Mass Balance Notes: CSTR and PFR (Environmental Pollutants)

Mass balance concepts

• Mass conservation: mass cannot be produced or destroyed; pollutant mass in a body of water changes due to inflow, outflow, and chemical production/consumption.
• If a pollutant enters a lake, its concentration may rise, implying sources or in-situ production/consumption.
• Goal: write equations to track pollutant mass and concentration over time in inflows, outflows, and in the lake.

Governing mass-balance equation (derivation, intuition)

• Start from a mass balance over a short interval t to t+Δt: M(t+\Delta t) = M(t) + M{in}(t,\Delta t) - M{out}(t,\Delta t) + M_{reaction}(t,\Delta t)
  • Mass in = mass flow in; mass out = mass flow out; reaction term = generation/consumption via chemical reactions.
• Rearranged and divided by Δt, then taking the limit Δt → 0 gives the differential form:
  \frac{dM}{dt} = \dot{M}{in} - \dot{M}{out} + \dot{M}_{reaction}
• Define flows in terms of flow rate and concentration:
  \dot{M}{in} = Q{in} C{in}, \quad \dot{M}{out} = Q{out} C{out}
• Substitute into the balance:
  \frac{dM}{dt} = Q{in} C{in} - Q{out} C{out} + \dot{M}_{reaction}
• Relationship between mass and concentration when volume V is present: M = V C
• If the volume is constant, then \frac{dM}{dt} = V \frac{dC}{dt} and the mass-balance becomes
  V \frac{dC}{dt} = Q{in} C{in} - Q{out} C{out} + \dot{M}_{reaction}
• For a well-mixed system (CSTR), the concentration inside is uniform, so C_{out} = C .

Well-mixed systems (CSTR) and constants

• Key definitions for a CSTR (continuous stirred tank reactor):
  • Inflow rate: Q{in} , Inflow concentration: C{in}
  • Outflow rate: Q{out} , Outflow concentration: C{out} (equal to the reactor concentration for a well-mixed tank)
  • Volume: V (often treated as constant in the class)
• Mass balance (well-mixed, constant V):
  V \frac{dC}{dt} = Q{in} C{in} - Q{out} C + \dot{M}{reaction}
• Reaction term for a conservative substance (no chemical transformation):
  \dot{M}_{reaction} = 0
• For a non-conservative substance with first-order decay:
  • Decay/conversion rate: \dot{M}_{reaction} = -k V C
  • Then the governing equation becomes
    V \frac{dC}{dt} = Q{in} C{in} - Q{out} C - k V C or equivalently \frac{dC}{dt} = \frac{Q{in} C{in}}{V} - \left(\frac{Q{out}}{V} + k\right) C
• Steady-state (dC/dt = 0):
  • Conservative: 0 = Q{in} C{in} - Q{out} C → with a single inflow, C = \frac{Q{in} C{in}}{Q{out}}
  • With decay (non-conservative): 0 = Q{in} C{in} - Q{out} C - k V C ⇒ C = \frac{Q{in} C{in}}{Q{out} + k V}
  • For multiple inflows, sum over inputs:
    C = \frac{\sumi Q{in,i} C{in,i}}{Q{out} + k V}
• Small summary of the big, general steady-state equation:
  \boxed{\; C = \frac{\sumi Q{in,i} C{in,i}}{Q{out} + k V} \; }

Multiple inflows (example worked)

• Problem setup (conservative substance, steady state): lake with two inflows, one pipe and one clean river.
• Given: inflows: 25 m^3/s with Cin=0 mg/L; 1 m^3/s with Cin=5 mg/L. Outflow equals total inflow for constant volume: Q_{out} = 25 + 1 = 26 \text{ m}^3\text{/s}
• Since the substance is conservative, k = 0 and steady state holds, so
  C_{out} = \frac{25 \cdot 0 + 1 \cdot 5}{26} = \frac{5}{26} \approx 0.19 \text{ mg/L}

