Direct heated rotary dryer

CALCULATION REPORT
TECHNOLOGICAL AND
CONSTRUCTION PROJECT
EQUIPMENT "DIRECT HEATED ROTARY
DRYER FOR SILICA SAND"
Nomenclature
1 Project initial data obtain from terms of reference (TOR)

DIRECT HEATED ROTARY DRYER FOR SILICA SAND
Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”
Nomenclature
2 Mass balance
2.1 Dry solid in the feed
2.2 Water rate in the feed
2.3 Water rate in the final product
2.4 Water rate evaporated
2.5 Final product weight
3 Heat balance
3.1 Heat for raising the temperature of material feedstock
3.2 Heat for raising temperature of moisture that remain in material
3.3 Heat for raising temperature of moisture that leave with air
3.4 Heat for evaporating moisture that leave with air
3.5 Heat for superheating moisture at 149*C, that leave with air
3.6 Heat loss from the surface of dryer
3.6.1 Pre determining the volume of dryer
3.6.2 Pre determining of diameter and length of dryer
3.6.3 Insulation thickness and dryer surface heat loss
3.6.3.1 Heat transfer coefficient
3.6.3.1.1 Convection heat transfer coefficient inside dryer
3.6.3.1.1.1 Mean temperature of gasses inside the dryer
3.6.3.1.1.2 Mean velocity of gasses inside the dryer
3.6.3.1.1.3 Air thermo-physical parameters, inside the dryer
3.6.3.1.1.4 Determining flow regime (Reynolds number)
3.6.3.1.1.5 Heat transfer coefficient due to forced convection
3.6.3.1.1.5.1 Nusselt number
3.6.3.1.1.6 Heat transfer coefficient due to free convection
3.6.3.1.1.6.1 Grashov number
3.6.3.1.1.7 Combined convection heat transfer coefficient
3.6.3.1.2 Convection heat transfer coefficient outside the dryer
3.6.3.1.2.1 Mean temperature of gasses outside the dryer
3.6.3.1.2.2 Mean velocity of gasses outside the dryer
3.6.3.1.2.3 Air thermo-physical parameters outside the dryer
3.6.3.1.2.4 Estimated dryer outer diameter
3.6.3.1.2.5 Prandtl number
3.6.3.1.2.6 Nusselt number
3.6.3.1.2.7 Heat transfer coefficient due to convection
3.6.3.1.3 Heat transfer coefficient due to radiation
3.6.3.1.4 Combined heat transfer coefficient
3.6.3.2 Determination of insulation thickness
3.6.3.2.1 Temperature inside dryer drum surfce
3.6.3.2.2 Temperature outside dryer drum surfce
3.6.4 Overall heat loss from drum surface
3.6.4.1 Calculation of mean log temperature
3.6.5 Absolut error (difrece betwen caculated heat losses and estimated heat losses)
3.6.6 Relative error (ratio betwen apolut error and estimated heat losses)
3.7 Heat loss from surface of combustion chamber
3.7.1 Estimation of diameter and length of combustion chamber

3.6.3.2.2 Temperature outside dryer drum surfce
3.6.5 Absolut error (difrece betwen caculated heat losses and estimated heat losses)
3.7.2 Insulation thickness and combustion chamber surface heat loss analysis
3.7.2.1.1 Convection heat transfer coefficient inside combustion chamber
3.7.2.1.1.1 Mean temperature of gasses inside combustion chamber
3.7.2.1.1.2 Mean velocity of gasses inside combustion chamber
3.7.2.1.1.3 Air thermo-physical parameters inside combustion chamber
3.7.2.1.1.4 Determining flow regime (Reynolds number)
3.7.2.1.2 Convection heat transfer coefficient outside combustion chamber
3.7.2.1.2.1 Mean temperature of gasses outside combustion chamber
3.7.2.1.2.2 Mean velocity of gasses outside combustion chamber
3.7.2.1.2.3 Air thermo-physical parameters outside combustion chamber
3.7.2.1.2.4 Estimated combustion chamber outer diameter
3.7.2.1.2.6 Nusselt numbber
3.7.2.2.1 Temperature inside combustion chamber surface
3.7.2.2.2 Temperature outside combustion chamber surface
3.7.3 Overall heat loss from combustion chamber surface
3.7.4 Absolute error from heat loss estimations
3.7.5 Relative error from heat loss estimation
3.8 Over all heat required for process equipment
3.9 Heat loss from exhaust gases leaving the dryer
3.9.1 Ratio between mass of air needed to make up for total heat loss from gases and total
mass of air needed
3.9.2 Total heat loss from gasses leaving the dryer
3.9.3 Mass of air needed to convey heat for the process
3.9.4 Mass of air needed to make up for total heat loss from gases
3.9.5 Total air mass
3.9.6 Sum of overall heat required (Q1 ) and total heat loss from gasses(Q2 )
3.9.7 Heat loss from exhaust gases leaving the dryer
3.10 Heat loss from moisture in exhaust gases leaving the dryer
3.10.1 Mass of moisture entering combustion chamber
3.11 Fuel analysis
3.11.1 Fuel data
3.11.1.1 Component data
3.11.2 Chemical reactions ocuring and energy released
3.11.3 Total energy released from combusting one kilogram of fuel
3.11.4 Total amount of air used for combusting one kilogram of fuel
3.11.4.1 Total amount of oxygen needed to combust one kilogram of fuel

3.11 Fuel analysis
3.11.1 Fuel data
3.11.4.2 Total amount of air needed to combust one kilogram of fuel
3.11.5 Total amount of gasses released after combusting one kilogram of fuel
3.11.5.1 Total amount of carbon dioxide released after combusting one kilogram of fuel
3.11.5.2 Total amount of steam released after combusting one kilogram of fuel
3.11.5.3 Total amount of sulphur dioxide released after combusting one kilogram of fuel
3.11.5.4 Total amount of gasses released as a by product of combustion for one kilogram of
fuel
3.11.6 Amount of fuel needed in an hour
3.11.6.1 Heat loss caused by combustion gases leaving the dryer
3.11.6.1.1 Total heat loss caused by combustion gases leaving the dryer
3.11.6.1.2 Ratio betwen heat loss caused by combustion gases for each unit of heat produced
3.11.6.2 Heat loss from impurities in the fuel
3.11.6.2.1 Ratio of heat loss from impurities in the fuel for each unit of heat produced
3.11.6.3 Heat loss as a cosequece of hydrogene in fuel
3.11.6.3.1 Ratio betwen heat loss as a cosequece of hydrogene in fuel for each unit of heat
produced
3.11.6.4 Total combustion energy needed
3.11.7 Amount of steam produced in an hour as a result of combustion
3.11.8 Total amount of heat loss from combustion
3.11.9 Process efficiency
3.11.10 Equipment efficiency
3.11.11 Combustion efficiency
3.11.12 Defining total air mass needed for combustion
4 Air mass balance
4.1.1 Air mass flow rate leaving the dryer
4.1.2 Water mass flow rate leaving the dryer
4.1.3 Air flow rate leaving the dryer
4.1.4 Moisture volume flow rate leaving the dryer
4.1.5 Total gas volume flow rate leaving the dryer
4.1.6 Density of gas leaving the dryer
4.1.7 Air humidity leaving the dryer
4.1.8 Mass of gasess and concentration entering combustion chamber, entering dryer and
leaving the dryer
5 Velocity of gasses
5.1 Dust data
5.2 Velocity of gasses
5.3 Correcting velocity of gasses in terms of density ratio
6 Dust load
6.1 Total volume of material leaving the dryer
6.2 Volume of material coveyored with air
6.3 Mass of material coveyored with air
7 Dryer diameter

5 Velocity of gasses
5.1 Dust data
6 Dust load
7 Dryer diameter
8 Design of lifters
8.1 Percentage of the loaded area
8.2 Depth of the loaded area
8.3 Lifter depth
8.4 Cross sectional area of material retained by each lifter
8.5 Number of lifters
9 Rotation speed
10 Time for a showering cycle
10.1 Particle falling time
10.2 Particle lifting time
10.3 Time of showering cycle
11 Showering load per unit length of dryer
11.1 Time ratio
11.2 Time of one revolution
12 Dryer length
12.1 Cross sectional surface area of dryer
12.2 Mean log temperature
12.3 Volumetric heat transfer coefficient
12.3.1 Heat transfer coefficient for unit of dried material
12.3.1.1 Mean diameter particle size
12.3.1.2 Air physical parameters
12.3.1.3 Mean velocity of the falling particle
12.3.1.4 Total velocity of the moving particle
12.3.1.5 Reynolds number
13 Bed load calculation
14 Effective slope
14.1 Showering load for unit of showering cycle
14.2 The effect of gas on the effective slope
15 Showering output
16 Retention time
17 Klin action load
18 Total bed load
19 Mean retention time
20 Percentage of loaded shell
21 Motor power calculation

16 Retention time
17 Klin action load
18 Total bed load
21.1 Showering power
21.2 Kling action power
21.3 Motor power
22 Dryer drum stress and strain analysis
22.1 Dryer drum weight
22.1.1 Total weight of material
22.1.2 Weight of lifters
22.1.3 Insulation weight
22.1.4 Dryer drum weights
22.1.5 Weight of girth gear
22.2 Determination of loads that act on dryer drum
22.2.1 Reaction forces
22.2.2 Maximal bending moment that is applied in the drum
22.2.3 Maximal torsional moment that is applied in the drum
22.3 Dryer drum section properties
22.3.1 Resistance momentum of dryer drum
22.3.2 Polar resistance momentum of dryer drum
22.4 Overall dryer drum stress
22.4.1 Dryer drum stress caused by bending moment
22.4.2 Dryer drum stress caused by torsional moment
22.5 Selection of dryer drum material
22.6 Checking dryer drum deflection
22.6.1 Moment of inertia of drum cross section
22.6.2 Maximal deflection of dryer shell in most critical positions
22.6.2.1 Deflection in the edge of dryer drum
22.6.2.2 Deflection in the middle of dryer drum
22.6.3 Relative deflection in the middle of dryer drum
23 Design calculations of riding ring and trunnion wheels
23.1 Determination of loads acting on dryer riding ring
23.1.1 Reaction force acting on riding ring and trunnion wheels
23.1.2 Angle between two near support palates with reference point the center of the drum
23.1.3 Number of riding ring support plates in a quadrant
23.1.4 Force acting on riding ring support plate that is located in the lowest point of the
vertical axis
23.1.5 Forces acting on riding ring support plates that are located near the lowest point of the
vertical axis
23.2 Riding ring geometrical parameters
23.2.1 Riding ring inner diameter
23.2.2 Riding ring outer diameter
23.2.3 Riding ring axis radius
23.3 Determination of internal forces actin on riding ring
23.3.1 Bending moments applied on top of the riding ring
23.3.1.1 Overall bending moment applied on top of the riding ring
23.3.2 Axial forces applied on top of the riding ring
23.3.2.1 Overall axial force applied on top of the riding ring
23.3.3 Calculation of bending moment for each section of the riding ring

vertical axis
23.3.4 Max bending moment applied on top of the riding ring
23.4 Determination of riding ring and trunnion wheels dimensions
23.4.1 Riding ring section dimensions
23.4.1.1 Riding ring width
23.4.1.2 Riding ring height
23.4.2 Gap between riding ring inner surface and support plates
23.4.3 Calculation of excess trunnion wheels width
23.4.4 Design of trust roller
23.4.4.1 Trust roller width
23.4.4.2 Trust roller height
23.4.4.3 Trust roller outer diameter
24 Design calculations of riding ring and trunnion rollers (second approach)
24.1 Longitudinal thermal expansion of dryer drum and calculation of trunnion roller width
24.2 Thermal expansion of dryer drum on vertical plane that is located in the middle of riding
ring width
24.3 Calculation of riding ring weight and width
24.4 Contac stress calculation according to Herx
24.5 Contact safety factor
24.6 Contac load calculation
24.7 Calculation of maximum bending moment applied on riding ring and trunnion wheel
contact point
24.8 Calculation of riding ring bending stress
24.9 Bending safety factor
24.10 Riding ring deflection
25 Design of dryer power transmission
25.1 Determine power transmissions ratios
25.1.1 Overall transmission ratio
25.1.2 Open transmission ratio
25.1.3 Second stage open transmission ratio
25.2 Design of first stage open transmission
25.2.1 Rotational speed of both dive and driven gear
25.2.2 Power and torsional moment in pinions shaft
25.2.3 Material selection
25.2.4 Determination of allowed bending and torsional stress and working factors
25.2.4.1 Allowed bending stress for drive and driven gear
25.2.4.2 Pinion total working cycles
25.2.4.3 Bending working condition factors for drive and driven gear
25.2.4.4 Allowed contact stress for drive and driven gear
25.2.4.5 Contact working condition factors
25.2.5 Design of gear teeth
25.2.5.1 Determination of gear toot modulus based on gear fatigue as a consequence of
bending stress
25.2.5.2 Number of teeth for each gear
25.2.5.3 Gear tooth shape coefficient
25.2.5.4 Gears durability comparison
25.2.5.5 Required gear modulus which is capable to withstand bending moment

