Question

A water-soaked \(10-\mathrm{cm} \times 10\)-cm-square sponge is experiencing parallel dry airflow over its surface. The average convection heat transfer coefficient of dry air at \(20^{\circ} \mathrm{C}\) flowing over the sponge surface is estimated to be \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the sponge surface is maintained at \(30^{\circ} \mathrm{C}\). If the sponge is placed under an array of radiant lamps, determine \((a)\) the rate of evaporation of water from the sponge and \((b)\) the net radiation heat transfer rate.

Short answer

a) The rate of evaporation of water from the sponge is approximately \(1.33\ \times 10^{-7}\ \mathrm{kg/s}\). b) To calculate the net radiation heat transfer rate (\(Q_\text{rad}\)), we would need the temperature of the radiant lamps. If that information was provided, we could use the Stefan-Boltzmann Law with the given formula to find the net radiation heat transfer rate.

Step-by-step solution

Calculate the convection heat transfer rate between sponge and air

To find the convection heat transfer rate (\(Q_\text{conv}\)) between the sponge and the air, we can use the following formula: \(Q_\text{conv} = hA(T_\text{s} - T_\text{air})\) Where: - \(h\) is the average convection heat transfer coefficient (\(30 \ \mathrm{W/m^2 \cdot K}\)) - \(A\) is the surface area of the sponge (\(10 \ \mathrm{cm} \times 10 \ \mathrm{cm} = 0.01\ \mathrm{m}\times 0.01 \ \mathrm{m} = 0.0001 \ \mathrm{m^2}\)) - \(T_\text{s}\) is the sponge surface temperature (\(30^{\circ} \ \mathrm{C}\)) - \(T_\text{air}\) is the air temperature (\(20^{\circ} \ \mathrm{C}\)). We can plug the known values to find the convection heat transfer rate: \(Q_\text{conv} = 30 \ \mathrm{W/m^2 \cdot K} \times 0.0001 \ \mathrm{m^2}(30^{\circ} \ \mathrm{C} - 20^{\circ} \ \mathrm{C}) \) \(Q_\text{conv} = 0.3 \ \mathrm{W}\)

Calculate the rate of evaporation of water

We know that \(Q_\text{conv}\) must be equal to the heat transfer rate due to the evaporation of water (\(Q_\text{evap}\)), which can be calculated using the following formula: \(Q_\text{evap} = m_\text{evap} L_\text{v}\) Where: - \(m_\text{evap}\) is the mass flow rate of evaporated water (the rate of evaporation, in \(\mathrm{kg/s}\)) - \(L_\text{v}\) is the latent heat of vaporization (\(\approx 2.25\ \times 10^6 \ \mathrm{J/kg}\) for water at \(100^{\circ} \ \mathrm{C}\)) We can rearrange this formula to find the rate of evaporation: \(m_\text{evap} = \frac{Q_\text{evap}}{L_\text{v}} = \frac{Q_\text{conv}}{L_\text{v}}\) Substitute the values: \(m_\text{evap}= \frac{0.3\ \mathrm{W}}{2.25\ \times 10^{6}\ \mathrm{J/kg}}\) \(m_\text{evap} \approx 1.33\ \times 10^{-7}\ \mathrm{kg/s}\) (a)

Calculate the net radiation heat transfer rate

From the given information, we know that the sponge is placed under an array of radiant lamps. To find the net radiation heat transfer rate (\(Q_\text{rad}\)), we will first calculate the radiative heat transfer from the lamps to the sponge (assuming it absorbs all incident radiation), and then subtract the radiation emitted by the sponge. Radiation heat transfer can be calculated using the Stefan-Boltzmann Law: \(Q_\text{rad} = \sigma A_\text{s} (T_\text{l}^4 - T_\text{s}^4)\) Where: - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \ \mathrm{W/m^2 \cdot K^4}\)) - \(A_\text{s}\) is the surface area of the sponge (\(0.0001 \ \mathrm{m^2}\)) - \(T_\text{l}\) is the temperature of the radiant lamps (unknown) - \(T_\text{s}\) is the sponge surface temperature (in Kelvin; \(30^{\circ} \ \mathrm{C} = 303 \ \mathrm{K}\)). The problem does not give the temperature of the radiant lamps, so we cannot calculate the exact value of \(Q_\text{rad}\). However, if the temperature of the lamps were provided, we could calculate the net radiation heat transfer rate using the above formula. (b)