Hydraulic Learning

Pollutant Propagation in Pressurized Networks

[Exercise] Pollutant Propagation in Hydraulic Networks (US Units)
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Pollutant Propagation in Pressurized Networks

Context: Environmental Hydraulics and Water Quality Monitoring.

In municipal water distribution systems, tracking the propagation of substances (chlorine, contaminants, or tracers) is critical for public safety. This exercise focuses on the AdvectionTransport of a substance by bulk motion of the fluid. and dilution of a pollutant at a pipe junction. We will analyze how a contaminant introduced in a secondary line mixes with the main flow and calculate its travel time downstream using US Customary Units.

Pedagogical Note: This exercise links hydraulic continuity (flow rates) with mass balance equations. It is essential for understanding how contaminants disperse in civil engineering infrastructure.


Learning Objectives

  • Apply the Law of Conservation of Mass to fluid flow (Continuity Equation).
  • Calculate the resulting concentration of a mixture using Mass Balance.
  • Determine flow velocity and travel time in a pressurized pipe.
  • Verify the flow regime using the Reynolds Number.

Study Data

A secondary pipe (Pipe B) carrying industrial runoff containing a tracer pollutant merges into a main water supply line (Pipe A) carrying clean water. The two streams mix instantly at the junction and flow into a common output pipe (Pipe C). We need to determine the characteristics of the flow in Pipe C.

Technical Sheet / Data
Pipe / Parameter Value
Pipe A (Main) - Flow Rate \(Q_{\text{A}} = 20 \, \text{cfs}\) (ft³/s)
Pipe A - Pollutant Concentration \(C_{\text{A}} = 0 \, \text{mg/L}\)
Pipe B (Runoff) - Flow Rate \(Q_{\text{B}} = 4 \, \text{cfs}\) (ft³/s)
Pipe B - Pollutant Concentration \(C_{\text{B}} = 600 \, \text{mg/L}\)
Pipe C (Output) - Diameter \(D_{\text{C}} = 2.5 \, \text{ft}\) (30 inches)
Pipe C - Length \(L_{\text{C}} = 8000 \, \text{ft}\)
Kinematic Viscosity (Water @ 68°F) \(\nu = 1.08 \times 10^{-5} \, \text{ft}^2/\text{s}\)
System Diagram: Junction Mixing
Pipe A (Clean) QA = 20 cfs Pipe B (Pollutant) QB = 4 cfs Pipe C (Mixture) L = 8000 ft, D = 2.5 ft
Questions to Address
  1. Calculate the total flow rate in Pipe C (\(Q_{\text{C}}\)).
  2. Calculate the resulting pollutant concentration in Pipe C (\(C_{\text{C}}\)).
  3. Calculate the flow velocity in Pipe C (\(v_{\text{C}}\)).
  4. Determine the Reynolds Number (\(\text{Re}\)) and the flow regime.
  5. Calculate the travel time (\(t\)) for the pollutant to traverse the length of Pipe C.

Theoretical Basics

Before solving, let's review the fundamental principles of fluid mechanics used in network analysis.

1. Continuity Equation (Conservation of Mass)
For an incompressible fluid (like water) at a junction, the sum of inflows equals the sum of outflows.

Flow Continuity

\[ \sum Q_{\text{in}} = \sum Q_{\text{out}} \]

2. Mass Balance for Constituents
Assuming complete mixing, the mass flux of the pollutant is conserved. Mass flux is the product of Flow Rate and Concentration.

Mass Balance

\[ Q_{\text{C}} \cdot C_{\text{C}} = Q_{\text{A}} \cdot C_{\text{A}} + Q_{\text{B}} \cdot C_{\text{B}} \]

3. Flow Velocity & Reynolds Number
Velocity is derived from flow rate and cross-sectional area. The Reynolds number predicts if flow is laminar or turbulent.

Velocity & Reynolds

\[ v = \frac{Q}{A} \quad ; \quad \text{Re} = \frac{v \cdot D}{\nu} \]

Where:

  • \(A = \pi \cdot D^2 / 4\) is the cross-sectional area.
  • \(\nu\) is the kinematic viscosity (\(\approx 1.08 \times 10^{-5} \, \text{ft}^2/\text{s}\)).
  • If \(\text{Re} > 4000\), flow is turbulent.

