Heat Transfer
Conduction, convection and radiation: thermal resistance and insulation, fins, transient cooling, forced and natural convection, heat exchangers, blackbody radiation, radiation shields, and heat pumps, with six labs. Free, self-paced, and tied to NASA’s view of Earth’s energy budget, Parker Solar Probe’s heat shield and efficient home heating.
Courses › AERO 440
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 senior heat transfer course for engineers. It is modeled on the topics and depth of the free online A Heat Transfer Textbook by Lienhard and Lienhard and MIT 2.51 Intermediate Heat and Mass Transfer (free on MIT OpenCourseWare), which we link where they help. None of those authors, schools or agencies endorses this page.
What we add: every unit ties to real work with a cited public source (NASA’s Earth energy budget, Parker Solar Probe and ENERGY STAR heat pumps), 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 spreadsheet.
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
Earth’s climate is a heat transfer problem: sunlight in, thermal infrared out, with convection and evaporation moving heat in between. The same physics decides how much energy our buildings, engines and electronics waste.
Insulation, heat pumps and efficient heat exchangers are some of the cheapest climate solutions. Each unit ends with a short, sourced section. Values that are not from a source are labeled as sample or textbook values.
Before you start
Prerequisites
- Thermodynamics: energy balances and the first and second laws.
- Fluid Mechanics: boundary layers, Reynolds number and pipe flow.
- Differential Equations: first-order ODEs and separation of variables.
- Calculus III: gradients and integrals over areas and volumes.
How to study
- Draw the system and every heat flow arrow before writing an equation.
- Use kelvin for radiation and temperature differences for conduction and convection.
- Turn every problem into a thermal-resistance network when you can.
- Check dimensionless numbers (Bi, Re, Pr, Nu, Ra) to pick the right correlation.
- Compare answers with everyday experience: a warm house, a cooling cup, a hot pan.
- 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 Energy Forms and Changes, Blackbody Spectrum and Greenhouse Effect simulations.
- Spreadsheet software (any free one works), or Python.
- The free online A Heat Transfer Textbook in a web browser.
- A calculator.
Suggested 15-week schedule
About 6 to 9 hours a week: reading, practice problems and, in lab weeks, the lab.
| Week | Unit | Topics | Reading |
|---|---|---|---|
| 1 | Modes of Heat Transfer and Energy Balances | Modes and energy balances | Lienhard AHTT |
| 2 | Steady Conduction and Thermal Resistance | Thermal resistance; Lab A | Lienhard AHTT |
| 3 | Cylinders, Spheres and Fins | Cylinders and critical radius | Lienhard AHTT |
| 4 | Cylinders, Spheres and Fins | Fins; Lab B | MIT 2.51 |
| 5 | Transient Conduction | Transient conduction; Lab C | Lienhard AHTT |
| 6 | External Forced Convection | Boundary layers | MIT 2.51 |
| 7 | External Forced Convection | Flat-plate correlations; Lab D; midterm practice | Lienhard AHTT |
| 8 | Internal Flow and Heat Exchangers | Internal flow | MIT 2.51 |
| 9 | Internal Flow and Heat Exchangers | Heat exchangers | Lienhard AHTT |
| 10 | Natural Convection and Boiling | Natural convection and boiling | Lienhard AHTT |
| 11 | Radiation Fundamentals | Radiation fundamentals; Lab E | NASA Earth Observatory |
| 12 | Radiation Exchange and Shields | View factors | Lienhard AHTT |
| 13 | Radiation Exchange and Shields | Shields and Parker Solar Probe | NASA Parker Solar Probe |
| 14 | Heat Pumps, Buildings and Earth’s Energy Balance | Heat pumps and Earth’s balance; Lab F | ENERGY STAR |
| 15 | Heat Pumps, Buildings and Earth’s Energy Balance | Review; final practice | MIT 2.51 |
Your progress
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.
Turn on JavaScript to track progress.
Unit 1 · Week 1
Modes of Heat Transfer and Energy Balances
Conduction, convection and radiation, their rate equations, and surface and control-volume energy balances.
Learning objectives
By the end of this unit you can:
- Name the three modes and their rate laws.
- Compute conduction, convection and radiation rates.
- Write a surface energy balance.
- Read Earth’s energy budget as a heat balance.
Key concepts
- Heat transfer is energy in transit because of a temperature difference; thermodynamics says how much, heat transfer says how fast.
- Conduction moves energy through a solid or still fluid by molecular contact (Fourier’s law).
- Convection moves energy between a surface and a moving fluid (Newton’s law of cooling, with a heat transfer coefficient h).
- Radiation moves energy by electromagnetic waves, needs no medium, and grows with T⁴ in kelvin (Stefan–Boltzmann law).
- A surface stores no energy, so what arrives by one mode must leave by the others.
Key equations
Worked examples, step by step
Example 1.1 A 0.2 m concrete wall (k = 1.4 W/(m·K), sample) of area 10 m² has a 15 K difference across it. Find q.
Show the steps
- q = 1.4 × 10 × 15/0.2.
Answer: 1050 W.
Example 1.2 A plate with ε = 0.9 at 400 K sits in 300 K surroundings. Net radiation per m²?
Show the steps
- 0.9σ(400⁴ − 300⁴) = 0.9 × 5.670374419 × 10⁻⁸ × 1.75 × 10¹⁰.
Answer: 893 W/m².
Example 1.3 NASA says about 29% of incoming sunlight is reflected and the global average is about 340 W/m². How much is absorbed?
Show the steps
- Reflected: 0.29 × 340 = 98.6 W/m².
- Absorbed: 340 − 98.6.
Answer: 241.4 W/m², matching NASA’s roughly 240 W/m².
Practice problems
Work each on paper first, then check.
Repeat the wall example with k = 0.04 W/(m·K).
Show answer
30 W.
Convection with h = 25 W/(m²·K), A = 2 m², ΔT = 40 K?
Show answer
2000 W.
Why must radiation use kelvin?
Show answer
T⁴ depends on absolute temperature.
Why is 1360/4 the global average sunlight?
