9.10. EQUATIONS FOR VARIOUS TYPES OF DRYERS

9.10A. Through-Circulation Drying in Packed Beds

For through-circulation drying, where the drying gas passes upward or downward through a bed of wet granular solids, both a constant-rate period and a falling-rate period of drying may result. Often the granular solids are arranged on a screen so that the gas passes through the screen and through the open spaces or voids between the solid particles.

1. Derivation of equations

To derive the equations for this case, no heat losses will be assumed, so the system is adiabatic. The drying will be for unbound moisture in the wet granular solids. We shall consider a bed of uniform cross-sectional area A m², where a gas flow of G kg dry gas/h · m² cross section enters with a humidity of H₁. By a material balance on the gas at a given time, the gas leaves the bed with a humidity H₂. The amount of water removed from the bed by the gas is equal to the rate of drying at this time:

Equation 9.10-1

where R = kg H₂O/h · m² cross section and G = kg dry air/h · m² cross section.

In Fig. 9.10-1 the gas enters at T₁ and H₁ and leaves at T₂ and H₂. Hence, the temperature T and humidity H both vary through the bed. Making a heat balance over the short section dz m of the bed,

Equation 9.10-2

where A = m² cross-sectional area, q is the heat-transfer rate in W (J/s), and c_S is the humid heat of the air–water vapor mixture in Eq. (9.3-6). Note that G in this equation is in kg/s · m². The heat-transfer equation gives

Equation 9.10-3

where T_W = wet bulb temperature of solid, h is the heat-transfer coefficient in W/m² · K, and a is m² surface area of solids/m³ bed volume. Equating Eq. (9.10-2) to (9.10-3), rearranging, and integrating,

Equation 9.10-4

Equation 9.10-5

where z = bed thickness = x₁ m.

For the constant-rate period of drying by air flowing parallel to a surface, Eq. (9.6-11) was derived:

Equation 9.6-11

Using Eq. (9.9-9) and the definition of a, we obtain

Equation 9.10-6

Substituting Eq. (9.10-6) into (9.6-11) and setting X₂ = X_C for drying to X_C, we obtain the equation for through-circulation drying in the constant-rate period:

Equation 9.10-7

In a similar manner, Eq. (9.7-8) for the falling-rate period, which assumes that R is proportional to X, becomes, for through-circulation drying,

Equation 9.10-8

Both Eqs. (9.10-7) and (9.10-8), however, hold for only one point in the bed in Fig. 9.10-1, since the temperature T of the gas varies throughout the bed. Hence, in a manner similar to the derivation in heat transfer, a log mean temperature difference can be used as an approximation for the whole bed in place of T − T_W in Eqs. (9.10-7) and (9.10-8):

Equation 9.10-9

Substituting Eq. (9.10-5) for the denominator of Eq. (9.10-9) and substituting the value of T₂ from Eq. (9.10-5) into (9.10-9),

Equation 9.10-10

Substituting Eq. (9.10-10) into (9.10-7) for the constant-rate period and setting x₁ = z,

Equation 9.10-11

Similarly, for the falling-rate period an approximate equation is obtained:

Equation 9.10-12

A major difficulty with the use of Eq. (9.10-12) is that the critical moisture content is not easily estimated. Different forms of Eqs. (9.10-11) and (9.10-12) can also be derived, using humidity instead of temperature (T1).

2. Heat-transfer coefficients

For through-circulation drying, where the gases pass through a bed of wet granular solids, the following equations for estimating h for adiabatic evaporation of water can be used (G1, W1):

Equation 9.10-13

Equation 9.10-14

where h is in W/m² · K, D_p is diameter in m of a sphere having the same surface area as the particle in the bed, G_t is the total mass velocity entering the bed in kg/h · m², and μ is viscosity in kg/m · h. In English units, h is btu/h · ft² · °F, D_p is ft, G_t is lb_m/h · ft², and μ is lb_m/ft · h.

