Convection Heat Transfer

Newton's law of cooling: the rate of convective heat transfer between a surface and a fluid is proportional to the surface area and to the temperature difference between the surface and the fluid:

$$Q = hA\,(T_s - T_\infty)$$

Convective heat transfer coefficient

$h$ (W/m$^2$ K) is the heat flow per unit area per unit temperature difference between the surface and the bulk fluid. It is not a property of the fluid alone; it depends on the whole flow situation. It is found from $h = -k_f\,(\partial T/\partial y)_{wall}/(T_s-T_\infty)$, which links it to the fluid conductivity and the temperature gradient at the wall.

Factors affecting $h$

• Type of flow: natural or forced, laminar or turbulent.
• Fluid properties: density, viscosity, thermal conductivity, specific heat, expansion coefficient.
• Fluid velocity and temperature difference.
• Geometry: shape, size, orientation and roughness of the surface.
• Phase change (boiling, condensation) raises $h$ very much.

Typical ranges

Hydrodynamic (velocity) boundary layer

When a fluid of free-stream velocity $u_\infty$ flows over a plate, the fluid in contact with the surface is at rest (no-slip) and velocity rises across a thin region to $0.99\,u_\infty$. This region, of thickness $\delta$, is the velocity boundary layer. All the viscous shear is confined to it; outside it the flow is practically inviscid.

Thermal boundary layer

If the plate temperature $T_s$ differs from the free-stream temperature $T_\infty$, the fluid temperature changes from $T_s$ at the wall to $T_\infty$. The thickness $\delta_t$ at which $(T-T_s)/(T_\infty - T_s) = 0.99$ is the thermal boundary layer. All convective heat transfer resistance lies within it.

Development

 u_inf --> . . . . . . . . . . . . . . . . .
                          _________ turbulent
                  _______/  ~~~~~~~  (buffer,
         _______/  transition        viscous
   _____/ laminar                    sublayer)
 -+--------------+-------------+------------> x
 leading edge    x_cr

• Laminar region near the leading edge: smooth layers, $\delta \propto \sqrt{x}$.
• Transition region: disturbances grow and the flow becomes unstable.
• Turbulent region: eddies mix the fluid, the layer grows faster ($\delta\propto x^{4/5}$), with a thin laminar sublayer at the wall. Both $\tau_w$ and $h$ increase sharply.

The transition is fixed by the critical Reynolds number

$$Re_{x,cr} = \frac{u_\infty x_{cr}}{\nu}\approx 5\times10^5$$

(it ranges from $10^5$ to $3\times10^6$ with roughness and turbulence level).

Effect of Prandtl number

$Pr = \nu/\alpha = \mu c_p/k$ compares momentum diffusion with heat diffusion. For laminar flow:

$$\frac{\delta}{\delta_t}\approx Pr^{1/3}$$

Film temperature $T_f = (100+20)/2 = 60$ °C, the temperature at which the given properties apply.

(a) Reynolds number

$$Re_L = \frac{u_\infty L}{\nu} = \frac{3\times1}{18.97\times10^{-6}} = 1.58\times10^5$$

This is less than $5\times10^5$, so the flow over the whole plate is laminar and the laminar correlation is valid.

(b) Average heat transfer coefficient

$$\begin{aligned} Nu_L &= 0.664\,Re_L^{0.5}\,Pr^{1/3}\\ &= 0.664\,(1.58\times10^5)^{0.5}\,(0.7202)^{1/3}\\ &= 0.664\,(397.7)(0.8962) = 236.7 \end{aligned}$$ $$\bar h = \frac{Nu_L\,k}{L} = \frac{236.7\times0.02896}{1} = 6.85\ \text{W/m}^2\text{K}$$

(c) Heat loss

$$Q = \bar h A (T_s-T_\infty) = 6.85\times(1\times0.6)\times(100-20) = 329\ \text{W}$$

(Heat is lost from one side. For both faces the value doubles.)

(d) Boundary layer thickness at $x = 1$ m

$$\delta = \frac{5x}{\sqrt{Re_x}} = \frac{5(1)}{397.7} = 0.01257\ \text{m} = 12.6\ \text{mm}$$ $$\delta_t = \frac{\delta}{Pr^{1/3}} = \frac{12.57}{0.8962} = 14.0\ \text{mm}$$

Since $Pr<1$ the thermal layer is slightly thicker than the velocity layer.

