Mass and Heat Balances-grinding

Grinding Technology Course

Mass and Heat Balances


MASS AND HEAT BALANCES TABLE OF CONTENTS

1.1

INTRODUCTION....................................................................................................... 3

1.2

MASS BALANCE ....................................................................................................... 4

1.2.1

The General System ................................................................................................ 4

1.2.2

Total Mass Balance ................................................................................................ 5

1.2.3

Partial Balances ..................................................................................................... 7

1.3

GAS BALANCES ........................................................................................................ 9

1.3.1

General Theory....................................................................................................... 9

1.3.2

Measurement and Calculation of Gas Flows ....................................................... 10

1.3.3

Determination of Gas Flows ................................................................................ 16

1.3.4

False Air Calculation ........................................................................................... 27

1.4

HEAT BALANCE ..................................................................................................... 33

1.4.1

Introduction .......................................................................................................... 33

1.4.2

Heat Balance for the Mill System ......................................................................... 33

1.5

CONCLUSION.......................................................................................................... 34

1.6

APPENDIX I .............................................................................................................. 35

1.7

APPENDIX II ............................................................................................................ 37

1.8

APPENDIX III........................................................................................................... 39

1.9

APPENDIX IV ........................................................................................................... 42

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1.1

INTRODUCTION

The mill installation is supplied with materials, gas flow and heat energy. The mass, gas and heat balances for the mill system should be in equilibrium. There are various reasons for which it is of interest to know the mass and heat balance. These are: A. Planning of the production B. Control of the material quality C. Control of the process D. Development of heat balances E. Optimisation of energy consumption F. Process optimisation In this module the significance of mass and heat balance and relevant factors identified for calculating mass and heat balances will be described. The subjects will be covered in the following order:  Mass Balance  Gas Balance  Heat Balance


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1.2

MASS BALANCE

The major part of planning the production, control of the quality and control of the process is based on mass balances. The mass balances can be calculated as total balances or partial balances. All of them can be divided in non-equilibrated and equilibrated balances (See Figure1). This section will deal with the following aspects of mass balance:  The general system  Total mass balance  Partial balances

1.2.1

The General System

The general system of the mass flow is shown on Figure 2. The total quantity of material A(t) in the system, which has CjA(t) concentration of component "j", is a sum of the different material flows In(t), which have different concentration of common component "j". The output of the system consist of different flows Ok(t), which have different concentration of component "j".

Figure 1 Non-equilibrated and Equilibrated Balances.

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Figure 2 The General System of the Mass Flow.

1.2.2

Total Mass Balance

The total mass balance is described as: INPUT = OUTPUT + ACCUMULATION and the general equation is: N

K

 I t   Q n 1

n

k 1

k



dA t  dt

There are two principal cases of the above mentioned equation:  Non-equilibrated (time-dependent) system.  Equilibrated or stabilised system (independent of time).


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The first one is very complicated, and is normally limited to considerations on large time horizons, f. ex. daily, weekly, and monthly balance of the production, contents of stocks, etc. The second case is less complex, and the equation is reduced to: N

K

 I t   Q n 1

n

k 1

k

This mass balance can represent output from mills, kilns, crushers etc. at a stabilised production. Figure3 shows the raw mill system, where the mill is fed with lime-stone, marl and sand, and supplied with drying air. Mill output consists of raw mix and humid air. A typical example is shown on the "Raw Mill Production Summary" (see Figure 4).

Quantity [tons] 874,2 134,5 51,0 1.059,7

Dry limestone Dry marl Dry sand Raw meal production

Figure 3 Raw Mill

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Figure 4 Raw Mill Production Summary.

Partial Balances The partial balances are divided into two types:  Component balances  Molecular balances The component balances are described as: N



n 1

I n t C

t    O k t C kO, j t  dA t  K

I n, j

dt

k 1

The component balances can also be divided in non-equilibrated and equilibrated systems. The non-equilibrated component balance requires a large amount of information f. ex. process control on filling of pre-homogenising stores, where a fixed quality is required. In cases where the flows and concentrations do not vary in relation to time (equilibrated component balances), the general equation is reduced to:

 I t C t    O t C t  N

n 1

K

n


I n, j

k 1

7

k

O k, j

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This equation is used for programming and proportioning purposes (assuming that the quality of the raw materials is known) such as raw material proportioning, proportioning of stockpiles, programming of quarry work and control of the dosage of gypsum to the cement mill. The last one is shown in Figure 5.

