Thermodynamics

Energy, entropy, cycles and engines: systems and properties, steam and ideal-gas tables, work and heat, the first law for closed systems and flowing devices, the second law and Carnot limits, entropy and isentropic efficiency, exergy, Otto, Diesel and Brayton cycles, and Rankine and refrigeration cycles, with six labs. Free, self-paced, and tied to power plants, heat pumps, combined heat and power, concentrating solar and geothermal energy.

NASA/Michael DeMocker (MAF20210116_GR0006)NASA/SDONASA/Kim Shiflett (KSC-20210927-PH-KLS04_0013, Sept. 27, 2021)NASA Blue Marble

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About this course

This is a free self-study course outline written by Saving Planets. It is not a university course, it carries no academic credit, and no school is involved.

Its scope follows a one-semester junior engineering thermodynamics course. It is modeled on the topics and depth of MIT 16.050 Thermal Energy and the thermodynamics parts of MIT Unified Engineering (both free on MIT OpenCourseWare), and on courses such as Purdue ME 200 and Georgia Tech ME 3322, using the free OpenStax University Physics Volume 2 chapters on heat and the laws of thermodynamics and NIST’s free fluid property data, which we link where they help. None of those authors or schools endorses this page.

What we add: every unit ties to real work with a cited public source (power-plant efficiency, nuclear and concentrating solar plants, heat pumps, combined heat and power and geothermal energy), every number on the page is re-computed by a checking script before it goes live, and every lab has a free simulation or a safe home version.

Work through it in order. Each unit has goals, key ideas, equations, step-by-step examples, practice with full answers, common mistakes, flashcards and a quiz. Your progress is saved only in this browser.

How this course helps save the planet

Almost all electricity from fuel, uranium, concentrated sunlight or Earth’s heat comes from a thermodynamic cycle, and almost all heating and cooling runs on one too. The first law says energy is conserved; the second law says how much of it can become useful work. Between them they decide how much fuel the world burns.

Better cycles, recovered waste heat and heat pumps are some of the fastest ways to cut emissions. Each unit ends with a short, sourced section. Values that are not from a source are labeled as sample values.

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Before you start

Prerequisites

  • Calculus II and III: integrals, partial derivatives and natural logarithms.
  • Physics I and General Chemistry I: energy, work, ideal gases and heat.
  • Differential Equations is helpful but not required.
  • Comfort with spreadsheets for tables and interpolation.

How to study

  • Start every problem by drawing the system or control volume and saying whether it is closed or open, steady or not.
  • List the known states and look up properties before writing any balance.
  • Write the energy balance in words first (energy in − energy out = change in stored energy), then cancel terms you can justify.
  • Keep units consistent: kJ, kPa, m³, kg and kelvin; never use °C inside a ratio of temperatures.
  • Sketch every process on a T–s or p–v diagram; cycles become much easier to follow.
  • Do every practice problem on paper before opening the answer, and retake the quizzes a few days later.

Lab equipment (for the wet labs)

  • A computer or tablet for the free PhET Gas Properties, Energy Forms and Changes and States of Matter simulations.
  • A kitchen thermometer (or a cheap digital probe), a measuring cup and a kitchen scale.
  • An electric kettle with a known power rating, ice cubes, an insulated mug or foam cup.
  • A phone stopwatch.
  • Spreadsheet software for the cycle labs.

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Suggested 15-week schedule

About 6 to 9 hours a week: reading, practice problems and, in lab weeks, the lab.

WeekUnitTopicsReading
1Systems, Properties, Energy and EfficiencySystems, properties, temperature, energy units, heat rateOpenStax UP2 1.1, 3.1; EIA
2Pure Substances, Phase Change and Property TablesPhases, saturation, quality; Lab AUP2 1.5, 2.1; NIST WebBook
3Pure Substances, Phase Change and Property TablesSteam tables, interpolation, ideal gasNIST WebBook
4Energy Transfer by Work and HeatBoundary, shaft and electrical work; heat transfer modes; Lab DUP2 3.2, 1.6
5The First Law for Closed SystemsFirst law for closed systems, specific heats, latent heat; Labs B and CUP2 3.3, 3.5
6The First Law for Control Volumes: Steady-Flow DevicesMass and energy for control volumes; nozzles, turbines, compressorsMIT 16.050
7The First Law for Control Volumes: Steady-Flow DevicesThrottles, mixing, heat exchangers; midterm practiceMIT 16.050
8The Second Law: Heat Engines, Refrigerators and Carnot LimitsSecond law, engines, heat pumps, Carnot; Lab EUP2 4.2–4.5
9Entropy and Isentropic EfficiencyEntropy, Clausius inequality, entropy generationUP2 4.6–4.7
10Entropy and Isentropic EfficiencyIsentropic processes and efficienciesUP2 3.6; NASA Glenn
11Exergy: How Much Energy Is UsefulExergy, irreversibility, second-law efficiencyMIT 16.050
12Gas Power Cycles: Otto, Diesel and BraytonOtto and Diesel cyclesNASA Glenn
13Gas Power Cycles: Otto, Diesel and BraytonBrayton cycle, regeneration, combined cyclesNASA Glenn; MIT Unified
14Vapor Power and Refrigeration CyclesRankine cycle and improvementsNIST WebBook
15Vapor Power and Refrigeration CyclesRefrigeration and heat pumps; Lab F; final practiceUP2 4.3

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Tick units and labs as you finish them, and take the quizzes. Progress and quiz scores are saved only in this browser, never sent to us.

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Unit 1 · Week 1

Systems, Properties, Energy and Efficiency

Closed systems, control volumes and surroundings, properties and states, equilibrium and quasi-equilibrium processes, temperature scales and the zeroth law, pressure, forms of energy, energy units, and efficiency as useful output over required input.

Learning objectives

By the end of this unit you can:

  • Choose a system or control volume and its boundary.
  • Tell intensive from extensive properties, and state from path.
  • Convert between kJ, kWh and Btu, and between °C and K.
  • Compute an efficiency and a power plant’s efficiency from its heat rate.

Key concepts

  • A system is the region we study; a closed system exchanges energy but not mass with its surroundings, while a control volume (open system) lets mass cross its boundary.
  • Properties such as pressure p, temperature T and specific volume v = V/m describe the state. Intensive properties (p, T, v) do not depend on size; extensive ones (V, m, E) do. Two independent intensive properties fix the state of a simple compressible substance.
  • The zeroth law: two bodies each in thermal equilibrium with a third are in equilibrium with each other; it is why thermometers work. Absolute temperature T(K) = T(°C) + 273.15 must be used in every ratio and every gas law.
  • Energy is stored as internal energy U, kinetic energy ½mV² and potential energy mgz, and it crosses a boundary only as heat Q or work W.
  • Efficiency = useful output ÷ required input. For power plants, the U.S. EIA uses heat rate: Btu of fuel per kWh of electricity, with 1 kWh = 3412 Btu.

Key equations

v=Vm,  ρ=1vv=\frac{V}{m},\ \ \rho=\frac1v
Specific volume and density
T(K)=T(∘C)+273.15T(\mathrm K)=T(^\circ\mathrm C)+273.15
Absolute temperature
E=U+12mV2+mgzE=U+\tfrac12mV^2+mgz
Total stored energy
η=3412 Btu/kWhheat rate\eta=\frac{3412\ \text{Btu/kWh}}{\text{heat rate}}
Power-plant efficiency from heat rate (EIA)

Worked examples, step by step

Example 1.1 EIA gives heat rates of 10,500 Btu/kWh and 7,500 Btu/kWh as examples. Find each efficiency.

Show the steps
  1. η = 3412/10,500.
  2. η = 3412/7500.

Answer: 32.5% and 45.5% (EIA rounds these to 33% and 45%).

Example 1.2 A tank holds 2 kg of gas in 0.5 m³. Find v and ρ.

Show the steps
  1. v = 0.5/2.
  2. ρ = 1/v.

Answer: v = 0.25 m³/kg, ρ = 4 kg/m³.

Example 1.3 A 1000 kg car at 30 m/s brakes to a stop. How much kinetic energy becomes internal energy of the brakes, in kJ and kWh?

Show the steps
  1. KE = ½ × 1000 × 30² = 450,000 J.
  2. 450 kJ ÷ 3600 kJ/kWh.

Answer: 450 kJ = 0.125 kWh.

Practice problems

Work each on paper first, then check.

  1. Convert 25 °C to kelvin.

    Show answer

    298.15 K.

  2. Is temperature intensive or extensive?

    Show answer

    Intensive.

  3. A plant has a heat rate of 9000 Btu/kWh. Find its efficiency.

    Show answer

    3412/9000 = 37.9%.

  4. Convert 1 MWh to GJ.

    Show answer

    3600 MJ = 3.6 GJ.

Common misconceptions

  • Myth: Heat and temperature are the same thing.

    Actually: Temperature is a property; heat is energy in transit.

  • Myth: A system “contains heat.”

    Actually: It contains internal energy; heat exists only crossing a boundary.

  • Myth: Ratios can use °C.

    Actually: Gas laws and efficiencies need kelvin.

Flashcards

  • System

    Region under study.

  • Control volume

    Mass can cross.

  • Specific volume

    v = V/m.

