2. Rotary kiln thermal load
• The thermal load of the kiln represent the heat quantity
introduced to the kiln tube as direct result of burning the
fuel in the kiln burner and the heat recuperated from the
cooler first stage and introduce to the kiln by the secondary
air.
• Therefore, the kiln burner is an instrument to produce a
flame, which have the characteristic that could make it able
to create this intense heat area so the kiln can produce its
designed capacity.
3. Therefore, the higher the thermal load toward the
upper limit of the kiln capacity i.e. the quantity of
heat in the kiln created by the flame of the kiln
burner is high enough.
4. How the kiln operator changes the thermal load in the
rotary kiln?
• If the kiln burner introduces 6000 litre/hour to the kiln-burning zone, the thermal
load is calculated as follow:
• The net heating value of the heavy oil = 9500 kcal/kg fuel. That means when we
fire one kilogram of oil in the kiln through the kiln burner. the net heat generated in
the kiln is 9500k.calorie.
• The density of the heavy oil = 0.875 kg/litre at operating temperature of 120ºC.
• Total heat in kcal /hour in the kiln =6000x9500x 0.875 =49875000
kcal/hour
• So the thermal load in the kiln per the hour is 49875000kcalorie i.e. 50 million
kcalorie/hour.
5. Relation between Thermal Load and Kiln System Productivity
• The relation between the thermal load and the kiln system
productivity can be expressed and defined in two different ways:
1-Kiln-cross section, defined as [kiln sectional thermal-load].
2-Kiln-volume,defined as [kiln volumetric load].
6. 1. Kiln Cross Section Thermal Load
This represent the thermal heat introduce to the kiln through its main burner per
hour distributed on each meter square of the kiln burning-zone cross-section.
Example
Kiln diameter = 5 meter
Kiln brick thickness = 220 mm.
The cross section of the kiln =3.14[kiln diameter –brick thickness on both sides of
the kiln diameter] ²÷4
Kiln cross section = 3.14[5-.440] ² ÷4
= 3.14[4.56] ² ÷4
=3.14x20.7936÷4
=16.32 m ² [this the active cross section of the kiln in which the
heat is generated
7. • If we assumed that the kiln is burning 6000 litre of heavy oil with the net calorific
value 9500 k calorie /kg of oil and density of .875 kg/litre at servicing temperature
of 120ºC
• Total heat in kcal /hour in the kiln =6000x9500x875 =49875000kcalorie
• So the thermal load in the kiln per the hour is 49875000kcalorie.
• When this thermal load is distributed on the total kiln cross section in meters, the
result will be the kiln cross sectional thermal load
• Kiln cross sectional thermal load = total thermal load in the kiln ÷kiln active cross
section
=49875000 kcal/hour÷16.32 =3056066 k calorie/hour/m²
In other words the thermal load in the kiln burning zone = 3x106 k calorie /hour/m²
From experience in cement industry the following has been observed
8. Kiln system burning zone thermal load
106 k calorie/hour/m²
Suspension preheater kiln
2.8----5.5
Suspension preheater kiln with secondary firing in the kiln riser
duct
2.8----6.0
Suspension preheater kiln with IN-LINE Precalciner
2.4----5.1
Suspension preheater kiln with Off-LINE Precalciner
2.4----5.1
Suspension Preheater, with SEPARATE-LINE Precalciner
2.4----5.1
Suspension Preheater, with SEPARATE-LINE Precalciner and
IN-LINE Precalciner attached to kiln riser duct
2.4----5.1
9. • The upper limit is representing the maximum heat-load that can be introduced
to a certain kiln with its specific diameter.
• This maximum thermal load is the direct results of the maximum fuel quantity
that can be introduced properly to the kiln cross section inside which the fuel
ignited and combusted.
• If we tried to introduce more fuel, the direct result will be direct failure of the
kiln bricks due to the break over-heating in the burning-zone that can reach in
some extreme cases the actual melting of the bricks.
• The lower limit represents the lowest limit of the thermal load that has been
produced from certain quantity of fuel ignited and combusted in the kiln burning
zone and still the kiln can be hot enough to produce sound cement clinker by
producing enough melted material.
10. Example:
This example is intended to illustrate the effect of diameter on the quantity of
fuel, which can be ignited in any kiln.
We will apply the higher and lower limits of cross sectional thermal load to dry
kiln with the same type of preheater and type of precalciner but of different
diameter and consequently of different cross section.
kiln is 4.4 meter in diameter.
five stages preheater equipped with off-line precalciner.
