# Solid Storage ( Dr. Mustafa BAYRAM )

# SOLID STORAGE (SILO DESIGN AND WAREHOUSE)

# FE 467 FOOD PLANT DESIGN

# PROF. DR. MUSTAFA BAYRAM UNIVERSITY OF GAZIANTEP FACULTY OF ENGINEERING DEPARTMENT OF FOOD ENGINEERING GAZIANTEP- TURKEY REV : FEB 2015 FE 467 Dr. Mustafa BAYRAM 1 MBAYRAM@GANTEP.EDU.TR

# TYPES OF SOLID STORAGE Open area • Over soil/ground • Under soil/ground By using PP, PE sheet By using pool, hill Closed area • Warehouse (concrete, steel, wood) • Silo (concrete, steel, wood)

# TYPES OF SOLID STORAGE BULK • Horizontal warehouse as bulk and packaged • Vertical silo (concrete, steel) • Cylindirical • Cubic • Hexagonal PACKAGED • In bag (25, 50 kg bags, box, bigbag «1 ton, 2 tons») with palette or not

# HORIZONTAL WAREHOUSE/STORAGE

# BULK / WAREHOUSE

# PACKAGED TYPE- WAREHOUSE

# Bag-Bigbag-Palette

# BULK SILO – Concrete – Wood – Steel/Iron/Galvanized (smooth surface or corrugated/hadveli) – Cylindirical – Conical bottom – Flat bottom – Square – Hexagonal

# TYPES OF SILOS – Steel/Iron/Galvanized (smooth surface or corrugated/hadveli) – Cylindirical – Conical bottom

# Unloading of silo with free flowing grain/flat silo Main slide electrical driven Auxiliary slide gate manually operated UNLOADING STEPS 1

# Flat bottom-Mass discharge “Non free flowing materials”: Soja cake, Rape cake, DDGS, etc Special reinforced silos and MASS DISCHARGE SYSTEMS Silo top view

# CONSTRUCTION OF SILO

# First roof, From top to bottom Top sheet thickness is less than the bottom

# COMPARISON OF WAREHOUSE AND SILO

# Thichness of wall Hygene Cleaning Aeration Fast construction Price Service life

# ADVANTAGES OF STEEL SILO

# COMPARISON OF CONICAL AND FLAT BOTTOM SILOS

# IMPOSTANCE OF GALZANIZATION

# STANDARDS OF SILO DESIGN

# SPECIAL ROOF STRUCTURES

# WALL SHEETS AND STIFFNERS IMPORTANCE

# SPECIAL EQUIPMENTS IN SILO

# SPEED REDUCER FOR FRAGILE PRODUCTS D Pistachio nut d1 Conical apparatus H angle d2 h FIGURE 3. HELEZONIC FIGURE 2. Conical flowing down of free (SCREW) type system for free flowing material (CONICAL speed reducer) flowing material (SCREWspeed reducer) FIGURE 4. Photo for screw system used in nut industry for filling silo to prevent 3 breaking kernel (screw type speed 5 reducer)

# Wall sliding speed reducer:

# Wrong discharge and silo damage

# If you prefer only two pipe for side discharging from wall; You can use the following procedures:

# FLAT BOOTOM SILO CONSTRUCTION General view for flat bottom silo

# BASIC INFORMATION ABOUT SILO SYSTEMS

# Discharging of flat bottom silo /tunnel

# SILO DESIGN

# TYPES OF BINS Conical Pyramidal Watch for in- flowing valleys in these bins!

# Chisel TYPES OF BINS Wedge/Plane Flow L B L>3B

# Figure 7: Hopper forms [9]

# THE FOUR BIG QUESTIONS What is the appropriate flow mode? What is the hopper angle? How large is the outlet for reliable flow? What type of discharger is required and what is the discharge rate?

# MATERIAL CONSIDERATIONS FOR HOPPER DESIGN Is it a fine powder (< 200 microns)? Is the material abrasive? Is the material elastic? Does the material deform under pressure?