  • Zeroth law

    Basis of thermometers.

  • Heat rate

    Btu per kWh.

  • 1 kWh

    3600 kJ = 3412 Btu.

Self-check quiz

1. A control volume allows…

Show answer

mass to cross its boundary. Open system.

2. Which is intensive?

Show answer

pressure. Independent of size.

3. A heat rate of 6824 Btu/kWh means efficiency of…

Show answer

50%. 3412/6824.

4. Energy crosses a closed-system boundary as…

Show answer

heat and work. Only two ways.

How this helps save the planet

The U.S. Energy Information Administration measures power-plant efficiency with the heat rate: the fuel energy a plant uses to deliver one kilowatthour to the grid, counting the electricity the plant itself uses for fuel feeding, boiler pumps, cooling and pollution control. Dividing 3,412 Btu (the energy in one kWh) by the heat rate gives efficiency; EIA’s examples are a 10,500 Btu plant at about 33% and a 7,500 Btu plant at about 45%.

That gap is the whole point of this course. For the same electricity, the 7,500 Btu plant burns 7,500/10,500 = 71.4% as much fuel, so about 28.6% less fuel and CO₂. The laws of thermodynamics explain where the rest of the energy goes and how engineers claw some of it back.

On Saving Planets: Solutions · Circuits

Sources: U.S. EIA: What is the efficiency of different types of power plants?

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Unit 2 · Weeks 2–3

Pure Substances, Phase Change and Property Tables

Phases and phase change, saturation temperature and pressure, compressed liquid, saturated mixture and superheated vapor, quality, T–v and p–v diagrams, steam tables and interpolation, the ideal-gas equation and when it applies, and the compressibility factor.

Learning objectives

By the end of this unit you can:

  • Identify the phase from p and T using saturation data.
  • Use quality to find v, u and h of a saturated mixture.
  • Interpolate in property tables.
  • Use the ideal-gas law and know when it fails.

Key concepts

  • A pure substance has a fixed chemical makeup. At a given pressure, water boils at one saturation temperature (99.61 °C at 100 kPa); while it boils, p and T stay fixed and the mixture’s quality x = m_vapor/m_total grows from 0 to 1.
  • For a saturated mixture any specific property is y = y_f + x(y_g − y_f) = y_f + x·y_fg, using saturated-liquid (f) and saturated-vapor (g) values from the tables.
  • If T is below the saturation temperature at that pressure, the water is a compressed liquid (approximately saturated liquid at T); if above, it is superheated vapor, found in separate tables.
  • Linear interpolation: y = y₁ + (x − x₁)(y₂ − y₁)/(x₂ − x₁). NIST’s free WebBook gives the same data for water and many refrigerants.
  • At low pressure and high temperature relative to the critical point, gases obey pv = RT with R = R_u/M; for air R = 0.287 kJ/(kg·K). Near saturation (steam in a boiler) use the tables instead.

Key equations

x=mgmf+mgx=\frac{m_g}{m_f+m_g}
Quality
y=yf+x yfgy=y_f+x\,y_{fg}
Saturated-mixture property
pv=RT,  R=RuMpv=RT,\ \ R=\frac{R_u}{M}
Ideal-gas law
y=y1+x−x1x2−x1(y2−y1)y=y_1+\frac{x-x_1}{x_2-x_1}(y_2-y_1)
Linear interpolation

Worked examples, step by step

Example 2.1 At 100 kPa, h_f = 417.51 and h_fg = 2257.5 kJ/kg, v_f = 0.001043 and v_g = 1.6941 m³/kg (steam tables). Find h and v at quality 0.6.

Show the steps
  1. h = 417.51 + 0.6 × 2257.5.
  2. v = 0.001043 + 0.6 × (1.6941 − 0.001043).

Answer: h = 1772.0 kJ/kg; v = 1.0169 m³/kg.

Example 2.2 Find the mass of air in a 0.5 m³ tank at 100 kPa and 300 K.

Show the steps
  1. m = pV/(RT) = 100 × 0.5/(0.287 × 300).

Answer: m = 0.581 kg.

Example 2.3 Find the pressure of 2 kg of air in 0.5 m³ at 400 K.

Show the steps
  1. p = mRT/V = 2 × 0.287 × 400/0.5.

Answer: p = 459.2 kPa.

Practice problems

Work each on paper first, then check.

  1. Water at 100 kPa and 80 °C is…

    Show answer

    Compressed liquid (80 °C < 99.61 °C).

  2. Water at 100 kPa and 150 °C is…

    Show answer

    Superheated vapor.

  3. Interpolate h at 25 °C if h = 83.9 at 20 °C and 125.7 at 30 °C (kJ/kg, saturated liquid).

    Show answer

    83.9 + 0.5 × 41.8 = 104.8 kJ/kg.

  4. Find R for helium (M = 4.003 kg/kmol).

    Show answer

    8.314/4.003 = 2.077 kJ/(kg·K).

Common misconceptions

  • Myth: Temperature rises while water boils.

    Actually: At constant pressure it stays at T_sat until all the liquid is gone.

  • Myth: Steam in a boiler is an ideal gas.

    Actually: Near saturation it is not; use steam tables.

  • Myth: Quality applies to superheated vapor.

    Actually: Only to saturated mixtures, 0 ≤ x ≤ 1.

Flashcards

  • Saturation

    Liquid and vapor coexist at T_sat(p).

  • Quality

    x = m_g/m.

  • Mixture property

    y_f + x y_fg.

  • Compressed liquid

    T < T_sat.

  • Superheated

    T > T_sat.

  • Ideal gas

    pv = RT.

Self-check quiz

1. Quality is…

Show answer

vapor mass over total mass. x = m_g/m.

2. Water at 100 kPa and 120 °C is…

Show answer

superheated vapor. Above T_sat.

3. R for air is about…

Show answer

0.287. kJ/(kg·K).

4. During boiling at constant p, T…

Show answer

stays constant. Saturation.

How this helps save the planet

The U.S. Department of Energy explains that nuclear reactors use fission heat to make steam that spins a turbine. All U.S. commercial reactors are light-water reactors: more than 65% are pressurized-water reactors, which keep core water under high pressure so it does not boil and make steam in a separate heat exchanger, and roughly a third are boiling-water reactors, which make steam directly in the vessel.

Both designs are exercises in this unit: raising the pressure raises the saturation temperature so water stays liquid, and the steam’s quality and enthalpy set how much work the turbine gets. DOE counts more than 400 commercial reactors worldwide, including 94 in the United States, all built on steam tables.

On Saving Planets: Solutions · General Chemistry I

Sources: U.S. DOE: Nuclear 101: How does a nuclear reactor work? (May 2025) · NIST Chemistry WebBook: Thermophysical properties of fluid systems

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Unit 3 · Week 4

Energy Transfer by Work and Heat

Sign conventions, boundary work and its area on a p–V diagram, constant-pressure, isothermal and polytropic processes, shaft and electrical work, and the three modes of heat transfer: conduction, convection and radiation.

Learning objectives

By the end of this unit you can:

  • Compute boundary work for common processes.
  • Show that work and heat are path functions.
  • Compute shaft and electrical work and power.
  • Estimate conduction and convection heat rates.

Key concepts

  • Sign convention: heat into the system and work done by the system are positive (Q_in, W_out). Both are path functions: they depend on the process, not just the end states.
  • Boundary work W_b = ∫p dV is the area under the process curve on a p–V diagram. At constant pressure W_b = p(V₂ − V₁).
  • For an ideal gas at constant temperature, W_b = mRT ln(V₂/V₁). For a polytropic process pVⁿ = constant (n ≠ 1), W_b = (p₂V₂ − p₁V₁)/(1 − n).
  • Shaft work: power = torque × angular speed, Ẇ = 2πṅT. Electrical work: W = VIΔt = PΔt.
  • Heat moves by conduction (Q̇ = kAΔT/L through a wall), convection (Q̇ = hAΔT to a moving fluid) and radiation (Q̇ = εσA(T_s⁴ − T_surr⁴)). Insulation lowers k; this is why building envelopes matter.

Key equations

Wb=∫12p dVW_b=\int_1^2 p\,dV
Boundary work
Wb=mRTln⁡V2V1W_b=mRT\ln\frac{V_2}{V_1}
Isothermal ideal gas
Wb=p2V2−p1V11−nW_b=\frac{p_2V_2-p_1V_1}{1-n}
Polytropic process
Q˙=kA ΔTL,  Q˙=hA ΔT\dot Q=\frac{kA\,\Delta T}{L},\ \ \dot Q=hA\,\Delta T
Conduction and convection

Worked examples, step by step

Example 3.1 1 kg of air expands isothermally at 300 K to twice its volume. Find the work.

Show the steps
  1. W = 1 × 0.287 × 300 × ln 2.

Answer: W = 59.7 kJ (done by the gas).

Example 3.2 Air is compressed polytropically (n = 1.3) from 100 kPa and 0.2 m³ to 0.1 m³. Find p₂ and the work.

Show the steps
  1. p₂ = 100 × 2^1.3 = 246.2 kPa.
  2. W = (246.2 × 0.1 − 100 × 0.2)/(1 − 1.3).

Answer: p₂ = 246.2 kPa; W = −15.4 kJ (work done on the gas).

