Mass and Energy Analysis of Control Volumes

Introduction

The first law of thermodynamics is a statement of conservation of energy. For a fixed quantity of matter, energy can cross the boundary as heat or work, but mass remains constant. Many engineering devices, however, operate with continuous flow: mass enters, changes state, and leaves. Turbines, compressors, pumps, nozzles, heat exchangers, boilers, condensers, mixers, and filling tanks are therefore best analyzed as control volumes.

A control volume is a selected region in space through which mass may cross a control surface. The control volume may be stationary, moving, fixed in shape, or changing in volume. The most useful engineering models are obtained by applying conservation of mass and the first law to this region.

This chapter develops general mass and energy balances and then specializes them to common steady-flow and transient-flow devices. The equations are applicable to real engineering systems provided that the assumptions used in simplifying them are stated and justified.

Learning Objectives

After studying this chapter, you should be able to:

1. Distinguish a system from a control volume.
2. Formulate and interpret the general mass balance.
3. Apply the first-law energy balance to a control volume.
4. Derive the steady-flow energy equation.
5. Analyze turbines, compressors, pumps, nozzles, diffusers, throttling devices, heat exchangers, mixing chambers, and storage vessels.
6. Identify appropriate assumptions and assess their effect on results.
7. Solve practical steady and transient flow problems using thermodynamic property data.

1. Control-Volume Concepts

A closed system contains a fixed amount of matter. No mass crosses its boundary. An open system, or control volume, permits mass transfer across its boundary. The boundary of a control volume is called the control surface.

Mass crossing a control surface carries energy in three principal forms:

• internal energy, represented by specific internal energy u;
• flow work, incorporated into specific enthalpy h = u + Pv;
• kinetic and potential energy.

Here, P is pressure, v is specific volume, and h is specific enthalpy. The use of enthalpy is especially convenient because the pressure–volume work required to push mass across a control surface is included in h.

For a stream, the specific total energy is

where θ is specific flow energy, h is specific enthalpy in J/kg, V is velocity in m/s, g is gravitational acceleration in m/s², and z is elevation in m. The terms V²/2 and gz are specific kinetic and potential energies, respectively, expressed in J/kg.

A flow stream may also carry energy associated with electrical, magnetic, nuclear, or chemical effects. In ordinary mechanical-engineering flow devices, these effects are often negligible unless explicitly stated.

2. Conservation of Mass

The general mass balance for a control volume is

where m_cv is the mass within the control volume in kg, t is time in s, and ṁ is mass flow rate in kg/s. The summations include all inlets and outlets.

The mass flow rate through a section is

where ρ is density in kg/m³, A is flow area in m², and V is the average velocity normal to the area in m/s. For a nonuniform velocity profile, the more exact expression is

where V_n is the velocity component normal to the area. The simpler equation is appropriate when density and velocity can be represented by suitable average values.

For a steady-flow control volume, the mass inside does not change with time:

For one inlet and one outlet, this reduces to ṁ₁ = ṁ₂. For several streams, the total inlet flow rate equals the total outlet flow rate.

Example 1: Mass flow through a pipe

Water with density 998 kg/m³ flows through a pipe of diameter 0.050 m at an average velocity of 2.0 m/s. The cross-sectional area is

A = πD²/4 = π(0.050)²/4 = 0.001963 m².

Therefore,

ṁ = ρAV = (998)(0.001963)(2.0) = 3.92 kg/s.

The result assumes one-dimensional average flow and incompressible water at approximately the stated density.

3. The Control-Volume Energy Balance

The first law for a control volume is

Here, E_cv is the total energy stored in the control volume, Q̇_cv is the net heat-transfer rate into the control volume, and Ẇ_cv is the net work-transfer rate from the control volume. The sign convention is therefore:

• Q̇_cv > 0 when heat enters;
• Ẇ_cv > 0 when work leaves the control volume.

The stored energy is commonly written as

where U_cv is internal energy, KE_cv is macroscopic kinetic energy, and PE_cv is macroscopic potential energy.

The general equation assumes that mass crossing the boundary carries enthalpy, kinetic energy, and potential energy. It also assumes that the only energy transfers not associated with mass flow are heat and work. If electrical or other energy transfers are important, they must be added explicitly.

For a steady-flow control volume, dE_cv/dt = 0. Thus,

For one inlet and one outlet,

Dividing by ṁ gives the specific steady-flow energy equation:

where q and w are heat and work transfers per unit mass, in J/kg. This equation is valid for a one-inlet, one-outlet steady-flow device under the stated sign convention.