# FLOW MODES Mass Flow – all the material in the hopper is in motion, but not necessarily at the same velocity Funnel Flow – centrally moving core, dead or non- moving annular region Expanded Flow – mass flow cone with funnel flow above it

# MASS FLOW D Does not imply plug flow with equal velocity Typically need 0.75 D to 1D to enforce mass flow Material in motion along the walls

# FUNNEL FLOW “Dead” or non- flowing region

# Funnel flow can result from : too rough and/or too shallow hopper walls, a feeder which discharges the bulk material only from a part of the outlet opening, a valve or gate which is not totally open, or from edges or welds protruding into the bulk solid.

# EXPANDED FLOW Funnel Flow upper section Mass Flow bottom section

# MASS FLOW (+/-) + flow is more consistent + reduces effects of radial segregation + stress field is more predictable + full bin capacity is utilized + first in/first out – wall wear is higher (esp. for abrasives) – higher stresses on walls – more height is required

# FUNNEL FLOW (+/-) + less height required – ratholing – a problem for segregating solids – first in/last out – time consolidation effects can be severe – silo collapse – flooding – reduction of effective storage capacity

# HOW TO INVESTIGATE THE FLOW PROFILE: Check the top of the silo. If homogen falling occurs (flat shape) àmass flow, or notàfunnel flow Use thin rod (short and long at different silo position, near or far position on te center of silo) Use different colored particles by controlling their discharge orders

# PROBLEMS WITH HOPPERS • Ratholing/Piping • Funnel Flow • Arching/Doming • Insufficient Flow • Flushing • Inadequate Emptying • Mechanical Arching • Time Consolidation – Caking

# PROBLEMS WITH HOPPERS Ratholing/Piping

# RATHOLING/PIPING •If only the bulk solid above the outlet is flowing out, and the remaining bulk solid – the dead zones -. •The reason: the strength (unconfined yield strength) of the bulk solid. Stable Annular •If the bulk solid consolidates Region increasingly with increasing period of storage at rest, the risk of ratholing increases. •If a funnel flow silo is not emptied completely in sufficiently small regular time intervals, the period of storage at rest can become very large thus causing a strong time

# PROBLEMS WITH HOPPERS Ratholing/Piping Funnel Flow

# FUNNEL FLOW -Segregation -Inadequate Emptying -Structural Issues e e s s r r a a o o C e C n i F

# PROBLEMS WITH HOPPERS Ratholing/Piping Funnel Flow Arching/Doming

# ARCHING/DOMING •If a stable arch is formed above the outlet so that the flow of the bulk solid is stopped. •In case of fine grained, cohesive bulk solid, the reason of arching is the strength (unconfined yield strength) of the bulk solid which is caused by the adhesion forces acting between the particles. •In case of coarse grained bulk Cohesive Arch preventing solid, arching is caused by material from exiting blocking of single particles. hopper •Arching can be prevented by sufficiently large outlets.

# PROBLEMS WITH HOPPERS Ratholing/Piping Funnel Flow Arching/Doming Insufficient Flow

# INSUFFICIENT FLOW – Outlet size too small – Material not sufficiently permeable to permit dilation in conical section -> “plop-plop” Material under flow compression in the cylinder section Material needs to dilate here

# PROBLEMS WITH HOPPERS Ratholing/Piping Funnel Flow Arching/Doming Insufficient Flow Flushing

# FLUSHING Uncontrolled flow from a hopper due to powder being in an aerated state – occurs only in fine powders (rough rule of thumb – Geldart group A and smaller) – causes –> improper use of aeration devices, collapse of a rathole

# PROBLEMS WITH HOPPERS Ratholing/Piping Funnel Flow Arching/Doming Insufficient Flow Flushing Inadequate Emptying

# INADEQUATE EMPTYING Usually occurs in funnel flow silos where the cone angle is insufficient to allow self draining of the bulk solid. Remaining bulk solid

