Cement chemistry; reactions and adsorption of water, volume of hydration products and

porosity of cement paste

Rak-82.3131 Concrete technology 2Exercise 3

Recap from last week:

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

Reactions and adsorption of water

Characteristics of hydration of the cement compounds

Amount of Contribution to cement

Compounds Reaction rate heat liberated Strength Heat liberationC3S Moderate Moderate High C2S Slow Low Low initially, Low

high later

C4AF + CH2 Fast Very high Low Very high C3A + CH2 Moderate Moderate Low Moderate

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

Tricalcium aluminate:

C3A + 3CH2 + 25H → C3A 3C∙ 31∙ H (ettringite) 1g + 1,91g + 1,67 → 4,58g

C3A + CH2 + 10H → C3A C∙ 12∙ H (monosulphite) 1g + 0,63g + 0,67g → 2,30g

C3A + 27H → C2AH8 + C4AH19

1g + 0,90g → 0,66g + 1,24g

C3A + 6H → C3AH6 1g + 0,40g → 1,40g

Tetracalcium aluminoferrite:

C4AF + 2CH + 10H → C3AH6 + C3FH6

1g + 0,30g + 0,37g → 0,78g + 0,90g

Water requirement of the hydration reactions

The total water requirement of the hydration reactions is the sum of the water requirements of the cement minerals.

Minerals Products Water requirementR1 R2 g water/ g R1

Silicate 2C3S 6H C3S2H3 + 3CH 0.24

phase 2C2S 4H C3S2H3 + CH 0.21

Aluminate C3A 6H C3AH6 0.40

phase 3C3A 36H C2AH8 + C4AH19 0.90

C3A CH + 18H C4AH19 1.20

C3A 3CŜH2 + 25H C3A·3CŜ·H31 1.67

C3A CŜH2 + 10H C3A·CŜ·H12 0.67

Ferrite C4AF 2CH + 10H C3AH6 + C3FH6 0.37

phase C4AF 4CH + 34H C4AH19 + C4FH19 1.26

Other CaO H2O Ca(OH)2 0.32

components MgO H2O Mg(OH)2 0.45

CaSO4 2H2O CaSO4 2H2O 0.26

Reacting elements

w = w(SP) + w(AP) + w(FP) + w(OC)

in which,SP = silicate phaseAP = aluminate phaseFP = ferrite phaseOC = other components

Water requirement w:

Clinker minerals Chemically bound water0

10

20

30

40

50

60

70

80

90

100

CaO + MgOC4AFC3AC2SC3S

In Exercise 2 concrete´s compound composition was calculated as: {C3S, C2S, C3A, C4AF, S E} = {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

Problem 1

First we have to change the given = SO3 amount (4,9 %) to equivalent gypsum amount (CH2)In order to do this, molecular weights are needed (see Exercise 2)

= 80,06 g/molC = 56,08 g/molH = 18,00 g/mol

In which case CH2 =56,08 + 80,06 + 2*18,00 = 172,14 g/mol

Transformed into equivalent gypsum amount:CH2 =4,9 % * 172,14/80,06 = 10,54 %

Water requirement of the hydration reactions

Minerals Products Water requirementR1 R2 g water/ g R1

Silicate 2C3S 6H C3S2H3 + 3CH 0,24

phase 2C2S 4H C3S2H3 + CH 0,21

Aluminate C3A 6H C3AH6 0,40

phase 3C3A 36H C2AH8 + C4AH19 0,90

C3A CH + 18H C4AH19 1,20

C3A 3CŜH2 + 25H C3A·3CŜ·H31 1,67

C3A CŜH2 + 10H C3A·CŜ·H12 0,67

Ferrite C4AF 2CH + 10H C3AH6 + C3FH6 0,37

phase C4AF 4CH + 34H C4AH19 + C4FH19 1,26

Other CaO H2O Ca(OH)2 0,32

components MgO H2O Mg(OH)2 0,45

CaSO4 2H2O CaSO4 2H2O 0,26

Reacting elements

MONOSULPHATEFrom the reactions it can be seen that 0,63g gypsum binds 1g C3A and 0,67g water.

Tricalcium aluminate:

C3A + 3CH2 + 25H → C3A 3C∙ 31∙ H (ettringite) 1g + 1,91g + 1,67 → 4,58g

C3A + CH2 + 10H → C3A C∙ 12∙ H (monosulphite) 1g + 0,63g + 0,67g → 2,30g

C3A + 27H → C2AH8 + C4AH19

1g + 0,90g → 0,66g + 1,24g

C3A + 6H → C3AH6 1g + 0,40g → 1,40g

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 C H∙ ∙ 12) = 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(C H∙ 12) = 0,26 * 0,04468 = 0,01162 g/g

C4AF (tetracalcium aluminoferrite)w(C4AF) = 0,37 * 0,082 = 0,03034 g/g

In addition we need to take into account the chemically bound water 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/g

w(C2S) = 0,21 * 0,21 = 0,0441 g/g

How does the water requirement of the cement in exercise 1 change when {S E} = {1,9} % ?

