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PROBLEMS OF FOOD PROCESS ENGINEERING
BLOCK I
Problem 1
The process to obtain oil from soya seeds is the following:
- By milling and pressing, a cake is obtained with an oil content of 6%
- In an extraction operation with hexane the cake becomes flour with 0.5 % of oil.
- Flour is dried till the amount of water in the final product is 8%
For 100 kg of beans:
a) How much oil and water will be separated in each operation? b) What composition will the final product have?
Data: Composition of soya beans: 35.0% proteins, 27.1% sugars, 18.0% oil, 10.5% water and 9.4 % fibre and inert
Solution: a) 12.77 kg of oil. 4.82 kg of oil. 4.25 kg of water. b) 44.78% protein; 34.67% sugars; 12.03% fibre; 8.00% water; 0.53% oil
Problem 2
In order to desalinize sea water, a process of reverse osmosis is used according to the following flow chart:
Sea water
Sea water has a salt concentration of 32 kg salt/m^3 and it is extracted with a flow of 900 L/h. The stream that is coming in the reverse osmotic unit has a salt concentration of 4% in weight. The flow of non- recycled brine is 500 kg/h.
Find out the flow of drinkable water if the maximum content of salt is 0.4 g/L What is the concentration of brine that goes out from the reverse osmotic unit? What percentage of brine is recycled? Data: density of sea water: 1020 kg/m^3 and density of drinkable water: 1000 kg/m^3
Solution: 418 kg/h drinkable water; salt mass fraction: 0.0573; 47.8%.
Problem 3
In a hot-air drying installation it is known that:
900 m^3 /h of air comes in the drying chamber with: Pressure=765 torr, temperature=60ºC, relative humidity=10% When air leaves the drying chamber it has: Pressure= 760 torr, temperature= 41ºC, relative humidity=40% The Raw material comes in the drier with a moisture content of 65% (wet basis) and the dehydrated product goes out with 35% (wet basis)
Drinkable water
Brine
Reverse Osmosis Unit
Calculate the capacity of the dryer (kg raw material/h) and its production (kg dehydrated product/h) Data: Steam pressure of liquid water at 60ºC=149.38 torr, at 41ºC=58.34 torr. Molecular weights: Water= 18; dry air=29. Consider that steam and dry air behave as ideal gases.
Solution: capacity: 14.92 kg/h; production: 8.03 kg/h
Problem 4
A product with a moisture content of 1.562 kg of water/kg of dry matter is taken to a drying installation and is dried up to a water content of 0.099 kg of water/ kg dry solid. 52.5 kg of air are introduced in the dryer per each kg of dry matter. Air coming out the drying chamber has an absolute humidity of 0. kg of water/kg dry air. How much air will be recycled if ambient air has an absolute humidity of 0.0152 kg of water/ kg of dry air in that place?
Solution: 0.235 kg recycled air/kg of air coming out
Problem 5
It is required to design an installation of continuous osmotic dehydration to prepare eel fillets. The production line will be fed with 100 kg/h of unsalted fillets with the following composition in weight: water 52.0%, fat 30.6%, protein 14.5%, inert 2.9%. The fillets will be in a salt solution (NaCl, 30 % (w/w). After this operation, the fillets will have a moisture content of 50% and 3% of salt content. The concentration and the amount of brine must be constant, so brine is continuously concentrated in an evaporator system and a pump injects salt solution with 35% NaCl. a) What must be the evaporation capacity of evaporator? b) What must be the pump flow? c) How much product will be obtained at the end of the process? d) What is the composition of the end product? Data: All percentages are in weight and referred to humid basis.
