Heat Transfer · Chapter 10
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Heat Transfer · Chapter Ten

Boiling and Condensation

When a fluid changes phase, it absorbs or releases its latent heat — enabling enormous heat fluxes at small temperature differences. This is why boiling and condensation power so much of thermal engineering.

This chapter treats phase-change convection: the regimes of pool boiling and the critical heat flux, flow boiling in tubes, and film and dropwise condensation — all governed by latent heat \(h_{fg}\) and surface tension.

Learning objectives

  • Distinguish pool boiling, flow boiling, film condensation, and dropwise condensation from physical cues.
  • Use saturation temperature, wall superheat or subcooling, and vapor quality with consistent signs.
  • Select boiling and condensation correlations only within their stated geometry and regime limits.
  • Evaluate phase-change properties at saturation and liquid-film properties at the specified film temperature.
  • Compute heat flux, phase-change rate, quality, and condensation coefficient.
  • Audit critical-heat-flux margin, units, energy balances, and physical limits.
10–1

Boiling Heat Transfer

Boiling is vapor formation at a liquid–solid interface when \(T_s>T_{\text{sat}}(p)\). It is not identified by temperature alone: pressure fixes saturation temperature, and a liquid may be locally boiling while its bulk remains subcooled. The signed wall superheat is \(\Delta T_e=T_s-T_{\text{sat}}>0\); use absolute temperature only where a correlation requires it.

Energy transport and nucleation

Each kilogram formed carries \(h_{fg}\) joules, while departing bubbles renew liquid at the wall. Active nucleation also depends on roughness, wettability, dissolved gas, and pressure; therefore a surface–fluid constant is not transferable without evidence.

Physical meaning

A large boiling coefficient does not mean the wall can accept unlimited heat. The boiling curve is non-monotonic, and the safe operating regime matters more than a single value of \(h=q''/\Delta T_e\).

Problem-solving strategy

  1. Identify pool versus forced-flow boiling and obtain \(T_{\text{sat}}\) at the local absolute pressure.
  2. Locate the likely regime from wall superheat, heat flux, and flow quality.
  3. Evaluate liquid/vapor properties at saturation unless the selected correlation specifies otherwise.
  4. Apply the minimum correlation that matches fluid, surface, geometry, and validity range.
  5. Audit energy conservation and compare operating heat flux with critical heat flux using a documented margin.
10–2

Pool Boiling

In pool boiling there is no imposed bulk flow. Increasing wall superheat produces natural convection, onset of nucleate boiling, vigorous nucleate boiling, critical heat flux (CHF), unstable transition boiling, a minimum-flux point, and stable film boiling. Transition boiling is generally unsuitable for controlled design because increasing wall temperature can reduce heat flux.

The Boiling Curve

Nucleate boiling combines latent transport and mixing. At CHF, vapor columns coalesce and interrupt rewetting; under imposed heat flux this can cause a sudden wall-temperature excursion. Stable film boiling has a continuous vapor layer and may require radiation as well as convection.

Rohsenow correlation (nucleate boiling)
\[ \dot{q}_{\text{nucleate}} = \mu_l h_{fg}\left[\frac{g(\rho_l-\rho_v)}{\sigma}\right]^{1/2}\left[\frac{c_{p,l}(T_s-T_{\text{sat}})}{C_{sf}\,h_{fg}\,Pr_l^{\,n}}\right]^3 \]
\(C_{sf}\) and \(n\) depend on the surface–fluid pairing; \(\sigma\) is surface tension. Heat flux scales with the cube of excess temperature.
Critical heat flux (Zuber)
\[ \dot{q}_{\max} = C_{cr}\,h_{fg}\left[\sigma g\,\rho_v^2(\rho_l - \rho_v)\right]^{1/4} \]
Use saturation properties at the operating pressure. This pool-boiling estimate is not a substitute for flow-boiling CHF maps or equipment qualification.
EXAMPLE 1Refrigerant production in a test evaporator
Problem. A laboratory evaporator delivers \(18.0\) kW directly to saturated R-134a liquid at its operating pressure. Supplied \(h_{fg}=201\) kJ/kg. Neglect heat loss and find the steady vapor production rate.

Strategy

Saturated liquid enters and saturated vapor leaves, so a latent-energy balance closes the problem without a boiling correlation.

Givens

  • \(\dot Q=18.0\) kW, \(h_{fg}=201\) kJ/kg; steady and adiabatic except for heater input.

Physical/model interpretation

The control volume is the refrigerant. Positive heat input creates vapor; kinetic, potential, and sensible-energy changes are neglected.

