5.3: Percentages
In Unit 1, we learned how to convert between fractions, decimals, and percentages. We also learned that a percent calculated
\[\Large{Percent}=\dfrac{Part}{Whole}\cdot 100\%\]
In this unit, we will use percentages in applications. We will concentrate on the following types of calculations which can be derived by rewriting the formula above (as learned in Unit 3):
- Calculating a percent of an amount: \(\left(percentage\right)\left(whole\right)=part\)
- Determine the percent based on the part and the whole: \(\frac{Part}{Whole}\cdot 100\%=Percent\)
- A value is a certain percent of a number: \(whole=\frac{part}{percent}\)
Examples:
- 52% of 560 is what number?
Since “of” means multiplication, we have \(\left(52\%\right)\left(560\right)=a\,\, number\), so let’s convert 52% to a decimal by dividing 52 by 100 or by moving the decimal place two units to the left to obtain 0.52, hence \(\left(52\%\right)\left(560\right)=\left(0.52\right)\left(560\right)=291.2\). As a result, 52% of 560 is 291.2.
- 32 is what percent of 420?
In this problem, we need to calculate the percent and we have the part, which is 32, and the whole, which is 420. Hence \(Percent=\frac{32}{420}\cdot 100\%=0.076\cdot 100\%=7.6\%\) (using two significant figures; see Unit 5).
- 23 is 12% of what number? Round the final answer to the nearest hundredth. (Unit 5)
In this problem, we are given the part and the percent and need to determine the whole, so we use: \(whole=\frac{part}{percent}=\frac{23}{12\%}=\frac{23}{0.12}=191.67\)
Percentages can be used in the Water Industry in many ways. Below are a few applications of percentages and ratios.
- The volatile content of raw sludge can be calculated as a rate, proportion, or a percent. Generally, we will use a percent to represent the volatile content of raw sludge by calculating the reduction of volatile matter using the following formula: \[\text{Volatile Solids Reduction (%) } = \frac{\left(Value\,\, IN\right)-(Value\,\, OUT)}{\left(Value\,\, IN\right)-[(Value\,\, IN)(Value\,\, OUT)]}\cdot 100\%\]
When the volatile solids reduction is between 50%-60%, it is a good or healthy anaerobic digester. Any amount less than 50% is not reducing the volatile solids enough. NOTE: The fraction portion of this formula represents the rate or proportion. By multiplying by 100% we obtain the percentage. Also, note that the Value IN and Value OUT are decimal values between 0 and 1.
Example: The volatile content of raw sludge is 68.2% while the digester sludge is 55.6%. Calculate the volatile solids reduction and determine if the reduction is a healthy amount.
\[\begin{align} \text{Volatile Solids Reduction (%) }&= \frac{\left(Value\,\, IN\right)-(Value\,\, OUT)}{\left(Value\,\, IN\right)-[(Value\,\, IN)(Value\,\, OUT)]}\cdot 100\% \\ &=\frac{0.682-0.556}{0.682-\left(0.682\right)\left(0.556\right)}\cdot 100\% \\ &=\frac{0.126}{0.682-0.379}\cdot 100\% \\ &=\frac{0.126}{0.303}\cdot 100\% \\ &=0.4158\cdot 100\%=41.6\% \end{align}\]
Since 41.6% is not between 50% and 60%, this is considered to not be a good reduction of waste solids.
- The efficiency of water can be calculated using the following formula:
\[Efficiency=\frac{\left(Value\,\, IN-Value\,\, OUT\right)}{Value\,\, IN}\cdot 100\%\]
where the Value IN corresponds to the water received from another water system through intertie and Value OUT corresponds to water supplied to another water system through an intertie. [Intertie as defined by Merriam- Webster Dictionary, “an interconnection permitting passage of current between two or more electric utility systems” such as in a share federal-state water system.
NOTE: The fraction portion of this formula represents the rate or proportion. By multiplying by 100% we obtain the percentage.
Example: Calculate the efficiency (as a percent) of removing BOD (Biological Oxygen Demand) at a wastewater treatment plant if the influx of BOD is 215 mg per liter and the outflow is 18 mg per liter.
\[\begin{align} \text{Efficiency}&=\frac{\left(Value\,\, IN-Value\,\, OUT\right)}{Value\,\, IN}\cdot 100\%\\ &= \frac{\left(215 \frac{mg}{L}-18\frac{mg}{L}\right)}{215\frac{mg}{L}}\cdot 100\%\\ &=\frac{197\frac{mg}{L}}{215\frac{mg}{L}}\cdot 100\%\\ &=0.916\cdot 100\% \\ &=91.6\% \end{align}\]
Hence, the efficiency of removing BOD at the wastewater treatment plant is 91.6%.