# PROBLEMS WITH HOPPERS Ratholing/Piping Funnel Flow Arching/Doming Insufficient Flow Flushing Inadequate Emptying Mechanical Arching 0

# MECHANICAL ARCHING a “traffic jam” at the outlet of bin – too many large particle competing for the small outlet 6 x dp,large is the minimum outlet size to prevent mechanical arching, 8-12 x is preferred 1

# PROBLEMS WITH HOPPERS • Ratholing/Piping • Funnel Flow • Arching/Doming • Insufficient Flow • Flushing • Inadequate Emptying • Mechanical Arching • Time Consolidation – Caking 2

# TIME CONSOLIDATION – CAKING Many powders will tend to cake as a function of time, humidity, pressure, temperature Particularly a problem for funnel flow silos which are infrequently emptied completely 3

# SEGREGATION •In case of centric filling, the larger particles accumulate close to the silo walls, while the smaller particles collect in the centre. •In case of funnel flow, the finer particles, which are placed close to the centre, are discharged first while the coarser particles are discharged at the end. .

# HOW IS A HOPPER DESIGNED? Measure – powder cohesion/interparticle friction – wall friction – compressibility/permeability Calculate – outlet size – hopper angle for mass flow – discharge rates Two steps are necessary for the design of mass flow silos: 1-The calculation of the required hopper slope which ensures mass flow, and the determination of the minimum outlet size to prevent arching. 5 2-

# WHAT ABOUT ANGLE OF REPOSE? Pile of bulk solids a a a 6

# ANGLE OF REPOSE Angle of repose is not an adequate indicator of bin design parameters “… In fact, it (the angle of repose) is only useful in the determination of the contour of a pile, and its popularity among engineers and investigators is due not to its usefulness but to the ease with which it is measured.” – Andrew W. Jenike Do not use angle of repose to design the angle on a hopper! (if you have data). Use inlined slope angle 7 FE 478 0 Dr. Mustafa BAYRAM 1

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# WALL FRICTION TESTING RESULTS t , Wall Yield Locus (WYL), s variable wall friction s e r t s r a e Wall Yield Locus, h s constant wall friction l l a W j’ Normal stress, s Powder Technologists usually express m as the “angle of wall friction”, j’ j’ = arctan m 8

# STRESSES IN HOPPERS/SILOS Cylindrical section – Janssen equation Conical section – radial stress field Stresses = Pressures a) Pressure in a silo filled with a fluid (imaginary); b) Vertical stress after filling the silo with a bulk solid; c) Vertical stress after the discharge of some bulk solid

# • If the height to diameter ratio of the silo is sufficiently large (usually: > 3), a constant vertical stress is attained. • This means that the vertical stress will not increase further even if the filling height is much larger. • The reason for this course are the shear stresses acting between the bulk solid and the silo walls even if the bulk solid is at rest. • Due to the shear stresses, the silo walls carry a part of the weight of the bulk solid.

# • The stresses acting in a hopper are different from those in the vertical section. • Just after filling an empty silo, the so called filling stress state (also: active stress state, figure b) prevails, where the vertical stress in the hopper decreases less in the upper part of the hopper and then more near the imaginary hopper apex. 1

# •In the emptying stress state the vertical stresses in the lower part of the hopper are nearly proportional to the distance to the imaginary hopper tip or, in other words, the stresses are proportional to the local hopper diameter. This linear course of stress is called the radial stress field •In principle, in the vertical section of the silo the stresses remain unchanged at discharge. 2

# STRESSES IN A CYLINDER Consider the equilibrium of forces on a differential element, dh, in a straight- sided silo P A = vertical pressure acting from v P A v h above h d rA g dh = weight of material in element D dh p t (P + dP ) A = support of material from v v below (P + dP ) A v v t p D dh = support from solid friction on rA g dh the wall D (P + dP ) A + t p D dh = P A + rA g dh v v v 3

