Chapter 7 Unit Operations Problems
1. Cabbage Drying
Initial composition
Final composition after drying
11% solids, 89% water.
100kg dry solids + 5kg water
= 4.8% water on wet basis.
Now 100kg initial cabbage yields 11kg dry solids + 0.55 kg water
Therefore remove 89-0.55 = 88.45 kg water
(a) heat energy per tonne of raw cabbage
(b) heat energy per tonne of dry cabbage
2. Spray dryer efficiency
Efficiency of spray dryer
Therefore

= (1000/100) x 88.45 x 2258
= 2 x 106 kJ
= (1000/11.55) x 88.45 x 2258
= 1.73 x 107 kJ
= (150-95)/(150-15) = 55/135
= 0.41
= 41%
Efficiency could be lifted by decreasing air outlet temperature, or increasing air inlet
temperature if the product could tolerate this.
3. Humidity of air
(a) Working from calculation and steam tables (Appendix 8):
Pressure of saturated steam at 65oC = 25.7kPa
= ps
Then if the mole fraction of the water is x
Then (x/18) / (1/29 + x/18)
= p/P
= p/100
And p/ps
Therefore
p
So
=
=
0.42
0.42 x 25.7
(100x/18) /(1/29 + x/18)
5.5x
x
= 10.794
= 10.794
= 0.372 + 0.600x
= 0.075 kgkg-1
(b) Reading from psychrometric chart (Appendix 9(b)), for temperature of 65 C and RH of
40%:
Absolute humidity
= humidity ratio
= 0.075kgkg-1
4. Cooling water
In an evaporative cooler, the lowest available temperature must be the wet-bulb temperature
of the ambient air: a real cooler may approach, but cannot get below this. So the problem is to
calculate the wet-bulb temperature of the air, with 18oC dry bulb and 42% RH.
From steam tables (Appendix 8) and 42%RH, the humidity of the air can be calculated,
assuming Y = 18p/29P
(a) At 42% RH, vapour pressure, ps = 2.06kPa
so p
= 2.06x 0.42 = 0.87kPa
p/P
= 0.0087
Y
= (18/29) x 0.0087 = 0.0054kgkg-1
Taking cs =1 from Appendix 6 and  =2534 kJkg-1 from Appendix 8
cs /  = 1/2534 = 0.0004 kgkg-1oC-1
cs /  = 4 x 10-4 kgkg-1oC-1
So moving along the temperature scale, we can estimate
o
C
18
16
14
Y kgkg-1
0.0054
0.0062
0.0070
12
0.0078
10
0.0086
Calculating the saturation humidities from Appendix 8 and Equation 7.2
Ys kgkg-1
0.1280
0.0120
0.0099
0.0087
0.0076
So the wet–bulb temperature lies between 10oC and 12oC
At 11oC
Y = 0.0082 kgkg-1
Ys = 0.0082 kgkg-1
So the wet-bulb temperature is very close to 11oC
This can be found very much more easily on the psychrometric chart, which agrees that the
wet-bulb temperature is 11oC.
(b) If the water approaches this by 5oC, then this implies the water cooling from 36oC to
16oC. Achieving this depends on the relative flow rates of air and water and the design of the
cooler.
5. Chiller store for fruit
Taking readings as accurately as you can determine at 5oC dry-bulb temperature from chart
(Appendix 9(a)), and interpolating between the (sloping) wet bulb lines at the constant
humidity conditions to read the wet-bulb temperatures we have:
%RH
90
Wet bulb depression 0.7
80
1.4
70
2.2
60
2.9
50
3.6
40
4.4
The chart is not easy to read, therefore imprecise and cannot be read below 40%RH.
Tabulated data should be sought if accuracy is important..
6. Heating air for dryer
(a) From the chart in Appendix 9(a):
Air at 16oC and 65%RH:
humidity 0.0064 kgkg-1; specific volume 0.829 m3 kg-1 ; enthalpy 35kJkg-1
volume 1300m3h-1
When heated to 150oC, the enthalpy increases to 170 kJkg-1 as shown in the Appendix 9(b)
chart.
