Etiket Arşivleri: FE 467

Food Processes ( Dr. Ali Coşkun DALGIÇ )

SECTION SIX FOOD PROCESSES

Principles of Food Process Design

Process Design is used in the design of entirely new (grass-roots) plants, but mostly in the expansion of existing production facilities, and in plant upgrading and modernization. New and improved processing facilities are necessary to meet the demands for new products, improved quality, process control, energy saving, and strict safety and environmental requirements. Economics and profitability must always be taken into account when designing new or upgraded processing plants.

Process and plant design are based on the following traditional engineering disciplines:

  • Chemical Engineering, which is concerned with the physical and engineering properties of materials, construction of process flowsheets, material and energy balances, equipment sizing, and plant utilities.

  • Mechanical Engineering, which deals with detailed engineering design of process and utility equipment, piping and material transport equipment, installation and maintenance of industrial equipment, and heating / air conditioning of industrial buildings.

  • Electrical Engineering is concerned with electrical power (e.g., motors), industrial lighting, process control, and automation of the processing plant.

  • Industrial Engineering is concerned with efficient plant operation, better utilization of material and labor resources, time-motion studies, and application of the various occupational and public health regulations at the local, state, and federal levels.

The design of food processes and food processing plants is based on the same principles of process (chemical) design, with the additional requirements for food safety and food quality. Food processes, processing equipment, and processing plants must comply with strict hygienic (sanitary) regulations.

Optimization of food processes is based on maximum preservation effect with minimum quality damage to the food product, and minimum health hazard to the consumers


Fish Oil Production ( MUS-BAY FISH OIL FACTORY )

  • FE 467 FOOD ENGINEERING DESIGN-2

  • Supervisor:

  • Prof. Dr. Mustafa BAYRAM

  • Fatma ATALAY

  • Mehmet Ali BACANAK

  • Tülay BAYIR

  • Ömer BEDDUR

  • Songül ÇÖREKÇİOĞLU

  • Rukiye ŞEN

  • MUS-BAY FISH OIL FACTORY

    Fish Oil Production

  • INTRODUCTION

  • ENTERING THE AMOUNT OF RAW MATERIALS:15,167 kg/h

  • THE RESULTING FISH OIL PRODUCT : 1500 kg/h

  • THE RESULTING FISH MEAL PRODUCT : 1127.52 kg/h

  • THE NUMBER OF EMPLOYEES : 102

  • THE NUMBER OF SHIFT: 2

  • RUNNING TIME : 300 DAY

  • SYSTEM : DAY/ CONTINUOUS

  • TOTAL CAPACITY : 23,017 TON/YEAR

  • CLOSED AREA : 10,030

  • TOTAL AREA : 23,150

  • AIM OF PROJECT

  • Production of fish oil and fish meal from fish

  • At a capacity 23,017 ton/year

  • High quality and healthy production fish oil

  • FLOW CHART OF FISH OIL

  • RAW MATERIAL

  • Fish caught for fish meal and fish oil production.

  • It must transport to factory immediately.

  • COOKING

    PRESSSING

The liquid is squeezed from the solid phase remove as much liquor higher oil yield .

  • PRESSSING

  • The liquid is squeezed from the solid phase remove as much liquor higher oil yield

  • DECANTER

  • Seperate the solids from the liquid in cooked fish.

  • CENTRIFUGE

  • The centrifuge is oil and water phase is seperated into oil and stick water.


Sunumunu bizlerle paylaşan Fatma Hanım özellikle öğrenci kardeşlerimize ışık tutması amacıyla paylaştığını belirtti. Bize de kendisinin bu notunu iletmek görevi düştü.

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
———————– Page 108———————–

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


Energy Management and Utilization in Food Plants ( Dr. Mustafa BAYRAM )

FE 467 FOOD PLANT DESIGN ENERGY MANAGEMENT AND UTILIZATION IN FOOD PLANTS

Prof. Dr. Mustafa BAYRAM University of Gaziantep Faculty of Engineering Department of Food Engineering Gaziantep/TURKEY

NOTE: From FE 366 DON’T FORGET • This semestry you have to prepare PID drawing • P&ID Diagram(YOU WILL DRAW THIS FOR YOUR PROJECT (Fe 467 Design II) The details of PID will be learned at FE 403 Food Process Control

ENERGY SOURCES • Liquids • Petrolium (Fuel oil-No:5 «Kalyak» or 6) • Waterdam (hydroelectric centralàelectiricty) • Ethly alcohol • Gas – LPG (Liq. petrolium gas: Propane+Butane «butane is heavy/high dense à red/ash fire)» – Natural gas – Hidrogen gas – Biogas (methane) • Solid • Coal • Solid waste • Wood

ALTERNATIVE ENEGRY SOURCES

Sun Wind Biomass Hydrogen Sea wave etc

Sun as an energy source • Sun collectors • Sun batteries • To produce; – Hot water – Steam – Electricity – Heated air Sun rock dryer (covered black dye or asphalt)

Wind • To produce electicity • Using wind turbine

Selection of energy sources • How can we select energy source? – Maintenance (coal needs more laboring, cleaning etc.) – Reparing – Calori value – Service life of heat converter machine – Economy (price, inflation etc.) – Availability – Constant price-constant supply – Waste and enviromental issue (coal ?) – Goverment policy and laws

