2.3 Unsteady-State Mass and Energy Balances

Unsteady-State Mass and Energy Balances

While steady-state operations dominate bulk continuous chemical manufacturing, unsteady-state (transient) processes are ubiquitous during plant startups, emergency shutdowns, batch reactor cycles, tank draining, vessel blowdowns, and process control upsets. On the NCEES PE Chemical exam, transient questions test your ability to formulate differential balance equations, separate variables, and integrate using proper boundary conditions.

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1. General Macroscopic Conservation Balance

For any conserved physical quantity $\Psi$ (mass, component mass, or energy) within a defined spatial control volume:

$\frac{d\Psi_{sys}}{dt} = \dot{\Psi}_{in} - \dot{\Psi}_{out} + \dot{\Psi}_{gen} - \dot{\Psi}_{cons}$

Transient Total Mass Balance

Because total mass cannot be created or destroyed ($\dot{m}{gen} = \dot{m}{cons} = 0$):

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

For an open liquid tank of constant surface area $A$ and constant liquid density $\rho$, $m_{sys} = \rho A h(t)$, where $h(t)$ is liquid level: $\rho A \frac{dh}{dt} = \sum \dot{m}_{in} - \sum \dot{m}_{out} \implies A \frac{dh}{dt} = \sum \dot{V}_{in} - \sum \dot{V}_{out}$

Transient Component (Species) Balance

For chemical species $i$ participating in reactions with volumetric rate $r_i$ ($\text{mol/(m}^3\cdot\text{s)}$):

$\frac{d n_i}{dt} = \sum_{in} \dot{n}_{i,in} - \sum_{out} \dot{n}_{i,out} + r_i V$

For a non-reactive, perfectly mixed continuous stirred tank of constant volume $V$ and volumetric throughput $\dot{V}$ ($n_i = V C_i(t)$ and $C_{i,out}(t) = C_i(t)$):

$V \frac{dC_i}{dt} = \dot{V} C_{i,in} - \dot{V} C_i(t)$

Dividing by $\dot{V}$ defines the hydraulic residence time (time constant $\tau = V / \dot{V}$): $\tau \frac{dC_i}{dt} + C_i(t) = C_{i,in}$

Integrating from initial concentration $C_0$ at $t = 0$ with a step change in inlet concentration to $C_{in}$ yields the classic first-order response: $C_i(t) = C_{in} + (C_0 - C_{in}) e^{-t / \tau} = C_{in}\left(1 - e^{-t / \tau}\right) + C_0 e^{-t / \tau}$

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2. Transient First Law Energy Balances

The general First Law of Thermodynamics for an unsteady open system is:

$\frac{d E_{sys}}{dt} = \dot{Q} - \dot{W}_s + \sum_{in} \dot{m}_{in} \left( \hat{h} + \frac{u^2}{2} + gz \right)_{in} - \sum_{out} \dot{m}_{out} \left( \hat{h} + \frac{u^2}{2} + gz \right)_{out}$

Neglecting kinetic and potential energies, total system energy equals internal energy ($E_{sys} = U_{sys} = m_{sys} \hat{u}_{sys}$):

$\frac{d(m_{sys} \hat{u}_{sys})}{dt} = \dot{Q} - \dot{W}_s + \sum_{in} \dot{m}_{in} \hat{h}_{in} - \sum_{out} \dot{m}_{out} \hat{h}_{out}$

The Fundamental Internal Energy vs. Enthalpy Asymmetry

Notice the profound thermodynamic distinction in the governing equation:

• Inside the control volume, energy is stored as internal energy ($\hat{u}$).
• Across the control surface boundaries, flowing streams transport enthalpy ($\hat{h} = \hat{u} + P\hat{v}$).

Why? Because fluid entering or exiting the boundary performs boundary work (flow work $P\hat{v}$) on or by the system fluid. Omitting this distinction is the single most pervasive source of error in transient thermodynamics.

Applying the product rule to the accumulation derivative: $\frac{d(m \hat{u})}{dt} = m \frac{d\hat{u}}{dt} + \hat{u} \frac{dm}{dt}$ For liquids and solids, $\hat{u} \approx \hat{h} \approx C_p(T - T_{ref})$ because specific volume is small. For gases, $\hat{u} = C_v(T - T_{ref})$ and $\hat{h} = C_p(T - T_{ref})$.