Solution: Pollutant Propagation in Pressurized Networks

Question 1: Total Flow Rate in Pipe C

Principle

We apply the Continuity Equation at the junction. Since water is incompressible, the volumetric flow rate entering the junction must equal the flow rate leaving it. In simple terms: what comes in must go out.

Key Concept

Continuity of Flow: In a steady state system, mass cannot be created or destroyed at a node. \(\sum Q_{\text{in}} = \sum Q_{\text{out}}\).

Pedagogical Note

This is the simplest form of mass balance because the fluid density is constant. If we were dealing with gases, we would need to account for density changes.

Standards

AWWA M31 (Distribution System Requirements) assumes steady-state flow for basic network sizing and modeling.

Formula(s)

Formulas Used

Sum of Flows

\[ Q_{\text{C}} = Q_{\text{A}} + Q_{\text{B}} \]
Assumptions

To apply this law, we make the following assumptions:

  • Steady-state flow (no changes over time).
  • Incompressible fluid (constant density).
  • No leakage at the junction.
Data
ParameterSymbolValueUnit
Flow A (Clean)\(Q_{\text{A}}\)20cfs
Flow B (Polluted)\(Q_{\text{B}}\)4cfs
Tips

Always check that your units match before adding. Here, both are in cubic feet per second (cfs), so direct addition is valid.

Flow Convergence
QA QB QC = ?
Calculations

Conversion(s)

While we work in cfs, it's useful to know the equivalent in Gallons Per Minute (GPM), common in US pump specs:

\[ \begin{aligned} 1 \, \text{cfs} &\approx 448.8 \, \text{GPM} \end{aligned} \]

This conversion helps visualize the scale of the flow in industrial terms.

Intermediate Calculation

Checking the magnitude of the main flow:

\[ \begin{aligned} Q_{\text{check}} &= 20 \times 448.8 \\ &= 8976 \, \text{GPM} \end{aligned} \]

This confirms we are dealing with a major supply line, not a small residential pipe.

Main Calculation

We substitute the variables with the known values from the Data table:

Step 1: Identify Inflows

\[ Q_{\text{A}} = 20 \, \text{cfs} \quad ; \quad Q_{\text{B}} = 4 \, \text{cfs} \]

Step 2: Add magnitudes

\[ \begin{aligned} Q_{\text{C}} &= 20 + 4 \\ &= 24 \, \text{cfs} \end{aligned} \]

This sum represents the total volume of water moving into Pipe C every second.

Diagram (After Calculations)

(Visualizing the combined flow magnitude)

Total Flow: 24 cfs
Reflections

The flow rate has increased by exactly 4 cfs. This represents a 20% increase over the original main flow (\(4/20 = 0.2\)).

Watch Out / Common Pitfalls

Do not confuse velocity (ft/s) with flow rate (ft³/s). You cannot add velocities directly (\(v_{\text{C}} \neq v_{\text{A}} + v_{\text{B}}\)). Only volumetric flow rates are additive.

Key Takeaways

Essential points to memorize:

  • Mass is conserved at a node.
  • For incompressible fluids, \(Q_{\text{in}} = Q_{\text{out}}\).
Did you know?

In the US, large scale water flow is often measured in MGD (Million Gallons per Day). 24 cfs is approximately 15.5 MGD.

FAQ
What if the pipes have different diameters?

The Continuity Equation (\(Q_{\text{total}} = \sum Q_i\)) holds true regardless of diameter changes. Diameter affects velocity, not the volumetric balance.

\(Q_{\text{C}} = 24 \, \text{ft}^3/\text{s}\)

Your Turn
If \(Q_{\text{A}}\) increases to 30 cfs, what is the new \(Q_{\text{C}}\)?

📝 Memo
Flow addition is linear: \(Q_{\text{total}} = Q_1 + Q_2 + ...\).


Question 2: Pollutant Concentration in Pipe C

Principle

We use the Mass Balance equation for the solute. The total mass of pollutant entering the junction per second must equal the mass leaving per second.

Key Concept

Dilution: Mixing a high concentration stream with a larger zero-concentration stream significantly lowers the final concentration.

