Fall from rest

Time for a solid spherical particle to reach 99 per cent of its terminal velocity when falling from rest in the Stokes regime... 

Consider a spherical particle of diameter dp and density pp falling from rest in a stationary fluid of density p and dynamic viscosity p.. The particle will accelerate until it reaches its terminal velocity a,. At any time t, let a be the particle s velocity. Recalling that the drag force acting on a sphere in the Stokes regime is of magnitude iirdppu, application of Newton s second law of motion can be written as... 

Show that the time to reach 50 per cent of the terminal velocity for a spherical particle falling from rest in laminar flow in a fluid is... 

Although a number of workers [e.g. (C5, Dl, D2)] have considered flow around particles started impulsively from rest at constant nonzero velocities, this situation is of little practical interest. Attention is concentrated on free fall from rest and oscillatory motion. 

Fig. 11.7 Variation of Re with dimensionless time for sphere falling from rest y — 1.22, Re,s = 3371, Rej = 364. Data are from Moorman (M10), run 33. (1) Eq. (11-33) and from Eqs. (11-31) and (11-32) (2) Eq. (11-33) A = Ah = 1 (3) Creeping how solution. Table 11.1 (4) Steady drag only (A = Aj, = 0) (5) Steady drag with A = 1. Bottom part of figure, giving drag components and coefficients, corresponds to curve 1.

Fig. 11.12 Fluid streamlines relative to sphere falling from rest showing development of wake, after (L5), y = 1.72, Re, = 145. Conditions as above.

As time progresses, the particle will attain a constant velocity given by xg. If a particle is initially given a velocity greater than this, it will decelerate until it has reached xg. If the particle s initial velocity is less, it will increase to a value of xg. If a particle falls from rest, it will accelerate until xg is attained. Thus xg represents the terminal settling velocity of the particle vt... 

Examples.—(1) In calculations involving mean values care must be taken not to take the wrong independent variable. Find the mean velocity of a particle falling from rest with a constant acceleration, the velocities being taken at equal distances of time. When a body falls from rest, V = gt,... 

Examples.—(1) A body falls from rest. Show that it travels 400 ft. in 6 sec. Hint. Use g = 32. 

Falling bodies n. For bodies falling from rest conditions are as for uniformly accelerated motion except that Vq = Q and g is the acceleration due to gravity. The formulate becomes - air resistance neglected,... 

A particle falling from rest in a fluid will initially experience a high acceleration as the shear stress drag, which increases with relative velocity, will be small. As the particle accelerates the drag force increases, causing the acceleration to reduce. Eventually a force balance is achieved when the acceleration is zero and a maximum or terminal relative velocity is reached. This is known as the single particle terminal velocity. 

Small particles in gases and all common particles in liquids quickly accelerate to their terminal velocity. As an example, a 100 fim particle falling from rest in water requires 1.5 ms to reach its terminal velocity of 2mm/s. Table 2.2 gives... 

Table 2.2 Sand particles falling from rest in air (particle density, 2600 kg/m )...

Assumption i If a body falls from rest, its velocity at any point is proportional to the distance already fallen. 

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