Question

The composition of an air-entrained concrete is given below:

$\text{Water}\to 184kg/m^3$

$\text{Ordinary Portland Cement (OPC)}\to 368kg/m^3$

$\text{Sand}\to 606kg/m^3$

$\text{Coarse aggregate}\to 1155kg/m^3$

Assume the specific gravity of OPC, sand and coarse aggregate to be 3.14, 2.67 and 2.74 respectively. The air content is_____$litres/m^3$

• A. 50.32

Answer 1

The correct option is A 50.32
$\frac{M_c}{{\rho }_c}+\frac{M_{fa}}{{\rho }_{fa}}+\frac{M_{ca}}{{\rho }_{ca}}+V_w+V_{air}=1$

$\frac{368}{3.14\times 1000}+\frac{606}{2.67\times 1000}+\frac{1155}{2.74\times 1000}+\frac{184}{1000}+V_{air}=1.0$


$0.11719+0.22696+0.42153+0.184+V_{air}=1.0$

$V_a=0.05032$


$V_a=0.05032\times 1000=50.32l/m^3$