Example 3.3 A wall of area 10 m², thickness 0.1 m and k = 0.04 W/(m·K) (sample insulation) has 20 K across it. Find the heat loss.

Show the steps
  1. Q̇ = 0.04 × 10 × 20/0.1.

Answer: Q̇ = 80 W.

Practice problems

Work each on paper first, then check.

  1. Gas expands at a constant 200 kPa from 0.5 to 0.8 m³. Work?

    Show answer

    200 × 0.3 = 60 kJ.

  2. A shaft delivers 50 N·m at 3000 rpm. Power?

    Show answer

    2π × 50 × 50 = 15.7 kW.

  3. A 1500 W heater runs 10 minutes. Electrical work?

    Show answer

    1500 × 600 = 900 kJ.

  4. Convection: h = 10 W/(m²·K), A = 2 m², ΔT = 30 K. Heat rate?

    Show answer

    600 W.

Common misconceptions

  • Myth: Work is a property of the state.

    Actually: It depends on the path.

  • Myth: A rigid tank can do boundary work.

    Actually: No; dV = 0.

  • Myth: Doubling insulation thickness halves the temperature.

    Actually: It halves the conduction heat rate at the same ΔT.

Flashcards

  • Sign convention

    Q in, W out positive.

  • Boundary work

    ∫p dV.

  • Isothermal

    mRT ln(V₂/V₁).

  • Polytropic

    (p₂V₂ − p₁V₁)/(1 − n).

  • Shaft power

    2πṅT.

  • Conduction

    kAΔT/L.

Self-check quiz

1. Boundary work equals…

Show answer

∫p dV. Area under p–V curve.

2. Isothermal ideal-gas work is…

Show answer

mRT ln(V₂/V₁). pV = const.

3. Heat and work are…

Show answer

path functions. Depend on the process.

4. Conduction heat rate is proportional to…

Show answer

ΔT/L. kAΔT/L.

How this helps save the planet

The U.S. Department of Energy explains that concentrating solar-thermal power uses mirrors to concentrate sunlight onto a receiver, heating a high-temperature fluid. That heat spins a turbine or powers an engine to make electricity, and it can also serve industry, from desalination to food processing and chemical production.

DOE notes that utility-scale plants use power towers or rows of linear mirrors, while single dish/engine systems produce 5 to 25 kilowatts each. Every one of them is a heat-transfer and work problem: absorb concentrated radiation, limit conduction and convection losses from the receiver, and turn the heat into shaft work.

On Saving Planets: Solutions · Physics II

Sources: U.S. DOE: Concentrating solar-thermal power basics

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Unit 4 · Week 5

The First Law for Closed Systems

Energy balance for closed systems, internal energy and enthalpy, specific heats of ideal gases, liquids and solids, constant-volume and constant-pressure heating, latent heat, and energy balances for cycles.

Learning objectives

By the end of this unit you can:

  • Write and apply Q − W = ΔU + ΔKE + ΔPE.
  • Use c_v and c_p for ideal gases, and c for liquids and solids.
  • Include latent heat in phase-change problems.
  • Show that for a cycle the net heat equals the net work.

Key concepts

  • First law for a closed system: Q − W = ΔE = ΔU + ΔKE + ΔPE. Usually ΔKE and ΔPE are negligible, so Q − W = ΔU.
  • Enthalpy h = u + pv. At constant pressure, the heat added equals the enthalpy change: Q = m(h₂ − h₁), because the boundary work is already included.
  • For ideal gases u and h depend only on T: Δu = c_vΔT, Δh = c_pΔT, with c_p − c_v = R and k = c_p/c_v (1.4 for air).
  • For liquids and solids c_p ≈ c_v = c, so Q = mcΔT. Melting or boiling at constant T needs latent heat: Q = m h_if or m h_fg.
  • Over a full cycle the system returns to its starting state, so ΔU = 0 and Q_net = W_net. This is the basis of every engine and refrigerator.

Key equations

Q−W=ΔU+ΔKE+ΔPEQ-W=\Delta U+\Delta KE+\Delta PE
First law, closed system
h=u+pvh=u+pv
Enthalpy
Δu=cvΔT, Δh=cpΔT, cp−cv=R\Delta u=c_v\Delta T,\ \Delta h=c_p\Delta T,\ c_p-c_v=R
Ideal-gas specific heats
Q=mcΔT,  Q=m hfgQ=mc\Delta T,\ \ Q=m\,h_{fg}
Sensible and latent heat

Worked examples, step by step

Example 4.1 2 kg of air in a rigid tank is heated from 300 K to 400 K. Find Q.

Show the steps
  1. W = 0 (rigid).
  2. Q = m c_v ΔT = 2 × 0.718 × 100.

Answer: Q = 143.6 kJ.

Example 4.2 The same air is heated at constant pressure (100 kPa) from 300 to 400 K. Find Q and the boundary work.

Show the steps
  1. Q = m c_p ΔT = 2 × 1.005 × 100.
  2. W = mRΔT = 2 × 0.287 × 100.

Answer: Q = 201.0 kJ; W = 57.4 kJ; ΔU = 143.6 kJ either way.

Example 4.3 Find the heat to warm 1 kg of water from 20 °C to 100 °C, and then to boil 0.1 kg of it away.

Show the steps
  1. Q₁ = 1 × 4.18 × 80.
  2. Q₂ = 0.1 × 2257.

Answer: Q₁ = 334.4 kJ; Q₂ = 225.7 kJ.

Practice problems

Work each on paper first, then check.

  1. Melt 0.2 kg of ice at 0 °C. Heat?

    Show answer

    0.2 × 334 = 66.8 kJ.

  2. Find c_p − c_v for air.

    Show answer

    1.005 − 0.718 = 0.287 kJ/(kg·K) = R.

  3. A cycle receives 500 kJ of heat and rejects 300 kJ. Net work?

    Show answer

    200 kJ.

  4. Find k for air.

    Show answer

    1.005/0.718 = 1.40.

Common misconceptions

  • Myth: Heat added always raises temperature.

    Actually: During phase change it does not.

  • Myth: Q = mc_pΔT works for a rigid tank.

    Actually: Use c_v when volume is constant.

  • Myth: Enthalpy is a form of heat.

    Actually: It is a property, u + pv.

Flashcards

  • First law

    Q − W = ΔU.

  • Enthalpy

    h = u + pv.

  • Ideal gas Δu

    c_vΔT.

  • Ideal gas Δh

    c_pΔT.

  • Latent heat

    Q = m h_fg.

  • Cycle

    Q_net = W_net.

Self-check quiz

1. For a rigid closed tank, Q equals…

Show answer

ΔU. No boundary work.

2. c_p − c_v for an ideal gas is…

Show answer

R. Mayer’s relation.

3. Over a cycle, ΔU is…

Show answer

zero. Same start and end state.

4. At constant pressure, Q equals…

Show answer

ΔH. h = u + pv.

How this helps save the planet

DOE reports that heat pumps are growing fast: as of 2020 over 17 million U.S. housing units had them, and they now account for more than half of all residential heating equipment sales in the United States. They work like a refrigerator, using electricity to move heat from a cool space to a warm one rather than making heat.

The first law is why that matters: an electric resistance heater turns each kWh of electricity into exactly one kWh of heat, but a heat pump delivers the electricity plus the heat it pulls from outside. DOE says modern air-source heat pumps can cut electricity use by 50% compared with furnaces and baseboard heaters, and that switching could save up to 7.6 tons of carbon emissions a year depending on the equipment replaced.

On Saving Planets: Solutions · Calculus II

Sources: U.S. DOE: Pump up your savings with heat pumps (May 30, 2023)

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Unit 5 · Weeks 6–7

The First Law for Control Volumes: Steady-Flow Devices

Mass and energy balances for control volumes, flow work and enthalpy, steady-flow devices (nozzles and diffusers, turbines and compressors, pumps, throttling valves, mixing chambers and heat exchangers), and an introduction to unsteady filling.

Learning objectives

By the end of this unit you can:

  • Write steady-flow mass and energy balances.
  • Explain why enthalpy appears for flowing streams.
  • Analyze nozzles, turbines, compressors, throttles and heat exchangers.
  • Choose which terms can be neglected for each device.

Key concepts

  • Mass balance: in steady flow, Σṁ_in = Σṁ_out, with ṁ = ρAV.
  • Pushing fluid across a boundary takes flow work pv per unit mass; adding it to u gives enthalpy h, which is why open-system balances use h.
  • Steady-flow energy equation, one inlet and one outlet: Q̇ − Ẇ = ṁ[(h₂ − h₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)].
  • Device shortcuts: nozzles trade enthalpy for speed (Q̇ = Ẇ = 0); turbines and compressors are near adiabatic with Ẇ = ṁ(h₁ − h₂); throttling valves keep h₁ = h₂; heat exchangers have Ẇ = 0 and the heat lost by one stream equals that gained by the other.
  • Remember the units: kinetic energy in J/kg must be divided by 1000 to add to kJ/kg enthalpy.