3. Geometry factors in a bed

To determine the value of a, m² surface area/m³ of bed, in a packed bed for spherical particles having a diameter D_p m,

Equation 9.10-15

where ε is the void fraction in the bed. For cylindrical particles,

Equation 9.10-16

where D_c is diameter of cylinder in m and h is length of cylinder in m. The value of D_p to use in Eqs. (9.10-13) and (9.10-14) for a cylinder is the diameter of a sphere having the same surface area as the cylinder, as follows:

Equation 9.10-17

4. Equations for very fine particles

The equations derived for the constant- and falling-rate periods in packed beds hold for particles of about 3–19 mm in diameter in shallow beds about 10–65 mm thick (T1, M1). For very fine particles of 10–200 mesh (1.66–0.079 mm) and bed depth greater than 11 mm, the interfacial area a varies with the moisture content. Empirical expressions are available for estimating a and the mass-transfer coefficient (T1, A1).

EXAMPLE 9.10-1. Through-Circulation Drying in a Bed A granular paste material is extruded into cylinders with a diameter of 6.35 mm and length of 25.4 mm. The initial total moisture content Xt1 = 1.0 kg H2O/kg dry solid and the equilibrium moisture is X* = 0.01. The density of the dry solid is 1602 kg/m3 (100 lbm/ft3). The cylinders are packed on a screen to a depth of x1 = 50.8 mm. The bulk density of the dry solid in the bed is ρS = 641 kg/m3. The inlet air has a humidity H1 = 0.04 kg H2O/kg dry air and a temperature T1 = 121.1°C. The gas superficial velocity is 0.811 m/s and the gas passes through the bed. The total critical moisture content is XtC = 0.50. Calculate the total time to dry the solids to Xt = 0.10 kg H2O/kg dry solid. Solution: For the solid, For the gas, T1 = 121.1°C and H1 = 0.04 kg H2O/kg dry air. The wet bulb temperature TW = 47.2°C and HW = 0.074. The solid temperature is at TW if radiation and conduction are neglected. The density of the entering air at 121.1°C and 1 atm is as follows: Equation 9.3-7 The mass velocity of the dry air is Since the inlet H1 = 0.040 and the outlet will be less than 0.074, an approximate average H of 0.05 will be used to calculate the total average mass velocity. The approximate average Gt is For the packed bed, the void fraction є is calculated as follows for 1 m3 of bed containing solids plus voids. A total of 641 kg dry solid is present. The density of the dry solid is 1602 kg dry solid/m3 solid. The volume of the solids in 1 m3 of bed is then 641/1602, or 0.40 m3 solid. Hence, є = 1 − 0.40 = 0.60. The solid cylinder length h = 0.0254 m. The diameter Dc = 0.00635 m. Substituting into Eq. (9.10-16), To calculate the diameter Dp of a sphere with the same area as the cylinder using Eq. (9.10-17), The bed thickness x1 = 50.8 mm = 0.0508 m. To calculate the heat-transfer coefficient, the Reynolds number is first calculated. Assuming an approximate average air temperature of 93.3°C, the viscosity of air is μ = 2.15 × 10−5 kg/m s = 2.15 × 10−5(3600) = 7.74 × 10−2 kg/m · h. The Reynolds number is Using Eq. (9.10-13), For TW = 47.2°C, λW = 2389 kJ/kg, or 2.389 × 106 J/kg (1027 btu/lbm), from steam tables. The average humid heat, from Eq. (9.3-6), is To calculate the time of drying for the constant-rate period using Eq. (9.10-11) and G = 2459/3600 = 0.6831 kg/s · m2, [View full size image] For the time of drying for the falling-rate period, using Eq. (9.10-12), [View full size image]

9.10B. Tray Drying with Varying Air Conditions

For drying in a compartment or tray dryer where the air passes in parallel flow over the surface of the tray, the air conditions do not remain constant. Heat and material balances similar to those for through circulation must be made to determine the exit-gas temperature and humidity.