Solution: a: 6.63 kg/h; b: 8.75 kg/h; c: 102.13 kg/h, d: 30.0% fat; 14.2% protein; 2.8% inert
Problem 6
Grape juice fermentation (10% Glucose and fructose, 6% fermentable solids and 84% water) provides bio-ethanol with 95 % weight in purity, with CO 2 generation and glucose and fructose fermentation. After fermentation, a distillation operation is necessary to remove non fermentable solids and a certain amount of water and thus obtain ethanol at 86% in weight and some water. Besides that, dehydration is performed in a land sieve reaching the final concentration of bio-ethanol. If 1 kg/s of grape juice comes in the production plant:
Osmotic Dehydrator
Tank with a solution of Evaporator 35% NaCl
Processed fillets
Fillets
Steam
Pump
Problem 10
There is a tank that is continuously receiving a flow of water at a flow rate of 75 kg/h. Simultaneously NaCl glasses are introduced continuously at a rate of 15 kg/h. The resulting homogenous solution is extracted from the tank at a flow rate of 60 kg/h. As the stirring is perfect, the concentration of the solution outside the tank is equal to the concentration of the solution inside the tank. At the beginning of this operation the deposit had 50 kg of pure water.
What is the concentration in the outlet stream after 1 hour?
Solution: 12.6% of salt in weight.
Problem 11
Work out the amount of heat that must be provided to an air flow of 30 Nm^3 /min to increase its temperature from 25 to 200 º C.
Data: Consider that the air behaves as a perfect gas, with a molecular weight of 29 kg/kmol. Variations of kinetic and potential energy are negligible. Specific air heat between 25 and 200 º C: 1.005 kJ/kg ºC
Solution: 176 kJ/kg; 114 kW
Problem 12
In a deodorizer plant, oil is pre-heated in a tubular heat exchanger, working in steady state and in counter flow. The heater agent is hot water obtained from the steam condensation of some waste products. The water flow (2500 Kg/h) comes through the exchanger at 100 ºC and goes out at 43 ºC. Raw oil comes in at 5000 kg/h at 21ºC.
Which will be the temperature of oil on the outside of the exchanger? Data:
- Specific heats of water and oil are respectively 1.0 and 0.5 kcal/kg º C
- The system is thermally isolated
- The inlet and outlet pipes of the exchanger are at the same height and they have the same size.
Solution: a: 78 ºC
Problem 13
An air-conditioner takes air at 31 ºC with a relative humidity of 60%, and it unloads it at 24 ºC with a relative humidity of 40 %. The atmospheric pressure is 1 atm. A flow of 3.822 kg/h of condensed water is collected at 24 ºC.
Calculate the flow of hot air taken by the machine (in m^3 /h) Calculate the flow of cold air thrown out (in Nm^3 /h) Find out the capacity of the air-conditioner to refrigerate (in Kcal/h)
Data: Vapour pressure of liquid water at 24 ºC=22.38 torr, at 31 º C = 33.7 torr Average specific heats (in kcal/kg º C): dry air = 0.24; steam = 0.46; liquid water = 1. Latent heat of vaporisation of water at 0º C = 595 kcal/kg Molecular weights: dry air = 29; water = 18 Reference states (with enthalpy 0): For steam: liquid water at 0 º C and 1 absolute atm. For air: dry air at 0 º C and 1 absolute atm.
Solution: a: 353 m^3 /h b: 312 Nm^3 /h; c: 2925 kcal/h
Problem 14
A centrifugal extractor extracts air from a room where temperature is 5 º C and the barometric pressure is the normal one (760 torr). The axis of the horizontal delivery pipe (D=300 mm) is 50 cm below the aspiring tube (D=400 mm). The air in the delivery pipe is at 30 mm of water column of water of gauge pressure. Then, the air flows through a flow meter which indicates a flow of 1.20 kg/s. The extractor supplies a power of 0.5 C.V
a) Neglecting the heat exchange with the room, calculate the temperature of air in the delivery pipe. b) Neglecting friction, work out the theoretical energy that is required for the fluid to flow. c) Calculate the yield of the extractor by supposing that all losses due to friction are located in it and the conduction losses are negligible.
Data: The air behaves as a perfect gas. Viscosity of air at 5ºC is 0.017 cP Specific heat is 0.24 cal/g º C Due to the small changes in temperature and pressure, the specific volume of air in the delivery pipe can be considered equal to the air of the room.
