Again, consider the plastic sheet of Example 6.5, and here allow for the surface-convection cooling by evaporation by exposing one side of the plastic (placed horizontally) sheet to a saturated water pool, as shown in Figure (a). At one atmosphere pressure, the boiling temperature of the water is found from Table and is T_{f,∞} = T_{lg} = 373.15 K. The plastic sheet is at T_{s} = 115^{\circ}C= 388.15 K. Assume that the surface superheat T_{s} − T_{lg} = 15^{\circ}C is sufficient to cause nucleate boiling of the plastic surface (which is a reasonable assumption).
(a) Draw the thermal circuit diagram.
(b) Use A_{ku}\left\langle R_{ku}\right\rangle_{L} =\frac{T_{s}-T_{lg}}{\left\langle q_{ku}\right\rangle _{L}} =a_{s}^{3}\frac{\Delta h_{lg}^{2}}{\mu _{l}c_{p,l}^{3}(T_{s}-T_{lg})^{2}} \left(\frac{\sigma }{g\Delta \rho _{lg}} \right) ^{1/2}Pr_{l}^{n} , n=3 (water) ,n=5.1(all other fluids) to determine the rate of surface-convection cooling.
(c) Again using A_{ku}\left\langle R_{ku}\right\rangle_{L} =\frac{T_{s}-T_{lg}}{\left\langle q_{ku}\right\rangle _{L}} =a_{s}^{3}\frac{\Delta h_{lg}^{2}}{\mu _{l}c_{p,l}^{3}(T_{s}-T_{lg})^{2}} \left(\frac{\sigma }{g\Delta \rho _{lg}} \right) ^{1/2}Pr_{l}^{n} , n=3 (water) ,n=5.1(all other fluids), determine the average surface-convection resistance \left\langle R_{ku}\right\rangle_{L}.
L = 0.2 m , p_{g} = 1.013 × 10^{5} Pa.
For lack of specific data, take a_{s} from Table 6.2, that corresponding to the water-copper pair.
