Question

Consider a 30 -cm-diameter pan filled with water at \(15^{\circ} \mathrm{C}\) in a room at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}\), and 30 percent relative humidity. Determine \((a)\) the rate of heat transfer by convection, (b) the rate of evaporation of water, and \((c)\) the rate of heat transfer to the water needed to maintain its temperature at \(15^{\circ} \mathrm{C}\). Disregard any radiation effects.

Short answer

Answer: (a) The rate of heat transfer by convection is \(-35.34\mathrm{~W}\). (b) The rate of evaporation of water is \(4.18 \times 10^{-11} \mathrm{kg/s}\). (c) The rate of heat transfer to the water needed to maintain its temperature at \(15^{\circ} \mathrm{C}\) is \(44.83\mathrm{~W}\).

Step-by-step solution

Calculate the convective heat transfer coefficient

We will assume the convective heat transfer coefficient (\(h\)) to be constant. For a natural convection problem like this, a reasonable value for \(h\) would be around \(10 \mathrm{~W/(m^2K)}\).

Calculate the rate of heat transfer by convection

Next, we need to find the rate of heat transfer by convection (\(q_{conv}\)). To do this, we will use Newton's Law of Cooling: \(q_{conv} = hA(T_s-T_\infty)\) In this equation, \(A\) is the surface area of the pan (\(\pi(0.15)^2 = 0.07069 \mathrm{~m^2}\)), \(T_s\) is the water temperature \((15^{\circ} \mathrm{C})\), and \(T_\infty\) is the room temperature \((20^{\circ} \mathrm{C})\): \(q_{conv} = 10 \cdot 0.07069 \cdot (15 - 20) = -35.34\mathrm{~W}\) Since the rate is negative, this means that heat is being transferred from the water to the room through convection.

Find the rate of evaporation due to the vapour pressure difference

Next, we need to find the evaporation rate (\(m_{e}\)) due to the difference in vapour pressure between the water surface and the air. We will use the following equation for mass transfer: \(m_{e} = k_m A (c_s - c_\infty)\) In this equation, \(k_m\) is the mass transfer coefficient, \(A\) is the surface area of the pan, \(c_s\) is the concentration at the water surface, and \(c_\infty\) is the concentration at the air. For simplicity, let's assume \(k_m = 5 \times 10^{-8} \mathrm{kg/(m^2~s)}\), \(c_s = 0.0173 \mathrm{kg/m^3}\) (saturated air with \(100\%\) moisture) and \(c_\infty = 0.30 \times 0.0173 = 0.00519 \mathrm{kg/m^3}\) (as given by the \(30 \%\) relative humidity): \(m_{e} = 5 \times 10^{-8} \cdot 0.07069 \cdot (0.0173 - 0.00519) = 4.18 \times 10^{-11} \mathrm{kg/s}\)

Calculate the rate of heat transfer due to evaporation

We can find the rate of heat transfer due to evaporation (\(q_{evap}\)) by using the heat of vaporization (\(H_v = 2.27 \times 10^6 \mathrm{W/kg}\)): \(q_{evap} = m_{e} H_v = 4.18 \times 10^{-11} \cdot 2.27 \times 10^6 = 9.49\mathrm{~W}\)

Calculate the rate of heat transfer required to maintain the water temperature

Finally, we can calculate the rate of heat transfer to the water (\(q_{in}\)) needed to maintain its temperature at \(15^{\circ} \mathrm{C}\) by combining the rates of heat transfer due to convection and evaporation. Apply the energy balance: \(q_{in} = -q_{conv} + q_{evap}\) \(q_{in} = -(-35.34\mathrm{~W}) + 9.49\mathrm{~W} = 44.83\mathrm{~W}\) So, the rate of heat transfer needed to maintain the water temperature at \(15^{\circ} \mathrm{C}\) is \(44.83\mathrm{~W}\). #Results# (a) The rate of heat transfer by convection is \(-35.34\mathrm{~W}\). (b) The rate of evaporation of water is \(4.18 \times 10^{-11} \mathrm{kg/s}\). (c) The rate of heat transfer to the water needed to maintain its temperature at \(15^{\circ} \mathrm{C}\) is \(44.83\mathrm{~W}\).