# STRESSES IN A CYLINDER (CONT’D) Two key substitutions t = m Pw (friction equation) (Pw:shear stress at wall to the lateral normal stress actingin the radialdirection at wall) Janssen’s key assumption: Pw = K Pv This is not strictly true but is good enough from an engineering view. Substituting and rearranging, A dPv = rA g dh – m K Pv p D dh Substituting A = (p/4) D2 and integrating between h=0, Pv = 0 and h=H and P = P v v Pv = (r g D/ 4 m K) (1 – exp(-4H mK/D)) This is the Janssen equation. 4

# STRESSES IN A CYLINDER (CONT’D) hydrostatic Bulk solids Notice that the asymptotic pressure depends only on D, not on H, hence this is why silos are tall and skinny, rather than short and squat. 5

# STRESSES – CONVERGING SECTION Over 40 years ago, the pioneer in bulk solids flow, Andrew W. Jenike, postulated that the magnitude of the stress in the converging section of a hopper was proportional to the distance s of the element from the hopper apex. s = s ( r, q) This is the radial stress field assumption. 6

# The Flow Function 7

# DETERMINATION OF OUTLET SIZE B = s H(q)/g c,i H(q) is a constant which is a function of hopper angle 9

# DETERMINATION OF OUTLET SIZE c Time flow function s sc,t Flow function sc,i Flow factor s1 0

# H(q) FUNCTION 3 ) q ( H 2 1 10 20 30 40 50 60 Cone angle from vertical 1

# EXPERIMENTAL SILO OUTLET SIZE DETERMINATION Generally, the rate of discharge (Q) is related to the orifice diameter (D) by an equation of the form: n Q = k D Log Q= log k + n . Log D where k is proportionality constant and n is a power of about 2.5 – 3.0. It is generally found that the head of material over the A straight line should result from orifice has no detectable effect on the rate which the values of k and n may of discharge. be determined and hence the Plot the calcu1ated discharge rates against equation relating Q and D the orifice diameter on a log-log basis: evaluated. 2

# DISCHARGE RATES Numerous methods to predict discharge rates from silos or hopper For coarse particles (>500 microns) Beverloo equation – funnel flow Johanson equation – mass flow For fine particles – one must consider influence of air upon discharge rate 3

# BEVERLOO EQUATION W = 0.58 r g0.5 (B – kd )2.5 b p where W is the discharge rate (kg/sec) 3 r is the bulk density (kg/m ) b g is the gravitational constant B is the outlet size (m) k is a constant (typically 1.4) dp is the particle size (m) Note: Units must be SI 4

# JOHANSON EQUATION Equation is derived from fundamental principles – not empirical W = r (p/4) B2 (gB/4 tan q )0.5 b c where qc is the angle of hopper from vertical This equation applies to circular outlets Units can be any dimensionally consistent set Note that both Beverloo and Johanson 5

# SILO DISCHARGING DEVICES Slide valve/Slide gate Rotary valve Vibrating Bin Bottoms/Vibrating Grates others 6

# ROTARY VALVES Quite commonly used to discharge materials from bins. 7

# SCREW FEEDERS Dead Region Better Solution 8

# VIBRATIONAL DISCHARGE EQUIPMENTS Air cannons Pneumatic Hammers Vibrators These devices should not be used in place of a properly designed hopper! They can be used to break up the effects of time consolidation.