Energy usage in the plant • For lighting (electiricty) • For motor power /Mechanical drive • Process Heating • Conditioning (indoor) (heating area etc.) • Air heating • Water heating • Steam generation (small scale) • Superheated steam • Material processing (cooking, pasteurization, sterilization, heatin of material, roasting etc.) • Cooling • Air • Water • Etc. • Space heating

Electricity sources in the plant • Energy Converter/Central •Natural gas central/converter •Coal central/converter •Water dam •Fuel/motorin central/convert

Electricity usage • In city • 220 V, 50 mHz (in Turkey, Europe etc.) • 110 V, 60 mHz (in USA) • 1-2 phases • Power (P, Watt): V (voltage) * I (ampere) • Energy consumption (E, kWh)= P * t (time) • In industry • 380 V , 50 mHz (in Turkey, Europe etc.) • 460 V, 60 mHz (in USA) • 3 phases • Power (P, Watt): V (voltage) * I (ampere) * √3 * Cos Ø • Energy consumption (E, kWh)= P * t (time)

Plug wires – colours Live – brown Neutral – blue Earth – green/yellow Fuse in live position

AC / DC Electricity So, in general, in industry, AC motor is used. Why??

In Turkey, 220 V in home

Three phases electricty

Utility Transmits at high voltage to reduce line losses 480, 600, 1K, 10K Vac Reduces voltage to useable levels Distributes current and protects circuits 120, 208, 277 Vac Heating cable Transformer Subpanel with circuit breakers generate heat

Decision and calculation: Energy/calori examination table: (Source: www.dosider.org)

Natural gas (pipeline in Turkey)-

Availabity of Natural gas in Cities

CONVERSION OF ENERGY

• For – Air heating – Water heating – Steam – Etc.

HEATING IN THE PLANT

Direct air heating Direct water heating Indirect air heating Indirect water heating

1- Direct air heating • A) Electricity (by using electrical resistance) – To process in duct/pipe + Cold air inlet Hot air outlet Resistant

B-Thermoblock (burner type) i-Ordinary thermoblock

ii-Turbo type thermoblock (back-pressure)

iii-Turbo type thermoblock (back-pressure)/ No Chimney

Indirect air heating

Circulating Air Systems • Heat distributed by an air stream through a heating unit to supply ducts • duct< 200’

2- WATER HEATING Direct and indirect 1. Direct

2.Indirect

Water Systems cooling tower off Heat is distributed by a pumped water system though a heating unit to supply piping boiler on Different piping arrangements are used – Hydronic Piping

INDUSTRIAL STEAM GENERATION

Sensible and Latent Heat • Sensible heat causes a change in temperature • Latent heat causes a change of state but no change in temperature

Calculation hints • Use Lower Heating Value (LHV) for energy source – The quantity known as lower heating value (LHV) (net calorific value (NCV) or lower calorific value (LCV)) is determined by subtracting the heat of vaporization of the water vapor from the higher heating value. This treats any H O formed as a vapor. The energy 2 required to vaporize the water therefore is not released as heat. – LHV calculations assume that the water component of a combustion process is in vapor state at the end of combustion, as opposed to the higher heating value (HHV) (a.k.a. gross calorific value or gross CV) which assumes that all of the water in a combustion process is in a liquid state after a combustion process. – The LHV assumes that the latent heat of vaporization of water in the fuel and the reaction products is not recovered. It is useful in comparing fuels where condensation of the combustion products is impractical, or heat at a temperature below 150°C cannot be put to use. – Higher heating value (HHV) (or gross energy or upper heating value or gross calorific value (GCV) or higher calorific value (HCV)) is determined by bringing all the products of combustion back to the original pre-combustion temperature, and in particular condensing any vapor produced. Such measurements often use a standard temperature of 25°C. This is the same as the thermodynamic heat of combustion since the enthalpy change for the reaction assumes a common temperature of the compounds before and after combustion, in which case the water produced by combustion is liquid.

Latent Heat of Evaporation 100 C 100 C • Evaporator: liquid changing to gas • Condenser: gas changing to liquid • It takes 180 Btu’s to raise the temperature of 1 lb. of water from 32 deg. to 212 deg. (100 C). • It takes 970 Btu’s to boil or vaporize 1 lb. of water at 212 deg. (100 C).

Training Agenda: Steam Introduction Steam distribution system Assessment of steam distribution system Energy efficiency opportunities

Introduction What is steam – Dryness fraction • Dry saturated steam: T = boiling point • Steam: mixture of water droplets and steam • Dryness fraction (x ) is 0.95 if water content of steam = 5% • Actual enthalpy of evaporation = dryness fraction X specific enthalpy hfg

Introduction • After obtaining saturated steam Saturated and super heated steam (in general it is used in all heating operation in food industry), saturated steam is heated with electirical resistant Superheated steam or hot oil or hot surface to obtain superheated steam (it is Sub-saturated water used for deodorization etc.)