Non-conservative substance with steady state (decay present)

• Example: single inflow, steady state, first-order decay.
• Given: Q{in}=50 \text{ m}^3/\text{d}, \; C{in}=100 \text{ mg/L}, \; Q_{out}=50 \text{ m}^3/\text{d}, \; V=500 \text{ m}^3, \; k=0.216 \text{ d}^{-1}
• Steady-state concentration in the outflow:
  C{out} = \frac{Q{in} C{in}}{Q{out} + k V} = \frac{50 \cdot 100}{50 + 0.216 \cdot 500} \approx \frac{5000}{50 + 108} \ = \frac{5000}{158} \approx 31.6 \text{ mg/L}

Non steady-state CSTR with decay (time-dependent)

• General solution (with decay) for a CSTR:
  • Define the steady-state concentration
    C{\infty} = \frac{\sumi Q{in,i} C{in,i}}{Q_{out} + k V}
  • The time-dependent concentration (well-mixed, constant V) solves to
    C(t) = C{\infty} + (C0 - C{\infty}) e^{ - \left( \frac{Q{out}}{V} + k \right) t }
• Example (pipe shut off, after the pipe was supplying pollutant):
  • Before shut-off: not needed here; after shut-off, inflow ceases so the inflow term vanishes: C_{\infty} = 0
  • Given: initial concentration C0 = 0.19 mg/L; V = 2 \times 10^5 m^3; Qout = 9 \times 10^4 m^3/yr; k = 0.12 yr^{-1}
  • The decay term total rate: \frac{Q_{out}}{V} + k = \frac{9\times 10^4}{2\times 10^5} + 0.12 = 0.45 + 0.12 = 0.57 \text{ yr}^{-1}
  • We want the time when C(t) = 0.5 C_0 = 0.095 mg/L.
  • Solve: 0.095 = 0.19 e^{-0.57 t} \Rightarrow e^{-0.57 t} = 0.5
  • So t = \frac{\ln(2)}{0.57} \approx 1.22 \text{ years}

Residence time (retention time)

• Definition: Roughly, how long the water (and pollutants) stay in a lake or reactor before leaving.
• Example numbers (illustrative): Lake Superior has very large volume and slow outflow → very long residence time (approx. 191 years in the lecture notes).
• Lake Erie has smaller volume and faster outflow → shorter residence time (approx. 2.6 years).
• Residence time for projects helps gauge cleanup times after pollution events.

Plug Flow Reactor (PFR) vs CSTR (well-mixed) — core idea

• Plug Flow Reactor (PFR): nonwell-mixed in the axial direction, but each “plug” of fluid is well mixed internally; no back-and-forth mixing between plugs.
• In environmental flows, PFR can model pipes, fast rivers, or air sheds between mountains.
• For a first-order decay in a PFR, the concentration decays along the flow path according to time spent in the system.
• PFR mass balance leads to an exponential decay with residence time T, where time through the system is T = Volume/Flow (V/q).
• Resulting PFR expression for outflow with first-order decay:
  C{out} = C{in} e^{-k T} = C_{in} e^{-k\frac{V}{Q}}
  with V = A \times L \quad (A = cross-section, L = length)
• Comparison: under many conditions, a PFR removes pollutants more efficiently than a CSTR with the same overall flow and volume credit.

Worked plug-flow example (revisited as a CSTR comparison)

• Earlier CSTR example (50 m^3/d, Cin = 100 mg/L, V = 500 m^3, k = 0.216 d^{-1}, Qin = Q_out = 50).
• Plug flow calculation for the same data:
  • C_in = 100 mg/L, V = 500 m^3, Q = 50 m^3/d, k = 0.216 d^{-1}
  • Residence time T = V / Q = 500 / 50 = 10 days (for this illustration, the lecture computed with a different time basis). The resulting C_out is lower in a PFR than in the CSTR case for the same input, illustrating that PFRs can be more efficient at pollutant removal in some setups.

