In extraterrestrial applications (e.g., in the space station), radiation cooling of liquids is made by forming droplets and injecting the droplets into open space (negligible surface-convection heat transfer), and after a short travel period collecting them. This is shown in Figure Ex. 4.14(a). We

Question

In extraterrestrial applications (e.g., in the space station), radiation cooling of liquids is made by forming droplets and injecting the droplets into open space (negligible surface-convection heat transfer), and after a short travel period collecting them. This is shown in Figure (a). We use water droplets of diameter D = 2 mm, emissivity [latex] \epsilon _{r,1} = 0.8[/latex], initial temperature [latex]T_{1}(t = 0) = 50^{\circ }C[/latex] ([latex]T_{1}[/latex] is assumed uniform throughout the droplet, [latex]N_{k} < 0.1)[/latex], and a blackbody surrounding at [latex]T_{∞} = 3 K[/latex].
(a) Draw the thermal circuit diagram.
(b) Determine the elapsed time required to cool the droplet to [latex]T_{1} = 20^{\circ}C[/latex].
Use the saturated water thermodynamic properties from Table and at T = 313 K.

Accepted Answer

(a) The thermal circuit diagram for this problem is given in Figure (b).
(b) From [latex]\frac{ 4\sigma _{SB}T_{\infty }^{3}}{( ho c_{p}V)_{1}R_{r,\sum }}t\equiv \frac{t}{ au _{1}} =\left(ln\left|\frac{T_{\infty }+T_{1}}{T_{\infty }-T_{1}} ight|+2tan^{-1}\frac{T_{1}}{T_{\infty}} ight) \mid ^{T_{1}}_{T_{1}(t=0)}=\left[ ln\left|\frac{T_{\infty }+T_{1}}{T_{\infty }-T_{1}} ight|-ln\left|\frac{T_{\infty }+T_{1}(t=0)}{T_{\infty }-T_{1}(t=0)} ight|+2tan^{-1}\frac{T_{1}}{T_{\infty }}-2tan^{-1}\frac{T_{1}(t=0)}{T_{\infty }} ight][/latex], we have

[latex] t=\frac{ R_{r,\sum }( ho c_{p}V)_{1}}{4\sigma _{SB}T_{\infty }^{3}}\left[ ln\mid\frac{T_{\infty }+T_{1}}{T_{\infty }-T_{1}} \mid-ln\mid\frac{T_{\infty }+T_{1}(t=0)}{T_{\infty }-T_{1}(t=0)} \mid+2tan^{-1}\frac{T_{1}}{T_{\infty }}-2tan^{-1}\frac{T_{1}(t=0)}{T_{\infty }} ight] [/latex]

where

[latex]R_{r,\sum{} }= (R_{r,\epsilon })_{1} + (R_{r,F} )_{1−∞} + (R_{r,\epsilon })_{∞}.[/latex]

The overall radiation resistance is

[latex] R_{r,\sum }=\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} ight)_{1}+\frac{1}{A_{r,1}F_{1-\infty }} +\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} ight)_{\infty } [/latex]

[latex] =\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} ight)_{1}+\frac{1}{A_{r,1}} +0 [/latex]

[latex] =\frac{1}{A_{r,1}\epsilon _{r,1}} [/latex]

because [latex] \epsilon_{r,∞} = 1[/latex] and[latex] F_{1−∞} = 1 [/latex] (neglecting the droplet-droplet and droplet-solid surface view factors).
We also note

[latex] A_{r,1} = 4\pi R^{2} = \pi D^{2} , V_{1} = \frac{4}{3} \pi R^{3} = \frac{1}{6} \pi D^{3} . [/latex]

The thermodynamic properties for water at [latex]T_{1} = 313 K[/latex] are

[latex] ho _{1} = 994.5 kg/m^{3}, c_{p,1} = 4.178 × 10^{3} J/kg-K [/latex] Table

We now use the numerical values where

[latex] A_{r,1} = \pi × (2 × 10^{−3}) ^{2} (m)^{2} = 1.257 × 10^{−5} m^{2} [/latex]

[latex] V_{1} =\pi imes \frac{(2 imes 10^{-3})^{3}(m)^{3}}{6}=4.189 imes 10^{-9}m^{3} [/latex]

[latex] R_{r,\sum }=\frac{1}{A_{r,1}\epsilon _{r,1}} =\frac{1}{1.257 imes 10^{-5}(m^{2}) imes 0.8} =9.944 imes 10^{4}1/m^{2} [/latex]

to determine t as

[latex]t=\frac{9.944 × 10^{4} (1/m^{2}) × 994.5(kg/m^{3}) × 4.178 × 10^{3} (J/kg-K) × 4.189 × 10^{−9} (m^{3} )}{4 × 5.670 × 10^{−8} (W/m^{2}-K^{4}) × (3)^{3} (K)^{3}}[/latex]

[latex]× \left[ln\mid \frac{3(K) + 293.15(K)}{3(K) - 293.15(K)} \mid -ln\mid \frac{3(K) + 323.15(K)}{3(K) - 323.15(K)}\mid +2tan^{-1}\frac{293.15(K)}{3(K)} -2tan^{-1}\frac{323.15(K)}{3(K)} ight] [/latex]

[latex]= 9.421 × 10^{7} (s) × (2.0468052 × 10^{−2} − 1.8567762 [/latex]

[latex] × 10^{−2} + 3.1211260 − 3.1230260)[/latex]

[latex]t = 9.421 × 10^{7}(s) × 2.9000000 × 10^{−7} = 27.31 s [/latex]

Question 4.14: In extraterrestrial applications (e.g., in the space station...

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