Calculation Guidelines for Rotary Drum Dryer

1. DESIGN PROCEDURE FOR A ROTARY DRYER:

2. ROTARY DRYER DESIGN
1. Heat and mass balance in a rotary dryer
1.1 Mass Balance
To calculate the heat duty, a mass balance must be established. General mass balance
around the dryer is given in the equation below.
(1.1)
(1.2)
Mass of feed is equivalent to the mass of solids and the moisture it contains. This can be
used in calculating the heat balance around the dryer.
(1.3)
(1.4)
1.2 Heat Duty of dryer
Heat transferred in direct-heat rotary dryer is expressed as follows:
(1.5)
Q is the rate of heat transfer, Ua is the volumetric heat transfer coefficient, V is the dryer
volume, and Δt is the true mean difference between the drying air and the material.
The heat supplied by the drying air is used for five different operations:
a. To heat the dry solid from its inlet temperature to its final temperature.
(1.6)
b. To heat moisture to vaporization temperature (inlet wet-bulb temperature).
(1.7)
c. Heat to evaporate moisture.
(1.8)
d. To heat residual moisture to final temperature.
(1.9)
e. To superheat the evaporated moisture
(1.10)
The overall heat transfer to the product is given by the equation.
𝑄 = (1 + 𝛼)( 𝑄1 + 𝑄2 + 𝑄3 + 𝑄4 + 𝑄5) (1.11)
Q=UaV(ΔT)lm
𝑄1 = 𝑚𝑠 × 𝐶𝑠 × (𝑇𝑓2 − 𝑇𝑓1)
𝑄2 = 𝑚𝑤 × 𝐶𝑤 × (𝑇𝑤 − 𝑇𝑓1)
𝑄3 = 𝑚𝑒 × λ𝑙𝑣
𝑄4 = 𝑚𝑟𝑤 × 𝐶𝑤 × (𝑇𝑓2 − 𝑇𝑤)
𝑄5 = 𝑚𝑒 × 𝐶𝑣 × (𝑇𝑓2 − 𝑇𝑤)
𝑚𝑓 = 𝑚𝑒 + 𝑚𝑝
𝑚𝑓 (1 − 𝑋1) = 𝑚𝑝 (1 − 𝑋2)
𝑚𝑠 = 𝑚𝑓 (1 − 𝑋1)
𝑚𝑤 = 𝑚𝑓 (𝑋1)

3. where 𝛼 is a factor that presents heat loss due to convection and radiation which ranger to 7-
10%.
1.3 Air mass rate
The air mass rate required to transfer sufficient heat for the drying is
(1.12)
Density of air can be estimated using the psychrometric chart or with the equation below
(1.13)
1.4 Gas mass velocity
Gas mass velocity must be determined before modelling a rotary dryer. Allowable mass
velocity, G, of the gas in a direct-contact rotary dryer depends on the dusting characteristic of the
material being dried and usually ranges from 2000-25000 kg/m2
h.
1.5 Air humidity
The humidity of the exit air must be checked to verify that it does not exceed the
maximum vapor the exit air can hold.
(1.13)
2. Number of Heat Transfer Unit and Log-Mean Temperature
NTU is the ratio of the overall thermal conduction to the smaller heat capacity. It is a
combination of overall heat transfer coefficient, transfer area, fluid flow rate, and heat capacity
parameters which is combined to form this one-dimensional parameter. This can be used to
calculate the wet-bulb temperature if exit gas temperature is given and vice versa.
(2.1)
Rotary dryers are operated most economically when Nt is between 1.5 to 2.5.
Log-mean temperature can be calculated using the equation
(2.2)
𝑚𝑎 =
𝑄
𝐶𝑎(𝑇𝑎2 − 𝑇𝑎1)
𝑌2 = 𝑌1 +
𝑚𝑒
𝑚𝑎
Nt=ln(Ta1-Tw/Ta2-Tw)
(ΔT)lm=
(𝑇𝑎1 − 𝑇𝑤) − (𝑇𝑎2 − 𝑇𝑤)
ln(Ta1-Tw/Ta2-Tw)
𝑣ℎ =
22.41𝑇
273.15
× (
1
28.97
+
𝐻
18.02
)