Figure 5 Dosage of Gypsum to a Cement Mill.

There are no chemical reactions; therefore, this is a component balance, whereby the general rule is valid: INPUT = OUTPUT + ACCUMULATION In this case there is no accumulation, and the result is: INPUT = OUTPUT Component balance for SO3: INPUT N

 I t C n

n 1

I n , SO 3

t  

0 ,5 46 ,1  6 th 6 100 100 0 , 43  2 , 77  0  0  3, 20

86

t

h 

t

h

0  2

t

h

0 

OUTPUT

 O k t C k , SO 3 t   100 K

k 1

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t

h

3, 20  3, 20 100 FLSmidth 2009, all rights reserved


The molecular balances are very difficult to handle, taking into consideration that chemical reactions take place in the system. The same is the case for non equilibrated molecular balances, where no chemical changes occur.

1.3

GAS BALANCES

This section will cover the following points:  General Theory.  Measurement and calculation of gas flow.  Determination of gas flow.  False Air Calculation.

1.3.1

General Theory

The major part of the process which takes place during the burning of cement clinker involve gases, for example: a)

Drying of raw materials and coal In general, the drying processes are affected by heat exchange between hot gases and moist solid materials. The heat transferred to the solid phase evaporates the moisture which subsequently is transferred as water vapour to the gas stream.

b)

Heating In the cement industry, heating takes place by combustion during which hot gases are formed and heat content exchanged with a solid phase.

c)

Cooling In the cement industry, two main types of cooling takes place; one is cooling by gases and the other is cooling by evaporation of water.

d)

Transport In various types of equipment, air or gases are used to transport powdery materials e.g. air slides, screw pumps, Flux pumps, air lifts etc.

As it was mentioned in section 3.1, there are various reasons for which it is of interest to know the gas flows which circulate between the different systems. The gas balances are formed on the basis of total balance in the system. During the process, it is possible that certain FLSmidth 2009, all rights reserved

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compounds change phases in such a manner that part of the solid material in the system leaves in the form of gases. The reasons for that are chemical reactions such as decarbonisation or physical-chemical reactions such as evaporation. Sometimes it is possible to separate the balance of water/vapour from the gas balance. In this way three balances are developed on basis of the total balance: A. Balance of dry solids B. Balance of moisture/water/vapour C. Balance of dry gases Based on the three balances, one balance of solids may be calculated. This balance comprises the balance of dry solids and the part of the moisture balance for the water or moisture which has not evaporated. In the same manner it is possible to calculate the total gas balance, i.e. the balance of dry gases plus the part of the moisture balance that comprises the vapour, i.e. the water evaporated during the process. The total gas balance in the system contains several unknown variables. These variables must be determined on the basis of specific measurements. In order to determine the total balance, the following specific measurements are performed: A. Determination of the gas composition by:  Gas analyzer.  Determination of the dew point temperature. B. Determination of gas quantities by:  Pitot tube measurements.  Fan performance curves.  Measurement of pressure losses. The most appropriate measurements depend on the particular situation and should be decided in each actual case. 1.3.2

Measurement and Calculation of Gas Flows

In cement factories, there is a great need to find out the different air and gas flows through mills, kilns, coolers and filters.

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To determine these flows several measurements and calculations exist, and they will be presented as follows:  Calculation of specific gravity of a gas.  Humidity and dew point.

1.3.2.1

Calculation of Specific Gravity of a Gas

The specific gravity of a gas depends on:  Chemical composition  Physical conditions. As a point of reference for the physical conditions used: 0oC and an atmospheric pressure of 10333 mm WG (pressure at sea level). The specific gravity, ρot, of a dry gas under these conditions, is calculated on the basis of the volumetric percentages of its content of CO 2, O2, CO and N2:

 ot  0,01  1,965  CO2  1,429  O2  1,250  CO  1,251 N 2



kg Nm3



If no gas analyses exist, the values from the following table can be used. Specific gravity and heat Specific gravity 3 kg/Nm * kg/m3 kg/m3 ρot (90oC) (100oC) 1,293 0,97 0,95 1,40 1,05 1,02 1,30 0,98 0,95 0,80 0,60 0,59

Specific heat cal/g/oC **

Atmospheric air 0,24 Kiln gases 0,25 Aux. furnace gases 0,26 Water vapour 0,45 Raw materials 0,22 Clinker 0,19 o *) Dry air at 10333 mm WG (760 mm Hg) and 0 C **) Valid for air and gases 0-200oC and materials and Clinker 0-400oC. If the physical conditions are different from the reference conditions (10333 mm WG, 0oC), the following formula is used to find the specific gravity of the actual condition.

 t   ot  FLSmidth 2009, all rights reserved

273 b  p  273 t 10333 11

  kg

m3

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where t is the actual temperature, measured in oC, p is the relative pressure (in duct), b is the barometric pressure which depends on the altitude h, in m, above sea level, and which is calculated as:

b  10333  e  0 , 0001255 h

mmWG 

The formula is shown graphically in Figure 6.

