FILTRATION AND COLD STORAGE CALCULATIONS
İSMAİL GENÇ

Filtration is made to remove the suspended pectin residues inside of the juice for clarification

Pectinase is used for break down of the pectins for clarification
α-(1-4)-linked D-galacturonic acid
(PECTIN)

Feed Rate= 1600 kg/h*(1m3/1050 kg)= 1.49 m3/h

CS= Mass of solids deposited on the medium per unit volume of filtrate

CS=(32 kg/h)/(1.49 m3/h)= 21.47 kg solid/ m3 filtrate

Operation time= 1h =3600 sec

Operation pressure= 5 bar =5*105 Pa

Assume radius of one plate = 0.25 m

Area of one plate= π* (0.25 m)2 = 0.196m2

If the volume of filtrate versus time per volume of filtrate graph is plotted according to the filtration data, intercept point will give B and slope will give Kp/2

KP=

Where;

µ: Viscosity in Pa*s

α: Specific cake resistance

CS: Mass of solids deposited on the medium per unit of volume of filtrate

A: Area of one plate

∆P: Pressure drop in Pa
B=

Where;

Rm: Medium Resistance

By using the data of filtration including the volume of filtrate and time period to collect that filtrate; graph is plotted to obtain the specific cake resistance and medium resistance

µ= 5.24*10-3 Pa.s

Equation of line : y = 0,106x + 1,308

Slope =0.106*106 s/m6 =KP/2

KP=212000 s/m6

KP=

α = 5.81*1011

B=

Rm= 97975.2 m-1

Where;

t: time of operation in seconds

V: volume of filtrate for operation time in m3

A= 6.32 m2 which is total required filtration area

Number of plates= Total area/Area of one plate
=32.24 plates

33 plates can be used for this operation

Power required=3.345 kW

Feed Rate= 1200 kg/h*(1m3/1061 kg)= 1.11 m3/h

CS=21.62 kg solid/ m3 filtrate

Operation time= 1h =3600 sec

Operation pressure= 5 bar =5*105 Pa

Assume radius of one plate = 0.25 m

Area of one plate= π* (0.25 m)2 = 0.196m2

µ= 6.58*10-3 Pa.s

Equation of line : y = 0,210x + 1,83

Slope =0.210*106 s/m6 =KP/2

KP=420000s/m6

KP=

α = 9.09*1011

B=

Rm= 109160.3 m-1

A= 6.64 m2 which is total required filtration area

Number of plates= Total area/Area of one plate

 = 33.87 plates

34 plates can be used for this operation

Cold Storage Calculations

Product storage in barrels

Barrel capacity=250 lt

392 kg pekmez production /h

Assume 30 days for transportation

392 kg pekmez/h*24h/day*30 days= 282240 kg maximum product storage should be mentioned.

282240 kg/(250 kg/barrel)=1129 barrels should be stored

1129barrels*250 lt=307250 about 307 m3 storage area required for product.

If we think about the space between barrels and movement spaces

Storage volume should be approximately 500 m3

Product temperature = 25 ˚C

Storage temperature= 5 ˚C

Average temperature for a year in Kilis= 17 ˚C

Dimensions of the storage room= 5*10*10 = 500 m3

According to the knowledge from Food Engineering Operations book, heat load from all causes other than heat generation, wall leakage and miscellaneous losses can be taken as approximately 0.003*V0.6 kj/s for each ˚C difference between the storage temperature and the outside temperature under standard storage conditions where V is the volume of the storage room in m3.

Conventional British practice leads to wall leakages of about 7-8 W/m2
Heat Losses
*Heat of product
*Wall Leakage
*Miscellaneous load
*Heat load from other causes such as from walls, roof, floor etc. which are related with the transfer between the storage room and the outside.
Heat of product
392 kg/h* 24h = 9408 kg for one day
Cp of pekmez =4.18*Xwater+2.5*Xsugar (kj/kg.K)
=4.18*0.28 + 2.5*0.72
=2.97 kj/kg.K
Qproduct=392 kg / h* (1h/3600s) *(2.97 kj/kg.K)*(25-5)K
=6.47 kW

Wall leakage

Area of walls= 4*5*10+10*10= 300 m2

Wall leakages= 7-8 W/ m2

 =8 W/ m2*300 m2

 =2.4 kW

Miscellaneous Load

Qforklift=500W*10 =5kW

QWorker=20* 270W=5.4 kW

Qlights=8*36 W=288 W for 4 hours of light opening= 1.15 kW

Qml=5+5.4+1.15=11.55 kW

Heat loss by other effects

Qother=0.003*V0.6

 =0.003*(500)0.6 =0.125 kW for each ˚C difference

12 ˚C difference

Qother=12*1.15 kW= 1.5 kW

Total loss=6.47+2.4+11.55+1.5

 =21.92 Kw

Apply 10% safety: 21.92 *1.1= 24.1 kW

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