5.3 Some steady-flow engineering devices

Steady-flow device principles

Steady-flow devices are the workhorses of engineering thermodynamics. They allow fluid to flow through continuously while energy is transferred between the fluid and the device. Turbines, compressors, nozzles, heat exchangers, and mixing chambers all fall into this category, and each one shows up constantly in real power and refrigeration systems.

To analyze any of these devices, you apply the steady-flow energy balance to a control volume drawn around the device. The key differences between devices come down to what crosses the control volume boundary: shaft work, heat, or just fluid streams with different properties.

Types of steady-flow devices

Turbines extract energy from a fluid and convert it into shaft work. The fluid drops in pressure and enthalpy as it passes through. Steam turbines in power plants, gas turbines in jet engines, and hydraulic turbines in dams are all common examples.

Compressors (and pumps, and fans) do the opposite: they use shaft work to increase the pressure of a fluid. The gas gains both pressure and temperature as it moves through. Reciprocating, centrifugal, and axial compressors each have different mechanical designs but follow the same thermodynamic principles.

Heat exchangers transfer thermal energy between two fluid streams without mixing them. The fluids flow through separate passages, and one stream heats up while the other cools down. Shell-and-tube and plate heat exchangers are the most common configurations.

Mixing chambers combine two or more fluid streams into a single outlet stream. The outlet mixture has properties (temperature, enthalpy, composition) determined by the properties and flow rates of the incoming streams. Examples include feedwater heaters in power plants and simple mixing valves.

Energy conversion in nozzles and diffusers

Nozzles and diffusers convert between kinetic energy and pressure energy in a flowing fluid.

• Nozzles accelerate the fluid: pressure drops and velocity increases. You'll find them in rocket engines, steam turbines, and garden hoses.
• Diffusers decelerate the fluid: velocity drops and pressure increases. Aircraft engine inlets and the diverging sections of wind tunnels are typical diffusers.

For both devices, there's typically no shaft work and negligible heat transfer. The energy balance simplifies to a trade-off between enthalpy and kinetic energy:

$h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}$

If the inlet velocity is small compared to the outlet (common for nozzles), you can often neglect $V_1$ and solve directly for the exit velocity.

Turbine and Compressor Efficiency

Energy balance and efficiency in turbines

For a turbine with one inlet and one outlet, negligible heat transfer, and negligible changes in kinetic and potential energy, the steady-flow energy balance reduces to:

$\dot{W}_{out} = \dot{m}(h_{in} - h_{out})$

where $\dot{W}_{out}$ is the rate of work output, $\dot{m}$ is the mass flow rate, and $h_{in}$ and $h_{out}$ are the specific enthalpies at the inlet and outlet.

The isentropic efficiency of a turbine compares the actual work output to the work you'd get from an ideal, reversible (isentropic) expansion between the same inlet state and the same exit pressure:

$\eta_{turbine} = \frac{w_{out,actual}}{w_{out,isentropic}} = \frac{h_{in} - h_{out,actual}}{h_{in} - h_{out,s}}$

Here $h_{out,s}$ is the exit enthalpy the fluid would have if the process were isentropic. Because real processes involve friction and other irreversibilities, the actual work output is always less than the isentropic work output, so $\eta_{turbine} < 1$. Modern steam and gas turbines typically achieve isentropic efficiencies of 70% to 90%.

Energy balance and efficiency in compressors

The energy balance for a compressor (again neglecting heat transfer, kinetic, and potential energy changes) is:

$\dot{W}_{in} = \dot{m}(h_{out} - h_{in})$

Notice the subscripts flip compared to the turbine equation, because now enthalpy increases through the device.

The isentropic efficiency of a compressor is defined with the ratio inverted relative to the turbine:

$\eta_{compressor} = \frac{w_{in,isentropic}}{w_{in,actual}} = \frac{h_{out,s} - h_{in}}{h_{out,actual} - h_{in}}$

The ideal work is in the numerator here because you want the compressor that requires the least work. Irreversibilities always make the actual work input greater than the isentropic work input, so $\eta_{compressor} < 1$ as well. Typical values for modern compressors range from 70% to 85%.

Energy Balance in Heat Exchangers

Heat transfer and effectiveness in heat exchangers

Heat exchangers involve no shaft work. If you also assume negligible kinetic and potential energy changes and no heat loss to the surroundings, the energy balance says that everything the hot fluid loses, the cold fluid gains:

$\dot{m}_{hot}c_{p,hot}(T_{hot,in} - T_{hot,out}) = \dot{m}_{cold}c_{p,cold}(T_{cold,out} - T_{cold,in})$

This equation assumes constant specific heats (a reasonable approximation for liquids and for gases over moderate temperature ranges). For fluids undergoing phase change, you'd use enthalpy differences instead.

The effectiveness $\varepsilon$ of a heat exchanger measures how close it comes to the thermodynamic maximum heat transfer:

$\varepsilon = \frac{\dot{Q}_{actual}}{\dot{Q}_{max}}$

The maximum possible heat transfer occurs when the fluid with the smaller heat capacity rate ($C_{min} = (\dot{m}c_p)_{min}$) undergoes the full temperature difference between the two inlet temperatures:

$\dot{Q}_{max} = C_{min}(T_{hot,in} - T_{cold,in})$

Effectiveness ranges from 0 to 1. Shell-and-tube heat exchangers typically achieve 0.6 to 0.8, while plate heat exchangers can reach 0.8 to 0.95.

Log mean temperature difference (LMTD) method

The LMTD method relates the heat transfer rate to the size and performance of the heat exchanger:

$\dot{Q} = UA \cdot \Delta T_{LMTD}$

where $U$ is the overall heat transfer coefficient, $A$ is the heat transfer surface area, and $\Delta T_{LMTD}$ is the log mean temperature difference.