Introduction Steam quality Steam should be available • In correct quantity • At correct temperature • Free from air and incondensable gases • Clean (no scale / dirt) • Dry

Classes of Steam • Low Pressure Heating Steam – 15 psig – Used Mainly for Space Heating Systems and Single Effect Absorption Chillers – Actual Code is More Restrictive • Medium Pressure Steam: 15-150 psig – Used in Hospitals, District Steam Systems, Some Industrial Heating • High Pressure: Above 150 psig – Strictly Industrial and Power Generating Applications • Each Class has Piping and Valve Requirements – Increase in Expense with Each Higher Class

Steam System Operation • Generation • Distribution • End Use • Condensate Recovery & Feed Water Systems

Steam Circle System Loads Radiators, Sterilizers Heat Exchangers, etc Steam Header Steam Traps BOILER Condensate Return Piping (Treated) Make up water Deaerator Condensate Return Tank

Generation • Boilers – Fire-tube or water-tube • Heat recovery generators – Turbine exhaust – Furnace exhaust Typical Boiler

Generation (2 types) • Water tube – fuel burned within combustion chamber – combustion gas surrounds water tubes within vessel • Fire tube – fuel burned in combustion chamber – combustion gases flow through tubes – water surrounds tubes – Scotch Marine – most popular Fire tube 44

Distribution & End Users • Distribution Systems – Distribution lines – Pressure reduction • Pressure reduction valve • Backpressure turbine • End Use Components – Heat exchangers – Mechanical drives – Steam sparging/injection equipment 45

Recovery & Feed Water • Condensate Recovery System – Steam traps – Collection tanks – Flash steam recovery – Pumps • Feed Water System – Deaerator – Economizer 46

BOILER EFFICIENCY HEAT LOSS METHOD BOILER EFFICIENCY = 100 – % AGE LOSSES 1. Heat Loss in Dry flue gas a. H = 0.24 w (T – T ) as percentage of heat input g g g a G.C.V a. H = K (T – T ) /1.8 K=0.32 for fuel oil g g a % CO2 in flue gas K=0.35 for Bituminous coal 2. Heat loss due to evaporation of moisture & H2 in fuel H = W +9H (100 – T ) + 540 – 4.6 (T – 100) %of heat input m m f g G.C.V 3. Heat loss due to moisture in air H = 0.26 W (T – T ) % of heat input a ma g a G.C.V 4. Heat loss due to Incomplete combustion to Co Hco = 2414 C x CO x 1 as % of heat input CO+CO2 G.C.V 5. Heat loss due to unburnt carbon “C” Hc = Wc x 7831 as % of heat input G.C.V

6. Heat loss due to Blow Down H = W (h – h ) as % of heat input bd b bw w G.C.V 7. Heat loss due to Radiation HR = Difficult to evaluate & thus take design values only In above Wg =Wt of dry flue gas W..G = [44.01 *CO2 + 32*O2 28.02 * N2 + 28.01*CO]*[Cb + 12.01 * S/32.07] 12.01 * (CO2 + CO) Tg = Tempt. Of flue gas at exit of Boiler Ta = Tempt. Of air at inlet (ambient) Tf = Tempt. Of fuel inlet h -h = Heat in blow down bw w Wm = Weight of moisture Wma = Wt of waterin Kg/Kg of air X Wt of air in Kg supplied / Kg of fuel Wc = Weight of unburnt “C” W = Wt of water blow down b All wts are / kg of fuel

Training Agenda: Steam Introduction Steam distribution system Assessment of steam distribution system Energy efficiency opportunities

Steam Distribution System Most important components 1. Pipes 7. Steam traps 2. Drain points 8. Air vents 3. Branch lines 9. Condensate 4. Strainers recovery 5. Filters system 6. Separators 10. Insulation

Steam Distribution System 1. Pipes • Pipe material: carbon steel or copper • Correct pipeline sizing is important • Oversized pipework: – Higher material and installation costs – Increased condensate formation • Undersized pipework: – Lower pressure at point of use – Risk of steam starvation – Risk of erosion, water hammer and noise • Size calculation: pressure drop or velocity

Steam Distribution System 1. Pipes • Pipeline layout: 1 m fall for every 100 m • Apply 5 degree angle for piping to remove/sliding condense (Spirax Sarco)

Steam Distribution System 2. Drain points Trap Pocket too small (Spirax Sarco)

Steam Distribution System 2. Drain points Properly Sized Trap Pocket (Spirax Sarco)

Steam Distribution System 3. Branch lines • Take steam away from steam main • Shorter than steam mains • Pressure drop no problem if branch line < 10 m A Branch Line (Spirax Sarco)

Steam Distribution System 3. Branch lines Branch line connections – Top: driest steam – Side or bottom: accept condensate and debris (Spirax Sarco)

Steam Distribution System 4. Strainers • Purpose – Stop scale, dirt and other solids – Protect equipment – Reduce downtime and maintenance • Fitted upstream of steam trap, flow meter, control valve • Two types: Y-type and basket type

Steam Distribution System 5. Filters • Consists of sintered stainless steel filter element • Remove smallest particles – Direct steam injection – e.g. food industry – Dirty stream may cause product rejection – e.g. paper machines – Minimal particle emission required from steam humidifiers – Reduction of steam water content

Steam Distribution System 6. Separators • Separators remove suspended water droplets from steam • Three types of separators Cyclonic type Coalescence type Baffle type

Steam Distribution System 7. Steam traps • What is a steam trap? – “Purges” condensate out of the steam system – Allows steam to reach destination as dry as possible

Steam Distribution System 8. Air vent – location • Within low lying steam trap opposite high level steam inlet • Opposite low level steam inlet • Opposite end of steam inlet

Steam Distribution System 9. Condensate recovery system • What is condensate – Distilled water with heat content – Discharged from steam plant and equipment through steam traps • Condensate recovery for – Reuse in boiler feed tank, deaerator or as hot process water – Heat recovery through heat exchanger

