torque is achieved when the fluid flows out in the radial direction, i.e. when cosβ2 = 0\n\n
- The significance of Euler’s turbine equation is that the details of the flow inside the turbine are irrelevant. All that matters is the total change in the angular momentum of the fluid between the inlet and the outlet. The maximum torque is achieved when the fluid flows out in the radial direction, i.e. when cosβ2 = 0\n\n
- The significance of Euler’s turbine equation is that the details of the flow inside the turbine are irrelevant. All that matters is the total change in the angular momentum of the fluid between the inlet and the outlet. The maximum torque is achieved when the fluid flows out in the radial direction, i.e. when cosβ2 = 0\n\n
- The significance of Euler’s turbine equation is that the details of the flow inside the turbine are irrelevant. All that matters is the total change in the angular momentum of the fluid between the inlet and the outlet. The maximum torque is achieved when the fluid flows out in the radial direction, i.e. when cosβ2 = 0\n\n
- The significance of Euler’s turbine equation is that the details of the flow inside the turbine are irrelevant. All that matters is the total change in the angular momentum of the fluid between the inlet and the outlet. The maximum torque is achieved when the fluid flows out in the radial direction, i.e. when cosβ2 = 0\n\n
- The significance of Euler’s turbine equation is that the details of the flow inside the turbine are irrelevant. All that matters is the total change in the angular momentum of the fluid between the inlet and the outlet. The maximum torque is achieved when the fluid flows out in the radial direction, i.e. when cosβ2 = 0\n\n
- The significance of Euler’s turbine equation is that the details of the flow inside the turbine are irrelevant. All that matters is the total change in the angular momentum of the fluid between the inlet and the outlet. The maximum torque is achieved when the fluid flows out in the radial direction, i.e. when cosβ2 = 0\n\n
- The significance of Euler’s turbine equation is that the details of the flow inside the turbine are irrelevant. All that matters is the total change in the angular momentum of the fluid between the inlet and the outlet. The maximum torque is achieved when the fluid flows out in the radial direction, i.e. when cosβ2 = 0\n\n
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