A tank holds a well-stirred solution, with a stream flowing in at one concentration and the same volume flowing out. The amount of salt y then obeys y′ = (rate in) − (rate out)·y/V, which is y′ = −k·(y − A) with the steady amount A equal to the inflow concentration times the volume. The lab gives the exact Euler column and the rounded closed form, and shows that the tank approaches A in the manner of a cooling cup and does not get there.
Core ideas
Three ideas: the balance that gives the equation, the steady amount it points at, and why the tank never gets there.
The salt in a tank changes by what the inflow brings minus what the outflow carries away. The inflow brings (flow)·(concentration) grams per minute. The outflow carries the same flow times the tank's current concentration y/V, so what leaves depends on how much is already there.
Setting rate in equal to rate out gives the amount at which y′ = 0. For equal flows that is the inflow concentration times the volume, A = c·V. Writing the equation as y′ = −(flow/V)·(y − A) puts it in the form of the last lesson.
The gap y − A is multiplied by 1 − kh at every Euler step and shrinks like e^(−kt) in the solution. Neither is ever zero, so the tank's concentration gets as close to the inflow concentration as you like and stays under it.
The material
The balance written once in general and once for a tank of one hundred litres, the steady amount, and the exact steps beside the closed form.
A tank holds 100 litres of water. Salt water at 2 grams per litre runs in at 5 litres per minute, and well-stirred solution runs out at the same 5 litres per minute, so the volume stays at 100. Let y be the grams of salt in the tank. The question the equation answers is how y changes.
A tank of constant volume V with a stream of concentration c flowing in at the flow rate r, and the same flow r leaving. The well-stirred assumption is that the concentration inside is y/V everywhere at once.
The balance is y′ = r·c − r·y/V: grams in per minute, minus grams out per minute.
The second line is the first rewritten by taking out −1/20: 10 − y/20 = −(1/20)·(y − 200). It is Newton's law with k = 1/20 and a steady amount A = 200 grams, and it is the equation of “Newton's Law of Cooling” with the salt's gap y − 200 in the place of the coffee's gap y − 20. The steady amount is where the rates balance: 10 = y/20 at y = 200, which is c·V = 2·100.
With the tank starting as pure water, y(0) = 0, the solution is y = 200 + (0 − 200)·e^(−t/20) = 200·(1 − e^(−t/20)). The lab steps it with h = 1 minute. The factor is 1 − 1/20 = 19/20, and the first two steps are y₁ = 200·(1 − 19/20) = 10 and y₂ = 200·(1 − 361/400) = 39/2. After ten steps the exact value is a fraction with a long denominator, which the lab prints in full beside ≈ 78.6939, the rounded closed form at t = 10.
The tank's concentration is y/V, which starts at 0 and has the inflow's concentration 2 as its limit. Does it get there? The gap 200·e^(−t/20) is positive at every time, so y is under 200 at every time, and the exact sequence has the same property, because (19/20)ⁿ is positive for every n. The gap halves every ln 2/(1/20) ≈ 13.8629 minutes, which is quick, and still never zero.
Fill the tank with 50 grams of salt and run pure water through it at 10 litres per minute into 100 litres: y′ = −y/10, steady amount 0. The factor with h = 1 is 9/10, so the exact amounts are 50, 45, 81/2, 729/20, …. They fall towards 0 and no step is 0. The third preset is a tank that already holds 100 grams and is fed by a stream whose steady amount is 300: y′ = 6 − y/50, which is 5 litres per minute at 1.2 grams per litre into 250 litres. It rises towards 300 and stays under it.
What the equation assumes is part of the problem. The volume does not change; the tank is perfectly mixed; the inflow concentration is constant. If the outflow differed from the inflow the volume would change, and V would become a function of time in the equation. That is a different problem and not one this course solves.
Interactive
A geometric sequence beside an exponential
Method
Five moves. The balance in the first two is where a mixing problem is won or lost.
Flow times inflow concentration, in grams per minute. Check the units before anything else.
Flow times the tank's current concentration, r·y/V. It contains y, which is what makes the equation differential.
Factor out the coefficient of y: y′ = −(r/V)·(y − A) with A = c·V. Read k and A off it.
y = A + (y₀ − A)·e^(−kt) rounded, and yₙ = A + (y₀ − A)·(1 − kh)ⁿ exact. Check the closed form by differentiating.
Constant volume, perfect stirring, a constant inflow. The steady amount is approached and not reached.
Flow 5 litres per minute both ways, inflow 2 grams per litre, volume 100 litres, an empty start. Exact until the closed-form line.
The gap to 200 is 200 grams at the start, and it is multiplied by 19/20 at every step. It halves, in the solution, every ≈ 13.8629 minutes, the half-life tile of this lab.
The concentration in the tank is y/100, and the inflow's concentration of 2 is its limit. The tank is below that limit at every time, in the exact sequence and in the closed form.
You should be able to write rate in minus rate out, rewrite it as −k·(y − A), read A as the inflow concentration times the volume, step it exactly, state the closed form, and say that the steady amount is approached and not reached.
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