8.3 Steady-State Energy Balances

Thermodynamic Principles and the First Law

Material balances track mass flows, whereas energy balances track the heat, work, and internal energy flows within a system. The thermodynamic foundation of energy balances is the First Law of Thermodynamics, which expresses the conservation of energy. For a closed system (no mass flow crossing the system boundary), the First Law is written as:

$$\Delta U + \Delta \text{KE} + \Delta \text{PE} = Q - W$$

where $\Delta U$ is the change in internal energy, $\Delta \text{KE}$ is the change in kinetic energy, $\Delta \text{PE}$ is the change in potential energy, $Q$ is the heat added to the system, and $W$ is the work done by the system. For most chemical process units, changes in kinetic and potential energy are negligible, simplifying the closed system balance to $\Delta U = Q - W$.

For an open system (control volume) operating at steady state, mass enters and leaves the system continuously. In this case, we must account for the work required to push fluid into and out of the system (flow work). Enthalpy ($H = U + PV$) is defined to combine internal energy and flow work. The steady-state open-system energy balance is:

$$\Delta \dot{H} + \Delta \dot{\text{KE}} + \Delta \dot{\text{PE}} = \dot{Q} - \dot{W}_s$$

where $\dot{Q}$ is the rate of heat input, $\dot{W}_s$ is the shaft work rate (work associated with pumps, turbines, or compressors), and $\Delta \dot{H}$ is the net rate of enthalpy flow leaving the system minus that entering. Neglecting kinetic and potential energy changes, the balance simplifies to the standard form found in the FE Reference Handbook:

$$\dot{Q} - \dot{W}_s = \sum \dot{m}_{out} \hat{H}_{out} - \sum \dot{m}_{in} \hat{H}_{in}$$

where $\hat{H}$ is the specific enthalpy (energy per unit mass). To evaluate these enthalpies, you must establish a reference state (e.g., liquid water at $0.01^\circ C$ and $1 \text{ bar}$) where specific enthalpy is defined as zero.

Enthalpy Changes in Non-Reactive Systems

For systems without chemical reactions, enthalpy changes occur due to temperature changes (sensible heat) or phase changes (latent heat). The change in specific enthalpy for a substance heated from $T_1$ to $T_2$ without phase change is calculated using its constant-pressure heat capacity ($C_p$):

$$\Delta \hat{H} = \int_{T_1}^{T_2} C_p(T) dT$$

If the heat capacity is constant or an average value $\bar{C}_p$ is used, this simplifies to $\Delta \hat{H} = \bar{C}_p(T_2 - T_1)$. When a phase change occurs (such as vaporization or melting), you must add the latent heat of vaporization ($\Delta H_v$) or latent heat of fusion ($\Delta H_m$) at the transition temperature:

$$\Delta \hat{H} = \int_{T_1}^{T_{transition}} C_{p,phase1} dT + \Delta H_{phase\_change} + \int_{T_{transition}}^{T_2} C_{p,phase2} dT$$

Thermochemistry and Heats of Reaction

When chemical reactions occur, chemical bonds are broken and formed, releasing or absorbing energy. The heat of reaction ($\Delta H_{rxn}$) is the enthalpy change when stoichiometric quantities of reactants react completely at a specified temperature and pressure. The standard heat of reaction at reference conditions ($T_{ref} = 298.15 \text{ K}$, $P = 1 \text{ bar}$) is calculated using standard heats of formation ($\Delta H_{f,i}^\circ$):

$$\Delta H_{rxn}^\circ = \sum \nu_i \Delta H_{f,i}^\circ = \sum \nu_j \Delta H_{f,j}^\circ \text{ (products)} - \sum \nu_k \Delta H_{f,k}^\circ \text{ (reactants)}$$

where $\nu_i$ is the stoichiometric coefficient of species $i$ (positive for products, negative for reactants). Standard heats of formation for common substances are tabulated in the chemistry and thermodynamics sections of the FE Reference Handbook. Note that the standard heat of formation of any elemental substance in its most stable physical state (e.g., $O_2(g)$, $N_2(g)$, $C(s)$) is zero.

To find the heat of reaction at a temperature $T$ other than the reference temperature, Kirchhoff's Law is applied:

$$\Delta H_{rxn}(T) = \Delta H_{rxn}^\circ(T_{ref}) + \int_{T_{ref}}^T \Delta C_p dT$$

where $\Delta C_p = \sum \nu_i C_{p,i}$. For a reactive system, the energy balance is evaluated by calculating the heat of reaction at a reference temperature and adding the sensible heat changes of the reactants and products relative to that reference temperature.

