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Chapter 9: Problem 97

Consider window blinds that are installed in the air space between the two panes of a vertical double-pane window. The window is \(H=0.5 \mathrm{~m}\) high and \(w=0.5 \mathrm{~m}\) wide, and includes \(N=19\) individual blinds that are each \(L=25 \mathrm{~mm}\) wide. When the blinds are open, 20 smaller, square enclosures are formed along the height of the window. In the closed position, the blinds form a nearly continuous sheet with two \(t=12.5 \mathrm{~mm}\) open gaps at the top and bottom of the enclosure. Determine the convection heat transfer rate between the inner pane, which is held at \(T_{s, i}=20^{\circ} \mathrm{C}\), and the outer pane, which is at \(T_{s, o}=-20^{\circ} \mathrm{C}\), when the blinds are in the open and closed positions, respectively. Explain why the closed blinds have little effect on the convection heat transfer rate across the cavity.

Short Answer

Expert verified
To determine the convection heat transfer rate between the inner and outer pane of the window when the blinds are in the open and closed positions, we need to first calculate the heat transfer coefficient for natural convection in each case. The heat transfer coefficient is given by the formula \(h = C \cdot (\Delta T/L)^\frac{1}{4}\), where C is a dimensionless constant, \(\Delta T\) is the temperature difference between the panes, and L is the characteristic length, which depends on the blinds' positions. Calculating the dimensionless constant C involves finding the Rayleigh, Grashof and Prandtl numbers, using the given properties of air and temperature values. Then, we can determine the heat transfer coefficients for both open and closed positions. Finally, the convection heat transfer rate across the cavity is calculated using Q = h路A路\(\Delta T\), where A is the area of the window. By comparing the heat transfer rates for the open and closed positions of the blinds, we find that there is little difference between them. The heat transfer rate remains nearly the same as the main factor affecting the natural convection heat transfer coefficient is the temperature difference, and the effect of reducing the gap between the blinds when they are closed is negligible.

Step by step solution

01

Declaration of universal constants

Let's declare some constants we'll need for our calculations: - Acceleration of gravity: \(g = 9.81 \frac{m}{s^2}\) - Boltzmann's constant: \(k = 1.38 \times 10^{-23} \frac{J}{K}\) - Fluid density: \(\rho = 1.2 \frac{kg}{m^3}\) - Average air viscosity: \(\mu = 1.85 \times 10^{-5} \frac{kg}{m \cdot s}\)
02

Given data

Let's list the given data: - Window height : \(H = 0.5 m\) - Window width : \(w = 0.5 m\) - Number of blinds : \(N = 19\) - Width of the blinds : \(L = 25 mm\) - Temperature of the inner pane: \(T_{s, i} = 20^\circ C\) - Temperature of the outer pane : \(T_{s, o} = -20^\circ C\) - Open gaps at the top and bottom when closed: \(t = 12.5 mm\)
03

Calculating the heat transfer coefficient for natural convection

To find heat transfer rate, we need to find heat transfer coefficient (h) for natural convection first. The formula for h is given by: \(h = C \cdot (\Delta T/L)^\frac{1}{4}\), where - C is a dimensionless constant which we need to determine first; - \(\Delta T\) is the temperature difference between the inner pane and the outer pane, which can be calculated as \((T_{s, i} - T_{s, o})\); - L is the characteristic length, which in this case will be the effective distance between the panes for both open and closed positions;
04

Calculating the dimensionless constant C

To compute the dimensionless constant C, we should use the Rayleigh number and Grashof and Prandtl numbers. With these dimensionless numbers, we can calculate C as follows: 1. The Grashof number (Gr) is given by: \[Gr = \frac{g \cdot \beta \cdot \Delta T \cdot L^3}{\nu^2}\] 2. The Prandtl number (Pr) is given by: \[Pr = \frac{\nu}{\alpha}\] 3. The Rayleigh number (Ra) is given by: \[Ra = Gr \times Pr\] 4. Finally, C is calculated as: \[C = 0.386 \times (1 + (\frac{0.492}{Pr})^\frac{9}{16})^\frac{-4}{9}\] So, we first need to find Gr, Pr, and then, Ra using the provided information, before we can determine the C constant.
05

Calculating the Grashof, Prandtl, and Rayleigh numbers

To calculate these dimensionless numbers, we need the following properties: - Coefficient of thermal expansion (\(\beta\)): \(\beta = \frac{1}{T_f}\), where \(T_f\) is the average air temperature in Kelvin; - Kinematic viscosity (\(\nu\)): \(\nu = \frac{\mu}{\rho}\); - Thermal diffusivity (\(\alpha\)): \(\alpha = \frac{k}{\rho \cdot C_p}\) The average air temperature inside the gap between the panes (\(T_f\)) can be calculated as: \(T_f = \frac{T_{s, i} + T_{s, o}}{2}\)
06

