Cement chemistry; reactions and
adsorption of water, volume of
hydration products and porosity of
cement paste
Rak-82.3131 Concrete technology 2
Exercise 3
Recap from last week:
C3 S
Tricalcium silicate
(alite)
Hydrates & hardens rapidly
Responsible for initial set and early strength
C2 S
Dicalcium silicate
(belite)
Hydrates & hardens slowly
Contributes to later age strength (beyond 7 days)
C3A
Tricalcium aluminate
Liberates a large amount of heat during first few days
Contributes slightly to early strength development
Cements with low %-ages are more resistant to
sulfates
C4AF
Tetracalcium
aluminoferrite
(ferrite)
Reduces clinkering temperature
Hydrates rapidly but contributes little to strength
Colour of hydrated cement (gray) due to ferrite
hydrates
Reactions and adsorption of
water
Characteristics of hydration of the cement
compounds
Compounds
Reaction rate
Amount of
heat liberated
Contribution to cement
Strength
Heat liberation
C3S
C2S
Moderate
Slow
Moderate
Low
C4AF + CSH2
C3A + CSH2
Fast
Very high
Low
Very high
Moderate
Moderate
Low
Moderate
Low initially,
high later
High
Low
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 + 3C𝐒H2 + 25H → C3A∙3C𝐒 ∙ 31H (ettringite)
1g + 1,91g
+ 1,67 →
4,58g
C3A + C𝑺H2 + 10H → C3A∙C𝑺 ∙ 12H (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
Minerals
Silicate
phase
Aluminate
phase
Ferrite
phase
Other
components
Reacting elements
R1
R2
2C3S
6H
2C2S
4H
C3A
6H
3C3A
36H
C3A
CH + 18H
C3A
3CŜH2 + 25H
C3A
CŜH2 + 10H
C4AF
2CH + 10H
C4AF
4CH + 34H
H2O
CaO
H2O
MgO
CaSO4
2H2O
Products
C3S2H3 + 3CH
C3S2H3 + CH
C3AH6
C2AH8 + C4AH19
C4AH19
C3A·3CŜ·H31
C3A·CŜ·H12
C3AH6 + C3FH6
C4AH19 + C4FH19
Ca(OH)2
Mg(OH)2
CaSO4 2H2O
Water requirement
g water/ g R1
0.24
0.21
0.40
0.90
1.20
1.67
0.67
0.37
1.26
0.32
0.45
0.26
The total water requirement of the hydration
reactions is the sum of the water requirements
of the cement minerals.
Water requirement w:
w = w(SP) + w(AP) + w(FP) + w(OC)
in which,
SP = silicate phase
AP = aluminate phase
FP = ferrite phase
OC = other components
100
90
80
70
CaO + MgO
60
C4AF
50
C3A
C2S
40
C3S
30
20
10
0
Clinker minerals
Chemically bound water
1
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 %).
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 (see
Exercise 2)
S = 80,06 g/mol
C = 56,08 g/mol
H = 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 %
Water requirement of the hydration
reactions
Minerals
Silicate
phase
Aluminate
phase
Ferrite
phase
Other
components
Reacting elements
R1
R2
2C3S
6H
2C2S
4H
C3A
6H
3C3A
36H
C3A
CH + 18H
C3A
3CŜH2 + 25H
C3A
CŜH2 + 10H
C4AF
2CH + 10H
C4AF
4CH + 34H
H2O
CaO
H2O
MgO
CaSO4
2H2O
Products
C3S2H3 + 3CH
C3S2H3 + CH
C3AH6
C2AH8 + C4AH19
C4AH19
C3A·3CŜ·H31
C3A·CŜ·H12
C3AH6 + C3FH6
C4AH19 + C4FH19
Ca(OH)2
Mg(OH)2
CaSO4 2H2O
Water requirement
g water/ g R1
0,24
0,21
0,40
0,90
1,20
1,67
0,67
0,37
1,26
0,32
0,45
0,26
MONOSULPHATE
From the reactions it can be seen that 0,63g gypsum binds 1g C3A and
0,67g water.