# DESIGN DATA

# AERATION OF SILO

# AERATION Correlation equations for airflow resistances through grain 1) The equation for the low velocity range of 0.004 to 0.05 m/s is: 2) The equation for the low velocity range of 0.05 to 0.35 m/s is:

# 3) Alternative formula:

# MASS AND FUNNEL FLOW REGIMES GRAPH FOR SOME SILOS skip Figure 4b: Design diagram for mass flow (conical hopper)

# skip Figure 4a: Design diagram for mass flow (wedge-shaped hopper)

# CONVEYING SYSTEM FOR SILO 8 08 Şubat 2015 Pazar4

# CONVEYOR SELECTION TABLE

# CHAIN CONVEYOR

# SCREW CONVEYOR Ms= A’ * Us * r * E b f A’= cross sectional area, available for flow (not occupied by screw and shaft) Ef= filling efficiency (within conveyor) (area occupied by the bed of solids) (compared to the cross-sectional area) Us= solids velocity inside a screw conveyor (Us) is the product of the blade pitch and revolutions per second) Us= Rev/s * pitch

# BELT CONVEYOR Work (ton/h)= 3600 f * v * r * f Power (PS)= (W * l * k) / 305 (except belt power) f= cross sectional area of product along width of belt, m2 V= belt velocity, m/s r= density of product ton/m3 f= distrubition coeffient of product (for dense: 1, loose:0.9, 0.8 ..) l= lenth of belt, m k: constant (frictional), 0.15-0.30

# Elevator: L:length of leg of elevator (~belt) X: distance between two dishs U belt= <0.5 m/s Ubelt: velocity of belt Width of dish= <0.5 m V= volume of each dish D:Diameter of pulley M= weight of material in a dish Number of dish= L/X Ubelt= motor drive (rev/sec) * 3.14 D/rev L= Ubelt * t Number of dish * X = U belt * t 4 3 2 FE 478 M= r * A * V = kg/m * m * m/s = kg/m * m/s

# VIBRO CONVEYOR spring Vs < 0.4 m/s conveyor velocity motor

# PNEUMATIC CONVEYOR

# EXAMPLES (HOME STUDY-EXAM QUESTIONS) L 1) a) Find h, L and volume of silo. (Bulk angle of repose: o 35 . o Sliding angle of repose: 45 . Diameter of silo: 2 m. Weight of product in silo: 7000 kg. Head space: 10%. Bulk density of 3 product: 700 kg/m . Volume of conical part: 1/3 2 *(3.14*r *height) h b) We want to discharge this silo within 30 mins. Calculate required discharge diamete n r) (Hint: Q=k.D ). Given; k Diameter of 0.1 0.3 0.5 discharger (m) Mass flow rate 1.80 1.88 1.97 (kg/sec) 8

# DISCHARGE RATE – EXAMPLE An engineer wants to know how fast a compartment on a railcar will fill with polyethylene pellets if the hopper is designed with a 6” Sch. 10 outlet. The car has 4 compartments and can carry 180000 lbs. The bulk solid is being discharged from mass flow silo and has a 65° angle from horizontal. Polyethylene has a bulk density of 35 lb/cu ft. 0

# DISCHARGE RATE EXAMPLE One compartment = 180000/4 = 45000 lbs. Since silo is mass flow, use Johanson equation. 6” Sch. 10 pipe is 6.36” in diameter = B 3 2 W = (35 lb/ft )(p/4)(6.36/12) (32.2x(6.36/12)/4 tan 25)0.5 W= 23.35 lb/sec Time required is 45000/23.35 = 1926 secs or ~32 min. In practice, this is too long – 8” or 10 “ would be a 1

# SILO CAPACITY AND SIZE DESIGN Note: In this example: Sliding slope angle is same with angle of repose. 2

# EXAMPLE: CALCULATION OF A HOPPER GEOMETRY FOR MASS FLOW An organic solid powder has a bulk density of 22 lb/cu ft. Jenike shear testing has determined the following characteristics given below. The hopper to be designed is conical. Wall friction angle (against SS plate) = j’ = 25º Bulk density = g = 22 lb/cu ft Angle of internal friction = d = 50º Flow function s = 0.3 s + 4.3 c 1 Using the design chart for conical hoppers, at j’ = 25º qc = 17º with 3º safety factor & ff = 1.27 5