For a counter-flow heat exchanger, the LMTD is calculated as:

$\Delta T_{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)}$

where $\Delta T_1 = T_{hot,in} - T_{cold,out}$ and $\Delta T_2 = T_{hot,out} - T_{cold,in}$.

The LMTD method assumes constant fluid properties and heat transfer coefficients with no phase change. For multi-pass or cross-flow arrangements, a correction factor $F$ is applied: $\dot{Q} = UAF \cdot \Delta T_{LMTD}$.

Energy and mass balance in mixing chambers

For an adiabatic mixing chamber with no work interactions, the mass and energy balances are:

Mass balance: $\dot{m}_{out} = \dot{m}_1 + \dot{m}_2 + \cdots$

Energy balance: $\dot{m}_{out}h_{out} = \dot{m}_1 h_1 + \dot{m}_2 h_2 + \cdots$

The outlet temperature will always fall between the temperatures of the incoming streams. You can solve for it using the energy balance along with either property tables or the relation $h = c_p T$ for ideal gases or incompressible liquids with constant specific heats.

Note that the outlet pressure of a mixing chamber is typically set by the lower of the inlet pressures (the streams must be at roughly the same pressure to mix). If the problem involves chemical reactions (like a combustion chamber), you also need to account for the heat of reaction in the energy balance and use stoichiometry in the mass balance.

Steady-flow device performance analysis

Problem-solving approach

Every steady-flow device problem follows the same general framework:

1. Sketch the device and define the control volume. Label all inlets, outlets, and any work or heat interactions crossing the boundary.
2. List known quantities (fluid properties, flow rates, temperatures, pressures, efficiencies).
3. Identify what you're solving for (work output, heat transfer rate, exit temperature, efficiency, etc.).
4. Write the mass balance ($\sum \dot{m}_{in} = \sum \dot{m}_{out}$) and the energy balance for the control volume, crossing out terms that are zero or negligible (e.g., no work for a heat exchanger, no heat transfer for an adiabatic device).
5. Look up fluid properties from steam tables, refrigerant tables, or ideal gas relations as needed. For isentropic efficiency problems, you'll need properties at both the actual and isentropic exit states.
6. Solve the equations and check that your answer makes physical sense (e.g., turbine exit enthalpy should be lower than inlet enthalpy; compressor exit temperature should be higher).

Turbine and compressor problems

These problems typically ask for work per unit mass, exit state properties, or isentropic efficiency.

Example: A steam turbine with an isentropic efficiency of 85% receives steam at 10 MPa and 500°C. The exit pressure is 50 kPa. Find the actual work output per unit mass and the actual exit enthalpy.

• Step 1: From steam tables, find $h_{in}$ at 10 MPa, 500°C.
• Step 2: Find $h_{out,s}$ by following an isentropic process from the inlet state to 50 kPa (use $s_{out} = s_{in}$ and the saturation/superheated tables at 50 kPa).
• Step 3: Calculate the isentropic work: $w_{s} = h_{in} - h_{out,s}$.
• Step 4: Apply the efficiency: $w_{actual} = \eta_{turbine} \times w_s = 0.85 \times w_s$.
• Step 5: Find the actual exit enthalpy: $h_{out,actual} = h_{in} - w_{actual}$.

Note that steam is not an ideal gas at these conditions, so you must use steam tables rather than ideal gas relations.

Heat exchanger problems

These problems typically ask for heat transfer rates, outlet temperatures, or required flow rates.

Example: A counter-flow heat exchanger cools oil ($c_p = 2.1$ kJ/kg·K) from 80°C to 50°C using water ($c_p = 4.18$ kJ/kg·K) entering at 20°C. The oil flow rate is 0.5 kg/s and the effectiveness is 0.8. Find the water flow rate and outlet temperature.

• Step 1: Calculate the actual heat transfer: $\dot{Q} = \dot{m}_{oil} c_{p,oil}(80 - 50) = 0.5 \times 2.1 \times 30 = 31.5$ kW.
• Step 2: Determine $C_{hot} = 0.5 \times 2.1 = 1.05$ kW/K. Use the effectiveness definition to find $C_{min}$ and then the water flow rate.
• Step 3: From the energy balance on the water side, solve for $T_{cold,out}$: $T_{cold,out} = T_{cold,in} + \frac{\dot{Q}}{\dot{m}_{water} c_{p,water}}$.

If fluid properties vary significantly with temperature, you may need to iterate.

Mixing chamber problems

These problems ask for the outlet mixture temperature, required inlet flow rates, or inlet conditions needed to achieve a target outlet state.

Example: Two air streams mix adiabatically. Stream 1 enters at 300 K with $\dot{m}_1 = 1$ kg/s; Stream 2 enters at 400 K with $\dot{m}_2 = 1.5$ kg/s. Find the mixture temperature (assume constant $c_p$).

• Step 1: Mass balance gives $\dot{m}_{out} = 1 + 1.5 = 2.5$ kg/s.
• Step 2: Energy balance (with constant $c_p$ canceling): $\dot{m}_{out} T_{out} = \dot{m}_1 T_1 + \dot{m}_2 T_2$.
• Step 3: Solve: $T_{out} = \frac{(1)(300) + (1.5)(400)}{2.5} = \frac{900}{2.5} = 360$ K.

The exit pressure in a mixing chamber is typically approximated as the lower of the two inlet pressures (both streams must be at compatible pressures to mix). If the pressures differ significantly, the higher-pressure stream throttles down to the mixing pressure, and you'd need to account for that separately.