Chapter 8, Problem 8.1P

To determine

The number of years to fill a football field that is $20\text{ft}$ deep.

Answer to Problem 8.1P

The time required to fill a football field that is $20\text{ft}$ deep is $7.85\text{years}$ .

Explanation of Solution

Calculation:

Assume the dimension of football field as $360\text{ft}\times \text{160}\text{ft}$ .

Calculate the volume of a $20\text{ft}$ deep football field.

  $V=l\times b\times h$   ...... (I)

Here, the volume is $V$ , the length of the field is $l$ , the width of the field is $b$ and the depth is $h$ .

Substitute $360\text{ft}$ for $l$ , $\text{160}\text{ft}$ for $b$ , and $20\text{ft}$ for $h$ in Equation (I).

  $\begin{array}{c}V=360\text{ft}\times 160\text{ft}\times 20\text{ft}\\ =1,152,000{\text{ft}}^{\text{3}}\\ =\left(1,152,000{\text{ft}}^{\text{3}}\right)\times {\left(\frac{\text{1}\text{yd}}{3\text{ft}}\right)}^3\\ =42666.67{\text{yd}}^{\text{3}}\end{array}$

Assume the dirt cover is $25\%$ in the $42666.67{\text{yd}}^{\text{3}}$ volume.

Write the expression to calculate the effective volume of football field.

  $V_{\text{e}}=V-\left(25\%\text{of}V\right)$   ...... (II)

Here, the effective volume is $V_{\text{e}}$ .

Substitute $42666.67{\text{yd}}^{\text{3}}$ for $V$ in Equation (II).

  $\begin{array}{c}{V}_{\text{e}}=42666.67{\text{yd}}^{\text{3}}-25\%\left(42666.67{\text{yd}}^{\text{3}}\right)\\ =42666.67{\text{yd}}^{\text{3}}-10666.67{\text{yd}}^{\text{3}}\\ =32000{\text{yd}}^{\text{3}}\end{array}$

Assume, in 2010 the university of Arizona produced $4075\text{tons}$ of waste. A total of about $7,000$ students live in the campus.

Calculate allowed waste per student.

  $\begin{array}{c}\text{Allowed}\text{waste}\text{per}\text{student}=\left(4075\text{ton}/\text{year}\right)\times \left(\frac{\text{2000}\text{lb}}{\text{1}\text{ton}}\right)\\ =8,150,000\text{lb}/\text{year}\end{array}$

Consider $70\%$ of the total allowed waste.

  $\begin{array}{c}\text{Allowed}\text{waste}\text{per}\text{student}=8,150,000\text{lb}/\text{year}\times 0.7\\ =5705000\text{lb}/\text{year}\end{array}$

Assume the density of the landfill as $1400\text{lb}/{\text{yd}}^3$ .

Write the expression to calculate the volume of the waste generated per year.

  $V_{\text{w}}=\frac{\text{Allowable}\text{waste}\text{per}\text{student}}{\text{Density}\text{of}\text{the}\text{landfill}}$   ...... (III)

Substitute $5705000\text{lb}/\text{year}$ for $\text{Allowable}\text{waste}\text{per}\text{student}$ and $1400\text{lb}/{\text{yd}}^3$ for $\text{Density}\text{of}\text{the}\text{landfill}$ in Equation (III).

  $\begin{array}{c}{V}_{\text{w}}=\frac{5,705,000\text{lb}/\text{year}}{1400\text{lb}/{\text{yd}}^{\text{3}}}\\ =4075{\text{yd}}^{\text{3}}/\text{year}\end{array}$

Write the expression to calculate the lifetime of the landfill.

  $T=\frac{V_{\text{e}}}{V_{\text{w}}}$   ...... (IV)

Substitute $32000{\text{yd}}^{\text{3}}$ for $V_{\text{e}}$ and $4075{\text{yd}}^{\text{3}}/\text{year}$ for $V_{\text{w}}$ in Equation (IV).

  $\begin{array}{c}T=\frac{32000{\text{yd}}^{\text{3}}}{4075{\text{yd}}^{\text{3}}/\text{year}}\\ =7.85\text{years}\end{array}$

Conclusion:

Thus, the lifetime of landfill is $7.85\text{years}$ .