Key equations

∑m˙in=∑m˙out\sum\dot m_{in}=\sum\dot m_{out}
Steady mass balance
Q˙−W˙=m˙[Δh+ΔV22+gΔz]\dot Q-\dot W=\dot m\left[\Delta h+\frac{\Delta V^2}{2}+g\Delta z\right]
Steady-flow energy equation
W˙turb=m˙(h1−h2)\dot W_{turb}=\dot m(h_1-h_2)
Adiabatic turbine
h1=h2h_1=h_2
Throttling valve

Worked examples, step by step

Example 5.1 Air enters an adiabatic nozzle slowly at 400 K and leaves at 350 K. Find the exit speed.

Show the steps
  1. Δh = c_pΔT = 1.005 × 50 = 50.25 kJ/kg = 50,250 J/kg.
  2. V₂ = √(2 × 50,250).

Answer: V₂ = 317.0 m/s.

Example 5.2 Steam flows at 5 kg/s through an adiabatic turbine from h₁ = 3116.1 to h₂ = 2403.0 kJ/kg. Find the power.

Show the steps
  1. Ẇ = 5 × (3116.1 − 2403.0).

Answer: Ẇ = 3565.5 kW ≈ 3.57 MW.

Example 5.3 An adiabatic compressor takes 1.2 kg/s of air from 300 K to 450 K. Find the power input.

Show the steps
  1. Ẇ_in = ṁc_pΔT = 1.2 × 1.005 × 150.

Answer: 180.9 kW.

Practice problems

Work each on paper first, then check.

  1. Water at 0.5 kg/s is heated from 15 to 45 °C in a heat exchanger. Heat rate?

    Show answer

    0.5 × 4.18 × 30 = 62.7 kW.

  2. Steam is throttled from h = 2800 kJ/kg. Exit h?

    Show answer

    2800 kJ/kg.

  3. Convert a speed change from 0 to 100 m/s into kJ/kg.

    Show answer

    100²/2 = 5000 J/kg = 5 kJ/kg.

  4. A pipe has ṁ = 2 kg/s at inlet. Outlet in steady flow?

    Show answer

    2 kg/s.

Common misconceptions

  • Myth: Turbine power uses u.

    Actually: Flowing streams use h, which includes flow work.

  • Myth: A throttle lowers enthalpy.

    Actually: h stays constant; p and usually T drop.

  • Myth: Kinetic energy is always negligible.

    Actually: Not in nozzles and diffusers.

Flashcards

  • Steady mass

    ṁ in = ṁ out.

  • Flow work

    pv per kg.

  • SFEE

    Q̇ − Ẇ = ṁΔ(h + V²/2 + gz).

  • Turbine

    ṁ(h₁ − h₂).

  • Throttle

    h₁ = h₂.

  • Heat exchanger

    Heat lost = heat gained.

Self-check quiz

1. For a throttling valve…

Show answer

h₁ = h₂. No work, no heat, small ΔKE.

2. Enthalpy appears in open systems because of…

Show answer

flow work. h = u + pv.

3. A nozzle converts…

Show answer

enthalpy into speed. Accelerates flow.

4. Adiabatic turbine power is…

Show answer

ṁ(h₁ − h₂). Enthalpy drop.

How this helps save the planet

DOE describes the core of a nuclear plant as a chain of steady-flow devices: fission heats water, a heat exchanger (in pressurized-water reactors) makes steam in a separate loop, the steam spins a turbine connected to a generator, and leftover steam is condensed back to water and pumped back to be reheated.

Each box is one steady-flow energy equation. Geothermal plants, which DOE says draw hot fluid up through wells and convert its heat to steam that drives turbines, use the same devices with Earth’s heat as the source. Getting a few more kJ per kg out of every turbine stage is how plant efficiency climbs.

On Saving Planets: Solutions · Aerodynamics I

Sources: U.S. DOE: Nuclear 101: How does a nuclear reactor work? (May 2025) · U.S. DOE: Geothermal basics

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Unit 6 · Week 8

The Second Law: Heat Engines, Refrigerators and Carnot Limits

Thermal reservoirs, heat engines and thermal efficiency, refrigerators, heat pumps and COP, the Kelvin–Planck and Clausius statements, reversible and irreversible processes, the Carnot cycle, and the Carnot limits on efficiency and COP.

Learning objectives

By the end of this unit you can:

  • Compute thermal efficiency and COP.
  • State the Kelvin–Planck and Clausius statements.
  • Identify sources of irreversibility.
  • Use Carnot limits to judge whether a claimed device is possible.

Key concepts

  • A heat engine takes Q_H from a hot reservoir, rejects Q_L to a cold one, and produces W = Q_H − Q_L. Thermal efficiency η = W/Q_H = 1 − Q_L/Q_H.
  • Kelvin–Planck: no cycle can turn all the heat from a single reservoir into work, so η < 100%. Clausius: heat does not flow from cold to hot by itself; a refrigerator or heat pump needs work input.
  • COP_R = Q_L/W for a refrigerator; COP_HP = Q_H/W = COP_R + 1 for a heat pump. COP can be well above 1, which is why heat pumps beat resistance heaters.
  • Irreversibilities (friction, heat transfer across a finite ΔT, unrestrained expansion, mixing) reduce performance. A reversible cycle between two reservoirs, the Carnot cycle, is the best possible.
  • Carnot limits use absolute temperatures: η_Carnot = 1 − T_L/T_H, COP_R,Carnot = T_L/(T_H − T_L), COP_HP,Carnot = T_H/(T_H − T_L).

Key equations

η=WQH=1−QLQH\eta=\frac{W}{Q_H}=1-\frac{Q_L}{Q_H}
Thermal efficiency
COPR=QLW, COPHP=QHWCOP_R=\frac{Q_L}{W},\ COP_{HP}=\frac{Q_H}{W}
Coefficients of performance
ηCarnot=1−TLTH\eta_{Carnot}=1-\frac{T_L}{T_H}
Carnot efficiency
COPHP,Carnot=THTH−TLCOP_{HP,Carnot}=\frac{T_H}{T_H-T_L}
Carnot heat pump

Worked examples, step by step

Example 6.1 An engine takes 1000 kJ of heat and produces 350 kJ of work. Find η and Q_L.

Show the steps
  1. η = 350/1000.
  2. Q_L = 1000 − 350.

Answer: η = 35%; Q_L = 650 kJ.

Example 6.2 Find the Carnot efficiency between 800 K and 300 K, and between 1073 K and 293 K.

Show the steps
  1. 1 − 300/800.
  2. 1 − 293/1073.

Answer: 62.5% and 72.7%.

Example 6.3 A heat pump keeps a house at 294 K when it is 273 K outside. Find the Carnot COP, and the heat delivered by a real unit with COP = 3 (sample value) using 3 kWh.

Show the steps
  1. COP = 294/(294 − 273).
  2. Q_H = 3 × 3.

Answer: COP_Carnot = 14.0; the real unit delivers 9 kWh of heat.

Practice problems

Work each on paper first, then check.

  1. A fridge has COP_R = 2.5 and removes 500 kJ. Work?

    Show answer

    200 kJ.

  2. Find COP_HP if COP_R = 2.5.

    Show answer

    3.5.

  3. A fridge works between 275 K and 298 K. Carnot COP_R?

    Show answer

    275/23 = 11.96.

  4. An inventor claims 70% efficiency between 600 K and 300 K. Possible?

    Show answer

    No: Carnot limit is 50%.

Common misconceptions

  • Myth: COP above 1 breaks energy conservation.

    Actually: The heat pump moves heat; it does not create it.

  • Myth: Carnot efficiency can use °C.

    Actually: It must use kelvin.

  • Myth: Better engineering can reach 100% efficiency.

    Actually: Kelvin–Planck forbids it.

Flashcards

  • Efficiency

    W/Q_H.

  • COP_R

    Q_L/W.

  • COP_HP

    Q_H/W.

  • Kelvin–Planck

    No 100% engine.

  • Clausius

    No cold-to-hot without work.

  • Carnot

    1 − T_L/T_H.

Self-check quiz

1. Carnot efficiency between 600 K and 300 K is…

Show answer

50%. 1 − 0.5.

2. COP_HP − COP_R equals…

Show answer

1. Q_H = Q_L + W.

3. Which statement forbids a 100% efficient engine?

Show answer

Kelvin–Planck. Single-reservoir statement.

4. Which is an irreversibility?

Show answer

friction. Dissipation.

How this helps save the planet

DOE describes heat pumps as moving heat rather than generating it, which makes them significantly more efficient than traditional heating. Modern air-source heat pumps can cut electricity use by 50% compared with furnaces and baseboard heaters, and geothermal heat pumps, which use the steady temperature of the shallow ground, can reduce energy use by 30% to 60%.

The Carnot COP explains both claims: the smaller the temperature lift T_H − T_L, the higher the COP. The ground stays milder than winter air, so a ground-source heat pump lifts heat across a smaller gap. DOE’s geothermal office sees potential for up to 28 million geothermal heat pumps by 2050.

On Saving Planets: Solutions · Oceans

Sources: U.S. DOE: Pump up your savings with heat pumps (May 30, 2023) · U.S. DOE: Geothermal basics

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Unit 7 · Weeks 9–10

Entropy and Isentropic Efficiency

The Clausius inequality, entropy as a property, the increase-of-entropy principle and entropy generation, the T ds relations, entropy change of liquids, solids and ideal gases, isentropic processes, T–s diagrams, and isentropic efficiencies of turbines, compressors and nozzles.

