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Thermodynamics · Chapter 5 of 10 · Intermediate

Mass and Energy Analysis of Control Volumes

Turbines, nozzles, compressors, and throttles all have mass flowing through them. Two balances, one for mass and one for energy, written for a region rather than a fixed mass, analyse every one.

Readiness check

This chapter applies the first law to flowing systems. Tick only what you can do closed-notes.

Define enthalpy h = u + Pv.
Write the energy balance Q − W = ΔE.
Read enthalpy from a steam table or compute it for an ideal gas.
Recall mass flow ṁ = ρAV.
Square a velocity and keep track of J versus kJ.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRevisit enthalpy and the first law in Chapter 4.

3 or more weak itemsPractise reading steam tables from Chapter 3.

The core idea

For a steady control volume, mass in equals mass out, and the energy each stream carries is its enthalpy plus kinetic and potential energy. The steady-flow energy equation balances heat and work against the change in that quantity.

ṁ_in = ṁ_outQ̇ − Ẇ = ṁ[(h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)]

A control volume is a fixed region that mass flows through. At steady state nothing inside accumulates, so the mass flow in equals the mass flow out. Each stream carries not just internal energy but also the flow work Pv needed to push it across the boundary, and the two combine into enthalpy h = u + Pv. The steady-flow energy equation then balances the net heat and shaft work against the change in enthalpy, kinetic energy, and potential energy of the stream.

The skill works when: you write both balances, then drop the terms a given device makes negligible.

The skill breaks down when: flow work is forgotten, or kinetic energy is kept where it does not matter and dropped where it does.

The concept. Mass enters and leaves the control volume carrying enthalpy and kinetic energy; heat and shaft work cross the boundary. At steady state the books balance with no accumulation inside.

The skills, taught in order

Five skills set the mass balance, explain why enthalpy appears, write the steady-flow energy equation, and reduce it for each standard device.

5.1 Conservation of mass

For a control volume, the rate of mass in minus the rate out equals the rate of accumulation. At steady state the accumulation is zero, so ṁ_in = ṁ_out. The mass flow through any cross-section is ṁ = ρAV = AV/v, which links velocity, area, and specific volume.

5.2 Flow work and enthalpy

Pushing a stream across the boundary takes flow work equal to Pv per unit mass. Adding it to internal energy gives enthalpy, h = u + Pv, the natural energy of a flowing fluid. This is why open-system balances use h where closed-system balances used u.

5.3 The steady-flow energy equation

For one inlet and one outlet at steady state, Q̇ − Ẇ = ṁ[(h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)]. The elevation term is almost always negligible for gases and vapors, and the kinetic term matters only where velocities are large, such as in a nozzle.

5.4 Device models

Each device is the same equation with terms removed. A turbine or compressor exchanges shaft work with negligible heat and kinetic change; a nozzle or diffuser exchanges kinetic energy with no work; a throttle does neither, so enthalpy is conserved.

Device	Simplification	Reduced balance
Nozzle / diffuser	adiabatic, no work	h₁ + V₁²/2 = h₂ + V₂²/2
Turbine	adiabatic, ΔKE and ΔPE small	ẇ = h₁ − h₂
Compressor / pump	adiabatic, ΔKE and ΔPE small	ẇ_in = h₂ − h₁
Throttle	adiabatic, no work, ΔKE small	h₁ = h₂
Heat exchanger	no work, ΔKE and ΔPE small	Σ ṁh (in) = Σ ṁh (out)

5.5 Choosing the boundary

The art is drawing the control volume so the streams you know cross it. For a heat exchanger, one boundary around both fluids gives a clean energy balance with no heat term; a boundary around just one fluid exposes the heat transferred between them.

Engineering connection: these device models are the building blocks assembled into the power and refrigeration cycles of Chapters 9 and 10.

Worked example 1: power from a steam turbine

Steam enters an adiabatic turbine at 3 MPa and 350 °C (h₁ = 3115.3 kJ/kg) and leaves at 10 kPa as a wet mixture of quality 0.90. Using the 10 kPa saturation data (h_f = 191.81 kJ/kg, h_fg = 2392.82 kJ/kg) and a mass flow of 2 kg/s, find the specific work and the power output.

Figure 1. An adiabatic turbine drops the steam enthalpy from inlet to outlet; that enthalpy difference, times the mass flow, is the shaft power.

ProblemFind the specific work and power of the turbine in Figure 1.
Given / findh₁ = 3115.3 kJ/kg, exit 10 kPa with x = 0.90, ṁ = 2 kg/s. Find ẇ and Ẇ.
AssumptionsSteady, adiabatic; changes in kinetic and potential energy are negligible.
ModelFind the exit enthalpy from quality, then apply the turbine balance ẇ = h₁ − h₂ and Ẇ = ṁẇ.
Equationsh₂ = h_f + x h_fgẇ = h₁ − h₂Ẇ = ṁ ẇ
Solveh₂ = 191.81 + 0.90 × 2392.82 = 2345.3 kJ/kg. ẇ = 3115.3 − 2345.3 = 769.9 kJ/kg. Ẇ = 2 × 769.9 = 1539.8 kW ≈ 1.54 MW.
CheckThe steam leaves wet, as real turbines do, and the work per kilogram is a sensible fraction of the inlet enthalpy. Units: (kg/s)(kJ/kg) = kW.
ConclusionA turbine simply turns an enthalpy drop into shaft work. The same balance, reversed in sign, sizes the compressor in the next example and the pump in Chapter 10.

Result. ẇ = 769.9 kJ/kg, so the 2 kg/s turbine delivers about 1.54 MW.

Worked example 2: accelerating air in a nozzle

Air enters an adiabatic nozzle at 300 kPa and 200 °C with negligible velocity and leaves at 100 °C. Treating air as an ideal gas with c_p = 1.005 kJ/kg·K, find the exit velocity.

