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During a hot summer day, a \(2-L\) bottle drink is to be cooled by wrapping it in a cloth kept wet continually and blowing air to it with a fan. If the environment conditions are \(1 \mathrm{~atm}, 80^{\circ} \mathrm{F}\), and 30 percent relative humidity, determine the temperature of the drink when steady conditions are reached.

Short Answer

Expert verified
Answer: The steady-state temperature of the drink will be approximately 72.9°F.

Step by step solution

01

Calculate the wet-bulb temperature

First, we need to find the wet-bulb temperature to determine the temperature to which the drink will be cooled. The wet-bulb temperature can be found using a psychrometric chart or equations. For simplicity, we will use the following empirical formula to estimate the wet-bulb temperature in Celsius: \(T_{wb} = T_a - 0.56(1 - RH)(T_a - 14.3) \) Where \(T_{wb}\) is the wet-bulb temperature in Celsius, \(T_a\) is the ambient temperature in Celsius, and \(RH\) is the relative humidity in decimals. Convert the given ambient temperature from Fahrenheit to Celsius: \(T_a (^{\circ}C) = \frac{80 - 32}{1.8} = 26.67 ^{\circ}C\) Now, we can calculate the wet-bulb temperature: \(T_{wb} = 26.67 - 0.56(1 - 0.3)(26.67 - 14.3) \) \(T_{wb} = 26.67 - 0.56(0.7)(12.37) \) \(T_{wb} ≈ 22.7 ^{\circ}C \)
02

Convert wet-bulb temperature to Fahrenheit

Now that we have the wet-bulb temperature in Celsius, we need to convert it back to Fahrenheit so that we can express the steady-state temperature of the drink in the original unit: \(T_{drink} (^{\circ}F) = 1.8(T_{wb}) + 32\) \(T_{drink} (^{\circ}F) = 1.8(22.7) + 32\) \(T_{drink} ≈ 72.9 ^{\circ}F\)
03

Determine steady-state temperature of the drink

The steady-state temperature of the 2-L bottle drink will be the wet-bulb temperature since the evaporative cooling process will cool the drink down to that temperature. Therefore, the drink will reach a temperature of: \(T_{drink} ≈ 72.9 ^{\circ}F\) When steady conditions are reached, the temperature of the drink will be approximately 72.9°F.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wet-Bulb Temperature
Wet-bulb temperature is a key concept when dealing with evaporative cooling. It tells us how cool an item can get if it is surrounded by moisture and exposed to a flow of air. Imagine dipping a thermometer in water and allowing air to blow over it. The temperature reading when no more evaporation occurs is the wet-bulb temperature. It's usually lower than regular air temperature due to the cooling effect of evaporation.
This measurement is integral in processes like air conditioning and meteorology. To calculate the wet-bulb temperature, we often refer to complex equations or tools like a psychrometric chart. However, for everyday uses, you might use simpler formulas. In the example exercise, an empirical formula was used to convert temperature readings. This allows for an efficient way to estimate cool temperatures, even with limited data. Once you find this temperature, you're discovering how cool a surface can theoretically become during evaporation under specific environmental conditions.
Psychrometric Chart
A psychrometric chart is a valuable tool in thermodynamics for illustrating the properties of moist air. By displaying various atmospheric temperature attributes, it helps you visualize how changes in one property affect another.
This essential chart can tell you a lot about the air's state, such as temperature, humidity, vapor content, and even the dew point. Reading the chart might look daunting at first, but once you understand it, it becomes an indispensable guide. It uses axes to mark dry-bulb temperature (the regular thermometer reading) and includes curves representing constant relative humidity levels. By finding the intersection of two known properties, you can determine others. For our problem, this chart could locate the wet-bulb temperature and help deduce the cooling potential through evaporative cooling, making it a central part of HVAC systems and environmental control.
Relative Humidity
Relative humidity represents how much moisture is in the air compared to the maximum amount it can hold at a given temperature. Understanding this concept is critical not only in weather forecasting but also for explaining how evaporative cooling works. When the relative humidity is low, evaporation occurs more effectively, leading to significant cooling effects.
High humidity reduces this efficiency since the air is already saturated with moisture. Consider the exercise scenario where the relative humidity was 30%. This implies that only 30% of the air's moisture capacity was filled, providing ample space for evaporation to occur. Relative humidity ranges from 0% (completely dry air) to 100% (fully saturated air). Operating at lower relative humidity levels facilitates processes such as drying clothes in the sun and, as seen in our example, cooling a drink using a wet cloth and fan. This underscores how crucial it is to factor in humidity when anticipating the cooling effect of evaporative techniques.

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Most popular questions from this chapter

Liquid methanol is accidentally spilt on a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) laboratory bench and covered the entire bench surface. A fan is providing a \(20 \mathrm{~m} / \mathrm{s}\) air flow parallel over the bench surface. The air is maintained at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), and the concentration of methanol in the free stream is negligible. If the methanol vapor at the air-methanol interface has a pressure of \(4000 \mathrm{~Pa}\) and a temperature of \(25^{\circ} \mathrm{C}\), determine the evaporation rate of methanol in molar basis.

The roof of a house is \(15 \mathrm{~m} \times 8 \mathrm{~m}\) and is made of a 20 -cm-thick concrete layer. The interior of the house is maintained at \(25^{\circ} \mathrm{C}\) and 50 percent relative humidity and the local atmospheric pressure is \(100 \mathrm{kPa}\). Determine the amount of water vapor that will migrate through the roof in \(24 \mathrm{~h}\) if the average outside conditions during that period are \(3^{\circ} \mathrm{C}\) and 30 percent relative humidity. The permeability of concrete to water vapor is \(24.7 \times 10^{-12} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m} \cdot \mathrm{Pa}\).

Helium gas is stored at \(293 \mathrm{~K}\) in a 3 -m-outer-diameter spherical container made of 5 -cm-thick Pyrex. The molar concentration of helium in the Pyrex is \(0.00073 \mathrm{kmol} / \mathrm{m}^{3}\) at the inner surface and negligible at the outer surface. Determine the mass flow rate of helium by diffusion through the Pyrex container.

Determine the mole fraction of dry air at the surface of a lake whose temperature is \(15^{\circ} \mathrm{C}\). Take the atmospheric pressure at lake level to be \(100 \mathrm{kPa}\).

Nitrogen gas at high pressure and \(298 \mathrm{~K}\) is contained in a \(2-\mathrm{m} \times 2-\mathrm{m} \times 2-\mathrm{m}\) cubical container made of natural rubber whose walls are \(4 \mathrm{~cm}\) thick. The concentration of nitrogen in the rubber at the inner and outer surfaces are \(0.067 \mathrm{~kg} / \mathrm{m}^{3}\) and \(0.009 \mathrm{~kg} / \mathrm{m}^{3}\), respectively. The diffusion coefficient of nitrogen through rubber is \(1.5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\). The mass flow rate of nitrogen by diffusion through the cubical container is (a) \(8.24 \times 10^{-10} \mathrm{~kg} / \mathrm{s}\) (b) \(1.35 \times 10^{-10} \mathrm{~kg} / \mathrm{s}\) (c) \(5.22 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (d) \(9.71 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (e) \(3.58 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\)

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