Chapter 1, Problem 1.89AP

Interpretation Introduction

Interpretation: The additional concentration of nitrogen (expressed in milligrams of nitrogen per liter) in the stream water due to the farmer’s yearly application of fertilizer is to be calculated.

Concept introduction: The additional concentration of nitrogen (expressed in milligrams of nitrogen per liter) in the stream water $\left({\text{A}}_{\text{c,}\text{N}}\right)$ per year is expressed as,

${\text{A}}_{\text{c,}\text{N}}=\frac{\text{Amount of nitrogen in stream water}\left(\text{mg}\right)}{\text{Average rate of stream flow}\text{per}\text{year}\left(\text{L}\right)}$

To determine: The additional concentration of nitrogen (expressed in milligrams of nitrogen per liter) in the stream water due to the farmer’s yearly application of fertilizer.

Answer to Problem 1.89AP

Answer

The additional concentration of nitrogen (expressed in milligrams of nitrogen per liter) in the stream per year is $\underset{\_}{0.031mg/L}$.

Explanation of Solution

Explanation

Given

The amount of fertilizer used by farmer in one year is $1500\text{kg}\text{.}$

The amount of nitrogen in farmer’s fertilizer is $10\%$.

The amount of fertilizer that is ran off by stream in one year is $15\%$.

The average rate of stream flow is $1.4{\text{m}}^3/\min$.

The amount of fertilizer in kilogram that is ran off by stream in one year is,

$\begin{array}{l}=1500\text{kg}\times \frac{15}{100}\\ =225\text{kg}\end{array}$

The amount of nitrogen in stream water is,

$\begin{array}{l}=225\text{kg}\times \frac{10}{100}\\ =22.5\text{kg}\end{array}$

The conversion of kilogram into milligram is,

$1\text{kg}={10}^6\text{mg}$

Convert the amount of nitrogen in stream water into gram.

$22.5\text{kg}=22.5\times {10}^6\text{mg}$ (1)

The average rate of stream flow is $1.4{\text{m}}^3/\min$.

The conversion of cubic meter into liter is,

$1{\text{m}}^3={10}^3\text{L}$

The conversion of minute into year is,

$1\min =\frac{1}{525600}\text{yr}$

Convert the average rate of stream flow into liter per year.

$\begin{array}{c}1.4{\text{m}}^3/\min =\frac{{10}^3\text{L}}{\frac{1}{525600}\text{yr}}\\ =735.84\times {10}^6\text{L}/\text{yr}\end{array}$

The above value indicates that the amount of stream water in one year is $735.84\times {10}^6\text{L}$.

The additional concentration of nitrogen (expressed in milligrams of nitrogen per liter) in the stream water $\left({\text{A}}_{\text{c,}\text{N}}\right)$ per year is expressed as,

${\text{A}}_{\text{c,}\text{N}}=\frac{\text{Amount of nitrogen in stream water}\left(\text{mg}\right)}{\text{Average rate of stream flow}\text{per}\text{year}\left(\text{L}\right)}$ (2)

Substitute the amounts of nitrogen in stream water $\left(\text{mg}\right)$ and rate of stream flow per year $\left(\text{L}\right)$ in equation (2).

$\begin{array}{c}{\text{A}}_{\text{c,}\text{N}}=\frac{22.5\times {10}^6\text{mg}}{735.84\times {10}^6\text{L}}\\ =0.031\text{mg}/\text{L}\end{array}$

Therefore, the additional concentration of nitrogen (expressed in milligrams of nitrogen per liter) in the stream per year is $\underset{\_}{0.031mg/L}$.

Conclusion

Conclusion

The additional concentration of nitrogen (expressed in milligrams of nitrogen per liter) in the stream per year is $\underset{\_}{0.031mg/L}$.