Pedagogical Note

Think of this as a "weighted average" calculation, where the "weights" are the flow rates of each pipe.

Standards

NPDES (National Pollutant Discharge Elimination System) permits often require such mixing zone calculations to ensure downstream compliance with EPA limits.

Formula(s)

Formulas Used

Concentration Mixture

\[ C_{\text{C}} = \frac{Q_{\text{A}} C_{\text{A}} + Q_{\text{B}} C_{\text{B}}}{Q_{\text{C}}} \]
Assumptions

Key assumptions for this calculation:

  • Instantaneous and complete mixing at the node (ideal CSTR model).
  • Conservative pollutant (no chemical decay or reaction).
Data
ParamValue
\(C_{\text{A}}\)0 mg/L (Clean)
\(C_{\text{B}}\)600 mg/L (Pollutant)
Tips

Since Pipe A has clean water, the \(Q_{\text{A}} C_{\text{A}}\) term becomes zero. This simplifies the math significantly!

Mixing Process
600 + 0 = ?
Calculations

Conversion(s)

Concentration is given in mg/L. No conversion is needed for the mixing ratio itself, as long as units are consistent across terms.

Unit Check

\[ \begin{aligned} \text{Unit}_{\text{mix}} &= \frac{\text{cfs} \cdot \text{mg/L}}{\text{cfs}} \\ &= \text{mg/L} \end{aligned} \]

The flow units (cfs) cancel out, leaving the result in mg/L.

Intermediate Calculation

Let's calculate the "Mass Flux" (Load) for each pipe individually before summing them:

Load from Pipe A (Clean)

\[ \begin{aligned} \text{Load}_{\text{A}} &= Q_{\text{A}} \times C_{\text{A}} \\ &= 20 \times 0 \\ &= 0 \end{aligned} \]

Load from Pipe B (Pollutant)

\[ \begin{aligned} \text{Load}_{\text{B}} &= Q_{\text{B}} \times C_{\text{B}} \\ &= 4 \times 600 \\ &= 2400 \, (\text{flow} \cdot \text{conc units}) \end{aligned} \]

We now have the total load entering the junction: \(0 + 2400 = 2400\). This represents the total amount of pollutant mass per time.

Main Calculation

Now we divide the Total Load by the Total Flow (\(Q_{\text{C}}\)) calculated in Question 1:

\[ \begin{aligned} C_{\text{C}} &= \frac{\text{Total Load}}{\text{Total Flow}} \\ &= \frac{2400}{24} \\ &= 100 \, \text{mg/L} \end{aligned} \]

The math simplifies perfectly to 100.

Diagram (After Calculations)

Resulting Mix Color

100 mg/L
Reflections

The concentration dropped from 600 to 100. This is a 6x dilution factor. This occurs because the clean water flow (20 cfs) is 5 times larger than the polluted flow (4 cfs), resulting in a 1 part polluted to 5 parts clean ratio (1:6 total ratio).

Watch Out / Common Pitfalls

Don't just average 600 and 0 to get 300! An unweighted average ignores the fact that there is much more clean water than polluted water.

Key Takeaways

Essential points to memorize:

  • \(C_{\text{mix}} = \sum(Q_i C_i) / \sum Q_i\)
  • More clean water = Lower final concentration.
Did you know?

1 mg/L is equivalent to 1 PPM (Part Per Million) in water density, because 1 Liter of water weighs approximately 1 million milligrams (1000g).

FAQ
Does temperature affect this calculation?

Only slightly by changing water density, but for standard engineering calculations, we ignore thermal expansion effects on concentration unless the temperature difference is extreme.

\(C_{\text{C}} = 100 \, \text{mg/L}\)

Your Turn
If \(C_{\text{B}}\) was only 300 mg/L, what would be \(C_{\text{C}}\)?

📝 Memo
Dilution is proportional to flow ratios.


Question 3: Flow Velocity in Pipe C

Principle

Velocity represents the speed of the fluid "plug" moving through the pipe. It relates the volumetric flow rate to the pipe's physical cross-sectional area. A smaller pipe for the same flow results in higher velocity.

Key Concept

Velocity Relationship: \(Q = v \times A\). Flow Rate = Area × Speed.

