Question 4.6: A double-wall, evacuated glass thermos is used for storing c...

A double-wall, evacuated glass thermos is used for storing cryogenic liquids (e.g., liquid nitrogen). The evacuation diminishes the conduction heat transfer between the walls, as depicted in Figure (a). However, heat transfer by radiation occurs between the glass wall surfaces. The glass walls can be reasonably assumed as very long parallel plates, i.e., F_{1-2} = 1 . The glass surface emissivity is \epsilon _{r,1} = \epsilon _{r,2} = 0.5 . The surface-radiation heat transfer q_{r,1-2} is reduced when a high reflectivity radiation shield, of emissivity \epsilon _{r,s} = 0.05 (e.g., aluminum foil, which is similar to highly polished aluminum, Table, is placed between the surfaces. For liquid nitrogen storage, at one atm pressure, from Table, T_{lg} = −195.9^{\circ }C . Here, surface 1, which is the cold surface, is at T_{1} = −190^{\circ }C , while the hot surface is at T_{2} = 2^{\circ }C .
Using the radiation shield model depicted in Figure (b), determine the radiation heat flux q_{r,1-2} , for the case of (a) no shield, (b) one shield, and (c) two shields present.

The shields have the same surface areas (on each side) as A_{r,1} and A_{r,2} .

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The thermal circuit model of the radiation heat transfer is shown in Figure (b). For F_{1-2} = 1   ,    Q_{r,1-2}=\frac{E_{b,1}(T_{1})-E_{b,2}(T_{2})}{(R_{r,\epsilon })_{1}+(R_{r,F })_{1-2}+(R_{r,\epsilon })_{2}+n[2(R_{r,\epsilon })_{s}+(R_{r,F})_{s}]} becomes

q_{r,1-2}=\frac{E_{b,1}(T_{1})-E_{b,2}(T_{2})}{A_{r,1}\left[(R_{r,\epsilon })_{1}+(R_{r,F })_{1}+(R_{r,\epsilon })_{2}\right] +A_{r,1}n[2(R_{r,\epsilon })_{s}+(R_{r,F})_{s}]}

where

A_{r,1}(R_{r,\epsilon })_{1}=\frac{1-\epsilon _{r,1}}{\epsilon _{r,1}}=\frac{1-0.5}{0.5} =1

 

A_{r,1}(R_{r,F })_{1}=1   since   F_{1-2}=1 

 

A_{r,1}(R_{r,\epsilon })_{2}=\frac{1-\epsilon _{r,2}}{\epsilon _{r,2}}=\frac{1-0.5}{0.5} =1    since      A_{r,1}=A_{r,2}

 

A_{r,1}(R_{r,\epsilon })_{s}=\frac{1-\epsilon _{r,s}}{\epsilon _{r,s}}=\frac{1-0.05}{0.05} =19         since      A_{r,s}=A_{r,1}

 

A_{r,1}(R_{r,F })_{s}=1           since      A_{r,s}=A_{r,1} 

Then for q_{r,1-2} we have the following.
(a) Without any shield:
q_{r,1-2}=\frac{\sigma _{SB}(T_{1}^{4}-T_{2}^{4})}{A_{r,1}[(R_{r,\epsilon })_{1}+(R_{r,F })_{1-2}+(R_{r,\epsilon })_{2}]}  

 

=\frac{5.670\times 10^{-8}(W/m^{2}-K^{4})[(83.15)^{4}(K)^{4}-(275.15)^{4}(K)^{4}]}{1+1+1}  

 

  q_{r,1-2}=\frac{-3.222\times 10^{2}(W/m^{2})}{3} =-1.074\times 10^{2}W/m^{2} 

(b) One shield:

q_{r,1-2}=\frac{\sigma _{SB}(T_{1}^{4}-T_{2}^{4})}{A_{r,1}[(R_{r,\epsilon })_{1}+(R_{r,F })_{1-s}+(R_{r,\epsilon })_{2}]+A_{r,1}[2(R_{r,\epsilon })_{s}+(R_{r,F})_{s-2}]}  

 

=\frac{-3.222\times 10^{2}(W/m^{2})}{1+1+1+(2\times 19+1)} 

 

q_{r,1-2}=-7.674  W/m^{2} 

(c) Two shields:

q_{r,1-2}=\frac{-3.222\times 10^{2}(W/m^{2})}{3+2\times 39}  

 

q_{r,1-2}=-3.979 W/m^{2}
b

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