Rhodes Solutions Ch2

SOLUTIONS TO CHAPTER 2: SINGLE PARTICLES IN FLUIDS EXERCISE 2.1: The settling chamber, shown schematically in Figure 2.E1.1, 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/m3 and viscosity 1.90 x 10-5 Pas. (a) Assuming Stokes Law applies, show that the efficiency of collection of particles of size x is given by the expression: x 2 g(ρp − ρ f )L collection efficiency, ηx = 18μHU where U 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 this expression applies. (c) When the volumetric flow rate of gas is 0.9 m3/s, and the dimensions of the chamber are those shown in Text-Figure 2.E1.1, determine the collection efficiency for spherical particles of diameter 30μm. SOLUTION TO EXERCISE 2.1: (a) Assuming plug flow of the gas and particles then the residence time of the L particles in the parallel-sided section of the separator is: U There is a critical particle diamter xcrit such that a particle of diameter xcrit falls at a L velocity Ucrit covering the height H in time . U HU i.e. U crit = L

All particles falling at a velocity greater than or equal to Ucrit will be collected no matter at which position in the cross section they start. Assuming particles of all sizes are evenly distributed across the cross section at the inlet to the parallel-sided section, then particle for which Ufall = 0.5Ucrit will be collected with an efficiency of 50% (since 50% of these particles will have too far to L fall in the time available ( ). U

SOLUTIONS TO CHAPTER 2 EXERCISES: SINGLE PARTICLES IN FLUIDS

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It follows that efficiency, η =

Ufall Ucrit

Assuming that all particles reach their terminal free fall velcocity in very short time and can be assumed to fall at this velocity, then

ηx =

UT , where UT is the single particle terminal velocity. Ucrit

(

x2 g ρp − ρf Assuming Stokes Law applies, then U T = 18μ

ηx =

(

x2g ρ p − ρf

)L

18μ

HU

)

, where η is the efficiency of collection of particles of size x.

(b) The upper limit of particle size for which this expression applies. The expression is limited to those particles for which Stokes Law applies, i.e. for Rep < 0.3 At the limiting Reynolds number,

(

U Tρ f x = 0.3 μ

x2 g ρp − ρf From Stokes Law, U T = 18μ

)

(2.1.1)

(2.1.2)

Solving Equations 2.1.1 and 2.1.2 simultaneously, x = 57.4 μm (not 50 μm as give in the book, which is calculated for Rep = 0.2) (c) Collection efficiency for spherical particles of diameter 30μm when volumetric flow rate of gas is 0.9 m3/s: 0.9 0.9 Superficial gas velocity in parallel-sided section, U = = = 0.15 m / s WH 2 × 3 From the equation derived for efficiency, 30 × 10 −6 ) × 9.81× (1500 − 0.7) ( = 2

η30

18 × 1.9 × 10

−5

10 = 0.86 3 × 0.15

Collection efficiency for 30μm particles is 86%.

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EXERCISE 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 x10-5 Pas. Estimate the terminal velocity reached by the particle. (Answer: 0.6 m/s) SOLUTION TO EXERCISE 2.2: 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 group C D Re2p : 3 4 x ρ f (ρ p − ρf )g 2 C D Re p = 3 μ2 ⎡ ⎤ 4 ⎢ (0.2 × 10 −3 )3 × 1.0 × (2500 − 1.0) × 9.81⎥ = 3⎢ −5 2 2 × 10 ⎥⎦ ⎣

(

)

= 653.7 This is the relationship between drag coefficient CD and single particle Reynolds number Rep for particles of size 0.2 mm and density 2500 kg/m3 falling in a fluid of density 1.0 kg/m3 and viscosity 2 x 10-5 Pas. Since C D Re2p is a constant, this relationship will give a straight line of slope -2 when plotted on the log-log coordinates of the standard drag curve. For plotting the relationship: Rep

CD

1

653.7

10

6.537

These values are plotted on the standard drag curves for particles of different sphericity (Text- Figure 2.3). The result is shown in Figure 2.2.1. Where the plotted line intersects the standard drag curve for a sphericity of 0.6 (ψ = 0.6), Rep = 6.0. The terminal velocity UT may be calculated from: ρ x U Re p = 6 = f v T μ Hence, terminal velocity, UT = 0.6 m/s

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EXERCISE 2.3: Spherical particles of density 2500 kg/m3 and in the size range 20 - 100 μm are fed

continuously into a stream of water (density, 1000 kg/m3 and viscosity, 0.001 Pas) flowing upwards in a vertical, large diameter pipe. What maximum water velocity is required to ensure that no particles of diameter greater than 60 μm are carried upwards with the water? SOLUTION TO EXERCISE 2.3: Assume that the upward velocity of the water if effectively uniform across the cross section of the large pipe and that the pipe walls have no effect [U∞ UD = 1.0].

