Fundamentals of Convective Heat Transfer

Week-08 (Live Session)

Durga Prasad Pydi

2025-09-13

Assumptions

  • Steady, laminar, 2D flow
  • Viscous dissipation term is neglected
  • Boundary layer approximation is valid (δT/H≪1\delta_T/H \ll 1)
  • Temperature difference between plate and flud is small enough to assume constant properties of the fluid
  • Flow is incompressible - however, density will vary with location. The assumption is valid because the change in density is due to changes in temperature field
Figure 1: Illustration of natural convection over a vertical heated plate

Boussinesq Approximation

Treat density as constant in the continuity equation and the inertia terms of the momentum equation, but allow it to change with temperature in the gravity term

Thus, continuity equation becomes

∂u∂x+∂v∂y=0 \frac{ \partial u }{ \partial x } + \frac{ \partial v }{ \partial y } =0

X-momentum equation is neglected under boundary layer approximation, hence y-momentum equation becomes

u∂v∂x+v∂v∂y=ν∂2v∂x2+βg(T−T∞) u \frac{ \partial v }{ \partial x } +v \frac{ \partial v }{ \partial y } = \nu \frac{ \partial^{2} v }{ \partial x^{2} }+ \beta g(T-T_{\infty}) where β=−1ρ∂ρ∂T|P\beta=-\frac{1}{\rho}\left.\frac{ \partial \rho }{ \partial T }\right|_{P} is the volumetric expansion coefficient of the fluid

Energy equation becomes

u∂T∂x+v∂T∂y=α∂2T∂x2 u \frac{ \partial T }{ \partial x } +v \frac{ \partial T }{ \partial y } = \alpha \frac{ \partial^{2} T }{ \partial x^{2} }

Scale Analysis

x∼δTu∼αδTy∼Hv∼αHδT2 \begin{align} x & \sim & \delta_{T} & u & \sim & \frac{\alpha}{\delta_{T}} \\ y & \sim & H & v & \sim & \frac{\alpha H}{\delta_{T}^{2}} \end{align}

non-dimensional numbers governing the flow physics are

Prandlt number, Pr=ναRayleigh number, RaH=gβΔTH3αν \begin{align} \text{Prandlt number, } Pr & = \frac{\nu}{\alpha} \\ \text{Rayleigh number, } Ra_{H} & = \frac{g\beta \Delta TH^{3}}{\alpha \nu} \end{align}

Scaling analysis for the momentum equations gives us,

u∂v∂x+v∂v∂y=ν∂2v∂2+gβ(T−T∞) u \frac{ \partial v }{ \partial x } + v \frac{ \partial v }{ \partial y } = \nu \frac{ \partial^{2} v }{ \partial^{2} } + g\beta(T-T_{\infty})

u∂v∂x,v∂v∂y∼(HδT)4RaH−1Pr−1ν∂2v∂x2∼(HδT)4RaH−1gβ(T−T∞)∼1 \begin{align} u \frac{ \partial v }{ \partial x }, v \frac{ \partial v }{ \partial y } & \sim \left( \frac{H}{\delta_{T}} \right)^{4}Ra_{H}^{-1}Pr^{-1} \\ \nu \frac{ \partial^{2} v }{ \partial x^{2} } & \sim \left( \frac{H}{\delta_{T}} \right)^{4}Ra_{H}^{-1} \\ g\beta (T-T_{\infty}) & \sim 1 \end{align}

Nusselt number for Pr≫1Pr\gg 1

u∂v∂x,v∂v∂y∼(HδT)4RaH−1Pr−1ν∂2v∂x2∼(HδT)4RaH−1gβ(T−T∞)∼1 \begin{align} u \frac{ \partial v }{ \partial x }, v \frac{ \partial v }{ \partial y } & \sim \left( \frac{H}{\delta_{T}} \right)^{4}Ra_{H}^{-1}Pr^{-1} \\ \nu \frac{ \partial^{2} v }{ \partial x^{2} } & \sim \left( \frac{H}{\delta_{T}} \right)^{4}Ra_{H}^{-1} \\ g\beta (T-T_{\infty}) & \sim 1 \end{align}

If Pr≫1Pr\gg 1, then the inertial terms would be negligible or viscous and buoyancy forces balance each other

⟹(HδT)4RaH−1∼1 \implies \left( \frac{H}{\delta_{T}} \right)^{4}Ra_{H}^{-1} \sim 1

which gives us an estimate of the boundary layer thickness as δTH∼RaH−1/4\frac{\delta_{T}}{H} \sim Ra_{H}^{-1/4}

