FAC1004 Tutorial 14 — Mixing Tank & Decayt Solutions
Unifying framework: All 6 problems are first-order ODEs. The two solution strategies are:
- Separable: $\dfrac{dy}{dx} = f(x)g(y) \implies \int \dfrac{dy}{g(y)} = \int f(x),dx$
- Linear (integrating factor): $\dfrac{dy}{dx} + P(x)y = Q(x) \implies \mu = e^{\int P,dx},; \dfrac{d}{dx}(\mu y) = \mu Q$
The table at the bottom maps each problem to its strategy.
Problem 1: Cylindrical Tank with Brine Solution
Setup: Tank holds $V = 1500$ L brine with $Q_0 = 90$ g salt. Pure water enters at $30$ L/s; mixed solution exits at $30$ L/s.
(a) Differential Equation
Inflow = outflow ($30 = 30$), so volume is constant at $1500$ L.
| Flow | Rate of Salt |
|---|---|
| In | $30 \cdot 0 = 0$ g/s (pure water) |
| Out | $30 \cdot \dfrac{Q(t)}{1500} = \dfrac{Q(t)}{50}$ g/s |
Balance: $$\frac{dQ}{dt} = 0 - \frac{Q}{50}$$
(a) Solution
This is separable: $$\frac{dQ}{dt} = -\frac{Q}{50} \implies \frac{dQ}{Q} = -\frac{dt}{50}$$
Integrate: $$\int \frac{dQ}{Q} = -\frac{1}{50}\int dt \implies \ln|Q| = -\frac{t}{50} + C$$
$$Q(t) = e^{-t/50 + C} = \tilde{C}e^{-t/50}$$
Apply $Q(0) = 90$: $90 = \tilde{C}e^0 = \tilde{C}$
$$\boxed{Q(t) = 90e^{-t/50}}$$
(b) After 25 Seconds
$$Q(25) = 90e^{-25/50} = 90e^{-0.5}$$
$$\boxed{Q(25) \approx 54.59\text{ g}}$$
(c) Time to Reach 35 g
$$35 = 90e^{-t/50} \implies e^{-t/50} = \frac{35}{90} \implies -\frac{t}{50} = \ln\left(\frac{35}{90}\right)$$
$$t = -50\ln\left(\frac{35}{90}\right)$$
$$\boxed{t \approx 47.23\text{ s}}$$
Problem 2: Chlorinated Container
Setup: $V = 300$ ft$^3$ container with $C_0 = 15$ m$^3$ chlorine. Mixed solution ($10$ m$^3$/ft$^3$) enters at $16$ ft$^3$/min; mixed solution exits at $16$ ft$^3$/min.
(a) Differential Equation
Volume is constant ($16 = 16$). Concentration at time $t$ is $C(t)/300$.
| Flow | Rate of Chlorine |
|---|---|
| In | $16 \cdot 10 = 160$ units/min |
| Out | $16 \cdot \dfrac{C(t)}{300} = \dfrac{4C}{75}$ units/min |
$$\frac{dC}{dt} = 160 - \frac{4C}{75}$$
(a) Solution
This is linear: $$\frac{dC}{dt} + \frac{4}{75}C = 160$$
Integrating factor: $$\mu = e^{\int \frac{4}{75},dt} = e^{4t/75}$$
Multiply through: $$e^{4t/75}\frac{dC}{dt} + \frac{4}{75}e^{4t/75}C = 160e^{4t/75}$$
$$\frac{d}{dt}\left(e^{4t/75}C\right) = 160e^{4t/75}$$
Integrate: $$e^{4t/75}C = 160 \cdot \frac{75}{4}e^{4t/75} + K = 3000e^{4t/75} + K$$
$$C(t) = 3000 + Ke^{-4t/75}$$
Apply $C(0) = 15$: $$15 = 3000 + K \implies K = -2985$$
$$\boxed{C(t) = 3000 - 2985e^{-4t/75}}$$
(b) After 60 Seconds (1 Minute)
$$C(1) = 3000 - 2985e^{-4/75}$$
$$\boxed{C(1) \approx 169.92\text{ units}}$$
Note: The steady-state limit as $t \to \infty$ is $C = 3000$ units, found by setting $\frac{dC}{dt} = 0$.
