Single Particles in a Fluid
2.6 WORKED EXAMPLES WORKED EXAMPLE 2.1
Calculate the upper limit of particle diameter xmax as a function of particle density rp for gravity sedimentation in the Stokes’ law regime. Plot the results as xmax
versus rp over the range 0 rp 8000 kg=m3 for settling in water and in air at ambient conditions. Assume that the particles are spherical and that Stokes’ law holds for Rep 0:3.
Solution
The upper limit of particle diameter in the Stokes’ regime is governed by the upper limit of single particle Reynolds number:
Rep¼rfxmaxU m ¼ 0:3
In gravity sedimentation in the Stokes’ regime particles accelerate rapidly to their terminal velocity. In the Stokes’ regime the terminal velocity is given by Equation (2.13):
UT¼x2ðrp rfÞg 18m Solving these two equations forxmaxwe have
xmax¼ 0:3 18m2 gðrp rfÞrf
" #1=3
¼ 0:82 m2 ðrp rfÞrf
" #1=3
Thus, for air (density 1:2 kg=m3and viscosity 1:84 105Pa s):
xmax¼ 5:37 104 1 ðrp 1:2Þ
" #1=3
And for water (density 1000 kg=m3and viscosity 0.001 Pa s):
xmax¼ 8:19 104 1 ðrp 1000Þ
" #1=3
WORKED EXAMPLES 35
These equations forxmaxas a function are plotted in Figure 2W1.1 for particle densities greater than and less than the fluid densities.
WORKED EXAMPLE 2.2
A gravity separator for the removal from water of oil droplets (assumed to behave as rigid spheres) consists of a rectangular chamber containing inclined baffles as shown schematically in Figure 2W2.1.
(a) Derive an expression for the ideal collection efficiency of this separator as a function of droplet size and properties, separator dimensions, fluid properties and fluid velocity (assumed uniform).
180.0 (a)
(b) 160.0 140.0 120.0 100.0 80.0 60.0 40.0 20.0 0.0
120.0
100.0
80.0
60.0
40.0
20.0
0.0
0 500 1000 1500 2000 2500 3000
0 500
Rising particles
Falling particles
Particle density (kg/m3) Particle density (kg/m3) Particle size, xmax (mm)Particle size, xmax (mm)
1000 1500 2000 2500 3000 3500 4000 4500 5000
Figure 2W1.1 (a) Limiting particle size for Stokes’ law in water. (b) Limiting particle size for Stokes’ law in air
(b) Hence, calculate the per cent change in collection efficiency when the throughput of water is increased by a factor of 1.2 and the density of the oil droplets changes from 750 to 800 kg=m3.
Solution
(a) Referring to Figure 2W2.1, we will assume that all particles rising to the under surface of a baffle will be collected. Therefore, any particle which can rise a distanceh or greater in the time required for it to travel the length of the separator will be collected.
Let the corresponding minimum vertical droplet velocity beUTmin.
Assuming uniform fluid velocity and negligible relative velocity between drops and fluid in the horizontal direction,
drop residence time; t ¼ L U= ThenUTmin¼ hU L=
Assuming that the droplets are small enough for Stokes’ law to apply and that the time and distance for acceleration to terminal velocity is negligible, then droplet velocity will be given by Equation (2.13):
UT¼x2ðrp rfÞg 18m
This is the minimum velocity for drops to be collected whatever their original position between the baffles. Thus
UTmin¼x2ðrp rfÞg
18m ¼hU
L
Assuming that drops of all sizes are uniformly distributed over the vertical height of the separator, then drops rising a distance less than h in the time required for them to travel the length of the separator will be fractionally collected depending on their original vertical position between two baffles. Thus for droplets rising at a velocity of 0:5UTmin
Water +
Figure 2W2.1 Schematic diagram of oil–water separator
WORKED EXAMPLES 37
only 50% will be collected, i.e. only those drops originally in the upper half of the space between adjacent baffles. For drops rising at a velocity of 0:25UTmin only 25% will be collected.