Steam Distribution System 10. Insulation • Insulator: low thermal conductor that keeps heat confined within or outside a system • Benefits – Reduced fuel consumption – Better process control – Corrosion prevention – Fire protection of equipment – Absorbing of vibration – Protects staff: hot surfaces, radiant heat

Steam Distribution System 10. Insulation Classification of insulators Temperature Application Materials o Low (<90 C) Refrigerators, cold / hot Cork, wood, 85% water systems, storage magnesia, mineral fibers, tanks polyurethane, expanded polystyrene Medium (90 – Low-temperature 85% magnesia, asbestos, o 325 C) heating and steam calcium silicate, mineral generating equipment, fibers steam lines, flue ducts, o High (>325 C) Boilers, super-heated Asbestos, calcium silicate, steam systems, oven, mineral fibre, mica, driers and furnaces vermiculite, fireclay, silica, ceramic fibre

Steam Distribution System Steam System Schematic

Example 1 Typical steam circuit

Example 2 –

Training Agenda: Steam Introduction Steam distribution system Assessment of steam distribution system Energy efficiency opportunities

Assessment of Steam Distribution System Three main areas of assessment • Stream traps • Heat loss from uninsulated surfaces • Condensate recovery

Training Agenda: Steam Introduction Steam distribution system Assessment of steam distribution system Energy efficiency opportunities

Efficient Steam Systems • Proper performance yields – Low operating costs – Minimal downtime – Reduced emissions – Effective process control • Effective maintenance is the best strategy!!

Energy Efficiency Opportunities 1. Manage steam traps 2. Avoid steam leaks 3. Provide dry steam for process 4. Utilize steam at lowest acceptable pressure 5. Proper utilization of directly injected steam 6. Minimize heat transfer barriers 7. Proper air venting 8. Minimize waterhammer 9. Insulate pipelines and equipment 10. Improve condensate recovery 11. Recover flash steam 12. Reuse low pressure steam

Why make a change? INEFFICIENT STEAM SYSTEM

Why make a change? EFFICIENT STEAM SYSTEM

Co-generator • To produce • Electricity • Hot water or steam From the engine

FOR PRESENTATION AND REPORTS • Show steam/hot water pipeline and ALL components • Calculate required steam/hot water for each machine separately • Calculate total steam/hot water quantity • Calculate total steam/hot water calories • Calculate total natural gas/electiricty/coal etc. «as energy source» quantity per hour, day, year and use them in feasibilty survey • Calculate boiler efficiency • Show in a table for; • Electricity • Energy

SOME IMPORTANT POWER CALCULATIONS FOR MOTOR/EQUIPMENT

EXAMPLES HOME STUDIES (SOME EXAM QUESTIONS) 1-Draw flow-diagram of a steam system, which includes the following components (use full page of your answer book, drawing is important, no partial grade will be given in this question) Pasteurizator (in parallel process) Jacketed tank (in parallel process) Steam distributor/head Steam traps Steam filters Condense tank Deaerator Water make-up/soft water system Pipes + Inclined pipes Air vents Valves Branch lines FLOW DIAGRAM !!!!!

2-In a food plant, overall steam requirement of the system is 5000 kg steam/hr at 120 C saturated steam. The efficiency of steam generator found in the plant is 85%. Condensed steam return at saturation temperature to the steam generator. According to following table, what quantity of the cheapest energy source is required (show all calculations)? Energy Source Heat value Unit cost Burning Efficiency Coal (C) 6500 kcal/kg 0.220 (TL/kg) 69% 3 3 Natural Gas (N) 8250 kcal/m 0.297 (TL/ m ) 93% Fuel-oil (No: 6) (F) 9200 kcal/kg 0.508 (TL/kg) 83%

Energy Burning TL/1000 Heat value Unit cost Source Efficiency kcal C 6500 0.220 0.69 0.0491 N 8250 0.297 0.93 CHEAPEST 0.0387 F 9200 0.508 0.83 0.0665 Natural gas is the cheapest: Required amount of natural gas: Real energy: 8250*0,93 = 7672,5 kcal/m3 3099620,04 kcal/h / 7672,5 kcal/m3 = 403,99 m3/h natural gas required.

(10 pts) Draw a steam generation and utilization system (in your drawing, show burner, steam generator, valves, termocouple, filters, piping, steam traps, one heating equipment/heat exchanger, condense tank, water treatment, flash etc),

1. (5 pts) Which energy source is the best according to following table? Energy source Energy value Unit price Burning yield (%) Diesel (Motorin) 10256 kcal/kg 2.805 TL/kg 84 Electricity 860 kcal/kWh 0.183 TL/kWh 99 Fuel oil (No:6) 9562 kcal/kg 0.95 TL/kg 80 Local Soma coal 4640 kcal/kg 0.348 TL/kg 65 LPG 11000 kcal/kg 3.100 TL/kg 92 3 3 Natural gas 8250 kcal/m 0.820 TL/m 93 Fe 478 Dr. Mustafa BAYRAM

