# Rhodes Solutions Ch10

October 13, 2017 | Author: Joson Chai | Category: Powder (Substance), Friction, Angle, Slope, Yield (Engineering)

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Rhodes Solutions Ch10...

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SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN EXERCISE 10.1: Shear cell tests on a powder show that its effective angle of internal friction is 40˚ and its powder flow function can be represented by the equation: σ y = σ 0.45 where σy is C the unconfined yield stress and σ is the compacting stress, both in kN/m2. The bulk C

density of the powder is 1000 kg/m3 and angle of friction on a mild steel plate is 16˚. It is proposed to store the powder in a mild steel conical hopper of semi-included angle 30˚ and having a circular discharge opening of 0.30 m diameter. What is the critical outlet diameter to give mass flow? Will mass flow occur? SOLUTION TO EXERCISE 10.1: a) With an effective angle of internal friction δ = 40˚ we refer to the flow factor chart in Text-Figure 10.18 (b), from which at Φw = 16˚ and with a safety margin of 3˚ we obtain the hopper flow factor, ff = 1.5 and hopper semi-included angle for mass flow, θ = 30° (see Solution-Manual-Figure 10.1.1). b)

For flow:

σC 〉 σy ff

(Text-Equation 10.3)

but for the powder in question σy and σC are related by the material flow function: σ y = σ 0.45 C Thus, the criterion for flow becomes: ⎡ σ 10.45 ⎤ ⎢ y ⎥ 〉 σ y ⎢ ff ⎥ ⎣⎢ ⎦⎥ ⎡ σ 10.45 ⎤ and so the critical value of unconfined yield stress σcrit is found when ⎢ y ⎥ = σ y ⎢ ff ⎥ ⎢⎣ ⎥⎦

hence, solving for σy gives σcrit = 1.393 kN/m2. From Text-Equation 10.5, H(θ ) = 2.0 +

θ . Hence, H(θ) = 2.5 when θ = 30˚ 60

Hence from Text-Equation 10.4: minimum diameter of circular outlet, B =

2.5 × 1.393 × 103 = 0.355 m. 1000 × 9.81

Summarising, mass flow without blockages is ensured by using a mild steel conical hopper with maximum semi-included cone angle 30˚ and a circular outlet diameter of

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN

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at least 35.5 cm. Although the hopper wall is steeper that required, the actual discharge diameter of the existing hopper is only 30 cm, and so blockages are likely to occur. EXERCISE 10.2: Describe how you would use shear cell tests to determine the effective angle of

internal friction of a powder. A powder has an effective angle of internal friction of 60˚ and has a powder flow function represented in the graph shown in Text-Figure 10E2.1. If the bulk density of the powder in 1500 kg/m3 and its angle of friction on mild steel plate is 24.5˚, determine, for a mild steel hopper, the maximum semi-included angle of cone required to safely ensure mass flow, and the minimum size of circular outlet to ensure flow when the outlet is opened. SOLUTION TO EXERCISE 10.2: a) With an effective angle of internal friction δ = 60˚ we refer to the flow factor chart in Text-Figure 10.18 (d), from which at Φw = 24.5˚ and with a safety margin of

3˚ we obtain the hopper flow factor, ff = 1.2 and hopper semi-included angle for mass flow, θ = 17.5˚ (see Solution-Manual-Figure 10.2.1). Critical conditions for flow b) The relationships between unconfined yield stress, σy and compacting stress, σC and

between stress developed in the powder σD and compacting stress, σC From TextEquation 10.3, the limiting condition for flow is: σC = σy ff This may be plotted on the same axes as the Powder Flow Function (unconfined yield stress, σy and compacting stress, σC) given in Text-Figure 10.E2.1 in order to reveal

the conditions under which flow will occur for this powder in the hopper. The limiting condition gives a straight line of slope 1/ff. Solution-Manual-Figure 10.2.2 shows such a plot. Where the powder has a yield stress greater than σC/ff, no flow occurs. Where the powder has a yield stress less than σC/ff, flow occurs. There is a critical condition, where unconfined yield stress, σy is equal to stress developed in the powder, σC/ff. This critical condition is given where the powder flow function and the straight line of slope 1/ff intersect (see Solution-Manual-Figure 10.2.2). This gives rise to a

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critical value of stress, σcrit which is the critical stress developed in the surface of the arch. From the figure we see that σcrit = 1.20 kN/m2. From Text-Equation 10.5, H(θ ) = 2.0 +

