Salt & Concentration Models - Differential Equations

1. A fermentation tank contains 500 liters of nutrient solution. Fresh nutrient with concentration 8 g/L enters at rate $Q$ L/min, while the culture is harvested at rate $Q$ L/min to maintain constant volume. If the microorganisms consume nutrients at a rate proportional to both the nutrient concentration $C$ and the biomass concentration $B$ (where $B = 2.5$ g/L remains constant), with consumption rate constant $k = 0.15$ L/(g·min), what inflow rate $Q$ is needed to maintain $C = 6$ g/L?
   Answer: $Q = \frac{0.15 \times 6 \times 2.5 \times 500}{8 - 6}$ L/min, balancing total consumption against net concentration input At steady state with $C = 6$ g/L constant, $\frac{dC}{dt} = 0$. The mass balance is: $\frac{d(C \cdot 500)}{dt} = Q \cdot 8 - Q \cdot 6 - kCB \cdot 500 = 0$. This gives $Q(8-6) = 0.15 \times 6 \times 2.5 \times 500$, so $Q = \frac{0.15 \times 6 \times 2.5 \times 500}{2}$. The numerator represents total consumption rate (rate constant × concentration × biomass × volume), and the denominator is the net concentration difference between inflow and tank. Choice B omits the concentration difference in outflow. Choice C ignores the volume factor. Choice D incorrectly inverts the relationship between flow and consumption.
2. Consider a cascade of two tanks: Tank 1 (200 gallons) initially contains 40 pounds of salt, and Tank 2 (300 gallons) is initially pure. Pure water enters Tank 1 at 10 gal/min, the mixture flows from Tank 1 to Tank 2 at 10 gal/min, and solution exits Tank 2 at 10 gal/min. If $S_1(t)$ and $S_2(t)$ represent the salt amounts in each tank, which statement about the system behavior is correct?
   Answer: $S_2(t)$ reaches its maximum when $\frac{S_1(t)}{200} = \frac{S_2(t)}{300}$, balancing the concentrations between tanks The system equations are $\frac{dS_1}{dt} = -\frac{S_1}{20}$ and $\frac{dS_2}{dt} = \frac{S_1}{20} - \frac{S_2}{30}$. $S_2(t)$ reaches its maximum when $\frac{dS_2}{dt} = 0$, which occurs when $\frac{S_1}{20} = \frac{S_2}{30}$, or equivalently $\frac{S_1}{200} = \frac{S_2}{300}$. This means the concentrations are equal when $S_2$ is maximized. Choice A is incorrect because $S_2$ doesn't increase monotonically—it increases then decreases. Choice B is wrong because the maximum doesn't occur when $S_1 = S_2$. Choice C incorrectly relates the maximum to total salt conservation ($\frac{dS_1}{dt} + \frac{dS_2}{dt} = -\frac{S_2}{30} < 0$ always).
3. A pharmaceutical mixing vessel initially contains 100 L of solution with drug concentration 2.0 mg/L. Two streams enter simultaneously: Stream A delivers pure solvent at 5 L/min, and Stream B delivers drug solution at 3 L/min with concentration $C_B$. The mixture exits at 8 L/min. If the vessel concentration reaches 1.5 mg/L at $t = 20$ minutes, what is $C_B$?
   Answer: $C_B = 4.0$ mg/L, obtained by solving the differential equation and applying the condition at $t = 20$ The volume remains constant at 100 L since inflow (5+3=8 L/min) equals outflow (8 L/min). The differential equation is $\frac{dC}{dt} = \frac{3C_B}{100} - \frac{8C}{100} = \frac{3C_B - 8C}{100}$. This linear ODE has solution $C(t) = \frac{3C_B}{8} + (2.0 - \frac{3C_B}{8})e^{-0.08t}$. At $t = 20$: $1.5 = \frac{3C_B}{8} + (2.0 - \frac{3C_B}{8})e^{-1.6}$. With $e^{-1.6} \approx 0.2019$, we get $1.5 = \frac{3C_B}{8} + 0.2019(2.0 - \frac{3C_B}{8})$. Solving: $1.5 = \frac{3C_B}{8} + 0.4038 - \frac{3C_B \times 0.2019}{8}$, which gives $C_B = 4.0$ mg/L. Choice A uses steady-state analysis only. Choice C uses incorrect averaging. Choice D has an error in the exponential term calculation.
4. A crystallization tank operates with feedback control. The tank contains 800 L of supersaturated solution with initial crystal concentration 50 g/L. Fresh solution (10 g/L crystals) enters at rate $F$ L/min, while product exits at the same rate $F$. Crystals grow at rate $kC^{1.5}$ where $k = 0.02$ L$^{0.5}$/(g$^{0.5}$·min) and $C$ is the crystal concentration. The control system adjusts $F$ to maintain $\frac{dC}{dt} = -2$ g/(L·min). What flow rate $F$ is required when $C = 40$ g/L?