So mass of air heated = 1300/0.829
= 1568kgh-1
Energy input
= 1568 (170 –35)
= 2.12 x 105 kJh-1
= 58.8kJs-1
= 58.8kW
(b) Air leaves at 90oC and 5%RH
Humidity 0.024kg kg-1
Quantity of water (total) removed per hour
= 1568 x 0.024
= 37.6kgh-1
= 1568 (0.024 – 0.0064)
= 27.6 kgh-1
(c) Water removed in drying
7. Drying food material
Dried in air at 90oC and 15% RH
From psychrometric chart Appendix 9(b)
wet-bulb temperature = 51oC
so T = (90 – 51)
= 39 oC
From Appendix 8, latent heat of evaporation = 2594 kJkg-1
Heat transfer coefficient 25Jm-2 s-1 oC-1 = 25 x 10-3 kJm-2 s-1 oC-1
Therefore rate of evaporation = [(25 x 10-3) x 39]/ 2594
= 0.376 x10-3 kg m-2 s-1
= 1.35 kg m-2 h-1
8. Loss weight of beef slices in chiller
(a) From Equation 5.27
h
= 5.7 + 3.9v
= 5.7 + 3.9x 0.5
= 7.7Jm-2 s-1 oC-1
From the psychrometric chart in Appendix 9(a), for air at 10oC and 50%RH
Wet-bulb depression = 4.5oC
From Appendix 8
Latent heat of evaporation of water = 2489 kJkg-1
So, rate of evaporation
1 day (24h)
Weight loss
= (7.7 x 10-3 x 4.5)/2489
= 0.014 x10-3kg m-2 s-1
= 0.014g m-2 s-1
= 8.64 x 104s
= 0.014 x 8.64 x 104
= 1.21x103 g m-2d-1
For each piece of steak, exposed surface
= top surface + 2 long sides + 2 short sides
= [5 x 15) + (2 x 30) + (2 x 10)]
= 155 cm2
= 1.55x10-2 m2
So weight loss per piece
= 1.21x103x 1.55x10-2 gd-1
= 18.8 gd-1
Specific weight of meat is 1050kgm-3
Volume of piece
= (15 x 5 x 2) cm3
= 150 cm3
= 150 x 10-6 m3
Weight of piece
= 150 x 10-6 x 1.050 x 103
= 0.158kg
= 158g
So fractional loss per day
= 18.8/158
= 0.12
= 12%
(b) If the RH were increased to 80%
Wet-bulb depression = 1.7 oC
Weight loss
= 18.8 x (1.7/4.5)
= 7.1 gd-1
Fractional loss per day
= 7.1/158
= 4.5%
(c) Net emissivity = 0.8
Radiation would add to heat input
From equation 5.13: Tm = 7.75 C = 281K.
qr
= 0.23 (Tm/100)3A T
qr
= 0.23 x 0.8x (281/100)3 x 1 x 4.5
= 18.4 Jm-2 s-1
and this would add to heat from air transfer, Example 8
(7.7 x 4.5)
= 34.7
Making an addition
= 18.4 /34.7
= 53%
So increasing rates from 18.8 g d-1per piece to 28.8g d-1
And the percentage from 12% to 18.4%
9. Drying of food material
Air 130oC, RH 1.6 %
Equilibrium moisture content 11% on dry basis
Dry from 400% to 16.3% on dry basis
Constant rate drying to a 100% moisture content on a dry basis.
From psychrometric chart,
Y = 0.028 kg kg-1
Ys = 0.066 kg kg-1
From the Lewis relationship:
kg
= 18gm-2s-1 = 0.018kgm-2s-1
Dry material = 100[1/(4+1)]= 20kg
Total water = (100-20)
= 80kg of which 20kg is retained.
Water to be removed in constant rate period:
= 100 - (20+20)
= 60kg
and so using Equation 7.3
where kg A(Ys –Y a)
= 0.018 x 12 x (0.066-0.028)
= 8.21 x 10-3 kgs-1
t
= 60/(8.21 x 10-3)
= 7308s
= 2.03h
For falling rate drying
Also from equn. 7.7
t
= w (Xo –Xf)/f kg'A(Ys -Ya)
Moisture content X 1.0
w(X1 - X2) kg
4
0.8
f (from Fig.7.7)
f kg'A(Ys -Ya) x 10-3
0.85
7
t
571
0.6
0.4
4
4
0.47
3.9
0.24
2
1025
0.3
2
0.12
0.99
2000
2020
0.25
0.20
0.18
1
1
0.4
0.163
0.34
0.07
0.57
0.05
0.41
0.03
0.25
0.02
0.16
1754
2439
 t = 13534s = 3.8h
Therefore total drying time = (2.03 + 3.8)h = 5.8h  6h.
1600
2125