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Fluid-flow theory
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Heat-transfer theory

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1. Heat and material balances are the basis for evaporator calculations.

2. The rate of boiling is governed by the heat transfer equations.

3. For multiple effect evaporators, i.e. two or more evaporators used in series,
with two evaporators, if   q1 = q2

                  U1A1DT1= U2A2DT2

and if A1 and A2 are equal     U2/U1 = DT1/DT2
(and this can be extended to more than two effects.)

4. Multiple-effect evaporators use less heat than single-effect evaporators. For an n-effect evaporator, the steam requirement is approximately 1/n, but requires more heat exchange surface; the heat-exchange surface required is approximately n times for the same output. From the energy viewpoint, evaporators can therefore be very much more efficient than dryers.

5. Condensers on an evaporator must provide sufficient cooling to condense the water vapour from the evaporator. Condenser calculations are based on the heat transfer equation.

6. Boiling-point elevation in evaporators can be estimated using Duhring's rule that the ratio of the temperatures at which two solutions exert the same pressure is constant.

7. Special provisions, including short residence times and low pressures to give low boiling points, are necessary if the maximum retention of volatile constituents is important, and to handle heat sensitive materials.


1. A single-effect evaporator is to produce a 35% solids tomato concentrate from a 6% solids raw juice entering at 18°C. The pressure in the evaporator is 20 kPa(absolute) and steam is available at 100 kPa gauge. The overall heat-transfer coefficient is 440 J m-2 s-1 °C-1, the boiling temperature of the tomato juice under the conditions in the evaporator is 60°C, and the area of the heat-transfer surface of the evaporator is 12 m2. Estimate the rate of raw juice feed that is required to supply the evaporator.
[ 536 kgh-1 ]

2. Estimate (a) the evaporating temperature in each effect, (b) the reirements of steam, and (c) the area of heat transfer surface for a two effect evaporator. Steam is available at 100 kPa gauge pressure and the pressure in the second effect is 20 kPa absolute. Assume an overall heat-transfer coefficient of 600 and 450 J m-2 s-1 °C-1 in the first and second effects respectively. The evaporator is to concentrate 15,000 kg h-1 of raw milk from 9.5 % solids to 35% solids.Assume the sensible heat effects can be ignored, and that there is no boiling-point elevation.
[ (a) 1st. effect 94°C, 2nd. effect 60°C, (b) 5,746 kgh-1, 0.53 kg steam/kg water (c) 450 m2 ]

3. A plate evaporator is concentrating milk from 10% solids to 30% solids at a feed rate of 1500 kg h-1. Heating is by steam at 200 kPa absolute and the evaporating temperature is 75°C. (a) Calculate the number of plates needed if the area of heating surface on each plate is 0.44 m2 and the overall heat-transfer coefficient 650 J m-2 s-1 °C-1. (b) If the plates, after several hours running become fouled by a film of thickness 0.1 mm, and of thermal conductivity 0.1 J m-1 s-1 °C-1, by how much would you expect the capacity of the evaporator to be reduced?
[ (a) 50 (b) 13% of its former value ]

4. (a) Calculate the evaporation in each effect of a triple-effect evaporator concentrating a solution from 5% to 25% total solids at a total input rate of 10,000 kg h-1. Steam is available at 200 kPa absolute pressure and the pressure in the evaporation space in the final effect is 55 kPa absolute. Heat-transfer coefficients in the effects are, from the first effect respectively, 600, 500 and 350 J m-2 s-1 °C-1. Neglect specific heats and boiling-point elevation. (b) Calculate also the quantity of input steam required per kg of water evaporated.
[ (a) 1 st. effect 2707 kgh-1 , 2 nd. effect 2669 kgh-1 , 3 rd. effect 2623 kgh-1; (b) 0.343 kgkg-1 ]

5. If in Problem 4 there were boiling point elevations of 0.60, 1.50 and 4°C respectively in the effects starting from the first, what would be the change in the requirement of input steam required per kg water evaporated?
[ 0.342 kgkg-1 compared with 0.343 kgkg-1; therefore no change ]

6. (a) Estimate the 12°C cooling water requirement for a jet condenser to condense the 70°C vapours from an evaporator which concentrates 4000 kg h-1 of milk from 9% solids to 30% solids in one effect. If the cooling water leaves at a maximumm of 25°C. (b) For the same evaporator estimate for a surface condenser, the quantity of 12°C cooling water needed and the necessary heat-transfer area if the cooling water leaves at a maximum of 25°C and the overall heat-transfer coefficient is 2200 J m-2 s-1 °C-1 in the condenser.
[ 130 x 103 kgh-1 Note in practice this is higher ; (b) 130 x 103 kgh-1, 17.3 m2 ]

7. If in the evaporator of Problem 6, one-half of the evaporated vapour were mechanically recompressed with an energy expenditure of 160 kJ kg-1, what effect would this have on the steam economy, assuming the steam supply was at 100 kPa absolute?
[ 46.6% steam energy saved )

8. A standard calandria type of evaporator with 100 tubes, each 1 m long, is used to evaporate fruit juice with approximately the same thermal properties as water. The pressure in the evaporator is 80 kPa absolute, and in the steam jacket 100 kPa absolute. Take the tube diameter as 5 cm. Estimate the rate of evaporation in the first evaporator. Assume the juice enters at 18°C and the overall heat transfer coefficient is 440 J m-2 s-1.
[ 72.5 kgh-1


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Unit Operations in Food Processing. Copyright © 1983, R. L. Earle. :: Published by NZIFST (Inc.)
NZIFST - The New Zealand Institute of Food Science & Technology