Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.0 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering 5. Silo and hopper pressure calculations 5.1 shaft and hopper pressures 5.2 Wall thickness …

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.1 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering 5.1 shaft and hopper pressures

Vertikal and Horizontal Pressures at Mass Flow Silo

height level of flattenedfree bulk surface

intersection betweenshaft and hopper

height level of discharge opening

fillingdischarging

silo

heig

ht H

silo pressures pv and ph

pv

pv

ph

ph

pv

ph

radial stress field r∝σ

F 5.1

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.2 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.3 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.4 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering

posit

ion

of ac

tive s

hear

pla

ne

Active and Passive Rankine's Stress State Limits

Yield locus: τ = tanϕi ·σ +τc = tanϕi ·(σ + σZ) or σR = sin ϕi ·(σM + σZ)

ph,a

activepassive

shea

r st

ress

τ

tensile strength σZ

cohesion τc

ϕi

σ2,aαa=

π ϕi4 2+

ϕiσR ,a

2·αa

pvσ1,a

yield locus

ph,p

σ1,p

αp =π ϕi

4 -

2

σM,p =σ1,p + σ2,p

2

position of passive shear plane

σR,p=

σ1,p σ2,p

2

-

active passive

principal stresses

lateral orhorizontal stress ratio

σ2,a = · σ1+ · τc1 - sin ϕi

1 + sin ϕi

2 · cos ϕi1 + sinϕi

σ1,p = · σ2,p+ · τc1 + sin ϕi

1 - sin ϕi

2 · cos ϕi

1 - sinϕi

lower limit for τc = 0 upper limit for τc = 0

ph,p

pv= = = tan21 + sin ϕi

1 - sin ϕi(π

4 +ϕi

2λp

ph,a

pv= = = tan21 - sin ϕi

1 + sin ϕi( π

4 -ϕi

2λa

·

·

·

active

π ϕi4 2+

pv

ph,aτ pv

ph,p

passive

π ϕi4 2

-τ

pv

ph,aτ

σ

pv ph,p

τσ

normal stress σ

given: pv = σ1 = ρb.g y τc, ϕi

searched: 2 stress states which meet the yield condition

y

x

σ2,p

F 5.4

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.5 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.6 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.7 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.8 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering

storage time tL = 22 hmFill = 11,5 tρb = 0,8 t/m3

ϕe = 33°ϕw = 23° steel sheet (CF)ϕw = 18° steel sheet (MF)

mass flow, switch load ∆pn = (c4 - c1) pn · cos Θ,TGL 32274/09

mass flow, switch load ∆pn = ρb· g · (H or D),DIN 1055 part 6

0 5 10 15 20 25 30 35 40

wall normal pressures ph, pn in kPa

0

1

2

3

4

5

heig

ht H

, H' i

n m

mass flow (MF)

core flowflow bondary

siloG 807

pn

Wall Normal Pressures ph, pn of Wheat

measured, filling measured, dischargingcalculated, fillingcalculated, discharging, load factors c1 = 1,4, c4 = 3,0calculated, DIN 1055 part 6

c3 = 1.8 (CF)

23°

30°

ph

D = 2,4 m

measurements according to Scholz

F 5.8

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.9 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering

Maximum Normal Pressures (Discharging) versus Effective Angle of Internal Friction

123456789

1011

ϕw= 10°

boundariesrough wall

ϕw=10°20°30° 40° 50°

0 10 20 30 40 50 60

p*h shaft

p*n hopper

20°

30°

40°50°60°

p*h ,

max

, p

* n,

max

12

ϕe in deg

Maximum Vertical Pressures (Filling) versus EffektiveAngle of Internal Friction

p*v,

max

0 10 20 30 40 50 60

12

345

678

9

1011

ϕw= 10°

20°

30°

40°

50°

boundaryrough wall

Θ = 10° mass flow =0,725

= 2,414

AUHD

shafthopper

ϕw=10°

20°30° 40° 50°

1213

14

ϕe in deg

Maximum Vertical Pressures (Filling) versus Wall Friction Angle

0 10 20 30 40 50 60

12

34

5

678

910

11

ϕe=30° 40° 50°

1213

isostatic pressure

boundary rough wall

p*v,

max

14

shafthopper

60°

ϕw in deg

Maximum Normal Pressures (Discharging) versus Wall Friction Angle

12

34

56789

10

11

30°40°

50°

0 10 20 30 40 50 60

boundariesrough wall

p*h ,

max

, p

* n,

max

12

1314

60°ϕe=30°

ϕw in deg

60°50°40°

p*h shaft

p*n hopper

Maximum Wall Friction Loads (Discharging)versus Effective Angle of Internal Friction

