Drilling in a stone drift 3 mm × 4 m in cross-section in a coal mine produces 0.5 gm of dust of 5 um size every minute. If the respirable dust has a free silica content of 8% and t... Drilling in a stone drift 3 mm × 4 m in cross-section in a coal mine produces 0.5 gm of dust of 5 um size every minute. If the respirable dust has a free silica content of 8% and the intake air has a respirable dust load of 0.8 mg/m3, what is the airflow rate required to dilute the dust to the TLV stipulated by CMR- 2017?
Understand the Problem
The question describes a scenario of dust production during drilling in a coal mine. We need to determine the airflow rate required to dilute the dust concentration to the Threshold Limit Value (TLV) stipulated by CMR-2017. This involves considering the dust generation rate, the silica content, the existing dust load in the intake air, and the TLV for respirable silica dust to calculate the required airflow.
Answer
$20 \times 10^6 \text{ m}^3/\text{min}$
Answer for screen readers
$20 \times 10^6 \text{ m}^3/\text{min}$
Steps to Solve
- Calculate the silica dust generation rate
First, find the amount of silica generated per minute. Since the dust generation rate is 10 kg/min and the dust contains 4% silica, the silica generation rate is:
$10 \text{ kg/min} \times 0.04 = 0.4 \text{ kg/min}$
Convert this to mg/min:
$0.4 \text{ kg/min} \times 10^6 \text{ mg/kg} = 4 \times 10^5 \text{ mg/min}$
- Calculate the silica dust load in intake air
The intake air has a dust concentration of 2 mg/m$^3$, and the dust contains 4% silica, the silica concentration in the intake air is:
$2 \text{ mg/m}^3 \times 0.04 = 0.08 \text{ mg/m}^3$
- Define variables and the target equation
Let $Q$ be the required airflow rate in m$^3$/min. Then, after mixing, the concentration of silica dust should be equal to the TLV. The equation is:
$\frac{\text{Silica generation rate} + (\text{Intake airflow rate} \times \text{Silica dust load in intake air})}{\text{Total airflow rate}} = \text{TLV}$
Plugging in the values:
$\frac{4 \times 10^5 \text{ mg/min} + (Q \text{ m}^3/\text{min} \times 0.08 \text{ mg/m}^3)}{Q \text{ m}^3/\text{min}} = 0.1 \text{ mg/m}^3$
- Solve for $Q$
Now, solve the equation for $Q$:
$4 \times 10^5 + 0.08Q = 0.1Q$
$4 \times 10^5 = 0.1Q - 0.08Q$
$4 \times 10^5 = 0.02Q$
$Q = \frac{4 \times 10^5}{0.02} = 20 \times 10^6 \text{ m}^3/\text{min}$
Therefore, the required airflow rate is $20 \times 10^6 \text{ m}^3/\text{min}$.
$20 \times 10^6 \text{ m}^3/\text{min}$
More Information
The required airflow rate to dilute the silica dust concentration to the TLV is extremely high, at 20 million cubic meters per minute. This highlights the importance of efficient dust control measures at the source to minimize dust generation rather than relying solely on ventilation for dilution. This also indicates that there is a risk in the workplace due to the high levels of dust, and more practical solutions such as dust suppression are needed.
Tips
A common mistake is forgetting to convert units properly. For example, failing to convert kg to mg can lead to errors. Another mistake is not considering the existing dust load in the intake air, which contributes to the overall dust concentration. Also, a frequent error involves incorrectly setting up or solving the concentration equation, particularly with units. It is necessary to keep track of the units for each step of the solution.
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