Sizing Of A Rotary Dryer

        Rotary dryers, called the “workhorse of chemical dryers”, belong to the most widely used class of continuous dryers in process industries. These dryers are suitable for relatively free-flowing, non-sticky and granular materials; for example, almost all types of crystals after crystallization and washing. Typical applications of rotary dryers are in drying o table salt, sodium sulphate, ammonium sulphate, and many other salts, drying of sand, minerals, organic solids, polymer resin beads, to mention a few. A rotary dryer consists of a slowly rotating slightly inclined cylindrical shell fed with the moist solid at the upper end. The material flows along the rotating shell, gets dried and leaves the dryer at the lower end.

         It is difficult, if not impossible, to design a rotary dryer on the basis of fundamental principles only. The available design correlations are a few in number and may not prove to be satisfactory for many systems. The design of a rotary dryer (and also of most other dryers) is better done by using the pilot plant test data or the full-scale operation data of a dryer of similar type together with the available correlations.  

Problem Statement & Given Data:

A moist non-hygroscopic granular solid at 26°C is to be dried rom 20% initial moisture to 0.3% final moisture (wet basis) in a rotary dryer at a rate of 1500 kg/h. Hot air enters at 135°C with a humidity of 0.015. The exit solid temperature must not exceed 110°C and the air velocity must not exceed 1.5 m/s in order to avoid dusting of the solid. Specific heat of the dry solid is c_ps = 0.85 kJ/kg.K. Suggest the diameter the length and the other parameters of the dryer.

Solution:

Basis of calculation is 1 hour operation.

Mass of dry solid, L_S = (1500) x (1 – 0.2) = 1200 kg/h; moisture in the wet solid

                 X₁ = 20/80 = 0.25; moisture in the dry solid, X₂ = 0.3/99.7 = 0.00301.

                  Water evaporated, m_S = L_S(X₁ – X₂) = 1200 x (0.25 – 0.00301) = 296.4 kg.

Figure 1 Solid and gas temperature profiles in a counter-current rotary dryer.

           Refer to figure 1. Given: T_S1 = 26°C; T_G2 = 135°C; Y₂ = 0.015

We assume that the exit temperature of the gas is T_G1 = 60°C and that of the solid is T_S2 = 100°C. These values are to be checked later on.

Calculation of enthalpy values of different streams (Reference temperature = 0°C)

H’_S1 = [c_ps + (4.187).X₁] (T_S1 – 0) = [0.85 + (4.187) x 0.25] x (26 – 0) 

        

        = 49.31 kJ/kg dry solid

H’_S2 = [c_ps + (4.187).X₂] (T_S2 – 0) = [0.85 + (4.187) x 0.00301] x (100 – 0) 

        

        = 86.2 kJ/kg dry solid

H’_G2 = [1.005 + 1.88.Y₂] (T_G2 – 0) + Y₂.λ₀ 

        = [1.005 + 1.88 x 0.015] x (135 – 0) + 0.015 x 2500 = 177 kJ/kg

H’_G1 = [1.005 + 1.88.Y₁] (T_G1 – 0) + Y₁.λ₀ 

        = [1.005 + 1.88 x Y₁] x (60 – 0) + Y₁ x 2500 = 60.3 + (2613 x Y₁)

Overall mass balance

G_S (Y₁ – Y₂) = L_S (X₁ - X₂) = G₂ (Y₁ – 0.015) = 296.4 = G_S = 296.4/(Y₁ – 0.015)

             L_S (H’_S2 – H’_S1) = G_S (H’_G2 – H’_G1) = 1200 x (86.2 – 49.31)

                       

                      = [296.4/(Y₁ – 0.015)] x (177 – 60.3 – 2613 x Y₁)

                 Y₁ = 0.04306 and G_S = 296.4/(Y₁ – 0.015) = 10,560 kg/h

Calculation of the shell diameter

Humid volume, V_H = [(1/28.97) + (Y/18.02)] x 22.4 x [(T_G + 273)/273]

Humid volume of the inlet gas (135°C, Y₂ = 0.015), V_H2 = 1.183 m³/kg dry air

Humid volume of the exit gas (60°C, Y₁ = 0.04306), V_H1 = 1.008 m³/kg dry air

The maximum volumetric gas flow rate (this occurs at the end 2 in figure 1) = G_S x V_H2

             

                                     = 10,560 x 1.183 = 12,490 m³/h = 3.47 m³/s

Take the maximum superficial air velocity to be 1.2 m/s (this is 20% less than the maximum allowable velocity since part of the dryer is filled with the moving solid, and the entire cross-section is not available for gas flow). If d is the diameter,

                (πd²/4) x (1.2) = 3.686 = d = 1.98. Select a 2 m diameter shell.

