Question 9.32: A steel tube fitted with transverse circular steel fins of c......

Assuming that the height of the fin is small compared with its circumference and that it may be treated as a straight fin of length (π/2)(d_{1} + d_{2}), then:
The perimeter: b={\frac{2\pi(d_{1}+d_{2})}{2}}=\pi(d_{1}+d_{2})

The area: A={\frac{\pi(d_{1}+d_{2})w}{2}}, i.e. the average area at right-angles to the heat flow

Then: m={\biggl(}{\frac{h b}{k A}}{\biggr)}^{0.5}={\biggl\{}{\frac{h\pi(d_{1}+d_{2})}{[k\pi(d_{1}+d_{2})w/2]}}{\biggr\}}^{0.5}

= \left({\frac{2h}{k w}}\right)^{0.5}

= \left({\frac{2\times30}{43\times0.002}}\right)^{0.5}

= 26.42\ \mathrm{m}^{-1}

From equation 9.254, the heat flow is given for case (b) as:

Q_{f}=-kAm\theta_{1}\frac{1-\mathrm{e}^{2m L}}{1+\mathrm{e}^{2m L}}=\sqrt{h b k A}\,\theta_{1}\,\mathrm{tanh}\,m L                   (9.254)

In this equation: A=\frac{[\pi(70.0+54.0)\times2.0]}{2} = 390 mm²     or     0.00039 m²

L={\frac{d_{1}-d_{2}}{2}} = 8.0 mm   or    0.008 m

mL = 26.42 x 0.008 = 0.211
\theta _{1} = 370 – 280 = 90 deg K

∴ Q_{f}=\frac{26.42\times43\times3.9\times10^{-4}\times90(\mathrm{e}^{0.422}-1)}{1+\mathrm{e}^{0.422}}

= {\frac{39.9\times0.525}{2.525}} = 8.29 W per fin

The heat loss per metre run of tube = 8.29 x 230 = 1907 W/m
or: \underline{\underline{1.91\ kW/m}}

In this case, the low value of mL indicates a fin efficiency of almost 1.0, though where mL tends to 1.0 the efficiency falls to about 0.8.