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3. Jacketed Batch and Semi-Batch Heating / Cooling

Consider a well-insulated, closed batch reactor of liquid mass $m$ and heat capacity $C_p$ being heated by an external jacket maintained at constant temperature $T_j$:

$\dot{m}_{in} = 0, \quad \dot{m}_{out} = 0, \quad \dot{W}_s \approx 0 \text{ (ignoring agitator friction)}$ $\frac{d(m \hat{u})}{dt} = m C_p \frac{dT}{dt} = \dot{Q}$

Heat transfer from the jacket follows Newton's Law of Cooling: $\dot{Q} = U A (T_j - T(t))$, where $U$ is the overall heat transfer coefficient and $A$ is heat transfer surface area:

$m C_p \frac{dT}{dt} = U A (T_j - T(t))$

Rearranging into standard first-order differential form: $\frac{dT}{dt} + \frac{U A}{m C_p} T(t) = \frac{U A}{m C_p} T_j$

Defining the thermal time constant: $\tau_{th} = \frac{m C_p}{U A}$

Separating variables from $t = 0$ ($T = T_0$) to time $t$: $\int_{T_0}^{T(t)} \frac{dT}{T_j - T} = \frac{1}{\tau_{th}} \int_0^t dt \implies -\ln\left(\frac{T_j - T(t)}{T_j - T_0}\right) = \frac{t}{\tau_{th}}$

Solving explicitly for vessel temperature $T(t)$:

$T(t) = T_j - (T_j - T_0) \exp\left(-\frac{t}{\tau_{th}}\right)$

The time $t$ required to reach a specific process temperature $T(t)$ is: $t = \tau_{th} \ln\left( \frac{T_j - T_0}{T_j - T(t)} \right) = \frac{m C_p}{U A} \ln\left( \frac{T_j - T_0}{T_j - T(t)} \right)$

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4. Gas Vessel Pressurization: The Evacuated Tank Paradox

Consider a rigid, thermally insulated tank of volume $V$ that is initially evacuated ($m_1 = 0$, $P_1 = 0$). A supply valve connected to a high-pressure header carrying an ideal gas at temperature $T_0$ is opened, filling the tank until flow stops. What is the final temperature $T_2$ in the tank?

Setting up the transient First Law balance:

• Rigid tank: $\dot{W}_s = 0$
• Insulated: $\dot{Q} = 0$
• No exit stream: $\dot{m}_{out} = 0$

$\frac{d(m \hat{u})}{dt} = \dot{m}_{in} \hat{h}_{in}$

Integrating from initial state ($m=0$) to final state ($m_2$): $m_2 \hat{u}_2 - 0 = \Delta m_{in} \hat{h}_{in} = m_2 \hat{h}_{in}$

Dividing by $m_2$: $\hat{u}_2 = \hat{h}_{in}$

For an ideal gas referenced to absolute zero ($T_{ref} = 0\text{ K}$), $\hat{u}2 = C_v T_2$ and $\hat{h}{in} = C_p T_0$:

$C_v T_2 = C_p T_0 \implies T_2 = \frac{C_p}{C_v} T_0 = \gamma T_0$

Where $\gamma = C_p / C_v$ is the heat capacity ratio ($1.40$ for diatomic gases such as $\text{N}_2$, $\text{O}_2$, and air).

Key Insight: If room-temperature air at $25^\circ\text{C}$ ($298.15\text{ K}$) fills an evacuated tank adiabatically, the final temperature inside the tank spikes to: $T_2 = 1.40 \times 298.15\text{ K} = 417.4\text{ K} = 144.3^\circ\text{C}!$ The gas heats up drastically without any external heat addition because the incoming fluid carries flowing enthalpy $P\hat{v}$ (work done by the supply header), which is converted entirely into internal thermal energy inside the rigid vessel.

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5. Gravity Draining Hydraulics: Torricelli's Law

When a liquid-filled vertical cylindrical tank drains by gravity through an open bottom orifice of area $A_o$, the discharge velocity is non-linear and governed by Torricelli's Law:

$u_{out}(t) = C_d \sqrt{2 g h(t)}$

Where $C_d$ is the discharge coefficient ($0.60$ to $0.62$ for sharp-edged orifices) and $h(t)$ is instantaneous liquid head above the orifice.

Setting up the transient total mass balance ($A \frac{dh}{dt} = -\dot{V}_{out}$):

$A \frac{dh}{dt} = -C_d A_o \sqrt{2 g h}$

Separating variables: $h^{-1/2} dh = -\frac{C_d A_o \sqrt{2g}}{A} dt$

Integrating from initial height $h_0$ at $t = 0$ to height $h_f$ at time $t$: $\int_{h_0}^{h_f} h^{-1/2} dh = 2\left(\sqrt{h_f} - \sqrt{h_0}\right) = -\frac{C_d A_o \sqrt{2g}}{A} t$

Solving for draining time $t$:

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

To drain the tank completely to $h_f = 0$: $t_{empty} = \frac{2 A \sqrt{h_0}}{C_d A_o \sqrt{2g}}$ Notice that draining from $h_0$ to $0$ takes exactly twice as long as it would if the initial discharge rate had remained constant!

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6. Summary: Canonical Transient Balance Archetypes

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7. Worked Numerical Example: Jacketed Polymerization Reactor Heating

Problem Statement

A batch chemical reactor holds $m = 2,500\text{ kg}$ of a liquid monomer mixture ($C_p = 3.500\text{ kJ/(kg}\cdot\text{K)}$) at an initial temperature of $T_0 = 20.0^\circ\text{C}$. To initiate polymerization, the charge must be preheated to $T(t) = 110.0^\circ\text{C}$.

