Chapter 8, Problem 8.3P

To determine

(a)

The total density of the refuse.

Answer to Problem 8.3P

The total density is $4.37\text{lb}/{\text{ft}}^{\text{3}}$ .

Explanation of Solution

Given data:

The following table shows the mass of materials.

Component	Percent by weight
Food waste	10
Yard waste	20
Mixed paper	40
Corrugated cardboard	100
Glass	10
Ferrous(cans)	10

Table (1)

Concept used:

Write the expression to calculate the total density.

  ${\rho }_{\text{total}}=\frac{M_{\text{A}}+M_{\text{B}}+M_{\text{C}}+M_{\text{D}}+M_{\text{E}}+M_{\text{F}}}{\frac{M_{\text{A}}}{{\rho }_{\text{A}}}+\frac{M_{\text{B}}}{{\rho }_{\text{B}}}+\frac{M_{\text{C}}}{{\rho }_{\text{C}}}+\frac{M_{\text{D}}}{{\rho }_{\text{D}}}+\frac{M_{\text{E}}}{{\rho }_{\text{E}}}+\frac{M_{\text{F}}}{{\rho }_{\text{F}}}}$   ..... (I)

Here, mass of material is $M$ and the density of material is $\rho$ .

The following table shows bulk density of given materials.

Component	Bulk Densities$\left(\text{lb}/{\text{ft}}^{\text{3}}\right)$
Food waste	23.04
Yard waste	4.45
Mixed paper	3.81
Corrugated cardboard	1.87
Glass	18.45
Ferrous (cans)	6.36

Table (2)

Calculation:

Calculate the total density.

Substitute $10\text{lb}$ for $M_{\text{A}}$ , $23.04\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{A}}$ , $20\text{lb}$ for $M_{\text{B}}$ , $4.45\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{B}}$ , $40\text{lb}$ for $M_{\text{C}}$ , $3.81\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{C}}$ , $10\text{lb}$ for $M_{\text{D}}$ , $1.87\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{D}}$ , $10\text{lb}$ for $M_{\text{E}}$ , $18.45\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{E}}$ , $10\text{lb}$ for $M_{\text{F}}$ , $6.36\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{F}}$ equation (I).

  $\begin{array}{c}{\rho }_{\text{total}}=\frac{10\text{lb}+20\text{lb}+40\text{lb}+10\text{lb}+10\text{lb}+10\text{lb}}{\left[\begin{array}{l}\frac{10\text{lb}}{23.04\text{lb}/{\text{ft}}^{\text{3}}}+\frac{20\text{lb}}{4.45\text{lb}/{\text{ft}}^{\text{3}}}+\frac{40\text{lb}}{3.81\text{lb}/{\text{ft}}^{\text{3}}}+\frac{10\text{lb}}{1.87\text{lb}/{\text{ft}}^{\text{3}}}+\frac{10\text{lb}}{18.45\text{lb}/{\text{ft}}^{\text{3}}}\\ +\frac{10\text{lb}}{6.36\text{lb}/{\text{ft}}^{\text{3}}}\end{array}\right]}\\ =\frac{100\text{lb}}{22.89{\text{ft}}^{\text{3}}}\\ =4.37\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Conclusion:

The total density is calculated by mass of materials and density and their components.

To determine

(b)

The bulk density of the refuse.

Answer to Problem 8.3P

The bulk density of refuse is $12.35\text{lb}/{\text{ft}}^{\text{3}}$ .

Explanation of Solution

Given data:

The following table shows the bulk density of the recovered refuse.

Component	Percentage recovered$\left(\text{lb}\right)$
Garbage	0
Yard waste	0
Mixed paper	50
Glass	60
Ferrous(cans)	95

Table (3)

Concept used:

Write the expression to calculate the not recovered mass of material.

  $M_{\text{not Recovered}}=100-\left(\%\text{Recovered}\right)$

Here, the not recovered mass of material is $M_{\text{non-Recovered}}$ .

Write the expression to calculate the total density of refuse.

  ${\rho }_{\text{total}}={\rho }_{\text{recovered}}+{\rho }_{\text{non-recovered}}$  ..... (II)

Here, the density of the recovered refuse is ${\rho }_{\text{recovered}}$ , the density of the non-recovered refuse is ${\rho }_{\text{non-recovered}}$ and the total density is ${\rho }_{\text{total}}$ .