Altitude and Barometric Pressure (b) b=10333*e-0,0001255*h

Barometric pressure b ( mmWG)

11000

10000

9000

8000

7000

6000 0

1000

2000

3000

4000

Altitude h (meter)

Figure 6 Altitude and Barometric Pressure.

In this way, it is possible to calculate the specific gravity of any dry gas, on condition that its chemical composition, temperature and barometric pressure are known.

1.3.2.2

Humidity and Dew Point

The dew point is an important value because a decrease in the gas temperature to a value below dew point will cause condensation of the water vapour and may result in problems during operation. The dry bulb temperature refers to the intensity of the heat content of substance as measured with a common thermometer in degrees.

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The wet bulb temperature is a method for determining the amount of moisture in air or gases. This is accomplished by using a standard, common thermometer with a wick or a sock on the sensing element (bulb). With the wick wet with clean water the thermometer is inserted into the gas stream or whirled through the air to create air movement over the bulb and wick. The resulting cooling effect from the evaporation process lowers the thermometer reading. The temperature will drop corresponding to the rate of evaporation, which will be determined by the amount of moisture in the air or gas stream. The more moisture there is present in the air, the slower will the rate of evaporation be, thus resulting in higher wet bulb reading.

The dew point temperature is defined as the temperature at which air is 100% saturated with moisture and the moisture begins to condense out of the air. When air or gas steam is cooled it becomes denser and, therefore, can hold less moisture. When the air is cooled to the temperature at which it cannot hold any more moisture, it is 100% saturated, and this represents the dew point temperature. The relation is shown graphically in Figure 7 (Dew Point Temperature). It should be pointed out that in lowering the temperature of the air until the dew point is reached, the moisture content of the air does not change. However, each change in the temperature changes the percentage of humidity in the air, since the ratio of moisture in the air to the amount of moisture the air can hold changes with the temperature.

Figure 7 Dew Point Temperature.

Humidity is a term used to describe the presence of moisture in the air. The amount of moisture that the air will hold depends upon its temperature. The warm air will hold more moisture than the cold air. FLSmidth 2009, all rights reserved

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Relative humidity refers to the amount of moisture in a given volume of air compared to the amount of moisture the air would hold if completely saturated at the temperature at which the comparison is being made. The relative humidity is expressed in percentage. Absolute humidity is the ratio between the amount of water and gases expressed as kg H2O per kg gas. If the units are moles, then the ratio expresses the absolute molar humidity. In most cases the molar ratio is more convenient to use, due to the fact that it is easy to make conversions between molar and volume ratios on basis of the ideal gas laws. Determination of the absolute humidity may be done graphically by using Figure 8 (Water Vapour in Air and Gases). The graph shows the ratio between water and dry gas amount wo and the ratio between the water and wet gas amount y = wo/(1+ wo).

1,0

0,9

0,9

0,8

0,8

y (kg water/kg wet gas) wo (kg water/kg drygasr)

0,7

0,7

0,6

0,6

0,5

0,5

0,4

0,4

0,3

0,3

0,2

0,2

0,1

0,1

0,0

kg water/kg wet gas y=wo/(1+wo)

kg water/kg dry gas wo

Absolute Humidity versus Dew Point Temperature 1,0

0,0 10

20

30

40

50

60

70

80

90

100

Dew point temperature (degC)

Figure 8. Water Vapour in Air and Gases

The determination of gas humidity can be made in the following way: a) Measure the temperature of the dry and the wet thermometers on the chosen gas and water vapour mixture. b) Look up the figure "Dew Point Temperature". The dew point is then found at the coordinates given by the wet and dry thermometer temperatures, respectively. The value of the dew point temperature is determined by interpolation between the curves ranging from 0 to 80oC. c) When the dew point has been determined as described in points a) and b) proceed to Figure 8. Water Vapour in Air and Gases. Select the dew point temperature on the abscissa and follow the corresponding vertical line to the interpolated curve for % Grasim Industries

