A horizontal \(1.5\)-m-wide, \(4.5\)-m-long double-pane window consists of two sheets of glass separated by a \(3.5-\mathrm{cm}\) gap filled with water. If the glass surface temperatures at the bottom and the top are measured to be \(60^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\), respectively, the rate of heat transfer through the window is (a) \(27.6 \mathrm{~kW}\) (b) \(39.4 \mathrm{~kW}\) (c) \(59.6 \mathrm{~kW}\) (d) \(66.4 \mathrm{~kW} \quad(e) 75.5 \mathrm{~kW}\) (For water, use \(k=0.644 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=3.55, v=\) \(0.554 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \beta=0.451 \times 10^{-3} \mathrm{~K}^{-1}\). Also, the applicable correlation is \(\mathrm{Nu}=0.069 \mathrm{Ra}^{1 / 3} \operatorname{Pr}^{0.074}\) ).

Short Answer

Expert verified
Question: Calculate the rate of heat transfer through a double-pane window with the following dimensions and properties: width = 1.5 m, length = 4.5 m, gap between glass panes = 0.035 m, glass surface temperatures: \(T_1 = 60^{\circ} \mathrm{C}\) and \(T_2 = 40^{\circ} \mathrm{C}\), thermal conductivity of water = 0.644 W/mK, Prandtl number of water = 3.55, kinematic viscosity of water = \(0.554 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), coefficient of thermal expansion of water = \(\beta = 0.451 \times 10^{-3} \mathrm{~K}^{-1}\), and correlation for Nusselt number: \(\mathrm{Nu}=0.069 \mathrm{Ra}^{1 / 3} \operatorname{Pr}^{0.074}\). Answer: The rate of heat transfer through the double-pane window is approximately \(50.655 \mathrm{~kW}\).

Step by step solution

01

Identifying given values.

First, let's note down all given values: - Window dimensions: width = 1.5 m, length = 4.5 m - Gap between glass panes: L = 0.035 m - Glass surface temperatures: \(T_1 = 60^{\circ} \mathrm{C}\), \(T_2 = 40^{\circ} \mathrm{C}\) - Thermal conductivity of water: k = 0.644 W/mK - Prandtl number of water: Pr = 3.55 - Kinematic viscosity of water: \(v = 0.554 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) - Coefficient of thermal expansion of water: \(\beta = 0.451 \times 10^{-3} \mathrm{~K}^{-1}\) - Correlation for Nusselt number: \(\mathrm{Nu}=0.069 \mathrm{Ra}^{1 / 3} \operatorname{Pr}^{0.074}\)
02

Calculate the temperature difference.

Calculate temperature difference between the bottom and top, which is used to derive the Rayleigh number. \(\Delta T = T_1 - T_2 = 60 - 40 = 20 \mathrm{~K}\)
03

Find the Rayleigh number (Ra).

Using the temperature difference, the gap length, and other given values, we can calculate the Rayleigh number as follows: \(\mathrm{Ra} = \frac{g \beta L^3 \Delta T}{v^2} = \frac{9.81 \times 0.451 \times 10^{-3} \times 0.035^3 \times 20}{(0.554 \times 10^{-6})^2}\) Calculating the Rayleigh number, we get: \(\mathrm{Ra} \approx 42.37 \times 10^8\)
04

Find the Nusselt number (Nu).

Using the Nusselt number correlation given in the problem, we can calculate the Nusselt number: \(\mathrm{Nu} = 0.069 \mathrm{Ra}^{1/3} \operatorname{Pr}^{0.074} = 0.069 \times (42.37 \times 10^8)^{1/3} \times 3.55^{0.074}\) Calculating the Nusselt number, we get: \(\mathrm{Nu} \approx 102.5\)
05

Calculate the heat transfer coefficient (h).

Using the Nusselt number and the thermal conductivity of the water, we can calculate the heat transfer coefficient as follows: \(h = \frac{\mathrm{Nu} \times k}{L} = \frac{102.5 \times 0.644}{0.035}\) Calculating the heat transfer coefficient, we get: \(h \approx 1886.5 \mathrm{~W} / \mathrm{m}^2 \mathrm{K}\)
06

Calculate the rate of heat transfer.

Finally, we can use the heat transfer coefficient, the surface area of the window, and the temperature difference to calculate the rate of heat transfer: \(Q = h \times A \times \Delta T = 1886.5 \times (1.5 \times 4.5) \times 20\) Calculating the rate of heat transfer, we get: \(Q \approx 50655 \mathrm{~W}\) or \(50.655 \mathrm{~kW}\) This rate of heat transfer is closest to option (e) \(75.5 \mathrm{~kW}\).

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.

Thermal Conductivity
The concept of thermal conductivity is pivotal when studying heat transfer, as it represents a material's ability to conduct heat. In the context of a double-pane window, we're typically dealing with two materials: glass and a filler substance, such as air or another gas, which in our case study is water.

Thermal conductivity, denoted by the symbol \( k \), is quantified as the amount of heat, in Watts (W), that passes through a material with a thickness of one meter (m) for each square meter of surface area and for each degree Kelvin (K) temperature difference across the material. For example, water has a thermal conductivity of \( 0.644 \frac{W}{m \times K} \). This property tells us that if we have a temperature difference across a layer of water, the material allows heat to transfer at a rate governed by its thermal conductivity.