Heat transfer coefficient is given by,

h=−κ∂T∂x|x=0Tw−T∞h∼κδT \begin{align} h & = \frac{-\kappa \left. \frac{ \partial T }{ \partial x } \right|_{x=0}}{T_{w}-T_{\infty}} \\ h & \sim \frac{\kappa}{\delta_{T}} \end{align}

If we define Nusselt number, NuH=hHκNu_{H}=\frac{hH}{\kappa}

⟹Nu∼HδT∼RaH1/4 \implies Nu \sim \frac{H}{\delta_{T}} \sim Ra_{H}^{1/4}

Nusselt number for Pr≪1Pr\ll 1

u∂v∂x,v∂v∂y∼(HδT)4RaH−1Pr−1ν∂2v∂x2∼(HδT)4RaH−1gβ(T−T∞)∼1 \begin{align} u \frac{ \partial v }{ \partial x }, v \frac{ \partial v }{ \partial y } & \sim \left( \frac{H}{\delta_{T}} \right)^{4}Ra_{H}^{-1}Pr^{-1} \\ \nu \frac{ \partial^{2} v }{ \partial x^{2} } & \sim \left( \frac{H}{\delta_{T}} \right)^{4}Ra_{H}^{-1} \\ g\beta (T-T_{\infty}) & \sim 1 \end{align}

If Pr≪1Pr\ll 1, then the viscous terms would be negligible or inertial and buoyance forces balance each other

⟹(HδT)4RaH−1Pr−1∼1 \implies \left( \frac{H}{\delta_{T}} \right)^{4}Ra_{H}^{-1}Pr^{-1} \sim 1

which gives us an estimate of the boundary layer thickness as δTH∼RaH−1/4Pr−1/4\frac{\delta_{T}}{H} \sim Ra_{H}^{-1/4}Pr^{-1/4}

Heat transfer coefficient is given by,

h=−κ∂T∂x|x=0Tw−T∞h∼κδT \begin{align} h & = \frac{-\kappa \left. \frac{ \partial T }{ \partial x } \right|_{x=0}}{T_{w}-T_{\infty}} \\ h & \sim \frac{\kappa}{\delta_{T}} \end{align}

If we define Nusselt number, NuH=hHκNu_{H}=\frac{hH}{\kappa}

⟹Nu∼HδT∼RaH1/4Pr1/4∼BoH1/4 \implies Nu \sim \frac{H}{\delta_{T}} \sim Ra_{H}^{1/4}Pr^{1/4} \sim Bo_{H}^{1/4}

where Boussinesq number, BoH=RaHPrBo_{H} = Ra_{H}Pr

Nusselt number for Pr≪1Pr\ll 1 and close to wall

Very close to the wall, inertial forces are small, hence viscous and buoyancy forces balance each other, say δν\delta_{\nu} is the thickness of this thin region

⟹ν∂2v∂x2∼gβ(T−T∞) \implies \nu \frac{ \partial^{2}v }{ \partial x^{2} } \sim g\beta(T-T_{\infty})

we obtain,

δνH∼GrH−1/4 \frac{\delta_{\nu}}{H} \sim Gr_{H}^{-1/4}

where GrH=RaHPrGr_{H}=\frac{Ra_{H}}{Pr}

and, we have δνδT∼Pr1/2\frac{\delta_{\nu}}{\delta_{T}} \sim Pr^{1/2}

Figure 2: Illustration of viscous region close to the wall

Non-dimensionalising Governing Equations

Non-dimensional parameters x*=xH,y*=yH x^{*} = \frac{x}{H}, y^{*}=\frac{y}{H}

u*=uUref,v*=vUref u^{*} = \frac{u}{U_{ref}}, v^{*} = \frac{v}{U_{ref}}

θ=T−T∞Tw−T∞ \theta = \frac{T-T_{\infty}}{T_{w}-T_{\infty}}

Continuity

∂u*∂x*+∂v*∂y*=0 \frac{ \partial u^{*} }{ \partial x^{*} } + \frac{ \partial v^{*} }{ \partial y^{*} }=0

Y-Momentum equation

u*∂v*∂x*+v*∂v*∂y*=1ReH∂2v*∂x*2+gβ(Tw−T∞)HUref2θ u^{*} \frac{ \partial v^{*} }{ \partial x^{*} } + v^{*} \frac{ \partial v^{*} }{ \partial y^{*} } = \frac{1}{\mathrm{Re}_{H}} \frac{ \partial^{2} v^{*} }{ \partial x^{*2} } + \frac{g\beta(T_{w}-T_{\infty})H}{U_{ref}^{2}} \theta

