Question

how did we get this answer (last pic)

Figure 8.35: View factor equation for parallel plates with midlines connected by per- pendicular Solar radiation Glass roof semi-circular cross-section Leaf surface 30 m 10 m Figure 8.36: Schematic of a greenhouse. 8.16 Greerhouse Energy Balance A greenhouse that has its bottom surface (floor) covered with leafy plants is shown as a schematic in Figure 8.36. The glass dome of the roof is semi-circular and the green- house can be considered long in the horizontal direction (actual length 30 m). Incident vertical solar radiation is 1000 W/m² Solar radiation can be assumed to be coming from a blackbody of 5800 K. Ignore any other atmospheric radiation. Transmissivity of the glass as a function of wavelength is given by Figure 8.37 and has no directional dependence. Reflectivity of the glass is considered zero at all wavelengths. The energy exchange between the leafy plants and the inside surface of the glass is given by the equation for a two-surface enclosure. For this particular exchange, the glass is effec- tively a blackbody. The leafy plants are arranged in such a way that the area of the plants can be approximated to be the same as the area of the greenhouse floor. The emissivity of the surface of the plants is 0.95. The inside surface of glass has a heat transfer coefficient of 5 W/m². K while the outside glass surface has a heat transfer coefficient of 10 W/m².K. The plant surface is at 30°C, the glass is at 45°C, and the outside air is at 25°C. 1) Calculate the amount of radiative heat exchange between these two bodies--glass and the leafy plant floor. Note that the glass property data means that it is effectively a blackbody for this exchange. 2) In order to find the green- house air temperature, first write the equation for energy balance for the glass surface, making sure all modes of heat transfer are included. 3) Calculate the greenhouse air temperature from the energy balance. Transmissivity of glass 1 Wavelength (jam) Figure 8.37: Transmissivity of the greenhouse glass

Answer 1

As you saw in the above step they are added all the required elements radiation to plants convective outside and convective inside

So for this also we should use the same formula and make appropriate changes so that we bring time of air inside to the left handed side and remaining all other terms to right handed side.

Then you will get the required answer