Question

Organic fraction of municipal solid waste (OFMSW) with bulk density of 315 kg/m$^3$ and water content of 30% is mixed with municipal sludge of bulk density 700 kg/m$^3$ and water content of 70%, such that the water content of the mixture is 40%. The amount (in kg) of sludge to be mixed per kg of OFMSW (rounded off to 2 decimal places) and the density of the mixture (in kg/m$^3$) (rounded off to the nearest integer) are calculated. Which of the following options is/are true:

• A. 0.33 kg of sludge added per kg of OFMSW.
• B. Density of the mixture is 365 kg/m$^3$.
• C. 0.66 kg of sludge added per kg of OFMSW.
• D. Density of the mixture is 450 kg/m$^3$.

Hint:

When mixing materials with different water contents, you can calculate the final water content by balancing the individual contributions from each material and then calculating the bulk density of the mixture based on their respective densities.

Answer 1

The Correct Option is A, B

Given Data:

• Bulk density of MSW: \( \rho_1 = 315 \, \text{kg/m}^3 \)
• Water content of MSW: 30%
• Bulk density of sludge: \( \rho_2 = 700 \, \text{kg/m}^3 \)
• Water content of sludge: 70%
• Water content of mixture: 40%

Step 1: Water Content Balance Equation

Let \( W_1 \) and \( W_2 \) be the weights of MSW and sludge, respectively. The water content equation for the mixture is:

\[ 0.3W_1 + 0.7W_2 = 0.4(W_1 + W_2) \]

Simplifying:

\[ 0.3W_1 + 0.7W_2 = 0.4W_1 + 0.4W_2 \]

\[ 0.3W_1 - 0.4W_1 = 0.4W_2 - 0.7W_2 \]

\[ -0.1W_1 = -0.3W_2 \]

\[ \frac{W_2}{W_1} = \frac{1}{3} \]

Step 2: Bulk Density Calculation

Bulk density of the mixture is given by:

\[ \rho_{\text{bulk}} = \frac{W_1 + W_2}{V_1 + V_2} \]

Since \( V = \frac{W}{\rho} \), we substitute:

\[ \rho_{\text{bulk}} = \frac{W_1 + W_2}{\frac{W_1}{\rho_1} + \frac{W_2}{\rho_2}} \]

Substituting \( \frac{W_2}{W_1} = \frac{1}{3} \), we solve and obtain:

\[ \rho_{\text{bulk}} \approx 365 \, \text{kg/m}^3 \]

Thus, the correct answers are (A) and (B).