4. Modeling Assumptions

Thermodynamic analysis is not merely substitution into equations. A sound solution begins by identifying the physical model.

Steady operation

Steady operation means that properties and flow rates at fixed locations do not vary with time. A device may operate continuously while fluid particles move through it, yet the control-volume contents remain constant on average.

Uniform properties at inlets and outlets

In engineering calculations, each inlet or outlet is often represented by average pressure, temperature, velocity, and enthalpy. This is reasonable when variations across the section are small or when measured values are already area-averaged.

Negligible kinetic energy

The kinetic-energy change per unit mass is V²/2. It may be neglected when it is much smaller than the enthalpy change or other energy terms. It cannot be neglected in high-speed nozzles, diffusers, or jet devices.

Negligible potential energy

The potential-energy change is g∆z. It is commonly negligible in compact devices but may be important in hydraulic systems, tall pipelines, and pumped-storage installations.

Adiabatic operation

An adiabatic control volume has Q̇ = 0. This does not mean that the device has the same temperature everywhere; it means that net heat transfer across the control surface is negligible.

5. Turbines

A turbine produces shaft work as a high-pressure or high-temperature fluid expands through it. For a steady turbine with one inlet and one outlet,

For an adiabatic turbine with negligible kinetic and potential-energy changes,

The enthalpy decreases because energy is extracted as shaft work. Real turbines also experience friction and other irreversibilities, so the actual work is less than the ideal isentropic work for the same inlet and exit pressures.

Turbine example

Steam enters an adiabatic turbine at a mass flow rate of 5 kg/s with h₁ = 3400 kJ/kg and exits with h₂ = 2500 kJ/kg. Neglect changes in kinetic and potential energy.

Ẇ_out = 5(3400 − 2500) = 4500 kW.

Thus, the turbine develops 4.50 MW of power. The result is positive because the fluid loses enthalpy while the control volume delivers work.

6. Compressors and Pumps

Compressors raise the pressure of gases, whereas pumps raise the pressure of liquids. Both normally require shaft work input.

For an adiabatic compressor with negligible kinetic and potential-energy changes,

The enthalpy increases because work is supplied to the fluid.

For an incompressible liquid pump, the enthalpy change is often approximated by the pressure-work relation. The ideal specific pump work is

where v is the approximately constant specific volume in m³/kg and pressures are in Pa, giving J/kg. If pump efficiency η_p is defined as ideal hydraulic power divided by actual shaft power,

so that

These equations assume incompressible flow and neglect changes in velocity and elevation unless they are included separately.

7. Nozzles and Diffusers

A nozzle accelerates a fluid by converting enthalpy into kinetic energy. A diffuser decelerates a fluid and converts kinetic energy into pressure and enthalpy-related energy.

For an adiabatic nozzle or diffuser with no shaft work and negligible elevation change,

Therefore,

When h is in kJ/kg, the velocity relation must include the conversion factor 1000:

where h is in kJ/kg and V is in m/s. In a nozzle, h₂ < h₁ and the velocity rises. In a diffuser, V₂ < V₁ and generally h₂ > h₁.

Nozzle example

Air enters an adiabatic nozzle at 100 m/s with h₁ = 300 kJ/kg. It exits with h₂ = 270 kJ/kg. Neglect elevation change.

V₂ = √[100² + 2000(300 − 270)]

V₂ = √70,000 = 264.6 m/s.

The increase in speed results from a 30 kJ/kg decrease in enthalpy.

8. Throttling Devices

A throttling valve, porous plug, capillary tube, or similar restriction causes a substantial pressure drop with no useful shaft work. In many applications the device is well insulated and changes in kinetic and potential energy are negligible. The steady-flow energy equation then gives

Thus, throttling is approximately isenthalpic, not isentropic. Entropy generally increases because the process is highly irreversible. Temperature may increase, decrease, or remain nearly unchanged depending on the fluid and its initial state. For an ideal gas, enthalpy depends only on temperature, so an ideal-gas throttling process has approximately constant temperature. Real gases may exhibit the Joule–Thomson effect.

9. Heat Exchangers

A heat exchanger transfers energy between two or more fluid streams, usually without mixing them. When the outer surface is well insulated and there is no shaft work, the total energy balance for a steady heat exchanger is

when kinetic and potential-energy changes are negligible.