Learning objectives

By the end of this unit you can:

  • Compute entropy changes for liquids, solids and ideal gases.
  • Find entropy generation and decide if a process is possible.
  • Use isentropic relations for ideal gases.
  • Apply isentropic efficiencies to real devices.

Key concepts

  • Entropy S is a property defined by dS = δQ_rev/T. For any process, S₂ − S₁ ≥ ∫δQ/T; the difference is the entropy generated, S_gen ≥ 0 (zero only if reversible).
  • Increase-of-entropy principle: the entropy of an isolated system (system plus surroundings) never decreases. A process with S_gen < 0 is impossible.
  • T ds relations: T ds = du + p dv = dh − v dp. For liquids and solids Δs = c ln(T₂/T₁); for ideal gases Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁).
  • An isentropic (reversible adiabatic) ideal-gas process with constant specific heats has T₂/T₁ = (p₂/p₁)^((k−1)/k). On a T–s diagram it is a vertical line, and the area under a reversible process is the heat transferred.
  • Real devices are compared with isentropic ones: η_T = (h₁ − h₂a)/(h₁ − h₂s) for turbines and η_C = (h₂s − h₁)/(h₂a − h₁) for compressors.

Key equations

dS=δQrevT,  Sgen≥0dS=\frac{\delta Q_{rev}}{T},\ \ S_{gen}\ge0
Entropy and the second law
Δs=cln⁡T2T1\Delta s=c\ln\frac{T_2}{T_1}
Liquids and solids
Δs=cpln⁡T2T1−Rln⁡p2p1\Delta s=c_p\ln\frac{T_2}{T_1}-R\ln\frac{p_2}{p_1}
Ideal gas
T2T1=(p2p1)(k−1)/k\frac{T_2}{T_1}=\left(\frac{p_2}{p_1}\right)^{(k-1)/k}
Isentropic ideal gas

Worked examples, step by step

Example 7.1 Find the entropy change of 2 kg of water heated from 293 K to 353 K.

Show the steps
  1. ΔS = 2 × 4.18 × ln(353/293).

Answer: ΔS = 1.557 kJ/K.

Example 7.2 Air goes from 300 K and 100 kPa to 500 K and 400 kPa. Find Δs.

Show the steps
  1. Δs = 1.005 ln(500/300) − 0.287 ln(400/100).

Answer: Δs = 0.1155 kJ/(kg·K).

Example 7.3 Air at 300 K is compressed through a pressure ratio of 8. Find the isentropic exit temperature and the actual one if η_C = 0.85.

Show the steps
  1. T₂s = 300 × 8^(0.4/1.4) = 543.4 K.
  2. T₂a = 300 + (543.4 − 300)/0.85.

Answer: T₂s = 543.4 K; T₂a = 586.4 K.

Practice problems

Work each on paper first, then check.

  1. 1000 kJ of heat flows from an 800 K reservoir to a 300 K one. S_gen?

    Show answer

    1000/300 − 1000/800 = 2.083 kJ/K.

  2. Isentropic T₂ for a pressure ratio of 10 from 300 K?

    Show answer

    579.2 K.

  3. Can S_gen be negative?

    Show answer

    No; that process is impossible.

  4. What does area under a reversible T–s curve represent?

    Show answer

    Heat transferred.

Common misconceptions

  • Myth: Entropy can never decrease.

    Actually: A system’s entropy can fall if it rejects heat; the total for an isolated system cannot.

  • Myth: Adiabatic means isentropic.

    Actually: Only if also reversible.

  • Myth: Isentropic efficiency can exceed 1.

    Actually: Not for turbines or compressors.

Flashcards

  • Entropy

    dS = δQ_rev/T.

  • S_gen

    ≥ 0.

  • Liquid Δs

    c ln(T₂/T₁).

  • Ideal gas Δs

    c_p ln(T₂/T₁) − R ln(p₂/p₁).

  • Isentropic T

    T₂/T₁ = (p₂/p₁)^((k−1)/k).

  • η_T

    (h₁ − h₂a)/(h₁ − h₂s).

Self-check quiz

1. S_gen for a reversible process is…

Show answer

zero. No irreversibility.

2. An isentropic process on a T–s diagram is…

Show answer

vertical. s constant.

3. Δs for a liquid heated is…

Show answer

c ln(T₂/T₁). Incompressible.

4. Compressor isentropic efficiency is…

Show answer

(h₂s − h₁)/(h₂a − h₁). Ideal work over actual.

How this helps save the planet

NASA Glenn’s Brayton-cycle page walks through a gas turbine on a T–s diagram: ideally the compressor raises pressure and temperature isentropically, a vertical line, but in reality the compression line leans to the right because entropy increases in the flow. Combustion happens at constant pressure, and the turbine, on the same shaft, drives the compressor.

NASA notes that the area enclosed on the T–s diagram is proportional to the useful work. Every bit of entropy generated in a compressor or turbine shrinks that area, so higher isentropic efficiencies mean more electricity or thrust from the same fuel, and less CO₂.

On Saving Planets: Aerodynamics I · Calculus III

Sources: NASA Glenn: Turbine engine thermodynamic cycle (Brayton cycle)

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Unit 8 · Week 11

Exergy: How Much Energy Is Useful

Dead state and surroundings, exergy (availability) of heat and of flowing streams, reversible work and irreversibility, exergy destruction I = T₀S_gen, second-law efficiency, and why matching energy quality to use matters.

Learning objectives

By the end of this unit you can:

  • Compute the exergy of heat from a reservoir.
  • Relate exergy destroyed to entropy generated.
  • Compute second-law efficiency.
  • Explain why combined heat and power and heat pumps use exergy well.

Key concepts

  • Exergy is the maximum useful work obtainable as a system comes to equilibrium with its surroundings (the dead state, T₀ and p₀). Energy is conserved; exergy is destroyed by irreversibility.
  • Heat Q from a reservoir at T has exergy X = Q(1 − T₀/T): heat at high temperature is valuable, heat near room temperature is almost worthless for work.
  • Exergy of a flowing stream per unit mass: ψ = (h − h₀) − T₀(s − s₀) + V²/2 + gz.
  • Exergy destroyed (irreversibility) I = T₀S_gen (Gouy–Stodola). Every irreversibility costs useful work.
  • Second-law efficiency η_II = η_th/η_rev for engines, or COP/COP_rev for heat pumps. Burning fuel at 2000 K just to heat water to 60 °C wastes most of the exergy; using the waste heat of a power plant for that job does not.

Key equations

XQ=Q(1−T0T)X_Q=Q\left(1-\frac{T_0}{T}\right)
Exergy of heat
ψ=(h−h0)−T0(s−s0)+V22+gz\psi=(h-h_0)-T_0(s-s_0)+\frac{V^2}{2}+gz
Flow exergy
I=T0SgenI=T_0S_{gen}
Exergy destroyed
ηII=ηthηrev\eta_{II}=\frac{\eta_{th}}{\eta_{rev}}
Second-law efficiency

Worked examples, step by step

Example 8.1 Find the exergy of 100 kJ of heat from a 600 K source with T₀ = 300 K.

Show the steps
  1. X = 100 × (1 − 300/600).

Answer: 50 kJ.

Example 8.2 1000 kJ flows from 800 K to 300 K (T₀ = 300 K). Find the exergy destroyed.

Show the steps
  1. S_gen = 1000/300 − 1000/800 = 2.083 kJ/K.
  2. I = 300 × 2.083.

Answer: I = 625 kJ, all the exergy the heat had at 800 K.

Example 8.3 An engine with η = 35% runs between 800 K and 300 K. Find η_II.

Show the steps
  1. η_rev = 62.5%.
  2. η_II = 0.35/0.625.

Answer: η_II = 56%.

Practice problems

Work each on paper first, then check.

  1. Exergy of 100 kJ at 330 K with T₀ = 300 K?

    Show answer

    100 × (1 − 300/330) = 9.09 kJ.

  2. A heat pump has COP 3; COP_rev = 14. η_II?

    Show answer

    3/14 = 21.4%.

  3. Why is room-temperature heat nearly worthless for work?

    Show answer

    Its exergy factor 1 − T₀/T is near zero.

  4. What destroys exergy?

    Show answer

    Every irreversibility: friction, mixing, heat transfer across finite ΔT.

Common misconceptions

  • Myth: Energy and exergy are the same.

    Actually: Energy is conserved; exergy is destroyed.

  • Myth: High first-law efficiency means a good process.

    Actually: A 95% efficient gas boiler can have a low second-law efficiency.

  • Myth: Exergy depends only on the system.

    Actually: It depends on the surroundings (T₀, p₀) too.

Flashcards

  • Exergy

    Maximum useful work.

  • Heat exergy

    Q(1 − T₀/T).

  • Dead state

    T₀, p₀.

  • Irreversibility

    I = T₀S_gen.

  • η_II

    η/η_rev.

  • Energy quality

    High T is valuable.

Self-check quiz

1. Exergy destroyed equals…

Show answer

T₀S_gen. Gouy–Stodola.

2. Exergy of heat at T = T₀ is…

Show answer

zero. 1 − T₀/T = 0.