0 10 20 30 40 50 60

ϕw=10°

20°

30°

40°

-50°

p*h shaftp*n hopper

20°

10°

30°

20°

40°

50°

-10°

ϕe=30°

1

2

1,5

2,5

0 0

0,5

0,6

0,7

0,4

0,3

0,2

0,1

p*w

, m

ax

boundariesrough wall

0,5

60°

40°50°

p*w

, m

ax

ϕe in deg

p*w

, m

ax

p*w

, m

ax

Maximum Wall Friction Loads (Discharging) versusWall Friction Angle

ϕe=30°40°

50°

50°40°

0

0,5

1

1,5

2

2,5

0

0,5

0,6

0,7

0,4

0,3

0,2

0,1

boundariesrough wall

0 10 20 30 40 50 60

60°

60°

ϕe=30°

ϕw in deg

p*w shaftp*w hopper

p* = p U ρb g A

.. .

F 5.9

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.10 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering

12

34

5

678

p*v,

max

9

1011

ϕw= 10°

20°

30°40°

50°

boundaryrough wall

Θ = 30° core flow =0,725

= 2,414

AUHD

shafthopper

ϕw=10°

20°30° 40° 50°

0 10 20 30 40 50 60ϕe in deg

Maximum Vertical Pressures (Filling) versus EffektiveAngle of Internal Friction

12

34

5

678

910

11

ϕe=30° 40° 50°

0 10 20 30 40 50 60

12

13

shafthopper

isistatic pressure

boundary rough wall

p*v,

max

60°

ϕw in deg

Maximum Vertical Pressures (Filling) versus Wall Friction Angle

12

345

678

9

10

11 ϕw= 10°

boundariesrough wall

ϕw=10°

20°

30°40° 50°

0 10 20 30 40 50 60

p*h shaft

p*n hopper

20°30°40°50°

60°

p*h ,

max

, p

* n,

max

60°

ϕe in deg

Maximum Normal Pressures (Discharging) versus Effective Angle of Internal Friction

12

345

678

910

11

ϕe=30° 40° 50°

0 10 20 30 40 50 60

ϕe=30°40°50°

60°

60°

boundariesrough wall

p*h ,

max

, p*

n, m

ax

ϕw in deg

p*h shaft

p*n hopper

Maximum Normal Pressures (Discharging) versus Wall Friction Angle

ϕe=30° 40°50°

50°

40°ϕe=30°

0 0

Ip* w

, m

ax

0,5

0,6

0,7

0,8

0,9

1,0

0,4

0,3

0,2

0,10,5

1

1,5

2

2,5

3

p*w

, m

ax boundariesrough wall

0 10 20 30 40 50 60

60°

60°

ϕw in deg

p*h shaft

p*n hopper

Maximum Wall Friction Loads (Discharging) versus Wall Friction Angle

3

ϕw=10° 20°

30°

40°

50°

20°

10°

30°

20°30°40°

50°

10°

ϕe=60°

1

2

1,5

2,5

00 10 20 30 40 50 60

0

0,5

0,6

0,7

0,8

0,9

1,0

0,4

0,3

0,2

0,1

Ip* w

, m

ax boundaries rough wall

0,5

ϕe in deg

p*w

, m

ax

p*h shaft

p*n hopper

Maximum Wall Friction Loads (Discharging)versus Effective Angle of Internal Friction

p* = p U ρb g A

.

. .

F 5.10

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.11 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering 5.2 Wall thickness

Necessary Steel Reinforcement of Concrete ASt

Di,1

pn

σϕ

ASt

H

Di

Di,2

Θ

pnσϕ

Di = Di,1 + Di,2 2

(1)

(2)

(3)

(4)

ΘcosF,Stσ2jcsiDnp

ASt⋅⋅

⋅ϑ⋅⋅=

HAσ2cosΘ/iDHnp0FSt

⋅⋅ϕ⋅−⋅⋅==∑

⋅ϕ⋅−−

⋅⋅ϕ⋅

⋅ϑ⋅⋅⋅=

iDH

wtanF4λexp1F,StσcosΘwtan8jcs

2iDgbρ

StA

load factors according to TGL 32 274/09

( ) Θcoscc1c 14or3j −+=cj = c1 oder c3 oder c4

with c1 = 1,2 ... 1,6 c3 = 1,7 ... 2,1 c4 = 2,1 ... 4,0

hoppershaftdischargingcore flowmass flow

F 5.11

Fig_PC&P_5 Lecture Product Characterization and Processing of Pharmaceutical Particulate Solids, Silo pressure calculations , Jürgen Tomas 29.05.2013

Figure 5.12 Prof. Dr.-Ing. habil. J. Tomas – chair for Mechanical Process Engineering