Solution: a: 5.43 ºC b: 287 J/kg; c: 58.6%
Problem 15
In an autoclave, 1500 cans of vegetable soup are sterilized by heating them up to 116 º C. Before taking them out, the autoclave and the cans cool down to 38 ºC by using cold water which comes at 24 º C and goes out at 30 º C. Each can has 500 g of soup. An empty can weights 70 g. A metallic basket is used to hold the cans inside the autoclave and it also cools to 38 ºC. The walls of the autoclave lose 2500 kcal in the cooling process. How much water is needed? Data: Specific heats in kcal/kg º C are: Soup: 0.94; can and basket: 0.12 and water: 1.
Solution: 9995.1 kg water
Problem 16
We want to freeze 1000 kg of beans from 30 º C to – 15 º C. Calculate the amount of heat that we must extract in kJ.
Data: Freezing point: - 4 º C Specific heat of beans above the freezing point: 0.9 kcal/kg º C Specific heat of beans below the freezing point: 0.4 kcal/ kg º C Latent heat of bean fusion: 70 kcal/kg
Solution: 439617 kJ
Problem 17
100 L/ minute of milk at 25 º C are added to a tank which already contains 2000 litres of milk at 35 º C. At the same time 100 L/minute of milk are taken out. An electrical resistance supplies 2000 kcal/ minute.
Exercise 22
Calculate the relative humidity, dew temperature and temperature of adiabatic saturation of air in a storage chamber that is kept at 21ºC by means of a heat generator if the ambient temperature outside is 15ºC and the absolute humidity of air is 0.006 kg w/kg d.a.
Solution: ϕ = 0.5; Tr = 6.5ºC; Ts = 13ºC
Exercise 223
In a summer day at 14:00 hours, in a store/warehouse the temperature is 32ºC and the ambient relative humidity is 30%.
a) How much mixture of air-water vapour is there in the store (in kg)? b) How many kg of water vapour would there be in a room of 6x9x3 m^3 having that air? c) Why do eggs exude when they are taken out of a chamber at 8ºC and then are left in a room like the one described above? d) If the same day the air of the room is humidified isothermally, what will be its absolute humidity after saturation?
Solution : a: 186.46 kg w.a.; b: 15.877 kg of water vapour; d: Xw = 0.031 kg H 2 O/kg d.a.
Exercise 24
Ambient air that is supplied to a drier has a temperature of 21.1ºC and the adiabatic saturation temperature is 15.6 ºC. By means of a heat exchanger it is heated till up to 93ºC, and then it is blown into the drier. When the air is getting through it, the air cools down and it is adiabatically saturated with humidity. a) What is the dew temperature of the ambient air at the beginning? b) What is the absolute humidity? c) What is the mass fraction of water? d) How much heat is required to heat 3 m^3 of air (measured at 21.1 ºC and P= 1atm) to 93ºC? e) How much water vapour from the substance that is being dried will be carried by 3 m^3 of air (measure at 21.1ºC and P=1 atm)? f) What temperature will the air going out the drier have?
Solution: a) Td≅ 12ºC. b) Xw = 0.009 kgw/kg (^) d.a. c) 0.00892 kgw/kg (^) humid air. d) 64 kcal. e) 9.5 g of water. f) T ≅ 32.5ºC.
Exercise 25
An air conditioner takes in 5 m^3 /h of external air (flow that a fan drives in) and introduces it in a room at 15ºC. It is known that external temperature is 30ºC and the temperature of the wet bulb is 24ºC.
a) Will the typical dropping of water be produced in the evaporator of the air conditioner? b) If it is so, how many L/h of water will be collected? c) What will be the power (kcal/h and kW) at which the air leaves when it cools in the air conditioner? d) In the case that the air coming in the room had a relative humidity of 70%, at what temperature should the outside of the air conditioner be heated?
Solution: b). 0.033 L/h. c). 40 kcal/h. d) 20ºC
Exercise 26
Find out the air conditions (property values in Mollier’s diagram) as a result of mixing two flows (A and B) with the following relation 0.25 m^3 A/m^3 B. Flow A:
Initially the wet bulb temperature is 30ºC and the dry bulb temperature is 50ºC. The air is cooling till reaching a dry bulb temperature of 20ºC and then it is mixed with Flow B. Flow B: Initially it has a dew temperature of 10ºC and 50% relative humidity. Before mixing with Flow A, Flow B is heated up to 60ºC. Draw on a Mollier’s diagram the path followed by each air flow till achieving the conditions produced in the mixture.