3. η_II compares actual performance with…

Show answer

the reversible limit. Second-law efficiency.

4. The dead state is…

Show answer

equilibrium with surroundings. T₀, p₀.

How this helps save the planet

DOE explains that combined heat and power (cogeneration) produces electricity and useful heating or cooling from a single fuel at or near where it is used, recovering heat that would otherwise be lost in power generation. DOE says CHP can reasonably operate at 65% to 75% efficiency, versus a national average of about 50% when electricity and heat are produced separately.

Exergy explains the win: the high-temperature part of the fuel’s exergy makes electricity, and the low-grade heat that a power plant would throw away meets heating needs that only ever required low-exergy energy. Going from 50% to 75% means the same services from two-thirds of the fuel.

On Saving Planets: Solutions · Space Weather

Sources: U.S. DOE: Combined heat and power basics

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Unit 9 · Weeks 12–13

Gas Power Cycles: Otto, Diesel and Brayton

Air-standard assumptions, mean effective pressure, the Otto cycle (spark ignition), the Diesel cycle (compression ignition), the Brayton cycle for gas turbines and jet engines, back-work ratio, regeneration, intercooling and reheat, and combined cycles.

Learning objectives

By the end of this unit you can:

  • Analyze ideal Otto and Diesel cycles.
  • Analyze the ideal Brayton cycle, including back-work ratio.
  • Explain how compression ratio and pressure ratio raise efficiency.
  • Describe regeneration and combined cycles.

Key concepts

  • Air-standard assumptions: the working fluid is air, an ideal gas, in a closed cycle; combustion is replaced by heat addition and exhaust by heat rejection. Cold-air-standard adds constant specific heats at room temperature.
  • Otto cycle: isentropic compression, constant-volume heat addition, isentropic expansion, constant-volume heat rejection. η_Otto = 1 − 1/r^(k−1), with compression ratio r = V_max/V_min.
  • Diesel cycle: heat is added at constant pressure; with cutoff ratio r_c, η_Diesel = 1 − (1/r^(k−1))·(r_c^k − 1)/(k(r_c − 1)). Diesels run at higher r, so they are more efficient in practice.
  • Brayton cycle (gas turbines): isentropic compression, constant-pressure heat addition, isentropic expansion, constant-pressure heat rejection. η_Brayton = 1 − 1/r_p^((k−1)/k). A large share of turbine work drives the compressor (the back-work ratio).
  • Regeneration uses hot exhaust to preheat compressed air; intercooling and reheat move the cycle toward Carnot. A combined cycle feeds gas-turbine exhaust to a steam (Rankine) cycle.

Key equations

ηOtto=1−1rk−1\eta_{Otto}=1-\frac{1}{r^{k-1}}
Otto cycle
ηDiesel=1−1rk−1rck−1k(rc−1)\eta_{Diesel}=1-\frac{1}{r^{k-1}}\frac{r_c^k-1}{k(r_c-1)}
Diesel cycle
ηBrayton=1−1rp(k−1)/k\eta_{Brayton}=1-\frac{1}{r_p^{(k-1)/k}}
Brayton cycle
rbw=wcompwturbr_{bw}=\frac{w_{comp}}{w_{turb}}
Back-work ratio

Worked examples, step by step

Example 9.1 Find the cold-air-standard efficiency of Otto cycles with r = 8 and r = 10.

Show the steps
  1. 1 − 8^(−0.4).
  2. 1 − 10^(−0.4).

Answer: 56.5% and 60.2%.

Example 9.2 A Diesel cycle has r = 18 and r_c = 2. Find η.

Show the steps
  1. 1/r^0.4 = 0.3147.
  2. (2^1.4 − 1)/(1.4 × 1) = 1.1703.
  3. η = 1 − 0.3147 × 1.1703.

Answer: η = 63.2%.

Example 9.3 An ideal Brayton cycle has r_p = 8, inlet 300 K and turbine inlet 1300 K. Find T₂, T₄, compressor and turbine work, back-work ratio, net work and η.

Show the steps
  1. T₂ = 300 × 8^0.2857 = 543.4 K; T₄ = 1300/1.8114 = 717.7 K.
  2. w_c = 1.005 × 243.4 = 244.7 kJ/kg; w_t = 1.005 × 582.3 = 585.3 kJ/kg.
  3. r_bw = 244.7/585.3; w_net = 340.6 kJ/kg; η = 1 − 8^(−0.2857).

Answer: r_bw = 0.418, w_net = 340.6 kJ/kg, η = 44.8%.

Practice problems

Work each on paper first, then check.

  1. Brayton efficiency at r_p = 12?

    Show answer

    1 − 12^(−0.2857) = 50.8%.

  2. Why are Diesel engines more efficient than gasoline ones in practice?

    Show answer

    They run at much higher compression ratios.

  3. Raising r in an Otto cycle from 8 to 10 adds how many points?

    Show answer

    60.2 − 56.5 = 3.7 points.

  4. What does regeneration do?

    Show answer

    Uses exhaust heat to preheat compressor air, cutting fuel needed.

Common misconceptions

  • Myth: Higher compression always means more efficient real engines.

    Actually: Knock limits spark-ignition engines.

  • Myth: Back work is negligible in gas turbines.

    Actually: Often 40–80% of turbine work.

  • Myth: Air-standard efficiency is the real efficiency.

    Actually: Real engines are much lower due to friction, heat loss and real gases.

Flashcards

  • Air standard

    Air, ideal gas, closed cycle.

  • Otto

    1 − 1/r^(k−1).

  • Diesel

    Constant-p heat addition.

  • Brayton

    1 − 1/r_p^((k−1)/k).

  • Back work

    w_c/w_t.

  • Regeneration

    Exhaust preheats air.

Self-check quiz

1. Otto efficiency depends on…

Show answer

compression ratio. 1 − 1/r^(k−1).

2. Brayton heat addition is at constant…

Show answer

pressure. Combustor.

3. Back-work ratio is…

Show answer

w_c/w_t. Compressor over turbine.

4. A combined cycle pairs Brayton with…

Show answer

Rankine. Steam bottoming cycle.

How this helps save the planet

NASA Glenn explains that the Brayton cycle is used in all gas turbine engines, whether turbojets, turbofans or turboprops, and that all share a compressor, a combustion section and a turbine that drives the compressor. NASA’s EngineSim program uses Brayton-cycle analysis to predict an engine’s thrust and fuel flow.

Ground-based gas turbines run the same cycle. EIA’s heat-rate examples show why cycle design matters: 7,500 Btu/kWh is about 45% efficient versus about 33% at 10,500 Btu/kWh. Raising pressure ratio and turbine temperature, and recovering exhaust heat, are how plants move toward the better number.

On Saving Planets: Aerodynamics I · Intro to Aerospace Engineering

Sources: NASA Glenn: Turbine engine thermodynamic cycle (Brayton cycle) · U.S. EIA: What is the efficiency of different types of power plants?

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Unit 10 · Weeks 14–15

Vapor Power and Refrigeration Cycles

The ideal Rankine cycle and its four devices, pump work, raising efficiency (higher boiler pressure and temperature, lower condenser pressure, reheat, regeneration), the vapor-compression refrigeration cycle and heat pumps, refrigerants, and cogeneration.

Learning objectives

By the end of this unit you can:

  • Analyze an ideal Rankine cycle with steam tables.
  • Explain how pressure, superheat and reheat change efficiency.
  • Analyze an ideal vapor-compression refrigeration cycle.
  • Relate cycle COP to heat pump performance.

Key concepts

  • Ideal Rankine cycle: pump (isentropic compression of liquid), boiler (constant-pressure heating to superheated steam), turbine (isentropic expansion), condenser (constant-pressure condensation).
  • Pump work is small because liquid has small v: w_p ≈ v(p₂ − p₁). Net work = w_turb − w_pump; η = w_net/q_in.
  • Efficiency rises with higher average heat-addition temperature (higher boiler pressure, more superheat, reheat) and lower condenser pressure. Regeneration with feedwater heaters preheats the water using steam bled from the turbine.
  • Vapor-compression refrigeration: compressor (to superheated vapor), condenser (rejects Q_H), expansion valve (throttle, h₃ = h₄), evaporator (absorbs Q_L). COP_R = (h₁ − h₄)/(h₂ − h₁).
  • A heat pump is the same cycle used for heating, with COP_HP = (h₂ − h₃)/(h₂ − h₁). Refrigerants are chosen for their saturation properties, safety and low global-warming potential.

Key equations

wp=v(p2−p1)w_p=v(p_2-p_1)
Pump work
η=wt−wpqin\eta=\frac{w_t-w_p}{q_{in}}
Rankine efficiency
COPR=h1−h4h2−h1COP_R=\frac{h_1-h_4}{h_2-h_1}
Refrigerator COP
h3=h4h_3=h_4
Expansion valve

Worked examples, step by step

Example 10.1 An ideal Rankine cycle runs between 3 MPa/350 °C and 75 kPa. Steam tables: h₁ = 384.44, v₁ = 0.001037 m³/kg, h₃ = 3116.1, h₄ = 2403.0 kJ/kg. Find w_p, q_in, w_net and η.

