91Ó°ÊÓ

We are given the evaporative flux, and we need to find the corresponding convection mass transfer coefficient. We can use the following equation: \(n_A'' = h_m (T_w - T_\infty)\) Rearrange the equation to solve for \(h_m\): \(h_m = \frac{n_A''}{(T_w - T_\infty)}\) Now, plug in the given values: \(h_m = \frac{0.030\, kg/s\cdot m^2}{(340 \, K - 300\, K)}\) \(h_m = 0.006\, kg/s\cdot m^2\cdot K\)

To find out how long it will take for the water to evaporate completely, we have to use the following formula: \(t = \frac{m_{water}}{n_A'' A}\) First, we need to calculate the mass of water in 1 m². We know that the water layer is 2 mm thick. The density of water is 1000 kg/m³. The mass of water per unit area can be calculated as follows: \(m_{water} =1000\, kg/m^3 \cdot 0.002\, m = 2\, kg/m^2\) Now, we can find the time it takes for the water to evaporate: \(t = \frac{2\, kg/m^2}{0.030\, kg/s\cdot m^2} = 66.67\, s\)

To find the convection heat transfer coefficient, we can use the following equation: \(q'' = h (T_w - T_\infty)\) Rearrange the equation to solve for \(h\): \(h = \frac{q''}{(T_w - T_\infty)}\) We know from step 1 that the mass transfer coefficient is: \(h_m = 0.006\, kg/s\cdot m^2\cdot K\) We can use the following relationship between the mass transfer and heat transfer coefficients: \(h = h_m \cdot L\) Where L is the latent heat of evaporation of water, approximately equal to \(2.257 \times 10^6\, J/kg\) Now, we can calculate the heat transfer coefficient: \(h = 0.006\, kg/s\cdot m^2\cdot K \cdot 2.257 \times 10^6\, J/kg = 13542\, W/m^2\cdot K\)

We have to find the electrical power supply per unit area required to maintain the prescribed temperature of the water. We know that the heat transfer includes both convective and radiative components: \(q'' = q_\text{supply} + q_\text{radiation}\) We can find the convective component by using the heat transfer coefficient we found in step 3: \(q_\text{convective} = h(T_w - T_\infty) = 13542\, W/m^2\cdot K \cdot (340\, K - 300\, K) = 541680\, W/m^2\) Now we need to calculate the radiative component. We know the emissivity of the water surface is \(\varepsilon_w = 0.95\), the Stefan-Boltzmann constant is \(5.67 \times 10^{-8}\, W/m^2\cdot K^4\), and the surrounding temperature is \(T_\infty = 300\, K\). We can calculate the radiative component as follows: \(q_\text{radiation} = \varepsilon_w \cdot \sigma \cdot (T_w^4 - T_\infty^4) = 0.95 \cdot 5.67 \times 10^{-8}\, W/m^2\cdot K^4 \cdot (340^4 - 300^4)\) \(q_\text{radiation} = 251989\, W/m^2\) Now we can find the electrical power supply per unit area: \(q_\text{supply} = q'' - q_\text{radiation} = 541680\, W/m^2 - 251989\, W/m^2 = 289691\, W/m^2\)

First, we need to determine the overall heat transfer from the plate and the surroundings: \(q_\text{radiation,p} = \varepsilon_p \cdot \sigma \cdot (T_p^4 - T_\infty^4)\) We know the emissivity of the plate, \(\varepsilon_p = 0.60\), and the electrical power found in step 4, \(q_\text{supply} = 289691\, W/m^2\) is now only lost through radiation: \(q_\text{supply} = \varepsilon_p \cdot \sigma\cdot(T_p^4 - T_\infty^4)\) Rearrange the equation for \(T_p^4\): \(T_p^4 = T_\infty^4 + \frac{q_\text{supply}}{\varepsilon_p \cdot \sigma}\) Plug in the given values: \(T_p^4 = 300^4 + \frac{289691\, W/m^2}{0.60 \cdot 5.67 \times 10^{-8}\, W/m^2\cdot K^4}\) \(T_p^4 = 9.135\times10^9\) Now, take the fourth root to find the plate temperature: \(T_p = \sqrt[4]{9.135\times10^9} = 410.79\, K\)