Consider a \(5-\mathrm{m} \times 5-\mathrm{m}\) wet concrete patio with an average water film thickness of \(0.3 \mathrm{~mm}\). Now wind at \(50 \mathrm{~km} / \mathrm{h}\) is blowing over the surface. If the air is at \(1 \mathrm{~atm}, 15^{\circ} \mathrm{C}\), and 35 percent relative humidity, determine how long it will take for the patio to dry completely.

Short Answer

Expert verified
Answer: No, we cannot determine the drying time for the patio due to the lack of information to calculate the rate of evaporation. Additional data or equations would be necessary to calculate the evaporation rate, taking into account the wind speed, air temperature, and humidity.

Step by step solution

01

Calculate the volume of water on the patio

To find the volume of water on the patio, we can multiply the patio area by the water film thickness. Make sure to convert the thickness from mm to meters. Patio area = \(5\,\text{m} \times 5\,\text{m} = 25\,\text{m}^2\) Water film thickness = \(0.3\,\text{mm} = 0.3 \times 10^{-3}\,\text{m}\) Volume of water = Patio area \(\times\) Water film thickness = \(25\,\text{m}^2 \times 0.3 \times 10^{-3}\,\text{m} = 7.5 \times 10^{-3}\,\text{m}^3\)
02

Calculate the rate of evaporation

The rate of evaporation depends on several factors, such as the wind speed, air temperature, and humidity. However, this exercise does not provide exact equations to determine the evaporation rate. As a result, we will assume the evaporation rate is directly proportional to the wind speed, air temperature, and relative humidity. To estimate the evaporation rate, we can use the following equation: Evaporation rate (\(\text{m}^3/\text{h}\)) = Constant \(\times\) Wind speed \(\times\) Temperature \(\times\) Humidity Since we don't have a value for the constant, we cannot calculate the evaporation rate. Unfortunately, without the evaporation rate, we cannot determine the drying time.
03

Conclusion

Due to the lack of information in the provided exercise, the drying time for the patio cannot be determined. To solve this problem, additional data or equations would be necessary to calculate the rate of evaporation, taking into account the wind speed, air temperature, and humidity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Heat Transfer
Heat transfer plays an essential role in the evaporation process. It involves the movement of thermal energy from one place to another through various means such as conduction, convection, and radiation.

Conduction is the transfer of heat between substances that are in direct contact with each other. Convection, on the other hand, involves the movement of heat by the physical movement of a fluid such as air or water. Finally, radiation is the transfer of heat through electromagnetic waves.

In the context of our problem, when wind blows over a wet concrete patio, it causes convective heat transfer from the air, which provides energy to the water film causing it to evaporate. The higher the temperature of the air and the greater the wind speed, the more energy is available for the water to convert from liquid to vapor, enhancing the evaporation rate.

This convective heat transfer is influenced by the air temperature and velocity; in our exercise, wind is blowing at 50 km/h, which increases the heat transfer to the water film and thus accelerates evaporation.
Mass Transfer
Mass transfer is another key concept integral to evaporation. It refers to the movement of mass from one location, usually within a fluid or between fluids, to another. Evaporation is a mass transfer process in which the molecules of a liquid escape into a gas phase, becoming vapor.

Factors Affecting Mass Transfer in Evaporation

Several factors impact the rate of mass transfer during the evaporation process:
  • Concentration Gradient: The difference in vapor concentration between the liquid interface and the surrounding air.
  • Wind Velocity: The higher the wind speed, the more quickly vapor is removed from the surface, enhancing the mass transfer.
  • Temperature: Higher temperatures typically increase the kinetic energy of molecules, making it easier for them to break into the gas phase.
  • Relative Humidity: Low humidity aids evaporation by maintaining a large concentration gradient; the air can accept more water vapor.
In our scenario, wind velocity and relative humidity greatly influence the mass transfer rate of water vapor from the patio surface into the air.
Evaporation Process
The final key concept to understand is the evaporation process itself. Evaporation is the transformation of water from its liquid state into a gas, also known as water vapor. It occurs when molecules at the surface of the liquid have enough kinetic energy to break free from the attraction of other liquid molecules and become a gas.

Evaporation in Open Air

In open-air conditions, like with our concrete patio, the evaporation process can be quite complex. It's not only governed by the temperature and humidity, but also by factors such as:
  • Wind speed which aids in dispersing the water vapor away from the surface, allowing more water to evaporate.
  • Sunlight, which can provide the heat energy required to increase the kinetic energy of the water molecules.
  • The area of the evaporating surface, affecting how much water can evaporate at once.
In the patio drying problem, we assume that heat and mass transfer combine to achieve evaporation. But without specific values or formulas, the rate of evaporation and therefore the time for the patio to dry cannot be accurately determined.

For practical problem solving, real-time measurements or empirical data would help to calculate the constant in the assumed evaporation rate equation, providing students with the ability to predict the drying time under varied conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Dry air whose molar analysis is \(78.1\) percent \(\mathrm{N}_{2}\), \(20.9\) percent \(\mathrm{O}_{2}\), and 1 percent Ar flows over a water body until it is saturated. If the pressure and temperature of air remain constant at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) during the process, determine (a) the molar analysis of the saturated air and \((b)\) the density of air before and after the process. What do you conclude from your results?

Benzene \((M=78.11 \mathrm{~kg} / \mathrm{kmol})\) is a carcinogen, and exposure to benzene increases the risk of cancer and other illnesses in humans. A truck transporting liquid benzene was involved in an accident that spilled the liquid on a flat highway. The liquid benzene forms a pool of approximately \(10 \mathrm{~m}\) in diameter on the highway. In this particular windy day at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) with an average wind velocity of \(10 \mathrm{~m} / \mathrm{s}\), the liquid benzene surface is experiencing mass transfer to air by convection. Nearby at the downstream of the wind is a residential area that could be affected by the benzene vapor. Local health officials have assessed that if the benzene level in the air reaches \(500 \mathrm{~kg}\) within the hour of the spillage, residents should be evacuated from the area. If the benzene vapor pressure is \(10 \mathrm{kPa}\), estimate the mass transfer rate of benzene being convected to the air, and determine whether the residents should be evacuated or not.

What is the physical significance of the Sherwood number? How is it defined? To what dimensionless number does it correspond in heat transfer? What does a Sherwood number of 1 indicate for a plane fluid layer?

Consider a glass of water in a room at \(20^{\circ} \mathrm{C}\) and \(97 \mathrm{kPa}\). If the relative humidity in the room is 100 percent and the water and the air are in thermal and phase equilibrium, determine (a) the mole fraction of the water vapor in the air and \((b)\) the mole fraction of air in the water.

Saturated water vapor at \(25^{\circ} \mathrm{C}\left(P_{\text {sat }}=3.17 \mathrm{kPa}\right)\) flows in a pipe that passes through air at \(25^{\circ} \mathrm{C}\) with a relative humidity of 40 percent. The vapor is vented to the atmosphere through a \(7-\mathrm{mm}\) internal-diameter tube that extends \(10 \mathrm{~m}\) into the air. The diffusion coefficient of vapor through air is \(2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). The amount of water vapor lost to the atmosphere through this individual tube by diffusion is (a) \(1.02 \times 10^{-6} \mathrm{~kg}\) (b) \(1.37 \times 10^{-6} \mathrm{~kg}\) (c) \(2.28 \times 10^{-6} \mathrm{~kg}\) (d) \(4.13 \times 10^{-6} \mathrm{~kg}\) (e) \(6.07 \times 10^{-6} \mathrm{~kg}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free