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CO2 (as determined by Orsat analysis). Then follow a horizontal line from the intercept to the ordinate axis, where the absolute humidity in kg vapour per kg dry gas can be read. If it is required to determine the molar humidity, then the above value should be converted to a volume ratio at the required conditions by using the corresponding water and gas densities. d) Knowing the absolute humidity of the gas wo, it is possible to calculate the specific gravity of the moist gas, - i.e. dry gas + water vapour, according to the following formula:

o 

1  wo 1 w  o  ot 0,8



kg Nm 3

moist gasses 

Examples of calculating gas humidity are available in Appendix I of this module.

1.3.2.3

Evaporation of water

Evaporation of water (transformation from liquid into steam) can happen not only when the water is boiled at 100 °C. The evaporation heat for the water heated up to 100 °C is 539 kcal/kg, Since the heat, used for heating the water to 100 °C, is 100 kcal ( specific heat 1 kcal/kg), the total heat consumption will be 639 kcal/kg. Knowing the specific heat for the steam between 0 °C and 100 °C, it is possible to calculate the evaporation heat at the temperature lower than 100 °C (see figure 9). If we assume that the average specific heat of the steam is 0.44 kcal/kg, the evaporation heat at 0 °C will be:

H e0  H e100  100  0,44 *100  595kcal / kg The total heat for evaporation of the water at 80 °C can be fide as:

He80  H e0  0,44 * 80  630,2 kcal / kg Attention: The values of the specific heat and heat of evaporation can vary, depending of the used basis sources.


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Evaporation of water 700 630,2 639

595 600

Heat content (kcal/kg)

500

400

300

200

100 100

80 0 0

10

20

30

40

50

60

70

80

90

100

Temperature (degC)

Evaporation at 0 degC



Figure 9 Evaporation of water

1.3.3

Determination of Gas Flows

Various methods can be used to determine a gas flow, such as: A) Measurements with Pitot tube B) Fan characteristics C) Heat balances. In this lesson, the methods A and B are treated, while C is treated separately.

1.3.3.1

Measurement with Pitot Tube

The measuring point is very important for the reliability of the measurement when D is the diameter of a duct. There should not be curves, dampers or other obstructions in a distance of minimum 5 x D before and 2 x D after the measuring point. If possible, a point with small dust load and maximum distance to cyclones, separator and fans should be chosen. The total pressure, p, is the sum of the dynamic pressure, pd, and the static pressure, ps. p = pd + ps Grasim Industries

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With the Pitot tube all three types of pressure can be measured by connecting simple manometers to the Pitot tube in the following way: p: ps: pd :

manometer connected to + manometer connected to manometer connected to + and -

To eliminate the effects of a non-laminar airflow, it is necessary to take measurements in the whole cross section of the duct to get a reliable average value. In practice, a set of measurements are made according to the duct diameter, as shown on page 19. The dynamic pressure is calculated as an average of the

p dm

 

p d of each point of measurement:

pd n

Knowing the dynamic pressure it is possible to calculate the air velocity in the duct, v, as:

v  4,43 

m sec 

f * p dm t

where f is dynamic pressure correction and ρ is the specific gravity of the gas. f = 1 for normal Pitot tube, f = 0,7 for S-type (normally marked on the tube) – see figures 10a and 10b) Knowing the air velocity the air or gas flow in the duct can be calculated as:

Q1  3600 

   D 2  v   t kg  hour  4

If it is necessary to find the flow in m 3 or Nm3 the following relations can be used:

Q  Q1   t

1

Q 0  Q1   o


1

17



m3



hour

Nm 3



hour



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Example of Air Flow Determination with Pitot Tube In the duct from R1S11 to the bottom of the separator, Pitot tube measurements have been carried our as follows: Data: Duct diameter = 2000 mm. pd(a) 1 12 85 120 125 70 20 2

0,97  D 0,90  D 0,81  D 0,68  D 0,32  D 0,19  D 0,10  D 0,03  D

pd(b) 2 18 75 115 120 80 15 2

mm WG mm WG mm WG mm WG mm WG mm WG mm WG mm WG

Calculate the air flow rate in m3/s, m3/min, Nm3/min and kg/min. ρt ρo

The gas density is:

Q

= 1,05 kg/m3 (ρo at 0oC and 1 atm) = 1,30 kg/Nm3

 3 3  2,0 2  27,02  84,9 m s  5093 m min 4 Q0  Q 

t 1,05 3  5093   4113 Nm min 0 1,30

Qm  Q   t  5093 1,05  5347 kg min

Qm  Q0   o  41131,30  5347 kg min The Pitot tube should be used as shown in figures 10a or 10b.