Tricalcium aluminate:
C3A + 3C𝐒H2 + 25H → C3A∙3C𝐒 ∙ 31H (ettringite)
1g + 1,91g
+ 1,67 →
4,58g
C3A + C𝑺H2 + 10H → C3A∙C𝑺 ∙ 12H (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
MONOSULPHATE
From 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 = 5,859
And 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
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/g
w(MgO) = 0,45 * 0,0108 = 0,00486 g/g
CALCIUMCILICATES
w(C3S) = 0,24 * 0,523 = 0,12552 g/g
w(C2S) = 0,21 * 0,21 = 0,0441 g/g
in which case wtot =
0,12552 + 0,0441 + 0,06231 + 0,01162+ 0,03034 + 0,00307 +
0,00486 = 0,25146 g/g
How does the water requirement of the cement in exercise 1
change when {S ̅} = {1,9} % ?
Exercise 2
Again we first change the given S = SO3 amount (1,9 %) to equivalent gypsum
amount (CSH2)
S = 80,06 g/mol
C = 56,08 g/mol
H = 18,00 g/mol
And then calculate the water requirements of the hydration reactions
Minerals
Silicate
phase
Aluminate
phase
Ferrite
phase
Other
components
Reacting elements
R1
R2
2C3S
6H
2C2S
4H
C3A
6H
3C3A
36H
C3A
CH + 18H
C3A
3CŜH2 + 25H
C3A
CŜH2 + 10H
C4AF
2CH + 10H
C4AF
4CH + 34H
H2O
CaO
H2O
MgO
CaSO4
2H2O
Products
C3S2H3 + 3CH
C3S2H3 + CH
C3AH6
C2AH8 + C4AH19
C4AH19
C3A·3CŜ·H31
C3A·CŜ·H12
C3AH6 + C3FH6
C4AH19 + C4FH19
Ca(OH)2
Mg(OH)2
CaSO4 2H2O
Water requirement
g water/ g R1
0,24
0,21
0,40
0,90
1,20
1,67
0,67
0,37
1,26
0,32
0,45
0,26
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 bound
4,09 % * 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 C3A
w(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 by
0,08197 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
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
Total volume of the cement paste Vpaste = Vc + Vo + L
Unhydrated cement (Vcnhyd) =
When ρc = 3100 kg/m³
• Vcnhyd = 0,32*C(1- α)
C−Ch
ρc
=
C−α∙C
ρc
=
C(1−α)
ρc
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
= Cρhc + 0,75 WρvN
h )
= Cρhc + 0,75( 0,25∗C
ρv
because Ch = α * C
→
Ch = α *
1
C(
ρC
+
0,19
)
ρV
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
=
=
=
WN
0,25 *
ρV
0,25×Ch
0,25 * (
)
ρV
α×C
0,0625 * (
)
ρV
(Chemically bound water is 25 % of the amount of hydrated cement )
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
(V is the solid part of the cement gel)
gs
→
Vg.h =
Vg.h =
0,28
0,72
0,28
0,72
Vg.s
α* C * (
1
𝜌𝐶
+
0,19
𝜌𝑉
)
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 ( ) are
weight 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
=
1
𝜌𝑉
( w/c * C – 0,25 *α*C – 0,2*α*C )
Vkap = C (w/c – 0,45 α)
Problem 3
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.
Chemically bound water is 25 % of the amount of
hydrated cement
WN = 0,25* α * 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 calculated
at 0,25148 g/g
In which case
WN = 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 =
0,247
Because WN was calculated at 75,44 kg/m3 in part a, we get:
WN = Wgh = 3* 0,247*75,44 = 55,93 kg/m³
3c
Excess water from the initial water amount (Wo),
forms the capillary water
Original 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
Problem 4
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?
Weight of the batch
= 835 + 150 + 57 = 1042 kg
Loss 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 =
C
ρc
+
WO
ρV
=
150
57
+
3,10
1
= 105,4 l
(In this type of
examination, the
amount of air can be
excluded)
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
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
VN = 0,25*α*C
[dm³]
[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 = 29,2
Vkapv = 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
1
capillary 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
0,28
0,72 Vgs
Vgh =
Vgs = Vch + VN – Vcon
=
αC
ρC
= 0,51 l
+
WN
ρV
−
WN
0,25
ρV
=
1∗1
3,1
+ 0,75 × 0,25/1
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