Assume that the particle accelerate so quickly to their terminal velocity so that the relative velocity between the particles and the water is equal to the single particle terminal velocity, UT. Thus, if the upward water velocity is less that UT for the particle, the particle will fall and if the upward water velocity if greater than UT, the particle will rise. In the limiting case: water velocity = UT

(

x2 g ρp − ρf Assuming Stokes Law applies for the 60μm particles, U T = 18μ

)

60 × 10 −6 ) × 9.81 × (2500 − 1000 ) ( = = 2.943 × 10− 3 m / s 2

hence, U T

18 × 0.001

ρx U 2.943 × 10 −3 × 1000 × 60 × 10 −6 = 0.177 Check Reynolds number, Re p = f v T = μ 0.001 Rep is less than 0.3, and so the assumption of Stokes Law is valid. Hence, maximum water velocity = 2.94 mm/s EXERCISE 2.4: Spherical particles of density 2000 kg/m3 and in the size range 20 - 100 μm are fed

continuously into a stream of water (density, 1000 kg/m3 and viscosity, 0.001 Pas) flowing upwards in a vertical, large diameter pipe. What maximum water velocity is required to ensure that no particles of diameter greater than 50 μm are carried upwards with the water? SOLUTION TO EXERCISE 2.4:

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Assume that the upward velocity of the water if effectively uniform across the cross section of the large pipe and that the pipe walls have no effect [U∞ UD = 1.0]. Assume that the particle accelerate so quickly to their terminal velocity so that the relative velocity between the particles and the water is equal to the single particle terminal velocity, UT. Thus, if the upward water velocity is less that UT for the particle, the particle will fall and if the upward water velocity if greater than UT, the particle will rise. In the limiting case: water velocity = UT

(

x2 g ρp − ρf Assuming Stokes Law applies for the 50μm particles, U T = 18μ

)

50 × 10 −6 ) × 9.81 × (2000 − 1000 ) ( = = 1.36 × 10 −3 m / s 2

hence, U T

18 × 0.001

ρx U 1.36 × 10−3 × 1000 × 50 × 10−6 = 0.068 Check Reynolds number, Re p = f v T = μ 0.001 Rep is less than 0.3, and so the assumption of Stokes law is valid. Hence, maximum water velocity = 1.36 mm/s EXERCISE 2.5: A particle of equivalent volume diameter 0.3 mm, density 2000 kg/m3 and sphericity 0.6 falls freely under gravity in a fluid of density 1.2 kg/m3 and viscosity 2 x10-5 Pas. Estimate the terminal velocity reached by the particle. SOLUTION TO EXERCISE 2.5: 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 group C D Re2p : 3 4 x ρ f (ρ p − ρf )g 3 μ2 ⎡ ⎤ 4 ⎢ (0.3 × 10 −3 )3 × 1.2 × (2000 − 1.2) × 9.81⎥ = 3⎢ −5 2 2 × 10 ⎣ ⎦⎥

C D Re2p =

(

)

= 2117

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This is the relationship between drag coefficient CD and single particle Reynolds number Rep for particles of size 0.3 mm and density 2000 kg/m3 falling in a fluid of density 1.2 kg/m3 and viscosity 2 x 10-5 Pas. Since C D Re2p is a constant, this relationship will give a straight line of slope -2 when plotted on the log-log coordinates of the standard drag curve. For plotting the relationship: Rep

CD

1

2117

10

21.17

100

0.2117

These values are plotted on the standard drag curves for particles of different sphericity (Text-Figure 2.3). The result is shown in Figure 2.5.1. Where the plotted line intersects the standard drag curve for a sphericity of 0.6 (ψ = 0.6), Rep = 12. The terminal velocity UT may be calculated from: ρx U Re p = 12 = f v T μ Hence, terminal velocity, UT = 0.667 m/s EXERCISE 2.6: (Cambridge University) Assuming that a car is equivalent to a flat plate 1.5 m square, moving normal to the air-stream, and with a drag coefficient, CD = 1.1, calculate the power required for

steady motion at 100 km/h on level ground. What is the Reynolds number? For air assume a density of 1.2 kg/m3 and a viscosity of 1.71 x 10-5 Pas. SOLUTION TO EXERCISE 2.6: R′ Drag coefficient, C D = 1 , where R ′ is the fluid drag force per unit projected 2 ρU 2 f area and U is the relative velocity of the "particle" and the fluid of density ρf.