In Fig. 9.10-2 air is shown passing over a tray. It enters having a temperature of T₁ and humidity H₁ and leaves at T₂ and H₂. The spacing between the trays is b m and dry air flow is G kg dry air/s m² cross-sectional area. Writing a heat balance over a length dL_t of tray for a section 1 m wide,

Equation 9.10-18

The heat-transfer equation is

Equation 9.10-19

Rearranging and integrating,

Equation 9.10-20

Defining a log mean temperature difference similar to Eq. (9.10-10) and substituting into Eqs. (9.6-11) and (9.7-8), we obtain the following. For the constant-rate period,

Equation 9.10-21

For the falling-rate period, an approximate equation is obtained:

Equation 9.10-22

9.10C. Material and Heat Balances for Continuous Dryers

1. Simple heat and material balances

In Fig. 9.10-3 a flow diagram is given for a continuous-type dryer where the drying gas flows countercurrently to the solids flow. The solid enters at a rate of L_S kg dry solid/h, having a free moisture content X₁ and a temperature T_S1. It leaves at X₂ and T_S2. The gas enters at a rate G kg dry air/h, having a humidity H₂ kg H₂O/kg dry air and a temperature of T_G2. The gas leaves at T_G1 and H₁.

For a material balance on the moisture,

Equation 9.10-23

For a heat balance a datum of T₀°C is selected. A convenient temperature is 0°C (32°F). The enthalpy of the wet solid is composed of the enthalpy of the dry solid plus that of the liquid as free moisture. The heat of wetting is usually neglected. The enthalpy of the gas in kJ/kg dry air is

Equation 9.10-24

where λ₀ is the latent heat of water at T₀°C, 2501 kJ/kg (1075.4 btu/lb_m) at 0°C, and c_S is the humid heat, given as kJ/kg dry air · K:

Equation 9.3-6

The enthalpy of the wet solid in kJ/kg dry solid, where (T_S – T₀)°C = (T_S – T₀) K, is

Equation 9.10-25

where c_pS is the heat capacity of the dry solid in kJ/kg dry solid · K and c_pA is the heat capacity of liquid moisture in kJ/kg H₂O · K. The heat of wetting or adsorption is neglected.

A heat balance on the dryer is

Equation 9.10-26

where Q is the heat loss in the dryer in kJ/h. For an adiabatic process Q = 0, and if heat is added, Q is negative.

EXAMPLE 9.10-2. Heat Balance on a Dryer A continuous countercurrent dryer is being used to dry 453.6 kg dry solid/h containing 0.04 kg total moisture/kg dry solid to a value of 0.002 kg total moisture/kg dry solid. The granular solid enters at 26.7°C and is to be discharged at 62.8°C. The dry solid has a heat capacity of 1.465 kJ/kg K, which is assumed constant. Heating air enters at 93.3°C, having a humidity of 0.010 kg H2O/kg dry air, and is to leave at 37.8°C. Calculate the air flow rate and the outlet humidity, assuming no heat losses in the dryer. Solution: The flow diagram is given in Fig. 9.10-3. For the solid, LS = 453.6 kg/h dry solid, cpS = 1.465 kJ/kg dry solid · K, X1 = 0.040 kg H2O/kg dry solid, cpA = 4.187 kJ/kg H2OK, TS1 = 26.7°C, TS2 = 62.8°C, X2 = 0.002. (Note that X values used are Xt values.) For the gas, TG2 = 93.3°C, H2 = 0.010 kg H2O/kg dry air, and TG1 = 37.8°C. Making a material balance on the moisture using Eq. (9.10-23), Equation 9.10-27 For the heat balance, the enthalpy of the entering gas at 93.3°C, using 0°C as a datum, is, by Eq. (9.10-24), ΔT°C = ΔT K and λ0 = 2501 kJ/kg, from Table A.2-9, For the exit gas, For the entering solid, using Eq. (9.10-25), Substituting into Eq. (9.10-26) for the heat balance with Q = 0 for no heat loss, Equation 9.10-28 Solving Eqs. (9.10-27) and (9.10-28) simultaneously,