4. 3. Diameter
Dryer diameter is a function of the amount of materials that will be subject to drying. The
gas mass velocity is also a factor in determining the ideal dryer diameter.
(3.1)
To calculate the diameter of the dryer
(3.2)
4. Overall heat transfer coefficient
The volumetric heat-transfer coefficient itself consists of a heat-transfer coefficient Uv
based on the effective area of contact between the gas and the solids, and the ratio a of this
area to the dryer volume. An empirical equation is to calculate volumetric heat transfer is given
as follow:
(4.1)
Optimum value for the constant k by AICHE ranges from 3.75-5.25 and is dependent on the
materials, flight geometry, rotational speed, and dryer holdup. Values for n are suggested by
various authors.
Friedman and Marshall (1949) 0.16
Aiken and Polsak (1982) 0.37
Miller et al. (1942) 0.60
McCormick (1962) 0.67
Myklestad (1963) 0.80
5. Length of Dryer
Dryer length is a function of the calculated values above. It highly affects the residence time of
the material and one of the bases of effective dryer. Length is calculated based on the heat transfer
coefficient (equation 1.5).
(5.1)
Volume can be calculated after length is determined.
(5.2)
𝐴 =
𝑚𝑎
𝐺
𝑑 = √
4𝑚𝑎
𝜋𝐺
𝑈𝑎 =
𝑘𝐺𝑛
𝐷
𝐿 =
4𝑄
𝜋𝐷2𝑈𝑎 ∙ (ΔT)lm
V=A×L

5. 6. Peripheral Speed and Rotational Speed
Peripheral Speed P is the distance travelled by a point in a perpendicularly rotating body over
a period of time. Peripheral speed ranges from 0.1 – 0.5 m/s.
Rotational speed is calculated
(6.1)
7. Flights design
Flights help to increase the surface area of contact between the material and drying air. Its
dimension is based on the percent loading of the solid in the dryer. An ideal solids loading is said
to be optimum between 10-15%. With this, the number and height of flights can be calculated
using the equations below.
(7.1)
(7.2)
For more accurate design procedure, see flights designing.
8. Residence Time
Residence time equation is given by Friedmann and Marshall. The second terms is the drag
friction caused by the gas.
(8.1)
where S is the slope in rad, F is the solids feed rate per minute. B is dependent on particle diameter
in microns.
9. Dryer Hold-up volume
Hold-up volume is volume occupied by the solids inside the dryer within the residence time. This
is important since it can identify if the dryer is underloaded or overloaded. It is also used for load
computations.
(9.1)
(9.2)
N=P/πD
Fd
=D/8
Nf
=3D
τ=
0.3344𝐿
S𝑁0.9D
+
0.608BLG
F
B=5𝑑𝑝−0.5
%𝐻𝑉 =
𝑉𝑠𝑜𝑙𝑖𝑑𝑠
𝑉𝑑𝑟𝑦𝑒𝑟
𝑉𝑠𝑜𝑙𝑖𝑑𝑠 =
𝐹 × 𝑅
ρb

6. Design procedure for rotary dryer

7. Case Study: Solids drying at AFC Granulation Plant 1
Step 1: Specification of Duty
Fertilizer from a granulator with a temperature of 80 0
C and at a rate of 30000 kg/hr containing
moisture of 5% (w.b.) will be dried in a rotary dryer to a final product containing 1.5% moisture. Drying air
from the burner is available at 400 0
C and 0.0188 kg/kg dA humidity. Material must not be heated above
130 0
C. Maximum air mass velocity is set at 9000 kg/hr-m2.
Step 2: Collection of physical and thermophysical data
Data for Process Design Calculation
Product Specification:
Parameter Data Symbol Unit
Feed Rate 30000 F Kg/hr
Initial Moisture content wb 5% X1 Kg/kg wb
Final Moisture content wb 1.5% X2 Kg/kg wb
Inlet Temperature feed 333 Tf1 K
Final Temperature feed 370 Tf2 K
Product average diameter 4 dp mm
Thermophysical Properties:
Parameter Data Symbol Unit
Specific heat of product 1.188 Cs Kg/hr
Specific heat of water 4.18 Cw Kg/kg wb
Specific heat of vapor 1.89 Cv Kg/kg wb
Specific heat of air 1.048 Ca K
Latent heat of vaporization 2300 λl kJ/kg
Bulk Density feed 970 ρb Kg/m3
Properties of Air
Parameter Data Symbol Unit
Gas Mass flow rate 9000 G Kg/hr m2
Atmospheric air
temperature
303 Ta K
Air humidity 0.0188 Y1 Kg H2O/kg dA
Temperature of air after
burner
673 Ta1 K

