/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Consider cross flow of gas \(\ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider cross flow of gas \(\mathrm{X}\) over an object having a characteristic length of \(L=0.1 \mathrm{~m}\). For a Reynolds number of \(1 \times 10^{4}\), the average heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The same object is then impregnated with liquid \(Y\) and subjected to the same flow conditions. Given the following thermophysical properties, what is the average convection mass transfer coefficient?

Short Answer

Expert verified
The average convection mass transfer coefficient (k_c) can be calculated using the Chilton-Colburn analogy and the given information about the flow and heat transfer properties. However, as the problem does not provide any information about the thermophysical properties of liquid Y, such as its diffusion coefficient (D) and thermal conductivity (k), we cannot complete the calculations. To solve the problem, these properties or correlations for them in terms of Reynolds number (\(\mathrm{Re}\)) or Prandtl number (\(\mathrm{Pr}\)) would be needed.

Step by step solution

01

List the given information

The exercise gives the following information: - Characteristic length, \(L = 0.1 \mathrm{~m}\) - Reynolds number, \(\mathrm{Re} = 1 \times 10^{4}\) - Average heat transfer coefficient, \(h = 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)
02

Determine the Prandtl number

The Chilton-Colburn analogy requires the Prandtl number (\(\mathrm{Pr}\)), which is not given in the problem. However, we know the Reynolds number (\(\mathrm{Re}\)) and heat transfer coefficient (h), and we can use the definition of Nusselt number (\(\mathrm{Nu}\)) to find the Prandtl number: \[\mathrm{Nu} = \frac{h L}{k} = \mathrm{Re} \cdot \mathrm{Pr} \cdot f(\mathrm{Re}, \mathrm{Pr})\] Here, \(k\) is the thermal conductivity of the fluid, and \(f(\mathrm{Re}, \mathrm{Pr})\) is a function that depends on the flow conditions and geometry. Different correlations exist for this function in the literature.
03

Find the function \(f(\mathrm{Re}, \mathrm{Pr})\)

In this exercise, it is not specified which correlation should be used to find the function \(f(\mathrm{Re}, \mathrm{Pr})\). We will assume a simple correlation in the form of: \[f(\mathrm{Re}, \mathrm{Pr}) = \mathrm{Re}^{-n} \mathrm{Pr}^{m}\] where n and m are constants. In such a case, the Nusselt number can be written as: \[\mathrm{Nu} = \mathrm{Re}^{1-n} \mathrm{Pr}^{1+m}\] Now we can solve for the Prandtl number in terms of the Nusselt number and Reynolds number: \[\mathrm{Pr} = \left(\frac{\mathrm{Nu}}{\mathrm{Re}^{1-n}}\right)^{1/(1+m)}\]
04

Apply the Chilton-Colburn analogy

The Chilton-Colburn analogy relates the mass transfer (Sherwood number, \(\mathrm{Sh}\)) and heat transfer (Nusselt number, \(\mathrm{Nu}\)): \(\mathrm{Sh} = \mathrm{Nu}\), where Sherwood number is defined as \(\mathrm{Sh} = \frac{k_c L}{D}\), and \(k_c\) is the mass transfer coefficient, and \(D\) is the diffusion coefficient of the impregnating liquid Y.
05

Calculate the average convection mass transfer coefficient

Now we can use the fact that \(\mathrm{Sh}=\mathrm{Nu}\) to find the mass transfer coefficient: \[k_c = \frac{h D}{k}\] We need the diffusion coefficient (D) and the thermal conductivity (k) of the impregnating liquid Y to calculate the average convection mass transfer coefficient (k_c). However, the exercise does not provide any information about the thermophysical properties of liquid Y, so we cannot proceed further without that data. To solve the exercise, we will need complete information about liquid Y's thermophysical properties or correlations for these properties in terms of the Reynolds number (\(\mathrm{Re}\)) or Prandtl number (\(\mathrm{Pr}\)).

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 91Ó°ÊÓ!

Key Concepts

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

Reynolds number
The Reynolds number (\( \mathrm{Re} \)) is a dimensionless quantity in fluid mechanics that helps predict flow patterns in different fluid flow situations. It is used to determine whether the flow is laminar or turbulent. A low Reynolds number indicates laminar flow, where fluid moves in parallel layers with minimal disruption. A high Reynolds number suggests turbulent flow, characterized by chaotic changes in pressure and flow velocity.

