If the moisture content of sludge is reduced sludge is reduced from 98% to 96%, the volume of sludge will have decreased by ______.
50%
This question asks how the volume of sludge changes when its moisture content decreases. Sludge is a mixture of solids and water. When water is removed, the amount of solids remains constant, while the total volume of the sludge decreases because the volume of water decreases.
To solve this, we can consider the solid content of the sludge, which stays the same regardless of the moisture content. Let's assume we have a certain amount of solids. The total volume of the sludge is made up of the volume of solids and the volume of water.
Let:
Moisture content is the percentage of water in the sludge by weight or volume (often assumed volume for simplicity in these types of problems if density isn't specified). The solid content is $100\% - \text{moisture content}$.
At 98% moisture, the solid content is $100\% - 98\% = 2\%$.
This means the volume of solids ($S$) is 2% of the initial total volume ($V_1$).
So, $S = 0.02 \times V_1$.
We can express the initial volume $V_1$ in terms of the constant solid volume $S$: $V_1 = \frac{S}{0.02}$.
At 96% moisture, the solid content is $100\% - 96\% = 4\%$.
This means the volume of solids ($S$) is 4% of the final total volume ($V_2$).
So, $S = 0.04 \times V_2$.
We can express the final volume $V_2$ in terms of the constant solid volume $S$: $V_2 = \frac{S}{0.04}$.
The decrease in volume is the difference between the initial volume and the final volume ($V_1 - V_2$). The percentage decrease is calculated with respect to the initial volume.
Percentage Decrease $= \frac{V_1 - V_2}{V_1} \times 100\%$
Substitute the expressions for $V_1$ and $V_2$ in terms of $S$:
Percentage Decrease $= \frac{\frac{S}{0.02} - \frac{S}{0.04}}{\frac{S}{0.02}} \times 100\%$
We can factor out $S$ from the numerator and denominator:
Percentage Decrease $= \frac{S \left(\frac{1}{0.02} - \frac{1}{0.04}\right)}{S \left(\frac{1}{0.02}\right)} \times 100\%$
The $S$ terms cancel out:
Percentage Decrease $= \frac{\frac{1}{0.02} - \frac{1}{0.04}}{\frac{1}{0.02}} \times 100\%$
Calculate the values $\frac{1}{0.02}$ and $\frac{1}{0.04}$:
Substitute these values back into the formula:
Percentage Decrease $= \frac{50 - 25}{50} \times 100\%$
Percentage Decrease $= \frac{25}{50} \times 100\%$
Percentage Decrease $= 0.5 \times 100\%$
Percentage Decrease $= 50\%$
Therefore, when the moisture content of the sludge is reduced from 98% to 96%, the volume of the sludge decreases by 50%.
| Parameter | Initial (98% Moisture) | Final (96% Moisture) |
|---|---|---|
| Moisture Content | 98% | 96% |
| Solid Content | $100\% - 98\% = 2\%$ | $100\% - 96\% = 4\%$ |
| Volume of Solids ($S$) | $0.02 \times V_1$ | $0.04 \times V_2$ |
| Total Sludge Volume | $V_1 = \frac{S}{0.02} = 50S$ | $V_2 = \frac{S}{0.04} = 25S$ |
| Volume Decrease | $V_1 - V_2 = 50S - 25S = 25S$ | |
| Percentage Decrease | $\frac{V_1 - V_2}{V_1} \times 100\% = \frac{25S}{50S} \times 100\% = 50\%$ | |
Here is a quick summary of the key concepts used in this sludge volume calculation:
Reducing the moisture content of sludge is a process called dewatering. It's a crucial step in wastewater treatment and sludge management for several reasons:
Common dewatering methods include:
The effectiveness of these methods is often measured by the final solid concentration achieved, which is directly related to the remaining moisture content.
The activated sludge process is an
The capacity of a septic tank is usually taken as:
Study the given statements with respect to soak pits and choose the correct option.
1. Soak pits are preferable at locations, where the water table level is high.
2. Soak pits are preferable at locations where soil is porous.A structure built up underground, focusing on receiving human waste in various forms is called:
The greasy and other substances floating on the surface of sewage is termed as