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Figure 10 Measurements with the Pitot Tube.


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D

- + pd

Figure 10a Normal Pitot tube

D +

-

pd

Figure 10b. S-type Pitot tube is used, if the gas stream is dust filled

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Fan Characteristics To determine the air flow in a system by means of fan characteristics the following rules and procedures should be used: Rule No. 1: Fan characteristics are valid only at the conditions given together with the curves (fan revolutions, temperature, pressure and specific gravity). Rule No. 2: Because of rule No. 1, it is necessary to distinguish between diagram and actual values:

Diagram value Actual value

1.3.3.2

Specific gravity (kg/m3)

Fan velocity (rpm)

Power consumption (kW)

Air flow (m3/s)

Total pressure (mbar)

ρ1

n1

PA1

Q1

Pt1

ρt

n

PA

Q

Pt

Procedure for Using Fan Characteristics

1) Calculate the specific gravity of the gas/air at actual conditions, using the equation: (ρ0 at 0oC and 1 atm.)

t o 

273 b  p  273 t 10333

  kg

m3

where p – relative pressure (in duct) in mmWG

2) Based on the actual power consumption of the fan, PA, the diagram value can be calculated as:

n   PA1  PA   1   1  n  t 3


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3) With PA1 calculated, use the fan characteristics to find Q1 and Pt1. 4) To transform Q1 and Pt1 to actual values use the following formulas:

Q  Q1

n n1 2

n  Pt  Pt1     t  n1  1

  m3

s

 mbar

pt is the total pressure at the pressure side minus the total pressure at the suction side, or the static pressure at the pressure side minus the static pressure at the suction side, providing that the dynamic pressure is the same on both sides. pt = (ps,p + pd,p) - (ps,s + pd,s) = ps,p - ps,s

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1.3.3.3

Recalculation of Fan Performance

Fan performance calculation at changed speed: (see Figure 11)

Q v1  Q v

n1 n

n  p t1  p t   1  n

2

n  PA1  PA   1   n 

3

Figure 11 Fan Performance Calculation at Changed Speed.


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Fan performance calculation at changed impeller diameter: (see Figure 12)

Q v1  Q v

d1 d

d  pt1  pt   1  d

2

d  PA1  PA   1  d 

3

Figure 12 Fan Performance Calculation at Changed Impeller Diameter.

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Fan performance calculation at changed density: (see Figure 13)

Q v1  Q v

  pt1  pt   1   t    PA1  PA   1   t 

Figure 13 Fan Performance Calculation at Changed Density.


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Specific gravity calculation at changed temperature and atmospheric pressure: (see Figure 14)

 273  t   PA       1   t    273  t1   PA1  The following symbols are used for all calculations: Qv pt PA ρt d n t b

-

volume flow, total pressure, power consumption, gas density, impeller diameter, fan speed, working temperature, atmospheric pressure,

(m³/s) (mbar or mm WG ) (kW) (kg/m³) (mm) (rpm) (oC) (mmWG)

The index "1" refers to new operating conditions.

Figure 14. Performance Curve for Fan HT-B.BR. 2000/2000.

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1.3.4

False Air Calculation

When considering the air flow participating in the overall clinker burning process, which are schematically represented in Fig 15, an important factor must always be taken into consideration. This factor is the so-called false air (also known as parasite, in-leaking or infiltrating air). False air enters through badly closed openings, e.g. mill inlet seal, inspection doors, dampers, etc.

Figure 15 Schematic Representation of Gas Flows in the Kiln System.

This in-leakage should at all times be kept at a minimum, since it cuts down the efficiency of the system fan. The cold false air may also cause an undesirable temperature drop to below the dew point which may cause condensation and eventually result in corrosion problems. The extra volume of air which the system fan has to transport requires extra power to drive this fan. It can be calculated that a false air quantity of 40 % may result in an extra power consumption FLSmidth 2009, all rights reserved

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consumption of the system fan of about 40 %, which is a significant inefficiency of the system. It is thus of importance at all times to maintain an effective sealing of all unnecessary openings. The quantity of false air is usually determined on the basis of O2 or CO2 analyses by using the Orsat apparatus at any two given points in the system. False air is usually expressed in per cent of unmixed gases at the inlet measuring point. In the calculations, atmospheric air is reckoned to contain 0 % CO2 and 21 % O2. In the calculations the term "gas" signifies the unmixed gases and the term "mixture" signifies the mixture of gases and false air. At the measuring point I is found: QI O2,I

Nm3 smoke gases containing % oxygen

corresponding to a total oxygen quantity of:

QI  O2, I 100

Nm

3

O2



At the measuring point II is found: QII O2,II

Nm3 (smoke gases + false air) containing % oxygen.

