·       Energy balance:

C_P (PTA) = 0.2871Btu/lb ° F; C_P (Water) = 1 Btu/lb°F

Product discharge temperature = (313 + 209)/2 = 261°F

Temperature of feed = 176°F

Heat required to raise the product to discharged temp.

= 23705 x 0.2871(261-176) + 24 (261-176) = 5.8143 x 10⁵ Btu/Hr

Heat required to remove the water = 1226 [(180-176) + 0.45 (209-180) + 550]

= 6.952 x10⁵ Btu/Hr

Total Heat = 1.27 x 10⁶ Btu/Hr

Air Required:

S_H­ -Humid Heat of inlet air = 0.24 + 0.45 x 0.002 = 0.2409

Use average humid heat = 0.242

G_G ' . S × Humid heat of air × Temperature= Total Heat , here S = cross sectional area, sq ft

G_G ' . S × (0.242) × (313-209) = 1.27 × 10⁶

G_G ' . S = 50723.27 lb/Hr

Humid heat = 0.24 + 0.45 × 0.02617 = 0.2517 and S_Havg = (0.2409 + 0.2517)/2 = 0.2463

Therefore the average humidity taken above is valid

Mean temperature difference across the rotary drier can be calculated by using following formulae

Let q_p = heat required to preheat the feed from inlet to wet bulb temperature.

q_s = heat required to heat product from wet bulb temperature to discharge temperature.

q_v = heat required to evaporate water at wet bulb temperature.