Cement Chapter 12

Technical Information Sheets

C E M E N T T E C H N O L O G Y N O T E S 2 0 0 5 138

1 . G y p s u m A n a l y s i s

2 . G y p s u m A d d i t i o n L e v e l

3 . L S F a n d B o g u e C o m p o s i t i o n

4 . Wa t e r - S o l u b l e A l k a l i s

5 . C e m e n t S S A , R e s i d u e a n d R o s i n - R a m m l e r P S D S l o p e

6 . C e m e n t F i n e n e s s

7 . Vo l u m e L o a d i n g

8 . M i l l C r i t i c a l S p e e d

9 . M i l l P o w e r C a l c u l a t i o n

1 0 . P o w d e r Vo i d F i l l i n g

1 1 . M i l l A i r f l o w

1 2 . M i l l E x h a u s t D e w P o i n t

1 3 . C i r c u l a t i n g L o a d

1 4 . C e m e n t G r i n d a b i l i t y

1 5 . S e p a r a t o r P e r f o r m a n c e

1 6 . A i r S e t Te n d e n c y

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GYPSUM ANALYSIS

1. Analyses required are: SO3, losses at 50C, 250C and 950C

2. Calculate the gypsum content (CaSO4.2H2O) from the loss between 50C and 250C. This is assumed to represent the water of crystallisation. At 50C it is assumed to be free moisture. The water of crystallisation is:2H2O/CaSO4.2H2O i.e. 36/172.14 = 20.9%

3. Calculate the SO3 accounted for by this level of gypsum, i.e. multiply bySO3 /CaSO4.2H2O i.e. 80.06/172.14 = 46.5%

4. Allocate the remaining SO3 (Total SO3 gypsum SO3) to anhydrite.

5. Calculate the amount of anhydrite present as SO3, i.e.SO3 /CaSO4 i.e. 80.06/136.14 = 58.8%

6. The loss between 250C and 950C can be allocated to calcite

EXAMPLE:

(i) Analysis: SO3 = 43.9%loss at 50C = 0.55%loss at 250C = 13.77%loss at 950C = 16.12%

(ii) Gypsum = (13.77 0.55)/20.91 = 63.2%Therefore the SO3 of this = 63.2 * 0.465 = 29.4%

(iii) SO3 of anhydrite = 43.9 29.4 = 14.5Therefore the anhydrite = 14.5/0.588 = 24.7%

(iv) Loss between 250C and 950C = 16.12 13.77 = 2.35Therefore the calcite (possibly) = 2.35/0.44 = 5.3%

SUMMARY:

Gypsum = 63.2%Anhydrite = 24.7%Calcite = 5.3%

TECHNICAL INFORMATION TIS MS001

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GYPSUM ADDITION LEVEL

1. Analysis required are: Clinker SO3Gypsum SO3 (i.e. SO3 of gypsum additive) Cement SO3

2. % Gypsum = (Cement SO3 Clinker SO3)/(Gypsum SO3 Clinker SO3) * 100

3. EXAMPLE:

Clinker SO3 = 0.88%Gypsum SO3 = 43.9%Cement SO3 = 3.20%

Therefore % gypsum = (3.20 0.88)/(43.9 0.88) *100 = 5.39%

4. For the presence of non-clinker components it is necessary to calculate a new target for the SO3 or a new effective Clinker SO, i.e.

(i) 5% limestone (of 0% SO3) in the cement.New target (for the clinker/gypsum)Cement SO3 = (limestone SO3 * limestone %) + (new target SO3 * (1-limestone %))

i.e. 3.20 = 0 * 0.05 + SO3 target * (1 - 0.05)Therefore New SO3 target = 3.20/0.95 = 3.37%

Therefore in the clinker/gypsum have 5.78% gypsum and 94.22% clinkerTherefore in the cement have

Clinker = 89.5%Gypsum = 5.5%Limestone = 5.0%

(ii) For 5% limestone on a clinker basis, calculate a new clinker/limestone SO3, i.e.

