A tube arranged on the wall of a furnace as indicated in Fig. 5.33 has an inner diameter of 25 mm and an outer diameter of 32 mm. The tube is made of a material of thermal conductivity 45 W/m^◦C. The heat flux incident on the exposed part is given to be 106 W/m^2. The inner wall temperature of the

Question

A tube arranged on the wall of a furnace as indicated in Fig. 5.33 has an inner diameter of 25 mm and an outer diameter of 32 mm. The tube is made of a material of thermal conductivity 45 W/m°C. The heat flux incident on the exposed part is given to be 106 W/m2. The inner wall temperature of the tube is 100°C. Make a plot of the temperature variation on the outer surface of the tube as a function of angle. Also, plot the temperature variation across the thickness of the tube at [latex]θ = \frac{π}{2}[/latex].

Accepted Answer

Step 1 We make use of the nomenclature introduced in Fig. 5.33 and write down the given data:

Inner radius of tube: [latex]R_{1}= \frac{0.025 }{2}= 0.0125 m[/latex]
Outer radius of tube: [latex]R_{2} = \frac{0.032 }{2} = 0.016 m[/latex]
Thermal conductivity of tube material: [latex]k = 45 W/m°C[/latex]
Datum temperature: [latex]T_{1} = 100°C[/latex]
Heat flux parameter:[latex]\frac{qR_{2} }{k} =\frac{10^{6} imes 0.016 }{45} =355.56 °C[/latex]

Step 2 In order to make a plot of the angular variation of temperature of the tube outer surface, we make use of Eq. 5.157 with [latex]r = R_{2}[/latex]. The data is calculated at 5° angular intervals from 0 to 360°. The Fourier summation needs a large number of terms and the data is generated using a computer program with 1000 terms. The resulting plot is shown in Fig. 5.36a. It is observed that the maximum tube wall temperature of 231.7 °C occurs, as expected, at θ = 90°.

[latex]z(r, heta )=\frac{2qR_{2}}{k\pi } \sum\limits_{1}^{\infty }{\frac{\sin [(2n-1) heta ] }{(2n-1)^{2}} } imes [/latex]

[latex] imes \left[\frac{\left(\frac{r}{R_{1}} ight) ^{(2n-1)}-\left(\frac{r}{R_{1}} ight) ^{-(2n-1)} }{\left(\frac{R_{2}}{R_{1}} ight) ^{(2n-1)} +{\left(\frac{R_{2}}{R_{1}} ight) ^{-(2n-1)}}} ight] +\frac{qR_{2} }{2k} \ln \left(\frac{r}{R_{1}} ight)[/latex] (5.157)

[latex]Note that 2n − 1 is 1, 3, 5, . . . for n = 0, 1, 2 . . ..[/latex]

Step 3 The second plot we make is actually for this angular location. Equation 5.157 is used to generate the data by taking the angle to be 90° and varying the radial position from [latex]R_{1} to R_{2}[/latex]. The resulting plot is shown in Fig. 5.36b). It is observed that the temperature varies very steeply within the tube wall. The radial temperature gradient is roughly 38000 °C/m!

Question 5.12: A tube arranged on the wall of a furnace as indicated in Fig...

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