Butter Production

FE 467
FOOD ENGINEERING DESIGN

BALIM BUTTER
Submitted by: ESRA KURTOĞLU
:MEHMET LÜTFİ GÜZEL

ADVISER : DR. MEDENİ MASKAN
process descrIptIon
Milk reception
Heat Pre-Treatment
Separation
Pasteurization
Ripening
Churning
Packaging
Storage
MİLK RECEPTION UNIT
Qualıty control of raw milk
color
odor
flavor
fat content
soluble solid content
ph
Microbial inspection
antibiotic
Milk filtration
Heat pre-treatment
SEPaRATION
Milk is separated into the cream and skimmed milk through the centrifugal seperator device
50 C for 1 hour at 5000 rpm
Pasteurization
RIPeNING
Churning prosess
Cream is converted into butter and butter milk
Churning process is done at 10 C and for 40 minutes
After 30 minutes % 1 by mass salt is added and churning is continued for 10 minutes(working )
PACKAGING
STORAGE
Butter is stored at 0 C between 2 and 4 weeks
Skimmed can be stored for several months without opening
Plant lay out
EquIpment lay out
MASS BALANCE
v Mass balanca around seperator:
A=200000L milk B=cream milk
Xa= 0,035 Xb=0,45
C=skimmed milk
Xc=0,0005
Total mass balance around seperator:
A = B + C
200000L*1,025KG/L = B + C
205000KG = B + C
Fat balance around seperator:
A*Xa=B*Xb+C*Xc
205000*0,035 = B*0,45 + (205000-B)*0,0005
B=15734,15 KG
C=189265,9 KG
Mass balance around churning:
B=15734,15 kg D=butter
Xb=0,45 Xd=0,82
E=butter milk
Xe = 0,0069
Mass balance
Total mass balance around churning:
B = D + E
15734,15= D + E
E = 15734,15 – D
Fat balance around churning:
B*Xb = D*Xd + E*Xe
15734,15*0,45 = D*0,82 + (15734,15-D)*(0,0069)
D=8574,35 KG
E=7159,8 KG
TANK DESIGN
Design Requirements
Volume of the tank
Height and diameter of the tank
Working pressure
Shell thickness
Head thickness
Bottom thickness
Balance tank (InsuLATED)
Keeps the product at a constant level above the pump inlet.
The head on the suction side is kept constant.
Volume of the tank :
V (volume of the milk) = 200000 L
1m3 = 1000 L
V (milk) =200 m3
safety factor is taken 20%
V tank = (200) * (1,2) =240 m3 è milk in 3 storage tank
ÒVolume of each tank: 240/3=80 m3
ÒHeight of each the tank :
V tank = π/4 *D2 * H
H = 4/3 D
V tank = π/3 * D3
80 m3 = π/3 * D3 è D = 4,24 m = 166,84 in
H =5,65 m
Working pressure :
P total = P optimum + H * g *ρ
= 1 atm + 5,65 m * 9.8 m/s2 * 1030 kg/m3 * 1atm * (1/101,3 * 103N/m2) * (1kg m /s2 N)
= 1 + 0,56
P total = 1,56 atm = 23 psi
Shell thickness :
ts = (PD + C) / (2Se – P)
S = Su * Fa * Fr * Fs * Fm
(Su = 9000psi for stainles steel type of 304)
S = 9000psi * 1 * 1 * 0,25 * 1
= 2250 psi = 153 atm
e = 0,80 è for double-buft joint
C = 1/16 inch = 1,58 * 10-3 m
Douter = 4,24+ 2* 0,0287
=4,2974 m
= 169,05 in
ÒHead thickness :
th = (P * L * W) / (2 * S * e)
W = 1,80
L (crown radius)= Di – 6 in
=166,84 – 6
=160,84 in =4,08 m
kr (knuckle radius)= 0.06*Do
=0.06 *169,05 in
=10,143 in
R= kr /L = 10,143/160,84 = 0,06
W = 1,8 (Values of Factor W for Dished heads table)
th = (1,56*4,08*1,80)/(2*153*0,8)
th = 0,0468 m = 46,8 mm
ÒBottom thickness : is equal to head thickness.
tb = 0,0468 m = 46,8 mm
PIPING DESIGN
PUMP POWER CALCULATION
HEAT BALANCE
PIPING DESIGN
•Determination of optimum diameter
•Evaluation of frictional losses
•Calculation of pump power
PUMP POWER CALCULATION
•CENTRIFUGAL PUMP ;
•In dairy process, centrifugal pump is used.
•Centrifugal pumps are commonly used to move liquids through a piping system.
•Centrifugal pumps are used for large discharge through smaller heads.
•POSITIVE PUMP ;
•UHT- steam injection section
•Define net amount of milk is pumped
•Flow of pump is controlled by regulating the speed