8. Step 3: Determination of temperature profiles
With the given properties of air, identify wet-bulb temperature using psychrometric chart. Calculate the
outlet air temperature using equation 2.1. At 673K and 0.0188 kg/kg dA, wet-bulb temperature is 343 K.
Using equation 2.1 with Nt say 2, estimate the outlet air.
The outlet air temperature is estimated to be 387 K. Calculate the log mean temperature using equation
2.2.
Step 4. Perform Mass balance
Step 5: Perform Heat Duty Calculation using equation (1.5-11)
Nt=ln(Ta1-Tw/Ta2-Tw)
2=ln(673-343/Ta2-343)
Ta2 = 387
(ΔT)lm=
(𝑇𝑎1 − 𝑇𝑤) − (𝑇𝑎2 − 𝑇𝑤)
ln(Ta1-Tw/Ta2-Tw)
(ΔT)lm=
(673 − 343) − (387 − 343)
ln(673-343/387-343)
= 142.669
𝑚𝑓 = 𝑚𝑒 + 𝑚𝑝
𝑚𝑓 (1 − 𝑋1) = 𝑚𝑝 (1 − 𝑋2)
30000 = 𝑚𝑒 + 𝑚𝑝
30000(1 − .05) = 𝑚𝑝 (1 − .015)
𝑚𝑝 = 28934.02
me = 1065.98
𝑄1 = 𝑚𝑠 × 𝐶𝑠 × (𝑇𝑓2 − 𝑇𝑓1)
𝑄2 = 𝑚𝑤 × 𝐶𝑤 × (𝑇𝑤 − 𝑇𝑓1)
𝑄3 = 𝑚𝑒 × λ𝑙𝑣
= 30000 × 0.95 × 1.188 (370 − 333) = 1252746 𝑘𝐽/ℎ𝑟
= 30000 × .05 × 4.184(343 − 333) = 62760 𝑘𝐽/ℎ𝑟
= 1065.98 × 2400 = 2558375.635 𝑘𝐽/ℎ𝑟

9. 𝑄 = (1 + 𝛼)( 𝑄1 + 𝑄2 + 𝑄3 + 𝑄4 + 𝑄5)
At an allowance of 10% for heat loss
𝑄 = (1 + .1)( 𝑄1 + 𝑄2 + 𝑄3 + 𝑄4 + 𝑄5)
𝑄 = 4414178.044 𝑘𝐽/ℎ𝑟
Step 6: Calculate air requirements
Using equation 1.12
Calculate volumetric flow of air using equation 1.13
𝑉 = 𝑚𝑎 ∗ 𝑣ℎ
𝑄4 = 𝑚𝑟𝑤 × 𝐶𝑤 × (𝑇𝑓2 − 𝑇𝑤)
𝑄5 = 𝑚𝑒 × 𝐶𝑣 × (𝑇𝑓2 − 𝑇𝑤)
= 28934.02 × .015 × 4.184(370 − 343) = 49029.26 𝑘𝐽/ℎ𝑟
= 1065.98 × 1.89 × (370 − 343) = 89978.24 𝑘𝐽/ℎ𝑟
𝑚𝑎 =
𝑄
𝐶𝑎(𝑇𝑎2 − 𝑇𝑎1)
𝑚𝑎 =
4414178.044
1.048(673 − 387)
= 14761.36488 𝑘𝑔/ℎ𝑟
𝑣ℎ =
22.41𝑇
273.15
× (
1
28.97
+
𝐻
18.02
)
=
22.41 ∗ 673
273.15
× (
1
28.97
+
0.0188
18.02
) = 1.9635
𝑚3
𝑘𝑔 𝑑𝐴
𝑌2 = 14761.365 × 1.9635 = 28984.4714 𝑚3
/ℎ𝑟

