Question : 0.126 kg/s of a solid product containing 4 per cent water is...

From the humidity chart, Figure 13.4 , air at 294 K and of 40 per cent relative humidity has a humidity of 0.006 kg/kg. This remains unchanged on heating to 366 K. At the dryer inlet, the wet bulb temperature of the air is 306 K. In the dryer, the cooling takes place along the adiabatic cooling line until 60 per cent relative humidity is reached.

At this point:

the humidity = 0.028 kg/kg and

the dry bulb temperature = 312 K.

The water picked up by 1 kg of dry air = (0.028 – 0.006) = 0.22 kg water/kg dry air.
The wet feed contains:
42 kg water/100 kg feed or 42/(100 – 42) = 0.725 kg water/kg dry solids
The product contains:
4 kg water/100 kg product or 4/(100 – 4) = 0.0417 kg water/kg dry solids.
Thus:

water evaporated = (0.725 – 0.0417) = 0.6833 kg/kg dry solids

The throughput of dry solids is:

0.126(100 – 4)/100 = 0.121 kg/s

and the water evaporated is:

(0.121 × 0.6833) = 0.0825 kg/s

The required air throughput is then:

0.0825/(0.078 – 0.006) = 3.76 kg/s

At 294 K, from Figure 13.4:

specific volume of the air = 0.84 m^{3}/kg

and the volume of air required is:

(0.84 × 3.76) = 3.16 m^{3}/s

At a humidity of 0.006 kg water/kg dry air, from Figure 13.4, the humid heat is 1.02 kJ/kg deg K and hence the heat required in the preheater is:

3.76 × 1.02(366 – 294) = 276 kW

In the second case, air both enters and leaves at 340 K, the outlet humidity is 0.111 kg water/kg dry air and the water picked up by the air is:

(0.111 – 0.006) = 0.105 kg/kg dry air

The air requirements are then:

(0.0825/0.105) = 0.786 kg/s or (0.786 × 0.84) = 0.66 m^{3} /s

The heat to be added in the preheater is:

0.66 × 1.02(340 – 294) = 31.0 kW

The heat to be supplied within the dryer is that required to heat the water to 340 K plus its latent heat at 340 K.

That is: 0.0825[4.18(340 – 290) + 2345] = 211 kW

Thus: Total heat to be supplied = (81 + 211) = 242 kW .