Calculation of the number of heat transfer units

The dryer is considered to consist of three zones as shown in figure 1. The temperature and humidity or moisture content of the streams can be obtained by material and energy balance.

Zone III:  Only heating of the solid occurs in this zone; there is little water left for vaporization. At the boundary between zones III and II, the solid is at T_SB (= T_SA) = 41°C (this value is to be checked and modified later if necessary).

Enthalpy of the solid at the inlet to zone III,

         H’_SB = [0.85 + (4.187) x 0.00301] x (41 – 0) = 35.37 kJ/kg dry solid

Humid heat of the gas entering zone III, c_HB = [1.005 + (1.88) x (0.015)] = 1.033 kJ/kg.K (this remains constant in zone III, since the humidity does not change in this section).

Heat balance over zone III:   L_S (H’_S2 – H’_SB) = G_S (c_HB)_III (T_G2 - T_GB)

  

               = 1200 x (86.2 – 35.37) = 10,560 x 1.033 x (135 – T_GB) = T_GB = 129°C

Adiabatic saturation temperature of air entering zone II (129°C and humidity of 0.015) is 41.3°C. This is fairly close to the guess value of 41°C and T_SA = T_SB = 41°C is not changed.

At the boundary B, ∆T_B = 129 – 41 = 88°C, at end 2, ∆T₂ = 135 – 100 = 35°C

Log mean temperature in zone III (∆T)_m = (88 – 35)/[ln(88 – 35)] = 57.5°C

Number of heat transfer units, (N_tG)_III = (T₂ – T_GB)/(∆T)_m = (135 – 129)/57.5 = 0.104

Zone II:  In order to calculate (N_tG)_II, we need the value of T_GA. This can be obtained by heat balance.

H’_GB = [1.005 + 1.88 x Y_B] x (129 – 0) + 2500 x Y_B = 170.8 kJ/kg. (since Y_B = 0.015)

H’_SA = [0.85 + c_PS x X₁] x (T_SA – 0) = [0.85 + 4.187 x 0.25] x (41 – 0) = 77.77 kJ/kg dry solid

Enthalpy balance: L_S x (H’_SB – H’_SA) = G_S x (H’_GB – H’_GA)

                   

                       = 1200 x (35.37 – 77.77) = 10,560 x (170.8 – H’_GA)

H’_GA = 175.6 = [1.005 + 0.04306 x 1.88] x (T_GA – 0) + 0.04306 x 2500 = T_GA = 63°C

Temperature differences; At section A, (∆T)_A = 63 – 41 = 22°C; (∆T)_B = 88°C

                                 (∆T)_m = (88 – 22)/[ln(88/22)] = 47.6

Number of heat transfer units, (N_tG)_II = (T_GB - T_GA)/(∆T)_m = (129 – 63)/47.6 = 1.386

Before calculating (N_tG)_I, let us check the validity of the assumed value of the exit gas temperature, T_G1 = 60°C, by making an energy balance over zone I.

                                       

                                     G_S (H’_G2 – H’_G1) = L_S (H’_S2 - H’_S1)

               = 10,560 (175.6 – H’_G1) = 1200 (77.77 – 49.31) = H’_G1 = T_G1 = 59.6°C, matches the assumed value.

Zone I:    (∆T)_I = 60 – 26 = 34°C; (∆T)_A = 22°C; (∆T)_m = (34 – 22)/[ln(34/22)] = 27.5

Number of heat transfer units, (N_tG)_I = (T_GA – T_G1)/(∆T)_m = (63 – 60)/27.5 = 0.109

Total number of heat transfer units N_tG = 0.104 + 1.386 + 0.109 = 1.53 (this lies within the usual range).

Length of a transfer unit

Average gas mass flow rate = [(10,560) x (1.015) + (10,560) x (1.04306)]/2 = 10,867 kg/h

The gas mass flow rate, G’ = (10,867/3600)/[(π/4) x 2²] = 0.961 kg/m² s

Volumetric heat transfer coefficient, U_a = [237 x (G’)^0.67]/d =[237 x (0.961)^0.67]/2 

                                                               

                                                         U_a = 115 W/m³ K

         Humid heat at the ends: c_H2 = 1.005 + 1.88 x 0.015 = 1.033

                                    

                                                c_H1 = 1.005 + 1.88 x 0.04306 = 1.083

Average humid heat, c_H = (1.033 + 1.083)/2 = 1.058 kJ/kg K = 1058 J/kg dry air.K

Length of a heat transfer unit, L_T = G’_cH/U_a = (0.961 x 1058)/115 = L_T = 8.84 m

Length of the dryer, L = (N_tG) x (L_T) = 1.56 x 8.84 = L = 13.8 m. Select 14 m

                               Select a  2 m diameter, 14 m long dryer