Heat Transfer over a Stretching Sheet

1. The energy equation and its reduction

For steady boundary-layer flow with constant properties and negligible dissipation, energy balance on a fluid element gives u∂T/∂x + v∂T/∂y = α∂²T/∂y², with α = k/ρc_p the thermal diffusivity. Take the sheet at constant T_w, ambient at T∞, and define θ(η) = (T − T∞)/(T_w − T∞) with the same η as the flow problem. Because T depends on η alone, the equation collapses to

Linear in θ — the nonlinearity lives entirely in f, which is already known from the momentum problem. This one-way coupling (flow drives heat, not vice versa) is called a forced convection regime.

2. Solving it

Integrating twice: θ′(η) = θ′(0) exp(−Pr∫₀^η f dξ), and with f = (1−e^(−βη))/β the inner integral is explicit, giving a solution in terms of the incomplete gamma function. The single number that matters is the wall gradient:

−θ′(0) is the dimensionless cooling rate every paper tabulates.

3. Reading the parameters

Prandtl number: Pr = ν/α compares momentum and thermal diffusion. Large Pr (oils, ~100) → thermal layer much thinner than momentum layer → steep wall gradient → high Nu. Small Pr (liquid metals, ~0.01) → the reverse. Magnetic parameter: raising M brakes the flow, weakens convection near the wall, thickens the thermal layer and reduces −θ′(0) — the field cools the cooling. Nanoparticle loading raises k_nf and typically raises Nu, the engineering motivation for the whole nanofluid programme.

Key takeaway

The thermal problem inherits the flow's similarity structure for free. One function f in, one number −θ′(0) out — and every radiation, dissipation or nanofluid extension in the literature is a modification of this equation.

4. Extended worked examples

Example 1 — from θ′(0) to hardware numbers. Suppose a computation returns θ′(0) = −0.60 for water at a station where Re_x = 10⁴. Then Nu_x = −√(Re_x)·θ′(0) = 100×0.60 = 60, and the convection coefficient is h = Nu_x·k/x: at x = 0.1 m with k = 0.613, h = 60×0.613/0.1 = 368 W/m²K. A sheet 20 K above ambient sheds q″ = hΔT = 7.4 kW/m² there. Every dimensionless table in the literature converts to watts by exactly this two-step.

Example 2 — Prandtl numbers set the picture. Water: Pr = ν/α = 1.004×10⁻⁶/1.47×10⁻⁷ ≈ 6.8. Air: ≈ 0.71. Engine oil: 10³–10⁴. Mercury: 0.025. For oil the thermal layer is buried deep inside the momentum layer (ratio ~ Pr^{−1/3} to ^{−1/2}); for mercury it pokes far outside, so liquid-metal heat transfer barely notices the velocity field's details — two different asymptotic worlds bracketing the same equation.

Example 3 — a large-Pr formula you can derive. Near the wall f ≈ η (since f(0) = 0, f′(0) = 1), so θ′(η) ≈ θ′(0)exp(−Pr η²/2), giving 1 = −θ′(0)∫₀^∞e^{−Prξ²/2}dξ = −θ′(0)√(π/2Pr). Hence −θ′(0) ≈ √(2Pr/π) ≈ 0.798√Pr for large Pr: at Pr = 7 this estimates 2.11; at Pr = 100, 7.98. A one-line asymptotic that reproduces the trend of every published table and provides the "does my code scale right?" check before any table is consulted.

Example 4 — the magnetic field's fingerprint. With f = (1−e^{−βη})/β, larger M (larger β) means f grows toward its plateau 1/β more slowly in integrated terms: ∫₀^η f dξ shrinks, the exponential damping of θ′ weakens, the thermal layer fattens and −θ′(0) falls. Physically: braked flow carries less warm fluid away per second. In tables this is the monotone Nu-versus-M column; now you can predict its sign before computing it.

5. Common misconceptions

"Nu and h are the same thing." h (W/m²K) is dimensional hardware; Nu = hx/k is its dimensionless shadow, and Nu_x here grows like √Re_x ∝ √x even while h itself decreases downstream (h ∝ x^{−1/2}) — quoting the wrong one reverses conclusions about where the sheet cools fastest (answer: near the die, where the layer is thinnest).

"The magnetic field heats the wall region, so it must raise heat transfer." Joule heating exists, but in this constant-property formulation the dominant M-effect is convective braking, which lowers −θ′(0). Papers including viscous/Ohmic dissipation add an Eckert number to carry the heating honestly.

"Pr is a property of the flow." It is a property of the fluid (ν/α), fixed before any flowing begins. The flow chooses Re; the fluid brings Pr.

