Septic Tank Unsteady-State Organic-Material Balance: 04-BS-13, May 2016, Q3

Question 2 (alternative): Septic Tank Unsteady-State Organic-Material Balance (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Tank capacity	100 000 L
Initial fill	75% → $V_0=75\,000$ L
Initial organic material	$M_0=15\,000$ kg
Water inflow (no organic material)	$F_{in}=20\,000$ L/h
Suspension outflow	$F_{out}=15\,000$ L/h
Suspension density	1 kg/L (constant)

Find. (a) organic material mass $M(3\text{ h})$; (b) time to fill the tank, and the organic-material concentration at that time.

Approach. The tank fills because $F_{in}>F_{out}$, so an unsteady total balance first gives $V(t)$. An unsteady species balance on organic material (incoming water carries none; outgoing suspension leaves at the tank's own, well-mixed concentration) then gives a separable ODE for $M(t)$, solved once $V(t)$ is known.

Figure 3. Holding tank filling faster than it drains (net +5000 L/h); well-mixed outflow carries the tank's own instantaneous concentration.

Total (volume) balance. $\dfrac{dV}{dt}=F_{in}-F_{out}=20\,000-15\,000=5000$ L/h, so $$V(t)=V_0+5000t=75\,000+5000t\ \text{L}.$$
Species (organic material) balance. Incoming water carries no organic material; outgoing suspension leaves at the tank's instantaneous, well-mixed concentration $C=M/V$: $$\frac{dM}{dt}=0-F_{out}\,C=-F_{out}\frac{M}{V(t)}.$$
Separate and integrate. $$\frac{dM}{M}=-\frac{F_{out}}{V_0+5000t}\,dt \;\Rightarrow\; \ln M=-\frac{F_{out}}{5000}\ln(V_0+5000t)+C',$$ so $M(t)=M_0\big(V(t)/V_0\big)^{-F_{out}/5000}$. With $F_{out}/5000=15\,000/5000=3$: $$\boxed{M(t)=M_0\left(\frac{V(t)}{V_0}\right)^{-3}=15\,000\left(\frac{75\,000+5000t}{75\,000}\right)^{-3}}\ \text{kg}.$$
Part (a): evaluate at $t=3$ h. $V(3)=75\,000+5000(3)=90\,000$ L, ratio $=90\,000/75\,000=1.2$: $$M(3)=15\,000(1.2)^{-3}=\boxed{8681\ \text{kg organic material}}.$$
Part (b): time to fill the tank. Tank is full at $V=100\,000$ L: $$100\,000=75\,000+5000t \;\Rightarrow\; \boxed{t_{full}=5.0\ \text{h}}.$$ At that time, ratio $=100\,000/75\,000=1.3333$: $$M(5)=15\,000(1.3333)^{-3}=\boxed{6328\ \text{kg}},\qquad C(5)=\frac{M(5)}{V(5)}=\frac{6328}{100\,000}=\boxed{0.0633\ \text{kg/L}\ (6.33\%\text{ w/w})}.$$

Quantity	Result
$V(t)$	$75\,000+5000t$ L
(a) Organic material at $t=3$ h	8681 kg
(b) Time to fill	5.0 h
(b) Organic material / concentration at $t_{full}$	6328 kg / 0.0633 kg/L (6.33%)

Topic: Unsteady-state (variable-volume) mass balance with a well-mixed outflow.

Key relations: total balance $dV/dt=F_{in}-F_{out}$; species balance $dM/dt=F_{in}C_{in}-F_{out}(M/V)$, integrable in closed form as a power-law decay once $V(t)$ is linear in $t$.

Why this works: because the outflow concentration equals the tank's own (well-mixed) concentration rather than a fixed feed value, the species balance becomes a variable-coefficient linear ODE in $M$; substituting the already-solved $V(t)$ converts it to a separable equation with a clean power-law solution $M\propto V^{-F_{out}/(F_{in}-F_{out})}$.

Common pitfall: treating this as a fixed-volume CSTR and using $dM/dt=-F_{out}M/V_0$ with constant $V_0$ — the volume is explicitly changing here, and ignoring that growth understates how quickly the concentration (and hence the removal rate) falls as the tank dilutes.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 6 (unsteady-state balances, variable-volume systems).

Question 3 of 9: Septic Tank Unsteady-State Organic-Material Balance

Question 2 (alternative): Septic Tank Unsteady-State Organic-Material Balance (20 marks)