Figure 2. A nozzle has no shaft work and little heat, so the drop in enthalpy as the air cools reappears as kinetic energy, accelerating the stream.

ProblemFind the exit velocity of the air in Figure 2.
Given / findT₁ = 473.15 K, T₂ = 373.15 K, V₁ ≈ 0, c_p = 1005 J/kg·K. Find V₂.
AssumptionsSteady, adiabatic, no shaft work; ideal gas with constant c_p; inlet velocity negligible.
ModelThe nozzle balance h₁ + V₁²/2 = h₂ + V₂²/2 gives V₂ from the enthalpy drop c_p(T₁ − T₂).
Equationsh₁ − h₂ = c_p(T₁ − T₂)V₂ = √(2 c_p(T₁ − T₂))
Solvec_p(T₁ − T₂) = 1005 × (473.15 − 373.15) = 1005 × 100 = 100 500 J/kg. V₂ = √(2 × 100 500) = √201 000 = 448 m/s.
CheckUsing c_p in J/kg·K keeps velocity in m/s. A 100 K cooling producing roughly 450 m/s is typical of nozzle flows approaching the speed of sound.
ConclusionA nozzle trades enthalpy for speed. The same balance run backwards describes a diffuser, which slows a stream and raises its pressure.

Result. The air leaves the nozzle at about 448 m/s.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Using u instead of h	Flow work left out of the balance	"Is mass crossing the boundary here?"	Flowing streams carry enthalpy h = u + Pv, not just u.
Keeping kinetic energy everywhere	Turbine work distorted by tiny velocities	"Are the velocities actually large?"	Drop ΔKE unless a nozzle or high-speed flow makes it significant.
Mixing J and kJ	Velocity off by a factor near 32	"Is c_p in J or kJ for this V² term?"	Use J/kg·K with V² in m²/s² to get m/s.
Throttle changes temperature predictably	Assuming a throttle always cools an ideal gas	"Is enthalpy constant, and is this an ideal gas?"	For an ideal gas h depends only on T, so an ideal gas throttle keeps T; real gases may cool or warm.

Practice ladder

Level 1 · Direct skill

An adiabatic turbine takes steam at h₁ = 3240 kJ/kg and exhausts it at h₂ = 2400 kJ/kg. Find the work per kilogram.

Show answer

ẇ = h₁ − h₂ = 3240 − 2400 = 840 kJ/kg. The work is just the enthalpy drop because the turbine is adiabatic with small kinetic and potential changes.

Level 2 · Mixed concept

The turbine of Worked Example 1 runs at 5 kg/s instead of 2 kg/s, same states. What is the new power?

Show answer

Power scales with mass flow: Ẇ = 5 × 769.9 = 3849.5 kW ≈ 3.85 MW. The specific work per kilogram is unchanged.

Level 3 · Independent problem

An air compressor raises air from 20 °C to 180 °C adiabatically. With c_p = 1.005 kJ/kg·K and ṁ = 0.4 kg/s, find the power input.

Show answer

ẇ_in = h₂ − h₁ = c_p(T₂ − T₁) = 1.005 × 160 = 160.8 kJ/kg. Ẇ = ṁẇ = 0.4 × 160.8 = 64.3 kW. A compressor is a turbine run in reverse, so work goes in.

Level 4 · Transfer to real engineering

Pick a flow device you can see (a jet engine nozzle, a hair-dryer fan, a pressure regulator). Draw a control volume, name which steady-flow terms survive, and state the reduced balance.

What good work looks like

A clear control volume, an honest list of which terms are negligible, and the correct reduced form (work for a turbine, kinetic energy for a nozzle, constant enthalpy for a throttle).

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check which steady-flow terms I dropped and whether each is justified."

"Give me five devices; I will write the reduced energy balance for each."

"What is the turbine power?" Taking the enthalpy drop times mass flow yourself is the skill.

"Solve this nozzle." Converting the enthalpy drop into velocity is the point.

Portfolio task

Analyse one real flow device with a control volume: draw the boundary, write the mass and energy balances, justify the terms you drop, and compute one output (work, velocity, or exit state).

Must include: a control-volume sketch, a reduced steady-flow balance, and one computed result with units.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. State conservation of mass for a steady control volume.

ṁ_in = ṁ_out, with ṁ = ρAV = AV/v through any section.

2. Why does enthalpy appear in open systems?

A flowing stream carries internal energy plus flow work Pv, and h = u + Pv bundles them together.

3. Reduce the energy equation for a turbine.

Adiabatic with small kinetic and potential changes gives ẇ = h₁ − h₂.

4. What is conserved across a throttle?

Enthalpy: h₁ = h₂, because there is no work and little kinetic change.

5. When does the kinetic term matter?

Where velocities are large, such as a nozzle or diffuser; elsewhere it is usually negligible.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the turbine and nozzle results from a blank page.

+3 daysReduce the energy equation for three new devices.

+7 daysCarry control volumes into the second law, Chapter 6.

+30 daysReuse the device models to build the cycles of Chapters 9 and 10.

Textbook mapping

Item	Mapping
Primary source	Borgnakke and Sonntag, Fundamentals of Thermodynamics, Chapter 6 (First-Law Analysis for a Control Volume)
Cross-reference	Çengel and Boles, Ch. 5 · Moran, Ch. 4
Core topics	5.1 Conservation of mass · 5.2 Flow work and enthalpy · 5.3 Steady-flow energy equation · 5.4 Device models · 5.5 Choosing the boundary
Engineering connection	Turbines, compressors, nozzles, and throttles are the parts of every cycle.
Read next	Chapter 6: The Second Law of Thermodynamics.