Chapter 2, Problem 2.18P

To determine

The composition of waste when some percentage of waste was recycled and to determine the change in energy content.

Answer to Problem 2.18P

The following table shows the composition of waste when some percentage of waste was recycled.

Component	Energy(Btu/lb)	Initial weight(lb)	Final weight(lb)
Food	2000	10	9
Paper	7200	33	16.5
Cardboard	7000	8	8
Plastics	14000	5	3.75
Textile	7500	3	3
Rubber	10000	4	4
Yard Waste	2800	18	18
Metal	300	10	7
Miscellaneous	3000	9	9

Thus, the change in energy content is $387300\text{Btu}$ .

Explanation of Solution

Calculation:

The following table shows the change in the composition of waste.

Component	Energy   $\left(E\right)$   $\left(\text{Btu}/\text{lb}\right)$	Initial weight$\left(\text{lb}\right)$   $\left(W_1\right)$	Change in weight$\left(\text{lb}\right)$   $\left(W_2\right)$	Final weight$\left(\text{lb}\right)$   $W_3=W_1-W_2$
Food	2000	10	$10\times \frac{10}{100}=1$	9
Paper	7200	33	$33\times \frac{50}{100}=16.5$	16.5
Cardboard	7000	8	-	8
Plastics	14000	5	$5\times \frac{25}{100}=1.25$	3.75
Textile	7500	3	-	3
Rubber	10000	4	-	4
Yard Waste	2800	18	-	18
Metals	300	10	$10\times \frac{30}{100}=3$	7
Miscellaneous	3000	9	-	9

Calculate the change in energy content.

  $E_C={\displaystyle \sum \left(E\times W_3\right)}$

Here, change in energy content is $E_C$ , the initial energy is $E$ and the final weight is $W_3$ .

Substitute the values from the table.

  $\begin{array}{c}{E}_C=\left[\begin{array}{l}\left(2000\text{Btu}/\text{lb}\times 9\text{lb}\right)+\left(7200\text{Btu}/\text{lb}\times 16.5\text{lb}\right)+\left(7000\text{Btu}/\text{lb}\times 8\text{lb}\right)\\ +\left(14000\text{Btu}/\text{lb}\times 3.75\text{lb}\right)+\left(7500\text{Btu}/\text{lb}\times 3\text{lb}\right)+\left(10000\text{Btu}/\text{lb}\times 4\text{lb}\right)\\ +\left(2800\text{Btu}/\text{lb}\times 18\text{lb}\right)+\left(300\text{Btu}/\text{lb}\times 7\text{lb}\right)+\left(3000\text{Btu}/\text{lb}\times 9\text{lb}\right)\end{array}\right]\\ =\left[\begin{array}{l}18000\text{Btu}+118800\text{Btu}+56000\text{Btu}+52500\text{Btu}+22500\text{Btu}\\ +40000\text{Btu}+50400\text{Btu}+2100\text{Btu}+27000\text{Btu}\end{array}\right]\\ =387300\text{Btu}\end{array}$

Conclusion:

The following table shows the composition of waste when some percentage of waste was recycled.

Component	Energy(Btu/lb)	Initial weight(lb)	Final weight(lb)
Food	2000	10	9
Paper	7200	33	16.5
Cardboard	7000	8	8
Plastics	14000	5	3.75
Textile	7500	3	3
Rubber	10000	4	4
Yard Waste	2800	18	18
Metal	300	10	7
Miscellaneous	3000	9	9

Thus, the change in energy content is $387300\text{Btu}$ .