Dryers Problem and Solution

1. Calculate the the flux (kg/sq.m sec) if mass of dry solid = 8 kg; Wet surface area = 4 sq.m; Change in moisture content with time is 0.4 /sec; Flux = 8/4 * 0.4 = 0.8 2. Find the gas phase mass transfer co-efficient co-effici ent for a unbound moisture to remove if the flux is 5 kg/sq.m sec and the difference in humidity of the liquid and the main stream is 0.5 units. Explanation: Flux= Mass transfer co-efficient * (Ys-Y) Mass transfer co-efficient= 5/0.5 = 10 3. Find the weight of the wet solid if dry solid is 2 kg and the moisture is 0.5 kg.

Wet solid= dry solid+ moisture = 2+0.5= 2.5 kg 4. Find the moisture content content in dry basis if the weight of dry solid is 5 kg and the moisture is 2 kg.

 As it is a dry basis, the moisture content = 2/5 =0.4 =0.4 5. Find the moisture content in wet basis is the weight of the dry solid is 3 kg and the weight of the moisture is 2 kg.

Weight of the solid = weight of moisture/ weight of moisture + dry solid = 2/5 = 0.4 6. Find the moisture to be evaporated per kg of dried product if initial and final moisture content per kg of dry solid is 85% and 15%.

Moisture to dry = Initial moisture – final moisture content = 0.85  – 0.15 =0.7 per kg of dry solid 7. Find the time taken to dry 50% to 40% of moisture (dry basis) if critical moisture content is 30%. If the weight of the wet solid is 5 kg and the product of area and constant rate is 0.5 kg/min.

Time for drying = ( wet solid weight * change in moisture )/ Area * constant rate = (5 * 0.1/0.5) min = 1 min = 60 sec. 8. Find the change in moisture content from the given data Given data: Time for drying= 6 sec  Area * Constant rate = 8 kg per sec Weight of wet solid = 100 kg

Difference in moisture = (time *area* rate )/Weight of the solid = 0.48. 9. A wet solid is dried from 36% to 8% moisture in 5 hours under constant drying conditions. The critical moisture is 14% and the equilibrium moisture is 4%. Caclulate the time needed to dr y from 8 to 5.5% moisture Figure 1.1 10. A wet solid is dried from 36% to 8% moisture in 5 hours under constant drying conditions. The critical moisture is 14% and the equilibrium moisture is 4%. Caclulate the time needed to dry from 30% to 15% moisture Figure 1.1

11. A wet solid is dried from 36% to 8% moisture in 5 hours under constant drying conditions. The critical moisture is 14% and the equilibrium moisture is 4%. Caclulate the time needed to dry from 36% to 5.5% moisture. Figure 1.1

Figure 1

12. A sheet material measuring 0.8 m square and 5 cm thick is dried from both sides from 20% to 2% moisture under constant drying conditions. The dry density of the mate rial is 500kg/m^3 and its equilibrium moisture is negligible. An experiment showed that the constant drying rate is 4.5 kg/s^2-hr and the falling rate begins at 25% moisture. How long is the drying time. What is the % moisture after 1 hour of drying?

13. Air at 90°C is being used to dry a solid food in a tunnel dryer. The product, with 1 cm thickness and a 5 cm by 10 cm surface, is exposed to the heated air with convective mass transfer coefficient of 0.1 m/s. Estimate the constant-rate drying time, when the initial moisture content is 85% and the critical moisture content is 42%. The air has been heated from 25°C and 50% RH. The product density is 875 kg/m3. Answer: 14-20 From Sta. Maria 21. Time for Drying in Constant-Rate Period A batch of wet solid was dried on a tray dryer using constant drying conditions and a thickness of material on the tray o f 25.4 mm. Only the top surface was exposed. The drying rate during the co nstant-rate period was R = 2.05 kg H_2O/h middot m^2 (0.42 lb_m H_2 O/h middot ft^2). The ratio Ls/A used was 24.4 kg dry solid/m^2 exposed surface (5.0 lbm dry solid/ft^2). The initial free moisture was = 0.55 and the critical moisture content Xc = 0.22 kg free moisture/kg dry solid. Calculate the time to dry a batch of this material from X-1 = 0.45 to X_2 = 0.30 using the same drying conditions but a thickness of 50.8 mm, with drying from the top and bottom surfaces.

22. A batch of wet solid was dried on a tray dryer. Drying air is blowing perpendicular to the surface of wet materials. The constant rate drying period was from X1 = 0.45 to XC = 0.18. It is known that when hot air flow rate is 5000 kg/h/m2, the drying time for decreasing moisture content from 0.30 to 0.20 is 2 h. When hot air flow rate increases to 15000 kg/h/m2, please calculate the new drying time for decreasing moisture content from 0.30 to 0.2

23.

Data is obtained from a batch drying experiment, with the weight monitored as a function of time. The material being dried contains water and 12.5 pounds of dry solids. The dry bulb temperature of

the air is 230°F, and the wet bulb temperature of the air is 110°F. The equilibrium moisture content of the material is 0.04 pounds of water per pound of dry solids. Time (hr)

0.0

0.2

0.4

0.6

0.7

0.9

1.1

1.3

1.4

Weight (lb)

40.0

32.9

25.8

18.7

16.4

14.3

13.4

13.0

13.0

24. a) From the data, determine the rate of drying as pounds of water removed per hour when the free moisture content is equal to 0.11 pounds of water per pound of dry solid. 25. b) Estimate the critical free moisture content of the material for the given conditions. 26. c) Determine the rate of drying during the constant rate period (pounds of water removed per hour per square foot of exposed surface area, assuming the total exposed surface area is 10 ft . 2

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