Thus; efficiency of collection; Z ¼actualUTfor droplet UTmin
(b) Comparing collection efficiencies when the throughput of water is increased by a factor of 1.2 and the density of the oil droplets changes from 750 to 850 kg=m3.
Let original and new conditions be denoted by subscripts 1 and 2, respectively.
Increasing throughput of water by a factor of 1.2 means thatU2=U1¼ 1:2.
Therefore from the expression for collection efficiency derived above:
Z2 The decrease in collection efficiency is therefore 50%.
WORKED EXAMPLE 2.3
A sphere of diameter 10 mm and density 7700 kg=m3 falls under gravity at terminal conditions through a liquid of density 900 kg=m3 in a tube of diameter 12 mm. The measured terminal velocity of the particle is 1:6 mm=s. Calculate the viscosity of the fluid. Verify that Stokes’ law applies.
Solution
To solve this problem, we first convert the measured terminal velocity to the equivalent velocity which would be achieved by the sphere in a fluid of infinite extent. Assuming Stokes’ law we can determine the fluid viscosity. Finally we check the validity of Stokes’ law.
Using the Francis wall factor expression [Equation (2.17)]:
UT1
UTD ¼ 1
ð1 x=DÞ2:25¼ 56:34
Thus, terminal velocity for the particle in a fluid of infinite extent, UT1¼ UTD 56:34 ¼ 0:0901 m=s
Equating this value to the expression for UT1 in the Stokes’ regime [Equa-tion (2.13)]:
UT1¼ð10 103Þ2 ð7700 900Þ 9:81 18m
Hence, fluid viscosity, m ¼ 4:11 Pa s.
Checking the validity of Stokes’ law:
Single particle Reynolds number; Rep¼xrfU m ¼ 0:197 Repis less than 0.3 and so the assumption that Stokes’ law holds is valid.
WORKED EXAMPLE 2.4
A mixture of spherical particles of two materials A and B is to be separated using a rising stream of liquid. The size range of both materials is 15–40 mm. (a) Show that a complete separation is not possible using water as the liquid. The particle densities for materials A and B are 7700 and 2400 kg=m3, respectively. (b) Which fluid property must be changed to achieve complete separation? Assume Stokes’ law applies.
Solution
(a) First, consider what happens to a single particle introduced into the centre of a pipe in which a fluid is flowing upwards at a velocityU which is uniform across the pipe cross-section. We will assume that the particle is small enough to consider the time and distance for its acceleration to terminal velocity to be negligible. Referring to Figure 2W4.1(a), if the fluid velocity is greater than the terminal velocity of the particleUT, then the particle will move upwards; if the fluid velocity is less thanUT
the particle will fall and if the fluid velocity is equal toUTthe particle will remain at the same vertical position. In each case the velocity of the particle relative to the pipe wall is ðU UTÞ. Now consider introducing two particles of different size and density having terminal velocitiesUT1 andUT2. Referring to Figure 2W4.1(b), at low fluid velocities ðU < UT2 < UT1Þ, both particles will fall. At high fluid velocities (U > UT1 > UT2), both particles will be carried upwards. At intermediate fluid velocities ðUT1 > U > UT2Þ, particle 1 will fall and particle 2 will rise. Thus we have the basis of a separator according to particle size and density. From the analysis above we see that to be able to completely separate particles A and B, there must be no overlap between the ranges of terminal velocity for the particles; i.e. all sizes of the denser material A must have terminal velocities which are greater than all sizes of the less dense material B.
Assuming Stokes’ law applies, Equation (2.13), with fluid density and viscosity 1000 kg=m3and 0.001 Pa s, respectively, gives
UT¼ 545x2ðrp 1000Þ
WORKED EXAMPLES 39
Based on this equation, the terminal velocities of the extreme sizes of particles A and B are:
We see that there is overlap of the ranges of terminal velocities. We can therefore select no fluid velocity which would completely separate particles A and B.
(b) Inspecting the expression for terminal velocity in the Stokes’ regime [Equation (2.13)]
we see that changing the fluid viscosity will have no effect on our ability to separate the particles, since change in viscosity will change the terminal velocities of all particles in the same proportion. However, changing the fluid density will have a different effect on particles of different density and this is the effect we are looking for.
