Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 117

In natural convection mass transfer, the Grashof number is evaluated using density difference instead of temperature difference. Can the Grashof number evaluated this way be used in heat transfer calculations also?

Short Answer

Expert verified
In conclusion, the Grashof number calculated using density differences, denoted as Gr', is equivalent to the Grashof number calculated using temperature differences, denoted as Gr. Therefore, Gr' can also be used in heat transfer calculations involving natural convection. The temperature difference remains the fundamental driving force in this scenario, and the modified Grashof number represents the same dependency using fluid densities while still relying on temperature differences.

Step by step solution

01

Understand natural convection mass transfer and Grashof number

Natural convection mass transfer occurs due to the differences in fluid density, which is induced by concentration or temperature gradients. Grashof number is a dimensionless number that characterizes the significance of buoyancy-induced flow in natural convection. The Grashof number (Gr) in heat transfer is defined as: Gr = \(\frac{g \beta (T_s - T_\infty) L^3}{\nu^2}\) where g is the acceleration due to gravity, \(\beta\) is the coefficient of thermal expansion, \(T_s\) is the surface temperature, \(T_\infty\) is the temperature of the undisturbed fluid, L is the characteristic length, and \(\nu\) is the kinematic viscosity of the fluid.
02

Derive the modified Grashof number using density differences

Let's denote the density of the fluid in the undisturbed state as \(\rho_\infty\) and the density of the fluid at the surface as \(\rho_s\). The density difference can be related to the temperature difference using the ideal gas law and the coefficient of thermal expansion: \(\rho_\infty - \rho_s = \rho_\infty \beta (T_s - T_\infty)\) Now we can define a modified Grashof number (Gr') using the density difference: Gr' = \(\frac{g (\rho_\infty - \rho_s) L^3}{\rho_\infty \nu^2}\) Substituting the expression for the density difference, we obtain: Gr' = \(\frac{g \rho_\infty \beta (T_s - T_\infty) L^3}{\rho_\infty \nu^2}\) We see that Gr' and Gr are identical in terms of temperature differences: Gr' = Gr
03

Discuss the applicability of the modified Grashof number in heat transfer calculations

Since the modified Grashof number (Gr') calculated using density differences is equivalent to the Grashof number (Gr) calculated using temperature differences, it can also be used in heat transfer calculations involving natural convection. The temperature difference still remains the fundamental driving force; the expression for Gr' just represents the same dependency using fluid densities, but it still relies on temperature differences.

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.

Start your free trial

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

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

What is the relation \(h_{\text {heat }}=\rho c_{p} h_{\text {mass }}\) known as? For what kind of mixtures is it valid? What is the practical importance of it?

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) \(6.07 \times 10^{-6} \mathrm{~kg}\)

A 2-mm-thick 5-L vessel made of nickel is used to store hydrogen gas at \(358 \mathrm{~K}\) and \(300 \mathrm{kPa}\). If the total inner surface area of the vessel is \(1600 \mathrm{~cm}^{2}\), determine the rate of gas loss from the nickel vessel via mass diffusion. Also, determine the fraction of the hydrogen lost by mass diffusion after one year of storage.

What is diffusion velocity? How does it affect the mass-average velocity? Can the velocity of a species in a moving medium relative to a fixed reference point be zero in a moving medium? Explain.

A glass bottle washing facility uses a well agi(Es) tated hot water bath at \(50^{\circ} \mathrm{C}\) with an open top that is placed on the ground. The bathtub is \(1 \mathrm{~m}\) high, \(2 \mathrm{~m}\) wide, and \(4 \mathrm{~m}\) long and is made of sheet metal so that the outer side surfaces are also at about \(50^{\circ} \mathrm{C}\). The bottles enter at a rate of 800 per minute at ambient temperature and leave at the water temperature. Each bottle has a mass of \(150 \mathrm{~g}\) and removes \(0.6 \mathrm{~g}\) of water as it leaves the bath wet. Makeup water is supplied at \(15^{\circ} \mathrm{C}\). If the average conditions in the plant are \(1 \mathrm{~atm}, 25^{\circ} \mathrm{C}\), and 50 percent relative humidity, and the average temperature of the surrounding surfaces is \(15^{\circ} \mathrm{C}\), determine (a) the amount of heat and water removed by the bottles themselves per second, \((b)\) the rate of heat loss from the top surface of the water bath by radiation, natural convection, and evaporation, \((c)\) the rate of heat loss from the side surfaces by natural convection and radiation, and \((d)\) the rate at which heat and water must be supplied to maintain steady operating conditions. Disregard heat loss through the bottom surface of the bath and take the emissivities of sheet metal and water to be \(0.61\) and \(0.95\), respectively.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Fields in Physics

Read Explanation

Oscillations

Read Explanation

Kinematics Physics

Read Explanation

Translational Dynamics

Read Explanation

Energy Physics

Read Explanation
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