Properties of steel pipe ( Taken from Perry’s Chemical Engineering Handbook)
Mass of the milk = 13733,33 kg/h
ID = 0,05 m
Viscosity of milk, μ = 2,12 cp = 2,12 *10-3 Pa.s
Density of milk, ρ = 1030 kg/m3
Process time= 15 hour
Area of pipe = 1,96 * 10-3 m2
Volumetric flow rate, Q = (200m3 ) (1h / 3600s)/(15 h) = 5,6 * 10-2 m3/s
Velocity, v = Q / A = 1,89 m/s
•Reynold’s Number, Nre = (D*v*ρ)/μ
Nre =0.05 *1.89 *1030 /2.12*10-3 =45912,7 è Turbulent flow
•From Geankoplis, ɛ=4.6*10-5 for commercial steel pipe (figure2.10-3)
• ɛ/D = 4,6*10-5 / 0,05 = 9,2*10-4
•By using the NRe and ɛ/D we can find friction factor, f
f = 0.006 (figure 2.10-3 , Geankoplis)
Friction loss due to:
Straight pipe
Elbow
Valve
Expansion
Contraction
1) Friction losses in contraction
A2/A1=0(negligible)
α=1 for turbulent flow
hc=0.982J/kg
2) Friction losses in enlargement
V1 = 1.89 m/s
hex=1.78 J/kg
3-)Losses in fittings and valves
2 elbow 90 ° are used.
Le/D =Equivalent length of straightpipe in pipe diameter
Le/D = 35 (FromGeankoplis)
Le = 2 * 35 * 0.05 = 3.5 m
1 gatevalve (wideopen)
Le/D = 9 (FromGeankoplis)
Le = 1 * 9 * 0,05 = 0,45 m
Total length = ΔL
ΔL = 4 m + 3.5 m + 0,45 m =7.95 m
hf = Kf* V12/2α
Kf =0.75 α= 1 from table 2.10-1
hf = 0,75*3,57/2=1,33
(For valve and elbow)
ÒFf = 13,63 J/kg

∑F = 13,63 J/kg+1,33J/KG+0.982J/kg+1,78J/KG

∑F = 17,72J/kg
Δz = 3 m
Ws = 3 m * 9,8 m/s2 + 17,72J/kg =47,12J/kg
Ws = η * Wp η=0,80
Wp = 58,9 J/kg
58,9J/kg * 13733,33 kg/h * (1/3600 sec)* ( 1 kW/ 1000 W)
Wp = 0,224kW
Power of pump = 0,224 kW * (1 hp / 0,745)
=0,3 hP
HEAT BALANCES
Plate heat exchanger
• The plate heat exchanger consist of a series of paralel plates which are held together within a rigid frame.
•The plates are seperated by rubber gaskets to form narrow chambers between each pair of plates and exchange of heat takes place between chambers through each plate
FOR cream PASTEURIZATION
Mass flow rate = 13733,33 kg/h
Cp(hot water) = 4200 J / kg.K
Cp(cold water) = 4180 J / kg.K
Cp(milk) = 3900 J / kg.K
ρ milk (4°C) = 1030 kg / m3
ρwater(95°C) = 975.3 kg / m3
ρ water(2°C) = 1000 kg / m3
Heat supplied in regeneration = 75 % of total heat
Calculation of overall heat transfer coefficient ( U )
L= 0.87 m
W = 0.3 m
Δx = 0.003 m è Technology of milk book
A = 0.146 m2
Gap between plates = 0.0126 m
Deq = 2*a*b/(a+b)
Deq = 2*0.3*0.0126/ 0.3+ 0.0126 = 0.024 m
Thermal Conductivity of stainless steel (k) =16.3 W/mK
Q= 13733,33 kg/h *(1 hr/3600 sec)*(1/1030 kg/m3) = 3.7*10-3 m3/sec
Q=V*A 3.7*10-3 m3 /s= V*π (0.024²/4) ; V=8,18 m/s
ÒNre:DVρ /µ = 45912.7
Npr: Cp µ/k =16.5
Nu=h*De/k=0.2536(NRe )0.65*(Npr)0.4
Heat transfer coefficient of milk (hmilk) = 2906 W/m2K
Heat transfer coefficient of water (hwater) = 3000 W/m2K
Ureg. = 5700 W / m2.K
Uheating = 5814 W / m2.K
Ucooling = 5650 W / m2.K
•Qtotal =mmilk *Cpmilk *(Tholding – Traw milk)
•Qtotal = 13733,33 kg/hr * 1 hr/ 3600 sec*3900 J/kg.K *(90-40ᵒC) =7,4*105W
•Qreg. =0.75*Qtotal =5,55*105 W
•Qheating = 0.25*Qtotal =1,85 *105 W
For regeneration section :
Qreg. = m*Cp*ΔT
5,55*105 W = 3,81 kg/s * 3900 * (T1 – 40)
T1 = 77,4°C