For two streams,

where subscripts h and c denote hot and cold streams. The heat lost by the hot stream equals the heat gained by the cold stream under the adiabatic-external assumption.

A heat exchanger may be classified as parallel-flow, counter-flow, or cross-flow. Counter-flow arrangements often provide a larger average temperature difference and better thermal performance than parallel-flow arrangements, but the first-law energy balance remains the same.

10. Mixing Chambers

A mixing chamber combines two or more streams. For steady operation with one outlet, the mass balance is

If the chamber is adiabatic, has no shaft work, and kinetic and potential-energy changes are negligible,

For liquids with nearly constant specific heat, enthalpy can often be related to temperature by h ≈ c_pT over a moderate range. This approximation must be used cautiously when phase changes occur or when property variation is significant.

Mixing example

Water streams of 2 kg/s at 20 °C and 3 kg/s at 80 °C mix adiabatically. Assume constant c_p and negligible kinetic and potential-energy changes. The outlet temperature is

T₃ = (ṁ₁T₁ + ṁ₂T₂)/(ṁ₁ + ṁ₂)

T₃ = [2(20) + 3(80)]/5 = 56 °C.

The result lies between the inlet temperatures, as required by the energy balance for this idealized situation.

11. Transient Control-Volume Analysis

A transient process occurs when conditions within the control volume change with time. Examples include filling a tank, emptying a reservoir, charging a compressed-air receiver, and startup or shutdown of a turbine.

The integrated mass balance between times 1 and 2 is

where m_in and m_out are the total masses that cross the boundary during the interval.

The integrated energy balance is

Here Q and W are total heat and work transfers over the time interval. The properties associated with entering and leaving mass may vary with time, in which case the flow-energy terms must be integrated rather than treated as constant.

Rigid tank filling

Consider a rigid, insulated tank initially evacuated and filled through one inlet. There is no boundary work because the tank volume is fixed, and no heat transfer because the tank is insulated. If inlet velocity and elevation effects are negligible, the energy balance becomes

Since the tank is initially empty, m₂ = m_in. Therefore,

This important result does not imply that the inlet and final temperatures are equal. Internal energy in the tank equals the enthalpy carried in by the entering stream because the stream enthalpy includes the flow work required to force fluid into the tank.

Rigid tank emptying

For an insulated rigid tank discharging through one outlet, with negligible outlet kinetic energy and no heat or work transfer,

The state inside the tank changes as mass and energy leave. A complete solution generally requires an equation of state and property relations, together with a relation for the outlet flow rate.

12. Detailed Applied Case: Compressed-Air Receiver

Background

Compressed-air receivers are used in manufacturing plants to smooth compressor delivery, supply short-duration demand peaks, and reduce compressor cycling. During operation, the receiver may undergo filling and discharge, so a steady-flow model is inappropriate.

Problem

An initially empty, rigid, insulated receiver is filled from a large compressed-air line whose specific enthalpy remains approximately constant at h_in. Determine the final receiver temperature conceptually and identify the governing balance.

Analysis and approach

Select the receiver as the control volume. The receiver is rigid, so boundary work is zero. It is insulated, so Q = 0. There is one inlet and no outlet. Neglect inlet kinetic and potential-energy changes. The transient balances are

m₂ − m₁ = m_in

and

E₂ − E₁ = m_inh_in.

Initially, m₁ = 0 and E₁ = 0. At the final state, if kinetic and potential energies inside the receiver are negligible, E₂ = m₂u₂. Since m₂ = m_in,

For an ideal gas with constant specific heats, u = c_vT and h = c_pT. Hence,

or

where γ = c_p/c_v. This idealized result predicts a final temperature higher than the inlet temperature. Real receivers exchange heat with their surroundings during filling, and the inlet state may change; therefore, measured temperatures may differ from the ideal adiabatic prediction.

Outcome and engineering significance

The receiver stores both mass and energy. Treating it as a steady-flow device would incorrectly set the stored-energy accumulation to zero. The analysis demonstrates why transient terms are essential during filling and why the choice of control volume determines the correct work and energy terms.

Learning points

1. A rigid vessel can have zero boundary work while still receiving flow work through its inlet.
2. Transient energy storage may dominate the process.
3. Initial conditions are required for a transient problem.
4. Heat transfer to the environment can substantially alter the final state.
5. Property models must be consistent with the pressure and temperature range.

