5.2 Energy analysis of steady-flow systems

Energy balance for steady-flow systems

General energy balance equation

The energy balance for a steady-flow system comes directly from the first law of thermodynamics combined with conservation of mass. It tracks every way energy can enter or leave a control volume: heat, work, and the energy carried by the fluid itself.

The general form is:

$\dot{Q} + \sum \dot{m}_i \left( h_i + \frac{V_i^2}{2} + g z_i \right) = \dot{W} + \sum \dot{m}_e \left( h_e + \frac{V_e^2}{2} + g z_e \right)$

where:

• $\dot{Q}$ = net rate of heat transfer into the system
• $\dot{W}$ = net rate of work done by the system
• $\dot{m}$ = mass flow rate
• $h$ = specific enthalpy
• $V$ = velocity of the fluid
• $g$ = gravitational acceleration
• $z$ = elevation above a reference datum

The left side represents everything coming in (heat plus the total energy carried by each inlet stream). The right side represents everything going out (work plus the total energy carried by each exit stream). If you rearrange this, you're really just saying "energy in = energy out" for a system at steady state, where nothing accumulates inside the control volume over time.

Simplifications and assumptions

Because the system is at steady state, the mass flow rate in equals the mass flow rate out: $\dot{m}_i = \dot{m}_e$. For a single-inlet, single-exit device, you can factor out $\dot{m}$ and work on a per-unit-mass basis:

$\dot{Q} - \dot{W} = \dot{m} \left[ (h_e - h_i) + \frac{V_e^2 - V_i^2}{2} + g(z_e - z_i) \right]$

From here, you simplify further depending on the device:

• Negligible kinetic and potential energy changes: Drop the $\frac{V^2}{2}$ and $gz$ terms. This is valid for most turbines, compressors, and heat exchangers where velocity and elevation changes are small compared to enthalpy changes.
• Adiabatic conditions: Set $\dot{Q} = 0$. Applies when the device is well-insulated or the fluid passes through too quickly for significant heat transfer (nozzles, turbines, throttling valves).
• No work interaction: Set $\dot{W} = 0$. Applies to nozzles, diffusers, heat exchangers, and throttling valves since they have no shaft, electrical, or other work output.

Knowing which terms to drop is half the battle on exam problems. Always justify your simplifications based on the physics of the device.

Energy transfer in steady-flow processes

Forms of energy transfer

Heat transfer ($\dot{Q}$) occurs whenever there's a temperature difference between the system and its surroundings. By convention in most thermodynamics texts, heat into the system is positive. A heat exchanger is the classic example: the whole point of the device is to transfer thermal energy between two fluid streams.

Work ($\dot{W}$) is energy transfer driven by a force through a displacement. In steady-flow systems, the most common type is shaft work, which shows up in turbines (work out) and compressors/pumps (work in). By the usual sign convention, work done by the system is positive. Note that flow work (the $PV$ part of pushing fluid into and out of the control volume) is already baked into the enthalpy term, so you don't account for it separately.

Enthalpy ($h$) is defined as:

$h = u + Pv$

where $u$ is specific internal energy, $P$ is pressure, and $v$ is specific volume. Enthalpy isn't a separate "transfer" mechanism. It's the property that naturally appears in the energy balance because it bundles the internal energy of the fluid together with the flow work needed to push that fluid through the control volume. The change in enthalpy, $\Delta h = h_e - h_i$, is usually the largest energy term in steady-flow analyses.

Significance of energy transfer terms

Which terms dominate depends entirely on the device:

• Heat exchangers: $\dot{Q}$ is the dominant term; $\dot{W} = 0$, and kinetic/potential energy changes are typically negligible. The energy balance reduces to $\dot{Q} = \dot{m}(h_e - h_i)$.
• Turbines and compressors: $\dot{W}$ is the dominant term. Many are modeled as adiabatic ($\dot{Q} \approx 0$), giving $\dot{W} = \dot{m}(h_i - h_e)$ for a turbine.
• Nozzles and diffusers: Neither heat nor work is significant. The energy exchange is between enthalpy and kinetic energy.
• Throttling valves: $\dot{Q} = 0$, $\dot{W} = 0$, and kinetic/potential energy changes are negligible, so $h_i \approx h_e$. The process is isenthalpic.