Show answer
A sphere’s area is 4 times its shadow disk: 1360/4 = 340 W/m².
Common misconceptions
Myth: Heat and temperature are the same.
Actually: Temperature is a state; heat is energy in transit.
Myth: Radiation needs air.
Actually: It travels through vacuum.
Myth: h is a material property.
Actually: It depends on flow and geometry.
Flashcards
Conduction
Molecular energy transfer through matter.
Convection
Surface-to-moving-fluid transfer.
Radiation
Electromagnetic energy exchange.
h
Convection coefficient, W/(m²·K).
k
Thermal conductivity, W/(m·K).
Surface balance
In = out at a surface.
Self-check quiz
How this helps save the planet
NASA’s Earth Observatory says total solar irradiance is about 1,360 W/m², averaging about 340 W/m² over the globe; about 29% is reflected, 23% is absorbed by the atmosphere and 48% by the surface, so Earth absorbs about 240 W/m².
For a stable temperature, Earth must radiate the same 240 W/m² back to space as thermal infrared, the heat balance this course returns to in Unit 10.
On Saving Planets: Earth Observation · Thermodynamics
Sources: NASA Earth Observatory: Climate and Earth’s Energy Budget
Free learning resources
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free textbook, Chapter 1.
- NASA Earth Observatory: Climate and Earth’s Energy Budget: NASA explainer.
Unit 2 · Week 2
Steady Conduction and Thermal Resistance
The heat equation, one-dimensional steady conduction, thermal-resistance networks, composite walls and insulation.
Learning objectives
By the end of this unit you can:
- State the heat equation and its 1D steady form.
- Build series and parallel resistance networks.
- Compute heat loss through a composite wall.
- Quantify the savings from insulation.
Key concepts
- The heat equation balances conduction, generation and storage; at steady state with no generation, a plane wall has a linear temperature profile.
- Each layer acts like an electrical resistor: R = L/(kA) for conduction, 1/(hA) for convection.
- Series resistances add; heat flow equals overall ΔT divided by total resistance, like current.
- The largest resistance controls the heat flow, which is why insulation (low k) matters most.
- Contact resistance between layers can add surprising resistance in electronics and joints.
Key equations
Worked examples, step by step
Example 2.1 A wall has inside air (h = 8), 0.1 m insulation (k = 0.04), 0.1 m brick (k = 0.7) and outside air (h = 25), all sample values. Find the resistance per m².
Show the steps
- 1/8 + 0.1/0.04 + 0.1/0.7 + 1/25 = 0.125 + 2.5 + 0.1429 + 0.04.
Answer: 2.808 m²·K/W.
Example 2.2 With 20 K between inside and outside, find the heat flux and the loss through 100 m².
Show the steps
- q″ = 20/2.808 = 7.12 W/m².
- q = 100 × 7.12.
Answer: 712 W.
Example 2.3 Remove the insulation. How much does the loss grow?
Show the steps
- R″ = 0.3079 m²·K/W.
- q″ = 20/0.3079 = 64.97 W/m².
Answer: About 9.1 times; insulation cuts the loss by 89.0%.
Practice problems
Work each on paper first, then check.
Double the insulation to 0.2 m. New q″?
Show answer
R″ = 5.308, q″ = 3.77 W/m².
Which layer dominates the original wall?
Show answer
The insulation, 2.5 of 2.808 m²·K/W.
Write R for two layers in parallel.
Show answer
1/R = 1/R₁ + 1/R₂.
Shape of T(x) in a plane wall at steady state?
Show answer
Linear.
Common misconceptions
Myth: Thin layers never matter.
Actually: A low-k thin layer can dominate.
Myth: Resistances in series multiply.
Actually: They add.
Myth: Air films add nothing.
Actually: Convection resistance is real.
Flashcards
Thermal resistance
ΔT per unit heat flow.
Heat equation
Energy balance in a solid.
Composite wall
Layers in series.
Insulation
Low-k material.
Contact resistance
Resistance at an interface.
U-value
Overall coefficient, 1/R″.
Self-check quiz
How this helps save the planet
ENERGY STAR says a typical U.S. household energy bill is around $1,900 a year and almost half of that goes to heating and cooling.
This unit’s sample wall shows why sealing and insulating comes first: adding insulation cut the conduction loss by about 89%, so any heating system has far less heat to replace.
On Saving Planets: Solutions · Thermodynamics
Sources: ENERGY STAR: Air-Source Heat Pumps
Free learning resources
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free textbook, Chapter 2.
- MIT OpenCourseWare 2.51 Intermediate Heat and Mass Transfer (Fall 2008): Free MIT lectures.
Unit 3 · Weeks 3–4
Cylinders, Spheres and Fins
Radial conduction, pipe insulation and the critical radius, extended surfaces, fin efficiency and electronics cooling.
Learning objectives
By the end of this unit you can:
- Compute radial conduction resistance.
- Find the critical insulation radius.
- Compute fin parameter m and fin efficiency.
- Estimate heat from a fin.
Key concepts
- In a cylinder, area grows with radius, so the temperature profile is logarithmic and R = ln(r₂/r₁)/(2πkL).
- Adding insulation to a small wire can increase heat loss until the outer radius passes the critical radius k/h.
- Fins add area where h is low (usually gases), as on engine cylinders, heat sinks and radiators.
- The fin parameter m = √(hP/(kA_c)) sets how fast the fin temperature decays from its base.
- Fin efficiency compares actual heat to the heat if the whole fin were at the base temperature.
Key equations
Worked examples, step by step
Example 3.1 Pipe insulation from r = 0.05 to 0.1 m, k = 0.04 W/(m·K), 1 m long, 100 K across it (sample values). Find R and q.
Show the steps
- R = ln 2/(2π × 0.04) = 2.758 K/W.
- q = 100/2.758.
Answer: 36.3 W per meter.
Example 3.2 An aluminum pin fin (k = 200) is 5 mm across with h = 50 W/(m²·K) and a 20 mm effective length. Find η.
Show the steps
- m = √(4h/(kD)) = √(200/1) = 14.14 m⁻¹.
- mL = 0.2828.
- η = tanh(0.2828)/0.2828.
