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: where y is the unconfined yield stress and C is the compacting stress, both in kN/m2. The bulk 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?

y = C0.45

SOLUTION TO EXERCISE 10.1: a) With an effective angle of internal friction = 40 we refer to the flow factor chart in Text-Figure 10.18 (b), from which at w = 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, = 30 (see Solution-Manual-Figure 10.1.1).

b) For flow: Cff

y (Text-Equation 10.3) but for the powder in question y and C are related by the material flow function:

y = C0.45Thus, the criterion for flow becomes:

y

45.01

y ff

and so the critical value of unconfined yield stress crit is found when y45.0

1

y ff

=

hence, solving for y gives crit = 1.393 kN/m2.

From Text-Equation 10.5, H ( )= 2.0 + 60

. Hence, H() = 2.5 when = 30 Hence from Text-Equation 10.4:

minimum diameter of circular outlet, B = 2.5 1.393 103

1000 9.81 = 0.355 m. 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 Page 10.1

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 = 60 we refer to the flow factor chart in Text-Figure 10.18 (d), from which at w = 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, = 17.5 (see Solution-Manual-Figure 10.2.1). b) Critical conditions for flow The relationships between unconfined yield stress, y and compacting stress, C and between stress developed in the powder D and compacting stress, C From Text-Equation 10.3, the limiting condition for flow is: Cff

= y This may be plotted on the same axes as the Powder Flow Function (unconfined yield stress, y and compacting stress, C) 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 C/ff, no flow occurs. Where the powder has a yield stress less than C/ff, flow occurs. There is a critical condition, where unconfined yield stress, y is equal to stress developed in the powder, C/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

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.2

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

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

Hence from Text-Equation 10.4

minimum diameter of circular outlet, B = 2.292 1.20 103

1500 9.81 = 0.187 m. 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 graduates reputation is established. Given the information below, what was the graduates 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, = 60 Angle of wall friction on mild steel, w = 8 Bulk density, B = 1250 kg/m3 Powder flow function C

0.55 (y and C in :N/m2)

y = k

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.3

SOLUTION TO EXERCISE 10.3: a) With an effective angle of internal friction = 60 we refer to the flow factor chart in Text-Figure 10.18 (d), from which at w = 8 and with a safety margin of 3we obtain the hopper flow factor, ff = 1.3 and hopper sfl

emi-included angle for mass

ow, = 35.5 (see Solution-Manual-Figure 10.3.1).

) For flow: C

ff yb (Text-Equation 10.3)

But for the powder in question, y and C are related by the material flow function:

es:

y = C0.55 Thus, the criterion for flow becom

y

55.01

y ff

and so the critical value of unconfined yield stress crit is found when y55.0

1

y ff

=

Hence, solving for y gives crit = 1.378 kN/m2.

rom Text-Equation 10.5,

H ( )= 2.0 + 60

. Hence, H() = 2.592 when = 35.5

inimum diameter of circular outlet,

F

Hence from Text-Equation 10.4

B = 2.592 1.378 103

1250 9.81 = 0.291 m. m 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 ircular opening diameter of at least 29.1 cm to be fitted. c

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.4

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 f the powder is 900 kg/m3.

a) , deduce the of internal friction of the powder.

safely nsuring mass flow.

c) r

e flow when required. (Note: extrapolation is ecessary here)

y angle of wall friction and effective angle of internal friction?

) effective yield locus. Slope = 1.85. (see Solution-Manual-Figure 10.4.1).

ence, the effective angle of internal friction,

10.4.1) the airs of values of C and y necessary to plot the powder flow function.

o

From the shear cell results of Text-Figure 10E4.1effective angle

b) Determine: (i) the semi included hopper angle e (ii) the Hopper Flow Factor, ff.

Combine this information with further information gathered from Text-Figure 10E4.1 in order to determine the minimum diameteof outlet to ensurn

d) What do you understand b SOLUTION TO EXERCISE 10.4 (i From Text-Figure 10E4.1, determine the slope of the

== 6.61)85.1(tan 1 H (ii) From Text-Figure 10E4.1, determine (see Solution-Manual-Figurep

y 0.4 0.45 0.5 c 0.65 1.0 1.45

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

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.5

factor, ff = 1.29 (approximately) and s(s

emi-included angle of hopper case, = 32.5 ee Solution-Manual-Figure 10.4.2).

re 10.4.3). ote that extrapolation is required in order to obtain the critical condition.

rom Text-Equation 10.5, H() = 2.54 when = 32.5

on 10.4, the minimum outlet diameter for mass flow, B is:

(iii) The relationship y = C/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, crit = 0.38 kN/m2 (Solution-Manual-FiguN F And from Text-Equati

B = 2.54 0.38 103

900 9.81 = 0.110 m. 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 ngle of case, 32.5 and a circular outlet with a diameter of at least 11 cm.

en the powder and aluminium is 16 and that the relevant ulk density is 900 kg/m3.

a) determine the effective angle of internal

c) ure

equired. Note: Extrapolation of these experimental results may be necessary.

) effective yield locus. gure 10.5.1).

ence, the effective angle of internal friction,

a 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 betweb

From Text-Figure 10E5.1friction of the powder.

b) 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-Fig10E5.1 in order to determine the minimum diameter of circular outlet to ensure flow when r

SOLUTION TO EXERCISE 10.5: (i From Text-Figure 10E5.1, determine the slope of the Slope = 0.833. (see Solution-Manual-Fi

40)833.0(tan 1 == H

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.6

(ii) From Text-Figure 10E5.1, determine (see Solution-Manual-Figure 10.5.1) the pairs alues o es to p e powder flow function.

y 2.0 2.18 2.4 2.58

of v f C and y nec sary lot th

c 5.0 7.0 9.0 9.9 Using the flow factor chart for = 40 (Text-Figure 10.18 (b)3

) with w = 16 and a margin of safety, gives a hopper flow factor, ff = 1.5 and semi-included angle of

a value of ritical unconfined yield stress, crit = 1.82 kN/m2 (Solution-Manual-Figure 10.5.3).

hopper case, = 29.5 (see Solution-Manual-Figure 10.5.2). (iii) The relationship y = C/ff is plotted on the same axes as the powder flow function and where this line intercepts the powder flow function we find c

Note that extrapolation is required in order to obtain the critical condition. From Text-Equation 10.5, H ( )= 2.0 + , and so H() = 2.492 when = 29.5

60

and from Text-Equation 10.4, the minimum outlet diameter for mass flow, B is:

B =900

2.492 1.82 103 9.81 = 0.514 m.

The value for crit varies depending on the extrapolation and so we quote B as

er in uestion 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%.

0.5m7% . Summarising then, to achieve mass flow without risk of blockage using the powdq

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.7

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

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.8

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

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

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.9

Figure 10.2.2: Estimation of crit from Powder Flow Function (Exercise-10.2).

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

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.10

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

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.11

Figure 10.4.2: Obtaining ff and from hopper flow factor chart ( = 60)

for Exercise 10.4.

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.12

Figure 10.4.3: Plotting of powder flow function and estimation of crit (Exercise 10.4).

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

SOLUTIONS TO CHAPTER 10 EXERCISES: HOPPER DESIGN Page 10.14

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

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

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