First the open position

Calculating dimensionless parameter C for open position: - Characteristic length (L): In the open position, L is the length of the square enclosure formed, which will be the same as the width of the blinds: \(L_{open} = 25 mm = 0.025 m\) - Calculate the Grashof, Prandtl, and Rayleigh numbers for the open position. - Calculate the dimensionless constant C for the open position. - Calculate the heat transfer coefficient (h) for the open position. - Calculate the heat transfer rate (Q) for the open position, using Q = h路A路\(\Delta T\), where A is the area of the window: \(A = H \cdot w\).
07

Now the closed position

Calculating dimensionless parameter C for closed position: - Characteristic length (L): In the closed position, L is the distance between the two open gaps, which is the window height minus the gap height: \(L_{closed} = H - 2t = 0.5 m - 2(0.0125 m)\) - Calculate the Grashof, Prandtl, and Rayleigh numbers for the closed position. - Calculate the dimensionless constant C for the closed position. - Calculate the heat transfer coefficient (h) for the closed position. - Calculate the heat transfer rate (Q) for the closed position, using the same formula Q = h路A路\(\Delta T\).
08

Result and explanation

Compare the heat transfer rates for the open and closed positions of the blinds. The result will show that the heat transfer rate does not change significantly when the blinds are closed. The reason behind this is that since the natural convection heat transfer coefficient is mostly dependent on the temperature difference, the effect of reducing the gap when the blinds are closed is negligible, and the convection heat transfer rate across the cavity remains nearly the same.

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Key Concepts

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

Natural Convection
Natural convection is a process where fluid motion is generated by temperature differences within the fluid itself. In the context of the double-pane window described in the exercise, natural convection occurs between the inner and outer window panes due to the temperature difference between them.
When the blinds are opened, small square enclosures form, allowing air to circulate freely due to the natural convection currents created by the 40-degree Celsius temperature difference. When the blinds are closed, they form a continuous barrier that slightly alters the air currents but does not significantly impact the overall heat transfer rate.
The movement of fluid in natural convection relies on the warmer air rising and cooler air sinking, creating a convection current that facilitates heat transfer from the warmer to the cooler surface. The key factor regulating this movement is gravity acting on the temperature-induced density differences within the fluid. In simpler terms, the warm air from the inner pane is lighter and rises towards the outer pane, where it cools and sinks, creating a continuous loop of heat transport across the cavity.
Grashof Number
The Grashof number, often denoted as \(Gr\), is a dimensionless quantity that is crucial in analyzing natural convection scenarios. It predicts the relative strength of natural convection compared to conductive heat transfer in a fluid.
In mathematical terms, the Grashof number is given by the formula:
  • \[Gr = \frac{g \cdot \beta \cdot \Delta T \cdot L^3}{u^2}\]
  • Here, \(g\) stands for the acceleration due to gravity, \(\beta\) is the thermal expansion coefficient, \(\Delta T\) is the temperature difference, \(L\) is the characteristic length, and \(u\) is the kinematic viscosity of the fluid.
A high Grashof number typically indicates that convection will be the dominant mode of heat transfer compared to conduction. In the exercise, the Grashof number is used to quantify the buoyancy-driven flow that takes place because of the temperature gradient between the two window panes. It helps evaluate how effective the convection process will be in transferring heat in both the open and closed blind positions.
Prandtl Number
The Prandtl number, denoted as \(Pr\), is another dimensionless number important for characterizing fluid flow, specifically defining the relationship between momentum diffusivity (viscous diffusion) and thermal diffusivity. It is calculated using the formula:
  • \[Pr = \frac{u}{\alpha}\]
  • where \(u\) is the kinematic viscosity, and \(\alpha\) is the thermal diffusivity of the fluid.
A smaller Prandtl number suggests that heat diffuses quickly compared to the velocity of fluid movement, essential for predictively analyzing convection currents. In the context of the window exercise, knowing the Prandtl number aids in determining how efficiently heat is being transferred relative to the movement of air between the panes. It characterizes the flow behavior and heat distribution pattern, directly influencing the convection heat transfer rate.
Rayleigh Number
Combining the influences of both the Grashof and Prandtl numbers, the Rayleigh number \(Ra\) provides an overall measure of the convection heat transfer capability. It is defined as:
  • \[Ra = Gr \times Pr\]
The Rayleigh number integrates the effects of buoyancy-driven flow and thermal response of the fluid, acting as a predictor for the onset of turbulence in natural convection flow.
In the exercise of the window blinds, calculating the Rayleigh number helps determine whether convection will remain laminar or become turbulent. Higher Rayleigh numbers indicate more complex flow structures and increased heat transfer rates. This is crucial in assessing the effectiveness of opening vs. closing the blinds in terms of altering the convection heat transfer dynamics across the window, offering insights into energy efficiency strategies.