Answer: 0.974.
Example 3.3 Heat from that fin with a 50 K base excess?
Show the steps
- A_f = πDL = 3.142 × 10⁻⁴ m².
- q = ηhA_fθ = 0.974 × 50 × 3.142 × 10⁻⁴ × 50.
Answer: 0.765 W.
Practice problems
Work each on paper first, then check.
Critical radius for k = 0.04 and h = 8?
Show answer
5 mm.
Fin efficiency at L = 40 mm?
Show answer
mL = 0.566, η = 0.905.
Why do fins suit air more than water?
Show answer
Air’s low h makes area the bottleneck.
Profile shape in a cylinder wall?
Show answer
Logarithmic.
Common misconceptions
Myth: More insulation always helps.
Actually: Not below the critical radius.
Myth: Longer fins always help a lot.
Actually: Efficiency drops with length.
Myth: Fins work in any medium.
Actually: They pay off where h is low.
Flashcards
Critical radius
k/h for cylinders.
Fin
Extended surface.
Fin efficiency
Actual over ideal fin heat.
m
Fin decay parameter.
Heat sink
Array of fins.
Log profile
Cylinder steady temperature.
Self-check quiz
How this helps save the planet
Heat moves the same way in a pipe wall as in a planet’s crust: from hot to cold, through resistances in series.
Insulating hot-water and steam pipes is one of the simplest efficiency steps in buildings and industry. All values in this unit are sample textbook values.
On Saving Planets: Solutions · Circuits
Sources: J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition)
Free learning resources
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free textbook, Chapter 4.
- MIT OpenCourseWare 2.51 Intermediate Heat and Mass Transfer (Fall 2008): Free MIT lectures.
Unit 4 · Week 5
Transient Conduction
The Biot number, the lumped-capacitance model, time constants, and when spatial temperature gradients matter.
Learning objectives
By the end of this unit you can:
- Compute the Biot number.
- Use the lumped model when Bi < 0.1.
- Compute a thermal time constant.
- Find the time to reach a target temperature.
Key concepts
- The Biot number compares internal conduction resistance to external convection resistance.
- When Bi < 0.1, the body is nearly uniform in temperature and the lumped model works.
- In the lumped model, the temperature excess decays exponentially with time constant τ = ρcV/(hA).
- For larger Bi, use one-term series solutions or charts (Heisler) or a numerical method.
- High heat capacity slows response: oceans and thick walls are thermal flywheels.
Key equations
Worked examples, step by step
Example 4.1 A 10 mm aluminum sphere (k = 237, ρ = 2700, c = 900) cools in air with h = 50. Find Bi.
Show the steps
- L_c = D/6 = 1.667 × 10⁻³ m.
- Bi = 50 × 1.667 × 10⁻³/237.
Answer: 3.52 × 10⁻⁴, so the lumped model works.
Example 4.2 Find τ for that sphere.
Show the steps
- τ = 2700 × 900 × 1.667 × 10⁻³/50.
Answer: 81 s.
Example 4.3 How long to cool from 300 °C to 50 °C in 20 °C air?
Show the steps
- θ_i = 280 K, θ = 30 K.
- t = 81 ln(280/30) = 81 × 2.234.
Answer: 181 s.
Practice problems
Work each on paper first, then check.
Fraction of the excess left after one τ?
Show answer
e⁻¹ = 0.368.
Time to halve the excess?
Show answer
τ ln 2 = 56.1 s.
Bi = 0.5. Is lumped valid?
Show answer
No; use series or numerical solutions.
Double h. What happens to τ?
Show answer
It halves to 40.5 s.
Common misconceptions
Myth: Bi is the Nusselt number.
Actually: Bi uses the solid’s k; Nu uses the fluid’s.
Myth: Lumped works for any size.
Actually: Only when Bi < 0.1.
Myth: Cooling is linear in time.
Actually: It is exponential.
Flashcards
Biot number
Internal over external resistance.
Lumped capacitance
Uniform-temperature model.
Time constant
ρcV/(hA).
Heisler chart
One-term transient solution chart.
Thermal mass
Heat capacity that slows change.
L_c
V/A characteristic length.
Self-check quiz
How this helps save the planet
NASA’s Earth Observatory notes that Earth’s surface averages about 15 °C, more than 30 °C warmer than it would be with no atmosphere.
In this unit’s language, a large heat capacity means a long time constant; the oceans’ heat capacity gives Earth’s climate a slow response (a textbook idea, not a NASA figure).
On Saving Planets: Oceans · Earth Observation
Sources: NASA Earth Observatory: Climate and Earth’s Energy Budget
Free learning resources
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free textbook, Chapter 5.
- MIT OpenCourseWare 2.51 Intermediate Heat and Mass Transfer (Fall 2008): Free MIT lectures.
Unit 5 · Weeks 6–7
External Forced Convection
Boundary layers, Reynolds, Prandtl and Nusselt numbers, flat-plate correlations and flow over cylinders.
Learning objectives
By the end of this unit you can:
- Compute Re, Pr and Nu.
- Use the laminar flat-plate correlation.
- Find h and the heat rate.
- Check transition at Re ≈ 5 × 10⁵.
Key concepts
- Velocity and thermal boundary layers form on a surface; heat crosses the thin layer of slow fluid by conduction.
- The Nusselt number Nu = hL/k is the dimensionless heat transfer coefficient.
- The Prandtl number compares momentum and thermal diffusivity (about 0.7 for air, about 7 for water near room temperature, textbook values).
- On a flat plate, flow is usually laminar below Re ≈ 5 × 10⁵; turbulence raises h.
- Correlations are fits to experiments; use them inside their stated ranges.
Key equations
Worked examples, step by step
Example 5.1 Air (ν = 1.5 × 10⁻⁵, k = 0.026, Pr = 0.71) flows at 5 m/s over a 0.5 m plate. Find Re.
Show the steps
- Re = 5 × 0.5/1.5 × 10⁻⁵.
Answer: 166,667, laminar.
Example 5.2 Find the average Nu and h.
