Energy Balance Fundamentals

The First Law of Thermodynamics states that energy can neither be created nor destroyed; it can only change forms. In engineering, we apply this principle using an energy balance across a defined system boundary.

Closed Systems

A closed system can exchange heat and work with its surroundings, but no mass crosses its boundary. For a closed system undergoing a process from state 1 to state 2, the energy balance is:

$\Delta U + \Delta E_K + \Delta E_P = Q - W$

where:

• $\Delta U$ = change in internal energy [J]
• $\Delta E_K$ = change in kinetic energy [J]
• $\Delta E_P$ = change in potential energy [J]
• $Q$ = heat transferred to the system [J]
• $W$ = work done by the system [J]

(Note: The sign convention $Q - W$ assumes heat IN is positive and work OUT is positive. Some textbooks use $Q + W$ with work defined as work done ON the system. Always verify the convention).

Often, changes in kinetic and potential energy are negligible for closed systems, simplifying the equation to $\Delta U = Q - W$.

Open Systems (Steady-Flow)

An open system involves mass flowing across the system boundaries (e.g., pumps, turbines, heat exchangers). For steady-state open systems, energy entering the system equals energy leaving the system.

Because mass carries energy into and out of the system, we introduce Enthalpy ($H$), which combines internal energy and flow work ($U + PV$). The macroscopic steady-flow energy balance on a rate basis is:

$\Delta \dot{H} + \Delta \dot{E}_K + \Delta \dot{E}_P = \dot{Q} - \dot{W}_s$

where:

• $\Delta \dot{H}$ = difference in enthalpy flow rate between outlets and inlets [W or J/s]
• $\dot{Q}$ = rate of heat transfer to the system [W]
• $\dot{W}_s$ = rate of shaft work done by the system [W]

Worked Example

Problem: Steam flows steadily through an adiabatic turbine (no heat transfer, $\dot{Q} = 0$). The inlet enthalpy is $3200 \text{ kJ/kg}$ and the outlet enthalpy is $2400 \text{ kJ/kg}$. Changes in kinetic and potential energy are negligible. The mass flow rate of steam is $5 \text{ kg/s}$. Calculate the power output (shaft work rate) of the turbine.

Solution:

1. State the steady-flow energy balance: $\Delta \dot{H} = \dot{Q} - \dot{W}_s$
2. Note that $\dot{Q} = 0$ (adiabatic). Therefore: $\dot{W}_s = -\Delta \dot{H}$
3. Expand $\Delta \dot{H}$ using the mass flow rate $\dot{m}$ and specific enthalpy $h$: $\Delta \dot{H} = \dot{m} (h_{out} - h_{in})$
4. Calculate the work rate: $\dot{W}_s = -5 \text{ kg/s} \times (2400 \text{ kJ/kg} - 3200 \text{ kJ/kg})$ $\dot{W}_s = -5 \times (-800) = 4000 \text{ kW} = 4 \text{ MW}$

Engineering Check

In the example above, the positive sign for $\dot{W}_s$ confirms that 4 MW of power is produced by the system (the turbine), matching our chosen sign convention. If this were a compressor, $\dot{W}_s$ would be negative, indicating work must be supplied to the system.\n