Solution: Xw = 0.0095 kgw/kga.s.; T = 52ºC.
Problem 27
Estimate the heat flow rate and the heat flux through a cork wall, in a cooling plant. The cork wall is 4. m long, 2.7 m high and 1.2 thick. The temperature on both sides of the wall is -0.5 and -15.5 ºC.
Data: the thermal conductivity of cork at -8ºC: 0.372 cal / h.cm2.^ (ºC/cm).
Solution: 6.0 kcal/h; 0.54 W/m^2.
SOLUTION
8ºC
k m f(Tm) Tm =−
A 0.^5399 W/m
q =
6. 9973 W 6.0264kcal/h
q 4.8 2.7 = =
Problem 28
The wall of a freezing chamber is built with bricks (11cm thick) on the external side, cement (7.5cm thick) in the central part of the wall, and 10 cm of cork on the inside wall of the chamber. The average temperature on the upper surface is 18 ºC and the internal temperature on the cork surface is -18ºC. Estimate the heat flux through the wall and the temperature distribution between the materials Data: The thermal conductivities in W/m.K are brick: 0.69; cement: 0.76; cork: 0.
Solution: 13.93 W/m^2. Temp. brick-cement: 15.8º C. Temp. cement-cork: 14.4ºC.
SOLUTION
A 13.93W/m
q =
18 T
q A 13.93^1
= = then Tbrick-cement^ = 15.8ºC
18 T
q A 13.93^2
= = then Tcement-cork^ = 14.4ºC
11cm 7.5cm 10cm
Text^ = 18ºC Tint^ = -18ºC
- 15.5ºC - 0.5ºC
2
1
2
air 2 1
ml 0.3204m
r
r
ln
r r
A 2 ππ =
= ; A insml = 0.4663m^2 ; then
∆T
k A
e
k
e
∆T
q
ins ml
ins air
air
b
air
Aml
Energy saving: 0. 18667
q
q q
a
a b
Problem 31
A cold air stream is flowing inside a rectangular pipe (300 x 600 mm) of an air conditioner. The internal wall surface temperature is 5ºC and the external wall surface temperature is lower than 35ºC. The thermal conductivity of the material is 0.5 kcal / h m ºC. Estimate the minimum thickness of this pipe to obtain a heat flow rate along the wall pipe lower than 300 kcal / h and linear meter of tube. Estimate the thermal conductivity of the pipe material for the same heat flow rate and 20 mm of thickness. Answer : 111 mm 0.106 kcal/h m ºC
A = 2 ⋅ ( 0. 3 ⋅ 1 + 2 x ) + 2 ⋅( 0. 6 ⋅ 1 + 2 x ) = 1. 8 + 8 x
dx
dT
q = 300 =− 0. 5 ⋅ 1. 8 + 8 x ⋅
∫ (^) ( ) ∫
2
T T
th dT
x
dx
+^ th
Ln
300
- 5308
⋅ ⋅
e
th
; then th = 0. 11066 m
If the thickness is 0.02 m, then the thermal conductivity is :
+^ k
Ln
k = Ln kcal/h m ºC
Problem 32
It is necessary to install a thermal insulator on a roof in order to avoid water condensation as a consequence of cold. The average minimum temperature in winter is 4ºC in the zone, and the average temperature and relative humidity inside the house are 21 ºC and 39% respectively. The structure is composed of 5 cm of cover, one insulator and 10 cm of concrete. If air flows through the structure, getting progressively colder as it goes out, when the dew temperature is reached some water vapour in the air can condensate. If there is condensation on the concrete, some water will pass through the ceiling of the house and consequently there would be “humidity” in the house. Therefore, it is necessary to calculate the thickness of the insulator for dew temperature to be reached between the isolator and the cover. Data:
Data: kcover: 0.69 W/m.K, kisolater (high density polystyrene): 0.029 W/m.K, kconcrete: 0.76 W/m.K, hinternal: 500 W/m2.K, hexternal: 50 W/m2.K.