Show the steps
- Nu = 0.664 × 166,667^½ × 0.71^⅓ = 0.664 × 408.2 × 0.892 = 241.8.
- h = 241.8 × 0.026/0.5.
Answer: 12.6 W/(m²·K).
Example 5.3 Heat from a 0.5 × 1 m plate 30 K above the air?
Show the steps
- q = 12.58 × 0.5 × 30.
Answer: 189 W.
Practice problems
Work each on paper first, then check.
Repeat h at 10 m/s.
Show answer
Re = 333,333, h = 17.8 W/(m²·K) (√2 times higher).
Re at 1 m/s?
Show answer
33,333.
Name the dimensionless h.
Show answer
Nusselt number.
Why does turbulence raise h?
Show answer
Mixing thins the slow near-wall layer.
Common misconceptions
Myth: Nu uses the solid’s k.
Actually: It uses the fluid’s k.
Myth: One correlation fits all flows.
Actually: Check range and geometry.
Myth: h is constant along a plate.
Actually: Local h falls as the layer grows.
Flashcards
Reynolds number
Inertia over viscosity.
Prandtl number
ν/α.
Nusselt number
hL/k.
Boundary layer
Thin near-wall region.
Correlation
Fit to experiments.
Transition
Laminar to turbulent.
Self-check quiz
How this helps save the planet
NASA’s Earth Observatory says convection carries about 5% of incoming solar energy from the surface into the atmosphere, while evaporation carries about 25%.
Wind speed sets h over oceans and land just as velocity sets h on this unit’s plate, which is why winds shape how fast the surface sheds heat.
On Saving Planets: Fluid Mechanics · Oceans
Sources: NASA Earth Observatory: Climate and Earth’s Energy Budget
Free learning resources
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free textbook, Chapters 6–7.
- MIT OpenCourseWare 2.51 Intermediate Heat and Mass Transfer (Fall 2008): Free MIT lectures.
Unit 6 · Weeks 8–9
Internal Flow and Heat Exchangers
Pipe-flow convection, mean temperature, the Dittus–Boelter correlation, heat exchanger types, LMTD and effectiveness–NTU.
Learning objectives
By the end of this unit you can:
- Use laminar and turbulent pipe correlations.
- Compute a log-mean temperature difference.
- Size a heat exchanger with UA and LMTD.
- Use effectiveness–NTU.
Key concepts
- In a pipe, the bulk (mean) temperature changes along the length as heat enters or leaves.
- Fully developed laminar flow with constant wall temperature has Nu = 3.66; turbulent flow uses Dittus–Boelter (textbook correlations).
- Counterflow exchangers beat parallel flow because the temperature difference stays more even.
- The LMTD method suits sizing when all four temperatures are known.
- The ε-NTU method suits rating when outlet temperatures are unknown.
Key equations
Worked examples, step by step
Example 6.1 Water (k = 0.6, Pr = 7, textbook values) flows at Re = 20,000 in a 20 mm tube. Find h.
Show the steps
- Nu = 0.023 × 20,000^0.8 × 7^0.4 = 138.2.
- h = 138.2 × 0.6/0.02.
Answer: About 4150 W/(m²·K).
Example 6.2 Counterflow: hot 90 → 50 °C, cold 20 → 40 °C. Find the LMTD.
Show the steps
- ΔT₁ = 90 − 40 = 50 K; ΔT₂ = 50 − 20 = 30 K.
- LMTD = 20/ln(50/30).
Answer: 39.15 K.
Example 6.3 With U = 500 W/(m²·K) and A = 2 m², find q.
Show the steps
- q = 500 × 2 × 39.15.
Answer: 39.2 kW.
Practice problems
Work each on paper first, then check.
ε for NTU = 2 when one stream condenses (C_r = 0)?
Show answer
1 − e⁻² = 0.865.
Laminar Nu for constant wall temperature?
Show answer
3.66.
Why prefer counterflow?
Show answer
More even ΔT, more heat per area.
Dittus–Boelter Nu for Pr = 0.7, Re = 20,000?
Show answer
55.0.
Common misconceptions
Myth: Use the arithmetic mean ΔT.
Actually: Use the log mean.
Myth: ε can exceed 1.
Actually: It cannot.
Myth: Laminar correlations work for turbulent flow.
Actually: Check Re first.
Flashcards
LMTD
Log-mean ΔT.
NTU
UA/C_min.
Effectiveness
Actual over maximum heat.
Counterflow
Opposite stream directions.
Fouling
Deposits that add resistance.
Dittus–Boelter
Turbulent pipe correlation.
Self-check quiz
How this helps save the planet
ENERGY STAR explains that an air-source heat pump moves heat between indoor and outdoor air instead of making it from fuel, using heat exchangers (coils) on both sides.
Better exchangers need a smaller temperature difference to move the same heat, which is one reason modern heat pumps deliver more heat per unit of electricity.
On Saving Planets: Solutions · Thermodynamics
Sources: ENERGY STAR: Air-Source Heat Pumps
Free learning resources
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free textbook, Chapters 3 and 7.
- MIT OpenCourseWare 2.51 Intermediate Heat and Mass Transfer (Fall 2008): Free MIT lectures.
Unit 7 · Week 10
Natural Convection and Boiling
Buoyancy-driven flow, Grashof and Rayleigh numbers, vertical-plate correlations, and an introduction to boiling and condensation.
Learning objectives
By the end of this unit you can:
- Compute the Rayleigh number.
- Use a vertical-plate correlation.
- Compare natural and forced h.
- Describe the boiling curve.
Key concepts
- Warm fluid is lighter and rises; buoyancy, not a fan, drives the flow.
- The Rayleigh number Ra = gβΔTL³/(να) plays the role Re plays in forced flow.
- On a vertical plate, flow is laminar below about Ra = 10⁹ (textbook value).
- Natural-convection h in air is only a few W/(m²·K), so radiation is often comparable.
- Boiling and condensation move large heat at small ΔT because of latent heat.
Key equations
Worked examples, step by step
Example 7.1 A 0.3 m vertical plate is 30 K above air (β = 1/300 K⁻¹, ν = 1.6 × 10⁻⁵, α = 2.2 × 10⁻⁵, sample values). Find Ra.
