Chapter 15
... pressure are created on opposite sides of the foil, the result is a lift force directed perpendicular to the foil from the high pressure zone toward the low pressure zone. ...
... pressure are created on opposite sides of the foil, the result is a lift force directed perpendicular to the foil from the high pressure zone toward the low pressure zone. ...
buoyant force
... • Stokes’ Law will not work if the object is not spherical • Assume the resistive force has a magnitude given by Fr = k v – k is a coefficient to be determined experimentally ...
... • Stokes’ Law will not work if the object is not spherical • Assume the resistive force has a magnitude given by Fr = k v – k is a coefficient to be determined experimentally ...
Chapter 3
... Chapter 3 The linear momentum of a mass (m) with a velocity v vxi v y j vzk is defined to be mv . Important: linear momentum is a vector. ...
... Chapter 3 The linear momentum of a mass (m) with a velocity v vxi v y j vzk is defined to be mv . Important: linear momentum is a vector. ...
Lecture 13
... The flow is laminar (steady, streamline flow) – all particles passing through a point have the same velocity at any time. The fluid is incompressible – the density remains constant The flow is irrotational – the fluid has no angular momentum about any point. A small paddle wheel placed anywhere does ...
... The flow is laminar (steady, streamline flow) – all particles passing through a point have the same velocity at any time. The fluid is incompressible – the density remains constant The flow is irrotational – the fluid has no angular momentum about any point. A small paddle wheel placed anywhere does ...
chapter14
... Rearranging and expressing in terms of density: P1 + ½ rv12 + ρgy1 = P2 + ½ rv22 + ρgy2 This is Bernoulli’s Equation as applied to an ideal fluid and is often expressed as P + ½ r v2 + r g y = constant When the fluid is at rest, this becomes P1 – P2 = r g h which is consistent with the pressure vari ...
... Rearranging and expressing in terms of density: P1 + ½ rv12 + ρgy1 = P2 + ½ rv22 + ρgy2 This is Bernoulli’s Equation as applied to an ideal fluid and is often expressed as P + ½ r v2 + r g y = constant When the fluid is at rest, this becomes P1 – P2 = r g h which is consistent with the pressure vari ...
Rooney AP Physics - Ch 9 Solids and Fluids
... – Otherwise, the fluid would not be in equilibrium – The fluid would flow from the higher pressure region to the lower pressure region – This does not happen, so our assumption must be true Section 9.4 ...
... – Otherwise, the fluid would not be in equilibrium – The fluid would flow from the higher pressure region to the lower pressure region – This does not happen, so our assumption must be true Section 9.4 ...
AT Physics II. Air Resistance The motion of
... where L is a characteristic length for the object moving through a fluid (say the radius or the diameter of a sphere), v its speed, ρ the density of the liquid and η its viscosity. Generally, high Reynolds number (anything much bigger than 1) means that viscosity is negligible; low Reynolds number ( ...
... where L is a characteristic length for the object moving through a fluid (say the radius or the diameter of a sphere), v its speed, ρ the density of the liquid and η its viscosity. Generally, high Reynolds number (anything much bigger than 1) means that viscosity is negligible; low Reynolds number ( ...
Physics Annotated Formula Sheet
... r = distance between centers in m Force of gravity is centripetal v = perimeter velocity in m/s GMm/r2 = mv2/r cm = center of mass in m Center of mass cm = m1r1 + m2r2 + ... m = mass in kg (m1 + m2 + ...) r = distance from 0 position in m Non-accelerating force problems where forces act through cm. ...
... r = distance between centers in m Force of gravity is centripetal v = perimeter velocity in m/s GMm/r2 = mv2/r cm = center of mass in m Center of mass cm = m1r1 + m2r2 + ... m = mass in kg (m1 + m2 + ...) r = distance from 0 position in m Non-accelerating force problems where forces act through cm. ...
pdf file - Wayne State University Physics and Astronomy
... Example: Depressing the brake pedal in a car pushes on a piston with crosssectional area 3.0 cm2. The piston applies pressure to the brake fluid, which is connected to two pistons, each with area 12.0 cm2. Each of these pistons presses a brake pad against one side of a rotor attached to one of the r ...
