LTR: Convective Heat Transfer

As we already discussed, convective heat transfer refers to the transport of heat due to a moving fluid (and in the engineering sense, applies to the case of a fluid "carrying" heat away from a solid boundary). In either sense of the term, it is clear that the rate of heat transfer will depend on the character of the fluid flow.

The rate expression for convection was suggested by Newton in 1701, commonly called Newton's "Law" of Cooling, and has the form:

EQUATION:

$\displaystyle{\frac{q}{A} = h\Delta T}$

where $h$ is the convective heat transfer coefficient. It is sometimes also referred to as a "film coefficient" since in some strong flows the temperature changes are confined to a relatively thin film (see below).

"Law" is in quotes above because it is perhaps more correct to think of this expression as an empirical (data fitting) definition of $h$ rather than a law. It will depend on:

• the geometry of the solid boundaries (this will typically be known)
• the nature of the fluid (conductivity, heat capacity, density, etc. : we can look these up)
• the nature of the flow (fluid mechanics!)
NOTE:

Determining the parameter, $h$, will often be the bulk of the work (or at least the only hard part) in a given convection problem.

OUTCOME:

Perform convective heat transfer calculations

EXAMPLE:

Lets look at an example:

Let us compare two cases of flow in a 1 m diameter pipe (a forced, internal, convection problem). In both cases the fluid is water ($\rho = 1000 kg/m^3$; $\nu = 1.25x10^{-6}m^2/s$). In both cases a suitable "bulk" temperature is 20C and the wall temperature is 40C. In the first case the fluid velocity is 1 mm/s, and in the second it is 1 $cm/s$. How does the flux of heat differ in these cases?

As in the conduction examples, we are simply interested in solving for the heat flow (not the temperature profile), so using Newton's Law may be sufficient (provided we know $h$).

Assuming we know $h$, the governing equation is:

$\displaystyle{\frac{q}{A} = h (T_s-T_b)}$

Since we know the two temperatures, as mentioned before, the real problem is determining h. In this case, we can look up relations for flow in a pipe which tell us that:

for fully developed laminar flow, there is an analytical solution:

$\displaystyle{h = 3.66 \frac{k}{D}}$

where $D$ is the pipe diameter. Here it is intuitive that the heat transfer coefficient goes up with higher fluid conductivity and goes down for a larger pipe. However, the convection coefficient does not depend on the velocity of the flow! (Which is weird to me anyhow. It is sufficient to say that the flow is laminar.)

for turbulent flow, there is an empirical relation:

$\displaystyle{h = 0.023\frac{V^{0.8}k^{0.6}(\rho c)^{0.4}}{D^{0.2}\nu^{0.4}}}$

where $V$ is the fluid velocity. Again the heat transfer coefficient goes up with higher fluid conductivity and goes down (much less!) for a larger pipe, but now it does depend on the velocity of the flow as well as other properties of the fluid.

So, all we need to do now is determine the nature of the flow in our two cases. One important way to characterize fluid flows is to calculate the Reynolds number. The Reynolds number is a measure of the relative importance of the inertial and viscous forces in a flow.

$\displaystyle{Re = \frac{VD}{\nu}}$

In a pipe flow (and only in a pipe flow!), the flow is considered laminar for Re<2300 and turbulent for Re>5000.

Let us calculate Re's for our flows...

$\displaystyle{Re = \frac{(0.001m/s)(1m)}{1.25x10^{-6}m^2/s} = 800}$ (laminar)

$\displaystyle{Re = \frac{(0.01m/s)(1m)}{1.25x10^{-6}m^2/s} = 8000}$ (turbulent)

The heat transfer coefficients in each case are:

$\displaystyle{h = 3.66 \frac{(0.56W/mK)}{(1m)} = 2.05 W/m^2K}$

$\displaystyle{h = 0.023 \frac{(0.01m/s)^{0.8}(0.56W/mK)^{0.6}((1000kg/m^3)(4200J/kgK))^{0.4}}{(1m)^{0.2}(1.25x10^{-6}m^2/s)^{0.4}} = 41.8 W/m^2K}$

So,

$\displaystyle{\frac{q}{A} = (2.05W/m^2K)(40^oC-20^oC) = 41W/m^2}$

$\displaystyle{\frac{q}{A} = (41.8W/m^2K)(40^oC-20^oC) = 836W/m^2}$

What would happen if we did this same calculation for air rather than water?

Note:

Often you will not be given any details of the fluid, the flow, (or even the geometry). Instead, you will simply be told what the "bulk" temperature in the fluid is as well as what the heat transfer coefficient (h) is. This is enough info to solve this type of convection problem.