Consider a horizontal \(0.7\)-m-wide and \(0.85\)-m-long plate in a room at \(30^{\circ} \mathrm{C}\). Top side of the plate is insulated while the bottom side is maintained at \(0^{\circ} \mathrm{C}\). The rate of heat transfer from the room air to the plate by natural convection is (a) \(36.8 \mathrm{~W}\) (b) \(43.7 \mathrm{~W}\) (c) \(128.5 \mathrm{~W}\) (d) \(92.7 \mathrm{~W}\) (e) \(69.7 \mathrm{~W}\) (For air, use \(k=0.02476 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7323, v=\) \(\left.1.470 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\)

Short Answer

Expert verified
Answer: The rate of heat transfer (Q) is approximately 36.8 W.

Step by step solution

01

Calculate Grashof number

To calculate the Grashof number (Gr), we use the following formula: $$ Gr = \frac{g \beta (T_s - T_{\infty}) L^3}{\nu^2} $$ where: - \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)) - \(\beta\) is the thermal expansion coefficient of the fluid (For air, \(\beta \approx \frac{1}{T_{\infty}}\)) - \(T_s\) is the bottom side temperature of the plate (\(0^{\circ} \mathrm{C}\)) - \(T_{\infty}\) is the room temperature (\(30^{\circ} \mathrm{C}\)) - \(L\) is the characteristic length of the plate (\(0.85 \mathrm{~m}\)) - \(\nu\) is the kinematic viscosity of air (\(1.470 \times 10^{-5} \mathrm{~m^2/s}\)) First, we need to convert the temperatures to Kelvin: \(T_s = 273.15 \mathrm{~K}\) and \(T_{\infty} = 303.15 \mathrm{~K}\) Now, we can find \(\beta\) and Gr: $$ \beta = \frac{1}{T_{\infty}} = \frac{1}{303.15} \approx 3.298 \times 10^{-3} \mathrm{~K^{-1}} $$ $$ Gr = \frac{9.81 \times 3.298 \times 10^{-3} \times (273.15 - 303.15) \times 0.85^3}{(1.470 \times 10^{-5})^2} \approx 4.643 \times 10^7 $$
02

Calculate Rayleigh number

The Rayleigh number (Ra) is the product of Grashof number (Gr) and Prandtl number (Pr): $$ Ra = Gr \times Pr $$ Using the given value for Pr (\(0.7323\)) and the calculated Gr (\(4.643 \times 10^7\)), we can find Ra: $$ Ra = 4.643 \times 10^7 \times 0.7323 \approx 3.400 \times 10^7 $$
03

Calculate Nusselt number

Since the top side of the plate is insulated, it's considered a one-sided heating problem. Based on the Rayleigh number calculated in step 2, we can use the following correlation for horizontal plates to find the Nusselt number (Nu): $$ Nu = 0.15 \times Ra^{1/3} $$ Now, we can calculate Nu: $$ Nu = 0.15 \times (3.400 \times 10^7)^{1/3} \approx 13.53 $$
04

Calculate heat transfer rate Q

With the Nusselt number (Nu), we can calculate the rate of heat transfer (Q) using the following formula: $$ Q = Nu \times k \times \frac{T_{\infty} - T_s}{L} $$ where: - \(k\) is the thermal conductivity of air (\(0.02476 \mathrm{~W/m \cdot K}\)) Now, we can find Q: $$ Q = 13.53 \times 0.02476 \times \frac{303.15 - 273.15}{0.85} \approx 36.8 \mathrm{~W} $$ So, the correct answer is (a) \(36.8 \mathrm{~W}\).

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.

Understanding Grashof Number
The Grashof number (Gr) is a dimensionless value in fluid dynamics and heat transfer that is essential to the analysis of natural convection scenarios. Natural convection occurs when a fluid moves due to density differences that result from temperature gradients within the fluid. In the context of the plate example, the Grashof number helps us predict how much the difference in temperature between the cold plate and the warmer surrounding air will drive the air motion due to buoyancy forces.

To calculate Gr, an understanding of several factors is necessary: the acceleration due to gravity, the characteristic length of the object, the temperature difference between the object's surface and the ambient fluid, and the fluid's kinematic viscosity. To help students grasp this concept, it's pivotal to emphasize the importance of using a consistent set of units, such as the International System of Units (SI), converting all temperatures to Kelvin, and expressing all numerical values to an appropriate number of significant digits.