Show the steps
  1. w_p = 0.001037 × (3000 − 75) = 3.03 kJ/kg; h₂ = 387.47.
  2. q_in = 3116.1 − 387.47 = 2728.6 kJ/kg.
  3. w_net = (3116.1 − 2403.0) − 3.03 = 710.1 kJ/kg; η = 710.1/2728.6.

Answer: w_p = 3.03 kJ/kg, q_in = 2728.6 kJ/kg, w_net = 710.1 kJ/kg, η = 26.0%.

Example 10.2 A vapor-compression cycle has h₁ = 244.5, h₂ = 275.4 and h₃ = h₄ = 95.5 kJ/kg (sample values). Find q_L, w_in and COP_R.

Show the steps
  1. q_L = 244.5 − 95.5 = 149.0 kJ/kg.
  2. w = 275.4 − 244.5 = 30.9 kJ/kg.
  3. COP = 149.0/30.9.

Answer: q_L = 149.0 kJ/kg, w = 30.9 kJ/kg, COP_R = 4.82.

Example 10.3 For that cycle used as a heat pump, find COP_HP.

Show the steps
  1. q_H = 275.4 − 95.5 = 179.9 kJ/kg.
  2. COP_HP = 179.9/30.9.

Answer: COP_HP = 5.82 (= COP_R + 1).

Practice problems

Work each on paper first, then check.

  1. In the Rankine example, what is the heat rejected?

    Show answer

    2403.0 − 384.44 = 2018.6 kJ/kg.

  2. What fraction of turbine work does the pump use?

    Show answer

    3.03/713.1 = 0.425%.

  3. Find the steam flow for 100 MW net in the example.

    Show answer

    100,000/710.1 = 140.8 kg/s.

  4. Why does lowering condenser pressure help?

    Show answer

    It lowers the heat-rejection temperature.

Common misconceptions

  • Myth: Pump work in a Rankine cycle is large.

    Actually: It is tiny because liquid is nearly incompressible.

  • Myth: The expansion valve does work.

    Actually: It is a throttle: h constant, no work.

  • Myth: More superheat always helps with no limit.

    Actually: Turbine materials limit temperature.

Flashcards

  • Rankine

    Pump, boiler, turbine, condenser.

  • Pump work

    v(p₂ − p₁).

  • Reheat

    Raises mean heat-addition T, drier exhaust.

  • Vapor compression

    Compressor, condenser, valve, evaporator.

  • COP_R

    (h₁ − h₄)/(h₂ − h₁).

  • Throttle

    h₃ = h₄.

Self-check quiz

1. In a Rankine cycle, heat is added in the…

Show answer

boiler. Constant pressure.

2. Pump work is about…

Show answer

v Δp. Incompressible liquid.

3. In refrigeration, Q_L is absorbed in the…

Show answer

evaporator. Cold side.

4. COP_HP minus COP_R is…

Show answer

1. q_H = q_L + w.

How this helps save the planet

DOE’s geothermal office explains that geothermal plants bring hot fluid up through wells and convert its heat to steam that drives turbines, a Rankine-family cycle with Earth as the boiler. Geothermal plants can run essentially 24 hours a day regardless of weather, and DOE’s analysis sees potential for 90 gigawatts of U.S. geothermal electricity by 2050 if enhanced geothermal costs fall.

Nuclear plants use the same Rankine cycle, and DOE’s heat pump article describes the vapor-compression cycle in reverse: moving heat into homes in winter and out in summer. Better Rankine and refrigeration cycles are where much of the world’s clean electricity and efficient heating will come from.

On Saving Planets: Solutions · Earth Observation

Sources: U.S. DOE: Geothermal basics · U.S. DOE: Nuclear 101: How does a nuclear reactor work? (May 2025) · U.S. DOE: Pump up your savings with heat pumps (May 30, 2023)

Free learning resources

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The lab (Thermodynamics lab)

The lab is the hands-on part of the course. Here it is free and for practice only, with no credit.

Every lab can be done in a free simulation or with kitchen items. Hot water and steam scald: use oven mitts, keep the kettle on a stable counter, pour away from yourself, and keep children and pets clear. Never heat a sealed container.

Safety reference: NFPA: Burn awareness information

Lab A: The ideal-gas law in PhET Gas Properties

Week 2

Goal: Show that pressure is proportional to absolute temperature at fixed volume and amount, and that pV/T stays constant.

p1T1=p2T2  (V, m fixed)\frac{p_1}{T_1}=\frac{p_2}{T_2}\ \ (V,\ m\ \text{fixed})
Ideal-gas law at constant volume

Procedure

  1. Open PhET Gas Properties, Ideal tab; pump in gas and hold volume constant.
  2. Heat the gas in steps and record T and p.
  3. Plot p against T in kelvin; extend the line to p = 0.
  4. Repeat at constant pressure and record volume.

Data table

T (K)p (kPa), starting 100 kPa at 300 K
300100
400133.3
600200

Ideal-gas values; the simulator’s numbers fluctuate a little because it counts particle collisions.

Sample calculation

  1. p₂ = 100 × 400/300 = 133.3 kPa.
  2. The line through the data passes through p = 0 at T = 0 K.

Safety

  • Computer only.

Home or virtual alternative

  • Gas Properties runs in a browser.

PhET: Gas Properties · PhET: States of Matter

Planet link: DOE says reactor designers pressurize water so it does not boil; the same state relationships set every boiler and tank design. U.S. DOE: Nuclear 101: How does a nuclear reactor work? (May 2025)

Lab B: Kettle energy balance and efficiency

Week 5

Goal: Apply the first law to heating water and find how much of a kettle’s electrical energy reaches the water.

η=mcΔTP t\eta=\frac{mc\Delta T}{P\,t}
Kettle efficiency

Procedure

  1. Weigh 1.00 kg (1 L) of tap water and measure its temperature.
  2. Read the kettle’s power rating from its label.
  3. Time how long it takes to reach boiling (the kettle switches off).
  4. Compute the heat absorbed and the electrical energy used.

Data table

QuantityValue
Water heat, 1 kg 20→100 °C (kJ)334.4
Electrical energy, 1500 W × 250 s (kJ)375.0
Efficiency89.2%

Sample values; your kettle’s time depends on its power, the water’s starting temperature and your altitude.

Sample calculation

  1. Q = 1 × 4.18 × 80 = 334.4 kJ.
  2. η = 334.4/375 = 89.2%.

Safety

  • Steam and boiling water scald: keep the lid closed, pour away from yourself, and keep children and pets away.
  • Do not touch the kettle body while hot.

Home or virtual alternative

  • This is a home lab.

PhET: Energy Forms and Changes

Planet link: DOE notes resistance heating turns 1 kWh into 1 kWh of heat, while heat pumps cut electricity use by about 50%. U.S. DOE: Pump up your savings with heat pumps (May 30, 2023)

Lab C: Melting ice and latent heat

Week 5

Goal: Use a heat balance to predict the final temperature when ice melts in warm water, and compare with measurement.

mwc(Tw−Tf)=mihif+mic Tfm_wc(T_w-T_f)=m_i h_{if}+m_ic\,T_f
Heat lost by water = heat gained by ice

Procedure

  1. Put 300 g of 40 °C water in an insulated cup.
  2. Add 50 g of ice at 0 °C (pat it dry).
  3. Stir gently until all the ice has melted and record the lowest temperature.
  4. Compare with the predicted final temperature.

Data table

QuantityValue
Water0.300 kg at 40 °C
Ice0.050 kg at 0 °C
Predicted final T22.9 °C

Sample values; heat gained from the cup and air makes the measured value a little higher.

Sample calculation

  1. 0.3 × 4.18 × (40 − T) = 0.05 × 334 + 0.05 × 4.18 × T.
  2. T = 33.46/1.463 = 22.9 °C.

Safety

  • Use warm tap water, not boiling.

Home or virtual alternative

  • This is a home lab.

PhET: States of Matter · PhET: Energy Forms and Changes

Planet link: Latent heat is how thermal storage works; DOE notes geothermal energy can also support thermal energy storage. U.S. DOE: Geothermal basics

Lab D: Insulation and cooling rate

Week 4

Goal: Compare how fast hot water cools in a plain and an insulated mug, and find each cooling constant.

T−T∞T0−T∞=e−kt\frac{T-T_\infty}{T_0-T_\infty}=e^{-kt}
Newton’s law of cooling

Procedure

  1. Fill two mugs, one plain and one insulated (or wrapped in a towel), with water at about 80 °C.
  2. Record temperature every 2 minutes for 20 minutes; note room temperature.
  3. Compute k = ln[(T₀ − T∞)/(T − T∞)]/t for each.
  4. Compare the heat-loss rates.

Data table

MugT after 20 min (°C)k (1/min)
Plain600.0203
Insulated700.00912

Sample values with room at 20 °C.

Sample calculation

  1. Plain: k = ln(60/40)/20 = 0.0203 per minute.
  2. Insulated: k = ln(60/50)/20 = 0.00912 per minute, less than half.

Safety

  • Hot water scalds: use oven mitts and keep mugs on a stable surface away from edges.

Home or virtual alternative

  • This is a home lab.