Show the steps
- Ra = 9.80665 × (1/300) × 30 × 0.3³/(1.6 × 10⁻⁵ × 2.2 × 10⁻⁵).
Answer: 7.52 × 10⁷, laminar.
Example 7.2 Find Nu and h with k = 0.026.
Show the steps
- Nu = 0.59 × (7.52 × 10⁷)^¼ = 54.9.
- h = 54.9 × 0.026/0.3.
Answer: 4.76 W/(m²·K).
Example 7.3 Heat from a 0.3 × 1 m plate?
Show the steps
- q = 4.76 × 0.3 × 30.
Answer: 42.9 W.
Practice problems
Work each on paper first, then check.
Compare with Unit 5’s forced h of 12.6.
Show answer
Forced is about 2.64 times higher.
Laminar limit on Ra?
Show answer
About 10⁹.
Why is boiling so effective?
Show answer
Latent heat moves energy at small ΔT.
β for an ideal gas at 300 K?
Show answer
1/300 K⁻¹ = 3.33 × 10⁻³.
Common misconceptions
Myth: Natural convection is negligible.
Actually: It often matches radiation.
Myth: Ra uses velocity.
Actually: It uses buoyancy.
Myth: Boiling h is low.
Actually: It is among the highest.
Flashcards
Rayleigh number
Buoyancy strength.
Grashof number
Buoyancy over viscosity.
β
Expansion coefficient.
Boiling curve
Heat flux vs wall superheat.
Critical heat flux
Peak nucleate boiling flux.
Condensation
Vapor to liquid on a cold wall.
Self-check quiz
How this helps save the planet
NASA’s Earth Observatory says evaporation removes about 25% of incoming solar energy from the surface and convection about 5%; the latent heat is released when water vapor condenses into clouds.
At 340 W/m² average sunlight, those shares are about 85 and 17 W/m², the same phase-change and buoyancy processes this unit treats.
On Saving Planets: Oceans · Earth Observation
Sources: NASA Earth Observatory: Climate and Earth’s Energy Budget
Free learning resources
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free textbook, Chapters 8–9.
- MIT OpenCourseWare 2.51 Intermediate Heat and Mass Transfer (Fall 2008): Free MIT lectures.
Unit 8 · Week 11
Radiation Fundamentals
Blackbodies, Planck’s law, Wien’s law, the Stefan–Boltzmann law, emissivity, absorptivity and Kirchhoff’s law.
Learning objectives
By the end of this unit you can:
- Find a blackbody’s peak wavelength.
- Compute emissive power.
- Use emissivity and absorptivity.
- Find a planet’s effective temperature.
Key concepts
- A blackbody absorbs all radiation and emits the most possible at each temperature and wavelength.
- Wien’s law: peak wavelength is inversely proportional to absolute temperature.
- Total emissive power grows as σT⁴, so doubling temperature multiplies emission by 16.
- Real surfaces emit a fraction ε of blackbody power; for a gray surface α = ε (Kirchhoff).
- Solar radiation is short-wave; Earth’s emission is long-wave infrared, which is why gases like CO₂ and water vapor matter.
Key equations
Worked examples, step by step
Example 8.1 NASA gives the Sun about 5,800 K. Find its peak wavelength.
Show the steps
- 2898/5800.
Answer: 0.500 µm, visible light.
Example 8.2 Earth absorbs about 240 W/m² (NASA). Find its effective temperature.
Show the steps
- T = (240/5.670374419 × 10⁻⁸)^¼.
Answer: 255.1 K, about −18 °C (NASA rounds to about −20 °C).
Example 8.3 Blackbody emission from a 288 K surface?
Show the steps
- σ × 288⁴.
Answer: 390 W/m².
Practice problems
Work each on paper first, then check.
Peak wavelength at 288 K?
Show answer
10.06 µm, thermal infrared.
Emission ratio when T doubles?
Show answer
2⁴ = 16.
Peak for a 3000 K lamp filament?
Show answer
0.966 µm, near infrared.
Gray surface with ε = 0.8: α?
Show answer
0.8.
Common misconceptions
Myth: Hot means visible.
Actually: Room-temperature objects glow in infrared.
Myth: Use °C in σT⁴.
Actually: Use kelvin.
Myth: Shiny means cold.
Actually: Low ε means it emits and absorbs little.
Flashcards
Blackbody
Perfect absorber and emitter.
Wien’s law
λ_max T = 2898 µm·K.
Emissivity
Emission relative to blackbody.
Kirchhoff’s law
α = ε for gray surfaces.
Planck’s law
Spectral blackbody emission.
Effective temperature
Blackbody-equivalent T.
Self-check quiz
How this helps save the planet
NASA’s Earth Observatory says the Sun’s surface is about 5,800 K, Earth’s effective radiating temperature is about −20 °C, and doubling an object’s temperature multiplies its radiated energy by 16.
NASA also notes that water vapor lets most infrared escape between 8 and 14 µm, while carbon dioxide absorbs at wavelengths beyond about 12–13 µm, near the edge of that window.
On Saving Planets: Earth Observation · Solutions
Sources: NASA Earth Observatory: Climate and Earth’s Energy Budget
Free learning resources
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free textbook, Chapter 10.
- PhET: Blackbody Spectrum: Free simulation.
Unit 9 · Weeks 12–13
Radiation Exchange and Shields
View factors, reciprocity, gray-surface networks, parallel plates, radiation shields and thermal protection in space.
Learning objectives
By the end of this unit you can:
- Use view-factor reciprocity.
- Compute net exchange between parallel plates.
- Show how a shield cuts heat flow.
- Estimate radiation from a hot shield.
Key concepts
- The view factor F₁₂ is the fraction of radiation leaving surface 1 that reaches surface 2.
- Reciprocity: A₁F₁₂ = A₂F₂₁; summation: the view factors from a surface add to 1.
- Gray surfaces exchange heat through surface resistances (1 − ε)/(εA) and space resistances 1/(A₁F₁₂).
- A thin low-ε shield between two plates adds resistances and cuts exchange sharply; multilayer insulation stacks many.
