Reactive Energy Balance and Steam Tables in Chemical Engineering

Reactive Energy Balances

When a chemical reaction occurs within the system, the energy balance must account for the energy released or absorbed by the reaction itself. This is known as the heat of reaction ($\Delta H_R^\circ$).

If we choose reference conditions such that the heat of reaction data is based on those same conditions, the energy balance for a steady-state process can be written as: $\sum \dot{H}_{\text{output streams}} - \sum \dot{H}_{\text{input streams}} + \alpha \Delta H_R^\circ = \dot{Q}$ In this equation, $\alpha$ represents the extent of the reaction. In a non-reactive system, the term $\alpha \Delta H_R^\circ$ simply becomes zero.

Exam Tip: Always check if the problem mentions a chemical reaction. If it does, you cannot simply use $\dot{Q} = \Delta \dot{H}$; you must include the heat of reaction term to account for the chemical energy change.

Using Steam Tables and Specific Enthalpy

To solve these balances, you need to know the energy content of the fluids. For water and steam, we use specific enthalpy ($h$), which is the enthalpy per unit mass. Because the properties of water change drastically with temperature and pressure, we rely on Steam Tables.

Saturated Steam

Saturated tables are used when water and steam are in equilibrium at a specific temperature or pressure. There are two critical states to identify:

1. Saturated Liquid ($'$): The state of water at its boiling point just before any vapor begins to form.
2. Saturated Vapour ($''$): The state of steam at its condensation point where no liquid remains.

Superheated Steam

If the steam is heated to a temperature higher than its saturation temperature for a given pressure, it enters the superheated region. In this state, the steam exists entirely as a single phase and is no longer in equilibrium with liquid water.

Worked Example: The Boiler Problem

To see these concepts in action, let's look at a common industrial scenario: a boiler.

Problem Statement: Two streams of water are mixed to form the feed to a boiler.

• Stream 1: $120\text{ kg/min}$ at $30^\circ\text{C}$
• Stream 2: $175\text{ kg/min}$ at $65^\circ\text{C}$
• The boiler operates at $17\text{ bar}$ (absolute).
• The emerging steam is saturated at the boiler pressure and exits through a $6\text{-cm}$ ID pipe.
• Calculate the required heat input ($\dot{Q}$) if we include kinetic energy.

Step 1: Mass Balance First, determine the total mass flow rate of the output. Since it is a steady-state process: $\dot{m}_{\text{out}} = \dot{m}_{\text{in,1}} + \dot{m}_{\text{in,2}} = 120 + 175 = 295\text{ kg/min}$

Step 2: Determine Specific Enthalpies ($h$) Using steam tables:

• $h_{\text{in,1}}$ (liquid at $30^\circ\text{C}$) $\approx 125.7\text{ kJ/kg}$
• $h_{\text{in,2}}$ (liquid at $65^\circ\text{C}$) $\approx 271\text{ kJ/kg}$
• $h_{\text{out}}$ (saturated steam at $17\text{ bar}$) $\approx 2793\text{ kJ/kg}$

Step 3: Enthalpy Change ($\Delta \dot{H}$) $\Delta \dot{H} = \sum \dot{m}_{\text{out}}h_{\text{out}} - \sum \dot{m}_{\text{in}}h_{\text{in}}$ $\Delta \dot{H} = (295 \times 2793) - (120 \times 125.7 + 175 \times 271) = 7.61 \times 10^5\text{ kJ/min}$

Step 4: Kinetic Energy ($\Delta \dot{E}_K$) To find the velocity ($u$) of the steam in the $6\text{-cm}$ pipe, we use the mass flow rate and the density (or specific volume $V$): $u = \frac{\dot{V}}{A} = \frac{\dot{m}V}{A}$ After calculating $u \approx 202\text{ m/s}$, we find: $\dot{E}_{K,\text{out}} = \frac{1}{2} \dot{m} u^2 \approx 6.02 \times 10^3\text{ kJ/min}$

Step 5: Final Energy Balance Assuming $\dot{W} = 0$ and $\Delta \dot{E}_P = 0$: $\dot{Q} = \Delta \dot{H} + \Delta \dot{E}_K$ $\dot{Q} = 7.61 \times 10^5 + 6.02 \times 10^3 = 7.67 \times 10^5\text{ kJ/min}$

By combining mass balances, enthalpy calculations from steam tables, and accounting for kinetic energy, we arrive at a complete energy profile for the system.