Steady-State Energy Balances

Concept

An energy balance is an application of the First Law of Thermodynamics, which states that energy cannot be created or destroyed. For continuous chemical processes, we analyze open systems (control volumes) where mass flows across boundaries. At steady state, the total energy entering the system equals the total energy leaving. For most process equipment (like heaters, coolers, compressors), changes in kinetic and potential energy are negligible compared to thermal and mechanical work terms. Thus, the energy balance revolves around Enthalpy ( $H$ ), a property combining internal energy and flow work.

Formula & Method

The general steady-state energy balance for an open system with one inlet and one outlet is:

$\Delta H + \Delta E_k + \Delta E_p = Q - W_s$

Where $\Delta$ represents (Output - Input). Neglecting kinetic ( $\Delta E_k$ ) and potential ( $\Delta E_p$ ) energy changes, the equation simplifies to:

$\Delta H = Q - W_s$ $\dot{m}(h_{out} - h_{in}) = \dot{Q} - \dot{W}_s$

For heat exchangers or vessels without moving parts, shaft work $W_s = 0$ , leading to: $\dot{Q} = \Delta \dot{H}$

Variables & Units

$\Delta H, \Delta \dot{H}$ = Change in enthalpy, or rate of enthalpy change (Joules, kJ/h, Watts).
$h$ = Specific enthalpy of the fluid stream (kJ/kg or J/mol).
$Q, \dot{Q}$ = Heat added to the system (or rate of heat transfer). Positive if heat enters.
$W_s, \dot{W}_s$ = Shaft work done by the system on the surroundings. Positive if system does work (e.g., turbine), negative if work is done on the system (e.g., pump, compressor).
$\dot{m}$ = Mass flow rate (kg/s).

Worked Example

Problem: Water flows continuously through a boiler. The feed water enters at $10 \text{ kg/s}$ with a specific enthalpy of $200 \text{ kJ/kg}$ . Steam leaves the boiler with a specific enthalpy of $2800 \text{ kJ/kg}$ . The boiler does no shaft work, and kinetic/potential energy changes are negligible. Calculate the rate of heat input ( $\dot{Q}$ ) required in kW.

Calculation:

Identify the variables: $\dot{m} = 10 \text{ kg/s}$ , $h_{in} = 200 \text{ kJ/kg}$ , $h_{out} = 2800 \text{ kJ/kg}$ , $\dot{W}_s = 0$ .
Apply the simplified steady-state energy balance: $\dot{Q} = \dot{m}(h_{out} - h_{in})$
Substitute the values: $\dot{Q} = 10 \text{ kg/s} \times (2800 - 200) \text{ kJ/kg}$ $\dot{Q} = 10 \times 2600 = 26,000 \text{ kJ/s}$
Since $1 \text{ kJ/s} = 1 \text{ kW}$ : $\dot{Q} = 26,000 \text{ kW} = 26 \text{ MW}$

Engineering Meaning

The steady-state energy balance allows engineers to determine the utility requirements of a plant. By calculating $\dot{Q}$ , engineers can size the necessary boilers, furnaces, or cooling towers. By calculating $\dot{W}_s$ , they can determine the electrical power required for compressors and pumps, or the power generated by turbines. Enthalpy is the critical bridge connecting mass flow to thermal requirements.

Engineering Check

Always carefully define the sign convention for heat and work. A common error is mixing up work done on the system vs. work done by the system. Ensure that the specific enthalpy values used are referenced to the same thermodynamic datum state. When a phase change occurs (e.g., boiling), remember that $h$ incorporates the latent heat of vaporization implicitly if pulled from a steam table.

Explicit Exclusions

This article covers non-reactive energy balances. It explicitly excludes energy balances on reactive systems (heats of reaction, heats of formation), transient energy balances, and entropy/availability (exergy) analysis.