Minerals Products Water requirementR1 R2 g water/ g R1

Silicate 2C3S 6H C3S2H3 + 3CH 0,24

phase 2C2S 4H C3S2H3 + CH 0,21

Aluminate C3A 6H C3AH6 0,40

phase 3C3A 36H C2AH8 + C4AH19 0,90

C3A CH + 18H C4AH19 1,20

C3A 3CŜH2 + 25H C3A·3CŜ·H31 1,67

C3A CŜH2 + 10H C3A·CŜ·H12 0,67

Ferrite C4AF 2CH + 10H C3AH6 + C3FH6 0,37

phase C4AF 4CH + 34H C4AH19 + C4FH19 1,26

Other CaO H2O Ca(OH)2 0,32

components MgO H2O Mg(OH)2 0,45

CaSO4 2H2O CaSO4 2H2O 0,26

Reacting elements

Again we first change the given = SO3 amount (1,9 %) to equivalent gypsum amount (CH2)

= 80,06 g/molC = 56,08 g/molH = 18,00 g/mol

And then calculate the water requirements of the hydration reactions

Exercise 2

When gypsum and C3A react, water is bound: w(C3A C H∙ ∙ 12) = 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?

1st, the amount of gypsum added to cement clinker is expressed as the mass of SO3 present; 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.

Volume of hydration products and porosity of cement paste

Properties of concrete, p.31

Calculating the volumes of hydration products

Total volume of the cement paste Vpaste = Vc + Vo + L

Unhydrated cement (Vcnhyd) =

When ρc = 3100 kg/m³

• Vcnhyd = 0,32*C(1- α)

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 α• Air L

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 = α * C→ Ch = α * C()

when ρC =3100kg/m³ ja ρV =1000kg/m³

Vg.s = 0,51 * α * C [dm³]

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 ρV =1000 kg/m³

Vcon = 0,0625 * α * C [dm³]

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 = α* C * ( )

Insert values → Vg.h = 0,2 * α * 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 / ρV

The volume of the capillary pores Vcap

= total water density Vo – chemically bound water VN – gel water Vg.h

= Vo – VN – Vg.h

= ( w/c * C – 0,25 *α*C – 0,2*α*C )

Vkap = C (w/c – 0,45 α)

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 α = 0,80. Calculate how much (l/m3) there is a) chemically bound water, b) gel water and d) capillary water.

Problem 3

Chemically bound water is 25 % of the amount of hydrated cement WN = 0,25* α * C = 0,25*0,8*375 = 75 kg/m

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 calculated at 0,25148 g/gIn which caseWN = 0,25148*α * C = 0,25148*0,8*375 = 75,44 kg/m3

3b

Vgh = 0,28*(Vgs+Vgh) = 0,28/0,72 Vgs

Vgs = Vch + VN – Vcon

= α*C/ρC + 0,25*α*C/ρV – 0,25*0,25*α*C/ρV

= αC(1/ρC+ 0,1875/ρV)→ Vgh = 0,28/0,72* α C(1/ρC+ 0,1875/ρV)

= 0,2αC = 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 used In 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

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

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

Weight of the batch = 835 + 150 + 57 = 1042 kgLoss of water = 0,028*1042 = 29,176 kg

Thus, from the added water over 50 % has evaporated.

In order to demonstrate the effect of this water loss we can calculate the maximum hydration degree (αMAX) which could be achieved with this remaining amount of water.

Total volume of the cement paste Vpaste

Vpaste = = = 105,4 l

Hydration reactions end when the amount of free water = 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)

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*α*C [dm³]VN = 0,25*α*C [dm³]

The maximum degree of hydration of the original mix design (without evaporation) can be calculated as:Vcapv = 0 = Vo – VN –Vgh

= 57 – 0,25*α*C - 0,2*α*C = 57 – 0,45*α*C C=150 = 57 – 67,5* α

→ αMAX = 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*α*150 = 27,8 – 67,5α

→ αMAX = 0,41

5

The composition of a concrete is such that it can theoretically achieve full hydration. Calculate the w/c ratio of such concrete.

You can use a simplified calculation tehnique: amount of cement 1kg

degree of hydration α 1capillary porosity 0

CALCULATE the amounts of chemically bound and gel water!

Chemically bound water:WN = 0,25*α*C = 0,25*1*1 = 250g

Gel water: Vgh/(Vgh+Vgs) = 0,28

Vgh = Vgs

Vgs = Vch + VN – Vcon

= = = 0,51 l

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