8.4 Unsteady-State Energy Balances

Dynamic Energy Balances

Unsteady-state or transient energy balances describe processes where temperature, pressure, or internal energy change with time. These models are crucial for analyzing the heat-up or cool-down of vessels, the startup and shutdown of heat exchangers, and the dynamic behavior of chemical reactors. The general transient energy balance is formulated from the First Law of Thermodynamics. For an open system, it is written as:

$$\frac{dU_{sys}}{dt} = \sum \dot{m}_{in} \hat{H}_{in} - \sum \dot{m}_{out} \hat{H}_{out} + \dot{Q} - \dot{W}_s$$

For liquid-phase systems, we assume that the difference between internal energy and enthalpy is negligible ($U \approx H$) and that the liquid is incompressible with a constant heat capacity $C_p$. The accumulation term becomes $\frac{dU_{sys}}{dt} \approx \frac{dH_{sys}}{dt} = \frac{d(m_{sys} \hat{H}_{sys})}{dt}$. Utilizing the definition of specific enthalpy relative to a reference temperature, we can write the accumulation in terms of temperature:

$$\frac{d(m_{sys} C_p T)}{dt} = m_{sys} C_p \frac{dT}{dt} + C_p T \frac{dm_{sys}}{dt}$$

For closed systems or systems with constant mass ($m_{sys} = m$), the mass derivative is zero, and the transient energy balance simplifies to:

$$m C_p \frac{dT}{dt} = \dot{Q} - \dot{W}_s$$

Unsteady Heat-Up or Cool-Down of Batch Vessels

Consider a closed, well-mixed batch vessel containing a mass $m$ of liquid with heat capacity $C_p$. The liquid is heated by a utility fluid (such as steam) circulating through a heating jacket or internal coil. The utility fluid is assumed to remain at a constant temperature $T_j$ (e.g., condensing steam). The rate of heat transfer $\dot{Q}$ from the jacket to the vessel liquid is given by:

$$\dot{Q} = U A (T_j - T)$$

where $U$ is the overall heat transfer coefficient, $A$ is the heat transfer area, and $T(t)$ is the liquid temperature in the vessel. Substituting this rate into the closed-system energy balance (with $\dot{W}_s = 0$) yields:

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

To solve this differential equation, separate the variables $T$ and $t$:

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

Integrating from the initial liquid temperature $T(0) = T_0$ to the temperature $T(t)$ at time $t$ gives:

$$-\ln\left(\frac{T_j - T(t)}{T_j - T_0}\right) = \frac{U A}{m C_p} t \implies \frac{T_j - T(t)}{T_j - T_0} = e^{-\frac{U A}{m C_p} t}$$

Rearranging this expression gives the liquid temperature as a function of time:

$$T(t) = T_j + (T_0 - T_j) e^{-t/\tau}$$

where $\tau = \frac{m C_p}{U A}$ is defined as the thermal time constant of the system. The thermal time constant represents the time required for the temperature to change by $63.2\%$ of the difference between its initial value and the jacket temperature. A smaller time constant (resulting from a larger heat transfer coefficient, larger area, or smaller mass) leads to a faster thermal response.

Unsteady Energy Balances in Jacketed Reacting Systems

In a jacketed reactor, the energy balance must account for the heat of reaction ($\Delta H_{rxn}$). The reaction rate $r$ ($\text{mol}/(\text{L} \cdot \text{s})$) inside a reactor of volume $V$ generates or consumes heat. The rate of heat generation is:

$$\dot{Q}_{gen} = (-\Delta H_{rxn}) r V$$

Note that for an exothermic reaction, $\Delta H_{rxn}$ is negative, so $-\Delta H_{rxn}$ is positive, meaning heat is generated. For an endothermic reaction, heat is consumed. Incorporating heat generation and heat removal via a jacket into a batch reactor energy balance yields:

$$m C_p \frac{dT}{dt} = U A (T_j - T) + (-\Delta H_{rxn}) r V$$

This equation is coupled to the material balance because the reaction rate $r$ depends on the reactant concentrations, which decrease over time. Solving these coupled equations typically requires numerical methods. However, the FE Chemical exam often tests qualitative understandings of thermal runaway. Thermal runaway occurs in exothermic reactors when the rate of heat generation increases exponentially with temperature (following the Arrhenius rate law, $k = A e^{-E_a/RT}$), while the rate of heat removal via the jacket increases only linearly with temperature ($UA(T - T_j)$). If heat generation exceeds the maximum possible heat removal, the reactor temperature escalates rapidly, leading to dangerous overpressurization or equipment failure.