25.2.4.5 Contact working condition factors
bending stress
25.2.6 Gear geometrical parameters
25.2.7 Gear linear velocity
25.2.8 Recalculating gear module with the new coefficients
25.2.9 Gear transmission Center distance
25.2.10 Contac stress analysis
25.2.11 Recalculation of gears pitch diameters and contact stress
25.2.12 Determination of loads that are caused by transmission
25.2.12.1 Peripheral force
25.2.12.2 Radial force
25.3 Design of second stage open transmission
25.3.1 Rotational speed of both dive and driven gear
25.3.2 Power and torsional moment in pinions shaft
25.3.3 Material selection
25.3.4 Determination of allowed bending and torsional stress and working factors
25.3.4.5 Contac working condition factors
bending stress
25.3.6 Gear geometrical parameters
25.3.7 Gear linear velocity
25.3.8 Recalculating gear module with the new coefficients
25.3.9 Gear transmission Center distance
25.3.10 Contac stress analysis
25.3.11 Determination of loads that are caused by transmission
25.3.11.1 Peripheral force
25.3.11.2 Radial force
25.4 Design of open transmission shafts
25.4.1 Design of the first shaft
25.4.1.1 Material selection
25.4.1.2 Loads transmitted to the shaft from first open transmission
25.4.1.3 Determination of all loads acting on the shaft
25.4.1.4 Shaft smallest diameter in section A subjected to torsional forces
25.4.1.5 Shaft diameter in section M subjected to torsional and bending forces
25.4.1.6 Final design of the shaft
25.4.1.7 Shafts final solidity calculation
25.4.1.8 Checking shaft rigidity
25.4.2 Design of the second shaft

25.4.1.8 Checking shaft rigidity
25.4.2 Design of the second shaft
25.4.2.8Checking shaft rigidity
In this report we are going to show a step by step design calculations of sand rotary dryer.
Necessary data:

In this report we are going to show a step by step design calculations of sand rotary dryer.
Necessary data:
≔Qin ⋅30000 ―
kg
hr
Input capacity: weight of material to be processed in an hour
≔φin %14 Input Moisture: percentage of moisture content for unit of input
product in mass.
≔φout %3 Output Moisture: percentage of moisture content for unit of output
product in mass.
≔γMwet 1760 ――
kg
m3
Bulk density (wet): Mass of material for each unit of volume whith
moistue level as imput
≔γMdry 1440 ――
kg
m3
Bulk density (dry): Mass of material for each unit of volume with moisture
level as output
≔cpM 840 ――
J
⋅kg K
Specific heat of material: heat needed to raise the temperature of one
kilogram of material with one degree celsius at constant pressure.
≔cpA 1005 ――
J
⋅kg K
Specific heat of air: heat needed to raise the temperature of one
kilogram of air with one degree celsius at constant pressure.
% of material Mesh size Mean diameter particle Granulometry
≔ψ1 %11.8 -10 100 ≔χ1 0.42 mm
≔ψ2 %46.2 -100 140 ≔χ2 0.118 mm
≔ψ3 %29.8 -140 200 ≔χ3 0.103 mm
≔ψ4 %8.7 -200 270 ≔χ4 0.0742 mm
≔ψ5 %3.5 -270 ≔χ5 0.0612 mm
≔TMin 16 °C ≔TAin 1090 °C Inlet temperatures
≔TMout 100 °C ≔TAout 149 °C Outlet temperatures
The effectiveness of any form of dryer varies with the nature of material (physical and
chemical properties) and size of the feedstock and before the design can proceed, it is
necessary to know the safety range of temperature of drying medium. This can be obtained
by experience with similar materials or by conducting a pilot plant tests. For the present
purpose it is assumed that the material can be heated by direct heating (chemical and the
structural integrity of the material is preserved in high temperature environments) in parallel
flow. From pilot tests we determined that the material can be heated at 1090 *C and output
temperature of the material must be 100 *C (because of the boiling point of water) and gas
temperatures will be 1090 *C inlet and 149 *C outlet (must be at least 20 *C higher from the
material because of saturation).

- Figure 1: Direct heated rotary dryer, process description diagram
2 Mass balance
2.1 Dry solid in the feed(mass of material with 0% moisture)

2 Mass balance
2.1 Dry solid in the feed(mass of material with 0% moisture)
≔QDrySol =⋅Qin
⎛⎝ -1 φin
⎞⎠ ⎛⎝ ⋅2.58 104 ⎞⎠ ―
kg
hr
2.2 Water rate in the feed (total mass of water found in material)
≔QWin =⋅Qin φin
⎛⎝ ⋅4.2 103 ⎞⎠ ―
kg
hr
2.3 Water rate in the final product (mass of water found in material, after being dyed)
≔QWMout =――――
⋅φout QDrySol
-1 φout
797.938 ―
kg
hr
2.4 Water rate evaporated (mass of water to be removed from material)
≔QWAout =-QWin QWMout
⎛⎝ ⋅3.402 103 ⎞⎠ ―
kg
hr
2.5 Final product weight (mass of output product)
≔Qout =+QDrySol QWMout
⎛⎝ ⋅2.66 104 ⎞⎠ ―
kg
hr
Based on the rate of moisture to be removed and input mass of product we select the tension
of the dryer as:
≔AV 70 ―――
kg
⋅m3
hr
Moisture removed for each unit volume of the dryer, in each unit of time
3 Heat balance
- Heat requirements to be determined
a)Heat for raising the temperature of material feedstock
b)Heat for raising temperature of moisture that remain in material
c)Heat for raising temperature of moisture that leave with air
d)Heat for evaporation moisture that leave with air
e)Heat for superheating moisture at 149*C, that leave with air
f)Heat loss from the surface of the dryer
g)Heat loss from the surface of combustion chamber
h)Heat loss from exhaust gases leaving the dryer
i)Heat loss from moisture in exhaust gases leaving the dryer
j)Heat loss from burning fuel

- Figure 2: Direct heated rotary dryer, heat diagram
3 Heat balance

3 Heat balance
≔QM =⋅⋅QDrySol cpM
⎛⎝ -TMout TMin
⎞⎠ 505.68 kW
- Figure 3: Direct heated rotary dryer, heat for raising the temperature of material feedstock
3.2 Heat for raising temperature of moisture that remain in material
From thermodynamic tables of water and steam we find:
≔HW16 ⋅65300 ―
J
kg
Enthalpy of water at 16 *C
≔HW100 ⋅419000 ―
J
kg
Enthalpy of water at 100 *C
With the data that we found we can define:
≔QWM =⋅QWMout
⎛⎝ -HW100 HW16
⎞⎠ 78.397 kW
- Figure 4: Direct heated rotary dryer, heat for raising temperature of moisture that remain in
material

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”- Figure 4: Direct heated rotary dryer, heat for raising temperature of moisture that remain in
material
≔QWA =⋅QWAout
⎛⎝ -HW100 HW16
⎞⎠ 334.253 kW
- Figure 5: Direct heated rotary dryer, heat for raising temperature of moisture that leave with
air
3.4 Heat for evaporating moisture that leave with air
≔HWL100 ⋅2256680 ―
J
kg
Latent heat at 100*C
≔QWAev =⋅QWAout HWL100
⎛⎝ ⋅2.133 103 ⎞⎠ kW
- Figure 6: Direct heated rotary dryer, heat for evaporation moisture that leave with air

≔HWS100 ⋅2675830 ―
J
kg
Enthalpy of steam at 100 *C
≔HWS149 ⋅2774680 ―
J
kg
≔QWMSH =⋅QWAout
⎛⎝ -HWS149 HWS100
⎞⎠ 93.415 kW
- Figure 7: Direct heated rotary dryer, heat for superheating moisture at 149*C, that leave
with air
3.6 Heat loss from the surface of dryer
Heat loss from the surface of the dryer can be estimated from chart 1 (the chart below) by
entering values of variables as being presented in this chart. We assume that the temperature
difference between ambient air and dryer drum surface is 65*C and wind velocity 3.5 m/s
than heat loss for square meters of surface is 2208 w/m2. From these data, we are going to
determine insulation thickness which will make sure that heat loss from the drum of the dryer
will be closer to values that we previously determined.

- Figure 8: Direct heated rotary dryer, heat loss from the surface of the dryer
≔Ud 2208 ――
W
m2
＝QDLos ⋅⋅⋅Ud π D L
3.6.1 Pre determining the volume of dryer
=QWAout
⎛⎝ ⋅3.402 103 ⎞⎠ ―
kg
hr
Water rate to be removed
=AV 70 ―――
kg
⋅m3
hr
Tension of the dryer
＝AV ―――
QWAout
VD
From this we determine ≔VD =―――
QWAout
AV
48.601 m3
3.6.2 Pre determining of diameter and length of dryer
With little to no mistake we approximate the ratio between dryer diameter and dryer length is
equals to 10.
＝＝Ξd ―
L
D
10 ≔Ξd 10
Also we know: ＝Sd ⋅⋅π D L ＝VD ―――
⋅⋅π D2
L
4
After a few transformation we can write
≔D =
‾‾‾‾‾3
――
⋅4 VD
10 π
⎛⎝ ⋅1.836 103 ⎞⎠ mm ≔L =
‾‾‾‾‾‾‾3
―――
⋅400 VD
π
⎛⎝ ⋅1.836 104 ⎞⎠ mm
Finally we can determine heat loss from surface of dryer:
≔QDLos =⋅⋅⋅Ud π D L 233.804 kW

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔QDLos =⋅⋅⋅Ud π D L 233.804 kW
- Figure 9: Direct heated rotary dryer, heat loss graph from the surface of the dryer
Detailed analysis of heat loss from dryer surface and the range of operating temperatures of
dryer drum.
To deal with such complex problem like heat transfer from dryer drum we must take a few
assumption or hypothesis into consideration in order to simplify it.

Detailed analysis of heat loss from dryer surface and the range of operating temperatures of
dryer drum.
To deal with such complex problem like heat transfer from dryer drum we must take a few
assumption or hypothesis into consideration in order to simplify it.
- Mean velocity of gases
- Mean temperature differences
- Ignoring seasons effects on ambient conditions
- Ignoring influence of material
- Figure 10: Direct heated rotary dryer, heat transfer from inside to outside of dryer
3.6.3.1.1.1 Mean temperature of gasses inside the dryer
≔TMeanin =――――
+TAout TAin
2
619.5 °C
Ambjent temperature
≔TA 25 °C
3.6.3.1.1.2 Mean velocity of gasses inside the dryer
(We calculate this value in the following analysis)
≔VMean 3 ―
m
s

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔VMean 3 ―
m
s
=TMeanin 619.5 °C Gas temperature
≔CPMeanin 1120 ――
J
⋅kg K
Specific heat in constant pressure
≔CVMeanin 834 ――
J
⋅kg K
Specific heat in constant volume
≔μMeanin ⋅⋅3.897 10-5
――
kg
⋅m s
Dynamic viscosity of gases
≔νMeanin ⋅⋅9.936 10-5
――
m2
s
Kinematic viscosity of gases
≔ρMeanin 0.3922 ――
kg
m3
Density of gases
≔λMeanin ⋅⋅6.276 10-5
――
kW
⋅m K
Thermal conductivity of gases
3.6.3.1.1 Convection heat transfer coefficient inside dryer
3.6.3.1.1.4 Determining flow regime
Renolds number
≔Rein =―――
⋅VMean D
νMeanin
⋅5.543 104
＝Nuin ⋅⋅0.018 Re0.8
εl
For: ＝＝Ξd ―
L
D
10 ≔Ξd 10 and =Rein ⋅5.543 104
≔εl 1.518
≔Nuin =⋅⋅0.018 Rein
0.8
εl 170.433
Forced convection heat transfer coefficient

