cement chemistry; reactions and adsorption of water ... · cement chemistry; reactions and...
TRANSCRIPT
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Cement chemistry; reactions and adsorption of water, volume of
hydration products and porosity of cement paste
Rak-82.3131 Concrete technology 2Exercise 3
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Recap from last week:
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C3S Tricalcium silicate Hydrates & hardens rapidly(alite) Responsible for initial set and early strength
C2S Dicalcium silicate Hydrates & hardens slowly(belite) Contributes to later age strength (beyond 7 days)
C3A Tricalcium aluminate Liberates a large amount of heat during first few daysContributes slightly to early strength developmentCements with low %-ages are more resistant tosulfates
C4AF Tetracalcium Reduces clinkering temperaturealuminoferrite Hydrates rapidly but contributes little to strength(ferrite) Colour of hydrated cement (gray) due to ferrite
hydrates
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Reactions and adsorption of water
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Characteristics of hydration of the cement compounds
Amount of Contribution to cementCompounds Reaction rate heat liberated Strength Heat liberationC3S Moderate Moderate HighC2S Slow Low Low initially, Low
high later
C4AF + CSH2 Fast Very high Low Very highC3A + CSH2 Moderate Moderate Low Moderate
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Hydration
Tricalcium silicate (alite):2C3S + 6H C3S2H3 + 3CH
1g + 0,24g 0,75g + 0,49g
Dicalcium silicate (belite):2C2S + 4H C3S2H3 + CH
1g + 0,21g 0,99g + 0,22g
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Tricalcium aluminate:
C3A + 3C H2 + 25H C3A·3C · 31H (ettringite)1g + 1,91g + 1,67 4,58g
C3A + C H2 + 10H C3A·C · 12H (monosulphite)1g + 0,63g + 0,67g 2,30g
C3A + 27H C2AH8 + C4AH19
1g + 0,90g 0,66g + 1,24g
C3A + 6H C3AH61g + 0,40g 1,40g
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Tetracalcium aluminoferrite:
C4AF + 2CH + 10H C3AH6 + C3FH61g + 0,30g + 0,37g 0,78g + 0,90g
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Water requirement of the hydration reactions
The total water requirement of the hydration reactions is the sum of the water requirementsof the cement minerals.
Minerals Products Water requirementR1 R2 g water/ g R1
Silicate 2C3S 6H C3S2H3 + 3CH 0.24phase 2C2S 4H C3S2H3 + CH 0.21Aluminate C3A 6H C3AH6 0.40phase 3C3A 36H C2AH8 + C4AH19 0.90
C3A CH + 18H C4AH19 1.20C3A 3C H2 + 25H C3A·3C ·H31 1.67C3A H2 + 10H C3A·C ·H12 0.67
Ferrite C4AF 2CH + 10H C3AH6 + C3FH6 0.37phase C4AF 4CH + 34H C4AH19 + C4FH19 1.26Other CaO H2O Ca(OH)2 0.32components MgO H2O Mg(OH)2 0.45
CaSO4 2H2O CaSO4 2H2O 0.26
Reacting elements
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w = w(SP) + w(AP) + w(FP) + w(OC)
in which,SP = silicate phaseAP = aluminate phaseFP = ferrite phaseOC = other components
Water requirement w:
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Clinker minerals Chemically bound water
CaO + MgO
C4AF
C3A
C2S
C3S
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In Exercise 2 concrete´s compound composition was calculated as: {C3S, C2S, C3A, C4AF, S } = {52.3, 21.0, 9.3, 8.2, 4.9} %. Calculate the water requirement of the cement (for complete hydration). Take into account the amounts of CaOfree (0.96 %) and MgO (1.08 %).