When analyzing the rate of heat transfer through a window, we need to consider the thermal conductivity of the filler material in combination with the window's dimensions and temperature difference to estimate how much heat is being lost or gained.
Nusselt Number Calculation
While thermal conductivity tells us about the material's intrinsic ability to conduct heat, the Nusselt number offers insight into the convective heat transfer occurring in a fluid. The Nusselt number, \( Nu \), is a dimensionless parameter that correlates the convective to the conductive heat transfer at a boundary in a fluid. In our double-pane window example, the water between the panes serves as our fluid.

To calculate the Nusselt number, we employ empirical correlations, which are developed through experimental data and theoretical analysis. The correlation provided for our double-pane window problem is \( Nu = 0.069 Ra^{1/3}Pr^{0.074} \), which ties the Nusselt number to both the Rayleigh and Prandtl numbers. This calculated Nusselt number is then utilized to figure out the convective heat transfer coefficient, essential for determining the total heat transfer through the window.
Rayleigh Number
The Rayleigh number \( (Ra) \) is another dimensionless quantity crucial in the study of convection. It essentially assesses the potential for convection given a specific set of conditions. Rayleigh number is influenced by a fluid's buoyancy-driven movement resulting from temperature differences, which in turn is affected by factors like the fluid's coefficient of thermal expansion \( (\beta) \), the gravitational acceleration \( (g) \), the temperature difference across the fluid \( (\Delta T) \), the characteristic length \( (L) \), and the kinematic viscosity of the fluid \( (v) \).

Using the formula \( Ra = \frac{g \beta L^3 \Delta T}{v^2} \), we include these variables to gain a Rayleigh number that helps estimate how the fluid will behave in response to the temperature gradient imposed by the window's hot and cold surfaces. A high Rayleigh number suggests a turbulent flow and therefore more effective convection, while a low number indicates a more stable, less convective flow.
Prandtl Number
Last but not least, we come to the Prandtl number \( (Pr) \), named after the German physicist Ludwig Prandtl. This dimensionless figure expresses the ratio of momentum diffusivity (kinematic viscosity, \( v \)) to thermal diffusivity. Essentially, the Prandtl number tells us about the fluid's thermal and velocity boundary layers.

The Prandtl number is calculated as \( Pr = \frac{\mu C_p}{k} \), where \( \mu \) is the fluid's dynamic viscosity, \( C_p \) is the specific heat capacity at constant pressure, and \( k \) is the previously discussed thermal conductivity. The Prandtl number plays a role in determining how a fluid transfers heat and is put to practical use when calculating the Nusselt number. For water, \( Pr \) typically falls around 3.55, indicating that the momentum and thermal diffusivity are of a similar order of magnitude, leading to comparable thicknesses of the velocity and thermal boundary layers.

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

Flat-plate solar collectors are often tilted up toward the sun in order to intercept a greater amount of direct solar radiation. The tilt angle from the horizontal also affects the rate of heat loss from the collector. Consider a \(1.5-\mathrm{m}\)-high and 3-m-wide solar collector that is tilted at an angle \(\theta\) from the horizontal. The back side of the absorber is heavily insulated. The absorber plate and the glass cover, which are spaced \(2.5 \mathrm{~cm}\) from each other, are maintained at temperatures of \(80^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\), respectively. Determine the rate of heat loss from the absorber plate by natural convection for \(\theta=0^{\circ}, 30^{\circ}\), and \(90^{\circ}\).

Under what conditions does natural convection enhance forced convection, and under what conditions does it hurt forced convection?

A \(12-\mathrm{cm}\)-high and 20-cm-wide circuit board houses 100 closely spaced logic chips on its surface, each dissipating \(0.05 \mathrm{~W}\). The board is cooled by a fan that blows air over the hot surface of the board at \(35^{\circ} \mathrm{C}\) at a velocity of \(0.5 \mathrm{~m} / \mathrm{s}\). The heat transfer from the back surface of the board is negligible. Determine the average temperature on the surface of the circuit board assuming the air flows vertically upward along the 12 -cm-long side by (a) ignoring natural convection and ( \(b\) ) considering the contribution of natural convection. Disregard any heat transfer by radiation. Evaluate air properties at a film temperature of \(47.5^{\circ} \mathrm{C}\) and 1 atm pressure. Is this a good assumption?

A \(0.2 \mathrm{~m} \times 0.2 \mathrm{~m}\) street sign surface has an absorptivity of \(0.6\) and an emissivity of \(0.7\). Solar radiation is incident on the street sign at a rate of \(200 \mathrm{~W} / \mathrm{m}^{2}\), and the surrounding quiescent air is at \(25^{\circ} \mathrm{C}\). Determine the surface temperature of the street sign. Assume the film temperature is \(30^{\circ} \mathrm{C}\).

What does the effective conductivity of an enclosure represent? How is the ratio of the effective conductivity to thermal conductivity related to the Nusselt number?

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