"θ is temperature." θ = (T−T∞)/(T_w−T∞) is a shape; multiply by the driving ΔT to recover degrees. Two experiments with different ΔT share one θ(η) exactly — the collapse that makes the similarity table universal.

6. Where this shows up

Polymer-film and fibre lines are the native habitat: the crystallinity, strength and clarity of drawn PET or nylon depend on the cooling curve through the glass transition, and that curve is −θ′(0) wearing units. Glass-fibre drawing (quench rates), continuous metal casting (mould heat flux), and paper drying all run the same mathematics with different Pr. In the research literature the equation of this chapter is the universal second act: whatever exotic momentum physics a paper proposes — Casson rheology, hybrid nanoparticles, porous media — its heat-transfer claims funnel through θ″ + Pr fθ′ (+ extras) = 0, and its headline deliverable is the −θ′(0) table you now know how to read, estimate, and audit.

7. Practice problems

P1. Convert θ′(0) = −1.9 at Re_x = 2.5×10⁵, x = 0.2 m, water: find Nu_x and h. Solution: Nu = 500×1.9 = 950; h = 950×0.613/0.2 = 2 912 W/m²K.

P2. Use Example 3's asymptotic to sanity-check a claimed −θ′(0) = 3.1 at Pr = 20. Solution: estimate 0.798×√20 = 3.57 — the claim is plausible (asymptotic overshoots slightly at moderate Pr since f < η away from the wall; a claimed 7 would be a red flag).

P3. Two coolants at equal Re: air (Pr 0.71) and water (Pr 6.8). Estimate the Nusselt ratio via the √Pr rule. Solution: √(6.8/0.71) ≈ 3.1 — water roughly triples the dimensionless cooling; with its far larger k, the dimensional advantage is bigger still.

P4. Show that with variable wall temperature T_w − T∞ = Axᵏ the similarity form survives, θ″ + Pr(fθ′ − k f′θ) = 0. Solution: substitute T = T∞ + Axᵏθ(η); the x-powers cancel for any k — recovering the Grubka–Bobba family, and explaining why published tables carry a wall-temperature exponent column.

8. Going deeper

Radiation enters most papers through the Rosseland approximation: for an optically thick medium the radiative flux behaves diffusively, q_r = −(16σ*T³/3k*)∂T/∂y, which (linearised about T∞) simply multiplies θ″ by (1 + 4Rd/3) — radiation as bonus conductivity, one parameter Rd. Viscous and Joule dissipation add Eckert-number source terms Ec·f″² and M·Ec·f′²; Buongiorno coupling adds Nbθ′φ′ + Ntθ′². Each extension is one more term in the same linear-in-θ scaffold, which is why a reader fluent in this chapter's bare equation can open any paper in the field and know, before the results section, which way every table must lean.

9. Historical context

The thermal problem rode in on the momentum solution's coat-tails. Fourier's Théorie analytique de la chaleur (1822) supplied the conduction law and the very idea of dimensionless heat flow that Nusselt formalised for convection in 1915. Once Crane (1970) had the exact stretching-sheet flow, the energy equation followed within the decade for constant wall temperature; Grubka and Bobba's 1985 paper solved the variable wall temperature family T_w − T∞ = Axᵏ (the exercise P4 reproduces its reduction) and published the −θ′(0) tables against Pr and k that remain standard citations — many later papers' "validation" rows are literally comparisons with Grubka–Bobba entries. The industrial motivation was explicit from the start: Sakiadis worked for a fibre company, and the polymer-drawing literature of the 1960s–80s (Tsou, Sparrow & Goldstein's combined analysis and experiment, 1967) confirmed that the moving-surface thermal boundary layer is a real, measurable object, not an analyst's toy.

10. Another way to see it: the reduction, every step shown

Claiming "the energy PDE becomes θ″ + Pr fθ′ = 0" deserves its receipts. Start from steady advection–diffusion with constant properties, u∂T/∂x + v∂T/∂y = α∂²T/∂y² (streamwise conduction dropped on boundary-layer scaling). Substitute T = T∞ + (T_w − T∞)θ(η), constant wall temperature, with η = y√(a/ν), u = axf′, v = −√(aν)f. Chain rule: ∂T/∂x = 0 (θ carries no x; ΔT constant); ∂T/∂y = ΔT θ′√(a/ν); ∂²T/∂y² = ΔT θ″(a/ν). Insert:

The left collapses to −aΔT fθ′; divide by aΔT and by α/ν = 1/Pr: θ″ + Pr fθ′ = 0, with θ(0) = 1, θ(∞) = 0. Every x cancelled without being asked — the definition of a successful similarity reduction — and the only fluid signature left standing is Pr. The first integral is also worth owning: θ′(η) = θ′(0)exp(−Pr∫₀^η f dξ), from which θ′(0) = −[∫₀^∞ exp(−Pr∫₀^η f)dη]^{−1}: the Nusselt number as an explicit functional of the flow — the formula behind Example 3's asymptotic and behind every numerical table's smooth dependence on Pr.

