Stationary Heat Transfer through Fins.
11-19-2016, 12:00 PM (This post was last modified: 11-22-2016 10:33 AM by Ángel Martin.)
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 Ángel Martin Senior Member Posts: 1,070 Joined: Dec 2013
Stationary Heat Transfer through Fins.
Stationary Heat Transfer through Fins. [ ANULAR, TRIANG, TRAPEZ ]
From the author’s Engineering Collection, included in the ETSII4 module.

1. Annular - Radial - Fins, with thickness w and r1, r2 the internal & external radius respectively.

Let n = sqrt( 2h / K.w ),

with h the heat transfer (film) coefficient and K the thermal conductivity.

Let T0 be the temperature difference between the base (r = r1) and the surrounding cooling fluid.

Assuming there’s no heat transfer at the fin’s tip, the expression for the corrected temperature at a distance r, (r1<= r <= r2) is given below:

T(r)/T0 = [ I0(n.r). K1(n.r2) + K0(n.r). I1(n.r2)] / [ I0(n.r1). K1(n.r2) + I1(n.r2). K0(n.r1)]

where I and K are the modified Bessel functions of first and second kind.

The expression for the dissipated heat is in this case:

Q = 2pi.n.K.w.r1.T0 [ I1(n.r2). K1(n.r1) - K1(n.r2). I1(n.r1)] / [ I0(n.r1). K1(n.r2) – K0(n.r1). I1(n.r2)]

2. Straight Fins with trapezoidal or triangular section profiles

With base thickness w and distance “d” to its (fictitious) triangular end point. Taking that end point as origin of coordinates, let "xe" be distance to the tip of the fin, and the base xb = d

• For a trapezoidal fin the actual length is: L = (d -Xe).
• For a triangular fin Xe =0 ; and its length is L = d

Let f = sqrt[ 1 + (w/2d)^2 ); and: p = 2 sqrt( 2f.h.d / K.w)

Let T0 be the temperature difference between the base and the surrounding fluid (air).

The expression for the corrected temperature (or difference) T(x) at a distance x >= xe is given below, and denoting x* = sqrt(x)

T(x)/T0= [ I0(p.x*).K1(p.xe*) + I1(p.xe*).K0(p.x*)] / [ I0(p.d*).K1(p.xe*)+ I1(p.xe*).K0(p.d*)]

Assuming there’s no heat transfer at the fin’s tip, the expression for the dissipated heat (per unit of depth) is in this case:

Q= -(A)[ I1(p.d*).K1(p.xe*) - I1(p.xe*).K1(p.d*)] / [ I0(p.d*).K1(p.xe*)+ I1(p.xe*).K0(p.d*)]

with A = K.w.p (Tb-T0) / sqrt(d)

Note: if you prefer using the base as origin of coordinates, simply replace x by (d–x) in the above expressions.

These programs use the Modified Bessel functions from the SandMath module, which needs to be plugged in the calculator as well.

Examples.

Calculate the temperature at the edge and the total dissipated heat for the following conditions: surrounding temperature Tinf = 30 deg C, base temperature Tb = 200 deg C. Physical properties: h = 34.89 [W/K.m^2] ; K = 53.498 [W/K.m]

a) an annular fin with r1=8 cm, r2=14 cm; w= 1 cm
b) a trapezoidal fin with w= 1cm, d = 14 cm; xe = 4 cm
c) a triangular fin with w= 1cm; d = 14 cm

The results for the corrected temperature (T(x)-Tinf) are given in the table below:

Code:
Fin type                Tc (deg C)   Q [J/s] --------------------------------------------- Annular, r = 0.14 m     131.1090     409.3259 Trapezoidal (x=4 cm)    81.5435      799.7711 Triangular  (x=0 cm)    29.6077      855.8098
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