Question

A mine void of dimension 100.0 m x 2.0 m x 1.2 m is to be filled in 3 hours by hydraulic stowing. The sand to slurry ratio is 0.4. If the hydraulic fill factor is 0.9, the hourly consumption of water for the operation, in m\(^3\), is ________ (rounded off to 2 decimal places).

Hint:

In hydraulic stowing, ensure the correct fill factor and slurry-to-sand ratio are applied to determine the volume of water required for the operation.

Answer 1

First, we calculate the total volume of the mine void ($V_{void}$): $$\mathbf{V_{void} = {length} \times {width} \times {height}}$$ $$\mathbf{V_{void} = 100.0 \, {m} \times 2.0 \, {m} \times 1.2 \, {m} = 240.0 \, {m}^3}$$ Next, we determine the volume of the sand required to fill the void ($V_{sand}$) using the hydraulic fill factor: $$\mathbf{V_{sand} = V_{void} \times {hydraulic fill factor}}$$ $$\mathbf{V_{sand} = 240.0 \, {m}^3 \times 0.9 = 216.0 \, {m}^3}$$ The sand to slurry ratio is given as 0.4. Let $V_{slurry}$ be the total volume of the slurry. We have: $$\mathbf{\frac{V_{sand}}{V_{slurry}} = 0.4}$$ $$\mathbf{V_{slurry} = \frac{V_{sand}}{0.4} = \frac{216.0 \, {m}^3}{0.4} = 540.0 \, {m}^3}$$ The volume of water in the slurry ($V_{water}$) is the difference between the volume of the slurry and the volume of the sand: $$\mathbf{V_{water} = V_{slurry} - V_{sand}}$$ $$\mathbf{V_{water} = 540.0 \, {m}^3 - 216.0 \, {m}^3 = 324.0 \, {m}^3}$$ This is the total volume of water required to fill the mine void in 3 hours. To find the hourly consumption of water, we divide the total volume of water by the time taken: $$\mathbf{{Hourly water consumption} = \frac{V_{water}}{{time}}}$$ $$\mathbf{{Hourly water consumption} = \frac{324.0 \, {m}^3}{3 \, {hours}} = 108.0 \, {m}^3/{hour}}$$ Rounding off to 2 decimal places, the hourly consumption of water for the operation is 108.00 m\(^3\). Final Answer: The hourly consumption of water is $\mathbf{108.00 \, {m}^3/{hour}}$.