A calculation of the false air which infiltrates the system between points I and II can then be determined according to: False Air = QII - QI 0,21 · (QII - QI)

Nm3 containing Nm3 oxygen,

Where:

Q1  O 2 , I O 2 , II 

100

 0,21  Q II  Q I  Q II

 100 %

from which:

Q II  Q I 

21  O 2 , I 21  O 2 , II

Nm 3

If the air or gas quantities are unknown, the percentage of false air may simply be calculated on the basis of the oxygen contents: Grasim Industries

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False air 

O 2 , II  O 2 , I 21  O 2 , II

 100 %

Example In the raw mill circuit shown in Figure 16, measurements of CO2 and O2 were carried out by Orsat analysis. The quantity of false air in the raw mill circuit is calculated according to the equation in Figure 17.

Figure 16 Determination of False Air in the Raw Mill Circuit


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Figure 17 Calculation of the Amount of False Air Into a System

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Position A to B:

CO 2 

O2 

29,0  27,4  100  5,84 % 27,4  0,0

4,75  4,33  100 2,58% 21 4,75

Average  4,2 % Position B to C:

CO 2  O2 

27,4  18,9  100  44,97 % 18,9  0,0 8,93  4,75  100  34,63 % 21  8,93

Average  39 ,8 % Position C to D:

CO 2 

O2 

18,9  16,8  100  12,50 % 16,8  0,0

9,97  8,93  100  9,43 % 21  9,97

Average  10 ,97 %


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Position A to D:

CO2  O2 

29,0  16,8  100 72,62 % 16,8  0

9,97  4,33  100  51,13 % 21  9,97

 61 ,88 %

Average

Too much false air at the system inlet lowers the temperature. False air increases the total air quantities, and thereby the load on the fan.

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1.4

HEAT BALANCE

1.4.1

Introduction

Now and then it is useful to establish a total material- and heat (energy) balance of the mill system. This is a rather elaborated calculation, however, it gives one of the most complete description of the operating system. The balance itself is based on the conservation of energy i.e. ENERGY (input) = ENERGY (output + accumulated) If we consider the mill as the system, it becomes basically a question of determining the heat content flowing into and out of the system. Apart from electrical power input all major energy exchanges are directly related to the heat transfer.

NOTE: The accumulated energy - i.e. the energy used to create new surface - is negligible according to the low efficiency of the grinding process (less than 1 %).

1.4.2

Heat Balance for the Mill System

To establish the heat balance an arbitrary standard condition of 0oC and 760 mm Hg are normally used. Still it is common to calculate in kcal/h. The system may consist of any subpart of the total mill system. Figure 18 shows a mill system with typical input and output components.

Figure 18 Mill System with Typical Input and Output Components.


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Input Material to the mill will be the heat content of the total dry feed. For close circuit the separator is normally not included, i.e. the return has to be considered as feed. Heat content of raw materials, clinker and other components ( e.g. gypsum) can be found from tables. 1.4.2.1

Air to the mill is an atmospheric air for the cement mill. For the raw mills it will often be solely kiln gases, or a mixture of kiln gases and auxiliary heat generator air. Coal mills use of atmospheric air or kiln gases will depend on the used system ( non-inert atmospheric air, inert kiln gases ). In the calculation a consideration of false air amount entering the system should be made. The water calculation should include moisture in the feed components and water injected into the mill (cement mills). The power supplied to the mill should be converted to kcal/h : 1 kWh = 860 kcal. Output The material can always be calculated by a simple mass balance. The air will contain a total air out of the system ( together with the false air). The water will include heat consumption of the residual moisture in the material and evaporated water as vapour. The heat losses are mainly surface losses from the mill shell, and depend on the surface temperature. Normally the surface loss for the ball mill is calculated as 600 kcal/h·m2. Examples of heat balance calculations are attached to this Module in Appendix IV. 1.4.2.2

1.5

CONCLUSION

Correct balances of mass, heat and gases in a ball mill grinding system are of considerable importance. Balances that are not properly checked and maintained can cause malfunctioning of the grinding system and also cause excessive power consumption. In this module problems of mass and heat calculations have been introduced and the proper solutions given. It is recommended to train additionally with problem solving by using the CBT pertaining to the module.

