Problem 4 - Advanced Water Mathematics

Question

Well A has a total dissolved solids (TDS) level of \(850 \mathrm{mg} / \mathrm{L}\). It is pumping \(1,500 \mathrm{gpm}\) which is \(40 \%\) of the total production from two wells. The other well (B) blends with well A to achieve a TDS level of \(375 \mathrm{mg} / \mathrm{L}\). What is the TDS level for Well B?

Step-by-Step Solution

Answer

The TDS level for Well B is 58.33 mg/L.

Step 1: Determine the total combined flow rate

Well A's flow rate is given as 1,500 gallons per minute (gpm) and this is 40% of the total flow rate. Solve for the total flow rate (FT): \[ FT = \frac{1,500}{0.40} = 3,750 \text{ gpm} \]

Step 2: Determine Well B's flow rate

Well B's flow rate (FB) is the remaining flow rate after accounting for Well A: \[ FB = FT - \text{ Well A flow rate} \] \[ FB = 3,750 - 1,500 = 2,250 \text{ gpm} \]

Step 3: Use TDS levels to set up an equation

Using the given TDS levels, set up the blend equation: \[ ( \text{Well A TDS} \times \text{Well A flow rate} ) + ( \text{Well B TDS} \times \text{Well B flow rate} ) = \text{Combined TDS} \times \text{Total flow rate} \] Plug the known values into the equation: \[ (850 \times 1,500) + ( \text{Well B TDS} \times 2,250 ) = 375 \times 3,750 \]

Step 4: Solve for Well B's TDS level

First, simplify and solve for Well B's TDS level: \[ 1,275,000 + 2,250 \times \text{Well B TDS} = 1,406,250 \]\[ 2,250 \times \text{Well B TDS} = 1,406,250 - 1,275,000 \]\[ 2,250 \times \text{Well B TDS} = 131,250 \]\[ \text{Well B TDS} = \frac{131,250}{2,250} = 58.33 \text{ mg/L} \]

Key Concepts

Total dissolved solids (TDS)

Total dissolved solids (TDS) represent the concentration of dissolved substances in water. They include organic and inorganic materials like salts, minerals, and metals. TDS levels are commonly measured in milligrams per liter (mg/L).
TDS levels are an important indicator of water quality. High TDS levels can affect the taste, odor, and appearance of the water and may pose a health risk. Monitoring TDS is crucial for ensuring safe drinking water.
In the exercise, Well A has a TDS level of 850 mg/L. This means that every liter of water from Well A contains 850 milligrams of dissolved substances. To find the TDS levels when blending water from different sources, we need to consider both the TDS levels and the flow rates of each source.

Flow rate calculations

Flow rate is the volume of fluid that passes through a point in a given time, typically measured in gallons per minute (gpm) for water. It's essential for calculating the combined properties of blended water sources.
Step 1: Determine Total Combined Flow Rate
Well A's flow rate is given as 1,500 gpm, which is 40% of the total flow rate. To find the total flow rate (FT), we use the formula:
\[ FT = \frac{1,500}{0.40} = 3,750 \text{ gpm} \]
This reveals that the combined flow rate of both wells is 3,750 gpm.
Step 2: Determine Well B's Flow Rate
Well B's flow rate (FB) is the remaining flow after Well A's contribution:
\[ FB = FT - \text{ Well A flow rate} \]
\[ FB = 3,750 - 1,500 = 2,250 \text{ gpm} \]
Thus, Well B flows at 2,250 gpm.

Blended water quality

Blended water quality involves combining water from different sources to achieve a desired quality. In this case, we blend water from Well A and Well B to reach a specific TDS level of 375 mg/L.
Step 3: Use TDS Levels to Set Up an Equation
We use the following blending equation:
\[ ( \text{Well A TDS} \times \text{Well A flow rate} ) + ( \text{Well B TDS} \times \text{Well B flow rate} ) = \text{Combined TDS} \times \text{Total flow rate} \]
When we plug in the known values, we get:
\[ (850 \times 1,500) + ( \text{Well B TDS} \times 2,250 ) = 375 \times 3,750 \]
Step 4: Solve for Well B's TDS Level
Simplify and solve for Well B's TDS level:
\[ 1,275,000 + 2,250 \times \text{Well B TDS} = 1,406,250 \]
\[ 2,250 \times \text{Well B TDS} = 1,406,250 - 1,275,000 \]
\[ 2,250 \times \text{Well B TDS} = 131,250 \]
\[ \text{Well B TDS} = \frac{131,250}{2,250} = 58.33 \text{ mg/L} \]
Therefore, Well B has a TDS level of 58.33 mg/L.