Question

A 2-m-deep 20-m \(\times 20-\mathrm{m}\) heated swimming pool is maintained at a constant temperature of \(30^{\circ} \mathrm{C}\) at a location where the atmospheric pressure is \(1 \mathrm{~atm}\). If the ambient air is at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity and the effective sky temperature is \(0^{\circ} \mathrm{C}\), determine the rate of heat loss from the top surface of the pool by ( \(a\) ) radiation, \((b)\) natural convection, and (c) evaporation. ( \(d\) ) Assuming the heat losses to the ground to be negligible, determine the size of the heater.

Short answer

To maintain the temperature of a heated swimming pool with dimensions of \(20\,\text{m} \times 20\,\text{m}\) and a surface temperature of \(30^{\circ}\mathrm{C}\), you would need a heater with a capacity of approximately \(101.5\,\text{kW}\). This calculation takes into account the heat loss due to radiation, natural convection, and evaporation.

Step-by-step solution

Calculate the surface area of the pool

The swimming pool dimensions are \(20\,\text{m} \times 20\,\text{m}\). Thus, the surface area of the swimming pool is: \(A_s = \text{length} *\text{width} = 20\,\text{m} * 20\,\text{m} = 400\,\text{m}^2\)

Compute the radiation heat loss rate

We are given the surface temperature of the pool (\(30^{\circ}\mathrm{C}\)), which we will first convert to Kelvin: \(T_{surf} = 30+273.15 = 303.15\,\text{K}\) Next, we will convert the effective sky temperature to Kelvin: \(T_{eff} = 0+273.15 = 273.15\,\text{K}\) Now, let's calculate the radiation heat loss rate using the given formula, assuming an emissivity of 0.95 for the water surface: \(q_{rad} = ε * σ * A_s * (T_surf^4- T_sky^4) = 0.95 * 5.67*10^{-8}\,\text{W/m}^2\mathrm{K}^4 * 400\,\text{m}^2 * (303.15^4 - 273.15^4)\,\mathrm{K}^4\) \(q_{rad} = 16402\,\text{W}\)

Compute the natural convection heat loss rate

We will use an average value for the convection heat transfer coefficient \(h_c\) as \(15\,\text{W/m}²\mathrm{K}\) (a reasonable value for natural convection). Then, we can calculate the natural convection heat loss rate: \(q_{conv} = h_c * A_s * (T_surf - T_{\infty}) = 15\,\text{W/m}^2\mathrm{K} * 400\,\text{m}^2 * (303.15-293.15)\,\mathrm{K}\) \(q_{conv} = 60000\,\text{W}\)

Compute the evaporation heat loss rate

To compute the evaporation heat loss rate, we need to determine the mass flow rate of evaporated water \(ṁ\). For this problem, we'll assume that all water evaporates at the pool's surface at a constant rate. The evaporation rate can be estimated as \(0.1\,\text{kg/h}\mathrm{m}^2\). So, the mass flow rate is: \(ṁ= A_s * \text{evaporation rate} = 400\,\text{m}^2 * 0.1\,\text{kg/h}\mathrm{m}^2 = 40\,\text{kg/h}\) Now, we can calculate the evaporation heat loss rate: \(q_{evap} = ṁ * L_v = \frac{40}{3600}\,\text{kg/s} * 2.26 * 10^6\,\text{J/kg}\) \(q_{evap} = 25111\,\text{W}\)

Determine the size of the heater

Assuming the heat losses to the ground are negligible, the heater needs to supply the combined heat loss of radiation, natural convection, and evaporation: \(q_{total} = q_{rad} + q_{conv} + q_{evap} = 16402\,\text{W} + 60000\,\text{W} + 25111\,\text{W} = 101513\,\text{W}\) The size of the heater needed is approximately \(101,513\,\text{W}\) or \(101.5\,\text{kW}\).