11. Frequently asked questions

Why does Pr appear but not Re? Re was consumed by the similarity scaling itself (it reappears only when converting back to dimensional Nu = −√Re_x θ′(0)). Inside the reduced world, the fluid's only remaining fingerprint is the diffusivity ratio.

Can θ overshoot 1 or undershoot 0? Not for this equation with these boundary conditions: the first-integral form shows θ′ keeps one sign, so θ descends monotonically from 1 to 0. Overshoots in the literature signal added physics — dissipation, heat sources, radiation nonlinearity — never the bare equation.

Which way does suction move Nu? Suction (f(0) = S > 0) fattens f, strengthens the exponential damping in the first integral, thins the thermal layer: Nu rises — the standard control knob when the field's braking must be compensated.

Is constant ΔT realistic for polymer lines? Often no — hence Grubka–Bobba's power-law family and the constant-heat-flux variant (θ′(0) prescribed instead of θ(0)); real quench analyses pick whichever idealisation brackets the plant data, and the k-exponent column in published tables exists precisely because the answer genuinely depends on it.

12. Further practice

P5. Evaluate the first integral exactly for M = 0 using f = 1 − e^{−η}: show −θ′(0) = e^{−Pr}Pr^{Pr}/γ(Pr, Pr)·(…) is messy but that Pr = 1 gives −θ′(0) = 1/(e−1)·e ≈ … — instead, verify numerically: with f known, a one-line quadrature gives −θ′(0)|_{Pr=1} ≈ 0.582. Solution: ∫₀^η f = η − 1 + e^{−η}; −θ′(0) = [∫₀^∞ e^{−(η−1+e^{−η})}dη]^{−1} — substituting t = e^{−η} turns it into an incomplete-gamma expression evaluating to ≈ 0.582; matching the literature's Pr = 1 entry validates both the formula and the tables at once.

P6. Show that adding uniform heat generation Q₀(T−T∞) adds +λθ (λ = Q₀/aρc_p) to the equation, and predict its sign effect on Nu. Solution: the source feeds θ, flattening its decay: −θ′(0) falls; internal generation always fights wall heat extraction — and λ > some threshold even reverses the wall flux, the "critical heat generation" rows of extended tables.

P7. A pilot line draws PET at a = 5 s⁻¹ through 40 °C water (Pr ≈ 4.3, k = 0.628). Using the √Pr asymptotic, estimate h at x = 0.05 m. Solution: −θ′(0) ≈ 0.798×2.07 = 1.65; Re_x = ax²/ν = 5×0.0025/10⁻⁶... careful: Re_x = u_w x/ν = ax²/ν = 1.25×10⁴; Nu = 1.65×112 = 185; h = 185×0.628/0.05 ≈ 2 320 W/m²K — pilot-plant arithmetic from a similarity table and one asymptotic.

13. Worked exam problem

Problem. A published table gives −θ′(0) = 1.42 at Pr = 5, M = 1 for water (k = 0.613) over a sheet stretching at a = 3 s⁻¹. At x = 8 cm find (a) Re_x, (b) Nu_x, (c) h, and (d) the heat flux for a sheet 25 K above ambient.

Solution. (a) Re_x = u_w x/ν = ax²/ν = 3×0.0064/1.004×10⁻⁶ = 1.91×10⁴. (b) Nu_x = √Re_x×1.42 = 138.3×1.42 = 196. (c) h = Nu_x k/x = 196×0.613/0.08 = 1 505 W/m²K. (d) q″ = hΔT = 1 505×25 = 37.6 kW/m² — a serious quench rate, extracted from one dimensionless table entry plus unit bookkeeping. This four-step conversion is the entire practical interface between the similarity literature and a factory floor; fluency in it is what "reading the field" means.

14. Key takeaways

The energy equation rides the flow: θ″ + Pr fθ′ = 0 with θ falling monotonically from 1 to 0, its wall slope −θ′(0) the whole story — Nu_x = −√Re_x θ′(0). Pr sets the thermal layer's thickness relative to the momentum layer (√(2Pr/π) asymptotic at large Pr); magnetic braking thickens it and cuts Nu; suction thins it and raises Nu; every literature extension (radiation, dissipation, Buongiorno terms) is one more auditable term in the same scaffold.

15. Where to go next

The velocity field f comes from the similarity chapter; the M-dependence traces to the Lorentz force; property shifts from loading enter via nanofluid models; and the numerics are shooting/bvp4c. The Hagen–Poiseuille calculator automates this chapter's table-to-watts conversion.