Continuous Transient Systems (CSTR Startup)

For a continuous stirred-tank reactor (CSTR) or a continuous heating tank during startup, fluid enters and leaves the vessel continuously. The transient energy balance must include flow enthalpy streams:

$$V \rho C_p \frac{dT}{dt} = F \rho C_p (T_{in} - T) + U A (T_j - T)$$

where $V$ is the tank volume, $F$ is the volumetric flow rate, and $T_{in}$ is the inlet stream temperature. Dividing by $V \rho C_p$ and rearranging yields:

$$\frac{dT}{dt} + \left(\frac{F}{V} + \frac{U A}{V \rho C_p}\right) T = \frac{F}{V} T_{in} + \frac{U A}{V \rho C_p} T_j$$

This is a first-order linear ODE with an effective time constant $\tau_{eff} = \left(\frac{1}{\tau_{flow}} + \frac{1}{\tau_{thermal}}\right)^{-1}$ where $\tau_{flow} = V/F$ and $\tau_{thermal} = \frac{V \rho C_p}{U A}$.

Worked Example: Steam-Jacketed Batch Heating

A well-insulated batch heating vessel contains $500 \text{ kg}$ of a liquid reactant (heat capacity $C_p = 3.50 \text{ kJ/(kg}\cdot^\circ C)$) initially at $20^\circ C$. The vessel is heated by steam condensing in a jacket at $120^\circ C$. The heat transfer area is $2.50 \text{ m}^2$, and the overall heat transfer coefficient is $400 \text{ W/(m}^2\cdot^\circ C)$ ($0.400 \text{ kW/(m}^2\cdot^\circ C)$). Calculate the time (in minutes) required to heat the liquid from $20^\circ C$ to $80^\circ C$.

Solution:

Identify the process parameters:

• $m = 500 \text{ kg}$
• $C_p = 3.50 \text{ kJ/(kg}\cdot^\circ C)$
• $T_0 = 20^\circ C$
• $T(t) = 80^\circ C$
• $T_j = 120^\circ C$
• $U = 0.400 \text{ kW/(m}^2\cdot^\circ C)$
• $A = 2.50 \text{ m}^2$

Calculate the product of $U$ and $A$:

$$U A = 0.400 \text{ kW/(m}^2\cdot^\circ C) \times 2.50 \text{ m}^2 = 1.00 \text{ kW/}^\circ C = 1.00 \text{ kJ/(s}\cdot^\circ C)$$

Calculate the thermal capacity of the liquid batch:

$$m C_p = 500 \text{ kg} \times 3.50 \text{ kJ/(kg}\cdot^\circ C) = 1,750 \text{ kJ/}^\circ C$$

Calculate the thermal time constant $\tau$:

$$\tau = \frac{m C_p}{U A} = \frac{1,750 \text{ kJ/}^\circ C}{1.00 \text{ kJ/(s}\cdot^\circ C)} = 1,750 \text{ seconds}$$

Next, use the analytical solution for temperature over time:

$$\frac{T_j - T(t)}{T_j - T_0} = e^{-t/\tau}$$ $$\frac{120 - 80}{120 - 20} = e^{-t/1750} \implies \frac{40}{100} = 0.40 = e^{-t/1750}$$

Take the natural logarithm of both sides:

$$\ln(0.40) = -\frac{t}{1,750} \implies -0.9163 = -\frac{t}{1,750}$$

Solve for time $t$ in seconds:

$$t = 0.9163 \times 1,750 = 1,603.5 \text{ seconds}$$

Convert the time to minutes:

$$t = \frac{1,603.5 \text{ seconds}}{60 \text{ seconds/minute}} = 26.7 \text{ minutes}$$

Thus, it takes $26.7$ minutes to heat the liquid from $20^\circ C$ to $80^\circ C$.