Application of Mathematics in Real-World Usage · Guide 24 · BTT Mathematics Hub
A concentration can fall while the total pollutant load rises. A treatment process can remove 90% of incoming mass without producing a 90% reduction in concentration if flow changes at the same time. Two sampling points can report the same concentration but represent very different total quantities because one stream is much larger. Environmental monitoring is therefore a problem of units, rates, mixtures and conservation before it is a problem of interpretation.
This guide uses fictional substances, flows, reservoirs, sampling schedules and emissions purely for Mathematics teaching. It is not environmental compliance advice, chemical handling guidance, exposure guidance or an operational monitoring protocol. Real environmental decisions require validated sampling methods, calibrated instruments, applicable regulations and domain specialists.
The central ledger has three layers: mass is an amount of substance, concentration is mass per volume or another stated basis, and load or emission rate is mass per time. These quantities are related but not interchangeable.
Concentration is a ratio with units
Suppose a fictional water sample contains 30 mg of a marker in 10 L. Concentration is 3 mg/L.
If the same 30 mg is distributed through 20 L, concentration falls to 1.5 mg/L even though total marker mass is unchanged.
A concentration change therefore does not by itself identify whether mass changed, volume changed or both.
Mass can be recovered from concentration and volume
Mass = concentration × volume when the units are compatible.
A tank with 200 L at 4 mg/L contains 800 mg under a uniform-mixing assumption.
If concentration varies within the tank, one point measurement cannot automatically represent the whole volume. The uniformity assumption is part of the calculation.
A dilution is a conservation problem
Suppose 10 L at 8 mg/L is mixed with 30 L of clean water in a fictional conservative-tracer model. Initial marker mass is 80 mg. Final volume is 40 L, so final concentration is 2 mg/L.
The concentration fell by 75%, but marker mass did not change. The denominator increased.
This simple dilution model excludes reactions, deposition, evaporation and other processes unless explicitly added.
Mixing two nonzero streams requires a weighted average
Stream A contributes 20 L at 2 mg/L, containing 40 mg. Stream B contributes 80 L at 5 mg/L, containing 400 mg.
Total concentration after complete mixing is 440 mg/100 L = 4.4 mg/L.
The simple mean of 2 and5 is 3.5 mg/L and is wrong because the stream volumes are unequal.
Flow-weighted concentration uses flow as the weight
Suppose two steady streams flow at 2 m³/s and 8 m³/s with concentrations 1 and4 mg/L respectively. If density conventions and units are handled consistently, the mixed concentration under a conservative steady model is (2×1+8×4)/(2+8)=3.4 mg/L.
The higher-flow stream dominates the mixture. Equal-weight averaging would understate its influence.
Whenever a mean combines streams, identify whether the weights are volume, mass, flow, time or something else.
Pollutant load combines concentration and flow
If concentration is C mg/L and flow is Q L/s, mass load rate is CQ mg/s.
For 3 mg/L at 200 L/s, load is 600 mg/s. If concentration falls to 2 mg/L while flow rises to 400 L/s, load becomes 800 mg/s.
Concentration improved downward, but total mass transported per second increased. A monitoring report needs both quantities when both matter.
Daily load requires a time conversion
At 600 mg/s for a full day, total mass is 600×86,400 = 51,840,000 mg = 51.84 kg.
This assumes the rate is constant for the entire day. If flow or concentration varies, the load must be accumulated over shorter intervals or integrated over time.
Rate multiplied by time gives quantity only when the rate model is appropriate.
Variable loads add as areas under a rate-time graph
A fictional emission runs at 2 kg/h for three hours and 5 kg/h for two hours. Total mass is 2×3+5×2=16 kg.
Time-weighted average rate is 16/5=3.2 kg/h. The simple mean of the two rates, 3.5 kg/h, incorrectly gives equal weight to unequal durations.
This is the same weighted-rate logic used in transport and data networks.
Mass balance tracks sources, sinks and storage
The US EPA APEX technical documentation describes mass balance in terms of inflow, outflow, removal and internal sources for a defined volume.
A generic discrete teaching balance is:
Ending mass = beginning mass + inflow + internal generation − outflow − removal.
If a reservoir begins with 100 kg, receives 30 kg, generates 5 kg, sends out 20 kg and removes 25 kg, ending mass is 90 kg.
A balance discrepancy is evidence of a missing term or measurement mismatch
Suppose the ledger predicts 90 kg ending mass but a measurement-based estimate gives 84 kg.
