Lecture 3: DESIGN CONSIDERATION OF DRIERS

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Lecture 3: DESIGN CONSIDERATION OF DRIERS 8. DESIGN OF DRYER Design of a rotary dryer only on the basis of fundamental principle is very difficult.

1 Lecture 3: DESIGN CONSIDERATION OF DRIERS 8. DESIGN OF DRYER Design of a rotary dryer only on the basis of fundamental principle is very difficult. Few of correlations that are available for design may not prove to be satisfactory for many systems. The design of a rotary dryer is better done by using pilot plant test data and the full scale operating data of dryer of similar type if available, together with the available design equations. A fairly large number of variables are involved such as solid to be dried per hour, the inlet and exit moisture contents of the solid, the critical and equilibrium moisture contents, temperature and humidity of the drying gas. The design procedure based on the basic principles and available correlations is discussed below. In this case we assume that the solid has only unbound moisture and as shown in fig 2.7 in stage II the solid is at the wet bulb temperature of the gas. Figure 2.7: Temperature profile for solid and gas in a counter current rotary dryer 1. Heat losses from dryer surfaces are neglected. 2. Once the capacity of the dryer is known, the drying gas flow rate, its temperature and humidity are decided considering a number of factors. And the following moisture & enthalpy balances need to be satisfied. Gs (Y 1 - Y 2 ) = Ms (X 1 X 2 ) Gs (Hg 2 Hg 1 ) = Ms (H S2 Hs 1 ) Here, G s = flow rate of air (dry basis, kg/h), Ms = flow rate of solid (kg/h, dry basis), H s = humidity of air (kg/h 2 O/kg dry air) Joint initiative of IITs and IISc Funded by MHRD Page 17 of 39

2 3. The gas and solid temperatures at the stage boundaries are obtained by moisture and energy (enthalpy) balances. The number of heat transfer unit for each zone is calculated. for the stage II. The number of heat transfer units is given by 4. The total length of dryer is given by (N tg ) h,ii T m = (T GB T GA ) L = (L T ) 1 (N tg ) 1 + (L T ) II (N tg ) II + (L T ) III (N tg ) III 5. The shell diameter is calculated from the dry gas flow rate (from step I) and suitable gas flow velocity or gas mass flow rate Some useful correlations for the design of a rotary dryer are given below. Volumetric gas-solid heat transfer coefficient. Ū a = (W/m 3.K) = 237 (G ) 0.67 /d Here, G = gas mass flow rate (kg/m 2.h) and d, dryer diameter Length of transfer unit L T = G CH / Ū a L T = CH. d G S Here, c H = average humid heat, and d = dryer diameter Solid retention time: ' 0.23 L B L G θ = S N d F Where, θ = retention time (min); (+ve sign is for counter flow; ve sign is for parallel flow of the gas and solid) L = dryer length (m) S = slope of the dryer (m/m); N = speed (rpm) G = gas mass flow rate (Kg/m 2.h) F = feed rate (Kg/m 2. h) dry basis B = 5 (d p ) -0.5 d p = weight average particle diameter (micron) d = dryer diameter (m) Joint initiative of IITs and IISc Funded by MHRD Page 18 of 39

3 Example 2.1: Size of the rotary dryer can be estimated for the following case. A moist non hygroscopic granular solid at 26 0 C is to be dried from 20% initial moisture to 0.3% final moisture in a rotary dryer at a rate of 1500 kg/h. The hot air enters the dryer at C with a humidity of With condition that the temperature of the solid leaving the dryer must not exceed C and the air velocity must not exceed 1.5 m/s in order to avoid dust carry over. C ps = 0.85 kj/kg.k. Recommend the diameter, length and other parameters of the dryer. Solution: Basis of calculation is 1 hr operation Solid contains 20% initial moisture Mass of dry solid = M S = 1500 (1-0.2) = 1200 kg/hr Moisture in the wet solid = X 1 = 20/80 = 0.25 Moisture in the dry solid = X 2 = 0.3/99.7 = Water evaporated, m S, evaporated = M S (X 1 X 2 ) = 1200 ( ) = Kg Given data: T S1 = 26 0 C; T G2 = C; Y 2 = Let us assume that the exit temperature of the gas is T G1 = 60 o C and for solid T S2 = 100 o C Now enthalpy of different streams (suppose ref temp = 0 o C) H S1 = [C PS + (4.187) X 1 ] [T S1 0] = [ (4.187) 0.25] [26 0] = KJ/kg dry air H S2 = [C PS + (4.187) X 1 ] [T S1 0] = [ (4.187) ] [100 0] = 86.2 KJ/kg dry solid H g2 = [ (1.88) 0.015] [135 0] + (0.015) (2500) = 177 KJ/kg H g1 = [ (1.88) Y 1 ] [60 0] + Y 1 (2500) = Y 1 Overall mass balance G S (Y 1 Y 2 ) = M S (X 1 X 2 ) G S (Y ) = G S = 296.4/(Y ) M S [H S2 H S1 ] = G S [H g2 H g1 ] 1200 [ ] = 296.4/(Y ) ( Y 1 ) Joint initiative of IITs and IISc Funded by MHRD Page 19 of 39