# EXAMPLE: CALCULATION OF A HOPPER GEOMETRY FOR MASS FLOW ff = s/s or s = (1/ff) s a a Condition for no arching => s > s a c (1/ff) s = 0.3 s + 4.3 (1/1.27) s = 0.3 s + 4.3 1 1 s = 8.82 s = 8.82/1.27 = 6.95 1 c B = 2.2 x 6.95/22 = 0.69 ft = 8.33 in 6

# EXAMPLE – (20 pts.) In a food plant, we want to transfer semolina from a machine at 2nd floor (upstair) to another machine at 1st floor (downstair) using a pipe. The bulk density of semolina is 700 3 kg/m . The diameter of pipe is 10 cm. The height between the floors is 5 m. 30% of pipe is only full at cross-section with semolina during transportation (see figure). The velocity of o product is 2 m/s in the pipe. The angle of repose of semolina is 45 . The product is out from the bottom of the machine found on 2nd floor and then inlet at the bottom of the machine found on 1st floor. a) (10 pts.) Calculate possible minimum horizontal distance between the machines and show the positions of both machines by drawing. b) (7 pts.)Calculate the mass flow rate of semolina in the pipe. c) (3 pts.)What is the capacity of the machine (Machine 2) under this condition at steady- state (no by-product, in=out)? Machine 2 2nd floor pipe Semolina level in pipe (%30) Machine 1 1st floor

# Machine 2 2nd floor pipe H=5 Semolina level in pipe (%30) Machine 1 45o 1st floor X=? Tan 45= H/X=1è X=5 m a) (7 pts.)Calculate the mass flow rate of semolina in the pipe. m=(bulk density)*(Cross sectional area of filled portion)*(velocity) 2 2 2 Cross section of whole pipe: 3.14 * (r )= 3.14 * (0.05 ) =0.00785 m . 2 Cross sectional area of filled portion “%30”= 0.00785 * 0.30= 0.002355 m . m=(bulk density)*(Cross sectional area of filled portion)*(velocity) m=700*0.002355*2 = 3.297 kg/s b) (3 pts.)What is the capacity of the machine (Machine 2) under this condition at steady- state (no by-product, in=out)? It is same with themass flow rate. The produced product flows in the pipe: 3.297 kg/s

# 3-In a food plant, the capacity is 30 tons/hr. You have a silo of 100 tons to feed the plant. The product found in 3 the silo is soybean and its particle size is 5 mm. Its bulk density is 800 kg/m . The flow in silo is funnel flow. o The angle of hopper from vertical is 45 . Gravitational constant=9.8. a)What is the silo discharge orifice diameter?, b)How long will it take to feed the plant by using this silo? Hints: Beverloo Equation; Johanson Equation; 0.5 2.5 2 0.5 W = 0.58 r g (B – k d ) W = r (p/4) B (g B/4 tan q ) b p b c Units must be in SI unit W is the discharge rate, r is the bulk density, g is the gravitational constant, B is the outlet size, k is a constant b (typically 1.4), d is the particle size, q is the angle of hopper from vertical. p c 9

# b) 100 t / (30 t/h) = 3.33 hr 0

# EXAMPLE Draw the flow-diagram of full silo system 1

# 1. (10 pts) Design a silo to storage 1000 kg bread wheat (h=6D). Use given table. (Compact, Columbus). Silo is cylindrical. 2 Volume of conical part: 1/3 * (3.14*r *h) 2from mass flow silo and has a 65° angle from horizontal. Lentil has a bulk density of 35 lb/cu ft. Use given Hint. Hints: Beverloo Equation; Johanson Equation; 0.5 2.5 2 0.5 W = 0.58 r g (B – k d ) W = r (p/4) B (g B/4 tan q ) b p b c Units must be in SI unit W is the discharge rate, r is the bulk density, g is the gravitational constant, B is the outlet b size, k is a constant (typically 1.4), d is the particle size, q is the angle of hopper from p c vertical. 3

# ANNEX ABOUT SILO STANDARD AND CALCULATION

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