Analytic Solutions Numerical Solutions

8/14/2019 Analytic Solutions Numerical Solutions

1/12

Lecture 16, Feb 9, 2004

Last Day

Lumped Capacitance

Today

Analytic SolutionsNumerical Solutions

When can we use the lumped capacitance method?

When the Biot number is small ( < 0.1). Otherwise there will be temperature

variations in the solid, and we will be getting the energy transfer by convection wrong,and hence the time it takes to heat/cool will be wrong.

appears in the exponential in the expression for Lumped Capacitance method. The

temperature of our part approaches the ambient temperature exponentially. If this term is

large, means that equilibrium will be reached quickly. If it is small this will take a longertime.

This is like a time constant for our system, and can be related to another electric circuitanalogy. The thermal energy storage is like a capacitance in an electric circuit hencethe name, the lumped capacitance method.

Or, We can express our problems in convenient non-dimensional forms

Where the length scale is as given Friday.

We can use the lumped capacitance method when Bi is small, but there are also other

approximations hidden in here

h is often only known within 20% (Next section)


2/12

h is never constant around the entire part (recall the fins)Therefore it will be approximate no matter what and Bi < 0.1 is a rough guideline.

Example using 2D unsteady conduction program developed in Matlab

Consider a cube of metal, initially at 100

o

C. The part is removed from an oven andplaced in air at T=30oC. The origin of our coordinate system is at the centre of the

cube, and because of symmetry we can examine the upper right quadrant only.

F

First, let us consider the case of Bi=0.005, where we expect the lumped capacitanceapproximation to hold. Shown below are temperature contours at four different times as

the cube cools.


3/12

Is is clear from the above figure that when Bi is low, the part cools uniformly, and our

lumped capacitance model will be valid. Shown below is a plot of the variation of thetemperature at the centre of the cube, and at the upper right corner of the cube. Also

plotted is the variation of the area averaged temperature of the entire part.

Note that all three of these curves are virtually identical, clearly portraying that it is

appropriate to neglect temperature variations within the part.

If we increase Bi, however to 5.34 the situation changes considerably.


4/12

There are very clearly temperature gradients in the solid. If we look now at the time

variation of the temperatures at the same location as well as the average temperature, we

clearly see that the centre is at a very different temperature than the outside edge forvirtually all of the cooling process. When we do a lumped capacitance analysis, the

temperature we are using is the average temperature, which if there is no variation in

space, is also the temperature at the surface. When we misapply the lumped capacitancemethod however, we are using the average temperature when we should be using the

surface temperature for the convection process. Consider the situation at t=500s for

example. The average temperature is approximately 50oC, while the surface temperature

is very close to 30oC. Using the average temperature would greatly over estimate the rate

of convective heat transfer, and this greatly accelerate the time to cool. As an exercise,

using the lumped capacitance method predict how long the part would take to cool, andcompare it to this numerical solution.


5/12

Analytic Solution (Section 5.5.1 in text)

We can also formulate analytic solutions to unsteady heat transfer problems which

involve temperature gradients within the solid.

Start with non-dimensionalization

The non-dimensional temperature will be a function of position, the Fourier number and

the Biot number.


6/12


7/12

We can see that the temperature distribution is initially a straight line at 30oC. As the

x/L=1 end is exposed to the hot air, the temperature at that end heats up. Initally howeverthe centre of the wall (x/L=0) does not experience any increase in temperature as it takes

time for the heat energy to be conducted through the wall to the centre. Eventually

however the centre begins to heat as well and subsequent profiles are at increasingly

higher average temperatures. Eventually, the wall would reach a steady state where thewall would be at a constant temperature of 100

oC, but this is not shown in the above

figure. As the x/L position heats up, there is less and less convective heat transfer, and

the heat flux will decrease to zero at steady state.

The thing to note is that at early times, the temperature distribution has a significant

section that is a straight line. It is very difficult to represent a straight line by summingup different sinusoidal functions, as is done in the Fourier series solution. The next figure

plots the first four terms in the analytical solution at some instant in time early in the

heating process. Each term in the Fourier series is plotted separately as well as the sum

of the first four terms which is our approximation to the non-dimensional temperature atthe time. It is clear that using enough terms we can come very close to a straight line, but

this will require at least these four terms.


8/12

Comparing to our numerical solution at two different times (Fo) we can see the problem.

In order to compare, we must first calculate the dimensional temperature from the non-dimensional temperature shown above.


9/12

At early times, even four terms of the Fourier series are not enough to represent thestraight line section. At later times however, there is very good agreement between thenumerical and analytic solutions.

While analytic methods have problems representing straight lines in this case, they are

very good for validating numerical methods at later times. We can then have the

confidence to apply the numerical methods to more complex problems that we cannot

solve analytically at all. And, we can also use many more terms in our Fourier series toimprove solution at early times. Using eight terms in the Fourier series, we see

considerable improvement. We could continue to add terms, and continue to improve the

analytic solution.


10/12

In addition, the analytic solutions also greatly help to develop our intuition about heattransfer problems.

Numerical Solutions

We have already seen how to develop the equations for numerical predictions of twodimensional steady conduction. Next we shall focus how to approximate unsteady

conduction problems.


11/12


12/12

Analytic Solutions Numerical Solutions

Documents

Transcript of Analytic Solutions Numerical Solutions