Question 8.12: The heat transfer rate from a very long fin of constant cros...

First, draw a sketch of the system (Figure 8.13).

The unknown is the entropy production rate for the fin. The material is aluminum.

Since there is no work mode entropy production here, Eq. (7.66) is used to find the fin’s entropy production rate by the direct method (note that we have insufficient information to use the more convenient indirect method or entropy balance here) as

 \dot{S}_P = (\dot{S}_P)_Q = – \int_{\sout{V}}[\frac{\dot{q}}{T^{2}}(\frac{dT}{dx} ) ]_\text{act}  d\sout{V}

Since this is a one-dimensional heat transfer problem, we can substitute A  dx for  d\sout{V}. Then,

(\dot{S}_P)_Q = – \int_{0}^{L} [\frac{\dot{q}}{T^{2}}(\frac{dT}{dx} ) ] A  dx = \int_{0}^{∞} [\frac{\dot{k_tA}}{T^{2}} (\frac{dT}{dx} )^{2}] dx

where we have used Fourier’s law, \dot{q} = −k_t(dT/dx), and have let L → ∞ for a very long fin. We can differentiate the fin’s temperature profile given previously to obtain

\frac{dT}{dx} = -m( T_f −T_∞)e^{−mx}

then

k_t(\frac{A}{T^{2}} )(\frac{dT}{dx})^{2} = \frac{\sqrt{hPk_tA} (m)(T_f −T)^{2}e^{−mx}}{[T_∞ + ( T_f −T_∞)e^{−mx}]^{2}}

and

(S_P)_Q = m( T_f −T_∞)^{2} \sqrt{hPk_tA} \int_{0}^{∞} \frac{e^{−2mx}dx}{[T_∞ + ( T_f −T_∞)e^{−mx}]^{2}}

This expression can be integrated using a table of integrals and a change of variables (e.g., let y = e^{−mx}) to obtain

(\dot{S}_P)_Q = \sqrt{hPk_tA} (\text{ln} \frac{T_f}{T_∞} + \frac{T_∞}{T_f} -1  )

In this problem, we have

h = 3.50  \text{W}/(\text{ m}^{2}.\text{K})          A = 1.00 × 10^{−4}\text{ m}^{2}

P = 0.0400  \text{m}         T_∞ = 20.0°\text{C} = 293.15  \text{ K}

k_t = 204  \text{W/(m.K)}         T_f = 95.0°\text{C} = 368  \text{ K}

Then,

\dot{S}_P = (\dot{S}_P)_Q

= \sqrt{[3.50  \text{W}/(\text{ m}^{2}.\text{K})](0.0400  \text{m})[204  \text{W/(m.K)}](1.00 × 10^{−4}\text{ m}^{2})}   (\text{ln} \frac{368}{293.15} + \frac{293.15}{368} -1)

= 0.00128  \text{W/K}

Note that the entropy production rate in this example is quite small.