Question

In the manufacture of an active pharmaceutical ingredient API, the API goes through a purification step in which it is crystallized, filtered and washed. The washed crystals contain 47% water. They are feed into a tunnel dryer and leave the tunnel dryer at 165kg/h containing 5% adhered mositure.Dry air enters the dryer at 145 degree Farenhiet(*F) and 1atm and the outlet air is 135*F and 1 atm with relative humidity of 50%.calculate the rate at which the API enters the dryer (kg/h) and the volumetric flow rate (ft^3/h) of the inlet air.

Answer 1

Crystal % in feed, F = 100- 47 = 53 %

Crystal % in product = 100- 5 = 95 %

Product mass flow = 165 kg/hr

Performing crystal mass balance we have,

F * 0.53 = 165 * 0.95

F = 295.75 kg/hr

Let flow rate of Dry air be S kg/hr

Outlet flow rate of air = (S+W) kg/hr where W is the water content in air

By Mass balance:

S + 295.75 = 165 + (S+W)

W = 130.75 kg/hr

Outlet temperature, T_out = 135 ^oF

T_out = (135-32)*5/9 + 273.15 K = 330.37 K

T_in = (145-32)*5/9 + 273.15 K = 335.93 K

P_sat at T = 16500 Pa (From saturation curves of water)

P_total = 1 atm = 101325 Pa

W/S = 0.5 * P_sat / P_total

S = W * 101325 / (0.5 * 16500) = 1605.9 kg/hr

Density of air at T_in = P M_w / RT = 1 * 28.97 / (0.0821 * 335.93) = 1.05 kg/m3

Volumetric flow rate = S / Density = 1605.9 / 1.05 = 1528.8 m³/hr = 1528.8 * 3.28³ ft³/hr

= 53990.65 ft³/hr