1
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Problem 1
First we have to change the given S = SO3 amount (4,9 %) to equivalent gypsum amount (CSH2)In order to do this, molecular weights are needed (seeExercise 2)
S = 80,06 g/molC = 56,08 g/molH = 18,00 g/mol
In which case CSH2 =56,08 + 80,06 + 2*18,00 = 172,14 g/mol
Transformed into equivalent gypsum amount:CSH2 =4,9 % * 172,14/80,06 = 10,54 %
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Water requirement of the hydration reactions
Minerals Products Water requirementR1 R2 g water/ g R1
Silicate 2C3S 6H C3S2H3 + 3CH 0,24phase 2C2S 4H C3S2H3 + CH 0,21Aluminate C3A 6H C3AH6 0,40phase 3C3A 36H C2AH8 + C4AH19 0,90
C3A CH + 18H C4AH19 1,20C3A 3C H2 + 25H C3A·3C ·H31 1,67C3A H2 + 10H C3A·C ·H12 0,67
Ferrite C4AF 2CH + 10H C3AH6 + C3FH6 0,37phase C4AF 4CH + 34H C4AH19 + C4FH19 1,26Other CaO H2O Ca(OH)2 0,32components MgO H2O Mg(OH)2 0,45
CaSO4 2H2O CaSO4 2H2O 0,26
Reacting elements
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MONOSULPHATEFrom the reactions it can be seen that 0,63g gypsum binds 1g C3A and 0,67g water.
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Tricalcium aluminate:
C3A + 3C H2 + 25H C3A·3C · 31H (ettringite)1g + 1,91g + 1,67 4,58g
C3A + C H2 + 10H C3A·C · 12H (monosulphite)1g + 0,63g + 0,67g 2,30g
C3A + 27H C2AH8 + C4AH19
1g + 0,90g 0,66g + 1,24g
C3A + 6H C3AH61g + 0,40g 1,40g
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MONOSULPHATEFrom the reactions it can be seen that 0,63g gypsum binds 1g C3A and 0,67g water.
Thus, gypsum is bound to the monosulphate:9,3 * 0,63 = 5,859And when gypsum and C3A react water is bound:w(C3A·CS·H12) = 0,05859 * 0,67 /0,63 = 0,06231 g/g
The amount of gypsum not bound to the monosulphate is: 10,54 – 5,859 = 4,681 %
Water requirement of the excess gypsum:w(CS·H12) = 0,26 * 0,04468 = 0,01162 g/g
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C4AF (tetracalcium aluminoferrite)w(C4AF) = 0,37 * 0,082 = 0,03034 g/g
In addition we need to take into account the chemically boundwater by MgO and CaOfree and thus we get:
w(CaOfree) = 0,32 * 0,0096 = 0,00307 g/gw(MgO) = 0,45 * 0,0108 = 0,00486 g/g
in which case wtot = 0,12552 + 0,0441 + 0,06231 + 0,01162+ 0,03034 + 0,00307 + 0,00486 = 0,25146 g/g
CALCIUMCILICATESw(C3S) = 0,24 * 0,523 = 0,12552 g/gw(C2S) = 0,21 * 0,21 = 0,0441 g/g
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How does the water requirement of the cement in exercise 1 change when {S } = {1,9} % ?
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Minerals Products Water requirementR1 R2 g water/ g R1
Silicate 2C3S 6H C3S2H3 + 3CH 0,24phase 2C2S 4H C3S2H3 + CH 0,21Aluminate C3A 6H C3AH6 0,40phase 3C3A 36H C2AH8 + C4AH19 0,90
C3A CH + 18H C4AH19 1,20C3A 3C H2 + 25H C3A·3C ·H31 1,67C3A H2 + 10H C3A·C ·H12 0,67
Ferrite C4AF 2CH + 10H C3AH6 + C3FH6 0,37phase C4AF 4CH + 34H C4AH19 + C4FH19 1,26Other CaO H2O Ca(OH)2 0,32components MgO H2O Mg(OH)2 0,45
CaSO4 2H2O CaSO4 2H2O 0,26
Reacting elements
Again we first change the given S = SO3 amount (1,9 %) to equivalent gypsum amount (CSH2)
S = 80,06 g/molC = 56,08 g/molH = 18,00 g/mol
And then calculate the water requirements of the hydration reactions
Exercise 2
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When gypsum and C3A react, water is bound: w(C3A·CS·H12) = 0,0409 * 0,67 /0,63 = 0,0435 g/g
And at the same time C3A is bound4,09 % * 1 / 0,63 = 6,49 %
The original C3A amount was 9,3 % thus the amount of C3A not bound to the monosulphate is 9,3 – 6,49 = 2,81 %
Water requirement of the C3Aw(C3A) = 0,40 * 0,0281 = 0,1124 g/g
wtot = w(SP) + w(AP) + w(FP) + w(OC)
0,12552 + 0,0441 +0,0435 +0,1124 + 0,03034 + 0,00307 + 0,00486= 0,33345 g/g
Water requirement is increased by0,08197 g/g
If the water requirement is increased, why excess gypsum is not added to the cement?