13. Practical Problem-Solving Procedure

A reliable control-volume solution can be organized as follows:

1. Define the control volume. Draw or imagine its boundary and identify every inlet, outlet, heat interaction, and work interaction.
2. Choose coordinates and signs. State that heat into the control volume and work out of it are positive, or use another convention consistently.
3. Write the general mass balance. Do not assume steady operation before checking it.
4. Write the general energy balance. Include enthalpy, kinetic energy, and potential energy initially.
5. State assumptions. Common assumptions include steady state, adiabatic operation, negligible elevation change, or negligible kinetic-energy change.
6. Use appropriate property data. Consult steam tables, refrigerant tables, compressed-liquid data, ideal-gas relations, or reliable software.
7. Check units. Enthalpy in kJ/kg combined with mass flow in kg/s produces kW. Pressure in kPa multiplied by specific volume in m³/kg also produces kJ/kg.
8. Check physical reasonableness. A turbine should generally produce work while fluid enthalpy decreases; a compressor should require work while enthalpy increases; a mixing temperature should normally lie between inlet temperatures for simple adiabatic mixing.

14. Limitations and Contemporary Practice

Control-volume balances express conservation laws, but they do not by themselves determine all unknown states. A complete analysis may require an equation of state, phase-equilibrium relations, efficiency definitions, or the second law of thermodynamics.

Real devices involve pressure losses, nonuniform velocity profiles, heat leakage, mechanical friction, leakage flows, and time-dependent operation. Computational fluid dynamics and digital monitoring can resolve local flow fields and transient behavior, but these tools do not replace conservation balances. In industrial practice, sensor data and automated control systems are often used to estimate mass flow, detect energy losses, and identify abnormal operation. AI-based monitoring may assist fault detection, but its outputs should remain subject to physical validation, sensor calibration, cybersecurity controls, and engineering responsibility.

Safety is particularly important in pressurized tanks, boilers, compressors, and refrigeration systems. Designs must comply with applicable pressure-vessel, process-safety, electrical, environmental, and occupational regulations. A mathematically correct balance is not a substitute for certified equipment design or hazard analysis.

Chapter Summary

A control volume is a region through which mass may cross a control surface. Conservation of mass is expressed by the rate of accumulation of mass inside the control volume equaling inlet mass flow minus outlet mass flow. For steady operation, total inlet mass flow equals total outlet mass flow.

The first-law control-volume energy balance accounts for heat transfer, work transfer, stored energy, and the enthalpy, kinetic energy, and potential energy carried by flowing mass. The steady-flow energy equation follows when stored energy does not change. Turbines generally convert enthalpy decrease into shaft work; compressors and pumps require work input; nozzles convert enthalpy into kinetic energy; diffusers reduce velocity; throttling devices are approximately isenthalpic; heat exchangers transfer energy between streams; and mixing chambers combine streams according to mass and energy balances.

Transient analysis is required whenever mass or energy stored in the control volume changes with time. Rigid filling and emptying tanks illustrate the importance of initial conditions and accumulation terms. Accurate engineering analysis depends on clearly stated assumptions, consistent properties and units, and physical checks on the result.

Key Terms

Control volume: A selected region in space through which mass and energy may cross.

Control surface: The boundary enclosing a control volume.

Open system: A thermodynamic system that permits mass transfer across its boundary.

Mass flow rate: The rate at which mass crosses a section, measured in kg/s.

Steady flow: Flow in which properties and flow rates at fixed locations do not change with time.

Transient flow: Flow in which conditions within the control volume change with time.

Enthalpy: The property h = u + Pv, especially useful for flowing fluids.

Flow work: Pressure–volume work associated with pushing mass across a control surface.

Kinetic energy: Energy associated with macroscopic velocity, V²/2 per unit mass.

Potential energy: Energy associated with elevation, gz per unit mass.

Adiabatic process: A process with zero net heat transfer across the boundary.

Turbine: A device that produces shaft work through fluid expansion.

Compressor: A device that raises gas pressure using work input.

Pump: A device that raises liquid pressure using work input.

Nozzle: A device that increases fluid velocity.

Diffuser: A device that decreases fluid velocity and commonly increases pressure.

Throttling device: A restriction through which flow undergoes an approximately constant-enthalpy process.

Heat exchanger: A device that transfers heat between fluid streams, usually without mixing them.

Mixing chamber: A control volume in which two or more streams combine.

Accumulation: The time rate of change of mass or energy stored in a control volume.

