# 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

…