The residual is observed minus predicted = −6 kg. It may reflect an unmodelled sink, measurement error, sampling mismatch or incorrect flow estimate.
The residual identifies inconsistency. It does not by itself identify which mechanism caused it.
Removal efficiency uses incoming mass as its base
If a treatment stage receives 100 kg and discharges 15 kg, removed mass is 85 kg. Removal efficiency is 85/100=85%.
If the incoming amount were 200 kg and outgoing 30 kg, efficiency is still 85% while removed mass doubles to 170 kg.
Percentage efficiency and absolute mass removed answer different questions.
Successive removal stages multiply retained fractions
Suppose stage 1 removes 80%, leaving 20%. Stage 2 removes 50% of what remains, leaving 50% of 20%=10% of the original.
Overall removal is 90%, not 130%. The stages act on changing bases.
The general retained fraction after several independent sequential percentage removals is the product of each retained fraction.
Concentration thresholds are classification rules
Suppose a classroom rule labels a sample “above threshold” at concentration greater than 5 mg/L. Values 4.9,5.0,5.1 produce classifications below, not above, above.
If the rule uses ≥5 instead, the middle classification changes. The inequality sign belongs to the definition.
This page does not assert any regulatory threshold; the numbers are fictional. Real environmental standards must be taken from the applicable authority and measurement method.
Detection limits create censored observations
Suppose a laboratory reports “<1 mg/L” rather than an exact value for one sample. Replacing it automatically with 0, 0.5 or 1 creates different averages.
For exact reported values 2 and4 plus one “<1,” the true three-sample mean under a hard 0–1 range for the censored result lies between 2.0 and 2.333… mg/L.
Censored-data analysis can require specialised statistical methods. The safe mathematical lesson is not to invent an exact observation where the measurement reports only a bound.
Sampling frequency changes which events can be seen
A concentration spike lasts one hour each day. A sample taken once every 24 hours at the same time could consistently miss it.
Hourly sampling would provide much more temporal coverage, but also more data and possibly different measurement costs.
Sampling design is therefore part of what the dataset can support. More decimal precision on infrequent samples cannot recover unobserved intervals.
Spatial sampling can be biased by convenient locations
Imagine a lake whose near-shore concentration is systematically higher than mid-lake concentration. Sampling only from an accessible pier can estimate pier conditions accurately while misrepresenting lake-wide average conditions.
A representative design may require multiple locations or area-based weighting. The mathematical issue is the relationship between sampled units and the target population.
Sampling bias is not fixed by increasing the number of repeated samples at the same unrepresentative site.
Emission factors are conditional multipliers
Suppose a fictional process uses a teaching emission factor of 0.4 kg per unit of activity. At 250 activity units, modelled emissions are 100 kg.
If the activity doubles and the same factor is assumed, emissions double. If technology changes the factor, the old proportionality no longer applies.
Real emission factors are source-specific empirical tools. This example demonstrates the Mathematics of a rate factor, not an approved factor for any process.
Mass fraction and concentration percentage need a defined base
A mixture contains 2 kg of component in 50 kg total mixture. Mass fraction is 2/50=0.04, or 4% by mass.
If 10 kg of clean carrier is added without changing component mass, new fraction is 2/60=3.333%.
The component mass stayed fixed while its percentage fell because the denominator increased.
Parts per million is a ratio scale
One part per million means a ratio of 1 to 1,000,000 on the stated basis. For a mass fraction, 25 ppm means 25 units of mass per million equal mass units, or fraction 25×10⁻⁶.
In water, shortcuts between mg/L and ppm depend on density assumptions and concentration regime. They should not be treated as exact universal identities without stating the approximation.
Dimensional discipline is especially important when different environmental datasets use mass/mass, mass/volume and volume/volume conventions.
First-order decay is an exponential model
If a conservative teaching system instead includes first-order removal rate constant k, a closed-system model can be M(t)=M₀e^(−kt).
With M₀=100 kg and k=0.1 per day, after 10 days M≈36.79 kg.
Half-life is ln2/k≈6.93 days. This is an abstract decay model, not a chemical claim about a real pollutant.
Steady state means inflow rate balances total removal rate
Suppose a well-mixed box receives pollutant at 12 g/h and total removal is proportional to stored mass at rate 0.3 per hour.
At steady state, 12=0.3M, so M=40 g.