2. Air recirculation in dryers

In many dryers it is desired to control the wet bulb temperature at which the drying of the solid occurs. Also, since steam costs are often important in heating the drying air, recirculation of the drying air is sometimes used to reduce costs and control humidity. Part of the moist, hot air leaving the dryer is recirculated (recycled) and combined with the fresh air. This is shown in Fig. 9.10-4. Fresh air having a temperature T_G1 and humidity H₁ is mixed with recirculated air at T_G2 and H₂ to give air at T_G3 and H₃. This mixture is heated to T_G4 with H₄ = H₃. After drying, the air leaves at a lower temperature T_G2 and a higher humidity H₂.

The following material balances on the water can be made. For a water balance on the heater, noting that H₆ = H₅ = H₂,

Equation 9.10-29

Making a water balance on the dryer,

Equation 9.10-30

In a similar manner heat balances can be made on the heater and dryer and on the overall system.

9.10D. Continuous Countercurrent Drying

1. Introduction and temperature profiles

Drying continuously offers a number of advantages over batch-drying. Smaller sizes of equipment can often be used, and the product has a more uniform moisture content. In a continuous dryer the solid is moved through the dryer while in contact with a moving gas stream that may flow parallel or countercurrent to the solid. In countercurrent adiabatic operation, the entering hot gas contacts the leaving solid, which has been dried. In parallel adiabatic operation, the entering hot gas contacts the entering wet solid.

In Fig. 9.10-5 typical temperature profiles of the gas T_G and the solid T_S are shown for a continuous countercurrent dryer. In the preheat zone, the solid is heated up to the wet bulb or adiabatic saturation temperature. Little evaporation occurs here, and for low-temperature drying this zone is usually ignored. In the constant-rate zone, I, unbound and surface moisture are evaporated, and the temperature of the solid remains essentially constant at the adiabatic saturation temperature if heat is transferred by convection. The rate of drying would be constant here but the gas temperature is changing, as well as the humidity. The moisture content falls to the critical value X_C at the end of this period.

In zone II, unsaturated surface and bound moisture are evaporated and the solid is dried to its final value X₂. The humidity of the gas entering zone II is H₂ and it rises to H_C. The material-balance equation (9.10-23) may be used to calculate H_C as follows:

Equation 9.10-31

where L_S is kg dry solid/h and G is kg dry gas/h.

2. Equation for constant-rate period

The rate of drying in the constant-rate region in zone I would be constant if it were not for the varying gas conditions. The rate of drying in this section is given by an equation similar to Eq. (9.6-7):

Equation 9.10-32

The time for drying is given by Eq. (9.6-1) using limits between X₁ and X_C:

Equation 9.10-33

where A/L_S is the exposed drying surface m²/kg dry solid. Substituting Eq. (9.10-32) into (9.10-33) and (G/L_S) dH for dX,

Equation 9.10-34

where G = kg dry air/h, L_S = kg dry solid/h, and A/L_S = m²/kg dry solid. This can be integrated graphically or numerically.

For the case where T_W or H_W is constant for adiabatic drying, Eq. (9.10-34) can be integrated:

Equation 9.10-35

The above can be modified by use of a log mean humidity difference:

Equation 9.10-36

Substituting Eq. (9.10-36) into (9.10-35), an alternative equation is obtained:

Equation 9.10-37

From Eq. (9.10-31), H_C can be calculated as follows:

Equation 9.10-38

3. Equation for falling-rate period

For the situation where unsaturated surface drying occurs, H_W is constant for adiabatic drying, the rate of drying is directly dependent upon X as in Eq. (9.7-9), and Eq. (9.10-32) applies:

Equation 9.10-39

Substituting Eq. (9.10-39) into (9.6-1),

Equation 9.10-40

Substituting G dH/L_S for dX and (H − H₂)G/L_S + X₂ for X,

Equation 9.10-41

Equation 9.10-42

Again, to calculate H_C, Eq. (9.10-38) can be used.

These equations for the two periods can also be derived using the last part of Eq. (9.10-32) and temperatures instead of humidities.