Relative velocity, U = 27.78 m/s. Power required for steady motion = force x velocity

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1 3 = R ′AU = CD ρ f AU 2 1 = 1.1 × × (1.5 × 1.5) × 1.2 × 27.783 = 31836 kW 2 = 31.8 kW. Reynolds number =

Uρ f x 27.78 × 1.5 × 1.2 = = 2.92 × 10 6 −5 μ 1.71 × 10

EXERCISE 2.7: (Cambridge University) A cricket ball is thrown with a Reynolds number such that the drag coefficient is 0.4 (Re ≈ 105).

(a) (b)

Find the percentage change in velocity of the ball after 100 m horizontal flight in air. 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. SOLUTION TO EXERCISE 2.7: (a) percentage change in velocity of the ball after 100 m horizontal flight in air: The kinetic energy of the cricket ball is dissipated by working against the drag force, F, which varies with relative velocity. Thus: 1 F × ds = − d ⎡ mu 2 ⎤ ⎣2 ⎦ 2 d ⎛u ⎞ F = −m ⎜ ⎟ ds ⎝ 2 ⎠ 2F ds = − d(u2 ) and so, m ⎛ πx2 ⎞ 1 ⎟ , where x is the diameter of the ball. Now drag force, F = CD ρ f u 2 ⎜ 2 ⎝ 4 ⎠ ⎛ π0.0672 ⎞ 1 ⎟ = 8.461× 10−4 u 2 Newton If CD = 0.4, then: F = 0.4 × × 1.2 × u 2 ⎜ 4 2 ⎝ ⎠

and with mass of ball, m = 0.15 kg, d(u2 ) 0.01128 ds = − 2 u

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integrating:

0. 01128 s = − ln(u2 ) + K

boundary conditions: when s = 0, u = u0; when s = 100, u = u100 hence: 0 = − ln(u 20 ) + K

2 1.128 = − ln(u100 )+K ⎛u ⎞ Eliminating K, 1.128 = −2 × ln⎜ 100 ⎟ ⎝ u0 ⎠ u100 −0.564 Therefore, =e = 0.569 u0

and

⎛ u ⎞ And so the percentage change in velocity, ⎜1 − 100 ⎟ × 100 = 43.1% ⎝ u0 ⎠ (b) Percentage change in velocity over 100 m horizontal flight a new ball, with a drag coefficient of 0.1: u With CD = 0.1, using the same procedure, 100 = e −0.141 = 0.868 u0 Percentage change in velocity of the new ball = 13.2% (The new cricket ball can therefore be delivered with greater pace to the batsman) EXERCISE 2.8: (Cambridge University) The resistance F of a sphere of diameter x, due to its motion with velocity u through a fluid of density ρ and viscosity μ varies with Reynolds number (Re = ρux/μ) as given

below: log10Re

F CD = 1 2 ⎛⎜ πx 2 ⎞⎟ ρu 2 ⎝ 4 ⎠

2.0

2.5

3.0

3.5

4.0

1.05

0.63

0.441

0.385

0.39

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/m3 and viscosity 0.0015 Pas.

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SOLUTION TO EXERCISE 2.8: At steady terminal velocity the weight of the sphere is balanced by the sum of the buoyancy force and the fluid drag force:

weight of sphere Mg = drag force + buoyancy force ⎛ πx 2 ⎞ π 3 1 ⎟ + x ρf g therefore, Mg = CD ρf u2 ⎜ 2 ⎝ 4 ⎠ 6

(2.8.1)

ρ xU 1000 × 0.013 × 0.6 Under the conditions, Reynolds number, Re = f = = 5200 μ 1.5 × 10 − 3 From the data given, plot CD versus log10Re and interpolate to find CD = 0.385 at Re = 5200. From Equation 2.8.1, mass of sphere, M = 0.00209 kg. EXERCISE 2.9 A particle of 2 mm in diameter and density of 2500 kg/m3 is settling in a stagnant fluid in the Stokes’ flow regime.

a) Calculate the viscosity of the fluid if the fluid density is 1000 kg/m3 and 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? c) What is the particle acceleration at these conditions? d) What is the apparent weight of the particle? SOLUTION TO EXERCISE 2.9 2 In the Stokes law region: U = x (ρ p − ρ f )g T

(EQ1.13)

18μ

Hence, with UT = 4 x 10-3 m/s, ρf = 1000 kg/m3, ρp = 2500 kg/m3 and x = 2 x 10-3 m: −3 (a) Viscosity, μ = (2 × 10 ) (2500 − 1000 ) × 9.81 = 0.8175 Pa.s −3 2

18 × 4 × 10

(EQ 1.3) (b) Drag force, FD = 3πμUx −3 So FD = 3π × 0.8175 × (4 × 10 )× (2 × 10 −3 ) = 6.164 × 10 −5 N (c) Drag coefficient CD = 24/Rep So: C D =

24μ 24 × 0.8175 = = 2452 .5 −3 Uρ f x 4 × 10 × 1000 × 2 × 10 −3

(

)

(

)

(d) At terminal velocity, acceleration is zero.

