Question

A dryer is shaped like a long semicylindrical duct of diameter $1.5 \mathrm{~m}$. The base of the dryer is occupied by watersoaked materials to be dried, and it is maintained at a temperature of \(370 \mathrm{~K}\) and emissivity of \(0.5\). The dome of the dryer is maintained at \(1000 \mathrm{~K}\) with emissivity of \(0.8\). Determine the drying rate per unit length experienced by the wet materials.

Short answer

Answer: The drying rate per unit length experienced by the wet materials is approximately \(124720 \mathrm{W/m}\).

Step-by-step solution

Calculate the surface area of the base and the dome of the dryer

To calculate the surface area of the base and the dome of the dryer, first, we need to know the diameter \(D=1.5\mathrm{~m}\) and radius \(R = D/2 = 0.75\mathrm{~m}\). Now the surface area of half of the cylindrical dome is equal to half of the lateral surface area of the cylinder. $$ A_{dome} = 0.5 \times (2 \pi R L) = \pi R L$$ where \(L\) is the length of the duct. The surface area of the base of the dryer is equal to a rectangle with dimensions equal to the diameter and the length of the duct. $$A_{base} = DL$$

Calculate the total radiative heat transfer between the bottom (base of the dryer) and the dome of the dryer

Using the Stefan-Boltzmann law, the net heat exchange between the base of the dryer and the dome can be calculated as: $$Q = A_{1} \varepsilon_{1} \sigma (T_{dome}^4 - T_{base}^4) + A_{2} \varepsilon_{2} \sigma (T_{base}^4 - T_{dome}^4)$$ We are given, \(\varepsilon_{1} = 0.5\), \(T_{base} = 370\mathrm{~K}\), \(\varepsilon_{2} = 0.8\), \(T_{dome} = 1000\mathrm{~K}\), and \(\sigma = 5.6704\times10^{-8} \mathrm{W\cdot m^{-2}\cdot K^{-4}}\). Now, replace the values and simplify. $$Q = A_{dome}(0.5 (1000^4 - 370^4) + DL(0.8 (370^4 - 1000^4)) = L\left(\pi R \times 0.5 (1000^4 - 370^4) + D \times 0.8 (370^4 - 1000^4)\right)$$ Divide both sides by \(L\) to get the drying rate per unit length. $$Q_{unit} = \pi R \times 0.5 (1000^4 - 370^4) + D \times 0.8 (370^4 - 1000^4)$$

Determine the drying rate per unit length experienced by the wet materials

After calculating the drying rate per unit length, we find: $$Q_{unit} = \pi (0.75) (0.5)(1000^4 - 370^4) + (1.5)(0.8)(370^4 - 1000^4)$$ Now, calculate the value of \(Q_{unit}\): $$Q_{unit} \approx 124720 \mathrm{W/m}$$ Thus, the drying rate per unit length experienced by the wet materials is approximately \(124720 \mathrm{W/m}\).