Problem Statement

DE for axisymmetric one-dimensional steady heat transfer through the fin-cylinder design of an air-cooled IC engine cylinder is

                on        a < r < b

where r is radius, k is thermal conductivity and T is temperature. The boundary conditions (BCs) for convective heat transfer on the fin and radiation to the cylinder wall, see the figure, are

                           on        (a + tc) £ r £ b

                       at         r = a

 
 

 

 

 

 

 

 

 

 

where h is heat transfer coefficient, Ta is ambient air temperature, Tr is the radiation temperature of the cylinder combustion gases, s is the Stefan-Boltzmann constant and e is emissivity.

The GWSh for the non-linear DE problem statement was fully discussed in the lecture FE discrete implementation of GWSN leads to

GWSN Þ GWSh = Se {WS}e = 0

Recalling the FE algorithm template syntax statement

{WS}e = (constant) (avg.data)e {dist.data}e (metric)e [FE matrix] {Q or data}e

the template pseudo-code for the IC engine problem statement GWSh is

{WS}e = (COND) (DZ) {R} (-1) [A3011] {Q}

 + (2) (H ) {R} (1) [A3000] {Q}

 + (-2) (H ) {R} (1) [A3000] {TA}

 + (DZ, RC, e, s) ( ) { } ( ) [ONE] {Q exp 4}

 + (-DZ, RC, e, s ) ( ) { } ( ) [ONE] {TR exp 4}

and recall an empty bracket implies no data of that type. 

      No convection occurs on the elements lying in the cylinder wall, hence the convection coefficient h, while constant, must be entered into the element average data bracket.   This requires generating an element length data array which contains (0,H) entries dependent on the number of elements M. 

      The radiation BC makes this GWSh non-linear, hence an iterative algebraic process is required to solve the matrix statement. The lecture detailed construction of the Newton algorithm  jacobian template as

JACe = (COND) (DZ) {R} (-1) [A3011] { }

+ (2) (H) {R} (1) [A3000] { }

+ (4, DZ, RC, e, s) ( ) { } ( ) [ONE] [Q exp 3]

The available error estimate for Th is from a linear problem asymptotic convergence theory. The resultant energy norm expression for smooth data is

|| errorh ||E  Cle2 || data ||W È W

Under uniform mesh refinement, i.e., regularly doubling the number M of equal length elements constituting Wh, the estimate of the FE solution error on mesh Wh/2 in energy is

|| errorh/2 || = D || Th ||E / 3

The lecture introduced the issue of genuine engineering problems being driven by non-smooth data, and stated the associated refinement of the asymptotic convergence theory as

            || errorh ||E  Cle2g || data ||W È W              g  max(k, r – 1)

Here, r is a measure of the required solution smoothness, as instilled by the data driving the problem.  You must interpret your data in this context.

You have assurance of a quality FE solution process when the error data exhibit agreement with the theory, as generalized. The FE solution energy norm for DE is computed as

The slope computation from the generated solution data is the post processing operation

      Finally, being a non-linear problem requires you to ascertain that the matrix iteration algorithm convergence character is acceptable.  The lecture stated that a Newton iteration will generate iterates that converge at a quadratic rate to a small value.  The corresponding mathematical expression, for p the iteration counter, is

max |dQ(p+1)^2| / max |dQ(p)| < constant

hence the sequence of iterates, when plotted on log-log scale, should exhibit a slope of two.  A quasi-Newton iteration procedure, if it converges, will generate data exhibiting a slope considerably less than two.  Your computational experiment data will include the maximum iterate sequence for p.

      The fin-cylinder is assumed to be the aluminum alloy 2024-T6. The suggested data specification for the problem statement is:

      Ta (Ambient temperature) = 70 F (294.26K)

      Tr (Reference temperature of the fluid) = 1822.7 F (1268K)

      h (heat transfer coefficient) = 20 Btu/hrft2F (113.56 W/m2K)

      k (thermal conductivity)= 107.5 Btu/hrftF (186 W/mK)

      a (cylinder inner radius)= 1.5 in. (2/39.3708m)

      tc (thickness of the cylinder)= 0.25 in. (1.5/39.3708m)

      b (cylinder outer radius)= 2.5 in. (4/39.3708m)

      s (Stefan-Boltzmann constant) = 0.1714Btu/hrft2R4 (5.67E-8 W/m2K4)

      e (emissivity) = 0.8

Assume the fin spacing is such that DZ = 0.33 in. for the cylinder section with basic fin thickness DZ = 0.15 in. if it is uniform.   However, your undergraduate heat and mass transfer text-book, [1], suggests that the maximum heat transfer per unit volume can be obtained for the fin being tapered of parabolic profile.  As an approximation you may use a linearly-tapered fin to analyze this effect in the second part of this experiment.    

[1]. Fundamentals of Heat and Mass Transfer; F. P. Incropera; D. P. Dewitt; John Wiley and Sons, New York.