A combustion system consists of a cylindrical annulus combustion chamber, as shown in Figure Ex. 3.10(a). The heat from the combustion is to flow mostlyto the outside surface designated by T2. Therefore, a high conductivity solid material is used for the outer cylindrical shell. Here we use silicon

Question

A combustion system consists of a cylindrical annulus combustion chamber, as shown in Figure (a). The heat from the combustion is to flow mostly to the outside surface designated by [latex]T_{2}[/latex]. Therefore, a high conductivity solid material is used for the outer cylindrical shell. Here we use silicon carbide SiC, a nonoxide ceramic, with k given in Figure (c) as a function of temperature. Silicon carbide has a relatively large k and remains inert at high temperatures.The inner cylindrical shell is made of a lower conductivity material and here we have selected zirconium oxide [latex]ZrO_{2}[/latex], an oxide ceramic, k also given in Figure. The inner radius for the [latex]ZrO_{2}[/latex] shell is selected as [latex] R_{1} = 25 mm[/latex]. The wall thickness for the [latex]ZrO_{2}[/latex] and SiC shells are selected as [latex]l_{1} = l_{2} = 3 mm[/latex].
(a) Draw the thermal circuit diagram.
(b) Determine the inner radius for the SiC shell [latex]R_{2}[/latex] such that the heat flowing through the SiC shell is 10 times that which leaves through the [latex]ZrO_{2}[/latex] shell.

Use [latex]\left\langle T_{g}\right\rangle = 2,000 K, T_{2} = 1,000 K, T_{1} = 800 K[/latex], and evaluate k’s at T = 1,200 K.

Accepted Answer

(a) The thermal circuit diagram for this energy conversion and heat transfer is shown in Figure (b). We have used nodes g, 1, and 2.

(b) The energy equation [latex] Q\mid _{A,i}=Q_{i}+\sum\limits_{j=i}^{n}{Q_{k,i-j}}=\dot{S}_{i} [/latex], with [latex]Q_{g} = 0 [/latex] (i.e., no prescribed heat transfer is given), gives

[latex] Q_{g}+\sum\limits_{j=i}^{2}{Q_{k,g-i}}= \frac{\left\langle T_{g} ight angle -T_{1}}{R_{k,g-1}}+ \frac{\left\langle T_{g} ight angle -T_{2}}{R_{k,g-2}} =\dot{S}_{r,c} [/latex]

The resistances are those given in Table i.e.,

[latex] R_{k,g-1}=\frac{\ln \frac{R_{1}+l_{1}}{R_{1}} }{2\pi Lk_{1}} [/latex]

[latex]R_{k,g-2}=\frac{\ln \frac{R_{2}+l_{2}}{R_{2}} }{2\pi Lk_{2}} [/latex]

The energy conversion rate is given as

[latex] \dot {S}_{r,c}=\dot{s}_{r,c}A_{u}L=\dot{s}_{r,c}\pi [R^{2}_{2}-(R_{1}+l_{1})^{2}]L [/latex]

The design requires that

[latex] \frac{Q_{k,g-2}}{Q_{k,g-1}}=\frac{\left\langle T_{g} ight angle -T_{2}}{\left\langle T_{g} ight angle -T_{1}}\frac{R_{k,g-1}}{R_{k,g-2}} =10 [/latex]

or

[latex] \frac{\left\langle T_{g} ight angle -T_{2}}{\left\langle T_{g} ight angle -T_{1}}\frac{\ln \frac{R_{1}+l_{1}}{R_{1}} }{\ln \frac{R_{2}+l_{2}}{R_{2}} } \frac{k_{2}}{k_{1}} =10 [/latex]

or

[latex] \ln \frac{R_{2}+l_{2}}{R_{2}}=\ln \frac{R_{1}+l_{1}}{R_{1}}\left(\frac{\left\langle T_{g} ight angle -T_{2}}{\left\langle T_{g} ight angle-T_{1} } ight) \frac{k_{2}}{k_{1}} imes \frac{1}{10} [/latex]

From Figure (c), we have for T = 1,200 K

[latex] ZrO_{2} : k_{1} \simeq 2.2 W/m-K [/latex] Figure (c)
[latex] SiC: k_{2} \simeq 20 W/m-K [/latex] Figure (c).

Then using the numerical values, we have

[latex] ln\left[\frac{R_{2}+0.003(m)}{R_{2}} ight] =ln\left[\frac{(0.025+0.003)(m)}{0.025(m)} ight] imes \frac{(2,000-1,000)(k)}{(2,000-800)(k)} imes \frac{20(W/m-k)}{2.2(W/m-k)} imes \frac{1}{10} [/latex]

The solution is found using a solver, and is

[latex]R_{2} = 0.03346 m = 33.46 mm.[/latex]

Question 3.10: A combustion system consists of a cylindrical annulus combus...

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