Question 10.3: A thermometer element, reflection coefficient 0.40, is expos...
A thermometer element, reflection coefficient 0.40, is exposed to a mean irradiance of 300 W m^{-2} of solar radiation in an environment where effective radiative temperature and air temperature are 20.0 ºC and windspeed is 1.0 m s^{-1}. If the resistance r_{H} to convective heat transfer of the thermometer is 80 s m^{-1}, what temperature will the thermometer indicate? Two methods of improving the thermometer were tested
(i) coating the element with white paint, reflection coefficient 0.90, and
(ii) surrounding the element with a radiation shield, emissivity unity, but which reduced the windspeed u around the bulb to 0.5 m s^{-1}. If the radiation shield reduced the mean solar irradiance of the bulb to 90 W m^{-2}, calculate which method gave the smallest temperature error, assuming that r_{H} was proportional to u^{-0.5}.
In equilibrium, the net radiation must balance convective heat loss. The net radiation R_{n} = (1-0.4)300+\sigma T^{4}_{a} – T^{4}_{t}. Writing the difference between air and thermometer temperature ( T_{a} -T_{t} ) as ∆T, it follows that, for small temperature differences, R_{n} = (1-0.4) 300+4\sigma T^{3}_{a} \Delta T . The rate of convective heat loss is given by C= -\rho c_{p} \Delta T/80 . Hence, equating the fluxes, -\rho c_{p} \Delta T/80 =180+4\sigma T^{3}_{a} \Delta T . Solving for ∆T gives ∆T = −8.7 °C, i.e. T_{t} = 28.7 °C.
i. Increasing the reflection coefficient gives ∆T = −1.4 °C, i.e. T_{t} = 21.4 °C.
ii. Adding a radiation shield that reduced ventilation increases r_{H} to 113 s m^{-1} and gives ∆T = −5.5 °C, i.e. T_{t} = 25.5 °C.