Question 13.7: A rotary dryer is fed with sand at the rate of 1 kg/s. The f...
A rotary dryer is fed with sand at the rate of 1 kg/s. The feed is 50% wet and the sand is discharged with 3% moisture. The entering air is at 380 K and has an absolute humidity of 0.007 kg/kg. The wet sand enters at 294 K and leaves at 309 K and the air leaves at 310 K. Calculate the mass flowrate of air passing through the dryer and the humidity of the air leaving the dryer. Allow for a radiation loss of 25 kJ/kg dry air. Latent heat of water at 294 K = 2450 kJ/kg. Specific heat capacity of sand = 0.88 kJ/kg K. Specific heat capacity of dry air = 0.99 kJ/kg k. Specific heat capacity of vapour = 2.01 kg K.
The feed rate of wet sand is 1 kg/s and it contains 50% moisture or (1.0 × 50/100) = 0.50 kg/s water.
∴ flow of dry sand = (1.0 – 0.5) = 0.50 kg/s
If the dried sand contains w kg/s water, then:
w/(w + 0.50) = (3.0/100) or w = 0.0155 kg/s
and: the water evaporated = (0.50 – 0.0155) = 0.4845 kg/s.
Assuming a flowrate of G kg/s dry air, then a heat balance may be made based on a datum temperature of 294 K.
Inlet streams:
(i) Sand: this enters at 294 K and hence the enthalpy = 0.
(ii) Air: G kg/s of dry air is associated with 0.007 kg/kg moisture.
∴ enthalpy = [(G × 0.99) + (0.007G × 2.01)](380 – 294) = 86.4G kW
and: the total heat into the system = 86.4G kW.
Outlet streams:
(i) Sand: 0.50 kg/s dry sand contains 0.0155 kg/s water and leaves the unit at 309 K.
∴ enthalpy = [(0.5 × 0.88) + (0.0155 × 4.18)](309 – 294) = 7.6 kW
(ii) Air: the air leaving contains 0.07 G kg/s water from the inlet air plus the water evaporated. It will be assumed that evaporation takes place at 294 K. Thus:
enthalpy of dry air = G × 0.99(310 – 294) = 15.8m kW
enthalpy of water from inlet air = 0.007G × 2.01(310 – 294) = 0.23G kW
enthalpy in the evaporated water = 0.4845[2430 + 2.01(310 – 294)] = 1192.9 kW, a total of (16.03G + 1192.9) kW
(iii) Radiation losses = 25 kJ/kg dry air or 25G kW and the total heat out = (41.03G + 1200.5) kW.
Mass balance:
86.4G = (41.03G + 1200.5) or G = 26.5 kg/s
Thus the flow of dry air through the dryer = 26.5 kg/s
and the flow of inlet air = (26.5 × 1.007) = 26.7 kg/s
As in Problem 13.5, water leaving with the air is: (26.5 × 0.007) + 0.4845 = 0.67 kg/s
and humidity of the outlet air = (0.67/26.5) = 0.025 kg/kg.