The critical condition for separation of particles A and B is when the terminal velocity of the smallest A particle is equal to the terminal velocity of the largest B particle.
UTB40¼ UTA15
(a)
(b)
Increasing fluid velocity
Increasing fluid velocity
1 2
(Relative velocity = U - UT) U
U < UT U = UT U > UT
U < UT1 U < UT2
U < UT1 U > UT2
U > UT1 U > UT2
U U
U U U
Figure 2W4.1 Relative motion of particles in a moving fluid
SizeðmmÞ ! 15 40
UTAðmm=sÞ 0.82 5.84
UTBðmm=sÞ 0.17 1.22
Hence,
545 x240 ð2400 rfÞ ¼ 545 x215 ð7700 rfÞ From which, critical minimum fluid density rf¼ 1533 kg=m3.
WORKED EXAMPLE 2.5
A sphere of density 2500 kg=m3falls freely under gravity in a fluid of density 700 kg=m3 and viscosity 0:5 103Pa s. Given that the terminal velocity of the sphere is 0:15 m=s, calculate its diameter. What would be the edge length of a cube of the same material falling in the same fluid at the same terminal velocity?
Solution
In this case we know the terminal velocity,UT, and need to find the particle size x. Since we do not know which regime is appropriate, we must first calculate the dimensionless groupCD=Rep[Equation (2.16)]:
CD
This is the relationship between drag coefficentCDand single particle Reynolds number Repfor particles of density 2500 kg=m3having a terminal velocity of 0:15 m=s in a fluid of density 700 kg=m3 and viscosity 0:5 103Pa s. Since CD=Rep is a constant, this relationship will give a straight line of slopeþ1 when plotted on the log–log coordinates of the standard drag curve.
For plotting the relationship:
These values are plotted on the standard drag curves for particles of different sphericity (Figure 2.3). The result is shown in Figure 2W5.1.
Rep CD
100 0.712
1000 7.12
10 000 71.2
WORKED EXAMPLES 41
Where the plotted line intersects the standard drag curve for a sphere ðc ¼ 1Þ, Rep¼ 130.
The diameter of the sphere may be calculated from:
Rep¼ 130 ¼rfxvUT
m Hence, sphere diameterxV¼ 619 mm.
For a cube having the same terminal velocity under the same conditions, the sameCD
versus Rep relationship applies, only the standard drag curve is that for a cube ðc ¼ 0:806Þ.
Cube sphericity
For a cube of side 1 unit, the volume is 1 cubic unit and the surface area is 6 square units. Ifxvis the diameter of a sphere having the same volume as the cube, then
px3v
6 ¼ 1:0 which gives xv¼ 1:24 units
Therefore; sphericity c ¼surface area of a sphere of volume equal to the particle surface area of the particle
c ¼4:836 6 ¼ 0:806 104
1000
100
10
1
0.10.001 0.01 0.1 1 10 100
Single particle Reynolds number, Rep
Drag coefficient, CD
sphere cube
1000 104 105 106
ψ = 0.125 ψ = 0.22
ψ = 0.6 ψ = 0.806
ψ = 1.0
Figure 2W5.1 Drag coefficientCD versus Reynolds numberRep.
At the intersection of this standard drag curve with the plotted line,Rep¼ 310.
Recalling that the Reynolds number in this plot uses the equivalent volume sphere diameter,
xv¼310 ð0:5 103Þ
0:15 700 ¼ 1:48 103m And so the volume of the particle ispx3v
6 ¼ 1:66 109m3
Giving a cube side length ofð1:66 109Þ1=3¼ 1:18 103mð1:18 mmÞ
WORKED EXAMPLE 2.6
A particle of equivalent volume diameter 0.5 mm, density 2000 kg=m3and sphericity 0.6 falls freely under gravity in a fluid of density 1:6 kg=m3 and viscosity 2 105Pa s.
Estimate the terminal velocity reached by the particle.