Answer: $Re_L = 1.58\times10^5$ (laminar); $\bar h = 6.85\ \text{W/m}^2$K; $Q = 329$ W; $\delta = 12.6$ mm, $\delta_t = 14.0$ mm.

Variables and dimensions

Fundamental dimensions: mass M, length L, time T, temperature $\theta$ (heat is expressed in mechanical units, so there are 4).

$n = 7$ variables, $m = 4$ fundamental dimensions, so the number of $\pi$ groups is $n - m = 3$.

Repeating variables

Choose $D$, $V$, $\mu$, $k$ (together they contain $L, T, M, \theta$ and are independent). Each non-repeating variable ($h$, $\rho$, $c_p$) forms one group.

Group 1 with $h$: $\pi_1 = h\,D^aV^b\mu^ck^d$. Equating exponents to zero:

• $M$: $1 + c + d = 0$
• $L$: $a + b - c + d = 0$
• $T$: $-3 - b - c - 3d = 0$
• $\theta$: $-1 - d = 0 \Rightarrow d = -1$

Then $c = 0$, $b = 0$ and $a = 1$, so $\pi_1 = \dfrac{hD}{k} = Nu$.

Group 2 with $\rho$: the same procedure gives $\pi_2 = \dfrac{\rho V D}{\mu} = Re$.

Group 3 with $c_p$: $\pi_3 = \dfrac{c_p\mu}{k} = Pr$.

Result

$$\pi_1 = f(\pi_2,\pi_3)\;\Rightarrow\; Nu = f(Re, Pr)$$

Experiments give the form $Nu = C\,Re^m Pr^n$, for example Dittus-Boelter $Nu = 0.023\,Re^{0.8}Pr^{n}$.

Physical meaning

(a) Reynolds number

$$Re = \frac{\rho V D}{\mu} = \frac{992.2\times1\times0.025}{653\times10^{-6}} = 3.80\times10^4$$

The flow is turbulent ($Re > 10\,000$) and the tube is long, so the Dittus-Boelter equation applies.

(b) Heat transfer coefficient

For heating of the fluid the exponent is $n = 0.4$:

$$\begin{aligned} Nu &= 0.023\,Re^{0.8}Pr^{0.4}\\ &= 0.023\,(37\,986)^{0.8}(4.32)^{0.4} = 190.4 \end{aligned}$$ $$h = \frac{Nu\,k}{D} = \frac{190.4\times0.631}{0.025} = 4805\ \text{W/m}^2\text{K}$$

(c) Heat gained

Mass flow rate:

$$\dot m = \rho V\frac{\pi D^2}{4} = 992.2\times1\times\frac{\pi(0.025)^2}{4} = 0.487\ \text{kg/s}$$ $$Q = \dot m c_p (T_{out}-T_{in}) = 0.487\times4179\times20 = 40\,700\ \text{W}$$

(d) Tube length

For constant wall temperature the mean temperature difference is the logarithmic mean:

$$\Delta T_{lm} = \frac{(90-30)-(90-50)}{\ln\left(\dfrac{90-30}{90-50}\right)} = \frac{60-40}{\ln1.5} = 49.3\ ^\circ\text{C}$$ $$Q = h\,(\pi D L)\,\Delta T_{lm}\;\Rightarrow\; L = \frac{40\,707}{4805\times\pi\times0.025\times49.3} = 2.19\ \text{m}$$

Check: $L/D = 87.5 > 60$, so the entrance effect is small and the correlation is acceptable.

Answer: $Re = 3.8\times10^4$; $h = 4805\ \text{W/m}^2$K; $Q = 40.7$ kW; $L \approx 2.2$ m.

Reynolds number

$$Re_D = \frac{VD}{\nu} = \frac{10\times0.05}{20.92\times10^{-6}} = 2.39\times10^4$$

$Re_D\,Pr = 1.72\times10^4 > 0.2$, so the Churchill-Bernstein equation is valid.

