Question

For many years, snow removed from the campus roads and parking lots has been located to the field next to the river. The plowed snow has a relatively high salt content due to the application of road salt (NaCl) during the winter to keep the campus roads safe. NaCl quickly dissociates in water, resulting in sodium (Na⁺) and chloride (Cl^-) ions. The US EPA recommends that the chloride content in remain below 230 mg/L to maintain healthy freshwater aquatic ecosystems.

Assume the following:

• Chloride is a non-reactive pollutant
• The Raquette River has a chloride concentration of 50 parts per million by mass

(note: 50 ppm chloride by mass = 50 grams chloride per 10⁶ grams water; the density of water is 1 g/cm³, 10³ cm³ = 1 L)

• The Raquette River has an average flowrate of 1500 ft³/s
• The runoff from the field has a chloride concentration of 5000 ppm by mass
• The flowrate of the runoff from field during an average storm is 20 ft³/s
• At the point of discharge from the field, the width of the river is 1000 ft, the depth is 5 ft, and there is near instantaneous mixing within 200 feet downstream of the field.

1. Draw a mass balance diagram for this system.
2. Calculate the chloride concentration of the river just downstream of the field one hour after an average rainstorm begins.
3. Is your estimate for the chloride concentration in the river just downstream of the field above the EPA limit of 230 mg/L chloride? Show your calculations for the comparison.
4. Name one fundamental assumption you needed to make to apply the mass balance approach to this problem. Is this a reasonable assumption? Briefly explain why or why not.

Answer 1

Before proceeding to solve the problem, let us relate all the units given:
1 ppm = 1 part per million = 1 gram chloride per million grams of water = 1 gram per 1000 kg of water = 1 mg per kg of water
Now, since density of water = 1g/cm³, 1 kg water = 1 litre water. Therefore,  1 ppm = 1 mg/litre
Also, 1 ft³ = 0.0283 m³

(a) Since, mixing is near instantaneous within 200 feet downstream of the field, we will consider this 1000 ft x 5 ft x 200 ft section of river as our system and apply the mass balance equation on it.
Now, for any hydrologic system the most basic mass balance equation is given as
Inflow = Outflow + Change in storage.
Here, the inflow includes various sources through which water reaches in to the system such as Precipitation, Surface runoff, Sub-surface flow, etc.
Similarly, the outflow includes Evaporation losses, percolation losses, water drawn for various usage from the reservoir, etc.
And change in storage simply tells us about the rise or fall in the volume of the storage.
In this given problem however, the inflow only includes surface runoff from the field and the outflow is simply the flow of the river. Also, since it's not a stagnant reservoir, there will not be any significant change in height/volume of the reservoir. The velocity of flow will increase instead to accommodate the extra inflow.
Here's the mass balance diagram for our system:

(b) To find out the chloride content downstream of the river, we will use the following equation:
Q₁C₁ + Q₂C₂ = (Q_mix)C_mix

Here, Q₁ is the discharge of the runoff from the field in ft³/s = 20 ft³/s,
C₁ = Chloride concentration in the field runoff in mg/l = 5000 mg/l,
Q₂ is the discharge of the Raquette river in ft³/s = 1500 ft³/s
C₂ = Initial chloride concentration in the Raquette river in mg/l = 50 mg/l
_Qmix = Discharge of the mixture 200 ft downstream of the field = 1500 + 20 = 1520 ft³/s
C_mix = Chloride concentration of the mixture at 200 ft downstream of the field

Putting all the value with their respective units in the equation,
20 ft³/s x 5000 mg/l + 1500 ft³/s x 50 mg/l = 1520 ft³/s x C_mix
Solving the above equation for C_mix, we get C_mix = 115.13 mg/l.

(c) From the above calculations, it is clear that the chloride concentration in the river just downstream of the field is well below the EPA limit of 230 mg/l.

(d) The fundamental assumption that I needed to make in order to apply the mass balance approach to this problem was there was no other inflow to the system except the field runoff or losses from the system. Now, this assumption can be reasonable for narrow, lined canals or narrow rivers with impermeable beds so that evaporation losses and infiltration losses are insignificant but for wide rivers with permeable beds, that may not be true because wide rivers have larger surfaces which gives more area for evaporation losses.