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: The rate of heat transfer needed to maintain the water temperature in the pan is 2230.84 W.

Step-by-step solution

Calculate the area of the water surface

To determine the area of the water surface, use the formula for the area of a circle: \(A=\pi r^2\) The diameter is given as 30 cm, which means the radius is 15 cm, or 0.15 m. The area is then: \(A=\pi (0.15)^2 = 0.0707 \, \mathrm{m^2}\)

Determine the heat transfer coefficient

The heat transfer coefficient can be estimated using the empirical correlation for natural convection over a horizontal surface in a room: \(h = 1.31 (\Delta T)^{1/3}\) Where \(h\) is the heat transfer coefficient, \(\Delta T = 20 - 15 = 5 \, ^\circ \mathrm{C}\) is the difference between room temperature and water temperature. \(h = 1.31 (5)^{1/3} = 2.214 \, \mathrm{W/(m^2 \cdot K)}\)

Calculate the rate of heat transfer by convection

Calculate the rate of heat transfer by convection using the following formula: \(q = h \cdot A \cdot \Delta T\) Plug in the values: \(q = 2.214 \, \mathrm{W/(m^2\cdot K)} \cdot 0.0707 \, \mathrm{m^2} \cdot 5 \, \mathrm{K}= 0.78 \,W\) So the rate of heat transfer by convection is 0.78 W.

Calculate the rate of evaporation

The rate of evaporation can be estimated using the following formula: \(E = h_v \cdot A \cdot [\rho_{sat} - \rho_{air}]\) Where \(h_v\) is the mass transfer coefficient which is assumed the half of heat transfer coefficient as \(h_v = 0.5h\) and \(\rho_{sat}\) and \(\rho_{air}\) are the humidity ratios at the water surface and air respectively. At the given conditions, we can assume: \(\rho_{sat} = 0.0164 \, \mathrm{kg_{water}/kg_{dry\,air}}\) \(\rho_{air} = 0.3 \cdot 0.0164 = 0.00492 \, \mathrm{kg_{water}/kg_{dry\,air}}\) Now, calculate the rate of evaporation: \(E = (0.5 \cdot 2.214) \, \mathrm{W/(m^2\cdot K)} \cdot 0.0707 \, \mathrm{m^2} \cdot (0.0164 - 0.00492) \, \mathrm{kg_{water}/kg_{dry\,air}} = 0.00098\, \mathrm{kg/s}\) So, the rate of evaporation is 0.00098 kg/s.

Find the rate of heat transfer needed to maintain the temperature

To maintain the temperature, the rate of heat transfer needed should be equal to the sum of the rate of heat transfer by convection and the rate of heat transfer due to evaporation. To calculate the rate of heat transfer due to evaporation, we use the following formula which utilizes the latent heat of evaporation, \(L_v\): \(q_{evap} = E \cdot L_v\) Using the latent heat of evaporation (\(L_v \approx 2.27 \times 10^6 \, \mathrm{J/kg}\)) and the rate of evaporation, we can calculate the rate of heat transfer due to evaporation: \(q_{evap} = 0.00098\, \mathrm{kg/s} \times 2.27 \times 10^6 \, \mathrm{J/kg} = 2230.06\, W\) Add the rate of heat transfer by convection to find the total rate of heat transfer needed: \(q_{total} = q + q_{evap} = 0.78\, W + 2230.06\, W = 2230.84\, W\) So, the rate of heat transfer needed to maintain the temperature is 2230.84 W. In summary: (a) The rate of heat transfer by convection is 0.78 W. (b) The rate of evaporation is 0.00098 kg/s. (c) The rate of heat transfer needed to maintain the temperature is 2230.84 W.

Key concepts

Natural Convection Coefficient

The natural convection coefficient is a critical factor in understanding how heat transfers through fluids such as air or water without the assistance of an external source like a fan or pump. It's commonly used in calculations involving objects cooling or heating merely due to temperature differences.

Natural convection occurs because of the density variations in a fluid owing to temperature gradients – warmer fluid rises while cooler fluid sinks. In the example problem, we estimate this coefficient using an empirical correlation designed for horizontal surfaces in still air, which is influenced by the temperature difference \( \Delta T \) between the fluid and its surroundings.

The formula \( h = 1.31 (\Delta T)^{1/3} \) calculates the heat transfer coefficient \( h \) in watts per square meter per Kelvin (\( \mathrm{W/(m^2 \cdot K)} \) ), which is a measure of the heat transfer rate from the water to the air due to natural convection.

Rate of Evaporation

The rate of evaporation is a measure of how quickly a liquid turns into a vapor. This process can significantly affect the rate at which a body of water loses heat in an environment, as in our example where water in a pan evaporates into the air.

In the given problem, we calculated the rate of evaporation using the mass transfer coefficient, \( h_v \) , which was assumed to be half of the heat transfer coefficient, and the difference between the humidity ratios of saturated air at the water surface and the air. The underlying principle is that evaporation is more potent when the air is dry as opposed to when it's humid since dry air can 'absorb' more water vapor.

By using \( E = h_v \cdot A \cdot [\rho_{sat} - \rho_{air}] \) , we determine the mass of water evaporated per second (in kilograms per second), which, in this scenario, is 0.00098 \( \mathrm{kg/s} \) .

Convective Heat Transfer

Convective heat transfer is the movement of heat between a solid surface and a fluid, or between layers within a fluid, due to the fluid's motion over the surface or internal flow. In the given textbook problem, the heat transfer by convection from the water's surface to the surrounding air is what we seek.

Using the formula \( q = h \cdot A \cdot \Delta T \) , we calculate the rate at which heat is transferred through convection in watts (W). The factors involved include the natural convection coefficient \( h \) we previously determined, the area of the water surface \( A \) , and again the temperature differential \( \Delta T \) .

The result is the rate of heat transfer by convection, which for our exercise, is calculated to be 0.78 W, indicating a relatively low rate of heat transfer for the scenario provided.

Latent Heat of Evaporation

Latent heat of evaporation is the amount of heat energy required to transform a unit mass of a liquid into a gas at constant temperature and pressure. It's a vital concept in understanding how substances like water absorb and release energy during phase changes.

In the context of our problem, the latent heat of evaporation comes into play when calculating the total heat transfer needed to maintain the water's temperature as it evaporates. The formula \( q_{evap} = E \cdot L_v \) involves the rate of evaporation \( E \) and the latent heat \( L_v \) - for water, this value is approximately 2.27 x 10^6 joules per kilogram (J/kg).

When liquid water evaporates, it absorbs this large quantity of energy from its environment which must be replaced to maintain a constant temperature of the remaining water, hence, the significant value of 2230.84 W for the heat transfer rate required to sustain the water temperature in the example.