- A heat shield facing the Sun gets hot, reradiates, and protects what lies in its shadow.
Key equations
Worked examples, step by step
Example 9.1 Large parallel plates at 400 K and 300 K, both ε = 0.8. Net flux?
Show the steps
- Denominator: 1.25 + 1.25 − 1 = 1.5.
- q″ = σ(400⁴ − 300⁴)/1.5.
Answer: 662 W/m².
Example 9.2 Insert a shield with ε = 0.05 on both sides.
Show the steps
- Resistance: 2 × (1.25 + 20 − 1) = 40.5.
- q″ = σ × 1.75 × 10¹⁰/40.5.
Answer: 24.5 W/m², 27 times less.
Example 9.3 A small sphere (A₁ = 1 m²) sits inside a shell of area 4 m². Find F₂₁.
Show the steps
- F₁₂ = 1.
- F₂₁ = A₁F₁₂/A₂.
Answer: 0.25.
Practice problems
Work each on paper first, then check.
Blackbody flux at Parker’s 1,377 °C shield limit?
Show answer
σ × 1650.15⁴ = 4.20 × 10⁵ W/m².
Why does one shield with ε = 0.05 work so well?
Show answer
Its two surface resistances add about 39.
Sum of view factors from an enclosure surface?
Show answer
1.
What is MLI?
Show answer
Many thin low-ε shields in a stack.
Common misconceptions
Myth: View factors are symmetric.
Actually: Only A₁F₁₂ = A₂F₂₁.
Myth: Shields must be thick.
Actually: Low ε matters, not thickness.
Myth: Black surfaces make good shields.
Actually: Shields need low ε.
Flashcards
View factor
Fraction of radiation intercepted.
Reciprocity
A₁F₁₂ = A₂F₂₁.
Radiosity
All radiation leaving a surface.
Radiation shield
Low-ε barrier.
MLI
Multilayer insulation.
Gray surface
ε independent of wavelength.
Self-check quiz
How this helps save the planet
NASA says Parker Solar Probe, launched August 12, 2018, is protected by a 4.5-inch (11.43 cm) carbon-composite shield that withstands nearly 2,500 °F (1,377 °C) as the spacecraft passes about 3.9 million miles (6.2 million km) from the Sun.
In 2021 Parker became the first spacecraft to fly through the Sun’s corona; its shield keeps the instruments behind it in shadow, the same idea as this unit’s radiation shield.
On Saving Planets: Spacecraft Design · Earth Observation
Sources: NASA Science: Parker Solar Probe
Free learning resources
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free textbook, Chapter 10.
- NASA Science: Parker Solar Probe: NASA mission page.
Unit 10 · Weeks 14–15
Heat Pumps, Buildings and Earth’s Energy Balance
Heat-pump COP, Carnot limits, whole-building heat loss, and the planetary energy imbalance measured from space.
Learning objectives
By the end of this unit you can:
- Compute heat-pump COP and its Carnot limit.
- Size electricity for a heating load.
- Convert a flux imbalance to total power.
- Connect all three modes in one system.
Key concepts
- A heat pump moves heat from cold to hot using work; its heating COP is heat delivered divided by work.
- The Carnot COP, T_h/(T_h − T_c), is an upper limit that falls as the outdoor temperature drops.
- A building’s heat loss is its UA times the indoor–outdoor difference; insulation lowers UA before you buy equipment.
- Earth’s energy imbalance is a small difference between two large flows, measured by satellites such as NASA’s CERES instruments.
- Every system in this course combines conduction, convection and radiation in a resistance network.
Key equations
Worked examples, step by step
Example 10.1 ENERGY STAR says a heat pump can deliver up to three times more heat than the electricity it uses. Electricity for 9 kW of heat?
Show the steps
- W = 9/3.
Answer: 3 kW.
Example 10.2 Carnot heating COP with 35 °C (308 K) delivery and 0 °C (273 K) outdoors?
Show the steps
- 308/(308 − 273).
Answer: 8.8; real units reach about a third of that.
Example 10.3 NASA says the imbalance is a little over 0.8 W/m². With Earth’s area 5.1 × 10¹⁴ m² (textbook value), total power?
Show the steps
- 0.8 × 5.1 × 10¹⁴.
Answer: About 4.1 × 10¹⁴ W.
Practice problems
Work each on paper first, then check.
Half of a $1,900 yearly energy bill?
Show answer
About $950 for heating and cooling.
Carnot COP at −10 °C (263 K) outdoors, 308 K delivery?
Show answer
6.84.
Why insulate before buying a heat pump?
Show answer
Lower UA means a smaller, cheaper unit.
What does CERES measure?
Show answer
Earth’s radiant energy from space.
Common misconceptions
Myth: COP above 1 breaks energy conservation.
Actually: Heat is moved, not created.
Myth: Turn heat pumps way down at night.
Actually: ENERGY STAR advises a steady setting.
Myth: 0.8 W/m² is negligible.
Actually: Over Earth it is about 4 × 10¹⁴ W.
Flashcards
COP
Heat delivered per unit work.
Carnot COP
Ideal limit.
UA
Overall conductance.
Energy imbalance
Absorbed minus emitted.
CERES
NASA radiation-budget instruments.
Heat pump
Moves heat with work.
Self-check quiz
How this helps save the planet
ENERGY STAR says an air-source heat pump can deliver up to three times more heat energy than the electrical energy it uses, because it moves heat instead of making it.
NASA’s Earth Observatory says Earth is absorbing a little more than 0.8 W/m² more than it emits; NASA’s CERES instruments measure the radiation budget from orbit.
On Saving Planets: Solutions · Earth Observation
Sources: ENERGY STAR: Air-Source Heat Pumps · NASA Earth Observatory: Climate and Earth’s Energy Budget · NASA Langley: CERES (Clouds and the Earth’s Radiant Energy System)
Free learning resources
- ENERGY STAR: Air-Source Heat Pumps: ENERGY STAR guide.
- NASA Langley: CERES (Clouds and the Earth’s Radiant Energy System): NASA CERES.