Forced convection heat transfer coefficient
≔α'in =―――――
⋅Nuin λMeanin
D
5.826 ―――
W
⋅m2
K
3.6.3.1.1.6.1 Nussel numbber
＝Nuin ⋅0.47 Gr0.25
≔Grin =―――――――
⋅⋅g D3
⎛⎝ -TAin TAout
⎞⎠
⋅νMeanin TAin
――
s
cm2
⋅4.216 109
≔Nuin =⋅0.47 Grin
0.25
119.764
Free convection heat transfer coefficient
≔α''in =―――――
⋅Nuin λMeanin
D
4.094 ―――
W
⋅m2
K
≔α =1.3 ⎛⎝ +α'in α''in
⎞⎠ 12.896 ―――
W
⋅m2
K
3.6.3.1.2 Convection heat transfer coefficient outside the dryer
3.6.3.1.2.1 Mean temperature of gasses outside the dryer
Temperature difference between ambient air and the dryer shell surface is 65*C. This means
that mean temperature difference is:
≔TSjM 65 °C Temperature difference between ambient
air and the dryer shell surface is 65*C
≔TA 25 °C Air temperature
≔TSS 90 °C Outside surface temerature
3.6.3.1.2.2 Mean velocity of gasses outside the dryer (wind velocity)
≔VW 3.5 ―
m
s
≔TMeanout =――――
++65 25 25
2
57.5
≔TMeanout ⋅330.5 K

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔TMeanout =――――
++65 25 25
2
57.5
≔TMeanout ⋅330.5 K
3.6.3.1.2.3 Air thermo-physical parameters outside the dryer
≔CPMeanout 1006 ――
J
⋅kg K
≔CVMeanout 719.8 ――
J
⋅kg K
≔μMeanout ⋅⋅1.962 10-5
――
kg
⋅m s
Dynamic viscosity of air
≔νMeanout ⋅⋅1.807 10-5
――
m2
s
Kinematic viscosity of air
≔ρMeanout 1.086 ――
kg
m3
Density of air
≔λMeanout ⋅⋅2.816 10-5
――
kW
⋅m K
Thermal conductivity of air
In order to calculate heat transfer coefficient we must estimate dryer outer diameter and
insulation thickness in order to calculate outer diameter of the dryer.
≔δins 50 mm Insulation thickness
≔δSell 20 mm Shell thickness
3.6.3.1.2.4 Estimated dryer outer diameter
≔DOUT =++D 2 δins 2 δSell
⎛⎝ ⋅1.976 103 ⎞⎠ mm
≔Grout =―――――――
⋅⋅g DOUT
3
⎛⎝ -TSjM TA
⎞⎠
⋅νMeanout TA
――
s
cm2
⋅5.617 109
≔Prout =―――――――――
⋅⋅νMeanout CPMeanout ρMeanout
λMeanout
0.701
Product of Prandtl and Grashof number
≔Η =⋅Prout Grout ⋅3.938 109

≔Η =⋅Prout Grout ⋅3.938 109
With this value we can find nusselt respective coefficient c and n:
For: =Η ⋅3.938 109
≔c 0.135 ≔n ―
1
3
3.6.3.1.2.6 Nusselt number
≔Nuout =⋅c Ηn
213.181
≔α'out =―――――
⋅Nuout λMeanout
D
3.27 ―――
W
⋅m2
K
＝α''out ――――――――
⋅⋅ε σ
⎛
⎜
⎝
-
⎛
⎜
⎝
――
TSjM
100
⎞
⎟
⎠
4
⎛
⎜
⎝
――
TA
100
⎞
⎟
⎠
4 ⎞
⎟
⎠
-TSjM TA
≔ε 0.95 Emissivity coefficient
≔σ 5.7 ―――
W
⋅m2
K4
Stefan-Boltzmann Constant
≔α''out =――――――――
⋅⋅ε σ
⎛
⎜
⎝
-
⎛
⎜
⎝
――
TSjM
100
⎞
⎟
⎠
4
⎛
⎜
⎝
――
TA
100
⎞
⎟
⎠
4 ⎞
⎟
⎠
-TSjM TA
7.003 ―――
W
⋅m2
K
≔αout =+α'out α''out 10.273 ―――
W
⋅m2
K
We take into account that thickness of layer holding insulation is approximately 1 mm with
thermal conductivity about 50 W/m*K, this means that the influence of this layer is negligible,
so we are going to ignore it in our calculation.
Also we don't want to lose more than 234 kW of heat so from this assumption we can calculate
the temperature inside surface of the shell.
≔δc 1 mm ≔λc 50 ――
W
⋅m K
=QDLos 233.804 kW
＝＝QDLos ⋅⋅⋅Ud π D L ⋅⋅⋅α π D L⎛⎝ -TMeanin TSinD
⎞⎠

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”=QDLos 233.804 kW
＝＝QDLos ⋅⋅⋅Ud π D L ⋅⋅⋅α π D L⎛⎝ -TMeanin TSinD
⎞⎠
After transformation we get: ≔TSinD =-TMeanin ――――
QDLos
⋅⋅⋅α π D L
448.289 °C
3.6.3.2.2Temperature outside dryer drum surfce
＝＝QDLos ⋅⋅⋅Ud π D L ―――――――――
⋅⋅⋅2 π λSS L ⎛⎝ -TSinD TSoutD
⎞⎠
ln
⎛
⎜
⎝
――――
+D 2 δSell
D
⎞
⎟
⎠≔λSS 16 ――
W
⋅m K
After transformation we get: ≔TSoutD =-TSinD ―――――――
⋅QDLos ln
⎛
⎜
⎝
――――
+D 2 δSell
D
⎞
⎟
⎠
⋅⋅2 π λSS L
445.558 °C
Finally we can determine insulation thickness
≔λins 0.25 ――
W
⋅m K
Thermal conductivity of insulation
＝＝QDLos ⋅⋅⋅Ud π D L ―――――――――
⋅⋅⋅2 π λins L ⎛⎝ -TSoutD TSS
⎞⎠
ln
⎛
⎜
⎝
――――――
++D 2 δSell 2 δins
+D 2 δSell
⎞
⎟
⎠
After a few transformation we get:
＝ln
⎛
⎜
⎝
――――――
+D 2 δSell
⎞
⎟
⎠
―――――――――
⎞⎠
QDLos
＝――――――
+D 2 δSell
e
―――――――――
⎞⎠
QDLos
＝δins ―――――――――――――――
-
⎛
⎜
⎝ ⋅e
―――――――――
⎞⎠
QDLos
⎛⎝ +D 2 δSell
⎞⎠
⎞
⎟
⎠ ⎛⎝ +D 2 δSell
⎞⎠
2

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”＝δins ―――――――――――――――
-
⎛
⎜
⎝ ⋅e
―――――――――
⎞⎠
QDLos
⎛⎝ +D 2 δSell
⎞⎠
⎞
⎟
⎞⎠
2
≔δins =―――――――――――――――
-
⎛
⎜
⎝ ⋅e
―――――――――
⎞⎠
QDLos
⎛⎝ +D 2 δSell
⎞⎠
⎞
⎟
⎞⎠
2
42.05 mm
We accept insulation thickness: ≔δins 40 mm
Overall heat transfer coefficient
≔λtot =―――――――――
1
++++―
1
α
――
δSell
λSS
――
δins
λins
―
δc
λc
――
1
αout
2.975 ―――
W
⋅m2
K
＝QDLos ⋅⋅⋅⋅λtot π D L ΔTm
≔ΔTm =+――――――――
-⎛⎝ -TAin TA
⎞⎠ ⎛⎝ -TAout TA
⎞⎠
ln
⎛
⎜
⎝
――――
-TAin TA
-TAout TA
⎞
⎟
⎠
273 K 437.433 °C
Overall heat loss
≔QDLos' =⋅⋅⋅⋅λtot π D L ΔTm 223.833 kW
3.6.5 Absolut error (difrece betwen caculated heat losses and estimated heat
≔Δ =-QDLos QDLos' 9.971 kW
≔ε =⋅――
Δ
QDLos
100 4.265
Error margin is acceptable
The same logistic is applied to combustion chamber
Heat loss from combustion chamber also can be estimated from chart 1 by entering values of
variables as being presented in this chart. We assume that temperature difference between
ambient air and combustion chamber shell surface is 177*C and wind velocity 3.5 m/s than
heat loss for square meter of combustion chamber is 6940 W/m2.

The same logistic is applied to combustion chamber
Heat loss from combustion chamber also can be estimated from chart 1 by entering values of
variables as being presented in this chart. We assume that temperature difference between
ambient air and combustion chamber shell surface is 177*C and wind velocity 3.5 m/s than
heat loss for square meter of combustion chamber is 6940 W/m2.
- Figure 11: Direct heated rotary dryer, heat loss from the surface of combustion chamber
≔Uc 6940 ――
W
m2
＝QCLos ⋅⋅⋅Ud π D L
≔Dc 1700 mm Combustion chamber diameter
≔Lc 4000 mm Combustion chamber length
≔QCLos =⋅⋅⋅Uc π Dc Lc 148.258 kW
3.7.2 Insulation thickness and combustion chamber surface heat loss analysis
Detailed analysis of heat loss from combustion chamber surface and the range of operating
temperatures of combustion chamber shell.
To deal with such complex problem like heat transfer in a combustion chamber shell, we must
take a few assumption or hypothesis into consideration in order to simplify it.
- Mean velocity of gases
- Mean temperature differences
- Ignoring seasons effects on ambient conditions

- Figure 12: Direct heated rotary dryer, heat transfer from inside to outside of combustion
chamber
3.7.2.1.1.1 Mean temperature of gasses inside combustion chamber
≔TCAin 1600 °C
≔TCAout 1090 °C
≔TCMeanin =―――――
+TCAin TCAout
2
⎛⎝ ⋅1.345 103 ⎞⎠ °C
Ambient temperature
≔TA 25 °C
3.7.2.1.1.2 Mean velocity of gasses inside combustion chamber
≔VCMean 5 ―
m
s
3.7.2.1.1.3 Air thermo-physical parameters inside combustion chamber
=TCMeanin
⎛⎝ ⋅1.345 103 ⎞⎠ °C Gas temperature

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”=TCMeanin
⎛⎝ ⋅1.345 103 ⎞⎠ °C
≔CCPMeanin 1220 ――
J
⋅kg K
≔CCVMeanin 933 ――
J
⋅kg K
≔μCMeanin ⋅⋅5.45 10-5
――
kg
⋅m s
Dynamic viscosity of gases
≔νCMeanin ⋅⋅24.74 10-5
――
m2
s
Kinematic viscosity of gases
≔ρCMeanin 0.22 ――
kg
m3
Density of gases
≔λCMeanin ⋅⋅9.2 10-5
――
kW
⋅m K
Thermal conductivity of gases
3.7.2.1.1 Convection heat transfer coefficient inside combustion chamber
3.7.2.1.1.4 Determining flow regime
Renolds number
≔ReCin =――――
⋅VCMean Dc
νCMeanin
⋅3.436 104
＝NuCin ⋅⋅0.018 ReCin
0.8
εl
For: ＝＝Ξd ――
LC
DC
2.35 ≔Ξd 2.35 and =Rein ⋅5.543 104
≔εl 2.75
≔NuCin =⋅⋅0.018 ReCin
0.8
εl 210.582
From this we can determine heat transfer coefficient due to forced convection
≔α'Cin =―――――
⋅NuCin λCMeanin
Dc
11.396 ―――
W
⋅m2
K
Nussel numbber

Nussel numbber
＝NuCin ⋅0.47 GrCin
0.25
≔GrCin =――――――――
⋅⋅g Dc
3
⎛⎝ +TCAin TCAout
⎞⎠
⋅νCMeanin TCAin
――
s
cm2
⋅3.365 109
≔NuCin =⋅0.47 GrCin
0.25
113.197
Heat transfer coefficient
≔α''Cin =―――――
⋅NuCin λCMeanin
Dc
6.126 ―――
W
⋅m2
K
≔αC =1.3 ⎛⎝ +α'Cin α''Cin
⎞⎠ 22.779 ―――
W
⋅m2
K
3.7.2.1.2 Convection heat transfer coefficient outside combustion chamber
3.7.2.1.2.1 Mean temperature of gasses outside combustion chamber
Temperature difference between ambient air and the combustion chamber shell surface is
177*C. This means that mean temperature difference is:
≔TCSjM 177 °C Temperature difference between
ambient air and the combustion
chamber shell surface is 177*C
≔TA 25 °C Air temperature
≔TSS 202 °C Outside surface temperature
3.7.2.1.2.2 Mean velocity of gasses outside combustion chamber
≔VW 3.5 ―
m
s
≔TCMeanout =―――
+TSS TA
2
113.5 °C
≔TCMeanout 113.5 °C