Review Questions

Short Answer Questions

1. What is the difference between a closed system and a control volume?
2. Why is enthalpy used rather than internal energy alone in flow-energy balances?
3. State the steady-flow mass balance for a control volume.
4. Under what assumptions is a throttling process modeled as isenthalpic?
5. Why must transient analysis include an accumulation term?

Long Answer Questions

1. Derive the one-inlet, one-outlet steady-flow energy equation from the general control-volume energy balance.
2. Explain the energy transformations in an adiabatic turbine, compressor, nozzle, and diffuser.
3. Develop the mass and energy balances for a two-stream adiabatic heat exchanger.
4. Explain the physical meaning of flow work and its inclusion in enthalpy.
5. Compare steady operation with transient filling of a rigid tank, including the assumptions used in each analysis.

Analytical / Application Questions

1. Air enters an adiabatic nozzle at 150 m/s with h₁ = 315 kJ/kg and leaves with h₂ = 285 kJ/kg. Neglect elevation change. Determine the exit velocity.
2. A pump raises water pressure from 100 kPa to 2,000 kPa. If v = 0.001 m³/kg and pump efficiency is 75%, determine the actual specific work input.
3. A steady adiabatic mixer receives 1 kg/s of water at 15 °C and 4 kg/s at 65 °C. Assuming constant specific heat and negligible kinetic and potential-energy changes, determine the outlet temperature.

Multiple Choice Questions with Answers

Question 1

Which equation represents the general mass balance for a control volume?

A. dE_cv/dt = Q̇ − Ẇ

B. dm_cv/dt = ∑ṁ_in − ∑ṁ_out

C. ∑ṁ_in = 0

D. m = ρV

Correct Answer: B. dm_cv/dt = ∑ṁ_in − ∑ṁ_out

Question 2

For a steady, one-inlet, one-outlet control volume, which condition applies?

A. ṁ_in = ṁ_out

B. ṁ_in > ṁ_out

C. ṁ_in < ṁ_out

D. ṁ_in = 0

Correct Answer: A. ṁ_in = ṁ_out

Question 3

An adiabatic turbine has negligible kinetic and potential-energy changes. Which expression gives its power output?

A. Ẇ_out = ṁ(h₂ − h₁)

B. Ẇ_out = ṁ(h₁ − h₂)

C. Ẇ_out = ṁ(h₁ + h₂)

D. Ẇ_out = 0

Correct Answer: B. Ẇ_out = ṁ(h₁ − h₂)

Question 4

For an idealized steady throttling process, which property remains approximately constant?

A. Temperature for every real fluid

B. Entropy

C. Enthalpy

D. Pressure

Correct Answer: C. Enthalpy

Question 5

A nozzle is adiabatic, has no shaft work, and has negligible elevation change. If enthalpy decreases, what generally happens to velocity?

A. It decreases

B. It increases

C. It becomes zero

D. It is unaffected by enthalpy

Correct Answer: B. It increases

Question 6

Water flows through a 0.050 m diameter pipe at 2.0 m/s. Using ρ = 998 kg/m³, what is the approximate mass flow rate?

A. 0.0392 kg/s

B. 0.392 kg/s

C. 3.92 kg/s

D. 39.2 kg/s

Correct Answer: C. 3.92 kg/s

Question 7

A pump raises the pressure of an incompressible liquid. Neglecting other effects, which quantity represents the ideal specific pump work?

A. v(P₂ − P₁)

B. (P₂ − P₁)/v

C. v/(P₂ − P₁)

D. P₁P₂v

Correct Answer: A. v(P₂ − P₁)

Question 8

For an externally adiabatic two-stream heat exchanger with negligible kinetic and potential-energy changes, what is conserved between inlet and outlet streams?

A. Total stream enthalpy flow

B. Temperature of every stream

C. Pressure of every stream

D. Entropy of every stream

Correct Answer: A. Total stream enthalpy flow

Question 9

An insulated rigid tank is initially evacuated and filled through one inlet, with negligible kinetic and potential-energy effects. Which relation results from the transient energy balance?

A. u₂ = 0

B. u₂ = h_in

C. h₂ = 0

D. u₂ = h₂/2

Correct Answer: B. u₂ = h_in

Question 10

Two water streams of 2 kg/s at 20 °C and 3 kg/s at 80 °C mix adiabatically with constant specific heat. What is the outlet temperature?

A. 32 °C

B. 44 °C

C. 56 °C

D. 68 °C

Correct Answer: C. 56 °C