Thermodynamics and Propulsion

16.3 Steady-State One-Dimensional Conduction

Figure 16.3: One-dimensional heat conduction

For one-dimensional heat conduction (temperature depending on one variable only), we can devise a basic description of the process. The first law in control volume form (steady flow energy equation) with no shaft work and no mass flow reduces to the statement that $\sum \dot{Q}$ for all surfaces $= 0$ (no heat transfer on top or bottom of Figure 16.3). From Equation (16.6), the heat transfer rate in at the left (at $x$ ) is

$$\dot{Q}(x) = -k\left(A\frac{dT}{dx}\right)_x.$$ (16..9)

The heat transfer rate on the right is

$$\dot{Q}(x +dx) = \dot{Q}(x) + \frac{d\dot{Q}}{dx}\biggr\vert _x dx + \dots.$$ (16..10)

Using the conditions on the overall heat flow and the expressions in (16.9) and (16.10)

$$\dot{Q}(x) - \left(\dot{Q}(x) + \frac{d\dot{Q}}{dx}(x) dx + \dots\right) =0.$$ (16..11)

Taking the limit as $dx$ approaches zero we obtain

$$\frac{d\dot{Q}(x)}{dx} =0,$$ (16..12)

or

$$\frac{d}{dx}\left(kA\frac{dT}{dx}\right)=0.$$ (16..13)

If $k$ is constant (i.e. if the properties of the bar are independent of temperature), this reduces to

$$\frac{d}{dx}\left(A\frac{dT}{dx}\right)=0,$$ (16..14)

or (using the chain rule)

$$\frac{d^2T}{dx^2} +\left(\frac{1}{A}\frac{dA}{dx}\right)\frac{dT}{dx}=0.$$ (16..15)

Equation (16.14) or (16.15) describes the temperature field for quasi-one-dimensional steady state (no time dependence) heat transfer. We now apply this to an example.

16.3.1 Example: Heat transfer through a plane slab

Figure 16.4: Temperature boundary conditions for a slab

For this configuration (Figure 16.4), the area is not a function of $x$ , i.e. $A =\textrm{ constant}$ . Equation (16.15) thus becomes

$$\frac{d^2T}{dx^2} = 0.$$ (16..16)

Equation (16.16) can be integrated immediately to yield

$$\frac{dT}{dx} = a$$ (16..17)

and

$$T = ax + b.$$ (16..18)

Equation (16.18) is an expression for the temperature field where $a$ and $b$ are constants of integration. For a second order equation, such as (16.16), we need two boundary conditions to determine $a$ and $b$ . One such set of boundary conditions can be the specification of the temperatures at both sides of the slab as shown in Figure 16.4, say $T(0) = T_1$ ; $T(L) = T_2$ .

The condition $T(0) = T_1$ implies that $b = T_1$ . The condition $T_2 = T(L)$ implies that $T_2 = aL + T_1$ , or $a =(T_2 - T_1)/L$ . With these expressions for $a$ and $b$ the temperature distribution can be written as

$$T(x)= T_1 + \left(\frac{T_2-T_1}{L}\right)x.$$ (16..19)

This linear variation in temperature is shown in Figure 16.5 for a situation in which $T_1 > T_2$ .

Figure 16.5: Temperature distribution through a slab

The heat flux $\dot{q}$ is also of interest. This is given by

$$\dot{q} = -k \frac{dT}{dx} = -k\frac{(T_2 - T_1)}{L} =\textrm{ constant}.$$ (16..20)

Muddy Points

How specific do we need to be about when the one-dimensional assumption is valid? Is it enough to say that $dA/dx$ is small? (MP 16.2)

Why is the thermal conductivity of light gases such as helium (monoatomic) or hydrogen (diatomic) much higher than heavier gases such as argon (monoatomic) or nitrogen (diatomic)? (MP 16.3)