Question 13.5: A rotary countercurrent dryer is fed with ammonium nitrate c...
A rotary countercurrent dryer is fed with ammonium nitrate containing 5% moisture at the rate of 1.5 kg/s, and discharges the nitrate with 0.2% moisture. The air enters at 405 K and leaves at 355 K; the humidity of the entering air being 0.007 kg moisture/kg dry air. The nitrate enters at 294 K and leaves at 339 K.
Neglecting radiation losses, calculate the mass of dry air passing through the dryer and the humidity of the air leaving the dryer. Latent heat of water at 294 K = 2450 kJ/kg. Specific heat capacity of ammonium nitrate = 1.88 kJ/kg K. Specific heat capacity of dry air = 0.99 kJ/kg K. Specific heat capacity of water vapour = 2.01 kJ/kg K.
The feed rate of wet nitrate is 1.5 kg/s containing 5.0% moisture or (1.5 × 5/100) =
0.075 kg/s water.
∴ flow of dry solids = (1.5 – 0.075) = 1.425 kg/s
If the product contains w kg/s water, then:
w/(w + 1.425) = (0.2/100) or w = 0.00286 kg/s
and: the water evaporated = (0.075 – 0.00286) = 0.07215 kg/s
The problem now consists of an enthalpy balance around the unit, and for this purpose a datum temperature of 294 K will be chosen. It will be assumed that the flow of dry air into the unit is m kg/s.
Considering the inlet streams:
(i) Nitrate: this enters at the datum of 294 K and hence the enthalpy = 0.
(ii) Air: G kg/s of dry air is associated with 0.007 kg moisture/kg dry air.
∴ enthalpy = [(G × 0.99) + (0.007G × 2.01)](405 – 294) = 111.5G kW
and the total heat into the system = 111.5G kW
Considering the outlet streams:
(i) Nitrate: 1.425 kg/s dry nitrate contains 0.00286 kg/s water and leaves the unit at 339 K.
∴ enthalpy = [(1.425 × 1.88) + (0.00286 × 4.18)](339 – 294) = 120.7 kW
(ii) Air: the air leaving contains 0.007 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(355 – 294) = 60.4G kW
enthalpy of water from inlet air = 0.007G × 2.01(355 – 294) = 0.86G kW
enthalpy in the evaporated water = 0.07215[2450 + 2.01(355 – 294)] = 185.6 kW
and the total heat out of the system, neglecting losses = (306.3 + 61.3G) kW.
Making a balance:
111.5G = (306.3 + 61.3G) or G = 6.10 kg/s dry air
Thus, including the moisture in the inlet air, moist air fed to the dryer is:
6.10(1 + 0.007) = 6.15 kg/s
Water entering with the air = (6.10 × 0.007) = 0.0427 kg/s.
Water evaporated = 0.07215 kg/s.
Water leaving with the air = (0.0427 + 0.07215) = 0.1149 kg/s
Humidity of outlet air = (0.1149/6.10) = 0.0188 kg/kg dry air.