Question 10.22: KNOWN: Concentric stainless steel tubes packed with dense bo......

ANALYSIS: Construct the thermal circuit shown above where \mathrm R_{23}^{\prime}~\text{and}~\mathrm R_{34}^{\prime} represent the resistances due to the boron nitride between \mathrm r_2~\text{and}~\mathrm r_3 and to the outer stainless steel tube, respectively. From an overall energy balance,

q _{\text {gen }}^{\prime}= q ^{\prime}_{\text{boil}},

\dot{ q } \pi\left( r _2^2  –  r _1^2\right)=\left(2 \pi r _4\right) C \left( T _{ s }  –  T _{\text{sat}}\right)^3.

With prescribed values for \mathrm T_{\text{sat}}, \mathrm T_{\mathrm s} and C, the required volumetric heating of the inner stainless steel tube is

\dot{ q }=\frac{2 r _4}{\left( r _2^2  ~- ~ r _1^2\right)} C \left( T _{ s }  –  T _{\text{sat}}\right)^3.

Using the thermal circuit, we can write expressions for the maximum temperature of the stainless steel (ss) and boron nitride (bn).

Stainless steel: \mathrm T_{\mathrm{ss},\max} occurs at \mathrm r_1. Using the results of Section 3.4.2, the temperature distribution in a radial tube of inner and outer radii \mathrm r_1~\text{and}~\mathrm r_2 is

T ( r )=-\frac{\dot{ q }}{2 k _{ ss }} r ^2+ C _1  \ln  r + C _2

for which the boundary conditions are

BC#1:    r = r _{ 1 } ~~\quad~~ \frac{ dT }{ dr }=0 ~~\quad~~ 0=-\frac{\dot{ q }}{2 k _{ ss }} 2 r _1+\frac{ C _1}{ r _1}+0 \rightarrow C _1=+\frac{\dot{ q } r _1^2}{ k _{ ss }}

BC#2:    r = r _2 ~~\quad~~ T \left( r _2\right)= T _2 ~~\quad~~ T _2=-\frac{\dot{ q }}{2 k _{ ss }} r _2^2+\frac{\dot{ q } r _1^2}{ k _{ ss }} \ln r _2+ C _2

C_2 = T_2 + \frac{\dot{ q }}{2 k _{ ss }} r _2^2  –  \frac{\dot{ q } r _1^2}{ k _{ ss }} \ln r _2

Hence,

T ( r )=-\frac{\dot{ q }}{2 k _{ ss }}\left( r ^2- r _2^2\right)+\frac{\dot{ q } r _1^2}{ k _{ ss }} \ln \left( r / r _2\right)+ T _2.

Using the thermal circuit, find \mathrm T_2 in terms of known parameters \mathrm T_{\mathrm s}, \mathrm T_{\text{sat}} and C.

\frac{ T _2~-~ T _{ s }}{ R _{23}^{\prime}~+~ R _{34}^{\prime}}=\left(2 \pi r _4\right) C \left( T _{ s }  –  T _{ sat }\right)^3.

Hence, the maximum temperature in the inner stainless steel tube (\mathrm r = \mathrm r_1) is

T _{ ss,\max}= T \left( r _1\right)=-\frac{\dot{ q }}{2 k _{ ss }}\left( r _1^2  –  r _2^2\right)+\frac{\dot{ q } r _1^2}{ k _{ ss }} \ln \left( r _1 / r _2\right)+ T _{ s} +\left( R _{23}^{\prime}+ R _{34}^{\prime}\right)\left(2 \pi r _4\right) C \left( T _{s }  –  T _{ sat }\right)^3

where from Eq. 3.27

R _{23}^{\prime}=\frac{\ln \left( r _3 / r _2\right)}{2 \pi k _{ bn }} ~~~\quad~~~ R _{34}^{\prime}=\frac{\ln \left( r _4 / r _3\right)}{2 \pi k _{ ss }}.

Boron nitride: \mathrm T_{\mathrm{bn},\max} occurs at \mathrm r_1. Hence

T_{\mathrm{bn},\max} = T(r_1)

as derived above for the inner stainless steel tube.