Question 13.6: Material is fed to a dryer at the rate of 0.3 kg/s and the m...
Material is fed to a dryer at the rate of 0.3 kg/s and the moisture removed is 35% of the wet charge. The stock enters and leaves the dryer at 324 K. The air temperature falls from 341 K to 310 K, its humidity rising from 0.01 to 0.02 kg/kg. Calculate the heat loss to the surroundings. Latent heat of water at 324 K = 2430 kJ/kg. Specific heat capacity of dry air = 0.99 kJ/kg K. Specific heat capacity of water vapour = 2.01 kJ/kg K.
The wet feed is 0.3 kg/s and the water removed is 35%, or: (0.3 × 35/100) = 0.105 kg/s
If the flowrate of dry air is G kg/s, the increase in humidity = (0.02 – 0.01) = 0.01 kg/kg
or: 0.01G = 0.105 and G = 10.5 kg/s
This completes the mass balance, and the next step is to make an enthalpy balance along the lines of Problem 13.5. As the stock enters and leaves at 324 K, no heat is transferred from the air and the heat lost by the air must represent the heat used for evaporation plus the heat losses, say L kW.
Thus heat lost by the inlet air and associated moisture is:
[(10.5 × 0.99) + (0.01 × 10.5 × 2.01)](341 – 310) = 328.8 kW
Heat leaving in the evaporated water = 0.105[2430 + 2.01 (310 – 324)] = 252.2 kW.
Making a balance:
328.8 = (252.2 + L) or L = 76.6 kW