Solution
In this case we know the particle size and are required to determine its terminal velocity without knowing which regime is appropriate. The first step is, therefore, to calculate the dimensionless groupCDRe2p:
CDRe2p¼4
This is the relationship between drag coefficientCDand single particle Reynolds number Rep for particles of size 0.5 mm and density 2000 kg=m3 falling in a fluid of density 1:6 kg=m3 and viscosity 2 105Pa s. SinceCDRe2p is a constant, this relationship will give a straight line of slope2 when plotted on the log–log coordinates of the standard drag curve.
For plotting the relationship:
These values are plotted on the standard drag curves for particles of different sphericity (Figure 2.3). The result is shown in Figure 2W6.1.
Rep CD
10 130.7
100 1.307
1000 0.013
WORKED EXAMPLES 43
Where the plotted line intersects the standard drag curve for a sphericity of 0:6ðc ¼ 0:6Þ, Rep¼ 40.
The terminal velocityUTmay be calculated from Rep¼ 40 ¼rfxvUT
m Hence, terminal velocity,UT¼ 1:0 m=s.
TEST YOURSELF
2.1 The total drag force acting on a particle is a function of:
(a) the particle drag coefficient;
(b) the projected area of the particle;
(c) both (a) and (b);
(d) none of the above.
2.2 The particle drag coefficient is a function of:
(a) the particle Reynolds number;
(b) the particle sphericity;
104
1000
100
10
1
0.10.001 0.01 0.1 1 10 100
Single particle Reynolds number, Rep
Drag coefficient, CD
1000 104 105 106 ψ = 0.125
ψ = 0.22 ψ = 0.6 ψ = 0.806
ψ = 1.0
Figure 2W6.1 Drag coefficientCD versus Reynolds numberRep.
(c) both (a) and (b);
(d) none of the above.
2.3 Stokes drag assumes:
(a) the drag coefficient is constant;
(b) the particle Reynolds number is constant;
(c) the drag force is constant;
(d) none of the above.
2.4 The particle Reynolds number is not a function of:
(a) the particle density;
(b) the pipe diameter;
(c) both (a) and (b);
(d) none of the above.
2.5 The force on a particle due to buoyancy depends on:
(a) the particle density;
(b) the particle size;
(c) both (a) and (b);
(d) none of the above.
2.6 The force on a particle due to gravity depends on:
(a) the particle density;
(b) the particle size;
(c) both (a) and (b);
(d) none of the above.
2.7 The particle Reynolds number is dependent on:
(a) the particle density;
(b) the fluid density;
TEST YOURSELF 45
(c) both (a) and (b);
(d) none of the above.
2.8 For particle in suspension in a fluid flowing in a pipe, as the pipe size increases (all other things held constant), the particle Reynolds number:
(a) increases;
(b) decreases;
(c) remains the same.
2.9 The particle Reynolds number does not depend on:
(a) particle density;
(b) velocity;
(c) viscosity;
(d) none of the above.
2.10 The units of viscosity are:
(a) length;
(b) mass/length;
(c) mass length/time;
(d) none of the above.
EXERCISES
2.1 The settling chamber, shown schematically in Figure 2E1.1, is used as a primary separation device in the removal of dust particles of density 1500 kg=m3 from a gas of density 0:7 kg=m3and viscosity 1:90 105Pa s.
(a) Assuming Stokes’ law applies, show that the efficiency of collection of particles of size x is given by the expression
collection efficiency; Zx¼x2gðrp rfÞL 18mHU
whereU is the uniform gas velocity through the parallel-sided section of the chamber.
State any other assumptions made.
(b) What is the upper limit of particle size for which Stokes’ law applies?
(c) When the volumetric flow rate of gas is 0:9 m3=s , and the dimensions of the chamber are those shown in Figure 2E1.1, determine the collection efficiency for spherical particles of diameter 30 mm [Answer: (b) 57 mm; (c) (86%).]