... Example: Depressing the brake pedal in a car pushes on a piston with crosssectional area 3.0 cm2. The piston applies pressure to the brake fluid, which is connected to two pistons, each with area 12.0 cm2. Each of these pistons presses a brake pad against one side of a rotor attached to one of the r ...
P 2
... • If the weight of the object is less than the weight of the fluid displaced, the object will float, and the magnitude of the buoyant force is equal to the magnitude of its weight (object < fluid). • If the weight of the object is the same as the weight of the fluid displaced, then the object will ...
... • If the weight of the object is less than the weight of the fluid displaced, the object will float, and the magnitude of the buoyant force is equal to the magnitude of its weight (object < fluid). • If the weight of the object is the same as the weight of the fluid displaced, then the object will ...
Lecture 28
... Two beakers are filled to the same level with water. One of them has a plastic ball floating in the water. If the beakers are placed on a scale, one at a time, which weighs more? A) The beaker without the plastic ball weighs more. B) The beaker with the plastic ball weighs more. C) The beakers have ...
... Two beakers are filled to the same level with water. One of them has a plastic ball floating in the water. If the beakers are placed on a scale, one at a time, which weighs more? A) The beaker without the plastic ball weighs more. B) The beaker with the plastic ball weighs more. C) The beakers have ...
3.0 bouyancy, archimedes` principles , surface tension
... When an object is submerged in a fluid, it appears to weigh less than they do when outside the fluid. For example, a large rock at the bottom of a stream would be easily lifted compare to lifting it from the ground. As the rock breaks through the surface o f the water, it becomes heavier. This pheno ...
... When an object is submerged in a fluid, it appears to weigh less than they do when outside the fluid. For example, a large rock at the bottom of a stream would be easily lifted compare to lifting it from the ground. As the rock breaks through the surface o f the water, it becomes heavier. This pheno ...
Phy_103_-3
... Considering the motion of real fluids, it is very complex and not fully understood. When fluid is in motion, its flow can be characterized as being one of 2 main types. Steady or laminar flow. The flow is steady if the overall flow pattern does not change with time. In this type of flow every elemen ...
... Considering the motion of real fluids, it is very complex and not fully understood. When fluid is in motion, its flow can be characterized as being one of 2 main types. Steady or laminar flow. The flow is steady if the overall flow pattern does not change with time. In this type of flow every elemen ...
Gauge pressure as a function of depth d below the surface of a fluid
... would show that the streamlines on the left and upper right would be unchanged, while the smoke pattern on the lower right would be quite different. ...
... would show that the streamlines on the left and upper right would be unchanged, while the smoke pattern on the lower right would be quite different. ...
Structure of the interstellar atomic gas Structure of molecular clouds?
... Thus the quantity Q is modified by exchanging flux at the surface of the fluid elements. ...
... Thus the quantity Q is modified by exchanging flux at the surface of the fluid elements. ...
AP Physics 2
... In the case of a car's brake pads, you have a small initial force applied by you on the brake pedal. This transfers via a brake line, which had a small cylindrical area. The brake fluid then enters a chamber with more AREA allowing a LARGE FORCE to be applied on the brake shoes, which in turn slow t ...
... In the case of a car's brake pads, you have a small initial force applied by you on the brake pedal. This transfers via a brake line, which had a small cylindrical area. The brake fluid then enters a chamber with more AREA allowing a LARGE FORCE to be applied on the brake shoes, which in turn slow t ...
ME33: Fluid Flow Lecture 1: Information and Introduction
... electric, and magnetic forces) and surface forces that act on the control surface (such as pressure and viscous forces, and reaction forces at points of contact). • Body forces act on each volumetric portion dV of the CV. • Surface forces act on each portion dA of the CS. ...
... electric, and magnetic forces) and surface forces that act on the control surface (such as pressure and viscous forces, and reaction forces at points of contact). • Body forces act on each volumetric portion dV of the CV. • Surface forces act on each portion dA of the CS. ...
Fluids
... where h1 is the height of the water in the tank, p1 is the pressure there, and v1 is the speed of the water there; h2 is the altitude of the hole, p2 is the pressure there, and v2 is the speed of the water there. is the density of water. The pressure at the top of the tank and at the hole is atm ...