Key concepts

Radiation Heat Loss

Radiation heat loss occurs when heat energy is emitted from a warm body to its cooler surroundings in the form of electromagnetic waves. In the context of a swimming pool, the pool surface radiates heat to the sky, which has a lower effective temperature. This process is characterized by several key components:

• Emissivity (\(\varepsilon\)), a measure of a material's ability to emit energy, is crucial. For water, it's around 0.95, meaning it radiates 95% of the energy it absorbs.
• The Stefan-Boltzmann constant (\(\sigma\)), used in the calculation of radiation heat loss, is \(5.67 \times 10^{-8} \,\mathrm{W/m}^2\mathrm{K}^4\).
• The temperature difference between the pool surface and the sky, calculated in Kelvin, affects the rate of heat loss significantly.

Using these parameters, the formula for radiation heat loss is given by:\[ q_{rad} = \varepsilon \cdot \sigma \cdot A_s \cdot (T_{\text{surf}}^4 - T_{\text{sky}}^4) \]where \(A_s\) is the surface area. Understanding these basics allows you to comprehend how energy transfer via radiation impacts thermal management in pools.

Natural Convection

Natural convection is the movement of fluid (in this case, air) that results from the fluid being heated. When the pool water heats the air just above it, this warmer, less dense air rises, while cooler air descends to take its place, setting up a convection current. This phenomenon results in heat being transferred from the pool surface to the surrounding air.

The rate of natural convection heat loss depends on:

• The surface area, which in our example is 400 m².
• The temperature difference between the pool surface and the ambient air, which dictates how vigorously the air circulates.
• The convective heat transfer coefficient (\(h_c\)), an empirical value reliant on conditions like surface roughness and air velocity. Here, it's approximated as 15 W/m²K.

The formula for natural convection heat loss is:\[ q_{conv} = h_c \cdot A_s \cdot (T_{\text{surf}} - T_{\infty}) \]where \(T_{\text{surf}}\) and \(T_{\infty}\) are the surface and air temperatures, respectively. This process is a key consideration in regulating pool temperature and energy consumption.

Evaporation Heat Loss

Evaporation heat loss describes the energy required to convert water from a liquid to a vapor, an endothermic process that occurs at the pool's surface as it is exposed to air. This cooling effect is actually a very efficient form of heat loss.

Key factors influencing evaporation include:

• Surface area: A larger surface means more water is exposed, leading to higher evaporation rates.
• The water temperature: Higher temperatures speed up molecular motion, increasing evaporation.
• Air humidity and temperature: Drier or warmer air can hold more water vapor, driving more evaporation from the pool.

The rate of evaporation can be calculated using an assumed rate (e.g., 0.1 kg/h per square meter for pools), and the latent heat of vaporization (\(L_v\)), which is around \(2.26 \times 10^6 \, \mathrm{J/kg}\) for water.
The formula to find evaporation heat loss is:\[ q_{evap} = \dot{m} \cdot L_v \]where \(\dot{m}\) is the mass flow rate of evaporated water. This highlights evaporation as a significant contributor to overall heat loss in swimming pools.

Thermodynamics

Thermodynamics is the science of energy transfer and transformation, helping us understand how heat flows within and between different systems. In our swimming pool example, thermodynamic principles enable us to calculate various heat losses and determine heater requirements.

Key concepts within thermodynamics related to this problem include:

• First Law of Thermodynamics: Often expressed as energy conservation, stating that energy cannot be created or destroyed, only transferred or changed in form.
• Heat Transfer Modes: The pool loses heat through radiation, convection, and evaporation, each governed by specific principles and relationships.
• Enthalpy: Particularly for evaporation, it defines the energy needed to turn liquid water to vapor, integrating directly into the energy balance.

To maintain the pool's temperature, a heater compensates for total heat losses. Consequently, calculating these losses precisely, as demonstrated in earlier sections, ensures efficient thermal management. By considering the pool's losses through radiation, convection, and evaporation, we integrate these concepts to specify the correct heater size, ensuring a comfortable swimming experience all year round.