θ . Hence, H(θ) = 2.292 when θ = 17.5˚ 60

Hence from Text-Equation 10.4 minimum diameter of circular outlet, B =

2.292 × 1.20 × 103 = 0.187 m. 1500 × 9.81

Summarising, mass flow without blockages is ensured by using a mild steel conical hopper with maximum semi-included cone angle 17.5˚ and a circular outlet diameter of at least 18.7 cm. EXERCISE 10.3: a) Summarise the philosophy used in the design of conical hoppers to ensure

flow from the outlet when the outlet valve is opened. b)

Explain how the Powder Flow Function and the effective angle of internal friction are extracted from the results of shear cell tests on a powder.

c) A company having serious hopper problems, takes on a Chemical Engineering graduate. The hopper in question feeds a conveyor belt and periodically blocks at the outlet and needs to be “encouraged” to restart. The graduate makes an investigation on the hopper, commissions shear cell tests on the powder and recommends a minor modification to the hopper. After the modification the hopper gives no further trouble and the graduate’s reputation is established. Given the information below, what was the graduate’s recommendation? Existing Design:

Material of wall - mild steel Semi-included angle of conical hopper - 33˚ Outlet - circular, fitted with 25 cm diameter slide valve

Shear Cell Test Data:

Effective angle of internal friction, δ = 60˚ Angle of wall friction on mild steel, Φw = 8˚ Bulk density, ρB = 1250 kg/m3 Powder flow function: σ y = σ 0.55 C (σy and σC in

kN/m2)

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SOLUTION TO EXERCISE 10.3: a) With an effective angle of internal friction δ = 60˚ we refer to the flow factor chart in Text-Figure 10.18 (d), from which at Φw = 8˚ and with a safety margin of 3˚

we obtain the hopper flow factor, ff = 1.3 and hopper semi-included angle for mass flow, θ = 35.5° (see Solution-Manual-Figure 10.3.1). b)

For flow:

σC 〉 σy ff

(Text-Equation 10.3)

But for the powder in question, σy and σC are related by the material flow function: σ y = σ 0.55 C Thus, the criterion for flow becomes: ⎡ σ 10.55 ⎤ ⎢ y ⎥ 〉 σ y ⎢ ff ⎥ ⎢⎣ ⎥⎦ ⎡ 0.55 ⎤ and so the critical value of unconfined yield stress σcrit is found when ⎢ σ y ⎥ = σ y 1

⎢ ⎢⎣

ff

⎥ ⎥⎦

Hence, solving for σy gives σcrit = 1.378 kN/m2. From Text-Equation 10.5, H(θ ) = 2.0 +

θ . Hence, H(θ) = 2.592 when θ = 35.5˚ 60

Hence from Text-Equation 10.4 minimum diameter of circular outlet, B =

2.592 × 1.378 × 10 3 = 0.291 m. 1250 × 9.81

Summarising, mass flow without blockages is ensured by using a mild steel conical hopper with maximum semi-included cone angle of 35.5˚ and a circular outlet diameter of at least 29.1 cm. The actual semi-included cone angle of 33˚, which means that the hopper wall is steep enough. However, the present discharge diameter is only 25 cm because of the restriction of the slide valve, which is the cause of the blockages. The solution offered by the new graduate is to remove the slide valve, cut a few centimetres off the lower end of the hopper to allow a slide valve with a circular opening diameter of at least 29.1 cm to be fitted.

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EXERCISE 10.4: Shear cell tests are carried out on a powder for which a stainless steel conical hopper

is to be designed. The results of the tests are shown graphically in Text-Figure 10E4.1. In addition it is found that the friction between the powder on stainless steel can be described by an angle of wall friction of 11˚, and that the relevant bulk density of the powder is 900 kg/m3. a)

From the shear cell results of Text-Figure 10E4.1, deduce the effective angle of internal friction δ of the powder.

b)

Determine: (i) the semi included hopper angle safely ensuring mass flow. (ii) the Hopper Flow Factor, ff. Combine this information with further information gathered from Text-Figure 10E4.1 in order to determine the minimum diameter of outlet to ensure flow when required. (Note: extrapolation is necessary here)

c)

d)

What do you understand by “angle of wall friction” and “effective angle of internal friction”?

SOLUTION TO EXERCISE 10.4 (i) From Text-Figure 10E4.1, determine the slope of the effective yield locus.