Heating is provided by condensing saturated steam inside a reactor jacket at $P_{sat} = 0.3614\text{ MPa}$, maintaining a constant inner wall temperature of $T_j = 140.0^\circ\text{C}$.

Vessel Specifications:

• Effective heat transfer area: $A = 5.20\text{ m}^2$
• Overall heat transfer coefficient: $U = 450.0\text{ W/(m}^2\cdot\text{K)}$
• Latent heat of steam condensation at $140.0^\circ\text{C}$: $\lambda_{vap} = 2,144.0\text{ kJ/kg}$

Calculate:

1. The thermal time constant of the reactor system ($\tau_{th}$).
2. The time required to reach the target temperature of $110.0^\circ\text{C}$ (in minutes).
3. The total thermal energy transferred to the batch ($\text{MJ}$).
4. The total mass of steam condensed during the heat-up cycle ($\text{kg}$).

Step-by-Step Solution

Step 1: Calculate the Thermal Time Constant

$\tau_{th} = \frac{m C_p}{U A}$ Convert $U$ and $C_p$ to consistent SI units:

• $m C_p = (2,500\text{ kg}) \times (3,500\text{ J/(kg}\cdot\text{K)}) = 8,750,000\text{ J/K}$
• $U A = (450.0\text{ W/(m}^2\cdot\text{K)}) \times (5.20\text{ m}^2) = 2,340.0\text{ W} = 2,340.0\text{ J/s}\cdot\text{K}$

$\tau_{th} = \frac{8,750,000\text{ J/K}}{2,340.0\text{ J/s}\cdot\text{K}} = 3,739.32\text{ seconds}$ In minutes: $\tau_{th} = 3,739.32 / 60 = 62.32\text{ minutes}$.

Step 2: Time to Reach $110.0^\circ\text{C}$

$\frac{T_j - T(t)}{T_j - T_0} = \frac{140.0 - 110.0}{140.0 - 20.0} = \frac{30.0}{120.0} = 0.250$ $t = \tau_{th} \ln\left( \frac{T_j - T_0}{T_j - T(t)} \right) = 3,739.32\text{ s} \times \ln\left( \frac{1}{0.250} \right) = 3,739.32 \times \ln(4.000)$ $t = 3,739.32\text{ s} \times 1.38629 = 5,183.80\text{ seconds}$ Converting to minutes: $t = \frac{5,183.80\text{ s}}{60\text{ s/min}} = 86.40\text{ minutes} \approx 1.44\text{ hours}$

Step 3: Total Heat Transferred to Monomer

$Q_{total} = m C_p (T_f - T_0) = (2,500\text{ kg}) \times (3.500\text{ kJ/(kg}\cdot\text{K)}) \times (110.0 - 20.0\text{ K})$ $Q_{total} = (2,500) \times (3.500) \times (90.0) = 787,500\text{ kJ} = 787.5\text{ MJ}$

Step 4: Total Steam Condensed

Assuming steam enters as saturated vapor and leaves as saturated condensate without subcooling: $m_{steam} = \frac{Q_{total}}{\lambda_{vap}} = \frac{787,500\text{ kJ}}{2,144.0\text{ kJ/kg}} = 367.30\text{ kg of steam}$ Average steam condensation rate over the heat-up cycle: $\dot{m}_{steam,avg} = \frac{367.30\text{ kg}}{1.44\text{ h}} = 255.1\text{ kg/h}$

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8. Common PE Exam Traps & Calculation Pitfalls

Trap 1: Confusing Flowing Enthalpy with System Internal Energy
For a closed batch vessel or a variable-mass tank, the accumulation term is $d(m\hat{u})/dt$, not $d(m\hat{h})/dt$. Writing $m C_p dT/dt$ for an ideal gas vessel results in an error of $\gamma = 1.4$, because the closed mass accumulation is governed by $C_v$, whereas boundary flow is governed by $C_p$.

Trap 2: Assuming Constant Gravity Draining Flow
In gravity-drained tanks, flow rate diminishes continuously as liquid head drops ($\dot{V}_{out} \propto \sqrt{h}$). Calculating emptying time by dividing initial volume by initial volumetric flow ($t = V / \dot{V}_0$) yields an answer that is exactly $50%$ of the true draining time! Always integrate Torricelli's equation.

Trap 3: Mixing Hydraulic and Thermal Time Constants
Do not confuse hydraulic residence time $\tau = V/\dot{V}$ with thermal time constant $\tau_{th} = (m C_p)/(UA)$. For a continuous stirred tank with both flow and jacket heat exchange, the total temperature response combines both terms: $\frac{1}{\tau_{total}} = \frac{1}{\tau_{flow}} + \frac{1}{\tau_{th}}$.