If the system begins away from 40 g, storage changes over time until the dynamic model determines whether it approaches that state. “Steady” refers to zero net rate of change, not absence of internal flow.
Per-capita emissions can move differently from total emissions
Total fictional emissions fall from 1,000 to 950 units while population falls from 100 to 90. Per-capita emissions rise from 10 to about 10.56.
A lower total does not guarantee a lower per-capita value. The denominator changed faster.
This is why environmental and demographic indicators often need to be read together rather than in isolation.
Normalising by area can also reverse rankings
Site A emits 100 units over 10 km², giving 10 units/km². Site B emits 150 over 30 km², giving 5 units/km².
B has the larger total but the smaller area-normalised rate.
Neither metric is universally superior. Total burden and spatial intensity answer different questions.
Trend slopes need their units and time base
Suppose annual average concentration values are 10,9.5,9.0,8.5 mg/L at equally spaced years. Linear slope is −0.5 mg/L per year.
Expressed as percentage of the first value, the total four-point endpoint change is −15% from 10 to8.5, but that is not the same object as the absolute annual slope.
Trend conclusions in real monitoring require attention to seasonality, sampling changes, autocorrelation and detection limits.
A complete environmental report labels the system boundary
State what volume, area or flow the concentration represents. State whether load is instantaneous, daily or annual. State whether mass balance assumes complete mixing, conservative transport, reactions or removal.
For observations, distinguish measured concentrations from derived loads and modelled emissions. For averages, state time and spatial weighting. For non-detects, preserve the censoring information.
The strongest mathematical habit is to make conservation and denominator choices visible enough that another reader can reconstruct the result.
Practice: twenty environmental Mathematics questions
- 30 mg is contained in 10 L. Find concentration.
- 30 mg is diluted to 20 L. Find new concentration.
- A 200 L tank is uniformly at 4 mg/L. Find total mass in mg.
- Mix 10 L at 8 mg/L with 30 L clean water. Find final concentration.
- Mix 20 L at 2 mg/L with 80 L at 5 mg/L. Find concentration.
- Flows 2 and8 m³/s carry concentrations 1 and4 mg/L. Find flow-weighted concentration.
- At 3 mg/L and 200 L/s, find load in mg/s.
- At 2 mg/L and 400 L/s, find load.
- Convert 600 mg/s for one day to kilograms.
- A rate is 2 kg/h for 3 h and 5 kg/h for 2 h. Find total mass and time-weighted mean rate.
- Beginning mass 100 kg, inflow 30, generation 5, outflow 20, removal 25. Find ending mass.
- If measured ending mass is 84 kg, find residual observed−predicted.
- Incoming mass 100 kg, outgoing 15 kg. Find removal efficiency.
- Stages remove 80% then 50% of remainder. Find overall removal.
- Two exact samples are 2 and4 mg/L and one censored sample is between 0 and1. Find possible range for the three-sample mean.
- An emission factor is 0.4 kg/activity unit for 250 units. Find modelled emissions.
- 2 kg component in 50 kg mixture: find mass percent.
- For M₀=100 and k=0.1/day, find M after 10 days.
- For source 12 g/h and proportional removal 0.3/h, find steady-state mass.
- Why can concentration fall while total pollutant load rises?
Worked answers
- 3 mg/L.
- 1.5 mg/L.
- 800 mg.
- 2 mg/L.
- 4.4 mg/L.
- 3.4 mg/L.
- 600 mg/s.
- 800 mg/s.
- 51.84 kg.
- 16 kg total; 3.2 kg/h average.
- 90 kg.
- −6 kg.
- 85%.
- 90% overall. Ten percent remains.
- 2.0 to 2.333… mg/L.
- 100 kg.
- 4%.
- About 36.79 kg.
- 40 g.
- Because load equals concentration multiplied by flow. A sufficiently large increase in flow can outweigh a decrease in concentration.
Sources and connected applications
For a general well-mixed source/sink mass-balance framework, see the US EPA APEX Technical Support Document, especially its mass-balance description of inflow, outflow, removal and internal sources. EPA also provides a distinct Chemical Mass Balance model for source apportionment; that specialised regulatory model should not be confused with the simple conservation exercises used here.
Continue with Weather, Climate, Averages, Anomalies and Forecast Uncertainty; Population, Demography, Growth and Dependency Ratios; and Astronomy, Scale, Orbits, Angles and Observation. Return to the BTT Mathematics Hub.