Confidential

Question 1

Filtration is a unit operation common in the food processing industry.

a) Briefly describe the application of filtration in food processing.

(2.5 marks)

Filtration is a process that removes suspended solids from a liquid. Filtration are used

to filter the water supply for food processing operations, boiler feed water, treatment of

liquid food processing waste. In the food processing industry fruit mash are filtered to obtain

fruit juices. Finer filters are used to obtain sparkling juice.

b) Name three types of filtration system, draw and label the schematic of the

equipment.

(7.5 marks)

The common types of filtration system are:

i) Bed or depth filter.

These are used to filter process water. In the simplest form it will consist of a

top layer of fine sand, a middle layer of coarser sand and a layer of stones.

ii) Filter press

The filter press consist of plates with flow channels covered by filter cloth.

These are arranged such that feed are pumped between the filter cloth

1

Confidential

causing solid particles to be trapped on the filter cloth. Once the spaces

between the filter cloth are filled with filter cake or if the pressure drop

exceeds the operating point the filtration is stopped and the filter cake

removed.

iii) Leaf filter

2

Confidential

iii) Centrifugal filter

iv) Rotary vacuum filter

v) Membrane filter

c) The mean pressure drop of a filtration system is 0.5 bar. Calculate the time to obtain

1000 kg/h filtrate from a suspension containing 0.1 kg solids/kg water. The filtration

area, A is 0.628 m2 the cake porosity ε is 0.5 and the cake resistance R is 1.42 x 10 5

(m/kg).. The water density is 1000 kg/m3 and the filtrate viscosity is 0.01 Pa.s.

(10 marks)

3

Confidential

t=( ηRC2 ΔP )(VA )2

η= filtrate viscosity (Pa.s)

R = filter resistance (m/kg)

C = solids concentration (kg/m3 of filtrate)

ΔP= pressure drop through filter (Pa)

V= filtrate volume (m3)

A = filter area (m2)

ΔP=5x104 ; η=0.01 ; R = 1.42x105 ; ρ=1000; F=1000/3600

V=F/ρ=1000/(1000*3600)=0.00027 m3/s

C=Fxc=1000*0.1=100

A=0.628

t=((0.01x1.42x105x100)/(2 x 5x104))(0.00027/0.628)2

t=2624.8 s = 0.73 hrs

Question 2

a) Name, draw and label the schematics of three types of heat exchangers as used in

the food industry.

(7.5 marks)

4

Confidential

i) Tubular heat exchanger

ii) Plate heat exchanger

iii) Scraped surface heat exchanger

b) 100 kg/h of a food sauce is to be pasteurised from a temperature of 30⁰C to 85⁰C in

a plate heat exchanger. Hot water flowing counter current to the sauce flow enters

at 98⁰C and leaves at a temperature of 85⁰C. If the overall heat transfer coefficient

5

Confidential

for the heat exchanger is 75 W/(m2.K) and the specific heats of the sauce is 3800

J/(kg.K) and the specific heats of the hot water is 4180 J/(kg.K), calculate

i) The hot water flow rate required, and

(5 marks)

Mf=100; cpf =3800; Tif= 30; Tio=85

Qf =100*3800*(85-30)=20900 kJ/h

Mw = Qf/(cpwxΔTw)

= 20900/(4.18x(98-80))

= 277.8 kg/h

ii) The heat transfer area required.

(7.5 marks)

Q = UxAxΔTlm

U = 75;

Q = 20900/3600 = 5800 W

ΔTlm = (ΔT1-ΔT2)/(ln(ΔT1/ΔT2))

ΔT1 = 98-85 =13

ΔT2 = 80-30 =50

ΔTlm = (13-50)/ln(13/50) = 27.5

A =Q/(UA ΔTlm) =5800/(75x27.5) =2.8 m2

Question 3

a) The jacketed stirred tank is a common piece of equipment used in the food

processing industry. Draw and label the schematic of a jacketed stirred tank and

briefly described some uses of the jacketed stirred tank.

(5 marks)

6

Confidential

b) A heating process using a 1000 litre jacketed stirred tank is used to cook a mixture

containing 200 kg of roselle calyx and 800 litre of sugar solution. The specific heats of

the mixture was estimated to be 3.2 kJ/(kg.K), the density 1100 kg/m3 and the

overall heat transfer coefficient was estimated at 200 W/(m2.K). Condensing steam

at 105⁰C is used to heat the jacketed tank.

i) Calculate the area for heat transfer available if the ratio of tank

diameter to height is 1:1.25

(7.5 marks)

H=1.25D; V=πD2H=1.25π x D3

Solving for D D=3√ V1.25 π

= 3√ 11.25 π

= 0.63 m

A = π x D2 x H = π x 1.25 x D3 = 1.25π x 0.633 = 9.82 m2

ii) Calculate the time required to heat the contents from 35⁰C to 95⁰C

assuming that the volume of the mixture is 1000 litres.