The Reynolds number is defined as:\[\mathrm{Re} = \frac{\rho v L}{\mu} = \frac{v L}{u}\]where:
  • \( \rho \) is the fluid density,
  • \( v \) is the flow velocity,
  • \( L \) is the characteristic length,
  • \( \mu \) is the dynamic viscosity,
  • \( u \) is the kinematic viscosity, the ratio of dynamic viscosity to density.
In the given problem, the Reynolds number helps in analyzing the flow conditions over the object, providing insight into the behavior of the gas flowing across the surface.
Chilton-Colburn analogy
The Chilton-Colburn analogy is a useful concept that connects heat, mass, and momentum transfer in fluid flow processes. It provides a relation between dimensionless numbers: the Nusselt number (\( \mathrm{Nu} \)), the Sherwood number (\( \mathrm{Sh} \)), and the friction factor. This analogy makes it easier to predict mass transfer from known heat transfer data and vice versa.

The analogy suggests that similar equations define both heat and mass transfer processes when dimensionless numbers are used:\[\mathrm{Nu} = \mathrm{Sh} = \left( \mathrm{Pr} \cdot \mathrm{Sc} \right)^{1/3}\]where Sc is the Schmidt number, another dimensionless number in fluid mechanics. This relationship assumes that heat and mass transfer occur similarly due to flow similarities, equating the Nusselt number (heat transfer) to the Sherwood number (mass transfer), leveraging known data to infer unknown properties.
Nusselt number
The Nusselt number (\( \mathrm{Nu} \)) is a key concept in heat transfer, signifying the ratio of convective to conductive heat transfer across a boundary. A higher Nusselt number indicates better convective heat transfer. It's particularly useful for understanding the effectiveness of heat exchangers.

The Nusselt number is defined by the equation:\[\mathrm{Nu} = \frac{hL}{k}\]where:
  • \( h \) is the convective heat transfer coefficient,
  • \( L \) is the characteristic length,
  • \( k \) is the thermal conductivity of the fluid.
In this problem, knowing the Nusselt number allows us to use the Chilton-Colburn analogy to connect heat transfer to mass transfer, facilitating the calculation of the mass transfer coefficient when certain values are unknown.
Prandtl number
The Prandtl number (\( \mathrm{Pr} \)) is a dimensionless number, essential in the study of heat transfer in flow processes. It represents the ratio of momentum diffusivity to thermal diffusivity, providing insights into the relative thickness of the velocity and thermal boundary layers.

The Prandtl number is calculated as follows:\[\mathrm{Pr} = \frac{\mu c_p}{k} = \frac{u}{\alpha}\]where:
  • \( \mu \) is the dynamic viscosity,
  • \( c_p \) is the specific heat at constant pressure,
  • \( k \) is the thermal conductivity,
  • \( u \) is the kinematic viscosity,
  • \( \alpha \) is the thermal diffusivity.
In the context of our exercise, the Prandtl number is vital for calculating the Nusselt number when using correlations, as it helps define the fluid's thermal behavior relative to its momentum. Understanding the Prandtl number's role aids in solving complex heat and mass transfer problems efficiently.

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

If laminar flow is induced at the surface of a disk due to rotation about its axis, the local convection coefficient is known to be a constant, \(h=C\), independent of radius. Consider conditions for which a disk of radius \(r_{o}=100 \mathrm{~mm}\) is rotating in stagnant air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and a value of \(C=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. If an embedded electric heater maintains a surface temperature of \(T_{x}=50^{\circ} \mathrm{C}\), what is the local heat flux at the top surface of the disk? What is the total electric power requirement? What can you say about the nature of boundary layer development on the disk?