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔TCMeanout 113.5 °C
≔CPCMeanout 1013 ――
J
⋅kg K
≔CVCMeanout 726 ――
J
⋅kg K
≔μCMeanout ⋅⋅2.28 10-5
――
kg
⋅m s
Dynamic viscosity of air
≔νCMeanout ⋅⋅2.591 10-5
――
m2
s
Kinematic viscosity of air
≔ρCMeanout 0.8824 ――
kg
m3
Density of air
≔λCMeanout ⋅⋅3.316 10-5
――
kW
⋅m K
Thermal conductivity of air
3.7.2.1.2.4 Estimated combustion chamber outer diameter
In order to calculate heat transfer coefficient we must estimate insulation and shell thickness
in order to calculate outer diameter of combustion chamber.
≔δins 30 mm insulation thickness
≔δSell 20 mm Shell thickness
Combustion chamber outer diameter
≔DcOUT =++Dc 2 δins 2 δSell
⎛⎝ ⋅1.8 103 ⎞⎠ mm
≔GrCout =――――――――
⋅⋅g DcOUT
3
⎛⎝ -TCSjM TA
⎞⎠
⋅νCMeanout TA
――
s
cm2
⋅1.125 1010
≔PrCout =――――――――――
⋅⋅νCMeanout CPCMeanout ρCMeanout
λCMeanout
0.698
Product of Prandtl and Grashof number
≔ΗC =⋅PrCout GrCout ⋅7.86 109

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔ΗC =⋅PrCout GrCout ⋅7.86 109
For: =ΗC ⋅7.86 109
≔c 0.452 ≔n ―
1
3
3.7.2.1.2.6 Nusselt numbber
≔NuCout =⋅c Ηn
713.761
≔α'Cout =――――――
⋅NuCout λCMeanout
Dc
13.923 ―――
W
⋅m2
K
＝α''Cout ―――――――――
⋅⋅ε σ
⎛
⎜
⎝
-
⎛
⎜
⎝
―――
TCSjM
100
⎞
⎟
⎠
4
⎛
⎜
⎝
――
TA
100
⎞
⎟
⎠
4 ⎞
⎟
⎠
-TCSjM TA
≔ε 0.95 Emissivity coefficient
≔σ 5.7 ―――
W
⋅m2
K4
Stefan-Boltzmann Constant
≔α''Cout =―――――――――
⋅⋅ε σ
⎛
⎜
⎝
-
⎛
⎜
⎝
―――
TCSjM
100
⎞
⎟
⎠
4
⎛
⎜
⎝
――
TA
100
⎞
⎟
⎠
4 ⎞
⎟
⎠
-TCSjM TA
11.813 ―――
W
⋅m2
K
≔αCout =+α'Cout α''Cout 25.735 ―――
W
⋅m2
K
We don't want to lose more than 149 kW of heat, so from this action we can calculate the
temperature inside surface of the insulation
≔δc 1 mm ≔λc 50 ――
W
⋅m K
3.7.2.2.1Temperature inside combustion chamber surface

3.7.2.2.1Temperature inside combustion chamber surface
=QCLos 148.258 kW
＝＝QCLos ⋅⋅⋅Uc π Dc Lc ⋅⋅⋅αC π Dc Lc
⎛⎝ -TCMeanin TCSinD
⎞⎠
After transformation we get: ≔TCSinD =-TCMeanin ―――――
QCLos
⋅⋅⋅αC π Dc Lc
⎛⎝ ⋅1.04 103 ⎞⎠ °C
3.7.2.2.2 Temperature outside combustion chamber surface
＝＝QDLos ⋅⋅⋅Uc π Dc Lc ――――――――――
⋅⋅⋅2 π λSS L ⎛⎝ -TCSinD TCSoutD
⎞⎠
ln
⎛
⎜
⎝
――――
+Dc 2 δSell
Dc
⎞
⎟
⎠≔λSS 16 ――
W
⋅m K
After transformation we get: ≔TCSoutD =-TCSinD ―――――――
⋅QCLos ln
⎛
⎜
⎝
――――
+Dc 2 δSell
Dc
⎞
⎟
⎠
⋅⋅2 π λSS Lc
⎛⎝ ⋅1.032 103 ⎞⎠ °C
Finally we can determine insulation thickness
≔λins 0.25 ――
W
⋅m K
Thermal conductivity of insulation
＝＝QCLos ⋅⋅⋅UC π Dc Lc ―――――――――
⋅⋅⋅2 π λins Lc
⎛⎝ -TCSoutD TSS
⎞⎠
ln
⎛
⎜
⎝
――――――
++Dc 2 δSell 2 δins
+Dc 2 δSell
⎞
⎟
⎠
＝ln
⎛
⎜
⎝
――――――
+Dc 2 δSell
⎞
⎟
⎠
―――――――――
⎛⎝ -TCSoutD TSS
⎞⎠
QCLos
＝――――――
+Dc 2 δSell
e
―――――――――
⎛⎝ -TCSoutD TSS
⎞⎠
QCLos

＝δins ――――――――――――――――
-
⎛
⎜
⎝ ⋅e
―――――――――
⎛⎝ -TCSoutD TSS
⎞⎠
QCLos
⎛⎝ +Dc 2 δSell
⎞⎠
⎞
⎟
⎠ ⎛⎝ +Dc 2 δSell
⎞⎠
2
≔δins =――――――――――――――――
-
⎛
⎜
⎝ ⋅e
―――――――――
⎛⎝ -TCSoutD TSS
⎞⎠
QCLos
⎛⎝ +Dc 2 δSell
⎞⎠
⎞
⎟
⎠ ⎛⎝ +Dc 2 δSell
⎞⎠
2
31.138 mm
We accept insulation thickness as it was estimated ≔δins 30 mm
Overall thermal conductivity coefficient
≔λtot =―――――――――
1
++++―
1
α
――
δSell
λSS
――
δins
λins
―
δc
λc
――
1
αout
3.377 ―――
W
⋅m2
K
3.7.3 Overall heat loss from combustion chamber surface
＝QDLos ⋅⋅⋅⋅λtot π D L ΔTm
≔ΔTm =+―――――――――
-⎛⎝ -TCAin TA
⎞⎠ ⎛⎝ -TCAout TA
⎞⎠
ln
⎛
⎜
⎝
――――
-TCAin TA
-TCAout TA
⎞
⎟
⎠
273 K ⎛⎝ ⋅1.303 103 ⎞⎠ °C
Overall heat loss
≔QCLos' =⋅⋅⋅⋅λtot π Dc Lc ΔTm 113.712 kW
3.7.4 Absolute error from heat loss estimations
≔Δ =-QCLos QCLos' 34.546 kW
3.7.5 Relative error from heat loss estimation
≔ε =⋅――
Δ
QDLos
100 14.776
Error margin is acceptable

≔Q1 =++++++QM QWM QWA QWAev QWMSH QDLos QCLos
⎛⎝ ⋅3.526 103 ⎞⎠ kW
3.9 Heat loss from exhaust gases leaving the dryer
This is the point that we ran into a solving problem, because to determine heat loss from
exhaust gases we must precisely know the quantity of gases leaving the dryer, which we are
not able to determine yet, because this variable is dependent on heat loss from exhaust gases
and heat loss from moisture in exhaust gases. To solve this problem we must take an
alternative root way which is different from current linear way that we are using.
By now we know that we have two unknown independent variables and five unknown
dependent variables, which are specified below:
- Heat loss from exhaust gases leaving the dryer (independent variable)QALoss
- Heat loss from moisture in exhaust gases leaving the dryer (independent variable)QAMLoss
- Total heat loss from gases leaving the dryer (dependent variable)Q2
-Sum of overall heat required ( )and total heat loss from gases( )(dependent variable)Q3 Q1 Q2
M1 - Mass of air needed to convey heat for the process ( )(dependent variable)Q1
M2 -Mass of air needed to make up for the total heat loss from gases( )(dependent variable)Q2
M - Total mass of air (dependent variable)
- Figure 13: Direct heated rotary dryer, heat loss from exhaust gases leaving the dryer
To solve this problem whit two independent variables we need a system with two equations.
We know that:

To solve this problem whit two independent variables we need a system with two equations.
We know that:
＝＝Q2 +QALoss QAMLoss +⎛⎝ ⋅⋅M cpA
⎛⎝ -TAout TMin
⎞⎠⎞⎠ ⎛⎝ ⋅⋅d M ⎛⎝ -HWS149 HW16
⎞⎠⎞⎠
Relative humidity at winter in the region that the equipment will be implemented is about 80%
and from psychrometric chart we determine:
≔d 9 ――
gm
kg
Moisture content
Also we know:
＝Q2 ⋅⋅M2 cpA
⎛⎝ -TAin TMin
⎞⎠
From our previous presupposition we can write:
＝+⎛⎝ ⋅⋅M cpA
⎛⎝ -TAout TMin
⎞⎠⎞⎠ ⎛⎝ ⋅⋅d M ⎛⎝ -HWS149 HW16
⎞⎠⎞⎠ ⋅⋅M2 cpA
⎛⎝ -TAin TMin
⎞⎠
3.9.1 Ratio between mass of air needed to make up for total heat loss from gases and total
mass of air needed
After a few mathematical transformation we can determine, ratio between mass of air needed
to make up for the total heat loss from gases and total mass of air needed:
＝＝Υ ――
M2
M
――――――――――――――
+⎛⎝ ⋅cpA
⎛⎝ -TAout TMin
⎞⎠⎞⎠ ⎛⎝ ⋅d ⎛⎝ -HWS149 HW16
⎞⎠⎞⎠
⋅cpA
⎛⎝ -TAin TMin
⎞⎠
≔Υ =――――――――――――――
+⎛⎝ ⋅cpA
⎛⎝ -TAout TMin
⎞⎠⎞⎠ ⎛⎝ ⋅d ⎛⎝ -HWS149 HW16
⎞⎠⎞⎠
⋅cpA
⎛⎝ -TAin TMin
⎞⎠
0.146
We know that the ratio of air masses is equal with the ratio of heat requirements:
＝＝＝Υ ――
M2
M
――
Q2
Q3
―――
Q2
+Q1 Q2
After a few mathematical transformation we can determine total heat loss from gases leaving
the dryer.

After a few mathematical transformation we can determine total heat loss from gases leaving
the dryer.
＝Q2 ―――
⋅Q1 Υ
(( -1 Υ))
≔Q2 =―――
⋅Q1 Υ
(( -1 Υ))
604.943 kW
3.9.3 Mass of air needed to convey heat for the process
≔M1 =――――――
Q1
⋅cpA
⎛⎝ -TAin TMin
⎞⎠
⎛⎝ ⋅1.176 104 ⎞⎠ ―
kg
hr
3.9.4 Mass of air needed to make up for total heat loss from gases
≔M2 =――――――
Q2
⋅cpA
⎛⎝ -TAin TMin
⎞⎠
⎛⎝ ⋅2.018 103 ⎞⎠ ―
kg
hr
≔M =+M1 M2
⎛⎝ ⋅1.378 104 ⎞⎠ ―
kg
hr
≔Q3 =+Q1 Q2
⎛⎝ ⋅4.131 103 ⎞⎠ kW
≔QALoss =⋅⋅M cpA
⎛⎝ -TAout TMin
⎞⎠ 511.611 kW
≔QAMLoss =⋅⋅d M ⎛⎝ -HWS149 HW16
⎞⎠ 93.333 kW
≔QMC =⋅d M 124.013 ―
kg
hr
3.11 Fuel analysis
Fuel is one of the most important factors in direct drying process and it cannot be ignored and
considered as separate from the system, because it have a large influence in our process both
termicaly and chemically:
1) Termically - influence of moisture produced by combustion fuel

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔QMC =⋅d M 124.013 ―
kg
hr
3.11 Fuel analysis
Fuel is one of the most important factors in direct drying process and it cannot be ignored and
considered as separate from the system, because it have a large influence in our process both
termicaly and chemically:
1) Termically - influence of moisture produced by combustion fuel
2) Chemically - influence of burning gases concentration in the air
For these reasons it is necessary to have a clear overview of combustion process
As fuel we are going to use furnace diesel.
- Figure 14: Direct heated rotary dryer, heat loss from burning fuel
3.11.1 Fuel data
Chemical composition of the fuel that we are going to use:
≔C %85.65 ≔O2 %0.7 ≔H2 %12.3 ≔S %0.5
≔N2 %0.2 ≔H2O %0.35 ≔Ash %0.05 ≔Sediment %0.25
Specific heat of each component participates in the process as an input or output:

Specific heat of each component participates in the process as an input or output:
≔CpO2 ⋅29.35 ―――
J
⋅mol K
Specific heat of oxygen
≔CpN2 ⋅29.125 ―――
J
⋅mol K
Specific heat of nitrogen
≔CpCO2 ⋅37.136 ―――
J
⋅mol K
Specific heat of carbon dioxide
≔CpSO2 ⋅39.87 ―――
J
⋅mol K
Specific heat of sulphur dioxide
≔CpH2O ⋅75.291 ―――
J
⋅mol K
Specific heat of steam
≔CpH2 ⋅28.825 ―――
J
⋅mol K
Specific heat of hydrogen
≔molCO2 ⋅44 ――
gm
mol
Molar weight of carbon dioxide
≔molSO2 ⋅64 ――
gm
mol
Molar weight of sulphur dioxide
≔molN2 28 ――
gm
mol
Molar weight of nitrogen
≔molO2 32 ――
gm
mol
Molar weight of oxygen
≔molH2O ⋅18 ――
gm
mol
Molar weight of water
≔molH2 ⋅1.002 ――
gm
mol

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔molH2 ⋅1.002 ――
gm
mol
I. ＝+C O2 CO2 ≔QCO2 8084 ――
kcal
kg
II. ＝+⋅2 H2 O2 H2O ≔QH2O 28922 ――
kcal
kg
III. ＝+S O2 SO2 ≔QSO2 2224 ――
kcal
kg
≔QCO2 =⋅QCO2 %85.65 ⎛⎝ ⋅6.924 103 ⎞⎠ ――
kcal
kg
≔QH2O =⋅QH2O %12.3 ⎛⎝ ⋅3.557 103 ⎞⎠ ――
kcal
kg
≔QSO2 =⋅QSO2 %0.5 11.12 ――
kcal
kg
≔Qkgld =++QCO2 QH2O QSO2
⎛⎝ ⋅1.049 104 ⎞⎠ ――
kcal
kg
The amount of oxygen required for combusting one kilogram of fuel and the amount of gases
released during combustion.
I. ＝+C O2 CO2
I. ＝+12 ―――
kg
⋅kg mol
32 ―――
kg
⋅kg mol
44 ―――
kg
⋅kg mol
For one kilogram of
combustible fuel
I. ＝+0.8565 ―――
kg
⋅kg mol
2.284 ―――
kg
⋅kg mol
3.1405 ―――
kg
⋅kg mol
II. ＝+⋅2 H2 O2 H2O

II. ＝+⋅2 H2 O2 H2O
II. ＝+⋅2 ―――
kg
⋅kg mol
16 ―――
kg
⋅kg mol
18 ―――
kg
⋅kg mol
For one kilogram of
combustible fuel
II. ＝+⋅0.123 ―――
kg
⋅kg mol
0.984 ―――
kg
⋅kg mol
1.107 ―――
kg
⋅kg mol
III. ＝+S O2 SO2
III. ＝+⋅32 ―――
kg
⋅kg mol
32 ―――
kg
⋅kg mol
64 ―――
kg
⋅kg mol
For one kilogram of
combustible fuel
III. ＝+⋅0.005 ―――
kg
⋅kg mol
0.005 ―――
kg
⋅kg mol
0.01 ―――
kg
⋅kg mol
≔MO2ld =++⋅2.284 kg 0.984 kg 0.005 kg 3.273 kg
≔O2 %23 Oxygene concentration
≔Majrit' =――
MO2ld
0.23
14.23 kg
Total amount of air used for combusting one kilogram of fuel
≔α 1.5 Coefficient of air surplus
≔Majrit =⋅Majrit' α 21.346 kg

≔MCO2ld 3.1405 kg
3.11.5.2 Total amount of steam released after combusting one kilogram of fuel
≔MH2Old =+1.107 kg 0.0035 kg 1.111 kg
3.11.5.3 Total amount of sulphur dioxide released after combusting one kilogram of fuel
≔MSO2ld 0.01 kg
3.11.5.4 Total amount of gasses released as a by product of combustion for one kilogram of fuel
≔MCO2ld 3.1405 kg Carbon dioxide
≔MH2Old 1.10 kg Steam
≔MSO2ld 0.01 kg Sulfur dioxide
≔MN2 =⋅Majrit 0.77 16.436 kg Nitrogen (we assume no presence of Nox)
≔MO2 =%23 ⎛⎝ -Majrit Majrit'⎞⎠ 1.637 kg Oxigene surplus
3.11.6 Amount of fuel needed in an hour
3.11.6.1 Heat loss caused by combustion gases leaving the dryer
≔QCO2ld =――――――――――
⋅⋅MCO2ld CpCO2
⎛⎝ -TAout TMin
⎞⎠
⋅molCO2 kg
⎛⎝ ⋅3.525 105 ⎞⎠ ―
J
kg
≔QH2Old =――――――――――
⋅⋅MH2Old CpH2O
⎛⎝ -TAout TMin
⎞⎠
⋅molH2O kg
⎛⎝ ⋅6.119 105 ⎞⎠ ―
J
kg
≔QSO2ld =――――――――――
⋅⋅MSO2ld CpSO2
⎛⎝ -TAout TMin
⎞⎠
⋅molSO2 kg
828.548 ―
J
kg
≔QN2 =―――――――――
⋅⋅MN2 CpN2
⎛⎝ -TAout TMin
⎞⎠
⋅molN2 kg
⎛⎝ ⋅2.274 106 ⎞⎠ ―
J
kg

≔QN2 =―――――――――
⋅⋅MN2 CpN2
⎛⎝ -TAout TMin
⎞⎠
⋅molN2 kg
⎛⎝ ⋅2.274 106 ⎞⎠ ―
J
kg
≔QO2 =―――――――――
⋅⋅MO2 CpO2
⎛⎝ -TAout TMin
⎞⎠
⋅molO2 kg
⎛⎝ ⋅1.996 105 ⎞⎠ ―
J
kg
3.11.6.1.1 Total heat loss caused by combustion gases leaving the dryer
≔Qldhngazet =++++QCO2ld QH2Old QSO2ld QN2 QO2
⎛⎝ ⋅3.439 106 ⎞⎠ ―
J
kg
3.11.6.1.2 Ratio betwen heat loss caused by combustion gases for each unit of heat produced
≔qldhngazet =―――
Qldhngazet
Qkgld
0.078
3.11.6.2 Heat loss from impurities in the fuel
≔M 0.15 kg
≔CpM ⋅1.3 ――
kcal
⋅kg K
≔QM =――――――――
⋅⋅M CpM
⎛⎝ -TAout TMin
⎞⎠
kg
⎛⎝ ⋅1.086 105 ⎞⎠ ―
J
kg
3.11.6.2.1 Ratio of heat loss from impurities in the fuel for each unit of heat produced
≔qM =――
QM
Qkgld
0.002
3.11.6.3 Heat loss as a cosequece of hydrogene in fuel
≔MH2 =⋅%12.3 1 kg 0.123 kg
≔HW1040 ⋅4642600 ―
J
kg
≔HW1600 ⋅6213000 ―
J
kg

≔HW1600 ⋅6213000 ―
J
kg
≔QH2 =――――――――
⋅9 MH2
⎛⎝ -HW1600 HW1040
⎞⎠
kg
⎛⎝ ⋅1.738 106 ⎞⎠ ―
J
kg
3.11.6.3.1 Ratio betwen heat loss as a cosequece of hydrogene in fuel for each unit of heat
produced
≔qH2 =――
QH2
Qkgld
0.04
3.11.6.4 Total combustion energy needed
≔Qtot =――――――――
+Q1 Q2
-1 ⎛⎝ ++qldhngazet qM qH2
⎞⎠
⎛⎝ ⋅4.696 103 ⎞⎠ kW
Amount of fuel needed in an hour
≔mld =――
Qtot
Qkgld
384.868 ―
kg
hr
3.11.7 Amount of steam produced in an hour as a result of combustion
≔QMBF =――――
⋅mld MH2Old
kg
423.355 ―
kg
hr
3.11.8 Total amount of heat loss from combustion
≔Qldhngazet =⋅⋅mld qldhngazet Qkgld 367.631 kW Total heat loss from combustion
gases leaving dryer
≔QM' =⋅⋅mld qM Qkgld 11.609 kW Total heat loss from impurities
in fuel
≔QH2' =⋅⋅mld qH2 Qkgld 185.852 kW Total heat loss from hydrogen
in fuel
Total heat loss from
combustion≔Qtoth =++Qldhngazet QM' QH2' 565.092 kW

≔ηpro =⋅――――――
--Q1 QDLos QCLos
Qtot
100 66.952
3.11.10 Equipment efficiency
≔ηeq =――
Q1
Qtot
100 75.087
3.11.11 Combustion efficiency
≔ηC =―――
+Q1 Q2
Qtot
0.88
3.11.12 Defining total air mass needed for combustion
≔Majritpercikel =――――
⋅Majrit mld
kg
⎛⎝ ⋅8.215 103 ⎞⎠ ―
kg
hr
4 Air mass balance

4 Air mass balance
≔QtotA =――――――
Qtot
⋅cpA
⎛⎝ -TAin TMin
⎞⎠
⎛⎝ ⋅1.566 104 ⎞⎠ ―
kg
hr
4.1.2 Water mass flow rate leaving the dryer
≔QtotW =++QWAout QMC QMBF
⎛⎝ ⋅3.949 103 ⎞⎠ ―
kg
hr
4.1.3 Air flow rate leaving the dryer
≔VtotA =⋅⋅⋅⋅QtotA 0.06243 ――
359
33.3
⎛
⎜
⎝
―――――
+273 °C TAout
273 °C
⎞
⎟
⎠
――
m3
kg
311.524 ――
m3
min
4.1.4 Moisture volume flow rate leaving the dryer
≔VtotW =⋅⋅⋅⋅QtotW 0.06243 ――
359
20
⎛
⎜
⎝
―――――
+273 °C TAout
273 °C
⎞
⎟
⎠
――
m3
kg
130.779 ――
m3
min
4.1.5 Total gas volume flow rate leaving the dryer
≔Vtot =+VtotA VtotW 442.303 ――
m3
min
4.1.6 Density of gas leaving the dryer
≔γAout =―――――
+QtotA QtotW
Vtot
0.739 ――
kg
m3
4.1.7 Air humidity leaving the dryer
≔Hum =――
QtotW
QtotA
0.252 ―
kg
kg
leaving the dryer

leaving the dryer
H2O
CO2
SO2
N2
O2
Tot
Mass_In_Ch
⎛
⎜
⎝
―
kg
hr
⎞
⎟
⎠
124.5
0
0
13230
3517
16750
Mass_In_Dr
⎛
⎜
⎝
―
kg
hr
⎞
⎟
⎠
547.314
1292
4.511
13230
3382
18510
Mass_Out_Dr
⎛
⎜
⎝
―
kg
hr
⎞
⎟
⎠
3949
1292
4.511
13230
3382
21910
Con_In_Ch
⎛
⎜
⎝
―
kg
kg
⎞
⎟
⎠
0.009
0
0
0.783
0.208
1
Con_In_Dr
⎛
⎜
⎝
―
kg
kg
⎞
⎟
⎠
0.033
0.07
0.00025
0.715
0.183
1
Con_Out_Dr
⎛
⎜
⎝
―
kg
kg
⎞
⎟
⎠
0.183
0.059
0.0002
0.604
0.154
1
(For purpose of simplifying the problem we neglect gases that have very low concentration)
As it is displayed in the table above the levels of carbon dioxide and sulfur dioxide
concentration are too low to have a notable effect in our gas thermal and chemical properties.
5. Velocity of gasses
Dust load leaving the dryer is proportional to the cost of dust collecting equipments. Taking
this fact into consideration we reason to determine the gas velocity in terms of the amount of
the particles which leave the dryer by being covered with air.
5.1 Dust data