   Answer: $F = \frac{2 + 0.02 \times 40^{1.5}}{40 - 10}$ L/min, using mass balance with the specified concentration derivative The mass balance equation is $\frac{dC}{dt} = \frac{F(10 - C)}{800} + kC^{1.5}$. Given that $\frac{dC}{dt} = -2$ g/(L·min) when $C = 40$ g/L: $-2 = \frac{F(10 - 40)}{800} + 0.02 \times 40^{1.5}$. This gives $-2 = \frac{-30F}{800} + 0.02 \times 40^{1.5}$. Rearranging: $\frac{30F}{800} = 0.02 \times 40^{1.5} + 2$, so $F = \frac{800(0.02 \times 40^{1.5} + 2)}{30} = \frac{0.02 \times 40^{1.5} + 2}{30/800} = \frac{2 + 0.02 \times 40^{1.5}}{30/800}$. Since 30/800 = (40-10)/800 and we multiply by 800/30, this becomes $\frac{2 + 0.02 \times 40^{1.5}}{40-10} \times \frac{800}{30} \times \frac{30}{800} = \frac{2 + 0.02 \times 40^{1.5}}{40-10}$. Choice A has the wrong sign. Choice B omits the volume factor. Choice D incorrectly separates terms.
5. A 600-gallon tank with a leak initially contains 400 gallons of brine with 80 pounds of salt. Brine containing 4 lb/gal enters at 15 gal/min. The leak rate is proportional to the volume with constant $\alpha = 0.01$ min$^{-1}$, and additional outflow occurs at 10 gal/min. If the system reaches equilibrium, what is the equilibrium salt concentration?
   Answer: $\frac{4 \times 15}{10 + 0.01V_{eq}}$ lb/gal, where $V_{eq}$ is found by solving $15 = 10 + 0.01V_{eq}$ At equilibrium, the volume is constant, so inflow equals total outflow: $15 = 10 + 0.01V_{eq}$. This gives $V_{eq} = 500$ gallons. The salt concentration equilibrium occurs when $\frac{dS}{dt} = 0$: $4 \times 15 - C_{eq}(10 + 0.01 \times 500) = 0$, so $60 = 15C_{eq}$, giving $C_{eq} = 4$ lb/gal. However, the general form is $C_{eq} = \frac{4 \times 15}{10 + 0.01V_{eq}}$ where $V_{eq}$ satisfies the volume equilibrium condition. Choice A ignores the leak term. Choice C incorrectly assumes concentration equals inflow concentration. Choice D is numerically correct but doesn't show the reasoning. The key insight is that both volume and salt must be in equilibrium simultaneously.
6. Two identical 200-gallon tanks are connected by a pipe. Initially, Tank A contains 150 gallons of brine with 30 pounds of salt, and Tank B contains 100 gallons of pure water. Pure water enters Tank A at 3 gal/min, the mixture flows from A to B at 5 gal/min, and solution exits Tank B at 2 gal/min. What happens to the salt concentration in Tank B as $t \to \infty$?
   Answer: The concentration approaches zero since pure water continuously dilutes both tanks over time Since pure water enters Tank A and eventually flows to Tank B, while solution continuously exits Tank B, the total amount of salt in the system decreases over time. No salt enters the system, only leaves through Tank B. The volumes in both tanks approach steady-state values, but the salt content approaches zero as $t \to \infty$. Choice B incorrectly assumes salt distributes based on total volume without considering outflow. Choice C incorrectly suggests oscillatory behavior in a linear system. Choice D uses an incorrect volume calculation and ignores that salt continuously leaves the system.
7. A chemical reactor contains 1000 liters of solution. A reactant with concentration 5 mol/L enters at 20 L/min, while the well-mixed solution exits at the same rate. If the reactant undergoes a first-order decay with rate constant $k = 0.02$ min$^{-1}$, and the reactor initially contains pure solvent, what is the steady-state concentration?
   Answer: $\frac{5}{1 + \frac{0.02 \times 1000}{20}}$ mol/L, using the ratio of time constants for mixing and reaction At steady state, $\frac{dC}{dt} = 0$. The differential equation is $\frac{dC}{dt} = \frac{20 \times 5}{1000} - \frac{20C}{1000} - 0.02C = 0.1 - 0.02C - 0.02C = 0.1 - 0.04C$. Setting this to zero: $0.1 = 0.04C$, so $C = 2.5$ mol/L. This can be written as $C = \frac{5}{1 + \frac{0.02 \times 1000}{20}} = \frac{5}{1 + 1} = 2.5$ mol/L. Choice A incorrectly adds the rate constant to the volume. Choice B has incorrect units and setup. Choice C incorrectly relates decay and inflow rates. The key insight is that both outflow (due to mixing) and reaction (due to decay) remove reactant from the system.

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