Combustion Calculations

Combustion is the rapid reaction of a fuel with oxygen to produce carbon dioxide, water, and heat. Combustion calculations are a major focus of the FE Chemical exam and require key definitions:

• Theoretical Oxygen (Stoichiometric $O_2$): The exact amount of oxygen required for complete combustion of all carbon to $CO_2$, hydrogen to $H_2 O$, and sulfur to $SO_2$.
• Excess Air: The amount of air fed to the burner in excess of the theoretical requirement. It is calculated as:

$$\% \text{ Excess Air} = \frac{(O_2)_{fed} - (O_2)_{stoich}}{(O_2)_{stoich}} \times 100\%$$

Since air is modeled as $21 \text{ mol\% } O_2$ and $79 \text{ mol\% } N_2$, the moles of nitrogen entering with air is $N_2 = O_2 \times (79/21) = 3.76 \times O_2$.

• Wet-Basis vs. Dry-Basis: Dry-basis composition lists the mole fractions of the flue gas components excluding water vapor. Wet-basis composition includes all components, including water vapor. Be sure to convert between wet and dry bases as specified in the exam problem.

Worked Example: Combustion Energy Balance

Propane gas ($C_3 H_8$) is burned completely with $20.0\%$ excess air. Both the propane and the air enter the burner at $25^\circ C$ ($298 \text{ K}$). The combustion products leave the chamber at $500^\circ C$ ($773 \text{ K}$). Calculate the heat removed from the burner ($\dot{Q}$, in $\text{kJ}$) per mole of propane burned. The standard heat of combustion of propane at $25^\circ C$ (with water as a gas) is $\Delta H_c^\circ = -2,044 \text{ kJ/mol}$. The average heat capacities ($C_p$) between $25^\circ C$ and $500^\circ C$ are: $C_{p, CO2} = 0.045 \text{ kJ/(mol}\cdot^\circ C)$, $C_{p, H2O} = 0.036 \text{ kJ/(mol}\cdot^\circ C)$, $C_{p, O2} = 0.032 \text{ kJ/(mol}\cdot^\circ C)$, and $C_{p, N2} = 0.030 \text{ kJ/(mol}\cdot^\circ C)$.

Solution:

First, write the balanced stoichiometric equation for propane combustion:

$$C_3 H_8 (g) + 5 O_2 (g) \to 3 CO_2 (g) + 4 H_2 O (g)$$

Based on $1.0 \text{ mole}$ of propane burned:

• Stoichiometric $O_2$ required = $5.0 \text{ mol}$
• Actual $O_2$ fed = $5.0 \times 1.20 = 6.0 \text{ mol}$ (representing $20\%$ excess)
• Actual $N_2$ fed = $6.0 \times (79/21) = 22.57 \text{ mol}$

The product stream composition is:

• $CO_2 = 3.0 \text{ mol}$
• $H_2 O = 4.0 \text{ mol}$
• $O_2 \text{ remaining} = 6.0 - 5.0 = 1.0 \text{ mol}$
• $N_2 \text{ remaining} = 22.57 \text{ mol}$

Since the reactants enter at the reference temperature ($25^\circ C$), their inlet enthalpy is zero. The heat removed $\dot{Q}$ is the sum of the heat of reaction at $25^\circ C$ and the sensible heat change of the products as they are heated from $25^\circ C$ to $500^\circ C$ ($\Delta T = 475^\circ C$):

$$\dot{Q} = \Delta H_c^\circ + \sum n_i C_{p,i} \Delta T$$

Calculate the sum of $n_i C_{p,i}$ for the products:

$$\sum n_i C_{p,i} = (3.0)(0.045) + (4.0)(0.036) + (1.0)(0.032) + (22.57)(0.030)$$ $$\sum n_i C_{p,i} = 0.135 + 0.144 + 0.032 + 0.6771 = 0.9881 \text{ kJ/}^\circ C$$

Now, calculate the sensible heat change:

$$\Delta H_{sensible} = (0.9881 \text{ kJ/}^\circ C)(475^\circ C) = 469.35 \text{ kJ}$$

Finally, calculate the heat removed:

$$\dot{Q} = -2,044 \text{ kJ} + 469.35 \text{ kJ} = -1,574.65 \text{ kJ}$$

Thus, approximately $1,575 \text{ kJ}$ of heat is released (removed) per mole of propane burned.