5. Velocity of gasses
Dust load leaving the dryer is proportional to the cost of dust collecting equipments. Taking
this fact into consideration we reason to determine the gas velocity in terms of the amount of
the particles which leave the dryer by being covered with air.
5.1 Dust data
Meshsize
10
20
35
48
65
100
150
200
325
micron
((mm))
1.65
0.83
0.42
0.3
0.22
0.15
0.11
0.074
0.044
Dust load is usually expressed in grains
＝0.454 kg 7000 grains
≔γMdry 1440 ――
kg
m3
Bulk density of dust
≔γW 1000 ――
kg
m3
Density of water
≔DP 0.074 mm Mean diameter particle
After a careful analysis, of the cost of dust collection equipments and volume flow rate of air
leaving dryer, we reasoned that it is a good techno-economical solution, to allow escaping
with the air, only particles that are under 200 mesh. From this we derive:
≔Vh =⋅―――――――
⋅⋅500 γMdry
‾‾‾‾‾‾‾5 ⎛
⎜
⎝
――
DP
mm
⎞
⎟
⎠
2
+γMdry γW
――
m
min
104.144 ――
m
min
Velocity of gases
Correcting at the design temperature in terms of density ratio.Vh
From psychrometric tables we find that the density of air in nominal condition is:

Correcting at the design temperature in terms of density ratio.Vh
From psychrometric tables we find that the density of air in nominal condition is:
≔γA 1.2 ――
kg
m3
≔VC =⋅Vh ――
γA
γAout
169.097 ――
m
min
6 Dust load
≔Vout =―――――
Qout
⎛⎝ -1 φout
⎞⎠ γMdry
19.042 ――
m3
hr
≔VDust_out =⋅Vout
⎛⎝ +ψ4 ψ5
⎞⎠ 2.323 ――
m3
hr
≔QDust_out =⋅VDust_out
⎛⎝ -1 φout
⎞⎠ γMdry
⎛⎝ ⋅3.245 103 ⎞⎠ ―
kg
hr
Dust load
≔FDust =―――――
⋅QDust_out 7000
⋅0.454 kg Vtot
⎛⎝ ⋅1.885 103 ⎞⎠ ――
1
m3
7 Dryer diameter
＝＝Vtot ⋅A VC ⋅―――
⋅π D2
4
VC From this we derive:
≔D =
‾‾‾‾‾‾2
―――
⋅4 Vtot
⋅π VC
⎛⎝ ⋅1.825 103 ⎞⎠ mm
We accept:
≔D 1800 mm
8 Design of lifters
We took as a reasonable value of percentage loaded area equal to 12%
(when we chose the percentage of the loaded area we must take into consideration the
percentage of moisture to be removed, if the percentage of moisture to be removed is height,

8 Design of lifters
We took as a reasonable value of percentage loaded area equal to 12%
(when we chose the percentage of the loaded area we must take into consideration the
percentage of moisture to be removed, if the percentage of moisture to be removed is height,
the percentage of loading area must be chosen at the lowest values and vice versa). From the
figure below we can determine the angle of the loaded area
- Figure 15: Direct heated rotary dryer, angle of loaded area
8.1 Percentage of the loaded area 12% - sin =0.76θ
≔sin((θ)) 0.76
=((asin((0.76)))) rad 49.464 °
≔θ 50 °
8.2 Depth of the loaded area
- Figure 16: Direct heated rotary dryer, depth of loaded area

- Figure 16: Direct heated rotary dryer, depth of loaded area

≔H =―――――
⋅D (( -1 cos((θ))))
2
321.491 mm
8.3 Lifter depth
Therefore, is reasonable to take lifter depths equal to 250mm, because it is recommended by
75 % of loaded depth.
≔dl 250 mm
- Figure 17: Direct heated rotary dryer, lifter depth
From table 3 we determine the angle response of material:
≔θR 32 °
8.4 Cross sectional area of material retained by each lifter
Cross sectional area of material retained by each lifter is determined as follows:
≔AM =――――
⋅dl
2
tan⎛⎝θR
⎞⎠
2
195.272 cm2
- Figure 18: Direct heated rotary dryer, material angle of response
Overlaping angle
The number of lifters in a dryer drum is corelated to overlaping angle, whch is the smallest
angle betwin two lifters with the condition that the above lifter dont puoer material on
upcoming lifter.

- Figure 18: Direct heated rotary dryer, material angle of response
Overlaping angle
The number of lifters in a dryer drum is corelated to overlaping angle, whch is the smallest
angle betwin two lifters with the condition that the above lifter dont puoer material on
upcoming lifter.
≔α =atan
⎛
⎜
⎝
―――――
⋅2 dl tan⎛⎝θR
⎞⎠
D
⎞
⎟
⎠
9.847 °
- Figure 19: Direct heated rotary dryer, overlapping angle
≔N1 =――
2 π
0.172
36.53
≔N1L 24 We decided to use only 24 lifters, for fabrication reasons
- Figure 20: Direct heated rotary dryer, lifters arrangements around dryer drum
9 Rotation speed
Speed of rotation times the dryer diameter empricaly lies between 9 to 12.
Keep in mind, when choosing this value, to take into consideration retention time which is
correlated with percentage of humidity to be removed and with temperature of gases.

- Figure 20: Direct heated rotary dryer, lifters arrangements around dryer drum
9 Rotation speed
Speed of rotation times the dryer diameter empricaly lies between 9 to 12.
Keep in mind, when choosing this value, to take into consideration retention time which is
correlated with percentage of humidity to be removed and with temperature of gases.
＝＝ξ ⋅n D ⋅12 rpm m
≔ξ ⋅12 rpm m
≔n =―
ξ
D
6.667 rpm
10 Time for a showering cycle
10.1 Particle falling time (the time needed for the paticle to all from the lifter to bed load).
≔tf =
‾‾‾‾‾‾‾‾‾‾‾2
―――――
⋅2 D sin⎛⎝θR
⎞⎠
g
0.441 s
- Figure 21: Direct heated rotary dryer, motion of falling particles over dryers drum cross section
10.2 Particle lifting time (time needed for a particle to be lifted from bed load to the hight that
starts to fall).
≔t1 =―――
⋅12 θR
⋅π n
3.056 s
10.3 Time of showering cycle (time needed for a particle to commplete a full cycle of being
lifted and falling)
≔tShC =+t1 tf 3.497 s

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔tShC =+t1 tf 3.497 s
- Figure 22: Direct heated rotary dryer, motion of particles over dryers drum cross section
11 Showering load per unit length of dryer
Number of lifters used per shawering cycle (number of lifters involved in a showering cycle).
＝NC ⋅Kc N1
11.1 Time ratio (time of shawering cycle devided by the time of one rotation or the time if a
full cycle)
＝Kc ――
tShC
tC
11.2 Time of one revolution (the time neeaded to completer one rotation)
≔tC =――
2 π
n
9 s
From this we can derive:
≔Kc =――
tShC
tC
0.389
And:

And:
≔NC =⋅Kc N1L 9.325
Showering load per unit length of dryer (weight of material showered in a cycle)
＝＝FM ⋅⋅NC AM γmes ――――――――
⋅⋅NC AM
⎛⎝ +γMwet γMdry
⎞⎠
2
≔FM =――――――――
⋅⋅NC AM
⎛⎝ +γMwet γMdry
⎞⎠
2
291.342 ―
kg
m
12 Dryer length
＝＝Qtot ⋅⋅Ua Sd ΔTm ⋅⋅⋅Ua D L ΔTm
＝L ――――
Qtot
⋅⋅Ua A ΔTm
=Qtot 4.696 MW Total heat necessary for ensuring the achievement of output product
requirements.
＝Ua ? Volumetric heat transfer coefficient
＝A ? Cross sectional area of dryer
＝ΔTm ? Mean log temperature
12.1 Cross sectional surface area of dryer
≔A =―――
⋅π D2
4
2.545 m2
12.2 Mean log temperature
＝ΔTm ――――――――――――
+-⎛⎝ -TAin TMin
⎞⎠ ⎛⎝ -TAout TMout
⎞⎠ ⋅273 K
ln
⎛
⎜
⎝
―――――
-TAin TMin
-TAout TMout
⎞
⎟
⎠
≔A. =-⎛⎝ -TAin TMin
⎞⎠ ⎛⎝ ⋅1.025 103 ⎞⎠ Δ°C
≔B. =ln
⎛
⎜
⎝
―――――
-TAin TMin
-TAout TMout
⎞
⎟
⎠
3.087

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔A. =-⎛⎝ -TAin TMin
⎞⎠ ⎛⎝ ⋅1.025 103 ⎞⎠ Δ°C
≔B. =ln
⎛
⎜
⎝
―――――
-TAin TMin
-TAout TMout
⎞
⎟
⎠
3.087
≔ΔTm =―
A.
B.
332.003 K
12.3 Volumetric heat transfer coefficient
＝Ua ⋅⋅hc ―――
⋅π Dp
2
4
――
Pnh
AL
＝hc ? Heat transfer coefficient for unit of dried material
＝Dp ? Mean diameter particle size
＝Re ? Flow regime, Reynolds number
＝Pr ? Prandtl number
＝kA ? Conductivity coefficient
12.3.1 Heat transfer coefficient for unit of dried material
＝hc ――
kA
Dp
⎛
⎝ +2 ⋅0.6 ‾‾‾
2
Re ‾‾‾
3
Pr
⎞
⎠
12.3.1.1 Mean diameter particle size
≔Dp =++++⋅ψ1 χ1 ⋅ψ2 χ2 ⋅ψ3 χ3 ⋅ψ4 χ4 ⋅ψ5 χ5 0.143 mm
12.3.1.2 Air physical parameters
From air parameters charts we find the physical parameters of air at T=149 *C and Gama =
0.735 kg/m3
=TAout 149 °C Temperature
=γAout 0.739 ――
kg
m3
Density
≔CpA 1025.8 ―――
J
⋅kg Δ°C
Specific heat of air

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔CpA 1025.8 ―――
J
⋅kg Δ°C
≔kA 0.0389 ―――
W
m2
Δ°C
Conductivity coefficient
≔μA ⋅2.566 10-5
――
kg
⋅m s
Viscosity of air
≔Pr 0.681 Prandl number
12.3.1.3 Mean velocity of the falling particle
≔VP =
‾‾‾‾2
――
⋅D g
2
2.971 ―
m
s
12.3.1.4 Total velocity of the moving particle
≔VRe =‾‾‾‾‾‾‾‾2
+VP
2
VC
2
4.095 ―
m
s
=VC 169.097 ――
m
min
Air velocity
- Figure 23: Direct heated rotary dryer, particle speed vectors decomposition
12.3.1.5 Reynolds number
≔Re =―――――
⋅⋅Dp VRe γAout
μA
16.909
From these findings we can determine heat transfer coefficient for unit of dried material
≔hc =⋅――
kA
Dp
⎛
⎝ +2 ⋅0.6 ‾‾‾
2
Re ‾‾‾
3
Pr
⎞
⎠ m ⎛⎝ ⋅1.132 103 ⎞⎠ ―――
W
m2
Δ°C
The term we are going to assume it to be equal to the showering area of⋅⋅π Dp
2
Pnh
material retained by the lifters and AL is the volume per unit of length of the dryer.