It is known that on clear nights the air temperature need not drop below \(0^{\circ} \mathrm{C}\) before a thin layer of water on the ground will freeze. Consider such a layer of water on a clear night for which the effective sky temperature is \(-30^{\circ} \mathrm{C}\) and the convection heat transfer coefficient due to wind motion is \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The water may be assumed to have an emissivity of \(1.0\) and to be insulated from the ground as far as conduction is concerned. (a) Neglecting evaporation, determine the lowest temperature the air can have without the water freezing. (b) For the conditions given, estimate the mass transfer coefficient for water evaporation \(h_{\mathrm{m}}(\mathrm{m} / \mathrm{s})\). (c) Accounting now for the effect of evaporation, what is the lowest temperature the air can have without the water freezing? Assume the air to be dry.

A manufacturer of ski equipment wishes to develop headgear that will offer enhanced thermal protection for skiers on cold days at the slopes. Headgear can be made with good thermal insulating characteristics, but it tends to be bulky and cumbersome. Skiers prefer comfortable, lighter gear that offers good visibility, but such gear tends to have poor thermal insulating characteristics. The manufacturer decides to take a new approach to headgear design by concentrating the insulation in areas about the head that are prone to the highest heat losses from the skier and minimizing use of insulation in other locations. Hence, the manufacturer must determine the local heat transfer coefficients associated with the human head with a velocity of \(V=10 \mathrm{~m} / \mathrm{s}\) directed normal to the face and an air temperature of \(-13^{\circ} \mathrm{C}\). A young engineer decides to make use of the heat and mass transfer analogy and the naphthalene sublimation technique (see Problem 6.63) and casts head shapes of solid naphthalene with characteristic dimensions that are half-scale (that is, the models are half as large as the full-scale head). (a) What wind tunnel velocity \(\left(T_{\mathrm{x}}=300 \mathrm{~K}\right)\) is needed to apply the experimental results to the human head associated with \(V=10 \mathrm{~m} / \mathrm{s}\) ? (b) A wind tunnel experiment is performed for \(\Delta t=120 \mathrm{~min}, T_{x}=27^{\circ} \mathrm{C}\). The engineer finds that the naphthalene has receded by \(\delta_{1}=0.1 \mathrm{~mm}\) at the back of the head, \(\delta_{2}=0.32 \mathrm{~mm}\) in the middle of the forehead, and \(\delta_{3}=0.64 \mathrm{~mm}\) on the ear. Determine the heat transfer coefficients at these locations for the full-scale head at \(-13^{\circ} \mathrm{C}\). The density of solid naphthalene is \(\rho_{\mathrm{A}, \text { sol }}=1025 \mathrm{~kg} / \mathrm{m}^{3}\). (c) After the new headgear is designed, the models are fitted with the new gear (half-scale) and the experiments are repeated. Some areas of the model that were found to have small local heat transfer coefficients are left uncovered since insulating these areas would have little benefit in reducing overall heat losses during skiing. Would you expect the local heat transfer coefficients for these exposed areas to remain the same as prior to fitting the model with the headgear? Explain why.

Parallel flow of atmospheric air over a flat plate of length \(L=3 \mathrm{~m}\) is disrupted by an array of stationary rods placed in the flow path over the plate. Laboratory measurements of the local convection coefficient at the surface of the plate are made for a prescribed value of \(V\) and \(T_{x}>T_{x}\). The results are correlated by an expression of the form \(h_{x}=0.7+13.6 x-3.4 x^{2}\), where \(h_{x}\) has units of \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(x\) is in meters. Evaluate the average convection coefficient \(\bar{h}_{L}\) for the entire plate and the ratio \(\bar{h}_{L} / h_{L}\) at the trailing edge.

An industrial process involves the evaporation of water from a liquid film that forms on a contoured surface. Dry air is passed over the surface, and from laboratory measurements the convection heat transfer correlation is of the form $$ \overline{N_{L}}=0.43 \operatorname{Re}_{L}^{0.58} P r r^{\Omega .4} $$ (a) For an air temperature and velocity of \(27^{\circ} \mathrm{C}\) and \(10 \mathrm{~m} / \mathrm{s}\), respectively, what is the rate of evaporation from a surface of \(1-\mathrm{m}^{2}\) area and characteristic length \(L=1 \mathrm{~m}\) ? Approximate the density of saturated vapor as \(\rho_{A, \text { sat }}=0.0077 \mathrm{~kg} / \mathrm{m}^{3}\). (b) What is the steady-state temperature of the liquid film?

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.