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1st, the amount of gypsum added to cement clinker is expressed as the mass of SO3present; this is limited by European Standards.
2nd, an excess of gypsum leads to an expansion and consequent disruption of the set of cement paste.
Optimum gypsum content leads to a desirable rate of early reaction and prevents local high concentration of products of hydration. In consequence the size of pores in hydrated cement paste is reduced and strength is increased.
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Volume of hydration products and porosity of cement paste
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Properties of concrete, p.31
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Calculating the volumes of hydration products
Total volume of the cement paste Vpaste = Vc + Vo + L
Unhydrated cement (Vcnhyd) = = = ( )
When c = 3100 kg/m³• Vcnhyd = 0,32*C(1- )
INITIAL DATA & ABBREVIATIONS (=lyhenteet)• The quantity and volume of cement C and Vc• The quantity and volume of water Wo and Vo• Water/cement ratio w/c = Wo/C• Degree of hydration • Air L
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Solid products of hydration Vg.s
= hydrated cement Vch + chemically bound water VN –contraction pores (supistumishuokoset) Vcon
= Vch + VN – Vcon Vcon = 0,25VN= Vch + VN – 0,25VN
= Vch + 0,75VN
= + 0,75
= + 0,75( , )because Ch = * C
Ch = * C( + , )
when C =3100kg/m³ ja V =1000kg/m³
Vg.s = 0,51 * * C [dm³]
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Volume of contraction pores Vcon
= Contraction pores are 25 % of the original volume of chemically bound water = 0,25* VN
= 0,25 *
= 0,25 * ( , × ) (Chemically bound water is 25 % of the amount of hydrated cement )
= 0,0625 * ( × )
when V =1000 kg/m³
Vcon = 0,0625 * * C [dm³]
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The volume of gel pores Vgh are 28 % of the total volume of the sement gel
Vg.h / (Vg.h + Vg.s) = 0,28 (Vgs is the solid part of the cement gel)
Vg.h = ,,
Vg.s
Vg.h = ,,
* C * ( + , )
Insert values Vg.h = 0,2 * * C [dm³]
A MORE PRECISE WAY:• Wg.h = 3*k*WN ,in which k 0,23(C3S) + 0,32(C2S) +
0,317(C3A) + 0,368(C4AF) , in which ( ) areweight per cent
• Vg.h = Wg.h / V
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The volume of the capillary pores Vcap
= total water density Vo – chemicallybound water VN – gel water Vg.h
= Vo – VN – Vg.h
= ( w/c * C – 0,25 * *C – 0,2* *C )
Vkap = C (w/c – 0,45 )
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The cement content of a concrete is 375 kg/m3 and the w/c ratio is 0,46. At a specific moment the degree of hydration = 0,80. Calculate how much (l/m3) there is a) chemically bound water, b) gel water and d) capillary water.