Slope = 1.85. (see Solution-Manual-Figure 10.4.1). Hence, the effective angle of internal friction, δ = tan −1 (1.85 ) = 61.6 ° (ii) From Text-Figure 10E4.1, determine (see Solution-Manual-Figure 10.4.1) the pairs of values of σC and σy necessary to plot the powder flow function. σy

0.4

0.45

0.5

σc

0.65

1.0

1.45

Using the flow factor chart for δ = 60˚ (the nearest to the estimated value of 61.6˚) (Text-Figure 10.18 (d)) with Φw = 11˚ and a 3˚ margin of safety, gives a hopper flow

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factor, ff = 1.29 (approximately) and semi-included angle of hopper case, θ = 32.5˚ (see Solution-Manual-Figure 10.4.2). (iii)

The relationship σy = σC/ff is plotted on the same axes as the powder flow

function and where this line intercepts the powder flow function we find a value of critical unconfined yield stress, σcrit = 0.38 kN/m2 (Solution-Manual-Figure 10.4.3). Note that extrapolation is required in order to obtain the critical condition. From Text-Equation 10.5, H(θ) = 2.54 when θ = 32.5˚ And from Text-Equation 10.4, the minimum outlet diameter for mass flow, B is: 2.54 × 0.38 × 10 3 B= = 0.110 m. 900 × 9.81 Summarising then, to achieve mass flow without risk of blockage using the powder in question we require a stainless steel conical hopper with a maximum semi-included angle of case, 32.5˚ and a circular outlet with a diameter of at least 11 cm. EXERCISE 10.5: The results of shear cell tests on a powder are given in Text-Figure 10E5.1. An

aluminium conical hopper is to be designed to suit this powder. It is known that the angle of wall friction between the powder and aluminium is 16˚ and that the relevant bulk density is 900 kg/m3. a) b) c)

From Text-Figure 10E5.1 determine the effective angle of internal friction of the powder. Also determine (i) the semi-included hopper angle safely ensuring mass flow (ii) the hopper flow factor, ff. Combine the information with further information gathered from Text-Figure 10E5.1 in order to determine the minimum diameter of circular outlet to ensure flow when required. Note: Extrapolation of these experimental results may be necessary.

SOLUTION TO EXERCISE 10.5: (i) From Text-Figure 10E5.1, determine the slope of the effective yield locus.

Slope = 0.833. (see Solution-Manual-Figure 10.5.1). Hence, the effective angle of internal friction, δ = tan −1(0.833 ) = 40

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(ii) From Text-Figure 10E5.1, determine (see Solution-Manual-Figure 10.5.1) the pairs of values of σC and σy necessary to plot the powder flow function. σy

2.0

2.18

2.4

2.58

σc

5.0

7.0

9.0

9.9

Using the flow factor chart for δ = 40˚ (Text-Figure 10.18 (b)) with Φw = 16˚ and a 3˚ margin of safety, gives a hopper flow factor, ff = 1.5 and semi-included angle of hopper case, θ = 29.5˚ (see Solution-Manual-Figure 10.5.2). The relationship σy = σC/ff is plotted on the same axes as the powder flow function and where this line intercepts the powder flow function we find a value of critical unconfined yield stress, σcrit = 1.82 kN/m2 (Solution-Manual-Figure 10.5.3). (iii)

Note that extrapolation is required in order to obtain the critical condition. From Text-Equation 10.5, H(θ ) = 2.0 +

θ , and so H(θ) = 2.492 when θ = 29.5˚ 60

and from Text-Equation 10.4, the minimum outlet diameter for mass flow, B is: 2.492 × 1.82 × 103 B= = 0.514 m. 900 × 9.81 The value for σcrit varies depending on the extrapolation and so we quote B as 0.5m±7% . Summarising then, to achieve mass flow without risk of blockage using the powder in question we require an aluminium conical hopper with a maximum semi-included angle of case, 29.5˚ and a circular outlet with a diameter of at least 50 cm +7%.

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN

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Figure 10.1.1: Obtaining ff and θ from hopper flow factor chart (δ = 40°) for Exercise 10.1.

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Figure 10E2.1: Powder flow function for Exercise 10.2.

Figure 10.2.1: Obtaining ff and θ from hopper flow factor chart (δ = 60°) for Exercise 10.2.

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN

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Figure 10.2.2: Estimation of σ crit from Powder Flow Function (Exercise10.2).

Figure 10.3.1: Obtaining ff and θ from hopper flow factor chart (δ = 60°) for Exercise 10.3.

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Figure 10.4.1: Obtaining the slope of the effective yield locus and pairs of values of σ c and σ y from the shear cell data (Exercise 10.4).

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN

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Figure 10.4.2: Obtaining ff and θ from hopper flow factor chart (δ = 60°) for Exercise 10.4.

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Figure 10.4.3: Plotting of powder flow function and estimation of σ crit (Exercise 10.4). SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN

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Figure 10.5.1: Obtaining the slope of the effective yield locus and pairs of values of σ c and σ y from the shear cell data (Exercise 10.5).

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Figure 10.5.2: Obtaining ff and θ from hopper flow factor chart (δ = 40°) for Exercise 10.5.

Figure 10.5.3: Plotting of powder flow function and estimation of σ crit (Exercise 10.5).

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