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1.6

APPENDIX I

GAS HUMIDITY CALCULATIONS Determination of ρot Orsat-analyses CO2 O2 CO N2

11,66 % 6,52 % 0,04 % 81,78 %

Specific gravities CO2

=

1,9643 kg/Nm3

O2

=

1,4289 kg/Nm3

CO

=

1,250 kg/Nm3

N2

=

1,251 kg/Nm3

Calculated specific gravity of gas:

CO 2

O2

Nm 3 CO 2 kg CO 2 0,1166  1 , 9643  0,2290 kg Nm3 gas 3 3 Nm gas Nm CO 2 Nm 3 O 2 kg O 2 kg 0,0652  1 , 4289  0 , 0932 gas Nm 3 Nm 3 gas Nm 3 O 2

Nm 3 CO kg CO CO 0,0004  1 , 250  0,0005 kg Nm 3 gas 3 3 Nm gas Nm CO N2

Nm 3 N 2 kg N 2 0,8178  1 , 251  1,0231 kg Nm 3 gas 3 3 Nm gas Nm N 2

 ot 1,3458 kg Nm3 gas dry  FLSmidth 2009, all rights reserved

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Determination of ρo Measurements:

tdry = 90oC and twet = 50oC

From Figure 7:

Dew point = 45oC

From Figure 8:

wo = 0,068 kg vapour/kg dry gas

Volume of 1,000 kg dry gas

1 1,3458

0,743 Nm3

Volume of 0,068 kg vapour

0 , 068 0 ,8

0,85 Nm3

0,743  0,085

0,828 Nm3

Volume of 1,068 kg moist gas

1, 068 0 ,828

Specific gravity of the moist gas, ρ0

1,290 kg/Nm3

Determination of ρ

static pressure, Ps = - 200 mm WG

Measurements:

H = 1400 m above sea level From Figure 6:

barometric pressure, b = 8640 mm WG

t o 

 t 1,290

Grasim Industries

273 b  ps  273  t dry 10333

  kg

m3

273 8640 200   0,792 273 90 10333

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  kg

m3



1.7

APPENDIX II

AIR FLOW CALCULATIONS Example of Air Flow Determination with Pitot Tube In the duct from R1S11 to the bottom of the separator (see Figure 19), Pitot tube measurements have been carried out as follows: Data: Duct diameter = 2000 mm. pd(a) 1 12 85 120 125 70 20 2

0,97  D 0,90  D 0,81  D 0,68  D 0,32  D 0,19  D 0,10  D 0,03  D

pd(b) 2 18 75 115 120 80 15 2

mm WG mm WG mm WG mm WG mm WG mm WG mm WG mm WG

Calculate the air flow rate in m3/s, m3/min, Nm3/min and kg/min. The gas density is:

Q

ρt ρ0

= =

1,05 kg/m3 1,30 kg/Nm3

 3 3  2,0 2  27,02  84,9 m s  5093 m min 4

Q0  Q 

t 1, 05  5093   4113 1,30 0

Nm 3

min

Q m  Q   t  5093  1,05  5347 kg min


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Figure 19 Pitot Tube Measurement in the Raw Mill Circuit

Solution to the Example of Air Flow Determination with Pitot Tube:

 pd

pd

pd

1,00 3,46 9,22 10,95 11,18 8,37 4,47 1,41 50,06

1,41 424 8,66 10,72 10,95 8,94 3,87 1,41 50,20

6,22

6,28

p d  6 , 26

Average

p d  39 ,3 mm WG

 t  1, 05 v  4, 43  Grasim Industries

kg m3

39 ,3  27 ,1 1,05 38

m s 



1.8

APPENDIX III

FAN PERFORMANCE CALCULATIONS Example of Fan Performance The power consumption of a fan of the type HT-B.BR 2000/2000 is: PA t p b

= = = =

230 kW 89oC -20 mbar = -204 mm WG 10072 mm WG

Use the performance curve for the fan to determine the quantity of air transported in: a) Nm3/s b) m3/s c) kg/s

The fan motor is replaced with a new motor with the speed: n = 1050 rpm. Calculate the quantity of air transported with the same data as above in: d) Nm3/s e) m3/s f) kg/s

