Mining Methods and Machinery · Mining Methods · 2 marks · Numerical answer
A 20×50×30 m mine void is filled with cemented paste backfill. Tailings:cement:water mass ratio is 1.0:0.1:0.2. Tailings and cement specific gravities are 2.8 and 2.4. Only 20% of original water volume remains in the final backfill. Find required cement in tonnes, to the nearest integer.
Let tailings mass be M tonnes. Cement mass is 0.1M and original water mass is 0.2M; retained water volume is 0.20×0.2M=0.04M m³. Total final volume is M/2.8+0.1M/2.4+0.04M=30000 m³. Hence cement = 0.1×30000/(1/2.8+0.1/2.4+0.04)=6836.679 tonnes, rounding to 6837 tonnes.
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