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔hc =⋅――
kA
Dp
⎛
⎝ +2 ⋅0.6 ‾‾‾
2
Re ‾‾‾
3
Pr
⎞
⎠ m ⎛⎝ ⋅1.132 103 ⎞⎠ ―――
W
m2
Δ°C
The term we are going to assume it to be equal to the showering area of⋅⋅π Dp
2
Pnh
material retained by the lifters and AL is the volume per unit of length of the dryer.
＝＝β ⋅⋅π Dp Pnh ⋅NC AM ≔β =⋅NC AM 0.182 m2
＝AL A' ≔A' 2.545 m3
- Figure 24: Direct heated rotary dryer, cross sectional area covered by the falling particles
Volumetric heat transfer coefficient
＝＝Ua ⋅⋅⋅hc π Dp
2
――
Pnh
AL
⋅hc ―
β
A
≔Ua =⋅3.3 hc ―
β
A'
267.187 ―――
W
m3
Δ°C
Finally, we can determine the length of the dryer
≔L =――――
+Q1 Q2
⋅⋅Ua A ΔTm
18.302 m
We acept: ≔L 20000 mm
Showering load (total weight of material showered in a cycle)

Showering load (total weight of material showered in a cycle)
≔FM' =⋅FM L ⎛⎝ ⋅5.827 103 ⎞⎠ kg
14 Effective slope
As a first trial we asume that the slope dryer must be 4.17 cm/m The effect of gas on effective
slope was determined experimentally to be followed by the following expression.
≔S 4.17 ――
cm
m
＝SC 5.2 ――
⋅A U
FM'
- Figure 25: Direct heated rotary dryer, slope of dryer drum
14.1 Showering load for unit of showering cycle
≔FM'' =⋅FM' ――
tf
tShC
734.943 kg
＝＝Divizor ⋅5000 log
⎛
⎜
⎝
――
4950
VC
50
⎞
⎟
⎠
4935
≔Divizor 4935
≔U =――――
⋅⋅VC γAout L
Divizor
0.506 ―――
kg
⋅m min
14.2 The effect of gas on the effective slope
≔SC =10.2 ――
⋅A U
FM'
――
s
mm
0.038 ―
m
m
From these findings we can determine the effective slope

From these findings we can determine the effective slope
≔Se =+S SC 79.302 ――
mm
m
15 Showering output
Showering output per minute is equal to showering load per meter of dryer length times the
advance rate of the sewing material.
≔Ar =―――――
⋅⋅D sin⎛⎝θR
⎞⎠ Se
tShC
1.298 ――
m
minShowering output
≔FMout =⋅FM Ar
⎛⎝ ⋅2.269 104 ⎞⎠ ―
kg
hr
16 Retention time
≔tR =――――――
⋅⋅3.094 ‾‾‾‾
2
32 ° L
⋅⋅n D S
14.708 min
Klin action output
≔KAout =-
⎛
⎜
⎝
――――
+Qin QDrySol
2
⎞
⎟
⎠
FMout
⎛⎝ ⋅5.212 103 ⎞⎠ ―
kg
hr
17 Klin action load
≔KALoad =⋅
⎛
⎜
⎝
-
⎛
⎜
⎝
――――
+Qin QDrySol
2
⎞
⎟
⎠
FMout
⎞
⎟
⎠
tR
⎛⎝ ⋅1.278 103 ⎞⎠ kg
18 Total bed load
≔Ltot =+FM' KALoad
⎛⎝ ⋅7.105 103 ⎞⎠ kg
≔tmr =――――
2 Ltot
+Qin QDrySol
15.279 min
≔X =―――――――
2 Ltot
⋅⋅A L ⎛⎝ +γMwet γMdry
⎞⎠
0.087
21.1 Showering power (enrergy needed for lifting material untill it falls, for an hour).

21.1 Showering power (enrergy needed for lifting material untill it falls, for an hour).
＝＝N1 ―――
⋅FM' B1
⋅K ηk
――――――
⋅FM' ――――
⋅D sin⎛⎝θR
⎞⎠
tShC
ηk
≔ηk 0.97 Process efficiency
≔N1 =――――――
⋅⋅⋅FM' D sin⎛⎝θR
⎞⎠ g
ηk tShC
16.069 kW
21.2 Kling action power
≔θo 18 ° =KALoad
⎛⎝ ⋅1.278 103 ⎞⎠ kg Conveying angle for this material
＝＝N2 ――――――
⋅⋅KALoad F sin⎛⎝θo
⎞⎠
⋅5 ηk
―――――――――
⋅⋅⋅⋅⋅KALoad π D n sin⎛⎝θo
⎞⎠ g
⋅5 ηk≔ηk 0.9
≔N2 =―――――――――
⋅⋅⋅⋅⋅KALoad π D n sin⎛⎝θo
⎞⎠ g
⋅5 ηk
3.397 kW
21.3 Motor power
≔ηOT 0.94 Gear one transmission efficiency
≔ηBe 0.99 Bearing efficiency
≔ηRe 0.95 Gearbox efficiency
≔η =⋅⋅ηOT ηBe
4
ηRe 0.858
3. Determine power consumption.
≔N =―――
+N1 N2
η
22.692 kW
4. Determine motor's power
≔β 1.1 ≔Nm =⋅β N 24.962 kW
5. Selection of Motor
≔ne 1440 rpm ≔Ne 25 kW
To simplify the problem we are considering the dryer drum as a beam supported in two pair
of trunnion wheels.
The main loads applied to drum shell are, weight of material, insulation weight, weight of
lifters, weight of girth gear, torsional moment caused by motor loads.

To simplify the problem we are considering the dryer drum as a beam supported in two pair
of trunnion wheels.
The main loads applied to drum shell are, weight of material, insulation weight, weight of
lifters, weight of girth gear, torsional moment caused by motor loads.
≔γs 7800 ――
kg
m3
Steel density
=KALoad
⎛⎝ ⋅1.278 103 ⎞⎠ kg Klin action load
=FM'
⎛⎝ ⋅5.827 103 ⎞⎠ kg Showering load
22.1 Dryer drum weight
22.1.1 Total weight of material
≔Kpara 1.2 Coefficient that take into consideration, the
changeing input capacity of dryer, due to
changing physical parameters of material in
entrance.
≔PMtot =Kpara
⎛⎝ +KALoad FM'
⎞⎠ ⎛⎝ ⋅8.525 103 ⎞⎠ kg
≔QMtot =Kpara ――――
+KALoad FM'
L
426.27 ―
kg
m
22.1.2 Weight of lifters
≔aPB 250 mm
≔δPB 10 mm
≔PLB =⋅⎛⎝ ⋅⋅⋅aPB δPB N1L L⎞⎠ γs
⎛⎝ ⋅9.36 103 ⎞⎠ kg
≔QLB =⋅⎛⎝ ⋅⋅aPB δPB N1L
⎞⎠ γs 468 ―
kg
m22.1.3 Insulation weight
≔γiz 1800 ――
kg
m3
Insulation density
≔δiz 40 mm Insulation thickness
≔δbar 20 mm Shell thickness
≔Piz =⋅γiz
⎛
⎜
⎜⎝
⋅―――――――――――――
⋅π
⎛
⎝ -⎛⎝ ++D ⋅2 δiz 2 δbar
⎞⎠
2
⎛⎝ +D 2 δbar
⎞⎠
2 ⎞
⎠
4
L
⎞
⎟
⎟⎠
⎛⎝ ⋅8.505 103 ⎞⎠ kg
≔Qiz =⋅γiz
⎛
⎜
⎜⎝
―――――――――――――
⋅π
⎛
⎝ -⎛⎝ ++D ⋅2 δiz 2 δbar
⎞⎠
2
⎛⎝ +D 2 δbar
⎞⎠
2 ⎞
⎠
4
⎞
⎟
⎟⎠
425.246 ―
kg
m
As it recommended:

As it recommended:
＝δbar
(( ‥0.007 0.02)) D
≔δbar 20 mm Dryer shell thickness
≔Pbar =⋅γs
⎛
⎜
⎜⎝
⋅――――――――
⋅π
⎛
⎝ -⎛⎝ +D 2 δbar
⎞⎠
2
D2 ⎞
⎠
4
L
⎞
⎟
⎟⎠
⎛⎝ ⋅1.784 104 ⎞⎠ kg
≔Qbar =⋅γs
⎛
⎜
⎜⎝
――――――――
⋅π
⎛
⎝ -⎛⎝ +D 2 δbar
⎞⎠
2
D2 ⎞
⎠
4
⎞
⎟
⎟⎠
891.961 ―
kg
mTotal weight
≔Ptot =+++PMtot PLB Piz Pbar
⎛⎝ ⋅4.423 104 ⎞⎠ kg
≔Qtot =+++QMtot QLB Qiz Qbar
⎛⎝ ⋅2.211 103 ⎞⎠ ―
kg
m
22.1.5 Weight of girth gear
From experience we decide:
≔DK 3000 mm Girth gear outer diameter
≔BK 200 mm Girth gear width
≔HK 200 mm Girth gear wall height
≔PK =⋅γs
⎛
⎜
⎜⎝
⋅――――――――
⋅π
⎛
⎝ -⎛⎝ +DK 2 HK
⎞⎠
2
DK
2 ⎞
⎠
4
BK
⎞
⎟
⎟⎠
⎛⎝ ⋅3.137 103 ⎞⎠ kg
- Figure 26: Direct heated rotary dryer, girth gear front representation
After a careful analysis we decided:
First pair of trunnion wheels-riding ring will be placed in distance

After a careful analysis we decided:
First pair of trunnion wheels-riding ring will be placed in distance
Second pair of trunnion wheels-riding ring will be placed in distance
Dryer girth gear will be placed in distance
≔L1 4000 mm
≔L2 16000 mm
≔LK 14000 mm
So the gap between the two pairs of trunnion wheels-riding
ring will be:
Gap between dryer girth gear and second pair of trunnion
wheels-riding ring will be placed in distance
≔L' =-L2 L1
⎛⎝ ⋅1.2 104 ⎞⎠ mm
≔L'' =-L2 LK
⎛⎝ ⋅2 103 ⎞⎠ mm

- Figure 27: Direct heated rotary dryer, loading, shear, bending moment, deflection, diagram

≔RA =+―――
⋅Qtot L
2
―――
⋅PK L''
L'
⎛⎝ ⋅2.264 104 ⎞⎠ kg
≔RB =+―――
⋅Qtot L
2
――――
⋅PK
(( -L' L''))
L'
⎛⎝ ⋅2.473 104 ⎞⎠ kg
22.2.2 Maximal bending moment that is applied in the drum
≔MB =+――――――
⋅⋅Qtot L (( -⋅2 L' L))
8
――――――
⋅⋅PK
(( -L' L'')) L''
L'
⎛⎝ ⋅2.734 104 ⎞⎠ ⋅kg m
22.2.3 Maximal torsional moment that is applied in the drum
≔MBpd =―――
⋅Ne η
⋅β n g
⎛⎝ ⋅2.848 103 ⎞⎠ ⋅kg m
22.3 Dryer drum section properties
22.3.1 Resistance momentum of dryer drum
≔Dmes =―――――
+⎛⎝ +D 2 δbar
⎞⎠ D
2
1.82 m
≔W =⋅―――
⋅π Dmes
2
4
δbar
⎛⎝ ⋅5.203 104 ⎞⎠ cm3
22.3.2 Polar resistance momentum of dryer drum
≔Jpol =――――――――
⋅π
⎛
⎝ -⎛⎝ +D 2 δbar
⎞⎠
4
D4 ⎞
⎠
32
⎛⎝ ⋅9.471 106 ⎞⎠ cm4
≔Wpol =――――――――
⋅π
⎛
⎝ -⎛⎝ +D 2 δbar
⎞⎠
4
D4 ⎞
⎠
⋅16 Dmes
⎛⎝ ⋅1.041 105 ⎞⎠ cm3

≔σpk =――
MB
W
52.55 ――
kg
cm2
22.4.2 Dryer drum stress caused by torsional moment
≔τpd =――
MBpd
Wpol
2.736 ――
kg
cm2
Overall dryer drum stress
=‾‾‾‾‾‾‾‾‾‾‾2
+σpk
2
⋅4 τpd
2
52.834 ――
kg
cm2
22.5 Selection of dryer drum material
Depending on thermo-physical condition we reasoned that it is appropriate to use as material
for the shell, stainless steel with mechanical properties as layed below
≔σrr 5800 ――
kg
cm2
Yield strength
≔σq 2900 ――
kg
cm2
Tensile strength
≔E ⋅2 106
――
kg
cm2
Elastic modulus
According to recommendation, maximum allowed stress in dryer shell must be:
≔σlej 100 ――
kg
cm2
Maximum allowed stress
22.6 Checking dryer drum deflection
22.6.1 Moment of inertia of drum cross section
≔Ix_x =――――――――
⋅π
⎛
⎝ -⎛⎝ +D 2 δbar
⎞⎠
4
D4 ⎞
⎠
64
⎛⎝ ⋅4.735 106 ⎞⎠ cm4
22.6.2 Maximal deflection of dryer shell in most critical positions
22.6.2.1 Deflection in the edge of dryer drum
≔Ymac0_L =―――――――――――
⋅⋅Qtot L1
⎛
⎝ -⋅L' ⎛
⎝ -L'2
⋅6 L1
2 ⎞
⎠ ⋅3 L1
3 ⎞
⎠
⋅⋅24 E Ix_x
0.149 mm

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔Ymac0_L =―――――――――――
⋅⋅Qtot L1
⎛
⎝ -⋅L' ⎛
⎝ -L'2
⋅6 L1
2 ⎞
⎠ ⋅3 L1
3 ⎞
⎠
⋅⋅24 E Ix_x
0.149 mm
≔YmacL2 =――――――――――――
⋅⎛
⎝ +⋅Qtot L'2
⋅PK L'⎞
⎠
⎛
⎝ -1.25 L'2
⋅6 L1
2 ⎞
⎠
⋅⋅96 E Ix_x
0.329 mm
22.6.3 Relative deflection in the middle of dryer drum
≔ε =―――
YmacL2
L
⋅1.645 10-5
Deflection is negligible
23 Design calculations of riding ring and trunnion wheels (first aproach)
Determination of riding rings dimensions, set freely in dryer drum shell based on maximum
allowed deformation caused by bending moment and contact pressure.