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(e) Apparent weight: at terminal velocity apparent weight = drag force = 6.164 x 10-5 N As a check, we calculate the apparent weight =

πx 3 (ρ p − ρ f )g = 6.164 × 10 −5 N 6

EXERCISE 2.10: Starting with the force balance on a single particle at terminal velocity, show that: 4 gx ⎡ ρp − ρ f ⎤ ⎢ ⎥ 3 U2 ⎣ ρ f ⎦ T where the symbols have their usual meaning. CD =

SOLUTION TO EXERCISE 2.10: See text EXERCISE 2.11: Using the drag coefficient-Reynolds number data given below, calculate the density of a sphere of diameter 10 mm which falls at a steady velocity of 0.25 m/s in large tank of water of density 1000 kg/m3 and viscosity 0.001 Pas.

Log10Rep CD

2.0 1.05

2.5 0.63

3.0 0.441

3.5 0.385

4.0 0.390

SOLUTION TO EXERCISE 2.11:

At terminal velocity: C D =

4 gx (ρ p − ρ f ) 3 U 2T ρf

Under the given conditions, Re p =

U T ρ f x 0.25 × 1000 × 10 × 10 −3 = = 2500 μ 0.001

Plotting the CD data given, we can interpolate to find CD at this value of Rep: Gives: CD = 0.40 1.2 1

CD

0.8 0.6 0.4 0.2 0 2

2.5

3

3.5

4

log10Re p

Hence, from C D =

4 gx (ρ p − ρ f ) , particle density, ρp = 1191 kg/m3 3 U 2T ρf

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EXERCISE 2.12: A spherical particle of density 1500 kg/m3 has a terminal velocity of 1 cm/s in a fluid of density 800 kg/m3 and viscosity 0.001 Pas. Estimate the diameter of the particle. SOLUTION TO EXERCISE 2.12: When UT is known and x unknown, we first calculate the dimensionless group: CD 4 gμ(ρ p − ρ f ) = (EQ1.16) Re p 3 U 3T ρ f2

So,

CD 4 9.81× 0.001× (1500 − 800) = × = 14.306 Re p 3 (0.01)3 × 800 2

Plotted on the drag curve (CD versus Rep) this gives a straight line of slope +1 (see Figure 2E12.1) This line intersects the curve for spherical particles (ψ = 1.0) at a Rep value of 1.4. Hence: Re p =

UT ρ f x = 1.4 , μ

giving particle size, x = 175 μm EXERCISE 2.13: Estimate the largest diameter of spherical particle of density 2000 kg/m3 which would be expected to obey Stokes's Law in air of density and viscosity, 1.2 kg/m3 and viscosity 18 x 10-6 Pas respectively. SOLUTION TO EXERCISE 2.13: The upper limit for Stokes Law is when Re p ≤ 0.3

Or:

Uρ f x ≤ 0.3 μ

The largest Reynolds number for given particle and fluid properties will be at U = UT (since this is the maximum relative velocity achieved by the particle) So:

UT ρ f x ≤ 0.3 μ

2 Now in Stokes Law region, U = x (ρ p − ρ f )g T

18μ

(EQ1.13)

⎛ x (ρ p − ρ f )g ⎞ ρ f x ⎟ So: ⎜ ≤ 0.3 2

⎜ ⎝

18μ

⎟ μ ⎠

With ρp = 2000 kg/m3, ρf = 1.2 kg/m3 and μ = 18 x 10-6 Pa.s: Particle size x ≤ 42.05 μm Largest diameter particle that will obey Stokes Law under these conditions is 42 μm

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Text-Figure 2.3: Drag coefficient CD versus Reynolds number Rep.

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Figure 2.E1.1: Schematic diagram of settling chamber (Exercise 2.1).

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Figure 2.2.1: Single particle drag curves: Solution to Exercise 2.2.

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Figure 2.5.1: Drag curves: Solution to Exercise 2.5

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Rep = 1.4

Figure 2E12.1: Drag curve plot for use in Exercise 2.12

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