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Chapter 14: Problem 152

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

Expert verified
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

01

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: \(A_s = \text{length} *\text{width} = 20\,\text{m} * 20\,\text{m} = 400\,\text{m}^2\)
02

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: \(q_{rad} = ε * σ * A_s * (T_surf^4- T_sky^4) = 0.95 * 5.67*10^{-8}\,\text{W/m}^2\mathrm{K}^4 * 400\,\text{m}^2 * (303.15^4 - 273.15^4)\,\mathrm{K}^4\) \(q_{rad} = 16402\,\text{W}\)
03

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: \(q_{conv} = h_c * A_s * (T_surf - T_{\infty}) = 15\,\text{W/m}^2\mathrm{K} * 400\,\text{m}^2 * (303.15-293.15)\,\mathrm{K}\) \(q_{conv} = 60000\,\text{W}\)
04

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: \(q_{evap} = ṁ * L_v = \frac{40}{3600}\,\text{kg/s} * 2.26 * 10^6\,\text{J/kg}\) \(q_{evap} = 25111\,\text{W}\)
05

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 \(101.5\,\text{kW}\).

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Most popular questions from this chapter

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