Thermodynamics · Chapter 5

Mass and Energy Analysis of Control Volumes

Most real devices have mass flowing through them. This chapter extends the conservation laws to control volumes and applies them to the workhorses of engineering: nozzles, turbines, compressors, valves, and heat exchangers.

Learning Objectives

Apply conservation of mass to steady- and unsteady-flow control volumes.
Account for flow work and define the energy of a flowing fluid.
Write the steady-flow energy equation.
Analyze nozzles, diffusers, turbines, compressors, throttles, and mixing chambers.
Handle unsteady (transient) charging and discharging processes.

Physical meaning

A control volume is a chosen region, not a packet of matter. Mass crossing its surface transports \(h=u+Pv\), kinetic energy, and potential energy. Steady means no accumulation inside; it does not mean fluid has stopped or that inlet and outlet states are equal.

Problem-solving strategy

Draw the control surface and label every inlet, outlet, heat path, shaft, velocity, and elevation.
Apply mass conservation first and decide from the wording whether accumulation is zero or transient.
Identify each thermodynamic state and phase, then obtain (h), density, or other properties from a declared ideal-gas model or authoritative table with justified interpolation.
Write the complete energy-rate equation before neglecting heat, work, kinetic energy, or potential energy.
Audit stream mass, energy, entropy generation, units, and physical direction; iterate when outlet properties depend on the answer.

5–1

Conservation of Mass

Mass balance

Mass cannot be created or destroyed, so the rate of mass entering a control volume minus the rate leaving equals the rate of mass accumulation inside it.

Mass flow rate

\[\dot m = \rho V_{\!avg} A = \frac{V_{\!avg} A}{v}\]

General mass balance

\[\sum \dot m_{\text{in}} - \sum \dot m_{\text{out}} = \frac{dm_{CV}}{dt}\]

Steady flow

\[\sum \dot m_{\text{in}} = \sum \dot m_{\text{out}}\]

Figure 5–1 At steady state the mass entering the control volume equals the mass leaving it.

5–2

Flow Work and the Energy of a Flowing Fluid

Flow work

The work needed to push a unit mass of fluid into or out of a control volume, equal to \(Pv\). Combining it with internal energy gives enthalpy, which is why enthalpy appears naturally in flow problems.

Flow work

\[w_{\text{flow}} = Pv\]

Energy of a flowing fluid

\[\theta = h + \tfrac{1}{2}V^2 + gz\]

5–3

Energy Analysis of Steady-Flow Systems

Steady-flow energy equation (per unit mass)

\[q - w = \left(h_2 + \tfrac{1}{2}V_2^2 + gz_2\right) - \left(h_1 + \tfrac{1}{2}V_1^2 + gz_1\right)\]

Rate form

\[\dot Q - \dot W = \sum_{\text{out}} \dot m\left(h + \tfrac{V^2}{2} + gz\right) - \sum_{\text{in}} \dot m\left(h + \tfrac{V^2}{2} + gz\right)\]

Key IdeaAt steady state nothing inside the control volume accumulates, so the energy carried out by the streams plus net work and heat must balance the energy carried in.

5–4

Some Steady-Flow Engineering Devices

Nozzles and diffusers

A nozzle speeds a fluid up at the expense of pressure; a diffuser does the reverse. Both are adiabatic with no work, so enthalpy converts to or from kinetic energy.

Turbines and compressors

A turbine extracts shaft work as fluid expands; a compressor, pump, or fan adds work to raise pressure. Both are usually modeled as adiabatic with negligible KE and PE change.

Throttling valves

A flow restriction that drops pressure with no work and essentially no heat, so enthalpy is conserved: \(h_1 \approx h_2\). Useful in refrigeration.

Mixing chambers and heat exchangers

A mixing chamber combines streams adiabatically; a heat exchanger transfers heat between streams that do not mix. Both reduce to mass and energy balances.

Figure 5–2 A converging–diverging nozzle accelerates a compressible flow; enthalpy is converted into kinetic energy.