11. Calculation of the cross –sectional –thermal load of both kilns:
a) Kiln cross section calculation:
• Kiln 4.4 meters diameter: brick height is 220mm.
Cross section of the kiln=3.14 (4.4-.44) 2 ÷4
Were .44 is the total brick height on both side of the kiln diameter.
Cross section of the kiln = 3.14 (4.4-.44) 2 ÷4
= 3.14x (3.96) 2 ÷4
=12.31 m2
12. b) Total thermal load in both kilns:
Expected thermal load in any kiln calculation:
Total thermal load in the kiln with maximum cross section= 5.1 x 10 6 x kiln cross-
section.
• Kiln 4.4meters diameter:
Total thermal load in the kiln burning zone = 5.1 x 10 6 x 12.31 m 2 = 62.78x10 6
Kcalorie /hour.
This is the heat available in the kiln-burning zone, which will create the intense heat
area we call burning zone or B.Z.
13. If kiln is consuming 750 kcalorie /kg clinker and the ratio of fuel in precalciner to kiln
is 63 in precalciner to 37 in the kiln main burner then the expected production will be
Kiln 4.4 meter in diameter
Kiln production / hour=62.78x10 6 ÷ [750x 37%] = 226234 kg clinker/hour
Kiln production /day = 226234 x 24 ÷ 1000 = 5429 ton / day
14. c) Quantity of fuel oil that by ignition in B.Z produced this
thermal load:
• Kiln 4.4 m diameter: fuel specific net heat value =9600 /kg fuel at service
temperature. Fuel density at service temp .875 kg / litre.
Quantity of fuel oil in the kiln burner = total thermal load in the kiln in B.Z ÷ fuel net
heat value
Quantity of fuel oil in the kiln burner = 62.78x106 ÷ 9600 =6539.5 kg fuel / hour in the
kiln main burner.
Quantity of fuel in litre= fuel in kg/ fuel density
= 6539 ÷ Quantity of fuel in litre= fuel in kg/ fuel density .875
= 7474 litre / hour.
15. 2. Kiln Volumetric Load
It is defined as the kiln clinker production in ton per 24 hour per meter cube of kiln
total volume inside lining [tpd/m3].
Example:
Kiln length = 78 meter.
Kiln diameter =5 meter
Assume brick thickness in the kiln 220mm.
Kiln volume= kiln length x kiln cross sectional area.
= Kiln, length x 3.14[kiln diameter-brick thickness on both sides] ² ÷4
Kiln cross section= 3.14[5 - .440] ² ÷4
= 3.14[4.56] ² ÷4
=3.14x20.7936÷4
=16.32 m ²
Kiln active volume=16.32x78=1273.2 m ³.
16. This volume is where the kiln thermal heat is produced.
Inside this volume the material moves and absorb the heat and melt and form the
clinker.
Therefore the higher this volume the bigger the material quantity that can be in the
kiln and the higher the thermal load that can be produced in this volume without
exceeding the proper cross-sectional thermal load for the kiln.
So from experience the following table is established:
17. Kiln system Volumetric load tpd/m3
Suspension preheater kiln 1.8----2.3
Suspension preheater kiln with secondary
firing in the kiln riser duct
1.8----2.5
In-line Precalciner kiln 3.6----5.0
Off-line Precalciner kiln 3.6----5.0
Suspension Preheater, with SEPARATE-
LINE Precalciner
3.6----5.0
Suspension Preheater, with SEPARATE-
LINE Precalciner and IN-LINE Precalciner
attached to kiln riser duct
3.6----5.0
18. The meaning of the maximum volumetric load is that the maximum production,
operator can produce from his kiln, is the kiln volume in m 3 x kiln type maximum
volumetric load tpd / m 3.
Kiln optimum production = the kiln volume in m 3 x kiln type maximum volumetric load
tpd / m 3
Example:
•Kiln 4.4 m diameter and its length 74 with 4 stages preheater
Kiln volume= kiln cross section x kiln length
Kiln volume= 12.31 x 74 = 910.94 m 3
Kiln expected optimum production capacity = 910.94 x 2.3 = 2095 ton / day
19. •2- Kiln 4.4 m diameter and its length 74 with precalciner
and 4- stage preheater
Kiln volume= kiln cross section x kiln length
Kiln volume= 12.31 x 74 = 910.94 m 3
Kiln expected optimum production = 910.94 x4.8 = 4372.5 t/pd
This kiln actually had been upgraded.
The average daily production after upgrading is 4200 ton / day. The original
nominal capacity was 3200 ton / day.
But surprisingly when the kiln was fed with easy to burn raw materials
produced for long periods of time 4400 ton /day i.e. reaching the maximum
possible production from such kiln volume, with acceptable cross thermal
load.