Calculation:

Calculate the density of the not recovered refuse.

Substitute $50\text{lb}$ for $M_{\text{A}}$ , $3.81\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{A}}$ , $40\text{lb}$ for $M_{\text{B}}$ , $18.45\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{B}}$ , $5\text{lb}$ for $M_{\text{C}}$ , $6.36\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{C}}$ , $5\text{lb}$ for $M_{\text{D}}$ , $4.45\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{D}}$ in Equation (I).

  $\begin{array}{c}{\rho }_{\text{Non-Recovered}}=\frac{50\text{lb}+40\text{lb}+5\text{lb}+5\text{lb}}{\left[\frac{50\text{lb}}{3.81\text{lb}/{\text{ft}}^{\text{3}}}+\frac{40\text{lb}}{18.45\text{lb}/{\text{ft}}^{\text{3}}}+\frac{5\text{lb}}{6.36\text{lb}/{\text{ft}}^{\text{3}}}+\frac{5\text{lb}}{4.45\text{lb}/{\text{ft}}^{\text{3}}}\right]}\\ =\frac{100\text{lb}}{17.20{\text{ft}}^3}\\ =5.81\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Here, $5\%$ of yard waste left in not recovered refuse.

Calculate the density of the recovered refuse.

Substitute $50\text{lb}$ for $M_{\text{A}}$ , $3.81\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{A}}$ , $60\text{lb}$ for $M_{\text{B}}$ , $18.45\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{B}}$ , $95\text{lb}$ for $M_{\text{C}}$ , $6.36\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{C}}$ in Equation (I).

  $\begin{array}{c}{\rho }_{\text{Recovered}}=\frac{50\text{lb}+60\text{lb}+95\text{lb}}{\frac{50\text{lb}}{3.81\text{lb}/{\text{ft}}^{\text{3}}}+\frac{60\text{lb}}{18.45\text{lb}/{\text{ft}}^{\text{3}}}+\frac{95\text{lb}}{6.36\text{lb}/{\text{ft}}^{\text{3}}}}\\ =\frac{205\text{lb}}{31.31{\text{ft}}^3}\\ =6.54\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Calculate the total density of the landfill.

Substitute $5.81\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{Non-Recovered}}$ , $6.54\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{Recovered}}$ in Equation(II).

  $\begin{array}{c}{\rho }_{\text{total}}=6.54\text{lb}/{\text{ft}}^{\text{3}}+5.81\text{lb}/{\text{ft}}^{\text{3}}\\ =12.35\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Conclusion:

The bulk density is calculated using the density of the recovered refuse, not recovered, recovered mass of material and not recovered mass of material.

To determine

(c)

Answer to Problem 8.3P

Explanation of Solution

Given data:

Original life of landfill is $10\text{years}$ .

Calculation:

Calculate the expected life of landfill.

  $\begin{array}{l}\left({\rho }_{\text{total}}\right)\times \left(N_{\text{Original}}\right)=\left({\rho }_{\text{non-Recovered}}\right)\times \left(N_{\text{Expected}}\right)\\ N_{\text{Expected}}=\frac{\left({\rho }_{\text{total}}\right)\times \left(N_{\text{Original}}\right)}{\left({\rho }_{\text{non-Recovered}}\right)}\end{array}$   ..... (III)

Here, the expected life of the landfill is $N_{\text{Expected}}$ and the original life of the landfill is $N_{\text{Original}}$ .

Substitute $12.35\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{Total}}$ , $5.81\text{lb}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{Non-Recovered}}$ and $10\text{years}$ for $N_{\text{Original}}$ in Equation (III).

  $\begin{array}{c}{N}_{\text{Expected}}=\frac{\left(12.35\text{lb}/{\text{ft}}^{\text{3}}\right)\times \left(10\text{years}\right)}{\left(5.81\text{lb}/{\text{ft}}^{\text{3}}\right)}\\ =\frac{123.5}{5.81}\text{years}\\ =21.26\text{years}\end{array}$

Conclusion:

Thus, the expected life for landfill is $21.26\text{years}$ .