Mixing at discharge: point-slice view of a mixing problem

• When an input pipe discharges into a natural stream, you can write a point-mixing balance at the discharge point:
  Q{stream} C{stream} + Q{pipe} C{pipe} = Q{down} C{down}
• If the upstream stream is clean (Cstream = 0) but the pipe brings pollutant, the mixing concentration just downstream of the discharge is C{down} = \frac{Q{pipe} C{pipe}}{Q{stream} + Q{pipe}}
• Example values from the lecture: upstream stream Qstream = 8.7 m^3/s with Cstream = 0 mg/L; pipe Qpipe = 0.9 m^3/s with Cpipe = 50 mg/L. Thus
  C_{0,down} = \frac{0.9 \times 50}{8.7 + 0.9} \approx \frac{45}{9.6} \approx 4.7 \text{ mg/L}
• Downstream concentration further downstream (50 km) requires travel time and handling decay along the path.

Travel time and decay downstream (PFR along a stream)

• Travel time for a downstream reach: if the cross-sectional area is A and length is L, then the volume is
  V = A L
• Downstream flow rate is the sum of inflows (assuming volume stays constant):
  Q{down} = Q{stream} + Q_{pipe} = 8.7 + 0.9 = 9.6 \text{ m}^3\text{/s}
• Time to travel 50 km:
  • Volume: V = A L = 10 \text{ m}^2 \times 50{,}000 \text{ m} = 5 \times 10^5 \text{ m}^3
  • Convert time scale to days or use days for decay constant; here, k = 0.2 day^{-1}
  • Time: T = \frac{V}{Q_{down}} = \frac{5 \times 10^5}{9.6 \text{ m}^3\text{/s}} \approx 5.21 \times 10^4 \text{s} \approx 0.603 \text{ days}
• Plug into PFR decay: with Cin at the downstream-mixing point Cin = 4.7 mg/L, we get
  C{out,down} = C{in} e^{-k T} = 4.7 \times e^{-0.2 \times 0.603} \approx 4.7 \times e^{-0.1206} \approx 4.17 \text{ mg/L}
• Insight: PFR travel and decay reduce concentration along the reach; the example illustrates the combined effect of mixing, travel time, and first-order decay.

Practical problem-solving steps (the class workflow)

• Step 1: Draw the system as a diagram (lake, river inflows, pipes, outflow, etc.).
• Step 2: Choose a control volume (usually the lake or the pipe in these problems).
• Step 3: Gather numerical information: flow rates, concentrations, volumes, and reaction (decay) rates.
• Step 4: Determine whether the system is well mixed or not (CSTR vs PFR assumptions).
• Step 5: Write the correct mass-balance equation for the chosen case.
• Step 6: Substitute the provided numbers and solve for the unknowns (Cout, Qout, time t, etc.).
• Step 7: If steady state, set dM/dt = 0; if transient, use the time-dependent forms and, if needed, solve for time using logs.

Common problem types covered in the lecture

• Conservative substance in a CSTR at steady state with multiple inflows: compute C_out from inflows.
• Non-conservative substance with first-order decay in a CSTR at steady state: compute C_out including decay term.
• Non-steady-state CSTR with decay: compute time to reach a target concentration using the time-dependent solution.
• Plug flow reactor (PFR) with decay: compute Cout from Cin using Cout = Cin e^{-k V/Q} or equivalently Cout = Cin e^{-k T} with T = V/Q.
• Mixing at a discharge point (two streams into a common downstream flow) to find the immediate downstream concentration.
• Residence time calculations for lakes or reactors to gauge pollutant clearance timelines.

