4.
Air is saturated with water vapour when the partial pressure of water
vapour in the air equals the saturation pressure of water vapour at the
same temperature.
5.
Humidity of air is the ratio of the weight of water vapour to the weight
of the dry air in the same volume.
6.
Relative humidity is the ratio of the actual partial pressure to the saturation
partial pressure of the water vapour at the air temperature.
7.
Water vapour/air humidity relationships are shown on the psychrometric
chart.
PROBLEMS
1.
Cabbage containing 89% of moisture is to be dried in air at 65°C down
to a moisture content on a dry basis of 5%. Calculate the heat energy
required per tonne of raw cabbage and per tonne of dried cabbage, for
the drying. Ignore the sensible heat.
[ 2x106 kJ ; 1.73x107 kJ ]
2. The efficiency of a spray dryer is given by the ratio of the heat energy
in the hot air supplied to the dryer and actually used for drying, divided
by the heat energy supplied to heat the air from its original ambient
temperature. Calculate the efficiency of a spray dryer with an inlet air
temperature of 150°C, an outlet temperature of 95°C, operating
under an ambient air temperature of 15°C. Suggest how the efficiency
of this dryer might be raised.
[ 41% ]
3.
Calculate the humidity of air at a temperature of 65°C and in which
the RH is 42% and check from a psychrometric chart.
[ 0.075 kgkg-1 ]
4.
Water at 36°C is to be cooled in an evaporative cooler by air which
is at a temperature of 18°C and in which the RH is measured to be
43%. Calculate the minimum temperature to which the air could be cooled,
and if the air is cooled to 5°C above this temperature, what is the
actual cooling effected. Check your results on a psychrometric chart.
[ 11°C ; 36°C to 16°C ]
5.
In a chiller store for fruit, which is to be maintained at 5°C, it
is important to maintain a daily record of the relative humidity. A wet-
and dry-bulb thermometer is available so prepare a chart giving the relative
humidity for the store in terms of the wet-bulb depression.
6.
A steady stream of 1300 m3 h-1 of room air at 16°C
and 65% RH is to be heated to 150°C to be used for drying. (a) Calculate
the heat input required to accomplish this. If the air leaves the dryer
at 92°C and at 98% RH (b) calculate the quantity of water removed
per hour by the dryer, and (c) the quantity of water removed per hopur
from the material being dried.
[ (a) 58.8 kW ; (b) 37.6 kg h-1 ; (c) 27.6 kg h-1 ]
7.
In a particular situation, the heat transfer coefficient from a food material
to air has been measured and found to be 25 J m-2 s-1
°C-1. If this material is to be dried in air at 90°C
and 15% RH, estimate the maximum rate of water removal.
[1.35 kg m-2 h-1 ]
8.
Food on exposure to unsaturated air at a higher temperature will dry if
the air is unsaturated. Steak slices are stored in a chiller at 10 °C.
(a) Estimate the maximum
weight loss of steak pieces, 15 cm x 5 cm x 2 cm in air at 10°C and
50% RH moving at 0.5 m s-1. The pieces are laid flat on shelves to
age. Assuming that the meat behaves as a free water surface, estimate
the percentage loss of weight in 1 day of exposure. Specific weight of
meat is 1050 kgm-3.
(b) if the H of the air were invcreased to 80% what would be the percentage
loss?
(c) it the meat pieces were also exposed to nearby surfaces at the temperature
of the air (dry bulb), what would then be the percentage loss? Assume
net emissivity is 0.8.
[ (a) 12% ; (b) 4.5% ; 18.4% ]
9.
Assume that the food material from worked Example
7.17 is to be dried in air at 130°C with a relative humidity of
1.6%. Under these conditions the equilibrium moisture content in the food
is 12% on a dry basis. Estimate the time required to dry it from 350%
down to 12 % on a dry basis. Constant-rate drying exists down to 100%
moisture content on a dry basis. All moisture contents on a dry basis.
[ 5.8h ; 2.03h constant rate, 3.8h falling rate ]
 CHAPTER
8: EVAPORATION
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