5,55*105 W = 3,81 kg/s * 3900 * (90– T3)
T3 =52,6°C
Ò 90 °C
52,6 °C 12,6°C
12,6 °C 77,4 °C
40 °C
ΔTlm = 12,6 °C
Q = U * A * ΔTlm
5,55*105 W = 5700W/m2 K* A* 12,6 °C
A = 7,72m2
Q = U * A * ΔTlm
5,1 * 105 = 5650 * A * 12,33
A = 7,32 m2
STEAM GENERATOR
A boiler or steam generator is a device used to create steam by applying heat energy to water
STEAM NEEDED FOR PASTEURIZATION PROCESS
Q=(m*Cp*ΔT)milk =(m*Cp*ΔT)water
Cp(milk) = 3900 J / kg.K
Cp(hot water) = 4200 J / kg.K
Mass flow rate = 13733,33 kg/hr
13733,33 kg/hr * 3900 J / kg.K*(40-18)K
=mwater * 4200j/kg.K *(95-86,2)K
mwater = 31881kg hot water/h is used
Assumed water at 10 °C
Amount of steam used to heat 31881 kg water at 10 °C to 95 °C
Q=(m*Cp*ΔT)water =ms * hfg
Steam is at 150 °C and 5 atm (from Dairy Handbook)
hfg = (2749-640) kj/kg =2109kj/kg
Q=(m*Cp*ΔT)water =ms * hfg
31881kg/h*4.2kj/kg.K*(95-10)K=1,1*107 kj/h
1,1*107 kj/h=ms*2109 kj/kg
ms = 5216 kgsteam/h
Amount of natural gas for steam :
Low energy value for natural gas : 8250kcal for 1m3 nat.gas
200m3/h nat. gas*8250 kcal/m3/h nat. Gas
=1.6*106 kcal/h
EFFICIENCY:93%
=1.4* 106 kcal/h
ÒSteam production and
distribution system
SEPERATOR
BELT CONVEYOR
COLD STORAGE
CREAM SEPERATION
A cream separator is a device used to separate cream from milk by centrifugal force
For this purpose, centrifugal separator is used for making it possible to separate cream from milk faster and more easily.
DESIGN OF CENTRIFUGE
•Process time:60 min
mass flow rate= 20600kg/60min * 1min/60s
=5,7 kg/s
•ώ:Angular velocity à ώ=(2πN)/60
•N=5000-6500 rpm (from tech.of milk book)
•N= 5000 rpm
•N= 5000rev/min * 1/60 = 83 rev/s
•ώ =(2*3.14*5000)/60=523 rev/s
•Radius of centrifuge à
•r c2 = (ρ1r12 – ρ2r22) / (ρ1 – ρ2)
•Ρmilk:1030 kg/m3 , Ρcream =950 kg/m3
•rout:0.15 m (outer radıus of the plates to the axıs of rotatıon)
•rin : 0.05 m (inner radıus of the plates to the axıs of rotation)
•r c2 =[1030 x (0.15)2 – 950 x (0.05)2] / (1030 – 950)
r c= 0.5 m ( radius of centrifuge )
V:Volume of centrifuge à
•V =A*h
•V= π (r22-r12) h
•h (height) = 0.2m rout:0.15 m rin : 0.05 m
•V=3.14*0.2*(0.152-0.052)
•V=0.012 m3 = 12 lt
Pressure of centrifuge : ∆P à
• ∆P=1/2(rcream* ώ 2(rout2-rin2))
•∆P=1/2(950kg/m3*(523 s-1)2 *(0.152-0.052)m²)
•∆P = 105 kg/m*s2 = 1*105 Pa
Fc:Centrifugal force à
Fc=m*rc * ώ 2
=5.7kg * 0.5m * (523s¯¹)2 =780 kN
POWER REQUIREMENT
P = (m/2 ) * (2* π*N*r)2
=(5.7/2) * (2*3.14*83*0.15) 2
=17422w =17,4kw =18 HP
Particle diameter à Dp
Q=(W2 (rl-rp) Dp2 V) / (18.µ. ln(r2/ r1))
Q=5.7kg/s *1/1030kg/m3 = 5.5*10-3 m3/s
5.5*10-3 = (5232)*(1030-950)* Dp2* 0.012 / 18 * 2.12 * 10-3 *ln (0.15/0.05)
Dp = 2.9*10-5 m=30 µm
Number of disc bowl à
•Q angle:60° (angle of the plates to the horızontal axıs (ranges from 45 to 60 degrees )
•5.5*10-3 =
((1030-950)*(3*10-5) 2*2*3.14*S*(83)2*(0.153-0.053)) / (3*18*0.00212*Tan60)
S = 109 plates
CONVEYOR BELT
•Carrying capacity:
lt=length = 3 m
b0=belt width = 1 m
V= belt speed = 1 m/s
Driving efficiency = 0.9
b = 1030 kg/m3
a=cross-section of material = b02 / 11 = 0.09 m2
T (carrying capacity) = a * b * V
= 0.09 * 1030 * 1 =92,7 kg / s
•Motor capacity:
•We : Power required to drive the empty conveyor
•We : Power required to move the material/load against friction of the rotating parts
•We : Power required to raise/elevate or lower the load
•WT = We + Wm ± Wr
•We = mi ( lt + 0.1*lt ) * g * µe * V
µe = friction coefficient= 0.03
mi = 60 * b0 = 60*1= 60 kg /m
•We = 60* (3 + 0.1 * 3) * 9.8 * 0.03 * 1
= 0.058kW = 0.079 hp
• Wm = T * lt * g * µm µm =0.04
= 92,7 * 3 * 9.8 * 0.04
=0.109 kW = 0.14 hp
• WT = We + Wm = 0.079 + 0.14 = 0.219 hp
MC (Motor Capacitiy) = 0.219/ 0.9 = 0.24 hp
COLD STORAGE ROOM
CALCULATIONS
PROPERTIES OF ROOM
•Tin=+4 °C
•Tout=30°C (average temperature of Balıkesir)
•Height of cold storage room = 6m
•Width of cold storage room =17m
•Length of storage room =51m
•Volume of cold storage = 51*17*6
= 5202m3
REFRIGERATION LOAD
1. Transmission load, which is heat conducted into the refrigerated space through its walls, floor, and ceiling;
2. Infiltration load, which is due to surrounding warm air entering the refrigerated space through the cracks and open doors
3. Product load, which is the heat re­moved from the food products as they are cooled to refrigeration temperature