Quick references (key formulas to memorize)

• Mass balance for a well-mixed system (CSTR), with volume V constant:
  V \frac{dC}{dt} = Q{in} C{in} - Q{out} C + \dot{M}{reaction}
• Conservative substance (no reaction): \dot{M}{reaction} = 0 \quad\Rightarrow\quad V \frac{dC}{dt} = Q{in} C{in} - Q{out} C
• Steady state for conservative case (single inflow):
  C = \frac{Q{in} C{in}}{Q_{out}}
• Steady state for non-conservative decay (single inflow):
  C = \frac{Q{in} C{in}}{Q_{out} + k V}
• Non-conservative, multiple inflows (steady state):
  C = \frac{\sumi Q{in,i} C{in,i}}{Q{out} + k V}
• Non-steady-state CSTR with decay (general):
  C(t) = C{\infty} + (C0 - C{\infty}) e^{ - (\frac{Q{out}}{V} + k) t }
  where
  C{\infty} = \frac{\sumi Q{in,i} C{in,i}}{Q_{out} + k V}
• Plug Flow Reactor (PFR) with first-order decay:
  C{out} = C{in} e^{-k T}, \quad T = \frac{V}{Q}, \quad V = A L
• Inflow mixing at a discharge into a stream: two inflows to a downstream reach
  Q{stream} C{stream} + Q{pipe} C{pipe} = Q{down} C{down}
• Downstream travel time for a reach: T = \frac{V}{Q_{down}}
• General travel time in a pipe segment: T = \frac{A L}{Q}
• Residence time examples (orders of magnitude vary by lake size and outflow): useful for intuition about cleanup timelines (e.g., Lake Superior vs Lake Erie).

Practical takeaways

• Mass balance provides a flexible framework to track pollutants, whether conservative or reactive, in well-mixed systems or along tubular paths.
• Constant-volume assumptions simplify equations; when volume changes, additional terms appear (dV/dt terms) and require more information.
• The choice between CSTR and PFR models depends on mixing: well-mixed bodies (lakes, reactors) vs. streams/pipes where axial mixing is limited.
• Steady-state solutions are often the simplest; transient problems require time-dependent solutions and sometimes exponential decay terms.
• Real-world relevance: residence times help assess how long pollutants linger in natural systems; mixing at discharge points affects immediate downstream concentrations; plug flow considerations matter for fast-flowing channels and pipes.

Quick reference problem outlines (from the lecture for practice)

• Problem 1 (Conservative, two inflows into a lake; steady-state):
  • Given: Qin1=25, Cin1=0; Qin2=1, Cin2=5; Qout=26; C_out ≈ \frac{25\cdot0 + 1\cdot5}{26} \approx 0.19 \text{ mg/L}
• Problem 2 (Non-conservative, steady-state, single inflow):
  • Given: Qin=50, Cin=100, Qout=50, V=500, k=0.216; C_out ≈ \frac{50\cdot100}{50 + 0.216\cdot 500} \approx 31.6 \text{ mg/L}
• Problem 3 (Non-steady-state CSTR with decay, pipe shut):
  • Cinfty = 0; C0 = 0.19 mg/L; V = 2 imes10^5 m^3; Q_out = 9 imes10^4 m^3/yr; k = 0.12 yr^{-1}
  • C(t) = C0 e^{ - (Q{out}/V + k) t } with Q_out/V = 0.45 yr^{-1} and total = 0.57 yr^{-1}; Solve for t when C(t) = 0.095 mg/L → t ≈ 1.22 yr
• Problem 4 (PFR: downstream concentration 50 km away):
  • Upstream mix: C_down,0 = 4.7 mg/L at discharge
  • Travel time: T ≈ 0.603 days; k = 0.2 day^{-1}
  • C_down(50 km) = 4.7 e^{-0.2×0.603} ≈ 4.17 mg/L

Note on exam expectations

• You will be asked to apply the appropriate equation to a given problem, not derive the equation on the exam.
• Practice with both well-mixed (CSTR) and plug-flow (PFR) scenarios, including conservative and non-conservative substances, steady-state and transient cases, to be ready for varied problems.