Numbered derivation
1. Write \(\dot Q=\dot m_v(h_g-h_f)=\dot m_v h_{fg}\).
2. Rearrange: \(\dot m_v=\dot Q/h_{fg}\).
3. Preserve compatible rate units: \(18.0\ {\rm kJ/s}/(201\ {\rm kJ/kg})\).
4. Evaluate \(\dot m_v=0.0896\ {\rm kg/s}\).

Unit/sign/limit checks

Units: \(({\rm kJ/s})/({\rm kJ/kg})={\rm kg/s}\).

Signs: positive heater input gives positive vapor generation.

Limit: as \(\dot Q\to0\), \(\dot m_v\to0\); doubling power doubles production under this model.

Final answer: \(\boxed{\dot m_v=0.0896\ {\rm kg/s}}\).

Engineering insight

This is an energy-capacity result, not a surface-temperature prediction; area, boiling regime, CHF margin, and losses must still be checked.

Course Note

The critical heat flux is the single most important number in boiling design. Operating just below it gives huge heat transfer; cross it and the surface temperature can jump by hundreds of degrees almost instantly — the "burnout" that destroys heating elements and fuel rods.

10–3

Flow Boiling

Flow boiling couples forced convection, nucleation, acceleration, and pressure drop. Equilibrium quality \(x\) is vapor mass fraction for a saturated mixture, \(0\le x\le1\); negative calculated quality denotes subcooled liquid and \(x>1\) denotes superheated vapor rather than a physical mixture quality.

Forced Convection Boiling

Local flow pattern may pass through bubbly, slug, annular, and mist regimes. A correlation must match channel size, orientation, mass flux, quality, pressure, and heat-flux range. Dryout and departure from nucleate boiling are distinct limits.

EXAMPLE 2Outlet quality of a heated water channel
Problem. Water at \(90^\circ{\rm C}\) enters a \(1\)-atm heated channel at \(0.080\) kg/s. It receives \(60.0\) kW. Using supplied \(c_p=4.20\) kJ/(kg·K), \(T_{\rm sat}=100^\circ{\rm C}\), and \(h_{fg}=2257\) kJ/kg, find outlet equilibrium quality.

Strategy

First pay the sensible-heating requirement to saturation; assign only the remaining energy to evaporation.

Givens

  • \(\dot m=0.080\) kg/s, \(\dot Q=60.0\) kW, \(T_i=90^\circ{\rm C}\), saturation properties supplied.

Physical/model interpretation

Steady, negligible losses and pressure drop; the outlet is an equilibrium saturated mixture if the computed quality lies from zero to one.

Numbered derivation
1. Sensible duty: \(\dot Q_s=\dot m c_p(T_{\rm sat}-T_i)\).
2. \(\dot Q_s=(0.080)(4.20)(10)=3.36\) kW.
3. Latent duty is \(60.0-3.36=56.64\) kW.
4. \(x_o=\dot Q_{\rm lat}/(\dot m h_{fg})=56.64/[0.080(2257)]=0.314\).

Unit/sign/limit checks

Units: kW divided by \(({\rm kg/s})({\rm kJ/kg})\) is dimensionless.

Signs: With heat into the control volume defined positive, \(60.0-3.36=56.64\) kW remains positive, so the computed quality increment must be positive; a negative result would instead indicate that the outlet remained subcooled.

Limit: \(0<x_o<1\), validating the mixture interpretation; the sensible load is only 5.6% of input.

Final answer: \(\boxed{x_o=0.314\ {\rm (31.4\%\,vapor\ by\ mass)}}\).

Engineering insight

Real design must include pressure-dependent saturation temperature, axial heat flux, dryout, and two-phase pressure drop.

10–4

Condensation Heat Transfer

Condensation occurs when vapor contacts a surface below its local saturation temperature, so \(\Delta T_c=T_{\rm sat}-T_s>0\). Noncondensable gas can build a diffusion layer and drastically lower performance; pure-vapor film correlations then overpredict heat transfer.

Film vs. Dropwise

Film condensation is predictable but the growing condensate layer adds conduction resistance. Dropwise condensation repeatedly exposes surface and can be much stronger, but its durability depends on surface chemistry and contamination.

10–5

Film Condensation

Nusselt's vertical-plate result assumes steady laminar film flow, negligible vapor shear and inertia, constant properties, a pure saturated vapor, and conduction across the film. Evaluate liquid properties at \(T_f=(T_{\rm sat}+T_s)/2\), while \(h_{fg}\) comes from saturation data.