4. Internal loads, which is heat generated by the lights, electric motors, and people in the refrigerated space
1. Transmission Load
Q=U*A*∆T
Q = heat loss
A= area of cold storage room
∆T= temperature difference between
outside and inside of room
U = overall heat transfer coefficient
wall
§For west wall , east wall and south wall
Δx polyurethane= 0.20 m
Δx ytong= 0.15 m
k polyurethane= 0.12 W/m.°K
k ytong= 0.13 W/m.°K
h outside= 22.7 W/m².°
h inside=9.37 W/m².°K
U= 0.3365 W/m².°K
Q= U*A*(To-Ti)
Q=0.3365*20*(30+3-4)
Q=195W
§For north wall
No solar effect
U=0.336 W/m².°K
Q=U*A*(To-Ti)
Q=0.336*20*(30-4)
Q=170W
§Roof
Ò2.Infiltration Load
Qf = (0.7*V + 2)ΔT
where Qf = heat flow, W
V = volume of room in m³
ΔT = temperature difference between room and ambient
Qf =(0.7*867+2) (30-4)
Qf = 15831,4W
3)Product Load
m=7555 kg/day * 1day/24*60*60 s
= 0,087 kg/s
Tin= +4 °C
Tproduct= 10 °C
Cp =3900 J/Kg.K
Q = m.Cp.∆T
Q = 0.087kg/s*3900 J/Kg.°K*(10-4)°K
Q = 2035 W
HYGIENE AND SANITATION
CIP
WASTE TREATMENT
COST REVISION
HYGIENE AND SANITATION
The arrangements for cleaning equipment that comes in contact with products are
an essential part of food processing plant.
HYGIENE AND SANITATION
Food manufacturers
should always keep İn mind the cleaning
obligations of equipment
and staff involved in
production to maintain
high hygienic standards.
TYPES OF DIRTS ON SURFACES
IN DAIRY MANUFACTURING
WASTE TREATMENT
Liquid waste = Waste water
sedimantion
Total Capital Investment
Total Product cost
Total Income
Gross & Net Profit
Cumulative Cash Position
Break Even Point
Profitability
Feasibility
Total Cap. Inv.= Fixed Cap. Inv.+Working Cap. Inv.
Fixed Cap. Inv.= Direct Cost + Indirect Cost
Purchased Equip. Cost = PEC + AEC + LEC
Where;
PEC – Process Equip. Cost
AEC – Auxiliary Equip. Cost
LEC – Lab. Equip. Cost
Process Equip. Cost
Auxiliary Equ. Cost
PE Cost = $ 4,180,354
Capital Estimation
WCI = 8,286,270 $
TCI = 24,754,892 $
Total Product Cost
=
General expenses + Manufacturing cost
Manufacturing cost = Direct Product Cost + Fixed charges + Plant Overhead Cost
General Expenses = Adm. Expenses + Dist. & Marketing + R & D
DIRECT PRODUCT COST
Raw material cost:-
-Raw milk: 200,000LPD *320 D/y*0.27$ = $ 17,280,000/y
-Salt: 180 kg/d * 320d/y *0.17$ = $ 9,792/ y
-Culture: 0.091 $/L * 16,221LPD *320d/y = 472,356 $/y
è Raw material cost = $ 17,762,148/y
Energy:
-Natural gas: 466,667 m^3/y * 0.447 $ = $ 208,732/y
-Electricity: 863 KW/d * 320 d/y = $ 276,160/ y
è Energy cost = $ 484,892/y
èTotal Product Cost
=
$ 33,145,083/y
Total Income
-Skim-milk : 58,801,087 L/y * 0.7 $/L =$ 41,160,760/y
-Butter : 2,720,000kg/y * 8.7$/L = $23,664,000/y
-Butter milk : 2,291,200 L/y *0.22$/L =$ 504,064/y
Total income = $ 65,328,824/y
Gross Earning = Total Income – Total Prod. Cost
Gross earning = $ 32,183,741/y
Net Profit
The required tax rate issued from the gov. is 20% of the gross earning.
According to the government’s incentive system for Balikesir, the gov. will pay back 60% of the 20% tax already has been paid after we -the company- earn half of the TCI.
The required tax to be paid regardless the incentive = $32,183,741/y * 0.2= $6,436,748/y
After earning ½ of the TCI the gov. will pay back 60% of the $ 6,436,748/y which is $3,862,050/y
è The obligatory tax is: 6,436,748 – 3,862,050
=
$ 2,574,698/y
èNet Profit = 32,183,741– 2,574,698
=
$29,609,043/y
Cumulative Cash Position
CCP = (Net profit+dep)X-TCI;
CCP= 0.83 y
Break Even Point
(Total Income/ton) (X) = (Total Product Cost/ton) (X) + Fixed Cost/y
997X = 506X + 588,755
èX = 1,197 Ton/y
1,197 Ton/65,576 Ton * 100= 1.82 % , i.e, the B.E.P is reached at 1.82 % of the present working capacity.
PROFITABILITY
Prof. = (total income – total prod. Cost)/total prod.cost *100

Profitability = 97 %
ÒFeasibility
S = TCI*(1+int.rate)n
The overall project is feasible

Sunflower Oil

FE 467

FOOD ENGINERING Design II

Name of company: İpek Yağ

Date of establısh: 2013

Type of process: SUNFLOWER OIL PRODUCTION

Area of the plant: 11.000 m² closed area 30.000 m² total area

Capacity

Operating time

NUMBER OF EMPLOYEES

760 tons (seed)/day

300 tons edible (refined) oil/day

24 hours/day
256

ENERGY REQUIREMENTS

COAL , STEAM, ELECTRICITY