Nusselt number

Compute the parts:

$$\begin{aligned} Re^{0.5} &= 154.6, \qquad Pr^{1/3} = 0.896\\ \text{numerator} &= 0.62\times154.6\times0.896 = 85.9\\ \left[1+(0.4/0.7202)^{2/3}\right]^{1/4} &= (1.676)^{0.25} = 1.138\\ \left[1+(Re/282000)^{5/8}\right]^{4/5} &= (1.2108)^{0.8} = 1.1663 \end{aligned}$$ $$Nu = 0.3 + \frac{85.9}{1.138}\times1.1663 = 0.3 + 88.0 \approx 88.5$$

Heat transfer coefficient

$$h = \frac{Nu\,k}{D} = \frac{88.5\times0.02953}{0.05} = 52.3\ \text{W/m}^2\text{K}$$

Heat loss per metre

$$\frac{Q}{L} = h\,(\pi D)(T_s-T_\infty) = 52.3\times\pi\times0.05\times100 = 821\ \text{W/m}$$

Answer: $h \approx 52.3\ \text{W/m}^2$K; $Q/L \approx 821\ \text{W/m}$.

Grashof number: physical meaning

$Gr = \dfrac{g\beta\,\Delta T\,L^3}{\nu^2}$ is the ratio of buoyancy force to viscous force acting on the fluid. In natural convection it plays the role that the Reynolds number plays in forced convection. A higher $Gr$ means stronger buoyancy-driven motion and higher $h$.

(a) Grashof and Rayleigh numbers

$T_f = 60$ °C $= 333.15$ K, so $\beta = 1/333.15 = 3.0\times10^{-3}\ \text{K}^{-1}$.

$$\begin{aligned} Gr_L &= \frac{9.81\times(1/333.15)\times(90-30)\times(0.5)^3}{(18.97\times10^{-6})^2} = 6.14\times10^{8}\\ Ra_L &= Gr_L\,Pr = 6.14\times10^8\times0.7202 = 4.42\times10^{8} \end{aligned}$$

$Ra < 10^9$, so the flow is laminar and the given correlation is valid.

(b) Nusselt number and $h$

$$Nu = 0.59\,Ra^{1/4} = 0.59\,(4.42\times10^8)^{0.25} = 0.59\times145.0 = 85.5$$ $$\bar h = \frac{Nu\,k}{L} = \frac{85.5\times0.02896}{0.5} = 4.95\ \text{W/m}^2\text{K}$$

(c) Heat loss

$$Q = \bar h A\,\Delta T = 4.95\times(0.5\times0.8)\times60 = 119\ \text{W}$$

Answer: $Gr = 6.14\times10^8$, $Ra = 4.42\times10^8$; $Nu = 85.5$; $\bar h = 4.95\ \text{W/m}^2$K; $Q \approx 119$ W.

Dimensionless numbers let results from one experiment be applied to similar situations of different size, fluid or velocity.

Relations

• Forced convection: $Nu = f(Re, Pr)$, e.g. $Nu = 0.023\,Re^{0.8}Pr^{0.4}$ for turbulent flow in tubes.
• Natural convection: $Nu = f(Gr, Pr)$, usually written $Nu = C\,(Gr\,Pr)^m$, where the product $Gr\,Pr$ is the Rayleigh number $Ra$.
• Mixed convection: $Gr/Re^2$ near 1 shows both effects are important.
• $St\,Pr^{2/3} = f/2$ is the Colburn analogy linking heat transfer with skin friction.

Differences

Mechanism on a heated vertical plate

Fluid next to the hot plate is heated, expands, becomes lighter and rises. Cooler fluid from the surroundings replaces it and a buoyancy-driven boundary layer forms.

      ^  ^  ^    rising warm
   |  |         turbulent
   |  |   -----------
   |  |    transition
   |  |   laminar layer
 hot plate --> grows with height

The layer starts laminar at the lower edge, grows in thickness, becomes turbulent when $Ra \gtrsim 10^9$, and $h$ then becomes nearly independent of height.

Dimensional analysis

Buoyancy force per unit mass is $g\beta\,\Delta T$, replacing the velocity $V$ of forced flow. Thus $h = f(L, \rho, \mu, k, c_p, g\beta\Delta T)$: 7 variables and 4 fundamental dimensions give 3 groups:

$$\frac{hL}{k}= f\left(\frac{\rho^2 g\beta\Delta T L^3}{\mu^2},\ \frac{\mu c_p}{k}\right)\;\Rightarrow\; Nu = f(Gr, Pr)$$

For a vertical plate, $Nu = 0.59\,(Gr\,Pr)^{1/4}$ for laminar flow and $Nu = 0.10\,(Gr\,Pr)^{1/3}$ for turbulent flow.