Solution: 12 mm
x L x dx x
5 ºC
35 ºC
Problem 33
A furnace is built with heat-resistance (ovenproof/fireproof) brick covered with a metallic external sheet. The internal walls of the furnace are at 100ºC and the external environment is at 20ºC. Security rules forbid a higher temperature than 50ºC in accessible surfaces. a) Does the furnace meet the security specifications? b) If not, what will be the thickness of the brick insulator (k: 0.25 W/mK) placed between the metallic sheet and the ovenproof material? c) What will be the temperature on the contact surface between the heat-resistance material and the insulator? Data: Thickness of fireproof brick: 9 inches Thermal conductivity of fireproof material: 2 W/mºC Thickness of the metallic sheet: 1 inch Thermal conductivity of the metallic sheet: 45.5 W/mºC Individual coefficient of heat transfer by air convection: 30 W/m^2 K Solution: a: No, b: 235 mm. c: 897.2 ºC
Problem 34
In the middle of a spacecraft there is a furnace with a cubic geometry of 1 m of edge. Work out the heat transferred through the furnace to the surroundings. Data: The furnace door is at 64ºC, while the other surfaces are at 34ºC Consider that the furnace base is properly insulated. Ambient temperature is 20ºC Individual heat transfer coefficients are:
- Horizontal flat surface at 34ºC: h 1 =4.03 W/m2.K
- Vertical flat surface at 34ºC: h 2 =3.74 W/m2.K
- Vertical flat surface at 64ºC: h 3 =5.30 W/m2.K Consider that the furnace behaves as a black body (ε = 1) Stefan-Boltzman’s constant: σ = 5.67.^10 -8^ W/m2.K^4 Solution: 1103 W.
Problem 35
A glass test tube 5 mm in internal diameter is full of chloroform (CHCl 3 , 119.5 kg/kmol) with 12 cm head space filled with air. The air velocity outside the test tube maintains the chloroform concentration in the neglected open test tube zone. All the system is at 20ºC and 1 atm. Estimate the mass flow rate of the chloroform through the gas phase considering a steady state system Calculate the expression of the gradient of the chloroform concentration by position in the headspace of the tube. Calculate the pressure of the chloroform at 5 cm of the liquid phase (7 cm to the open tube). Data: The molecular diffusion of the chloroform in air is 9.^10 -6^ m^2 /s. The saturated pressure of chloroform in air at 20ºC is 0.21 atm. The air is insoluble in liquid chloroform
Answer : a: 6.21 mg/h. b: gradient of molar fraction of chloroform: -1.55.^ exp(1.96.^ x). c: 0. atm. a)
dy
dx
RT
P
J Ay = xA ⋅ JA + JB − DAB ⋅ ⋅ A ; JB = 0 ;
( ) A
J
x
x
RT y y
P
J D
A
A
A AB =
1
2
ln
RT
P
cT kmol/m^3
2
A = π ⋅ r = ( )
3 2 5 2
2. 510 1. 9610 m
− −
pA1 = 0.
pA2 = 0
λ W = 2 ⋅ λ A , then, when one molecule of water is transferred, with this energy, two molecules of acetic
acid are also transferred. In this case, the water flux is half as much as that of the acetic flux:
→ 2 ⋅ JW =− JA
dy
dx
RT
P
J Ay = xA ⋅ JA − 0. 5 ⋅ JA − DAB ⋅ ⋅ A →
1
2
2 1 1 0.^5
ln
0. 5 A
A A AB
x
x
RT y y
P
J D
ln
J A = 8.706·
-5 (^) kmol/s m 2
MACETIC=58.9 kg/kmol; JA = 18.459 kg/h m^2
Problem 37
As a package for strawberries a rigid box has been chosen with a permeable plastic to CO 2 and O 2 0.3 mm thick and with airtight closing. When cut strawberries are packed they generate a O 2 flow of 1.09·10- mol/s and other CO 2 flow equal but in opposite sense. Once the steady state is reached, oxygen represents 12 % of the headspace volume in the box. a) What will be the value of O 2 diffusivity through plastic film if the O 2 external concentration is 21% in volume and the storage temperature is 8ºC? b) What will happen if CO 2 concentration was higher than 20%?