2.2 A particle of equivalent sphere volume diameter 0.2 mm, density 2500 kg=m3 and sphericity 0.6 falls freely under gravity in a fluid of density 1:0 kg=m3 and viscosity 2 105Pa s. Estimate the terminal velocity reached by the particle. (Answer: 0.6 m/s.) 2.3 Spherical particles of density 2500 kg=m3 and in the size range 20–100 mm are fed continuously into a stream of water (density, 1000 kg=m3and viscosity, 0.001 Pa s) flowing upwards in a vertical, large diameter pipe. What maximum water velocity is required to ensure that no particles of diameter greater than 60 mm are carried upwards with water?
(Answer: 2.9 mm/s.)
2.4 Spherical particles of density 2000 kg=m3 and in the size range 20–100 mm are fed continuously into a stream of water (density, 1000 kg=m3and viscosity, 0.001 Pa s) flowing upwards in a vertical, large diameter pipe. What maximum water velocity is required to ensure that no particles of diameter greater than 50 mm are carried upwards with the water?
(Answer: 1.4 mm/s.)
2.5 A particle of equivalent volume diameter 0.3 mm, density 2000 kg=m3and sphericity 0.6 falls freely under gravity in a fluid of density 1:2 kg=m3 and viscosity 2 105Pa s . Estimate the terminal velocity reached by the particle.
(Answer: 0.67 m/s.)
2.6 Assuming that a car is equivalent to a flat plate 1.5 m square, moving normal to the airstream, and with a drag coefficient,CD¼ 1:1, calculate the power required for steady motion
Gas
Figure 2E1.1 Schematic diagram of settling chamber
EXERCISES 47
at 100 km/h on level ground. What is the Reynolds number? For air assume a density of 1:2 kg=m3and a viscosity of 1:71 105Pa s. (Cambridge University)
(Answer: 32.9 kW; 2:95 106.)
2.7 A cricket ball is thrown with a Reynolds number such that the drag coefficient is 0.4 ðRe 105Þ.
(a) Find the percentage change in velocity of the ball after 100 m horizontal flight in air.
(b) With a higher Reynolds number and a new ball, the drag coefficient falls to 0.1. What is now the percentage change in velocity over 100 m horizontal flight?
(In both cases take the mass and diameter of the ball as 0.15 kg and 6.7 cm, respectively, and the density of air as 1:2 kg=m3.) Readers unfamiliar with the game of cricket may substitute a baseball. (Cambridge University) [Answer: (a) 43.1%; (b) 13.1%.]
2.8 The resistanceF of a sphere of diameter x, due to its motion with velocity u through a fluid of density r and viscosity m, varies with Reynolds number ðRe ¼ rux=mÞ as given below:
Find the mass of a sphere of 0.013 m diameter which falls with a steady velocity of 0.6 m/s in a large deep tank of water of density 1000 kg=m3and viscosity 0.0015 Pa s. (Cambridge University)
(Answer: 0.0021 kg.)
2.9 A particle of 2 mm in diameter and density of 2500 kg/m3is settling in a stagnant fluid in the Stokes’ flow regime.
(a) Calculate the viscosity of the fluid if the fluid density is 1000 kg/m3and the particle falls at a terminal velocity of 4 mm/s.
(b) What is the drag force on the particle at these conditions?
(c) What is the particle drag coefficient at these conditions?
(d) What is the particle acceleration at these conditions?
(e) What is the apparent weight of the particle?
2.10 Starting with the force balance on a single particle at terminal velocity, show that:
CD¼4 3
gx U2T
rp rf rf
" #
where the symbols have their usual meaning.
log10Re 2.0 2.5 3.0 3.5 4.0
CD¼ F
1
2ru2ðpx2=4Þ 1.05 0.63 0.441 0.385 0.39
2.11 A spherical particle of density 1500 kg=m3has a terminal velocity of 1 cm/s in a fluid of density 800 kg/m3and viscosity 0.001 Pa s. Estimate the diameter of the particle.
2.12 Estimate the largest diameter of spherical particle of density 2000 kg/m3 which would be expected to obey Stokes’ law in air of density 1:2 kg=m3 and viscosity 18 106Pa s.
EXERCISES 49