Chapter 7 Unit Operations Problems

Chapter 7 Unit Operations Problems

1. Cabbage Drying

Initial composition 11% solids, 89% water.

Final composition after drying100kg 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= (1000/100) x 88.45 x 2258

= 2 x 106 kJ

(b) heat energy per tonne of dry cabbage = (1000/11.55) x 88.45 x 2258

= 1.73 x 107 kJ

Spray dryer efficiency

Efficiency of spray dryer = (150-95)/(150-15)= 55/135

= 0.41

Therefore= 41%

Efficiency could be lifted by decreasing air outlet temperature, or increasing air inlet temperature if the product could tolerate this.

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=0.42

Thereforep=0.42 x 25.7= 10.794

So (100x/18) /(1/29 + x/18)= 10.794

5.5x= 0.372 + 0.600x

x= 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

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

oC1816141210

Y kgkg-10.00540.00620.00700.00780.0086

Calculating the saturation humidities from Appendix 8 and Equation 7.2

Ys kgkg-10.12800.01200.00990.00870.0076

So the wet–bulb temperature lies between 10oC and 12oC

At 11oCY = 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.

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:

%RH908070605040

Wet bulb depression0.71.42.22.93.64.4

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

= 27.6 kgh-1

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-1oC-1 = 25 x 10-3 kJm-2 s-1oC-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

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-1oC-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 = (7.7 x 10-3 x 4.5)/2489

= 0.014 x10-3kg m-2 s-1

= 0.014g m-2 s-1

1 day (24h)= 8.64 x 104s

Weight loss= 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%

Radiation would add to heat input

From equation 5.13: Tm = 7.75oC = 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-1Ys= 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)/fkg'A(Ys -Ya)

Moisture content X 1.0 0.8 0.6 0.4 0.3 0.25 0.20 0.18 0.163

w(X1 - X2) kg4 44 21 10.4 0.34

f (from Fig.7.7) 0.85 0.47 0.24 0.12 0.07 0.05 0.03 0.02

fkg'A(Ys -Ya) x 10-3 7 3.9 2 0.99 0.57 0.41 0.25 0.16

t 571 1025 2000 2020 1754 2439 1600 2125

t = 13534s = 3.8h

Therefore total drying time = (2.03 + 3.8)h = 5.8h  6h.