Data: Pressure is 1 atm both inside and outside the box. Lid dimensions are: 10x20 cm
Solution: a) 4.19·10-10^ m^2 /s ; b) 4.05·10-10^ m^2 /s
Problem 38
There is a fumes extraction bell above a cylindrical equipment to extract oil by means of liquid hexane at 30ºC. Hexane changes state and passes to the air, keeping a saturated concentration of hexane in the air very similar to liquid hexane. The bell is installed to avoid explosive air from going out of the extractor, considering there is no hexane in the bell. If liquid hexane is 20 cm below the outside of the extractor, what height will the bell have?
Data: Saturated air pressure in hexane at 30ºC: 52.6 kPa. Maximum partial pressure of hexane in order to avoid explosion: 7.5 kPa. Hexane diffusivity through air: 0.088 cm^2 /s. Consider air practically insoluble in liquid hexane.
Solution: 23.2 mm
Problem 39
A cylindrical copper stick 1 m long with its cylindrical surface thermally insulated is at 20ºC. The end of the stick is immersed in a boiling water bath at 100ºC. Estimate the time for 0.1 m of heating side to reach 60ºC. Data: The thermal diffusion of copper is 10-4^ m^2 /s The stick surface immediately arrives at the bath temperature Answer: 110 seconds
m=0, T 0 = 20ºC, T∞ = 100ºC rm = 1m, r = 0.1m, n = 0. 1 m
0.1 m
exp( ( v 0.5) π X)cos (( v-0.5)π n)
v 0.5 π
Y 2 2 2
v 1
v 1
∞
=
−
- 5 100 20
100 60
−
− Y = ,^2 r m
t X
α (^) = 10-4·t
Y = 0. 19385 + 0. 15095 + 0. 09141 + 0. 04292 + 0. 01554 + 0. 0043 = 0.
t = 109.9 s With error function:
∫
−
x x
erf x e dx
0
π
applying the integral
t
y
Y erf tables
2 α
( ) then t = 110 s
Problem 40
A piece of meat with a cylindrical geometry (diameter 10 cm, height 20cm ) initially at 25ºC is blanched at 95ºC in a water bath. a) Estimate the temperature in the middle of piece of meat after 6.5 minutes of blanching. b) Explain the way to calculate the time to arrive at 60ºC in the centre of the meat. Data Thermal conductivity of the meat: 5.19 W/m K Thermal diffusivity of the meat: 1.29.^10 -5^ m^2 /s Individual coefficient to transfer heat by convection around the meat surface: 51. W/m^2 K The physicochemical properties of the meat are constant during blanching time. Answer: a) 83.1ºC T 0 = 25ºC, T∞ = 95ºC a) t = 6.5 min = 390 s YCF = YC∞·YL∞ (Newman’s law)
rmC∞ = 0.1m X (^) C∞ = (^2)
r m
α ⋅ t
=2.0124 ; mC∞ = 2
YC∞ = 0.
rmL∞ = 0.05m X (^) L∞ = (^2)
r m
α ⋅ t
=0.5031; mL∞ = 1
YL∞ = 0.
YCF = YC∞·YL∞ = 0.
- 1712 95 25
95
−
−
T Y T = 83.013 ºC (plot)
b) Iteration:
YCF = YC∞·YL∞ ; 0. 5
YCF =
tsup^ = 180 s → X (^) C∞ =
Problem 41
A cylindrical can 100 mm in diameter and 118.5 mm in height is full of pressed meat with gelatine. The system is motionless with an average thermal diffusion of 1.75.^10 -7^ m^2 /s. The initial temperature of the can is 60ºC and it is introduced in an autoclave with water steam at 120ºC. The surface of the can immediately reaches 20ºC. Estimate the temperature in the centre of the can after 10 and 40 minutes of process Use the graphs and the equations Answer: using equations: 60ºC at 10 minutes; 86.9ºC at 40 minutes.
m = 0, n (middle) = 0
YCF = YC∞·YL∞ (Newman’s law)
rmC∞ = 0.1m rmL∞ = 0.05m
