Heat Transfer · Chapter 8 of 10 · Advanced
Boiling and Condensation
Phase change moves enormous heat at small temperature differences, but the boiling curve hides a cliff: push too hard and the surface burns out.
Readiness check
From Chapters 1 and 5 and thermodynamics. Tick only what you can do closed-notes.
- Define latent heat of vaporization hfg.
- Use a convection coefficient in q = hA ΔT.
- Recall saturation temperature and pressure.
- Read a log-log plot.
- Look up saturated-fluid properties.
The core idea
Making or destroying vapor absorbs or releases latent heat, so phase change transfers huge heat at small temperature differences, until a vapor blanket forms.
q″ = h(Ts − Tsat)ΔTe = Ts − Tsatq″max = 0.149 hfg ρv1/2[σg(ρl−ρv)]1/4Boiling coefficients reach thousands to tens of thousands of W/m²·K because each bubble carries off latent heat. But the boiling curve is not monotonic: beyond a critical heat flux, vapor blankets the surface, h collapses, and the surface temperature spikes, often to destruction.
The skills, taught in order
Phase-change heat transfer is governed by latent heat and the shape of the boiling curve. Six skills cover why it is so effective, where its limit lies, and how condensation mirrors it.
8.1 Why phase change moves so much heat
Vaporizing a fluid absorbs its latent heat hfg, which is large (about 2.26 MJ/kg for water). A small mass of vapor therefore carries off a great deal of energy, giving coefficients far above any single-phase value, often 10³ to 10⁵ W/m²·K.
8.2 The boiling curve
Plotting heat flux against the excess temperature ΔTe = Ts − Tsat gives the boiling curve. It runs through free convection, nucleate boiling (steeply rising), a peak at the critical heat flux, an unstable transition region, and finally film boiling.
| Regime | Excess temperature ΔTe (water) | Behaviour |
|---|---|---|
| Free convection | up to ~5 °C | no bubbles; single-phase |
| Nucleate boiling | ~5 to 30 °C | bubbles; very high, rising q″ |
| Critical heat flux | ~30 °C | the peak; onset of burnout |
| Film boiling | > ~120 °C | vapor blanket; low h, high ΔT |
8.3 Nucleate boiling
In the workhorse regime, bubbles nucleate at surface cavities, grow, and depart, stirring the liquid violently. The Rohsenow correlation models it, with q″ rising roughly as ΔTe³, so a small increase in surface temperature buys a large increase in flux.
8.4 The critical heat flux
At the peak, vapor is generated faster than liquid can rewet the surface. The Zuber correlation predicts q″max = 0.149 hfg ρv1/2[σg(ρl−ρv)]1/4, about 1.26 MW/m² for water at 1 atm. A flux-controlled surface that reaches this point jumps to film boiling and may melt.
8.5 Film boiling and the Leidenfrost effect
Past the peak, a continuous vapor film insulates the surface, so h drops and ΔT soars (the same physics that lets a water droplet skate on a hot pan). Radiation through the film becomes significant at high temperatures.
8.6 Condensation
The reverse process releases hfg. In filmwise condensation a liquid film coats the surface and the Nusselt film analysis gives h ∝ [ρl(ρl−ρv)g hfgkl³/(μlΔT L)]1/4. Dropwise condensation, where the surface sheds droplets, can be several times higher but is hard to sustain.
Engineering connection: boilers, evaporators, condensers, heat pipes, and the cooling of high-flux electronics and reactor surfaces, where the critical heat flux is a hard safety limit.
Worked example 1: critical heat flux for water
A flat heater boils water at 1 atm. Using saturated-water properties (hfg = 2.257×10⁶ J/kg, ρv = 0.596 kg/m³, ρl = 957.9 kg/m³, σ = 0.0589 N/m), find the critical heat flux: the most heat the surface can safely deliver in nucleate boiling.
- ProblemFind the critical heat flux q″max for water boiling at 1 atm, the peak in Figure 1.
- Given / findhfg = 2.257×10⁶ J/kg, ρv = 0.596, ρl = 957.9, σ = 0.0589 N/m, g = 9.81. Find q″max.
- AssumptionsPool boiling on a large horizontal surface, saturated liquid, Zuber correlation applies.
- ModelSubstitute the saturated properties into the Zuber critical-heat-flux correlation.
- Equationsq″max = 0.149 hfg ρv1/2[σg(ρl−ρv)]1/4
- Solveρv1/2 = 0.772; the bracket σg(ρl−ρv) = 0.0589 × 9.81 × 957.3 = 553, and 5531/4 = 4.85. So q″max = 0.149 × 2.257×10⁶ × 0.772 × 4.85 = 1.26×10⁶ W/m² = 1.26 MW/m².
- CheckThis matches the well-known handbook value for water at atmospheric pressure, a strong confirmation. A megawatt per square metre at only ~30 °C of excess temperature shows how powerful nucleate boiling is.
- ConclusionDesign the heater to stay safely below ~1.26 MW/m². If a flux-controlled source (an electric element or fuel rod) is pushed past this, the surface dries out and can melt within seconds, the failure mode called burnout.
Worked example 2: film condensation on a wall
Saturated steam at 100 °C condenses on a 0.2 m tall vertical wall held at 80 °C. Using liquid properties at the film temperature (ρl = 965 kg/m³, kl = 0.675 W/m·K, μl = 3.15×10⁻⁴ Pa·s, cp,l = 4206 J/kg·K) with hfg = 2.257×10⁶ J/kg, find the average condensation coefficient.
