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Lesson 3: Mixing Problems

Estimated time: 30-40 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Identify the key components of a mixing problem (tank, inflow, outflow, concentration)
Set up the ODE using the rate-in minus rate-out principle
Solve mixing problems where inflow rate equals outflow rate (constant volume)
Handle mixing problems where the volume changes over time
Interpret solutions in the physical context of the problem

The General Framework

A mixing problem involves a tank containing a well-stirred solution. Liquid flows in carrying some concentration of a dissolved substance, and liquid flows out. We want to track the amount of substance in the tank over time.

Fundamental Mixing Principle: dQ/dt = (rate in) - (rate out), where Q(t) is the amount of substance at time t.

Rate in = (inflow rate) × (inflow concentration)

Rate out = (outflow rate) × (concentration in tank) = (outflow rate) × Q(t)/V(t)

The concentration in the tank at time t is Q(t)/V(t), where V(t) is the volume of liquid in the tank. If inflow rate ≠ outflow rate, V changes with time.

Constant Volume Case

When inflow rate = outflow rate = F (liters per minute), the volume V stays constant. The ODE becomes:

dQ/dt = F · c_in - F · Q/V

This is a first-order linear ODE: dQ/dt + (F/V)Q = F · c_in.

Example 1: Basic Mixing (Equal Flow Rates)

A 200-liter tank is full of pure water. Brine containing 3 g/L salt enters at 5 L/min, and the well-mixed solution drains at 5 L/min. Find Q(t), the grams of salt at time t.

Step 1: Identify components.

V = 200 L (constant), F = 5 L/min, c_in = 3 g/L, Q(0) = 0.

Step 2: Write the ODE.

dQ/dt = 5(3) - 5(Q/200) = 15 - Q/40.

Step 3: Solve. This is linear: dQ/dt + Q/40 = 15.

Integrating factor: μ = e^t/40.

d/dt[e^t/40 Q] = 15e^t/40.

e^t/40 Q = 600e^t/40 + C.

Q(t) = 600 + Ce^-t/40. With Q(0) = 0: C = -600.

Answer: Q(t) = 600(1 - e^-t/40) grams.

Check: As t → ∞, Q → 600 g, which gives concentration 600/200 = 3 g/L (matches inflow).

Example 2: Initial Salt Present

Same setup as Example 1, but the tank initially contains 100 g of salt. Find Q(t).

Solution: Same ODE but Q(0) = 100.

Q(t) = 600 + Ce^-t/40. With Q(0) = 100: 100 = 600 + C, so C = -500.

Answer: Q(t) = 600 - 500e^-t/40 grams.

Variable Volume Case

When inflow rate ≠ outflow rate, the volume changes: V(t) = V₀ + (F_in - F_out)t.

Example 3: Tank Filling (Inflow > Outflow)

A 500-liter tank initially contains 200 L of pure water. Brine with 2 g/L enters at 6 L/min. Solution drains at 4 L/min. Find Q(t) and the concentration when the tank is full.

Step 1: V(t) = 200 + 2t. Tank is full when 200 + 2t = 500, i.e., t = 150 min.

Step 2: dQ/dt = 6(2) - 4 · Q/V(t) = 12 - 4Q/(200 + 2t).

Rewrite: dQ/dt + [4/(200 + 2t)]Q = 12. This is linear with P(t) = 4/(200 + 2t) = 2/(100 + t).

Step 3: Integrating factor: μ = e^{∫2/(100+t) dt} = (100 + t)².

d/dt[(100 + t)² Q] = 12(100 + t)².

(100 + t)² Q = 4(100 + t)³ + C.

Q(t) = 4(100 + t) + C(100 + t)^-2.

Step 4: Q(0) = 0: 0 = 4(100) + C(100)^-2 ⇒ C = -400 · 10000 = -4,000,000.

Q(t) = 4(100 + t) - 4,000,000/(100 + t)².

Step 5: At t = 150: Q(150) = 4(250) - 4,000,000/250² = 1000 - 64 = 936 g.

Concentration = 936/500 = 1.872 g/L.

Strategy for Setting Up Mixing Problems

Define Q(t) = amount of substance in the tank at time t.
Compute V(t) = initial volume + (inflow rate - outflow rate) × t.
Rate in = (inflow rate) × (inflow concentration).
Rate out = (outflow rate) × Q(t)/V(t).
Write the ODE: dQ/dt = rate in - rate out.
Solve using the integrating factor method (it is always linear!).
Apply the initial condition Q(0) = initial amount of substance.

Flushing a Contaminant

Example 4: Removing Pollution

A 1000-liter tank contains water with 50 g of pollutant. Fresh water (0 g/L) enters at 10 L/min and well-mixed solution exits at 10 L/min. How long until the pollutant drops to 5 g?

ODE: dQ/dt = 10(0) - 10(Q/1000) = -Q/100.

Solution: Q(t) = 50e^-t/100.

Find t: 5 = 50e^-t/100 ⇒ e^-t/100 = 0.1 ⇒ t = -100 ln(0.1) = 100 ln(10) ≈ 230.3 minutes.

Check Your Understanding

1. A 100-L tank of pure water receives brine at 4 g/L, flowing in at 2 L/min with outflow at 2 L/min. What is the ODE for Q(t)?

Answer: dQ/dt = 2(4) - 2(Q/100) = 8 - Q/50. This is linear: dQ/dt + Q/50 = 8.

2. For the problem above, what is the steady-state amount of salt?

Answer: At steady state, dQ/dt = 0, so Q/50 = 8, giving Q = 400 g. The concentration approaches 400/100 = 4 g/L (matching inflow concentration).

3. In a mixing problem, if the inflow rate is 3 L/min and outflow is 5 L/min starting with 400 L, what is V(t) and when is the tank empty?

Answer: V(t) = 400 + (3 - 5)t = 400 - 2t. Tank is empty when 400 - 2t = 0, i.e., t = 200 minutes.

Key Takeaways

The mixing ODE is always dQ/dt = rate in - rate out, where rate out = (outflow rate) × Q/V.
When inflow = outflow, volume is constant and the ODE is first-order linear with constant coefficients.
When inflow ≠ outflow, V(t) changes and the integrating factor involves variable coefficients.
The steady-state concentration always equals the inflow concentration (when volume is constant).
Always assume the solution is well-stirred (instantaneous mixing throughout the tank).

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