Pedagogical Note

Velocity is a crucial parameter for calculating travel time (Question 5) and for checking if shear stress is sufficient to clean the pipe.

Standards

Typical municipal water main velocities are kept between 2 and 5 ft/s. Velocities above 8-10 ft/s can cause excessive head loss and water hammer issues.

Formula(s)

Formulas Used

Velocity Formula

\[ v = \frac{Q}{A} \]

Area Formula (Circle)

\[ A = \pi \times r^2 = \pi \left(\frac{D}{2}\right)^2 \]
Assumptions

We assume:

  • The pipe is flowing full (pressurized).
  • The cross-section is perfectly circular.
Data
ParamValue
\(Q_{\text{C}}\)24 cfs
\(D_{\text{C}}\)2.5 ft
Tips

Remember that area scales with the square of the diameter. Doubling the diameter quadruples the area.

Pipe Cross Section
D = 2.5'
Calculations

Conversion(s)

The diameter is given in feet (2.5 ft). If it were in inches (30 inches), we would divide by 12:

Inches to Feet

\[ \begin{aligned} D &= 30 \, \text{in} \div 12 \\ &= 2.5 \, \text{ft} \end{aligned} \]

We use feet to remain consistent with cfs (cubic feet per second).

Intermediate Calculation

We cannot divide Flow by Diameter directly. We need the Area (A) in square feet first.

Step 1: Radius Calculation

\[ \begin{aligned} r &= \frac{D}{2} \\ &= \frac{2.5}{2} \\ &= 1.25 \, \text{ft} \end{aligned} \]

Step 2: Area Calculation

\[ \begin{aligned} A &= \pi \times r^2 \\ &= 3.14159 \times (1.25)^2 \\ &= 3.14159 \times 1.5625 \\ &\approx 4.9087 \, \text{ft}^2 \end{aligned} \]

This 4.9087 ft² is the effective cross-sectional area available for flow.

Main Calculation

Now we divide the Flow Rate (\(Q_{\text{C}}\)) by the calculated Area (\(A\)):

\[ \begin{aligned} v_{\text{C}} &= \frac{24 \, \text{ft}^3/\text{s}}{4.9087 \, \text{ft}^2} \\ &\approx 4.889 \, \text{ft/s} \end{aligned} \]

Rounding to two decimal places gives us 4.89 ft/s.

Diagram (After Calculations)

(Velocity Vector)

v = 4.89 ft/s
Reflections

The calculated velocity is ~4.89 ft/s. This falls perfectly within the acceptable engineering range (2-5 ft/s). It is fast enough to prevent sediment deposition but slow enough to avoid excessive friction losses.

Watch Out / Common Pitfalls

Common error: Using Diameter in the formula \(\pi r^2\). Always use radius (\(D/2\)) or the alternate formula \(A = \frac{\pi D^2}{4}\).

Key Takeaways

Essential points to memorize:

  • \(v = Q/A\).
  • Consistent units (ft, cfs) prevent errors.
Did you know?

High velocities (>10 ft/s) can cause "water hammer" (destructive pressure surges) if a valve is closed suddenly, potentially bursting pipes.

FAQ
Is velocity uniform across the pipe?

In turbulent flow, the velocity profile is relatively flat (uniform), meaning the average velocity is close to the max velocity. However, theoretically, it is zero at the very walls due to the "no-slip condition".

\(v_{\text{C}} \approx 4.89 \, \text{ft/s}\)

Your Turn
If diameter was decreased to 2.0 ft, would velocity go up or down?

(It would go UP, because Area decreases while Flow stays constant)

📝 Memo
Smaller pipe = Higher speed (for same flow).


Question 4: Reynolds Number & Regime

Principle

The Reynolds Number (Re) is a dimensionless quantity used to predict flow patterns. It represents the ratio of inertial forces (momentum) to viscous forces (friction/stickiness).

Key Concept

Turbulence: High Re (>4000) indicates chaotic, mixing flow. Low Re (<2000) indicates laminar, smooth flow. Between 2000 and 4000 is transitional.

Pedagogical Note

We need to know the regime to confirm if mixing is efficient. Laminar flow does not mix lateral pollutants well; turbulent flow mixes them excellent.

