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Capillary porosity with 75% hydration

Short answer

Water volume 1.575 cm³, paste volume 2.575 cm³. With 75% hydration, products 1.5 cm³, unhydrated cement 0.25 cm³; capillary pores 0.825 cm³, porosity 32.0%.

Assuming that 1 cm3^3 of cement produces 2 cm3^3 of hydrated products under the standard curing conditions (ASTM standard), calculate the percentage of capillary porosity in the hydrated cement paste after 28 days. Take w/c ratio as 0.5 and assume 75% hydration in 28 days.

Assumptions: specific gravity of cement 3.15; 1 cm3^3 of cement gives 2 cm3^3 of hydrated products (gel, with gel pores); capillary pores are the space not occupied by hydrated products or unhydrated cement. Take 1 cm3^3 of cement as the basis. Degree of hydration α=0.75\alpha = 0.75.

Volume of water=0.5×3.15=1.575 cm3Total paste volume=1+1.575=2.575 cm3Hydrated products=2×0.75=1.50 cm3Unhydrated cement=1−0.75=0.25 cm3Capillary pores=2.575−1.50−0.25=0.825 cm3Capillary porosity=0.8252.575×100=32.0%\begin{aligned} \text{Volume of water} &= 0.5\times3.15 = 1.575\ \text{cm}^3 \\ \text{Total paste volume} &= 1 + 1.575 = 2.575\ \text{cm}^3 \\ \text{Hydrated products} &= 2\times0.75 = 1.50\ \text{cm}^3 \\ \text{Unhydrated cement} &= 1 - 0.75 = 0.25\ \text{cm}^3 \\ \text{Capillary pores} &= 2.575 - 1.50 - 0.25 = 0.825\ \text{cm}^3 \\ \text{Capillary porosity} &= \frac{0.825}{2.575}\times100 = 32.0\% \end{aligned}

Answer: capillary porosity = 32.0% of the paste volume.

✓ Answer checked by Claude Sonnet 5.5

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