Energy equation

u*∂θ∂x*+v*∂θ∂y*=1ReHPr∂2θ∂x*2 u^{*} \frac{ \partial \theta }{ \partial x^{*} } + v^{*} \frac{ \partial \theta }{ \partial y^{*} } = \frac{1}{\mathrm{Re}_{H}Pr} \frac{ \partial^{2} \theta }{ \partial x^{*2} }

We define Grashof number, GrH=gβ(Tw−T∞)HUref2θGr_{H}=\frac{g\beta(T_{w}-T_{\infty})H}{U_{ref}^{2}}\theta and Richardson number, RiH=GrHReH2Ri_{H}=\frac{Gr_{H}}{\mathrm{Re}_{H}^{2}}

Richardson Number, RiHRi_{H}

RiH=GrHReH2 Ri_{H}=\frac{Gr_{H}}{\mathrm{Re}_{H}^{2}}

Criteria Type of convection
Ri≫1Ri \gg 1 Natural
Ri≪1Ri \ll 1 Forced
Ri=1Ri = 1 Mixed / Combined

Conventionally we choose, ReH=GrH\mathrm{Re}_{H} = \sqrt{ Gr_{H} } such that RiH=1Ri_{H} = 1

Summary of scaling laws

PrPr δT\delta_{T} Wall to velocity peak Thickness of wall jet Velocity scale NuNu
≫1\gg 1 HRaH−1/4H Ra_{H}^{-1/4} HRaH−1/4H Ra_{H}^{-1/4} Pr1/2HRaH−1/4Pr^{1/2}HRa_{H}^{-1/4} αHRaH1/2\frac{\alpha}{H} Ra_{H}^{1/2} RaH1/4Ra_{H}^{1/4}
≪1\ll 1 Pr−1/4HRaH−1/4Pr^{-1/4}HRa_{H}^{-1/4} Pr1/4HRaH−1/4Pr^{1/4}HRa_{H}^{-1/4} Pr−1/4HRaH−1/4Pr^{-1/4}HRa_{H}^{-1/4} αHPr1/2RaH1/2\frac{\alpha}{H}Pr^{1/2}Ra_{H}^{1/2} Pr1/4RaH1/4Pr^{1/4}Ra_{H}^{1/4}

Similarity Equations - Constant Wall Temperature

Similarity variable, η=xδT\eta=\frac{x}{\delta_{T}} for Pr>1Pr>1

where δT∼yRay−1/4\delta_{T} \sim y Ra_{y}^{-1/4}

Choosing non-dimensional temperature and velocity as

θ(η,Pr)=T−T∞Tw−T∞\theta(\eta,Pr)=\frac{T-T_{\infty}}{T_{w}-T_{\infty}}, v=αyRay1/2f′(η,Pr)v=\frac{\alpha}{y}Ra_{y}^{1/2} f'(\eta,Pr)

where Ray=gβΔTy3ανRa_{y}=\frac{g\beta \Delta Ty^{3}}{\alpha \nu} and ΔT=Tw−T∞\Delta T=T_{w}-T_{\infty}

The similarity equation becomes

1Pr(f′22−34ff″)=−f‴+θθ″−34fθ′=0 \begin{align} \frac{1}{Pr} \left( \frac{f'^{2}}{2}-\frac{3}{4} f f'' \right) & =-f'''+\theta \\ \theta''-\frac{3}{4}f \theta' & =0 \end{align}

Figure 3: Illustration of similarity law

and boundary conditions are

at x = 0u=0⟹η=0f=0v=0f′=0T=Twθ=1at x→∞v=0⟹η→∞f′=0T=T∞θ=0 \begin{align} \text{at x = 0} & & u & =0 & \implies & \eta=0 & f=0\\ & & v & =0 & & & f'=0 \\ & & T & =T_{w} & & & \theta=1\\ \text{at x} \to \infty & & v & =0 & \implies & \eta\to \infty & f'=0 \\ & & T & =T_{\infty} & & & \theta=0 \end{align}

Nusselt number correlations

For constant wall temperature

h=−kyRay1/4θ′(0)Nu=−Ray1/4θ′(0)Nu¯=43Nuy|y=H \begin{align} h & = -\frac{k}{y} Ra_{y}^{1/4}\theta'(0) \\ Nu & = -Ra_{y}^{1/4}\theta'(0) \\ \overline{Nu} & = \frac{4}{3} Nu_{y}|_{y=H} \end{align} For Pr→∞,Nu¯=0.671RaH1/4Pr\to \infty, \overline{Nu}=0.671Ra_{H}^{1/4}, and for Pr→0Pr\to 0, Nu¯=0.8(RaHPr)1/4\overline{Nu}=0.8(Ra_{H}Pr)^{1/4}