Standards

The Moody Chart uses the Reynolds Number to determine the friction factor for head loss calculations in pipe networks.

Formula(s)

Formulas Used

Reynolds Number

\[ \text{Re} = \frac{v \cdot D}{\nu} \]
Assumptions

Standard water properties:

  • Temperature @ 68°F (determines viscosity).
  • Newtonian fluid.
Data
ParamValueUnit
Velocity \(v\)4.89ft/s
Diameter \(D\)2.5ft
Viscosity \(\nu\)\(1.08 \times 10^{-5}\)ft²/s
Tips

Re has NO units. If your units don't cancel out, check your inputs (e.g., ensure viscosity is kinematic, not dynamic).

Viscosity vs Inertia
Inertia Viscosity
Calculations

Conversion(s)

Ensure kinematic viscosity is in ft²/s. Sometimes it is given in centistokes, which would require conversion.

Unit Check

\[ \begin{aligned} \text{Unit Check} &= \frac{(\text{ft/s}) \cdot (\text{ft})}{\text{ft}^2/\text{s}} \\ &= \frac{\text{ft}^2/\text{s}}{\text{ft}^2/\text{s}} \\ &= \text{dimensionless} \end{aligned} \]

The units cancel out perfectly.

Intermediate Calculation

Let's calculate the numerator (the inertial term) first:

\[ \begin{aligned} \text{Numerator} &= v \times D \\ &= 4.889 \times 2.5 \\ &\approx 12.2225 \, \text{ft}^2/\text{s} \end{aligned} \]

This value represents momentum diffusivity potential.

Main Calculation

Now divide the inertial term by the kinematic viscosity \(\nu\):

\[ \begin{aligned} \text{Re} &= \frac{12.2225}{1.08 \times 10^{-5}} \\ &= \frac{12.2225}{0.0000108} \\ &\approx 1,131,712 \end{aligned} \]

The result is over 1.1 million.

Diagram (After Calculations)

Flow Regime Visualization

TURBULENT (Re > 4000)
Reflections

Since \(\text{Re} \approx 1.13 \times 10^6\), which is much greater than 4000, the flow is Fully Turbulent. This guarantees that the pollutant will mix efficiently across the pipe's cross-section.

Watch Out / Common Pitfalls

Scientific notation errors are common here. Dividing by a negative exponent (\(10^{-5}\)) results in a huge number. Watch your calculator inputs.

Key Takeaways

Essential points to memorize:

  • Re > 4000 = Turbulent.
  • Re < 2000 = Laminar.
Did you know?

Osborne Reynolds introduced this number in 1883 by injecting dye into water pipes to visualize the transition from straight streamlines to chaotic eddies.

FAQ
Is Re usually this high?

Yes, for water transport in large municipal pipes (2.5 ft diameter), Re is almost always very high (turbulent). Laminar flow is mostly seen in tiny capillaries or very viscous fluids like oil.

\(\text{Re} \approx 1.13 \times 10^6\) (Turbulent)

Your Turn
If viscosity increased (e.g., colder water), would Re go up or down?

(It would go DOWN, because viscosity is in the denominator)

📝 Memo
Water mains = Turbulent flow.


Question 5: Travel Time Downstream

Principle

The Travel Time (or detention time) is calculated using simple kinematics. It represents how long the "plug" of pollutant takes to travel the length \(L\) of the pipe.

Key Concept

Advection Time: Time = Distance / Speed.

Pedagogical Note

This metric is critical for public safety. It tells operators how much time they have to shut down a valve or warn downstream users in case of a contamination event.

Standards

Emergency Response Plans (ERPs) for water utilities often use this method to calculate "Time to Shutdown" or "Time to Customer".

Formula(s)

Formulas Used

Time Formula

\[ t = \frac{L}{v} \]
Assumptions

We assume:

  • Plug Flow: The pollutant moves as a distinct block without longitudinal dispersion (spreading out ahead).
  • Uniform velocity along the entire length.
Data
ParamValue
Length \(L\)8000 ft
Velocity \(v\)4.89 ft/s
Tips

The result will initially be in seconds. Convert it to minutes for a human-readable answer.