0.05 * limestone SO3 + 0.95 * clinker SO3 = 0.836%

Then re-calculate as in 1. to find %gypsum = 5.49%.Therefore clinker = 100 5.49 * 0.95 = 89.8%

Therefore in the cement have

Clinker = 89.8%Gypsum = 5.49%Limestone = 4.72%


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LSF AND BOGUE COMPOSITION

LSF = (CaO 0.7 * SO3) / (2.80 * SiO2 + 1.18 * Al2O3 + 0.65 * Fe2O3)

Silica Ratio = SiO2 / (Al2O3 + Fe2O3)

Alumina Ratio = Al2O3 / Fe2O3

BOGUE Composition:

C3S = 4.07 CaO 7.6 SiO2 6.73 Al2O3 1.43 Fe2O3 - 2.85 SO3

C2S = 2.87 SiO2 0.754 C3S

C3A = 2.65 Al2O3 1.69 Fe2O3

C4AF = 3.04 Fe2O3

Note: For Bogue calculations (not potential) use total CaO free lime


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WATER SOLUBLE ALKALIS


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CEMENT SSA, RESIDUE AND ROSIN-RAMMLER PSD SLOPE

Note:

Above represent the Rosin-Rammler slope (n) of the psd.Based on neat OPC only, i.e. no non-clinker components present.Based on determination of psd by x-ray sedimentation.

Can provide a useful rapid guide to compare the sharpness of psds, e.g.

350m2/kg and 10% residue: Slope = 1.09380m2/kg and 9% residue: Slope = 1.06

Hence in the 2nd cement the reduction in residue is not compatible to the higher SSA. For thesame psd slope the residue would need to be 8%.


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CEMENT FINENESS


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VOLUME LOADING


Area of Segment = r2

Area of Triangle = 1/2 * r cos * 2 r sin = r2 cos sin

Area below level = r2 - r2 cos sin = r2 ( - cos sin )

Area of circle = r2

Volume occupied = r2 ( - cos sin ) / r2= ( - cos sin ) /

r cos = h - r = cos-1 {(h - r)/r}

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BALL MILL PARAMETERSMILL CRITICAL SPEED

Critical Speed when mv2/r = mgv2/r = g or v2 = rgv2 = D/2g or v = (D/2 * g)0.5

Also = (2 N)/60N = (60 ) / (2 )Also v = r = v/rNc = (60 v)/(2 r) Nc = (60 v) / ( D)Nc = {60 (D/2 * g)0.5 } / ( D)Nc = 42.3 / D0.5

Wherem = mass g = gravity D = mill diameterv = tangential velocity = angular velocityN = mill speed (rpm)Nc = mill critical speed (rpm)


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MILL POWER CALCULATION

1. The mill power (gross) can be measured from the kWh meter.

(i) Note the meter factor, usually in kWh/revolution(ii) Time 10 revolutions of the disk(iii) Calculate the revolutions per hour(iv) Calculate the mill power

Revs/hour * kWh/rev

EXAMPLE:

Meter factor = 3.210 kWh/revTime for 10 revolutions = 28.46 secsTherefore have = 1264.9 revolutions/hourMill Power = 1264.9 * 3.210 = 4060kW

Note: meter factor is also sometimes in revs/kWh or revs/MWh.

2. The mill net power can be estimated from the diameter, mill speed and the height above the charge (HAC) for each chamber.

Net Power (kW) = 0.2846 * D * A * W * N

Where D = effective chamber diameter (m)A = 1.073 V/100 (where V = the chamber volume loading%)W = the weight of grinding media in the chamber (tonnes)N = the mill speed (rpm)

Calculate for each chamber individually

EXAMPLE:

D = 4.45mV = 30.4%W = 100.8 tonnesN = 14.6rpm

Net power = 0.2846 * 4.45 * (1.073 30.4/100) *100.8 * 14.6 = 1433kW

3. The gross power/net ratio should be between 1.05 and 1.10.


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POWDER VOID FILLING

1. The powder (or material) void filling can be estimated during a mill internal inspection. Thefollowing data is required for each chamber:

Effective diameterHeight above charge (HAC)Height above powder (HAP)

Note: A realistic estimate can only be made for a crash stop situation.

2. Determine the volume loading from the diameter and HAC (see also TIS MS007).

3. Determine the volume % to the powder level using the diameter and HAP as above for thevolume loading (use HAP in place of HAC).

4. When the powder level is below the media:

Void Filling = powder vol% / media vol%

5. When the powder level is above the media:

Void Filling = (powder vol% - 0.55*media vol%) / (0.45*media vol%)

Note: 0.45 represents an average media voidage

EXAMPLE 1

D = 4.45HAC = 2.92HAP = 3.17

Volume loading (media) = 30.4%Volume loading to powder = 23.8%Therefore the void filling = 23.8/30.4 = 78%

EXAMPLE 2

D = 4.50HAC = 3.05HAP = 2.98

Volume loading (media) = 27.9%Volume loading to powder = 29.7%Therefore the void filling = (29.7 0.55*27.9) / (0.45* 27.9) = 114%


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MILL AIRFLOW

1. Mill design airflow rules:

2-3 volume changes per minute (volume of mill above media charge)0.4kg air per kg cement1 m/sec air velocity in mill (using mill diameter)

