Consider the chimney in Example 16.1. The chimney is made of a double wall construction with a t = 50.8mm thick Super Wool 607 Plus in the annular space. Thermal conductivity of the insulation is known to be k(Insul) = 0.05W/m°C. Assume that the chimney experiences a mild wind of 2m/s normal to its

Question

Consider the chimney in Example 16.1. The chimney is made of a double wall construction with a t = 50.8mm thick Super Wool 607 Plus in the annular space. Thermal conductivity of the insulation is known to be [latex]k_{ ext {Insul }}=0.05 W / m ^{\circ} C[/latex]. Assume that the chimney experiences a mild wind of 2m/s normal to its axis.

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This example will show how wemake use of thematerial already covered in previous chapters to model a fairly complex problem. We make use of the overall resistance concept in treating heat transfer radially across the chimney, assuming very little heat transfer to take place along the axis. Heat transfer on the outside surface of the chimney is modeled as that due to flow normal to the axis of a cylinder, using the Zhukaskas correlation. Resistance due to the insulation is considered as conduction through an annulus.The stainless sheet metal used inmaking the doublewall chimney is very thin and may be assumed to offer zero conduction resistance. Air properties required in the analysis are tabulated below.

T, K	[latex] ho[/latex] , [latex]kg / m ^{3}[/latex]	ν, [latex]m ^{2} / s[/latex]	k, [latex]W / m ^{\circ} C[/latex]	[latex]P r[/latex]
293	1.205	[latex]1.52 imes 10^{-5}[/latex]	0.0258	0.704
453	0.779	[latex]3.28 imes 10^{-5}[/latex]	0.0376	0.686

Step 1 We calculate all the resistances now.

(a) Conduction resistance due to the insulation is given by

[latex]R_{ ext {Insul }}=\frac{1}{2 \pi k_{ ext {Insul }}} \ln \left(\frac{I D+2 t}{I D} ight)[/latex] [latex]=\frac{1}{2 imes \pi imes 0.05} \ln \left(\frac{203.2+2 imes 50.8}{203.2} ight)=1.2906 m ^{\circ} C / W[/latex]

(b) Reynolds number for flow in the chimney is given, based on air properties at 453 K, by

[latex]R e_{I D}=\frac{5.612 imes 0.2032}{3.28 imes 10^{-5}}=34812.5[/latex]

Based on Dittus–Boelter equation the corresponding Nusselt number is given by

[latex]N u_{I D}=0.023 \operatorname{Re}_{I D}^{0.8} \operatorname{Pr}^{0.37}=0.023 imes 34812.5^{0.8} imes 0.686^{0.37}=88.3[/latex]

Corresponding to the above the inside heat transfer coefficient [latex]h_{i}[/latex] is given by

[latex]h_{i}=\frac{N u_{I D} k_{i}}{I D}=\frac{88.8 imes 0.0376}{0.2032}=16.34 W / m ^{2}{ }^{\circ} C[/latex]

Film resistance on the inside is then given by

[latex]R_{i}=\frac{1}{\pi I D h_{i}}=\frac{1}{\pi imes 0.2032 imes 16.34}=0.0959 m ^{\circ} C / W[/latex]

(c) Consider the cross flow across the chimney on the outside due to the prevailing wind. The wind velocity is specified as U = 2m/s. Air properties are taken at 293 K. Reynolds number based on OD is then given by

[latex]\operatorname{Re}_{O D}=\frac{2 imes 0.3048}{1.52 imes 10^{-5}}=39994.3[/latex]

The appropriate constants in the Zhukaskas correlation are C = 0.26, m = 0.6 and n = 0.37. The Nusselt number is then given by

[latex]N u_{O D}=0.26 imes 39994.3^{0.6} imes 0.704^{0.37}=131.8[/latex]

Then the external heat transfer coefficient is

[latex]h_{o}=\frac{N u_{O D} k_{o}}{O D}=\frac{131.8 imes 0.0258}{0.3048}=11.15 W / m ^{2 \circ} C[/latex]

Hence the outside film resistance is given by

[latex]R_{o}=\frac{1}{\pi O D h_{o}}=\frac{1}{\pi imes 0.3048 imes 11.15}=0.0936 m ^{\circ} C / W[/latex]

Hence the overall resistance is given by

[latex]R_{ ext {Overall }}=R_{i}+R_{ ext {Insul }}+R_{o}=0.0959+1.2906+0.0936=1.4801 m ^{\circ} C / W[/latex]

Step 2 We model the cooling of the gas as it moves up the chimney as a heat exchanger tube with constant wall temperature that corresponds to the outdoor air temperature. The analysis is akin to that presented in section 14.4.3. [latex]2 \pi R \bar{h}[/latex] in that analysis will correspond to [latex]\frac{1}{R_{ ext {Overall }}}[/latex] of the present case. Taking I D as the basis,wemay calculate the mean heat transfer coefficient as

[latex]\bar{h}=\frac{1}{\pi I D R_{ ext {Overall }}}=\frac{1}{\pi imes 0.2032 imes 1.4801}=1.0583 W / m ^{2}{ }^{\circ} C[/latex]

The mass flow rate of the flue gas is [latex]\dot{m}= ho_{i} Q=0.7942 imes 0.182=0.1445 kg/s[/latex]. We assume its specific heat to be [latex]C_{p}=1005 J / kg K[/latex]. With the heat transfer area being given by [latex]S_{H}=\pi I D H=\pi imes 0.2032 imes 20=12.767m^2[/latex], we then have

[latex]\frac{\bar{h} S_{H}}{\dot{m} C_{p}}=\frac{1.4801 imes 12.767}{0.1445 imes 1005}=0.0930[/latex]

The temperature difference between the gas leaving the chimney and the ambient is then obtained as

[latex]\Delta T_{H}=(180-20) e ^{-0.0930}=145.8^{\circ} C[/latex]

Hence the exit temperature of the gases from the chimney is

[latex]T_{H}=20+145.8=165.8^{\circ} C[/latex]

Thus the temperature drops by about [latex]14^{\circ} C[/latex] from entry to exit. Example 16.1 may be reworked by taking the temperature of the gases as the mean at entry and exit and rework all the numbers.

Question 16.2: Consider the chimney in Example 16.1. The chimney is made of...

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