Path Animation
Start End (8000')
Calculations

Conversion(s)

We will convert seconds to minutes at the end using the standard factor:

Time Conversion

\[ \text{min} = \frac{\text{sec}}{60} \]

Intermediate Calculation

Just to visualize the distance, let's see how long 8000 ft is in miles:

\[ \begin{aligned} L_{\text{miles}} &= 8000 \, \text{ft} \div 5280 \, \text{ft/mile} \\ &\approx 1.515 \, \text{miles} \end{aligned} \]

The pollutant has to travel about 1.5 miles.

Main Calculation

Dividing the length by the velocity:

\[ \begin{aligned} t_{\text{sec}} &= \frac{8000 \text{ ft}}{4.89 \text{ ft/s}} \\ &\approx 1635.99 \, \text{s} \end{aligned} \]

Now convert to minutes by dividing by 60:

\[ \begin{aligned} t_{\text{min}} &= \frac{1635.99}{60} \\ &\approx 27.27 \, \text{min} \end{aligned} \]
Diagram (After Calculations)

Result

~ 27 Minutes
Reflections

The operators have less than half an hour (27 mins) to react if a spill occurs at point B before it reaches the end of Pipe C. This implies automated sensors might be necessary, as manual intervention might be too slow.

Watch Out / Common Pitfalls

Don't mix miles and feet. Keep everything in feet for the calculation, then convert at the end if asked.

Key Takeaways

Essential points to memorize:

  • Time = L / v.
  • Always convert to meaningful units (min vs sec) for reporting.
Did you know?

In reality, tracers disperse longitudinally. The "leading edge" of the pollutant will arrive slightly faster than the average time calculated here due to the velocity profile center moving faster than the average.

FAQ
What is "Plug Flow"?

Plug flow is an ideal model where fluid moves like a solid piston or "plug", with no mixing along the length. In reality, some dispersion occurs, but for turbulent flow, plug flow is a good approximation for timing.

\(t \approx 27.3 \, \text{min}\)

Your Turn
If length was 4000 ft (half distance), what is the time?

📝 Memo
Distance / Speed = Time.


Exercise Summary Diagram

This diagram summarizes all calculated values, final states, and system configuration.

QA = 20 cfs QB = 4 cfs Conc = 600 mg/L Pipe C Results: • Q Total = 24 cfs • Conc = 100 mg/L • Velocity = 4.89 ft/s • Time = 27.3 min

📝 Grand Memo: What to absolutely remember

Here is the summary of key methodological and physical points covered in this exercise:

  • 🔑
    Key Point 1: Dilution Logic
    The final concentration is a flow-weighted average. A small polluted stream mixing with a large clean stream results in high dilution.
  • 📐
    Key Point 2: Mass Balance Formula
    Remember: \(Q_{\text{out}} C_{\text{out}} = \sum (Q_{\text{in}} C_{\text{in}})\). This applies to temperature mixing as well!
  • ⚠️
    Key Point 3: Units Matter
    Ensure flow rates match (cfs) and area calculations use Diameter (ft), not inches.
  • 💡
    Key Point 4: Turbulence
    High Reynolds numbers (>4000) indicate turbulent flow, which promotes good mixing of the pollutant.
"Dilution is the solution to pollution... until you run out of clean water."

🎛️ Interactive Simulator: Dilution Effect (US Units)

Adjust the flow rate of the clean water (Pipe A) and the concentration of the pollutant (Pipe B) to see the effect on the final concentration in Pipe C.

Parameters
Output Flow (Qc): -
Final Conc. (Cc): -

📝 Final Quiz: Test your knowledge

1. If the flow rate of the clean water (Pipe A) doubles, what happens to the pollutant concentration in Pipe C?

2. What physical principle states that \(\sum Q_{\text{in}} = \sum Q_{\text{out}}\)?

📚 Glossary

Advection
The transport of a substance or quantity by bulk motion of a fluid.
Mass Flux
The rate at which mass passes through a cross-section per unit time (Mass/Time).
Continuity
The principle that matter cannot be created or destroyed within a system; Flow In = Flow Out.
Turbulent Flow
Fluid motion characterized by chaotic changes in pressure and flow velocity (High Reynolds Number).
Tracer
A substance introduced into a fluid system to track motion or measure flow properties.
Exercise: Pollutant Propagation in Networks
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