PhET: Energy Forms and Changes

Planet link: DOE advises weatherizing before installing a heat pump, because insulation cuts the heat a home must replace. U.S. DOE: Pump up your savings with heat pumps (May 30, 2023)

Lab E: Carnot limits for a heat pump

Week 8

Goal: Build a spreadsheet of the Carnot COP of a heat pump against outdoor temperature and compare with a real unit’s rated COP.

COPHP,Carnot=THTH−TLCOP_{HP,Carnot}=\frac{T_H}{T_H-T_L}
Carnot heat pump

Procedure

  1. Set indoor T_H = 294 K.
  2. List outdoor temperatures from 253 K to 288 K.
  3. Compute the Carnot COP for each.
  4. Look up the rated efficiency of any heat pump on its label or ENERGY STAR listing and compare.

Data table

T_L (K)COP_Carnot
2639.48
27314.0
28326.7

Ideal limits; real heat pumps reach a fraction of these.

Sample calculation

  1. COP = 294/(294 − 263) = 9.48.
  2. Colder outdoor air means a lower limit.

Safety

  • Computer only.

Home or virtual alternative

  • Any spreadsheet works.

OpenStax UP2, 4.3 Refrigerators and heat pumps

Planet link: DOE says geothermal heat pumps, using steady ground temperatures, can cut energy use by 30% to 60%. U.S. DOE: Pump up your savings with heat pumps (May 30, 2023)

Lab F: Rankine cycle spreadsheet

Week 15

Goal: Compute an ideal Rankine cycle from steam-table values and see how the turbine, pump and boiler terms set efficiency.

η=(h3−h4)−v1(p2−p1)h3−h2\eta=\frac{(h_3-h_4)-v_1(p_2-p_1)}{h_3-h_2}
Ideal Rankine efficiency

Procedure

  1. Look up the four states in NIST’s WebBook (or use the values given).
  2. Compute pump work, heat in, turbine work, net work and η.
  3. Change the condenser pressure or boiler temperature using new table values and recompute.
  4. Find the steam flow for a 100 MW plant.

Data table

QuantityValue
w_pump (kJ/kg)3.03
q_in (kJ/kg)2728.6
w_net (kJ/kg)710.1
η26.0%

Steam-table values for 3 MPa/350 °C and 75 kPa.

Sample calculation

  1. η = 710.1/2728.6 = 26.0%.
  2. Steam flow for 100 MW = 100,000/710.1 = 140.8 kg/s.

Safety

  • Computer only.

Home or virtual alternative

  • Any spreadsheet works.

NIST Chemistry WebBook: Thermophysical properties of fluid systems

Planet link: DOE says nuclear and geothermal plants both turn heat into steam that spins turbines, the cycle in this lab. U.S. DOE: Nuclear 101: How does a nuclear reactor work? (May 2025) · U.S. DOE: Geothermal basics

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Exam prep practice sets

Time yourself: about 10 minutes per problem. Then compare with the full solutions.

Midterm practice set (Units 1–5)

  1. A plant has a heat rate of 8530 Btu/kWh. Find its efficiency.

    Show full solution
    1. 3412/8530.

    Answer: 40.0%.

  2. Find the mass of air in a 2 m³ tank at 200 kPa and 350 K.

    Show full solution
    1. m = 200 × 2/(0.287 × 350).

    Answer: 3.98 kg.

  3. Find h at 100 kPa and quality 0.9 (h_f = 417.51, h_fg = 2257.5 kJ/kg).

    Show full solution
    1. 417.51 + 0.9 × 2257.5.

    Answer: 2449.3 kJ/kg.

  4. 3 kg of water is heated from 15 °C to 65 °C. Find Q.

    Show full solution
    1. 3 × 4.18 × 50.

    Answer: 627 kJ.

  5. Steam at 10 kg/s drops 800 kJ/kg in an adiabatic turbine. Find the power.

    Show full solution
    1. 10 × 800.

    Answer: 8000 kW.

Final practice set (all units)

  1. Find the Carnot efficiency between 900 K and 300 K.

    Show full solution
    1. 1 − 300/900.

    Answer: 66.7%.

  2. A heat pump delivers 12 kW with COP 4. Find the electrical power and the heat taken from outside.

    Show full solution
    1. W = 12/4; Q_L = 12 − 3.

    Answer: 3 kW and 9 kW.

  3. Find Δs of 1 kg of air from 300 K, 100 kPa to 600 K, 100 kPa.

    Show full solution
    1. 1.005 ln 2.

    Answer: 0.697 kJ/(kg·K).

  4. Find the exergy of 500 kJ of heat at 1000 K with T₀ = 300 K.

    Show full solution
    1. 500 × 0.7.

    Answer: 350 kJ.

  5. Find the Otto efficiency at r = 9.

    Show full solution
    1. 1 − 9^(−0.4).

    Answer: 58.5%.

  6. A refrigeration cycle has q_L = 150 kJ/kg and w = 37.5 kJ/kg. Find COP_R and COP_HP.

    Show full solution
    1. 150/37.5.

    Answer: 4.0 and 5.0.

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Formula sheet

η=3412/heat rate, v=V/m\eta=3412/\text{heat rate},\ v=V/m
pv=RT, y=yf+x yfgpv=RT,\ y=y_f+x\,y_{fg}
Wb=∫p dV, W=mRTln⁡(V2/V1)W_b=\int p\,dV,\ W=mRT\ln(V_2/V_1)
Q−W=ΔU, h=u+pvQ-W=\Delta U,\ h=u+pv
Δu=cvΔT, Δh=cpΔT, Q=mcΔT\Delta u=c_v\Delta T,\ \Delta h=c_p\Delta T,\ Q=mc\Delta T
Q˙−W˙=m˙(Δh+ΔV2/2+gΔz)\dot Q-\dot W=\dot m(\Delta h+\Delta V^2/2+g\Delta z)
η=W/QH, COPR=QL/W, COPHP=QH/W\eta=W/Q_H,\ COP_R=Q_L/W,\ COP_{HP}=Q_H/W
ηCarnot=1−TL/TH\eta_{Carnot}=1-T_L/T_H
Δs=cpln⁡(T2/T1)−Rln⁡(p2/p1)\Delta s=c_p\ln(T_2/T_1)-R\ln(p_2/p_1)
T2/T1=(p2/p1)(k−1)/kT_2/T_1=(p_2/p_1)^{(k-1)/k}
XQ=Q(1−T0/T), I=T0SgenX_Q=Q(1-T_0/T),\ I=T_0S_{gen}
ηOtto=1−r1−k, ηBrayton=1−rp(1−k)/k\eta_{Otto}=1-r^{1-k},\ \eta_{Brayton}=1-r_p^{(1-k)/k}
wp=vΔp, COPR=(h1−h4)/(h2−h1)w_p=v\Delta p,\ COP_R=(h_1-h_4)/(h_2-h_1)

Constants

ConstantValue
Universal gas constantR_u = 8.314 kJ/(kmol·K)
Air (ideal gas, room temperature, textbook values)R = 0.287 kJ/(kg·K), c_p = 1.005, c_v = 0.718 kJ/(kg·K), k = 1.4
Liquid water (textbook value)c = 4.18 kJ/(kg·K)
Latent heats (textbook values)fusion of ice 334 kJ/kg; vaporization at 100 °C 2257 kJ/kg
Energy units1 kWh = 3600 kJ; 1 Btu = 1.055 kJ; 1 kWh = 3412 Btu
Absolute temperatureT(K) = T(°C) + 273.15

Source: NIST: CODATA value of the molar gas constant

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Free texts, data and simulations

Use these free lectures, property data and simulations alongside the course.

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Glossary

Adiabatic
No heat transfer.
Back-work ratio
Compressor work over turbine work.
COP
Coefficient of performance of a refrigerator or heat pump.
Carnot cycle
Reversible cycle between two reservoirs; the best possible.
Control volume
Region in space that mass can cross.
Dead state
Equilibrium with the surroundings.
Enthalpy
h = u + pv.
Entropy
Property with dS = δQ_rev/T; measures irreversibility.
Exergy
Maximum useful work obtainable.
Heat engine
Device that converts heat to work in a cycle.
Heat pump
Device that moves heat from cold to hot using work.
Heat rate
Fuel Btu per kWh generated.
Isentropic
Constant entropy; reversible and adiabatic.
Latent heat
Energy for phase change at constant temperature.
Quality
Vapor mass fraction of a saturated mixture.
Rankine cycle
Steam power cycle: pump, boiler, turbine, condenser.
Reversible process
Idealized process with no entropy generation.
Saturation
State where liquid and vapor coexist.
Specific heat
Energy to raise 1 kg by 1 K.
Throttling
Pressure drop at constant enthalpy.

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Self-assessment rubric

No grade or credit is given. Use this to judge honestly where you are in each unit.

LevelWhat it looks like
4 · MasteryDraws systems and diagrams, looks up states correctly, writes complete mass, energy and entropy balances, checks against Carnot limits, and runs labs safely.
3 · ProficientSolves the practice problems with occasional table or unit slips.
2 · DevelopingFollows worked examples but mixes up open and closed systems, or uses °C in ratios.
1 · BeginningPlugs numbers into formulas without a system sketch; needs to revisit the unit.

Suggested weighting if you want a score

PartWeight
Practice problems20%
Labs25%
Midterm practice set25%
Final practice set30%

Aim for level 3 or better in every unit before moving to the next course.

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