Problem 3
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Chemically bound water is 25 % of the amount of hydrated cement WN = 0,25* * C
= 0,25*0,8*375 = 75 kg/m³
A MORE SPECIFIC WAY:For example if the water demand is known (as calculated in previous exercises). Water demand for cement in exercise 1 was calculatedat 0,25148 g/gIn which caseWN = 0,25148* * C = 0,25148*0,8*375 = 75,44 kg/m3
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3bVgh = 0,28*(Vgs+Vgh) = 0,28/0,72 VgsVgs = Vch + VN – Vcon
= *C/ C + 0,25* *C/ V – 0,25*0,25* *C/ V
= C(1/ C+ 0,1875/ V) Vgh = 0,28/0,72* C(1/ C+ 0,1875/ V)
= 0,2 C = 0,2*0,8*375= 60 l/m³
A MORE SPECIFIC WAY :If the clinker mineral composition is known the formula Wg.h = 3*k*WN can be usedIn which k = 0,23*0,523+0,32*0,210+0,317*0,093+0,368*0,082 = 0,247Because WN was calculated at 75,44 kg/m3 in part a, we get: WN = Wgh = 3* 0,247*75,44 = 55,93 kg/m³
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3cExcess water from the initial water amount (Wo), forms the capillary waterOriginal water amount Wo = c*w/c
= 375*0,46 = 172,5 kg/m³
Vcap = Vo – VN – Vg.h
= 172,5 – 75 – 60 = 37,5 l/m³
A MORE PRECISE WAY Vcap = 172,5 - 75,44 - 55,93 = 41,13 l/m³
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Unreinforced concrete ducts were manufactured at a factory with a recipe:
• Aggregate 835 kg/m³• Cement 150 kg/m³• Water 57 kg/m³
A series of ducts were weighed right after manufacture and again 24 hours after. An average of 2,8 % weight loss caused by concrete drying was measured. What meaning does this water loss have?
Problem 4
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Weight of the batch= 835 + 150 + 57 = 1042 kgLoss of water = 0,028*1042 = 29,176 kg
Thus, from the added water over 50 % hasevaporated.
In order to demonstrate the effect of this water loss we can calculate the maximum hydration degree
MAX) which could be achieved with this remaining amount of water.
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Total volume of the cement paste Vpaste
Vpaste = + = ,
+ = 105,4 l
Hydration reactions end when the amount of freewater = capillary water runs out!!
The volume of capillary water:Vcapv = initial water amount Vo –evoporated water
V –chemically bound water VN – gel water Vgh
(In this type of examination, the amount of air can be excluded)
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The volume of the gel pores (gel water) are 28 % of the total volume of the sement gel
Vgh/ (Vgh + Vgs) = 0,28 Vgh = 0,28/0,72 * Vgs
Vgh = 0,2* *C [dm³]VN = 0,25* *C [dm³]
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The maximum degree of hydration of the original mix design (without evaporation) can be calculated as:Vcapv = 0 = Vo – VN –Vgh
= 57 – 0,25* *C - 0,2* *C= 57 – 0,45* *C C=150= 57 – 67,5*
MAX = 0,84
And the maximum degree of hydration for the evaporated batch: V = 29,2Vkapv = 0 = Vo V – VN – Vgh
= 57 – 29,2 – 0,45* *150= 27,8 – 67,5
MAX = 0,41
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5
The composition of a concrete is such that it can theoretically achieve full hydration. Calculate the w/c ratio of such concrete.
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You can use a simplified calculation tehnique:amount of cement 1kgdegree of hydration 1capillary porosity 0
CALCULATE the amounts of chemically bound and gel water!
Chemically bound water:WN = 0,25* *C = 0,25*1*1 = 250g
Gel water: Vgh/(Vgh+Vgs) = 0,28 Vgh = ,
, Vgs
Vgs = Vch + VN – Vcon
= + 0,25 = ,
+ 0,75 × 0,25/1= 0,51 l
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Vgh = 0,28/0,72*0,51 l = 0,198 l
TOTAL WATER AMOUNT IS THUS:250g + 198 g = 448g
AND THE WATER CEMENT RATIO:w/c = 0,448