The lab (Heat transfer lab)
The lab is the hands-on part of the course. Here it is free and for practice only, with no credit.
Every lab runs in a free simulation or a spreadsheet, so there are no burn or electrical hazards. Do not heat, boil or ignite anything for this course. Take screen breaks, sit with good posture, and keep cables out of walkways.
Safety reference: OSHA: Walking-working surfaces (slip, trip and fall hazards)
Lab A: Heat flow and a wall-insulation calculator
Week 2
Goal: Watch heat flow between objects in PhET Energy Forms and Changes, then compute wall heat flux for several insulation thicknesses.
Procedure
- In PhET Energy Forms and Changes, heat a block and watch energy flow to its neighbor and the air.
- In a spreadsheet, enter Unit 2’s wall without insulation: R″ = 0.3079 m²·K/W.
- Add insulation layers of 0.05, 0.1 and 0.2 m (k = 0.04) and compute q″ for ΔT = 20 K.
- Plot q″ against thickness and note the diminishing returns.
Data table
| Insulation (m) | R″ (m²·K/W) | q″ (W/m²) |
|---|---|---|
| 0 | 0.3079 | 64.96 |
| 0.05 | 1.558 | 12.84 |
| 0.1 | 2.808 | 7.12 |
| 0.2 | 5.308 | 3.77 |
Computed sample values.
Sample calculation
- R″(0.05) = 0.3079 + 0.05/0.04 = 1.558.
- q″ = 20/1.558 = 12.84 W/m².
Safety
- Computer only.
Home or virtual alternative
- Any spreadsheet works.
PhET: Energy Forms and Changes
Planet link: ENERGY STAR: almost half of a typical $1,900 household energy bill goes to heating and cooling. ENERGY STAR: Air-Source Heat Pumps
Lab B: Fin efficiency explorer
Week 4
Goal: Compute pin-fin efficiency for several lengths and see where adding length stops paying off.
Procedure
- Enter Unit 3’s fin: k = 200, D = 5 mm, h = 50, so m = 14.14 m⁻¹.
- Compute η for L = 10, 20 and 40 mm.
- Compute fin heat for a 50 K base excess.
- Change h to 200 and repeat.
Data table
| L (mm) | mL | η |
|---|---|---|
| 10 | 0.1414 | 0.993 |
| 20 | 0.2828 | 0.974 |
| 40 | 0.5657 | 0.905 |
Computed sample values.
Sample calculation
- m = √(4 × 50/(200 × 0.005)) = 14.14 m⁻¹.
- η(40 mm) = tanh(0.5657)/0.5657 = 0.905.
Safety
- Computer only.
Home or virtual alternative
- Any spreadsheet works.
J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition)
Planet link: Efficient fins and exchangers let heat pumps move more heat per unit of electricity (ENERGY STAR). ENERGY STAR: Air-Source Heat Pumps
Lab C: Lumped cooling curve
Week 5
Goal: Model Unit 4’s aluminum sphere cooling in a spreadsheet and compare an exact curve with a step-by-step (Euler) solution.
Procedure
- Enter T_i = 300 °C, T∞ = 20 °C, τ = 81 s.
- Compute T every τ out to 3τ.
- Add an Euler column with a 5 s step and compare.
- Double h and repeat.
Data table
| t (s) | T (°C) |
|---|---|
| 0 | 300 |
| 81 | 123.0 |
| 162 | 57.9 |
| 243 | 33.9 |
Computed sample values.
Sample calculation
- T(81) = 20 + 280e⁻¹ = 123.0 °C.
- T(243) = 20 + 280e⁻³ = 33.9 °C.
Safety
- Computer only.
Home or virtual alternative
- Any spreadsheet works.
PhET: Energy Forms and Changes
Planet link: NASA: Earth’s surface averages about 15 °C, more than 30 °C warmer than with no atmosphere. NASA Earth Observatory: Climate and Earth’s Energy Budget
Lab D: Flat-plate convection calculator
Week 7
Goal: Compute Re and average h for air over a 0.5 m plate at several speeds.
Procedure
- Enter air properties: ν = 1.5 × 10⁻⁵, k = 0.026, Pr = 0.71.
- Compute Re and h at 1, 5 and 10 m/s.
- Check each Re is below 5 × 10⁵.
- Plot h against √V.
Data table
| V (m/s) | Re | h (W/(m²·K)) |
|---|---|---|
| 1 | 33,333 | 5.62 |
| 5 | 166,667 | 12.58 |
| 10 | 333,333 | 17.78 |
Computed sample values.
Sample calculation
- Re(1) = 1 × 0.5/1.5 × 10⁻⁵ = 33,333.
- h(10) = 0.664 × 577.4 × 0.892 × 0.052 = 17.78.
Safety
- Computer only.
Home or virtual alternative
- Any spreadsheet works.
MIT OpenCourseWare 2.51 Intermediate Heat and Mass Transfer (Fall 2008)
Planet link: NASA: convection carries about 5% of incoming solar energy from the surface. NASA Earth Observatory: Climate and Earth’s Energy Budget
Lab E: Blackbody spectra
Week 11
Goal: Use PhET Blackbody Spectrum to see Wien’s law, then check peaks in a spreadsheet.
Procedure
- In PhET Blackbody Spectrum, set the Sun, a lamp and Earth.
- Read the peak wavelength from the graph.
- Compute 2898/T for 288, 3000 and 5800 K.
- Compute σT⁴ for each.
Data table
| T (K) | λ_max (µm) | σT⁴ (W/m²) |
|---|---|---|
| 288 | 10.06 | 390 |
| 3000 | 0.966 | 4.59 × 10⁶ |
| 5800 | 0.500 | 6.42 × 10⁷ |
Computed values.
Sample calculation
- 2898/288 = 10.06 µm.
- σ × 5800⁴ = 6.42 × 10⁷ W/m².
Safety
- Computer only.
Home or virtual alternative
- Runs in a browser.
Planet link: NASA: the Sun is about 5,800 K. NASA Earth Observatory: Climate and Earth’s Energy Budget
Lab F: Greenhouse effect and energy balance
Week 14
Goal: Explore PhET Greenhouse Effect, then compute Earth’s effective temperature for several albedos.