5–5

Energy Analysis of Unsteady-Flow Processes

Unsteady (transient) flow

Processes like filling or emptying a tank, where the control volume's mass and energy change with time. Both balances keep their accumulation terms.

Unsteady energy balance

\[Q - W = \sum_{\text{out}}\! m_e\theta_e - \sum_{\text{in}}\! m_i\theta_i + (m_2u_2-m_1u_1)_{CV}\]

EXAMPLE 1Vent nozzle velocity

Air enters an insulated test nozzle at 420 K with speed 35 m/s and exits at 330 K. Using the supplied constant ideal-gas value \(c_p=1.010\ \text{kJ/(kg}\!\cdot\!\text{K)}\), determine exit speed.

Recognition strategy

No shaft, insulated walls, and a large temperature decrease identify steady conversion of enthalpy into kinetic energy. A nozzle demands retention of the velocity terms.

Given information

\(T_1=420\ \text{K}\), \(T_2=330\ \text{K}\), \(V_1=35\ \text{m/s}\), supplied \(c_p=1.010\ \text{kJ/(kg}\!\cdot\!\text{K)}\); pressures are absolute but not needed.

Model and interpretation

The fixed control volume surrounds the nozzle with one inlet and outlet at steady state on a unit-mass basis. Heat into and shaft work out are positive. Air is an approximate constant-\(c_p\) ideal gas; pressure and temperature are absolute, elevation change is neglected, and no choking or shock branch is asserted because area and pressure data are absent.

Step-by-step solution

1. Apply steady mass conservation, so the single inlet and outlet carry the same mass flow rate.

2. Reduce energy conservation to \(h_1+V_1^2/2=h_2+V_2^2/2\) after setting heat, work, and elevation changes to zero.

3. Evaluate \(h_1-h_2=c_p(T_1-T_2)=1.010(90)=90.9\ \text{kJ/kg}\).

4. Solve \(V_2=\sqrt{V_1^2+2(90.9\times1000)}=427.8\ \text{m/s}\), or \(428\ \text{m/s}\).

Verification

Units: Multiplying kJ/kg by 1000 gives J/kg, identical to m²/s², so the square root correctly has velocity units.

Signs and direction: The enthalpy drop is positive and therefore raises speed; the computed exit direction matches the assumed outward stream.

Plausibility: Substitution returns \((427.8^2-35^2)/2=90.9\ \text{kJ/kg}\); energy and mass close, while entropy generation must remain nonnegative.

Final result: Computed nozzle exit speed: \(V_2=428\ \text{m/s}\).

Engineering interpretation

The high speed warns that compressibility, choking, and possible shocks may matter. Release analysis needs geometry, back pressure, variable properties, and a branch check against sonic conditions rather than extending this simple energy balance alone.

EXAMPLE 2Two-stream process-air mixer

Two ideal-gas air streams mix adiabatically at 250 kPa absolute: 0.80 kg/s at 310 K and 1.20 kg/s at 460 K. With the explicitly supplied common constant \(c_p=1.005\ \text{kJ/(kg}\!\cdot\!\text{K)}\), determine outlet flow and temperature.

Recognition strategy

Multiple inlets and one outlet identify simultaneous steady mass and energy balances. Equal constant specific heat allows enthalpy to be represented by \(c_pT\).

Given information

\(\dot m_1=0.80\ \text{kg/s}\), \(T_1=310\ \text{K}\), \(\dot m_2=1.20\ \text{kg/s}\), \(T_2=460\ \text{K}\), \(P=250\ \text{kPa absolute}\), supplied \(c_p=1.005\ \text{kJ/(kg}\!\cdot\!\text{K)}\).

Model and interpretation

The control surface encloses the mixer at steady state. Heat into and shaft work out are positive and both are zero; kinetic and potential energy changes are neglected. States are absolute-temperature ideal gases on a mass basis. The constant-property model is approximate and no phase quality applies.

