8.2 Unsteady-State Mass Balances

The Nature of Transient Systems

In contrast to steady-state operations, unsteady-state (or transient) systems exhibit process variables that change with time. Transient conditions are typical in batch processes, during startup and shutdown sequences, and when responding to process upsets or setpoint changes. For transient systems, the accumulation term in the general material balance equation cannot be neglected, meaning the rate of accumulation of mass within the system boundary is equal to the net rate of mass entering and leaving the system plus generation minus consumption. For a non-reactive system, this transient mass balance is written as:

$\frac{dm_{sys}}{dt} = \sum \dot{m}_{in} - \sum \dot{m}_{out}$

where $m_{sys}$ is the total mass within the system boundary at any time $t$, and $\dot{m}{in}$ and $\dot{m}{out}$ are the mass flow rates entering and leaving the system, respectively. If the system contains multiple components, a transient balance must be written for each individual species. For a component $i$ with mass fraction $w_i$, the transient species balance is:

$\frac{d(m_{sys} w_i)}{dt} = \sum \dot{m}_{in} w_{i,in} - \sum \dot{m}_{out} w_{i,out}$

To solve unsteady-state material balances, you must formulate the differential equation, separate the variables, and integrate over the specified time interval using the initial conditions of the system.

Liquid Level in Draining and Leaking Tanks

A classic FE Chemical exam problem involves modeling the liquid level in a storage vessel as it fills, drains, or leaks. Consider a tank with a constant cross-sectional area $A$ containing a liquid of constant density $\rho$. The volume of liquid in the tank is $V(t) = A h(t)$, where $h(t)$ is the liquid height. The mass of liquid is $m_{sys} = \rho A h$.

Substituting this into the transient mass balance gives:

$\frac{d(\rho A h)}{dt} = \rho F_{in} - \rho F_{out}$

where $F_{in}$ and $F_{out}$ are the volumetric flow rates entering and leaving the tank, respectively. Since density $\rho$ and area $A$ are constant, we can simplify this to a volumetric balance:

$A \frac{dh}{dt} = F_{in} - F_{out}$

If the tank is draining under the force of gravity through a leak or an orifice at the bottom, the outlet velocity is described by Torricelli's Law, which is derived from the Bernoulli equation. The outlet volumetric flow rate $F_{out}$ is:

$F_{out} = C_d A_o \sqrt{2gh}$

where $C_d$ is the discharge coefficient (typically between 0.60 and 0.98), $A_o$ is the cross-sectional area of the orifice, and $g$ is the acceleration due to gravity. The governing differential equation for the liquid height becomes:

$A \frac{dh}{dt} = F_{in} - C_d A_o \sqrt{2gh}$

This is a non-linear first-order ordinary differential equation. If there is no inlet flow ($F_{in} = 0$), the equation is separable and can be integrated directly to find the time required to drain the tank from an initial height $h_0$ to a final height $h_f$:

$\int_{h_0}^{h_f} \frac{dh}{\sqrt{h}} = -\frac{C_d A_o \sqrt{2g}}{A} \int_0^t dt$ $2(\sqrt{h_f} - \sqrt{h_0}) = -\frac{C_d A_o \sqrt{2g}}{A} t$

Rearranging this expression allows you to calculate the elapsed time $t$:

$t = \frac{2A}{C_d A_o \sqrt{2g}} (\sqrt{h_0} - \sqrt{h_f})$

Concentration Dynamics in Well-Mixed Dilution Vessels

Another high-yield transient problem is the dilution or mixing tank. Consider a well-mixed vessel of constant volume $V$ containing a solute of concentration $C(t)$. A liquid stream enters the vessel at a volumetric flow rate $F$ and solute concentration $C_{in}$, and an outlet stream leaves at the same flow rate $F$ to maintain constant volume ($F_{in} = F_{out} = F$).

Because the vessel is well-mixed, the solute concentration in the outlet stream is equal to the concentration inside the tank ($C_{out} = C$). The species balance for the solute is:

$\frac{d(V C)}{dt} = F C_{in} - F C$

Since $V$ is constant, this simplifies to:

$V \frac{dC}{dt} = F (C_{in} - C) \implies \frac{dC}{dt} + \frac{F}{V} C = \frac{F}{V} C_{in}$

This is a first-order linear ordinary differential equation. Defining the residence time or space time as $\tau = V/F$, we can write:

$\frac{dC}{dt} + \frac{1}{\tau} C = \frac{1}{\tau} C_{in}$

Integrating this equation from an initial concentration $C(0) = C_0$ yields the concentration as a function of time:

$C(t) = C_{in} + (C_0 - C_{in}) e^{-t/\tau}$

This equation shows that as $t \to \infty$, the concentration exponentially approaches the inlet concentration $C_{in}$. The quantity $e^{-t/\tau}$ represents the fraction of the initial fluid remaining in the vessel at time $t$.

For systems where the inlet and outlet flow rates are different ($F_{in} \neq F_{out}$), the volume is also a function of time: $V(t) = V_0 + (F_{in} - F_{out})t$. In this case, the product rule must be used to expand the accumulation term:

$\frac{d(V C)}{dt} = V \frac{dC}{dt} + C \frac{dV}{dt} = F_{in} C_{in} - F_{out} C$

Substitute $\frac{dV}{dt} = F_{in} - F_{out}$ into the equation and rearrange to isolate $\frac{dC}{dt}$:

$V \frac{dC}{dt} + C(F_{in} - F_{out}) = F_{in} C_{in} - F_{out} C \implies V(t) \frac{dC}{dt} = F_{in}(C_{in} - C)$

This must be solved by separating variables and integrating using the time-dependent volume function $V(t)$.

The table below compares the concentration dynamics equations for constant-volume and variable-volume well-mixed vessels:

Worked Example: Emptying a Leaking Storage Tank

A cylindrical storage tank with a diameter of $2.0 \text{ m}$ (cross-sectional area $A = 3.1416 \text{ m}^2$) is initially filled with water to a height of $4.0 \text{ m}$. A drain hole with a diameter of $0.05 \text{ m}$ ($A_o = 0.001963 \text{ m}^2$) at the bottom of the tank is opened. Assuming the discharge coefficient is $C_d = 0.60$ and the acceleration due to gravity is $g = 9.81 \text{ m/s}^2$, calculate the time (in minutes) required for the water level to drop from $4.0 \text{ m}$ to $1.0 \text{ m}$.

Solution:

We start with the derived integration formula for a draining tank with no inlet flow:

$t = \frac{2A}{C_d A_o \sqrt{2g}} (\sqrt{h_0} - \sqrt{h_f})$

First, calculate the constant coefficient:

$\text{Coefficient} = \frac{2(3.1416)}{(0.60)(0.001963) \sqrt{2(9.81)}} = \frac{6.2832}{(0.001178) (4.4294)} = \frac{6.2832}{0.005218} = 1,204.1 \text{ s/m}^{1/2}$

Next, substitute the initial and final heights:

$\sqrt{h_0} - \sqrt{h_f} = \sqrt{4.0} - \sqrt{1.0} = 2.0 - 1.0 = 1.0 \text{ m}^{1/2}$

Multiply the coefficient by the height difference:

$t = 1,204.1 \text{ s/m}^{1/2} \times 1.0 \text{ m}^{1/2} = 1,204.1 \text{ seconds}$

Convert the time into minutes:

$t = \frac{1,204.1 \text{ seconds}}{60 \text{ seconds/minute}} = 20.1 \text{ minutes}$

Therefore, it will take approximately $20.1$ minutes for the liquid level in the tank to drop from $4.0 \text{ m}$ to $1.0 \text{ m}$.