... where h1 is the height of the water in the tank, p1 is the pressure there, and v1 is the speed of the water there; h2 is the altitude of the hole, p2 is the pressure there, and v2 is the speed of the water there. is the density of water. The pressure at the top of the tank and at the hole is atm ...
Slide 1
... • If the weight of a system is greater than that of the displaced fluid, its density is greater than the fluid’s. • Since weight exceeds the buoyant force, the object will sink. ...
... • If the weight of a system is greater than that of the displaced fluid, its density is greater than the fluid’s. • Since weight exceeds the buoyant force, the object will sink. ...
Topic 9: The Impulse-Momentum Principle To summarize what we
... We first considered fluid statics, in which case a mass balance is of little value – it would simply tell us that the amount of mass in a static system remains constant. However, we did find a force balance useful to understand the relationship between pressure and depth in a liquid. Looking back, w ...
... We first considered fluid statics, in which case a mass balance is of little value – it would simply tell us that the amount of mass in a static system remains constant. However, we did find a force balance useful to understand the relationship between pressure and depth in a liquid. Looking back, w ...
Fluid Motion (ppt)
... • In hydrostatic equilibrium, pressure increases with depth due to gravity • The buoyant force is the weight of the displaced fluid • Fluid flow conserves mass (continuity eq.) and energy (Bernoulli’s equation) • A constriction in flow is accompanied by a velocity and pressure change. • Reread, Revi ...
... • In hydrostatic equilibrium, pressure increases with depth due to gravity • The buoyant force is the weight of the displaced fluid • Fluid flow conserves mass (continuity eq.) and energy (Bernoulli’s equation) • A constriction in flow is accompanied by a velocity and pressure change. • Reread, Revi ...
Powerpoint - UBC Computer Science
... – solves object + fluid motion simultaneously – handles tight coupling (eg. water balloons) – requires conforming (tet) mesh to avoid artifacts ...
... – solves object + fluid motion simultaneously – handles tight coupling (eg. water balloons) – requires conforming (tet) mesh to avoid artifacts ...
Fluid dynamics - Equation of continuity and Bernoulli`s principle.
... The radius of the aorta is ~ 10 mm and the blood flowing through it has a speed ~ 300 mm.s-1. A capillary has a radius ~ 4×10-3 mm but there are literally billions of them. The average speed of blood through the capillaries is ~ 5×10-4 m.s-1. Calculate the effective cross sectional area of the capil ...
... The radius of the aorta is ~ 10 mm and the blood flowing through it has a speed ~ 300 mm.s-1. A capillary has a radius ~ 4×10-3 mm but there are literally billions of them. The average speed of blood through the capillaries is ~ 5×10-4 m.s-1. Calculate the effective cross sectional area of the capil ...
Blade element momentum theory
Blade element momentum theory is a theory that combines both blade element theory and momentum theory. It is used to calculate the local forces on a propeller or wind-turbine blade. Blade element theory is combined with momentum theory to alleviate some of the difficulties in calculating the induced velocities at the rotor.This article emphasizes application of BEM to ground-based wind turbines, but the principles apply as well to propellers. Whereas the streamtube area is reduced by a propeller, it is expanded by a wind turbine. For either application, a highly simplified but useful approximation is the Rankine-Froude ""momentum"" or ""actuator disk"" model (1865,1889). Herein, we study the ""Betz Limit"" on the efficiency of a ground-based wind turbine.A development came in the form of Froude's Blade Element Momentum theory (1878), later refined by Glauert (1926). Betz (1921) provided an approximate correction to momentum ""Rankine-Froude Actuator-Disk"" theory to account for the sudden rotation imparted to the flow by the actuator disk (NACA TN 83, ""The Theory of the Screw Propeller"" and NACA TM 491, ""Propeller Problems""). In Blade Element Momentum theory, angular momentum is included in the model, meaning that the wake (the air after interaction with the rotor) has angular momentum. That is, the air begins to rotate about the z-axis immediately upon interaction with the rotor (see diagram below). Angular momentum must be taken into account since the rotor, which is the device that extracts the energy from the wind, is rotating as a result of the interaction with the wind.The following provides a background section on the Rankine-Froude model, followed by the Blade Element Momentum theory.