Kinetic and potential energy impact

When to include kinetic energy

Kinetic energy per unit mass is $\frac{V^2}{2}$. For most steady-flow devices, velocity changes are small enough that this term is negligible compared to enthalpy changes. The major exceptions are nozzles and diffusers, which exist specifically to change fluid velocity.

• Nozzle: Converts enthalpy into kinetic energy. The fluid speeds up, pressure drops, and $V_e \gg V_i$. You must keep the $\frac{V^2}{2}$ terms.
• Diffuser: Does the reverse. The fluid slows down, and kinetic energy converts back into enthalpy/pressure. Again, $\frac{V^2}{2}$ terms are essential.

A useful rule of thumb: if the velocity change is on the order of hundreds of m/s, the kinetic energy term (in kJ/kg) becomes comparable to typical enthalpy changes. For example, a velocity of 300 m/s gives $\frac{V^2}{2} = \frac{(300)^2}{2} = 45{,}000 \text{ J/kg} = 45 \text{ kJ/kg}$, which is definitely not negligible.

When to include potential energy

Potential energy per unit mass is $gz$. For most devices in a typical thermodynamics course, elevation changes are small and this term is dropped. The exceptions are systems with large elevation differences, such as hydraulic turbines and pumps in hydroelectric or piping systems.

• Pump raising fluid: Elevation increases, so potential energy increases. The pump must supply this energy in addition to any pressure rise.
• Hydraulic turbine: Fluid drops in elevation, converting potential energy into shaft work.

For a 100 m elevation change: $gz = (9.81)(100) = 981 \text{ J/kg} \approx 1 \text{ kJ/kg}$. Compare that to enthalpy changes in steam turbines that are often hundreds of kJ/kg. That's why $gz$ is almost always negligible unless the problem specifically involves large height differences.

Energy analysis of steady-flow devices

Problem-solving approach

Follow these steps for any steady-flow energy problem:

1. Sketch the system and identify the device. Draw the control volume with inlets and exits. Label known quantities (temperatures, pressures, velocities, elevations).
2. Write out the general energy balance. Start from the full equation before simplifying. This keeps you from accidentally dropping a term that matters.
3. Apply simplifications based on the device. Decide which terms are negligible and state why (e.g., "adiabatic because the nozzle is insulated," or "no work because there's no shaft").
4. Apply conservation of mass. For single-inlet, single-exit: $\dot{m}_i = \dot{m}_e = \dot{m}$. For mixing chambers or heat exchangers with multiple streams, write a separate mass balance.
5. Look up or calculate thermodynamic properties. Use steam tables, refrigerant tables, or ideal gas relations to find $h$, $v$, etc. at each state. This is often where mistakes happen, so double-check your state identification (superheated vs. saturated, etc.).
6. Substitute and solve. Plug everything into your simplified energy balance and solve for the unknown.
7. Check units and signs. Make sure your answer has the right sign (positive work out for a turbine, positive heat in for a boiler) and reasonable magnitude.

Interpreting results and energy transformations

Once you have a numerical answer, think about what's physically happening:

• Diffuser: Kinetic energy converts to pressure (enthalpy). Velocity drops, pressure rises. If your answer shows $h_e > h_i$ and $V_e < V_i$, that's consistent.
• Turbine: High-enthalpy fluid enters, delivers shaft work, and exits at lower enthalpy and pressure. A positive $\dot{W}$ confirms energy is leaving as work.
• Compressor: Work is done on the fluid, raising its enthalpy and pressure. You should see $h_e > h_i$ with work input.
• Throttling valve: Enthalpy stays roughly constant ($h_e \approx h_i$), but pressure drops significantly. For an ideal gas, temperature stays constant too. For real substances (like refrigerants), temperature usually drops, which is why throttling valves are used in refrigeration cycles.

If your results contradict the expected physics of the device, go back and check your simplifications and property lookups. A turbine that requires work input or a nozzle that slows the fluid down means something went wrong in the calculation.