(7.5 marks)

7

Confidential

T s−TT s−T 0

=expUAtMC

To=35 T=95 Ts=105 U=200

A=9.82 M=1100 C=3200

t= ln ((105-35)/(105-95))/((200 x 9.82)/(1100 x 3200)

t=3171 seconds

Question 4

a) Draw and label the schematic of a vapour compression refrigeration system.

(5 marks)

b) The refrigeration load of a cold store maintained at -5⁰C was estimated at 25 kW.

The temperature difference between the evaporator surface and the refrigerant in

the evaporator is 10⁰C. The refrigerant properties are as follows: BP -41⁰C, ΔH r =

166.6 kJ/kg , Cpl = 2.5 kJ/(kg K), and Tc = 40 ⁰C. If the refrigerant condensing

temperature is 40⁰C, calculate:

i) The refrigerant flow rate required for R-22.

(5 marks)

8

Confidential

Qe=[∆H r−C pl(Tc−Te )]F r

Qe=50 ;∆ H r=166.6 ;C pl=2.5;T c=313 ;T e=258

Fr = 25/((166.6-2.5(313-258)) = 0.86 kg/s

ii) The compressor power (kW)

(5 marks)

Qc=[∆H r(T c−T eT e )]F r

Qc=[166.6 ( 313−258258 )]0.86=24.15 kW

iii) The COP (coefficient of performance)

(5 marks)

COP=T e

(T c−T e )−

(C plT e )∆ H r

COP= 258313−258

−2.5×258166.6

=0.82

Question 5

a) Briefly describe the thermal process use for canning.

(5 marks)

b) The table below is the heat penetration data for the canning of tuna in curry.

Calculate the FO value for the process. The steam temperature in the retort is 121˚C.

The reference temperature is 121˚C and z=10.

Can Center Temperature 75 90 102 110 117 118 119 120 121 121 121

Time (mins) 5 7.5 10 12.5 15 17.5 20 22.5 25 27.5 30

9

Confidential

(10 marks)

Tr 121 z 11Time T L Δ t CL

(mins) ⁰C (T-Tr)/z 10^((T-Tr)/z)) 2.55 75

7.5 90 82.5 -3.85 0.000141 0.000353 0.00035310 102 96 -2.5 0.003162 0.007906 0.008259

12.5 110 106 -1.5 0.031623 0.079057 0.08731615 117 113.5 -0.75 0.177828 0.44457 0.531886

17.5 118 117.5 -0.35 0.446684 1.116709 1.64859520 119 118.5 -0.25 0.562341 1.405853 3.054448

22.5 120 119.5 -0.15 0.707946 1.769864 4.82431225 121 120.5 -0.05 0.891251 2.228127 7.05244

27.5 121 121 0 1 2.5 9.5524430 121 121 0 1 2.5 12.05244

Fo 12.05244

c) If Do for the organism is 1.5 mins, and m the reduction factor is 5,

i) calculate the F required for commercial sterility.

ii) Is the process given by the data in the table above adequate? State

your reasons.

d) Desired F=mD;‘m=5; D=1.5; F=5x 1.5 = 7.5

(2.5 marks)

The process yielded F of 12.05 whereas the value for commercial sterility is 7.5. Therefore the process is more than adequate. In fact steam should be stopped after 27.5 minutes of heating.

(2.5 marks)

Question 6

Tapioca chips enters a dryer at 1 tons/h having an initial moisture content of 90%

(wb) and is dried to a final moisture content of 9% (wb). The drying air enters at 30⁰C

and has a moisture content of 0.010 kg/kg db and enters the drying chamber at 65⁰C

and 25% relative humidity. The air leaving the dryer has a relative humidity of 70%.

The equilibrium moisture content is estimated at 0.08 kg/kg(db)

10

Confidential

a) Draw the psychrometric chart showing the inlet and outlet conditions of the air

leaving and entering the dryer and

(6 marks)

Point 1 Tdb=30, RH =38%, Y=0.01

Point 2 Tdb=65, RH =6.5%, Y=0.01

Point 3 Tdb=32; RH=70%; Y=0.025

b) From the chart establish the enthalpies of the drying air entering and leaving the

drying chamber.

(4marks)

Enthalpy at point 1 = 65 kJ/kg db

Enthalpy at point 2 = 94 kJ/kg db

c) If the equilibrium moisture content is 0.08 kg/kg(db) and drying time constant is

0.81 hours calculate the total drying time.

(5 marks)

Xo=90/10=9; Xe=0.08; X=9/91=0.099; tc = 0.81

t¿−t c ln [ X−X eXo−Xe ]

t=-(0.81)ln((0.099-0.08)/(9-0.08)=4.98 hrs

d) Calculate the volumetric air flow rate for the dryer.

(5 marks)

Mass balance: Wa(Ho-Hi)=Ws(Xi-Xo)

Hi=0.01; Ho= 0.035 ; Xi=9; Xo=0.099

11

Confidential

Mass flow of air: Wa=Ws(Xi-Xo)/(Ho-Hi)

= (1000*0.1)(9-0.099)/(0.035-0.01)

= 32004 kg/h

Volumetric air flow:

Specific volume of air v =0.875 m3/kg

F=32004 x 0.875 =28003.5 m3/h=7.78 m3/s

12