4 Y 1 = and G s = 296.4/(Y ) = Kg/h Shell Diameter Volume of humid inlet gas (135 0 C and Y 2 = 0.015) V H2 = m 3 /Kg dry air Volume of humid exit gas (60 0 C and Y 1 = ) V H1 = m 3 /Kg dry air The max. volumetric gas flow rate = G s.v H2 = = m 3 /h The working velocity i.e. superficial velocity = = 1.2 m/s / 4 d 2 (1.2) = d = 1.98 m, say 2.0 m Heat Transfer Unit Dryer is divided into three zones and therefore, the stage wise calculation of temperature and humidity of the stream can be obtained by material and energy balance. Stage III Very less water left for vaporization in stage III. Consider solid is at T SB, the wet bulb temperature of the air at location between III & II. assume T SB = T SA = 41 0 C Enthalpy of solid at the inlet to stage III H SB = [ ( ) (4.187)] (41-0) = KJ/kg dry solid Humid heat of gas entering stage III C HB = [ (1.88) (0.015)] = KJ/kg.K Heat balance over stage III M S [H S2 - H SB ] = G S (C HB ) III (135 T GB ) T GB = C Adiabatic saturation temperature of air entering stage II (129 0 C & humidity of 0.015) is C. At the boundary B, At end 2, T B = = 88 0 C T 2 = = 35 0 C Joint initiative of IITs and IISc Funded by MHRD Page 20 of 39

5 LMTD III = ( T) m = 88-35/ln(88/35) = C (N tg ) III = T 2 T GB /( T) m = /57.5 = Stage II Use heat balance equation over stage II to calculate the value of T GA. The calculated T GA value can be use to estimate the number of transfer units. Since Y B = H GB = [ Y B ] (129-0) (Y B ) = KJ/Kg H AS = [ C PS X 1 ] (T SA -0) = [ (4.187) (0.25)] (41) = KJ/(Kg dry solid) Enthalpy balance: M S (H SB H SA ) = G S (H GB H GA ) 1200 ( ) = ( H GA ) H GA = KJ/Kg Once H GA value is known then T GA can be calculated using the following equation H GA = = [ (1.88)] [T GA - 0] (2500) T GA = 63 0 C At section A temp diff. T A = = 22 0 C and T B = 88 0 C ( T) M = (88-22)/ ln(88/22) = C Number of transfer unit = (N tg ) II = T GB T GA /( T) M = (129 63)/47.6 = To validate the assumed value of exit gas temperature i.e. T G1 = 60 0 C, first do an energy balance over stage I. G S (H g2 H g1 ) = M S (H S2 H S2 ) (175.6 H g1 ) = 1200 ( ) H g1 = T G1 = C Joint initiative of IITs and IISc Funded by MHRD Page 21 of 39

6 Stage I ( T) 1 = = 34 0 C ( T) A = 22 0 C ( T) M = 34-22/ln (34/22) = 27.5 Number of transfer unit, N tg = = 1.53 Length of Transfer Unit: Avg. mass flow rate = [10560 (1.015) ( )]/2 = Kg/h The gas mass flow rate, G = (10867/3600)/ / 4 (2) 2 Volumetric heat transfer coeff. = Humid heat at the ends Avg. humid heat, = Kg/m 2.S U a = (237 (G ) 0.67 )/d U a = (237 (0.961) 0.67 )/2 = 115 W/m 3.K C H2 = (0.015) = C H1 = ( ) = C H = ( )/2 = KJ/Kg. K Length of transfer unit, L T = G. C H / U = ( )/115 = 8.84 m a Length of dryer, L = N tg. L T = = 13.8 m d = 2 m and L = 14 m Joint initiative of IITs and IISc Funded by MHRD Page 22 of 39

Principles of Food and Bioprocess Engineering (FS 231) Solutions to Example Problems on Psychrometrics

Principles of Food and Bioprocess Engineering (FS 21) Solutions to Example Problems on Psychrometrics 1. We begin by identifying the conditions of the two streams on the psychrometric chart as follows.