solids handouts 10

8
HANDOUT 10-1  xx σ  θ   xy σ   yx σ  yy σ   yy σ   xy σ   yx σ   xx σ  (a) Normal Shear Min Max (b) MAX NORMAL NORMAL AXIS    S    H    E    A    R    S    T    R    E    S    S    A    X    I    S MIN NORMAL MOHR CIRCLE (c)

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8/2/2019 Solids Handouts 10

http://slidepdf.com/reader/full/solids-handouts-10 1/8

HANDOUT 10-1

 xxσ   

θ   

 xyσ   

 yxσ 

 yyσ   

 yyσ   

 xyσ   

 yxσ   

 xxσ 

 (a)

Normal

Shear

Min

Max(b)

MAX NORMAL

NORMAL

AXIS

   S   H   E   A   R

   S   T   R   E   S

   S   A   X   I   S

MIN NORMAL

MOHR

CIRCLE

(c)

8/2/2019 Solids Handouts 10

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HANDOUT 10-2

Normal Stress

   S   h  e  a  r   S   t  r  e  s  s

Critical Point

(End Point)

Normal Stress

   S   h  e  a  r   S   t  r  e  s  s

c f   

Normal Stress

Critical Point

(End Point)

1σ   

JYL

   S   h  e  a  r   S   t  r  e  s  s

JYL

JYL

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HANDOUT 10-3

Normal Stress

   S   h  e  a  r   S   t  r  e  s  s

Critical Points

δ   

MFF

Non-cohesive

Less cohesive

More cohesive

Multiple JYL

curves

c f   

1σ   

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HANDOUT 10-4

0

10

20

30

40

50

0 10 20 30 40 50 60 70

Semi-included angle, degrees

   W  a   l   l   F  r   i  c   t   i  o  n ,   d

  e  g  r  e  e  s

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

δ=30

δ=40

δ=60

δ=50

δ=60

δ=40

δ=30

δ=50

δ=60

δ=50

δ=45

δ=40

δ=35

δ=30

   F   l  o  w   F  a

  c   t  o  r ,          f          f

Flow Factor Curves

Wall Friction Curves

 

Figure 10-17. Design chart for symmetrical slot outlet hoppers. For example (dashed arrows),

and gives and

o22=wδ o50=δ  o5.30=θ  19.1= ff  .

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HANDOUT 10-5

0

10

20

30

40

0 5 10 15 20 25 30 35 40 45

Semi-included angle, degrees

   W  a   l   l   F  r   i  c   t   i  o  n ,

   d  e  g

  r  e  e  s

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

δ=30

δ=40

δ=60

δ=50

δ=60

δ=40

δ=30

δ=50

δ=60

δ=50

δ=40

δ=35

δ=30

   F   l  o  w   F  a  c   t  o

  r ,          f          f

Flow Factor Curves

Wall Friction Curves

 

Figure 10-18. Design chart for conical outlet hoppers. For example, and gives

and .

o22=wδ o50=δ 

o5.20=θ  29.1= ff 

 

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HANDOUT 10-6

Steps on using the shear stress data to design a hopper.

1.  Rotating Shear Test

Plot SHEAR STRESS

vsNORMAL STRESS

Internal Friction

Get f c vs σ 1Get δ  

δ 

SHEARSTRESS

   f c  σ 1 

NORMAL STRESS

JYL

SHEAR

STRESS

 wδ 

2.  Rotating Shear Test

Wall material

Get wδ 

 

NORMAL STRESS

3.  Fit φ w and δ to hopper correlation

Get θ and ff .

(θ is the theoretical

angle of the hopper;in final design subtract

3 degrees for margin of 

safety)

4.  Plot 1/  ff on mff curve ( f c vs σ 1) to get CAS

Get CAS

δ 

δ 

wφ 

 ff 

θ  

MFFCAS

 f c

Slope 1/  ff  

σ 15.  Use CAS and θ in correlations to select opening size.

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HANDOUT 10-7

 D  

θ  Semi included angle

For conical hoppers, Figure 10-20, the opening

diameter, , is given by D

c

ogg

CAS H  D

 / )( ρ 

θ = (10-12)

602)(

θ θ  += H  (10-13)

Where θ  is in degrees, from the charts in

Figures 10-17 or 10-18. Typical values for H are

about 2.4.

Figure 10-20. Conical Hopper with outlet size D 

and semi included angle θ  .

 L

 W  

For symmetrical slot outlet hoppers the opening

size is determined from

c

ogg

CAS H W 

 / )( ρ 

θ = (10-14)

180

1)(θ 

θ  += H  (10-15)

W  L 3> (10-16)

Figure 10-21. Symmetrical slot outlet h

of opening size W x L.

opper

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HANDOUT 10-8

10.4.1 COARSE PARTICLES (particles > 500 microns in diameter)

MASS FLOW – JOHANSON EQUATION

)()1(2 θ  ρ 

Tanm

 Bg Am

o

+=& (10-17)

where θ  = semi included angle of the hopper

m& = discharge rate (kg/sec)o ρ  = bulk density (kg/m3)

g = gravity acceleration (9.807 m/s2)

Table 10-3. Parameters in the Johanson Equation, Eq.(10-18)

Parameter Conical hopper Symmetric slot hopper

B  D, diameter of outlet W  

A 2

4 D

π  

WL

m 1 0

FUNNEL FLOW – BEVERLOO EQUATION

5.25.0 )(58.0  p

okd  Dgm −= ρ & (10-18)

where d  p = particle diameter (m)

k = constant, typically 1.3 < k < 2.9 with k = 1.4 if not

discharge rate data are available.

)perimeteroutlet()areasectionalcross(4= D  

10.4.2 FINE PARTICLES (d  p < 500 microns)

CARLETON EQUATION

g

 B

 p p

oo=+

35

34

32

31

2

15sin4

 ρ 

µ  ρ θ (10-19)

o

o AV m ρ =&  

where = average velocity of solids discharging (m/s)oV  A, B = given in Table 10-3

µ  ρ , = air density and viscosity

 p ρ  = particle density

o ρ  = bulk density of the powder bed