Problem 3: Sugar Bowl (Variable Volume)
Setup: $V_0 = 120$ m$^3$ bowl with $S_0 = 595$ g sugar. Solution A ($5$ g/m$^3$) enters at $3$ m$^3$/s; mixed solution exits at $2$ m$^3$/s.
Volume Function
Inflow > outflow, so volume increases: $$V(t) = 120 + (3 - 2)t = 120 + t$$
Tank overflows when $V(t) = 400$: $$120 + t = 400 \implies t = 280\text{ s}$$
Differential Equation
| Flow | Rate of Sugar |
|---|---|
| In | $3 \cdot 5 = 15$ g/s |
| Out | $2 \cdot \dfrac{S(t)}{V(t)} = \dfrac{2S}{120 + t}$ g/s |
$$\frac{dS}{dt} = 15 - \frac{2S}{120 + t}$$
Rearrange to linear form: $$\frac{dS}{dt} + \frac{2}{120 + t}S = 15$$
Solution
Integrating factor: $$\mu = e^{\int \frac{2}{120+t},dt} = e^{2\ln(120+t)} = (120 + t)^2$$
Multiply: $$(120 + t)^2\frac{dS}{dt} + 2(120 + t)S = 15(120 + t)^2$$
$$\frac{d}{dt}\left[(120 + t)^2 S\right] = 15(120 + t)^2$$
Integrate: $$(120 + t)^2 S = 15 \cdot \frac{(120 + t)^3}{3} + C = 5(120 + t)^3 + C$$
$$S(t) = 5(120 + t) + \frac{C}{(120 + t)^2}$$
Apply $S(0) = 595$: $$595 = 5(120) + \frac{C}{120^2} = 600 + \frac{C}{14400}$$
$$C = (595 - 600) \cdot 14400 = -72000$$
$$\boxed{S(t) = 5(120 + t) - \frac{72000}{(120 + t)^2}}$$
After 2 Minutes
$t = 120$ s: $$S(120) = 5(240) - \frac{72000}{240^2} = 1200 - \frac{72000}{57600} = 1200 - 1.25$$
$$\boxed{S(120) = 1198.75\text{ g}}$$
At Overflow
$t = 280$ s: $$S(280) = 5(400) - \frac{72000}{400^2} = 2000 - \frac{72000}{160000} = 2000 - 0.45$$
$$\boxed{S(280) = 1999.55\text{ g}}$$
Problem 4: Variable Outflow Variations
Part (a): Outflow $1$ m$^3$/s
Volume: $V(t) = 120 + (3 - 1)t = 120 + 2t$
Overflow when $V = 400$: $120 + 2t = 400 \implies t = 140$ s.
ODE
$$\frac{dS}{dt} = 15 - \frac{S}{120 + 2t}$$
$$\frac{dS}{dt} + \frac{1}{120 + 2t}S = 15$$
Integrating Factor
$$\mu = e^{\int \frac{1}{120+2t},dt}$$
Let $u = 120 + 2t$, $du = 2,dt$: $$\int \frac{1}{120+2t},dt = \frac{1}{2}\ln(120 + 2t)$$
$$\mu = e^{\frac{1}{2}\ln(120+2t)} = (120 + 2t)^{1/2}$$
Solve
$$\frac{d}{dt}\left[(120 + 2t)^{1/2} S\right] = 15(120 + 2t)^{1/2}$$
$$(120 + 2t)^{1/2} S = 15 \int (120 + 2t)^{1/2},dt$$
Let $u = 120 + 2t$, $du = 2,dt$: $$\int u^{1/2},dt = \frac{1}{2}\int u^{1/2},du = \frac{1}{2} \cdot \frac{2}{3}u^{3/2} = \frac{1}{3}(120 + 2t)^{3/2}$$
$$(120 + 2t)^{1/2} S = 15 \cdot \frac{1}{3}(120 + 2t)^{3/2} + C = 5(120 + 2t)^{3/2} + C$$
