Transient (unsteady) Heat Conduction

Lecture 15, February 6, 2004 • Assignment 2 is posted

Transient (unsteady) Heat Conduction – Chapter 5 Introduction Recall conservation of energy,

Up to this point we have virtually always assumed that the storage term was zero, and we have limited the number of dimensions that we consider. In general, though, the temperature is

What does the energy storage term do to our temperature fields. The most obvious effect is that the temperature distribution changes with time Æ the problem is unsteady. Imagine a 1m long bar, heated at one end, with a steady heat flux imposed at one end and exposed to a air at T∞ = 30oC (h = 100 W/m2K) at the other end. If the bar is insulated alone its surface, then the temperature distribution will be 1 dimensional and will vary in time, T=T(x,t). Take k = 284 W/mK, and assume that initially the bar is in equilibrium with the air, T(x,0) = 30oC and apply a constant heat rate of 100 W to the x=0 face. In order to consider the energy storage term, we will need another material property, the thermal diffusivity introduced earlier,

which we will initially set at 97 x 10-6 m2/s. Applying conservation of energy using a finite volume method, we can calculate T(x,t), and examine the problem in a number of ways. First, lets look at the temperature of the two end points, and the heat rate out, q(L).

The above figure shows both of these quantities on two different time scales so that we can see what happens initially. On the left hand side of the figure, the first 500s are shown, and it is clear that while the left end of the bar is undergoing heating, the right end is practically not. Initially, then entire 100W applied to the bar is being stored in the first portion of the bar, and the heat energy has not had time to conduct all the way through to the other end. On the right hand scale, the temperatures and heat rates are shown over a much larger time scale. It is clear that the temperature of the right end (x=L) does begin to increases substantially, but more slowly than the left end which has the heat applied directly to it. As the temperature at the end does increase though, so does the heat rate out of the bar. Finally, after enough time the entire 100W is leaving the bar at x=L. Clearly there is no more energy being stored since all the input energy is passing right through the bar. At this point, the bar has reached steady state, and the temperature is no longer changing with time. Let’s look now at the temperature distribution across the entire length of the bar as a function of time.

The above figure shows the temperature distribution at the initial condition and at four subsequent times. Also shown is the heat rate out at x=L, as shown previously. At time 2, the effect of heating the bar has not propagated much past the centre of the bar, and the heat rate out practically zero. As time advances however T(L,t) increases as energy is conducted through the bar. Notice that for unsteady conduction, that the temperature profile is non-linear even when the cross sectional area is not changing. This clearly has to be the case since while energy is being stored, more energy is entering the bar at x=0 than is leaving at x=L. Fourier’s law still applies at these points, and therefore the temperature gradient must be larger at x=0 than at x=L. As time advances further, the temperatures increase further, and the distribution becomes more linear. Of course, when we reach a steady state, the distribution, must be linear since we have reached the steady case that we examined earlier, and the rate of energy storage is zero. This is clearly shown in the following figure.

In order to conserve energy, the heat rate out + the energy storage must equal the energy put in (100W in this case). The same information is presented above, using two different time scales. The top graph shows the physical time in [s], while the bottom graph shows the non-dimensional time given by the Fourier number

The top graph clearly shows that it takes approximately 250,000s to reach a steady state, and in this case that corresponds to a non dimensional time of Fo = 25. Why do we want to consider non-dimensional time? To answer this, lets consider what happens if we increase the thermal diffusivity. Remember, that the thermal diffusivity can be thought of as the ratio of the capacity of a material to conduct thermal energy to the materials capacity to store thermal energy. If we increase α, then the material will be relatively have a larger capacity to conduct thermal energy. This means that the effect of applying a heat flux will propagate more quickly from x=0 to x=L. The heat rate at x=L will then be larger sooner, because of the larger difference between the T(L,t) and T∞. Overall then we can expect that a steady state should be reached even sooner, and this is exactly what we see. Raising α, by an order of magnitude, and repeating the analysis yields, the following curve.

The bar now reaches a steady state before t=50000s, and the top plot looks quite different. If we plot the heat rate against the non-dimensional time though, it is identical to the previous case, reaching a steady state between Fo=15 and Fo=20! This example shows that it we properly non-dimensionalize our problem we can save ourselves a tremendous amount of time. If you use dimensional time, you have to solve the equations for every desired combination of material properties. If you use nondimensional time, then you can solve it once and interpret the solution for any other combination of properties! This may not seem like a big deal in this example, but consider some large thermo-fluid calculations where the computing time is on the order of weeks or even months for a single calculation (this is not uncommon). The person who uses dimensional time will take years to carry out a parametric study, while the the person who uses non-dimensional time will be done in weeks or months. In general, there are three ways to solve unsteady heat transfer problems • Simplified Analytical o Lumped Capacitance o Exact Solution of simple geometries

• •

Numerical (as in the previous example) Experimental

Lumped Capacitance Imagine a hot steel bolt just removed from a furnace. We can well imagine the bolt to be at a uniform T due to the relatively high conductivity of steel. In comparison, we know that a bowl of rice just removed from a microwave oven may well have hot regions and cold regions (it has a much lower effective conductivity). The lumped capacitance method is concerned with situations in which we can assume a uniform T throughout our system, and thus do not have to consider spatial variations of T. Either the system will begin to cool, or to heat, but either way we want to be able to analyze this situation. Conservation of Energy in this case reduces to,

The heat rate out is given by Newton’s law of cooling

Introduce the temperature difference at each time,

The energy storage is given by

And our expression for conservation of energy becomes

We can now easily integrate this expression to solve for the time rate of change of theta. At t=0, we have the largest temperature difference,

so that upon integration,

And, taking exponentials,

We can now determine the temperature of the bolt at any time, t, or we can determine the time it takes to reach a certain temperature. When is this analysis applicable? Recall the Biot number?

where Lc is a characteristic length scale. The appropriate length scale for three dimensional problems is

and

for two dimensional problems.

The Biot number represents the ratio of the resistance to conduction to the resistance to convection. This scenario is exactly analogous to the 1D approximation we made to solve for heat transfer in fins. If the Biot number is sufficiently small, then the temperature will be uniform in space and vary only in time. Otherwise we will need to consider the spatial variation of temperature as well.