10. Step 7: Calculate outside air humidity
Using equation
The outlet humidity is below the saturation humidity at outlet air properties.
Step 8: Calculate diameter
Assuming a Gas Mass Velocity of 3420 kg/hr m2, calculate the diameter using equation
Step 9: Calculate volumetric heat transfer coefficient
Using equation 4.1, setting k as 4 (optimum range 3.75-5.25) and n as 0.67 as McCormick suggested.
Step 10: Dryer Length and Volume Calculation
Using equation 5.1, calculate the length of the dryer. Equation 5.2 is used to calculate dryer volume.
𝑌2 = 𝑌1 +
𝑚𝑒
𝑚𝑎
𝑌2 = 0.0188 +
1065.98
14761.36488
= 0.0910 kg H2O/kg dA
𝐴 =
𝑚𝑎
𝐺
𝑑 = √
4𝑚𝑎
𝜋𝐺
𝐴 =
14761.36488
3420
= 4.316 𝑚2
𝑑 = √
4 × 14761.36
𝜋 × 3420
= 2.344 𝑚
𝑈𝑎 =
𝑘𝐺𝑛
𝐷
=
4 × 3420.67
2.344
=
397.97𝑘𝐽
ℎ𝑟 − 𝑚3 − 𝐾
=
4 × 4414178.044
𝜋 × 2.3442 × 397.97 ∙ 142.669
= 18.012 𝑚
V=A×L
𝐿 =
4𝑄
𝜋𝐷2𝑈𝑎 ∙ (ΔT)lm
= 4.316 × 18.012 = 77.7439 𝑚3

11. Step 11: Estimate Number of flights and flight design
(See flights design procedure)
Step 12: Estimate peripheral speed and Calculate rpm
Peripheral speed of dryers is typically between 0.1-0.5 m/s. Say the speed is 0.5 m/s, using equation 6.1,
calculate the rotational speed.
Step 13: Estimate slope and calculate residence time
Most dryers operate at a slope of 1ᵒ-5ᵒ, say for example that this dryer operates at 2ᵒ. General equation
for estimating the residence time of a material in a rotary dryer is given by Friedmann and Marshall.
Equation 8.1.
The drag of the gas in parallel operation pushes the material towards the outlet, thus, the negative sign
is used and results in a shorter residence time.
Step 14: Calculate hold up volume
Using equation 9.2
Holdup volume is calculated using 9.1.
N=P/πD
=0.5/π*2.344 = 4.07 rpm
B=5𝑑𝑝−0.5
τ=
0.3344𝐿
S𝑁0.9D
+
0.608BLG
F
=5*4000−0.5
= 0.07906
=
0.3344 ∗ 18.012
(
𝜋
180 × 2) 4.070.92.344
-
0.608*0.07906*18.012*28984.4714/60
30000/60
= 19.958 𝑚𝑖𝑛𝑠
𝑉𝑠𝑜𝑙𝑖𝑑𝑠 =
𝐹 × 𝑅
ρb
𝑉𝑠𝑜𝑙𝑖𝑑𝑠 =
30000 × 19.958 × 1/60
970
= 10.28762 𝑚3
%𝐻𝑉 =
𝑉𝑠𝑜𝑙𝑖𝑑𝑠
𝑉𝑑𝑟𝑦𝑒𝑟
%𝐻𝑉 =
10.28762
77.7489
× 100 = 13.23%

12. This value is within optimum holding volume value of 10-16%. Thus, design can be considered as
appropriate.
Step 15: Estimation of Mechanical works
Proceed to the calculation of insulation if applicable. Consult mechanical department for motor
specifications.