$$S(t) = 5(120 + 2t) + \frac{C}{(120 + 2t)^{1/2}}$$
Apply $S(0) = 595$: $$595 = 5(120) + \frac{C}{120^{1/2}} = 600 + \frac{C}{\sqrt{120}}$$
$$C = (595 - 600)\sqrt{120} = -5\sqrt{120}$$
$$S(t) = 5(120 + 2t) - \frac{5\sqrt{120}}{\sqrt{120 + 2t}}$$
After 2 Minutes ($t = 120$)
$$S(120) = 5(360) - \frac{5\sqrt{120}}{\sqrt{360}} = 1800 - 5\sqrt{\frac{120}{360}} = 1800 - 5\sqrt{\frac{1}{3}}$$
$$\boxed{S(120) = 1800 - \frac{5}{\sqrt{3}} \approx 1797.11\text{ g}}$$
At Overflow ($t = 140$)
$$S(140) = 5(400) - \frac{5\sqrt{120}}{\sqrt{400}} = 2000 - \frac{5\sqrt{120}}{20} = 2000 - \frac{\sqrt{120}}{4}$$
$$\boxed{S(140) = 2000 - \frac{\sqrt{30}}{2} \approx 1997.26\text{ g}}$$
Part (b): Outflow $6$ m$^3$/s
Volume: $V(t) = 120 + (3 - 6)t = 120 - 3t$
Tank empties when $V = 0$: $120 - 3t = 0 \implies t = 40$ s.
ODE
$$\frac{dS}{dt} = 15 - \frac{6S}{120 - 3t}$$
$$\frac{dS}{dt} + \frac{6}{120 - 3t}S = 15$$
Integrating Factor
$$\mu = e^{\int \frac{6}{120-3t},dt}$$
Let $u = 120 - 3t$, $du = -3,dt$: $$\int \frac{6}{120-3t},dt = \int \frac{6}{u}\cdot\frac{du}{-3} = -2\int\frac{du}{u} = -2\ln(120 - 3t)$$
$$\mu = e^{-2\ln(120-3t)} = (120 - 3t)^{-2}$$
Solve
$$\frac{d}{dt}\left[(120 - 3t)^{-2} S\right] = 15(120 - 3t)^{-2}$$
$$(120 - 3t)^{-2} S = 15 \int (120 - 3t)^{-2},dt$$
Let $u = 120 - 3t$, $du = -3,dt$: $$\int u^{-2},dt = -\frac{1}{3}\int u^{-2},du = -\frac{1}{3}(-u^{-1}) = \frac{1}{3u} = \frac{1}{3(120 - 3t)}$$
$$(120 - 3t)^{-2} S = 15 \cdot \frac{1}{3(120 - 3t)} + C = \frac{5}{120 - 3t} + C$$
$$S(t) = 5(120 - 3t) + C(120 - 3t)^2$$
Apply $S(0) = 595$: $$595 = 5(120) + C(120)^2 = 600 + 14400C$$
$$C = \frac{595 - 600}{14400} = -\frac{5}{14400} = -\frac{1}{2880}$$
$$S(t) = 5(120 - 3t) - \frac{(120 - 3t)^2}{2880}$$
After 20 Seconds ($t = 20$)
$$V(20) = 120 - 60 = 60\text{ m}^3$$
$$S(20) = 5(60) - \frac{60^2}{2880} = 300 - \frac{3600}{2880} = 300 - 1.25$$
$$\boxed{S(20) = 298.75\text{ g}}$$
When Tank Empties ($t = 40$)
$$S(40) = 5(0) - \frac{0^2}{2880} = 0$$
$$\boxed{S(40) = 0\text{ g}}$$
Problem 5: Mercury-194 Decay
Setup: $\dfrac{dP}{dt} = -0.728P$, $P(0) = P_0$, initial amount $650$ g.
(a) General Solution
This is separable: $$\frac{dP}{P} = -0.728,dt$$
$$\ln P = -0.728t + C \implies P(t) = P_0e^{-0.728t}$$
where $k = 0.728$ hr$^{-1}$ is the decay constant.