20. Therefore we now have the criteria to determine the following:
• The kiln optimum daily production that can be obtained from the kiln tube or
if there is a project of upgrading: what is the maximum production the kiln can
give with such project.
• This will help to determine the required changes in the other parts of the
plant such as quarry crusher, related transport system, stackers areas (clay
and limestone), raw mill department, kiln feed, kiln speed system and if it can
be increased, kiln preheater, related fans, kiln burner, kiln cooler, clinker
transport system after the cooler and all departments filters either bag filters
or electrostatic filters.
21. • The second condition we can now estimate with a high degree of accuracy
is the maximum fuel the operator can ignite through the kiln main burner
without damaging the kiln burning zone bricks and other kiln zones bricks
which will satisfy the operator needs so he can drive the kiln to it’s the
maximum possible production.
• There are other factors that have a great influence on the rotary kiln
production.
These factors can summarized in the following:
22. Loading percent of the kiln volume or Kiln volume loading
That factor is representing the quantity of the precalcined raw meal in the kiln tube,
in other wards the volume that this bed of material occupies inside kiln during the
operation of the kiln.
The kiln volume loading is in the range of 8 to 13 % of the kiln volume and the kiln
operator should not exceed the upper limit or the material will not tumble in side the
kiln and will just glide inside the kiln without mixing together during its movement in
side the kiln especially in the zones before the burning zone which will decrease the
heat transfer to the inside the material body, therefore the material when reaching
23. the burning zone will not ready for clinkerization process and will have easily free lime
and low litre weight.
There is a formula that is used normally in the plants to determine this factor.
The following formula is used to determine any kiln loading:
Kiln volume loading % = 1.5 x clinker production t/h ÷[active cross-section
area in m2].
Example:
Kiln production = 80 t/h, active cross-section = 10.173 m2
Kiln volume loading = 1.5 x 80 / 10.173 = 11.795 %
Another formula
Kiln charge or kiln volume loading = 3.2 x kiln production ÷[kiln diameter]3xkiln
inclination in percent x kiln number of rotation per minutes.
24. Example:
Dry kiln with preheater and precalciner and has the diameter of 4.4 m, use the
above-mentioned formula to calculate the expected kiln volume loading of the kiln
tube.
Kiln effective cross-section = 3.14 (4.4-.44) 2 /4
= 3.14x (3.96) 2 /4
=12.31 m2
Another formula is also used in determining the percent loading in the kiln:
Percent loading of the kiln = kiln output in kg/hour x raw meal factor for clinker x
retention time in minutes x 100÷ bulk density of the feed kg/m³ x 60minutes x
internal kiln volume m³.
25. Example:
Kiln production is 4200t/day the raw meal factor 1.85 kg raw meal / kg clinker,
Retention time in the kiln is 22 minutes, bulk density of the feed = 1.1 ton / m³ in
side the kiln, internal kiln volume 910 m³ what is the percent loading of the kiln.
Solution
Hourly clinker production = 4200 ÷24 =175 ton/ hour
Percent loading of the kiln = 175 x1000 x 1.85 x22 x100÷1.1 x1000 x 60x 910
= 712250000÷60060000
= 71225÷6006 =11.87 %
If we compared the first and second examples we will find that both are nearly the same.
26. Kiln retention-time
The correct determination of the hot calcined material inside short dry kiln with
external preheater and precalciner is depending mainly on kiln slope and the
number of rotation of the kiln but the accurate determination of retention time of the
material in the kiln should but into consideration the physical properties of the
material.
Main factor that affect the material retention time are:
•Kiln slope
Short dry kiln with preheater and precalciner has the slope range from 3 to 4. as an
average%.
27. 1.0º 1.20º 1.40º 1.60º 1.70º 1.80º 2.00º 2.20º 2.30º 2.40º 2.60º 2.8º
1.7 2.09 2.44 2.79 3.00 3.14 3.49 3.84 4.00 4.19 4.54 4.89 %
Kiln slope in % and equivalent slope of the kiln in degree:
•Kiln speed. Short dry kiln with preheater and precalciner has the speed range from
2.5 to 4 rpm [revolution per minute].
The following is the formula that is used to determine the kiln feed residue time i.e.
Retention-time in the dry kiln with SP and PC
Retention time = 11.2 x kiln length ÷ [kiln number of revolution x effective
diameter x kiln slop in degree].
Another formula
Kiln retention time = kiln length x 23÷kiln diameter x kiln number of revolution per
minute x kiln inclination in percent.