Step-by-step solution

1. Conserve mass: \(\dot m_3=\dot m_1+\dot m_2=0.80+1.20=2.00\ \text{kg/s}\).

2. Write energy conservation as \(\dot m_1h_1+\dot m_2h_2=\dot m_3h_3\).

3. Insert \(h=c_pT\); the common supplied \(c_p\) cancels, leaving \(T_3=(\dot m_1T_1+\dot m_2T_2)/\dot m_3\).

4. Calculate \(T_3=[0.80(310)+1.20(460)]/2.00=400\ \text{K}\).

Verification

Units: Each mass-flow times enthalpy term has kJ/s, and the weighted temperature expression leaves kelvin as required.

Signs and direction: Both inlet flows are positive toward the mixer and the outlet is positive outward; the energy-weighted result has no sign reversal.

Plausibility: The outlet temperature lies between 310 K and 460 K, mass residual is zero, energy closes, and real finite-temperature mixing has \(S_{gen}>0\).

Final result: \(\dot m_3=2.00\ \text{kg/s}\) and \(T_3=400\ \text{K}\).

Engineering interpretation

The hotter stream controls more of the outlet state because it carries greater mass flow. Production work would use calibrated flow rates and variable-property enthalpy data, propagate sensor uncertainty, and verify allowable mixer pressure loss.

EXAMPLE 3Emergency receiver charging

An evacuated rigid insulated receiver is charged from a large air line at 360 K and 700 kPa absolute until 5.00 kg has entered. Using supplied ideal-gas constants \(c_p=1.005\) and \(c_v=0.718\ \text{kJ/(kg}\!\cdot\!\text{K)}\), estimate the final uniform temperature after flow stops.

Recognition strategy

A tank whose mass changes is an unsteady control volume, not a steady device or closed system. Inlet flow carries enthalpy, while accumulated tank energy is internal energy.

Given information

Initially \(m_1=0\); finally \(m_2=5.00\ \text{kg}\); inlet \(T_i=360\ \text{K}\), \(P_i=700\ \text{kPa absolute}\); supplied constant \(c_p=1.005\), \(c_v=0.718\ \text{kJ/(kg}\!\cdot\!\text{K)}\).

Model and interpretation

The fixed boundary surrounds the rigid receiver over the complete filling process. Heat into and work out are positive and both vanish. Air is an approximate ideal gas on a mass basis; temperatures and pressure are absolute. Inlet kinetic and potential energy, stored kinetic and potential energy, and line-state variation are neglected.

Step-by-step solution

1. Integrate mass conservation: with no exit and an initially evacuated tank, \(m_i=m_2-m_1=5.00\ \text{kg}\).

2. Reduce the integrated energy balance to \(m_ih_i=m_2u_2\), because the tank is rigid and insulated.

3. Use the supplied constant-property ideal-gas relations \(h_i=c_pT_i\) and \(u_2=c_vT_2\).

4. Since \(m_i=m_2\), solve \(T_2=(c_p/c_v)T_i=(1.005/0.718)(360)=503.9\ \text{K}\).

Verification

Units: The ratio \(c_p/c_v\) is dimensionless, so multiplying by the absolute inlet temperature correctly returns kelvin.

Signs and direction: Incoming enthalpy raises stored internal energy; a final temperature above the line temperature is consistent with flow work entering the rigid receiver.

Plausibility: Mass closes at 5.00 kg and \(c_vT_2=c_pT_i\) closes energy; throttling and mixing are irreversible, so entropy generation is nonnegative.

Final result: Computed final receiver temperature: \(T_2=504\ \text{K}\).

Engineering interpretation

The counterintuitive temperature rise comes from flow work already embedded in inlet enthalpy. Release work requires transient heat transfer, real-gas or variable-property data, valve/choking analysis, and pressure-vessel and vendor review.

Chapter recap

Choose and label the control surface before writing balances.
Conserve mass first; steady flow removes accumulation but not stream energy.
Use enthalpy for crossing streams and internal energy for material stored inside a rigid control volume.
Before moving on, be able to distinguish steady and transient models, source properties, retain important kinetic and potential terms, and close mass, energy, and entropy audits.

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