(b) After 50 Minutes
Convert to hours: $50$ min $= \frac{50}{60} = \frac{5}{6}$ hr.
$$P\left(\frac{5}{6}\right) = 650e^{-0.728 \times 5/6} = 650e^{-0.6067}$$
$$\boxed{P \approx 354.24\text{ g}}$$
(c) Half-Life
$$t_{1/2} = \frac{\ln 2}{k} = \frac{\ln 2}{0.728}$$
$$\boxed{t_{1/2} \approx 0.952\text{ hr} \approx 57.1\text{ min}}$$
Problem 6: Plutonium-241 Decay
Setup: $P(0) = 400$ mg. After 2 years, 15% has decayed ($85%$ remains).
(a) Model
General decay: $P(t) = P_0e^{-kt} = 400e^{-kt}$
After 2 years, $P(2) = 0.85 \times 400 = 340$ mg:
$$340 = 400e^{-2k} \implies e^{-2k} = 0.85$$
$$-2k = \ln(0.85) \implies k = -\frac{\ln(0.85)}{2}$$
$$k \approx 0.08125\text{ yr}^{-1}$$
$$\boxed{P(t) = 400e^{-0.08125t}}$$
(b) After 6 Years
$$P(6) = 400e^{-0.08125 \times 6} = 400e^{-0.4875}$$
$$\boxed{P(6) \approx 245.68\text{ mg}}$$
(c) Half-Life
$$t_{1/2} = \frac{\ln 2}{k} = \frac{\ln 2}{0.08125}$$
$$\boxed{t_{1/2} \approx 8.53\text{ years}}$$
Strategy Reference
| # | Problem Type | Volume | ODE Form | Solution Method | Key Formula |
|---|---|---|---|---|---|
| 1 | Constant volume, pure water in | $V = 1500$ | $\dfrac{dQ}{dt} = -\dfrac{r}{V}Q$ | Separable | $Q(t) = Q_0e^{-rt/V}$ |
| 2 | Constant volume, solution in | $V = 300$ | $\dfrac{dC}{dt} = rC_{\text{in}} - \dfrac{r}{V}C$ | Linear IF | $C(t) = C_{\infty} + (C_0 - C_{\infty})e^{-rt/V}$ |
| 3 | Variable volume (in $>$ out) | $V(t) = V_0 + \Delta r \cdot t$ | $\dfrac{dS}{dt} = r_{\text{in}}c_{\text{in}} - \dfrac{r_{\text{out}}}{V(t)}S$ | Linear IF | $\mu = V(t)^{r_{\text{out}}/\Delta r}$ |
| 4a | Variable volume (slow outflow) | $V(t) = 120 + 2t$ | Same as #3 | Linear IF | $\mu = \sqrt{V(t)}$ |
| 4b | Variable volume (fast outflow) | $V(t) = 120 - 3t$ | Same as #3 | Linear IF | $\mu = V(t)^{-2}$ |
| 5 | Exponential decay (k given) | N/A | $\dfrac{dP}{dt} = -kP$ | Separable | $P(t) = P_0e^{-kt}$ |
| 6 | Exponential decay (data point) | N/A | $\dfrac{dP}{dt} = -kP$ | Separable | Find $k$ from $P(t_1) = P_1$ |
Decay vs. Mixing: Spotting the Pattern
- Decay problems (P5, P6): Rate is proportional to amount — always separable, always exponential.
- Constant-volume mixing (P1, P2): Rate = (constant inflow) − (proportional outflow). P1 has zero inflow so it's also separable; P2 is linear.
- Variable-volume mixing (P3, P4): Rate = (constant inflow) − (proportional outflow with time-dependent coefficient). Always linear IF, with $\mu = V(t)^{\text{outflow}/\Delta r}$.
Common Mistakes to Avoid
| Mistake | Why It's Wrong |
|---|---|
| Treating P4(a) like P3 | P3 has $\mu = V(t)^2$; P4(a) has $\mu = V(t)^{1/2}$ because outflow rate differs |
| Forgetting the minus sign in the IF exponent for P4(b) | Outflow $>$ inflow makes $\Delta r$ negative, which flips the IF exponent sign |
| Mixing up minutes, hours, and seconds | P5 units are hours but (b) asks for minutes — always convert first |
| Using the wrong $C$ sign | In linear IF solutions, $C$ can be positive or negative depending on initial vs. steady state |