Solution to the Example of Fan Performance From the performance curve it is found that:

 75  0 ,96

kg m3

 1   75  0,96 kg m3   0  0,96 


273  75 273 39

kg Nm3

1,22 kg Nm3

Grasim Industries


t  89 1,22 

273 10072 204  0,88 kg m3  273  89 10333

PA1  PA 

a)

b)

c)

1 0,96  230  251 kW t 0,88

Qv  Qv1  25,5 Qv  0,88  25,5

kg s

m3

s

 22,4 kg s

22 , 4 kg s Qv   18 ,4 kg 1,22 Nm 3

Nm 3

s

n  1050 RPM n 1  970 RPM

Q v  Q v1 

d)

n n1

24 ,3 kg s Qv   19 ,9 1, 22 kg Nm 3

Nm 3

1050  27,6 970

s

e)

Qv  25,5 

f)

Qv  0,88  27,6 kg s  24,3 kg s

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40

m

s



Figure 20 Performance Curve for Fan HT-B.BR. 2000/2000.


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1.9

APPENDIX IV

HEAT BALANCE CALCULATIONS Case 1 An open-circuit cement mill with two grinding compartments has the following characteristics:

Diameter of mill shell Length of mill shell Length of first compartment Power consumption, C1 Power consumption, C2

4,2 11,0 4,25 997 1669

m m M kW kW

Operational data: Production Feed temperature (clinker + gypsum) Material temperature, outlet C1 Material temperature, outlet C2 Air temperature, mill inlet Air temperature, outlet C1 Air temperature, outlet C2 Water temperature (injected water) Airflow at mill inlet

85 170 108 122 25 108 122 18 20000

tph o C o C o C o C o C o C o C kg/h

Fixed data to be used in heat balance: Specific heat of cement Specific heat of air Heat loss from mill surface 1 kWh consumed by mill Heat of evaporation at 100oC Specific heat of vapour (0 -120oC) Vapour in the air at mill inlet Specific gravity of dry air Specific gravity of vapour

0,19 0,24 600 860 539 0,44 0,015 1,293 0,80

kcal/kgo·C kcal/kgo·C kcal/m2·h kcal kcal/kg kcal/kgo·C kg/kg dry air kg/Nm3 kg/Nm3

a) Use a heat balance for the first compartment to calculate the amount of water that must be injected to maintain 108o in the outlet. b) Use a heat balance for the second compartment to calculate total amount of water that must be injected to maintain 122oC in the outlet. c) Calculate the total airflow in the mill outlet (airflow at inlet + vapour) in kg/h, Nm3/h and m3/h at 122oC. d) Calculate the air velocity in the free area above the ball charge at the mill outlet (q = 30%). e) Calculate the total amount of vapour in the airflow in kg vapour/kg dry air, and find the expected dew point and wet temperature in the mill outlet (Figures 3.7 and 3.8) Grasim Industries

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Case 2 For a TIRAX coal mill, 32 x 4,5 + 2,8, the following operational data are given: Production (incl. H2O) Water in product Water in feed Inlet temperature (gases) Outlet temperature Power consumption Specific heat of coal Feed temperature

20 2,2 14,2 400 75 450 0,25 20

tph % % o C o C kW kcal/kg·oC o C

a) Based on a heat balance for the coal mill operating at above conditions find the necessary amount of gases for drying, in kg/h and m3/h. b) The hot gas duct between the grate cooler and the coal mill inlet has a diameter of 1 m. Calculate the air velocity in this duct.


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Case 3 We will now treat a closed-circuit raw mill installation, as shown in the following diagram:

Actual operational data: Production Water in raw materials Water in product and return materials Temperature of raw materials Temperature of false air Mill inlet temperature Kiln gases available Temperature in mill outlet Temperature of return material Specific heat of raw meal Specific heat of kiln gases Specific heat of false air

124000 ? 0,5 10 10 325 110000 80 78 0,21 0,25 0,24

kg/h % % o C o C o C kg/h o C o C kcal/kg·oC kcal/kg·oC kcal/kg·oC

a) With a circulation factor of 2,0 calculate the maximum water % which can be dried in this mill. False air is 0%. b) Will the drying capacity increase with a circulation factor of 3,0? c) What will the drying capacity be with 25% false air in the mill? Circulation factor = 2,0 and the same amount of kiln gases.

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Mass and Heat Balances-grinding

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