Procedure
- In PhET Greenhouse Effect, add and remove clouds and greenhouse gases and watch the temperature.
- In a spreadsheet, use S = 1360 W/m² (NASA).
- Compute T_e for albedo 0.25, 0.29 and 0.35.
- Explain why T_e is below NASA’s 15 °C surface average.
Data table
| Albedo | T_e (K) |
|---|---|
| 0.25 | 259.0 |
| 0.29 | 255.4 |
| 0.35 | 249.9 |
Computed values.
Sample calculation
- (1360 × 0.71/(4σ))^¼ = 255.4 K.
- 255.4 − 273.15 = −17.7 °C.
Safety
- Computer only.
Home or virtual alternative
- Runs in a browser.
Planet link: NASA: about 29% of sunlight is reflected and Earth absorbs about 240 W/m². NASA Earth Observatory: Climate and Earth’s Energy Budget
Exam prep practice sets
Time yourself: about 10 minutes per problem. Then compare with the full solutions.
Midterm practice set (Units 1–5)
Convection from 3 m² at h = 15 W/(m²·K), ΔT = 20 K?
Show full solution
- 15 × 3 × 20.
Answer: 900 W.
Heat flux through 0.15 m of insulation (k = 0.04) plus 0.3079 m²·K/W of other layers at ΔT = 20 K?
Show full solution
- R″ = 0.3079 + 3.75 = 4.058.
- 20/4.058.
Answer: 4.93 W/m².
Critical radius for k = 0.1 and h = 10?
Show full solution
- k/h.
Answer: 10 mm.
τ for a 20 mm aluminum sphere, h = 50?
Show full solution
- 2700 × 900 × (0.02/6)/50.
Answer: 162 s.
Re for air at 2 m/s over 0.5 m?
Show full solution
- 2 × 0.5/1.5 × 10⁻⁵.
Answer: 66,667.
Final practice set (Units 1–10)
LMTD for ΔT₁ = 60 K and ΔT₂ = 20 K?
Show full solution
- 40/ln 3.
Answer: 36.4 K.
ε for NTU = 1, C_r = 0?
Show full solution
- 1 − e⁻¹.
Answer: 0.632.
Ra for Unit 7’s plate at L = 0.6 m?
Show full solution
- 7.52 × 10⁷ × 8.
Answer: 6.02 × 10⁸.
Peak wavelength at 1000 K?
Show full solution
- 2898/1000.
Answer: 2.90 µm.
Parallel plates 500 K and 300 K, ε = 0.9 both?
Show full solution
- σ(500⁴ − 300⁴)/(2/0.9 − 1).
Answer: 2524 W/m².
Carnot heating COP, 318 K delivery, 268 K outdoors?
Show full solution
- 318/50.
Answer: 6.36.
Formula sheet
Constants
| Constant | Value |
|---|---|
| Stefan–Boltzmann constant | σ = 5.670374419 × 10⁻⁸ W/(m²·K⁴) |
| Wien’s constant (textbook value) | 2898 µm·K |
| Standard gravity | g = 9.80665 m/s² |
| Air near 300 K (textbook values) | k = 0.026 W/(m·K), ν = 1.5 × 10⁻⁵ m²/s, Pr = 0.71 |
| Aluminum (textbook values) | k = 237 W/(m·K), ρ = 2700 kg/m³, c = 900 J/(kg·K) |
| Earth’s surface area (textbook value) | 5.1 × 10¹⁴ m² |
Free texts and simulations
Use these free texts, lectures and simulations alongside the course.
- J. H. Lienhard IV and J. H. Lienhard V, A Heat Transfer Textbook (free online edition): Free complete heat transfer textbook.
- MIT OpenCourseWare 2.51 Intermediate Heat and Mass Transfer (Fall 2008): Free MIT lectures and problems.
- NASA Earth Observatory: Climate and Earth’s Energy Budget: Free NASA explainer.
- PhET: Energy Forms and Changes: Heat flow interactively.
- PhET: Blackbody Spectrum: Thermal radiation.
- PhET: Greenhouse Effect: Radiation balance.
Glossary
- Albedo
- Fraction of sunlight reflected.
- Biot number
- Internal over external thermal resistance.
- Blackbody
- Perfect absorber and emitter.
- Boundary layer
- Thin near-wall flow region.
- COP
- Heat moved per unit work.
- Critical radius
- Radius of maximum heat loss from an insulated cylinder.
- Effectiveness
- Actual over maximum exchanger heat.
- Emissivity
- Emission relative to a blackbody.
- Energy imbalance
- Absorbed minus emitted planetary energy.
- Fin efficiency
- Actual over ideal fin heat.
- Fourier’s law
- Conduction rate law.
- Heat exchanger
- Device moving heat between fluids.
- LMTD
- Log-mean temperature difference.
- Lumped capacitance
- Uniform-temperature transient model.
- Nusselt number
- Dimensionless convection coefficient.
- Prandtl number
- Momentum over thermal diffusivity.
- Radiation shield
- Low-emissivity barrier.
- Rayleigh number
- Buoyancy-flow strength.
- Thermal resistance
- Temperature difference per heat flow.
- View factor
- Fraction of radiation one surface intercepts.
Self-assessment rubric
No grade or credit is given. Use this to judge honestly where you are in each unit.
| Level | What it looks like |
|---|---|
| 4 · Mastery | Draws the system and resistance network, checks dimensionless numbers before choosing a correlation, uses kelvin for radiation, and explains results in plain words. |
| 3 · Proficient | Solves the practice and exam sets with minor slips and checks units. |
| 2 · Developing | Follows worked examples but mixes up k of the solid and fluid, or uses °C in σT⁴. |
| 1 · Beginning | Plugs numbers into formulas without a sketch or energy balance; needs to revisit the unit. |
Suggested weighting if you want a score
| Part | Weight |
|---|---|
| Practice problems | 20% |
| Labs | 25% |
| Midterm practice set | 25% |
| Final practice set | 30% |
Aim for level 3 or better in every unit before moving to the next course.