Combined Conduction and Convection

We can now analyze problems in which both conduction and convection occur, starting with a wall cooled by flowing fluid on each side. As discussed, a description of the convective heat transfer can be given explicitly as

$\displaystyle \frac{\dot{Q}}{A}=\dot{q} = h(T_w-T_\infty).$

This could represent a model of a turbine blade with internal cooling. Figure shows the configuration.

**Figure :** Conducting wall with convective heat transfer

The heat transfer in fluid 1 is given by

$\displaystyle \frac{\dot{Q}}{A} = h_1(T_{w1}-T_1).$

which is the heat transfer per unit area to the fluid. The heat transfer in fluid 2 is similarly given by

$\displaystyle \frac{\dot{Q}}{A} = h_2 (T_2 -T_{w2}).$

Across the wall, we have

$\displaystyle \frac{\dot{Q}}{A} = \frac{k}{L}(T_{w2} -T_{w1}).$

The quantity $\dot{Q}/A$ is the same in all of these expressions. Putting them all together to write the known overall temperature drop yields a relation between heat transfer and overall temperature drop,

$\displaystyle T_2-T_1 = (T_2-T_{w2})+(T_{w2}-T_{w1})+(T_{w1}-T_1) =\frac{\dot{Q}}{A}\left[\frac{1}{h_1}+\frac{L}{k}+\frac{1}{h_2}\right].$

We can define a thermal resistance,

, as before, such that

$\displaystyle \dot{Q} =\frac{(T_2 -T_1)}{R},$

where

is given by

$\displaystyle R = \frac{1}{h_1A} + \frac{L}{Ak} + \frac{1}{h_2A}.$

Equation is the thermal resistance for a solid wall with convection heat transfer on each side.

For a turbine blade in a gas turbine engine, cooling is a critical consideration. In terms of Figure ,

is the combustor exit (turbine inlet) temperature and

is the temperature at the compressor exit. We wish to find $T_{w2}$ because this is the highest metal temperature. From , the wall temperature can be written as

$\displaystyle T_{w2} = T_2 - \frac{\dot{Q}}{Ah_2}= T_2 - \frac{T_2 - T_1}{R}\frac{1}{Ah_2}.$

Using the expression for the thermal resistance, the wall temperatures can be expressed in terms of heat transfer coefficients and wall properties as

$\displaystyle T_{w2} = T_2 - \cfrac{T_2 - T_1}{\cfrac{h_2}{h_1}+\cfrac{Lh_2}{k}+1}.$

Equation provides some basic design guidelines. The goal is to have a low value of $T_{w2}$ . This means

should be large,

should be large (but we may not have much flexibility in choice of material) and

should be small. One way to achieve the first of these is to have

low (for example, to flow cooling air out as in Figure 17.1 to shield the surface).

A second example of combined conduction and convection is given by a cylinder exposed to a flowing fluid. The geometry is shown in Figure 17.7.

**Figure :** Cylinder in a flowing fluid

For the cylinder the heat flux at the outer surface is given by

$\displaystyle \dot{q} = \frac{\dot{Q}}{A} = h(T_w-T_\infty)\quad \textrm{ at }r=r_2.$

The boundary condition at the inner surface could be either a heat flux condition or a temperature specification; we use the latter to simplify the algebra. Thus,

. This is a model for the heat transfer in a pipe of radius

surrounded by insulation of thickness

. The solution for a cylindrical region was given in Section 16.5.1 as

$\displaystyle T(r) =a\ln\left(\frac{r}{r_1}\right) +b.$

Use of the boundary condition

yields

At the interface between the cylinder and the fluid,

, the temperature and the heat flow are continuous. (Question: Why is this? How would you argue the point?)

$\displaystyle \dot{q} = \underbrace{-k\frac{dT}{dr}}_{\substack{\textrm{heat fl... ...2}{r_1}\right)+T_1\right)-T_\infty\right]} _\textrm{surface heat flux to fluid}$

Plugging the form of the temperature distribution in the cylinder into Equation yields

$\displaystyle -a\left(\frac{k}{r_2}+h\ln\left(\frac{r_2}{r_1}\right)\right)=h(T_1-T_\infty).$

The constant of integration,

, is

$\displaystyle a=\cfrac{-h(T_1-T_\infty)}{\cfrac{k}{r_2}+h\ln\left(\cfrac{r_2}{r... ...t)} = -\cfrac{(T_1-T_\infty)}{\cfrac{k}{hr_2}+\ln\left(\cfrac{r_2}{r_1}\right)}$

and the expression for the temperature is, in normalized non-dimensional form,

$\displaystyle \frac{T_1-T}{T_1-T_\infty} = \cfrac{\ln(r/r_1)}{\cfrac{k}{hr_2}+\ln(r_2/r_1)}.$

The heat flow per unit length, $\dot{Q}$ , is given by

$\displaystyle \dot{Q}=\cfrac{2\pi(T_1-T_\infty)k}{\cfrac{k}{hr_2}+\ln(r_2/r_1)}.$

The units in Equation are W/m-s.

A problem of interest is choosing the thickness of insulation to minimize the heat loss for a fixed temperature difference $T_1 -T_\infty$ between the inside of the pipe and the flowing fluid far away from the pipe. ( $T_1 -T_\infty$ is the driving temperature distribution for the pipe.) To understand the behavior of the heat transfer we examine the denominator in Equation as

varies. The thickness of insulation that gives maximum heat transfer is given by

$\displaystyle \frac{d}{dr_2}\left(\frac{k}{hr_2}+\ln\left(\frac{r_2}{r_1}\right)\right)=0.$

(Question: How do we know this is a maximum?)

From Equation , the value of

for maximum $\dot{Q}$ is thus

$\displaystyle (r_2)_{\textrm{maximum heat transfer}} = \frac{k}{h}.$

is less than this, we can add insulation and increase heat loss. To understand why this occurs, consider Figure , which shows a schematic of the thermal resistance and the heat transfer. As

increases from a value less than

, two effects take place. First, the thickness of the insulation increases, tending to drop the heat transfer because the temperature gradient decreases. Secondly, the area of the outside surface of the insulation increases, tending to increase the heat transfer. The second of these is (loosely) associated with the