EXAMPLE 1

Mill of 1000kW, effective diameter = 2.42m, volume = 60m3, volume above media = 42m3, 22t/hr

2-3 volume changes: mill airflow = 5,000 7,500m3/hr0.4kg air per kg cement: mill airflow = 11,400m3/hr1 m/sec air velocity in mill: mill airflow = 16,500m3/hr

EXAMPLE 2

Mill of 4400kW, effective diameter = 4.45m, volume = 210m3, volume above media = 147m3, 115t/hr

2-3 volume changes: mill airflow = 17,700 26,500m3/hr0.4kg air per kg cement: mill airflow = 59,000m3/hr1 m/sec air velocity in mill: mill airflow = 56,000m3/hr

BEWARE of in-leaks, i.e. a significant proportion of air passes the filter exhaust but not necessarily through the mill.

2. Temperature Balance.

Filter Exhaust, F = Mill Air, M + In-leaks, LTherefore M = 55,000 * (347-288)/(391-288) = 31,500 m3/hr


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MILL EXHAUST DEW POINT DEW-POINT TEMPERATURE ESTIMATION

EXAMPLE:

Water Input: T/hr Moisture % Water T/hr

Clinker 118.5 0 0Gypsum 3.4 5 0.17Other (e.g. limestone) 3.9 4 0.16Water Injection 1.22 100 1.22Air(1) 25 1 0.25Gypsum Dehydration 0.61

Total 2.41

(1) Airflow assumed to be 0.2kg/kg cement, i.e. 25,000kg/hr.Assumed to be 70% relative humidity and 20oc, humidity = 0.01

Humidity of Mill exhaust air = 2.41/25 = 0.096From diagram the dew-point temperature = 53C


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CIRCULATING LOAD

EXAMPLE 1, Sieve residues:

a = 47%, f = 4%, r = 65%Therefore A/F = (4 - 65)/(47 65) = 339%

EXAMPLE 2, Blaine SSA balance:A = 150, F = 50, R = 100 and f = 350, r = 200, a = ?

a = (Ff + Rr)/A = (50*350 + 100*200)/150 = 250


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CEMENT GRINDABILITY

AVERAGE GRINDABILITY CURVE. EXAMPLE:

Mill condition A: 66.4 t/hr at 370m2/kgMill condition B: 69.7 t/hr at 355m2/kg

Comparison of mill performance, i.e. is the reduction in Blaine SSA compatible with the higher mill output?

Assume constant mill power, say 2200kW

Condition A: kWh/t = 2200/66.4 = 33.13From graph, the kWh/t at 370 = 45.0

Condition B: kWh/t = 2200/69.7 = 31.56From graph, the kWh/t at 355 = 41.7

Therefore using the grindability curve a reduction in SSA from 370 to 355 would be expected to provide a 45.0/41.7 increase in mill output.

Hence expect t/hr to increase from 66.4 to 71.7. Therefore the increase was less than expected.


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SEPARATOR PERFORMANCE

1. Determine the psds for the separator feed, fines and rejects streams (a, f, r) and present as cumulative finer.

2. Sum each psd. Determine A, F, R. i.e. A = F + R, A = 100, F = A(a - r)/(f - r)

3. Using the calculated A, F, R, re-calculate the feed psd,i.e. a = (Ff + Rr)/A at each size interval.

4. Use the measured r and calculated a to derive the Tromp Efficiency at each size interval, i.e. Rr/Aa * 100%. Plot against the mid interval size.

5. Equiprobable size (effective cut-size) is interpolated at Tromp Efficiency = 50%.

6. Imperfection, I, = (D75 D50) / (2 x D50)

7. The By-Pass, S, is the minimum Grade Efficiency.

Coarse Grade Efficiency Tromp Curve

Sum a = 668.1Sum f = 814.4Sum r = 517.2

A = 100F = 50.8R = 49.2

Circulating Load (A/F) = 197%Equiprobable Size = 35 micronsImperfection = 0.32By-Pass = 19%


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AIR SETTING TENDENCY

1. The air setting tendency of cement is strongly influenced by the presence of free lime and K2O

2. Higher levels of K2O with gypsum and moisture can form SYNGENITE, which can lead to the formation of lumps in silos

3. Higher levels of free lime tend to act as a desiccant and thus can mitigate the problems of air setting

4. Air set propensity can be assessed by exposing a sample of cement to storage in closed bags at 27C, 95% RH at 4psi in a humidity cabinet for 3 days. The % retained at 600 microns isthen determined. This then represents the AIR SET TENDENCY.


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Cement Chapter 12

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