In the exercise, the Grashof number is computed based on the temperature of the insulated bottom side of the plate and the room's air temperature. The temperature difference provides the driving force for convection, while the kinematic viscosity suggests how easily the air moves. The resulting Grashof number indicates the relative strength of the buoyancy to viscous forces in the air surrounding the plate. A high Gr number signifies that buoyancy forces significantly influence the fluid flow, leading to more pronounced natural convection.
The Role of Rayleigh Number
The Rayleigh number (Ra) is another dimensionless quantity that combines the effects of buoyancy-driven flow and thermal diffusion, represented by the Grashof and Prandtl numbers, respectively. The Prandtl number (Pr) is a dimensionless number that represents the ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity. Thus, Ra plays a key role in characterizing the convection heat transfer in fluids.

Ra is particularly important as it tells us whether convection will occur. For instance, above a certain critical value of Ra, convection is likely to be the dominant heat transfer mode. In the provided solution, the Rayleigh number is first calculated by multiplying the Grashof number with the Prandtl number. Understanding the relationship between the Gr, Pr, and Ra can provide insights into how effectively heat is transferred through convection.

When educating students, it’s beneficial to articulate that the Rayleigh number is a measure of the driving force (due to temperature differences) for convection relative to the resisting forces (due to fluid viscosity and heat diffusion). Explaining the physical meaning behind dimensionless numbers like Ra makes them less abstract and helps students to better understand the mathematical steps involved in their calculation.
Calculating Heat Transfer with Nusselt Number
The Nusselt number (Nu) is the final piece of the puzzle that bridges the theoretical dimensionless analysis with the practical calculation of heat transfer rates in convection processes. Nu is a dimensionless number that represents the ratio of convective to conductive heat transfer across a fluid boundary. In essence, it quantifies the enhancement of heat transfer through a fluid layer as a result of convection relative to pure conduction.

For the case of the insulated plate, the Nusselt number is calculated using a correlation that takes into account the plate’s orientation (horizontal in this case) and the previously calculated Rayleigh number. It serves as a scaling factor in the formula that relates the temperature difference, thermal conductivity, and geometry of the plate to the actual heat transfer rate (Q). Additionally, when guiding students, it is important to explain that different empirical or semi-empirical correlations for Nu exist for various geometries and flow conditions.

By obtaining the Nusselt number and combining it with the thermal conductivity and temperature difference, we are able to calculate the heat transfer rate Q from the plate to the air. This calculated rate of heat transfer provides a tangible result that can validate the temperature maintenance, thermal design, or processing efficiency in numerous applications from industrial heat exchangers to maintaining the thermal comfort in living spaces.

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

Determine the \(U\)-factor for the center-of-glass section of a double-pane window with a 13-mm air space for winter design conditions. The glazings are made of clear glass having an emissivity of \(0.84\). Take the average air space temperature at design conditions to be \(10^{\circ} \mathrm{C}\) and the temperature difference across the air space to be \(15^{\circ} \mathrm{C}\).

A \(50-\mathrm{cm} \times 50-\mathrm{cm}\) circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at \(25^{\circ} \mathrm{C}\). Each chip dissipates \(0.18 \mathrm{~W}\) of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. Evaluate air properties at a film temperature of \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

Physically, what does the Grashof number represent? How does the Grashof number differ from the Reynolds number?

A room is to be heated by a coal-burning stove, which is a cylindrical cavity with an outer diameter of \(32 \mathrm{~cm}\) and a height of \(70 \mathrm{~cm}\). The rate of heat loss from the room is estimated to be \(1.5 \mathrm{~kW}\) when the air temperature in the room is maintained constant at \(24^{\circ} \mathrm{C}\). The emissivity of the stove surface is \(0.85\), and the average temperature of the surrounding wall surfaces is \(14^{\circ} \mathrm{C}\). Determine the surface temperature of the stove. Neglect the heat transfer from the bottom surface and take the heat transfer coefficient at the top surface to be the same as that on the side surface. The heating value of the coal is \(30,000 \mathrm{~kJ} / \mathrm{kg}\), and the combustion efficiency is 65 percent. Determine the amount of coal burned a day if the stove operates \(14 \mathrm{~h}\) a day. Evaluate air properties at a film temperature of \(77^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

Exhaust gases from a manufacturing plant are being discharged through a \(10-\mathrm{m}-\) tall exhaust stack with outer diameter of \(1 \mathrm{~m}\). The exhaust gases are discharged at a rate of \(0.125 \mathrm{~kg} / \mathrm{s}\), while temperature drop between inlet and exit of the exhaust stack is \(30^{\circ} \mathrm{C}\), and the constant pressure-specific heat of the exhaust gases is \(1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). On a particular calm day, the surrounding quiescent air temperature is \(33^{\circ} \mathrm{C}\). Solar radiation is incident on the exhaust stack outer surface at a rate of \(500 \mathrm{~W} / \mathrm{m}^{2}\), and both the emissivity and solar absorptivity of the outer surface are \(0.9\). Determine the exhaust stack outer surface temperature. Assume the film temperature is \(60^{\circ} \mathrm{C}\).

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