Nusselt film condensation on a vertical plate
\[ h = 0.943\left[\frac{g\rho_l(\rho_l - \rho_v)h_{fg}'\,k_l^3}{\mu_l(T_{\text{sat}} - T_s)L}\right]^{1/4} \]
\(h_{fg}' = h_{fg} + 0.68\,c_{p,l}(T_{\text{sat}}-T_s)\) is the modified latent heat correcting for subcooling of the film.
Horizontal tube
\[ h = 0.729\left[\frac{g\rho_l(\rho_l - \rho_v)h_{fg}'\,k_l^3}{\mu_l(T_{\text{sat}} - T_s)D}\right]^{1/4} \]
EXAMPLE 3Condensate load on a process panel
Problem. Saturated steam at \(60^\circ{\rm C}\) condenses on one side of a vertical panel \(0.60\) m high and \(0.30\) m wide held at \(50^\circ{\rm C}\). At the \(55^\circ{\rm C}\) film temperature, use \(\rho_l=983\) kg/m³, \(\rho_v=0.13\) kg/m³, \(k_l=0.653\) W/(m·K), \(\mu_l=4.67\times10^{-4}\) Pa·s, and modified \(h'_{fg}=2358\) kJ/kg. Find average \(h\), heat rate, and condensate rate.

Strategy

The pure vapor, vertical geometry, and supplied film properties cue the laminar Nusselt plate model.

Givens

  • \(L=0.60\) m, \(W=0.30\) m, \(\Delta T_c=10\) K, \(A=WL\); properties supplied at \(T_f=55^\circ{\rm C}\).

Physical/model interpretation

Gravity drives a downward film. Heat and condensate rates are reported positive from vapor to wall; edge effects and vapor shear are neglected.

Numbered derivation
1. Use \(\bar h=0.943[g\rho_l(\rho_l-\rho_v)h'_{fg}k_l^3/(\mu_l\Delta T_cL)]^{1/4}\).
2. Substitute SI values: \(\bar h=0.943[9.81(983)(982.87)(2.358\times10^6)(0.653)^3/(4.67\times10^{-4}(10)(0.60))]^{1/4}\).
3. Evaluate \(\bar h=6.47\times10^3\) W/(m²·K), then \(\dot Q=\bar h(WL)\Delta T_c=11.7\) kW.
4. \(\dot m_c=\dot Q/h'_{fg}=11.652/2358=4.94\times10^{-3}\) kg/s.

Unit/sign/limit checks

Units: \(hA\Delta T\) gives W and W/(J/kg) gives kg/s.

Signs: \(T_{\rm sat}>T_s\), so heat flows to the wall and condensation is positive.

Limit: the correlation gives \(h\propto(\Delta T_cL)^{-1/4}\), while \(\dot Q\propto\Delta T_c^{3/4}\), so zero subcooling gives zero heat rate despite divergent idealized \(h\).

Final answer: \(\boxed{\bar h=6.47\ {\rm kW/(m^2K)},\ \dot Q=11.7\ {\rm kW},\ \dot m_c=4.94\ {\rm g/s}}\).

Engineering insight

Verify film Reynolds number and remove air leakage before sizing; noncondensables can dominate the resistance.

10–6

Film Condensation Inside Horizontal Tubes

Inside horizontal tubes, gravity drains liquid around the circumference while vapor shear redistributes it. The Chato form is a low-vapor-velocity model; at higher mass flux use a validated internal-condensation correlation that includes quality, shear, and flow regime.

Low-vapor-velocity internal condensation (Chato)
\[ h_{\text{internal}} = 0.555\left[\frac{g\rho_l(\rho_l-\rho_v)k_l^3}{\mu_l(T_{\text{sat}}-T_s)D}\left(h_{fg} + \tfrac{3}{8}c_{p,l}(T_{\text{sat}}-T_s)\right)\right]^{1/4} \]
Applicable for low vapor Reynolds numbers (\(Re_v < 35{,}000\) at inlet).
10–7

Dropwise Condensation

Dropwise condensation may greatly outperform film condensation because droplets depart and renew the surface. It should not be credited as a permanent design condition unless coating life, fouling, chemistry, and inspection have been demonstrated.

Key Idea Dropwise condensation can yield coefficients up to ~10 times those of film condensation. The challenge is durability: special coatings or promoters that keep the surface non-wetting eventually degrade, and the process reverts to film condensation.

Chapter recap

  • Recognize the regime before selecting a correlation.
  • Anchor properties to saturation pressure and film temperature.
  • Use an energy balance for phase rate or quality before adding local heat-transfer detail.
  • Treat CHF, dryout, noncondensables, and coating durability as engineering limits.

Before moving on

You should be able to sketch the boiling curve, calculate quality and condensate rate, state each correlation's assumptions, and audit units, signs, and regime validity.

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