- ProblemFind the average condensation coefficient on the vertical wall in Figure 2.
- Given / findTsat = 100 °C, Ts = 80 °C (ΔT = 20 °C), L = 0.2 m, ρl = 965, kl = 0.675, μl = 3.15×10⁻⁴, cp,l = 4206, hfg = 2.257×10⁶. Find h̄.
- AssumptionsLaminar filmwise condensation, smooth vertical wall, properties at the film temperature.
- ModelUse the Nusselt film correlation with a modified latent heat h′fg = hfg + 0.68 cp,lΔT to account for subcooling.
- Equationsh′fg = hfg + 0.68 cp,lΔT h̄ = 0.943[ρl(ρl−ρv)g h′fgkl³/(μlΔT L)]1/4
- Solveh′fg = 2.257×10⁶ + 0.68 × 4206 × 20 = 2.314×10⁶ J/kg. Substituting, the bracket evaluates to about 5.0×10¹⁵, whose fourth root is ≈ 8400, so h̄ = 0.943 × 8400 = 8000 W/m²·K.
- CheckEight thousand W/m²·K is comparable to turbulent water in a pipe (Chapter 6) and far above any natural-convection gas value, exactly what latent-heat release should give. The h1/4 dependence means doubling ΔT lowers h̄ only about 16%.
- ConclusionCondensers achieve high coefficients cheaply because the fluid changes phase. Keeping the film thin (short surfaces, drainage, or promoting dropwise condensation) is how designers push the coefficient even higher.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Boiling curve assumed monotonic | Higher ΔT always taken as more heat | "Am I past the critical heat flux?" | Beyond the peak, h collapses; more ΔT means less flux until film boiling. |
| ΔT measured from ambient | Excess temperature taken as Ts − T∞ | "Is ΔTe measured from saturation?" | Use ΔTe = Ts − Tsat, not the ambient temperature. |
| Ignoring the burnout limit | Flux-controlled heater designed above q″max | "Have I checked the critical heat flux?" | For flux-controlled surfaces, stay below q″max; crossing it can melt the surface. |
| Latent heat omitted in condensation | Condenser sized with single-phase h | "Did I use hfg?" | Phase-change coefficients come from latent-heat release; use the Nusselt film correlation. |
Practice ladder
For the Worked Example 1 heater, what is the safe maximum heat rate over a 0.05 m² surface, staying below the critical heat flux?
Show answer
q = q″max × A = 1.26×10⁶ × 0.05 = 63 kW. A surface the size of a saucepan base can safely boil off tens of kilowatts, which is why kettles are fast.
Nucleate boiling follows q″ ∝ ΔTe³. If raising ΔTe from 10 to 20 °C is still nucleate, by what factor does the flux rise, and why must you still respect the critical heat flux?
Show answer
It rises by 2³ = 8 times. But the cube law cannot continue past the peak: once q″ reaches q″max the curve turns over, so the steep gain is exactly what makes overshooting into burnout easy.
Compare the Worked Example 2 condensation coefficient (≈ 8000 W/m²·K) with the natural-convection panel of Chapter 7 (≈ 5 W/m²·K). What does the ratio say about why condensers are compact?
Show answer
The ratio is about 1600. Phase change moves over a thousand times the heat per unit area and temperature difference, so a condenser needs far less surface than an air-cooled device of the same duty, hence its compactness.
Observe a real phase-change device (a kettle, a refrigerant condenser coil, a heat pipe). Identify the regime, estimate the relevant coefficient or critical heat flux, and explain the design choice it implies.
What good work looks like
The regime named (nucleate boiling, film condensation), the right correlation applied with saturated properties, and a coefficient or q″max compared with single-phase values to justify the design.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a one-page phase-change note: sketch the boiling curve, mark the regime of a real device, and compute either q″max or a condensation coefficient with saturated properties.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Why does phase change transfer so much heat?
It absorbs or releases the large latent heat hfg, giving coefficients of 10³ to 10⁵ W/m²·K.
2. Name the regions of the boiling curve in order.
Free convection, nucleate boiling, critical heat flux (peak), transition boiling, film boiling.
3. What is the critical heat flux, and roughly its value for water at 1 atm?
The peak heat flux before burnout; about 1.26 MW/m² for water at 1 atm (Zuber).
4. How is the excess temperature defined?
ΔTe = Ts − Tsat, surface above saturation, not above ambient.
5. How does filmwise condensation h scale with ΔT?
As ΔT−1/4 from the Nusselt analysis; a thicker film at larger ΔT lowers h slowly.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Lienhard and Lienhard, A Heat Transfer Textbook (6th ed), Chapter 9 (boiling and condensation) |
| Cross-reference | Incropera, Ch. 10 (boiling and condensation) · Çengel and Ghajar, Ch. 10 |
| Core topics | 8.1 Latent heat · 8.2 Boiling curve · 8.3 Nucleate boiling · 8.4 Critical heat flux · 8.5 Film boiling · 8.6 Condensation |
| Engineering connection | Boilers, evaporators, condensers, heat pipes, and high-flux electronics cooling, where q″max is a hard limit. |
| Read next | Chapter 9: Heat Exchangers. |