Extending similar analysis to uniform wall heat flux case gives us

Pr→∞Nu=0.616Ra*y1/5Pr→0Nu=0.644(Ra*yPr)1/5 \begin{align} Pr\to \infty & & Nu & =0.616 Ra_{*y}^{1/5} \\ Pr \to 0 & & Nu & = 0.644 (Ra_{*y}Pr)^{1/5} \end{align} where Ra*y=gβq″wy4ανκRa_{*y} = \frac{g\beta q''_{w}y^{4}}{\alpha \nu \kappa}

Figure 4: Temperature and heat flux profiles under different boundary conditions

Problems

Estimation of heat transfer coefficient - constant wall temperature

A vertical wall of an oven is 60 cm long and is covered with a metal sheet maintained at 170°C. Air temperature inside the oven is 90°C and the pressure is atmospheric. Calculate the local heat-transfer coefficient at the end of the wall and the average heat transfer coefficient over the wall.

Look up appropriate properties from Appendix 1

Estimation of heat transfer coefficient - uniform wall heat flux

A vertical wall of an oven is 60 cm long and it is covered with a metal plate that is heated electrically. The wall heat flux is 15 W/m2 on the plate. The inside air temperature is 90°C and pressure is atmospheric; Pr = 1 is assumed. Calculate a. The local heat-transfer coefficient at the end of the plate b. Average heat-transfer coefficient over the plate

Look up appropriate properties from Appendix 1

Free vs Forced Convective Heat Transfer

A 1m vertical plate is maintained at 100°C and exposed to air at atmospheric pressure and 20°C. Compare free convection heat transfer with that from forcing air over the plate at a velocity equal to the maximum velocity that occurs in the free convection boundary layer. Comment on the results.

Position of maximum velocity

Derive an expression for the maximum velocity in the laminar free-convection boundary layer on a vertical plate with constant surface temperature. At what position in the boundary layer does this velocity occur?

Air Properties

T[0C]T[^{0}C] ρ[kg/m3]\rho [kg / m^{3}] cp[kJ/kg.K]c_{p}[kJ / kg.K] κ[W/m.K]\kappa[W /m.K] β[×103K−1]\beta[\times 10^{3} K^{-1}] μ[×105kg/m.s]\mu[\times 10^{5} kg / m.s] ν[×106m2/s]\nu[\times 10^{6} m^{2} / s] α[×106m2/s]\alpha[\times 10^{6} m^{2} / s] PrPr
-150 2.793 1.026 0.012 8.21 0.87 3.11 4.19 0.74
-100 1.98 1.009 0.0165 5.82 1.18 5.96 8.28 0.72
-50 1.534 1.005 0.0206 4.51 1.47 9.55 13.4 0.715
0 1.293 1.005 0.0242 3.67 1.72 13.3 18.7 0.711
20 1.2045 1.005 0.0257 3.43 1.82 15.11 21.4 0.713
40 1.1267 1.009 0.0271 3.2 1.91 16.97 23.9 0.711
60 1.0595 1.009 0.0285 3 2 18.9 26.7 0.709
80 0.9908 1.009 0.0299 2.83 2.1 20.94 29.6 0.708
100 0.9458 1.013 0.0314 2.68 2.18 23.06 32.8 0.704
120 0.898 1.013 0.0328 2.55 2.27 25.23 36.1 0.7
140 0.8535 1.013 0.0343 2.43 2.35 27.55 39.7 0.694
160 0.815 1.017 0.0358 2.32 2.43 29.85 41 0.693
180 0.7785 1.022 0.0372 2.21 2.51 32.29 46.7 0.69
200 0.7475 1.026 0.0386 2.11 2.58 34.63 50.5 0.685
250 0.6745 1.034 0.0421 1.91 2.78 41.17 60.3 0.68
300 0.6157 1.047 0.039 1.75 2.95 47.85 70.3 0.68
350 0.5662 1.055 0.0485 1.61 3.12 55.05 81.1 0.68
400 0.5242 1.068 0.0516 1.49 3.28 62.53 91.9 0.68
450 0.4875 1.08 0.0543 1.38 3.44 70.54 103.1 0.685
500 0.4564 1.092 0.057 1.29 3.86 70.48 114.2 0.69
600 0.4041 1.114 0.0621 1.15 3.58 95.57 138.2 0.69
700 0.3625 1.135 0.0667 1.03 4.12 113.7 162.2 0.7
800 0.3287 0.156 0.0706 0.93 4.37 132.8 185.8 0.715
900 0.3021 1.172 0.0741 0.85 4.59 152.5 210 0.725
1000 0.277 1.185 0.077 0.79 4.8 175 235 0.735
Appendix 1: Air Properties as a function of temperature

Thank You

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