23 Design calculations of riding ring and trunnion wheels (first aproach)
Determination of riding rings dimensions, set freely in dryer drum shell based on maximum
allowed deformation caused by bending moment and contact pressure.
- Figure 28: Direct heated rotary dryer, reaction force acting on dryer riding rings
23.1 Determination of loads acting on dryer riding ring
≔Rm =max⎛⎝ ,RA RB
⎞⎠ ⎛⎝ ⋅2.473 104 ⎞⎠ kg Reaction force acting on dryer riding rings
≔α 30 ° Angle between roller and vertical access of dryer
drum with reference point the center of the drum
≔Np 14 Number of riding ring support plates
23.1.1 Reaction force acting on riding ring and trunnion wheels
≔Rr =――――
Rm
⋅2 cos((α))
⎛⎝ ⋅1.428 104 ⎞⎠ kg
23.1.2 Angle between two near support palates with reference point the center of the drum
≔α1 =――
2 π
Np
25.714 °
23.1.3 Number of riding ring support plates in a quadrant
≔Npk =―――
-Np 2
4
3

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔Npk =―――
-Np 2
4
3
- Figure 29: Direct heated rotary dryer, forces acting on riding ring support plates
23.1.4 Force acting on riding ring support plate that is located in the lowest point of the
vertical axis
≔P0 =――
⋅4 Rm
Np
⎛⎝ ⋅7.065 103 ⎞⎠ kg
vertical axis
≔P1 =⋅P0 cos⎛⎝α1
⎞⎠ ⎛⎝ ⋅6.366 103 ⎞⎠ kg
≔P2 =⋅P0 cos⎛⎝2 α1
⎞⎠ ⎛⎝ ⋅4.405 103 ⎞⎠ kg
The drum riding ring is a closed system,statically indeterminate, loaded with outside forces in
accordance with vertical acces. To solve this problem we have to do an imaginary cut on top
of the ring and and replace the nouns with a unit of bending moment M0 and a unit of axial
force N0. The action of each pare of forces will be examined separately and then there effect
will be gathered.
By knowing the orientation of each force that is applied on a riding ring support plants we can
determine the angle between the force and the vertical access of the drum, for each case
separately.

=P0
⎛⎝ ⋅7.065 103 ⎞⎠ kg ≔θ0 180 °
=P1
⎛⎝ ⋅6.366 103 ⎞⎠ kg ≔θ1 =-180 ° α1 154.286 °
=P2
⎛⎝ ⋅4.405 103 ⎞⎠ kg ≔θ2 =-180 ° 2 α1 128.571 °
≔β =-180 ° α 150 ° Angle between trunnion wheel and the imaginary cut on top
of the riding ring with reference point the center of the drum
Ring axis radius.
=D ⎛⎝ ⋅1.8 103 ⎞⎠ mm Dryer drum inside diameter
=δbar 20 mm Drum shell thickness
≔δPp =⋅1.5 δbar 30 mm Thickness of reinforcing plates
≔δPm =δbar 20 mm Times of riding ring support plates
≔δPsh =δbar 20 mm Times of fixing plates
≔δind 50 mm Insulation thickness
≔h 200 mm Estimated ring height
≔Dbb =+++++D 2 δind ⋅2 δbar ⋅2 δPp ⋅2 δPm ⋅2 δPsh
⎛⎝ ⋅2.08 103 ⎞⎠ mm
≔Djb =++++++D ⋅2 δiz ⋅2 δbar ⋅2 δPp ⋅2 δPm ⋅2 δPsh ⋅2 h ⎛⎝ ⋅2.46 103 ⎞⎠ mm
≔Rmes =―――
+Dbb Djb
4
⎛⎝ ⋅1.135 103 ⎞⎠ mm

≔M0_0 =⋅⋅P0 Rmes
⎛
⎜
⎝
-+1 ――――
1
cos(( -π α))
⋅α tan(( -π α))
⎞
⎟
⎠
⎛⎝ ⋅1.184 103 ⎞⎠ ⋅kg m
≔M0_1 =⋅⋅P1 Rmes
⎛
⎜
⎝
+-+1 ――――
cos⎛⎝θ1
⎞⎠
cos(( -π α))
⋅α1 sin⎛⎝θ1
⎞⎠ ⋅⋅α cos⎛⎝θ1
⎞⎠ tan(( -π α))
⎞
⎟
⎠
⎛⎝ ⋅1.424 104 ⎞⎠ ⋅kg m
≔M0 =+M0_0 M0_1
⎛⎝ ⋅1.543 104 ⎞⎠ ⋅kg m
≔N0_0 =――――――
⋅⋅P0 α tan(( -π α))
2 π
-339.93 kg
≔N0_1 =―――――――――――――
⋅P1
⎛⎝ -⋅α1 sin⎛⎝θ1
⎞⎠ ⋅⋅α cos⎛⎝θ1
⎞⎠ tan(( -π α))⎞⎠
2 π
69.626 kg
≔N0 =+N0_0 N0_1 -270.304 kg
- Figure 30: Direct heated rotary dryer, bending moment diagram

For: >θ β <<0 α2 β ≔α1' 0 °
≔Mα1' =+M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α1'
⎞⎠⎞⎠ ⎛⎝ ⋅1.543 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α2' 10 °
⎛⎝ -1 cos⎛⎝α2'
⎞⎠⎞⎠ ⎛⎝ ⋅1.542 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α3' 20 °
⎛⎝ -1 cos⎛⎝α3'
⎞⎠⎞⎠ ⎛⎝ ⋅1.541 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α4' 30 °
⎛⎝ -1 cos⎛⎝α4'
⎞⎠⎞⎠ ⎛⎝ ⋅1.539 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α5' 40 °
⎛⎝ -1 cos⎛⎝α5'
⎞⎠⎞⎠ ⎛⎝ ⋅1.536 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α6' 50 °
⎛⎝ -1 cos⎛⎝α6'
⎞⎠⎞⎠ ⎛⎝ ⋅1.532 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α7' 60 °
⎛⎝ -1 cos⎛⎝α7'
⎞⎠⎞⎠ ⎛⎝ ⋅1.528 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α8' 70 °
⎛⎝ -1 cos⎛⎝α8'
⎞⎠⎞⎠ ⎛⎝ ⋅1.523 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α9' 80 °
⎛⎝ -1 cos⎛⎝α9'
⎞⎠⎞⎠ ⎛⎝ ⋅1.518 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α10' 90 °
≔Mα10' =+M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α10'
⎞⎠⎞⎠ ⎛⎝ ⋅1.512 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α11' 100 °
≔Mα11' =+M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α11'
⎞⎠⎞⎠ ⎛⎝ ⋅1.507 104 ⎞⎠ ⋅kg m
<θ β ≤≤0 α2 β ≔α12' 110 °For:

Required: “CONFIDENTIAL” Sh.a. DESIGNED: Arberor MITA Design group“ELMAGERARD”≔Mα11' =+M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α11'
⎞⎠⎞⎠ ⎛⎝ ⋅1.507 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α12' 110 °
≔Mα12' =+M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α12'
⎞⎠⎞⎠ ⎛⎝ ⋅1.502 104 ⎞⎠ ⋅kg m
For: <θ β ≤≤0 α2 β ≔α13' 120 °
≔Mα13' =+M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α13'
⎞⎠⎞⎠ ⎛⎝ ⋅1.497 104 ⎞⎠ ⋅kg m
<θ β ≤≤0 α2 β ≔α14' 130 °
≔Mα14' =+M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α14'
⎞⎠⎞⎠ ⎛⎝ ⋅1.492 104 ⎞⎠ ⋅kg m
<θ β ≤≤0 α2 β ≔α15' 140 °
≔Mα15' =++M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α15'
⎞⎠⎞⎠ ⋅⋅P2 Rmes sin⎛⎝ -α15'
⎛⎝ -π α15'
⎞⎠⎞⎠ ⎛⎝ ⋅1.869 104 ⎞⎠ ⋅kg m
<θ β ≤≤0 α2 β ≔α16' 150 °
≔Mα16' =++M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α16'
⎞⎠⎞⎠ ⋅⋅P2 Rmes sin⎛⎝ -α16'
⎛⎝ -π α16'
⎞⎠⎞⎠ ⎛⎝ ⋅1.866 104 ⎞⎠ ⋅kg m
<θ β ≤≤0 α2 β ≔α17' 155 °
≔Mα17' -++M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α17'
⎞⎠⎞⎠ ⋅⋅P2 Rmes sin⎛⎝ -α17'
⎛⎝ -π α17'
⎞⎠⎞⎠ ⋅⋅Rr Rmes sin⎛⎝ -α17' β⎞⎠
=Mα17'
⎛⎝ ⋅6.328 103 ⎞⎠ ⋅kg m
<θ β ≤≤0 α2 β ≔α18' 160 °
≔Mα18'' -++M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α18'
⎞⎠⎞⎠ ⋅⋅P2 Rmes sin⎛⎝ -α18'
⎛⎝ -π α18'
⎞⎠⎞⎠ ⋅⋅Rr Rmes sin⎛⎝ -α18' β⎞⎠
≔Mα18' =+Mα18'' ⋅⋅P1 Rmes sin⎛⎝ -α18'
⎛⎝ -π α18'
⎞⎠⎞⎠ ⎛⎝ ⋅1.181 104 ⎞⎠ ⋅kg m
<θ β ≤≤0 α2 β ≔α19' 170 °
≔Mα19'' -++M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α19'
⎞⎠⎞⎠ ⋅⋅P2 Rmes sin⎛⎝ -α19'
⎛⎝ -π α19'
⎞⎠⎞⎠ ⋅⋅Rr Rmes sin⎛⎝ -α19' β⎞⎠
≔Mα19' =+Mα19'' ⋅⋅P1 Rmes sin⎛⎝ -α19'
⎛⎝ -π α19'
⎞⎠⎞⎠ ⎛⎝ ⋅1.18 104 ⎞⎠ ⋅kg m
<θ β ≤≤0 α2 β ≔α20' 180 °
≔Mα20'' -++M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α20'
⎞⎠⎞⎠ ⋅⋅P2 Rmes sin⎛⎝ -α20'
⎛⎝ -π α20'
⎞⎠⎞⎠ ⋅⋅Rr Rmes sin⎛⎝ -α20' β⎞⎠
≔Mα20' =+Mα20'' ⋅⋅P1 Rmes sin⎛⎝ -α20'
⎛⎝ -π α20'
⎞⎠⎞⎠ ⎛⎝ ⋅1.179 104 ⎞⎠ ⋅kg m

≔Mmax =+M0 ⋅⋅N0 Rmes
⎛⎝ -1 cos⎛⎝α1'
⎞⎠⎞⎠ ⎛⎝ ⋅1.543 104 ⎞⎠ ⋅kg m
23.4 Determination of riding ring and trunnion wheels dimensions
23.4.1 Riding ring section dimensions
23.4.1.1 Riding ring width
＝b ―――――――――――
⋅⋅⋅⋅⋅0.592
Rr E1 E2 2 ⎛⎝ +Djb Drmb
⎞⎠
⋅⋅⋅σlejk
2
⎛⎝ +E1 E2
⎞⎠ Djb Drmb
＝＝E1 E2 E Since we select as material for riding ring and trunnion wheels, casted medium
carbon steel, Elastic modulus will be the same.
≔σlejk ⋅5000 ――
kg
cm2
Maximum allowed contact stress
=Djb
⎛⎝ ⋅2.46 103 ⎞⎠ mm Riding ring outer diameter
＝Drmb ⋅(( ‥0.25 0.33)) Djb Pranojme: ≔Drmb ⋅0.33 Djb Trunnion wheel diameter
≔Drmb 800 mm
≔b =――――――――――
⋅⋅⋅⋅0.592
Rr E2
2 ⎛⎝ +Djb Drmb
⎞⎠
⋅⋅⋅⋅σlejk
2
2 E Djb Drmb
65.861 mm
Constructively we accept ≔b 300 mm
23.4.1.2 Riding ring height
From previous information we can derive the height of riding rings:
≔σlejpk ⋅1000 ――
kg
cm2
Maximum allowed bending stress
≔hb =
‾‾‾‾‾‾‾2
―――
⋅6 Mmax
⋅b σlejpk
175.662 mm
We accept: ≔hb 200 mm
=Djb
⎛⎝ ⋅2.46 103 ⎞⎠ mm Riding ring outer diameter

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