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Most popular questions from this chapter

A solid object is to be cooled by submerging it in a quiescent fluid, and the associated free convection coefficient is given by \(\bar{h}=C \Delta T^{1 / 4}\), where \(C\) is a constant and \(\Delta T=T-T_{\infty}\). (a) Using the results of Section \(5.3 .3\), obtain an expression for the time required for the object to cool from an initial temperature \(T_{i}\) to a final temperature \(T_{f}\). (b) Consider a highly polished, \(150-\mathrm{mm}\) square aluminum alloy (2024) plate of \(5-\mathrm{mm}\) thickness, initially at \(225^{\circ} \mathrm{C}\), and suspended in ambient air at \(25^{\circ} \mathrm{C}\). Using the appropriate approximate correlation from Problem 9.11, determine the time required for the plate to reach \(80^{\circ} \mathrm{C}\). (c) Plot the temperature-time history obtained from part (b) and compare with the results from a lumped capacitance analysis using a constant free convection coefficient, \(\bar{h}_{o}\). Evaluate \(\bar{h}_{o}\) from an appropriate correlation based on an average surface temperature of \(\bar{T}=\left(T_{i}+T_{f}\right) / 2\).

To assess the efficacy of different liquids for cooling by natural convection, it is convenient to introduce a fure of merit, \(F_{N}\), which combines the influence of all pertinent fluid properties on the convection coefficient. If the Nusselt number is governed by an expression of the form, \(N u_{L} \sim R a^{n}\), obtain the corresponding relationship between \(F_{N}\) and the fluid properties. For a representative value of \(n=0.33\), calculate values of \(F_{N}\) for air \((k=0.026\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}, \quad \beta=0.0035 \mathrm{~K}^{-1}, \quad \nu=1.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \quad P r=\) \(0.70)\), water \(\left(k=0.600 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=2.7 \times 10^{-4} \mathrm{~K}^{-1}\right.\), \(\nu=10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \quad P r=5.0\) ), and a dielectric liquid \(\left(k=0.064 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad \beta=0.0014 \mathrm{~K}^{-1}, \quad \nu=10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right.\), \(P r=25)\). What fluid is the most effective cooling agent?

Common practice in chemical processing plants is to clad pipe insulation with a durable, thick aluminum foil. The functions of the foil are to confine the batt insulation and to reduce heat transfer by radiation to the surroundings. Because of the presence of chlorine (at chlorine or seaside plants), the aluminum foil surface, which is initially bright, becomes etched with in- service time. Typically, the emissivity might change from \(0.12\) at installation to \(0.36\) with extended service. For a \(300-\mathrm{mm}\)-diameter foil-covered pipe whose surface temperature is \(90^{\circ} \mathrm{C}\), will this increase in emissivity due to degradation of the foil finish have a significant effect on heat loss from the pipe? Consider two cases with surroundings and ambient air at \(25^{\circ} \mathrm{C}\) : (a) quiescent air and (b) a cross-wind velocity of \(10 \mathrm{~m} / \mathrm{s}\).

A horizontal 100-mm-diameter pipe passing hot oil is to be used in the design of an industrial water heater. Based on a typical water draw rate, the velocity over the pipe is \(0.5 \mathrm{~m} / \mathrm{s}\). The hot oil maintains the outer surface temperature at \(85^{\circ} \mathrm{C}\) and the water temperature is \(37^{\circ} \mathrm{C}\). Investigate the effect of flow direction on the heat rate (W/m) for (a) horizontal, (b) downward, and (c) upward flow.

9.27 The vertical rear window of an automobile is of thickness \(L=8 \mathrm{~mm}\) and height \(H=0.5 \mathrm{~m}\) and contains fine-meshed heating wires that can induce nearly uniform volumetric heating, \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). (a) Consider steady-state conditions for which the interior surface of the window is exposed to quiescent air at \(10^{\circ} \mathrm{C}\), while the exterior surface is exposed to air at \(-10^{\circ} \mathrm{C}\) moving in parallel flow over the surface with a velocity of \(20 \mathrm{~m} / \mathrm{s}\). Determine the volumetric heating rate needed to maintain the interior window surface at \(T_{s, i}=15^{\circ} \mathrm{C}\). (b) The interior and exterior window temperatures, \(T_{s, i}\) and \(T_{s, \rho}\), depend on the compartment and ambient temperatures, \(T_{\infty, i}\) and \(T_{\infty, p}\), as well as on the velocity \(u_{\infty}\) of air flowing over the exterior surface and the volumetric heating rate \(\dot{q}\). Subject to the constraint that \(T_{s, i}\) is to be maintained at \(15^{\circ} \mathrm{C}\), we wish to develop guidelines for varying the heating rate in response to changes in \(T_{\infty,}, T_{\infty \rho,}\), and/or \(u_{\infty}\). If \(T_{\infty, i}\) is maintained at \(10^{\circ} \mathrm{C}\), how will \(\dot{q}\) and \(T_{s, \rho}\) vary with \(T_{\infty, o}\) for \(-25 \leq T_{\infty,,} \leq 5^{\circ} \mathrm{C}\) and \(u_{\infty}=10,20\), and \(30 \mathrm{~m} / \mathrm{s}\) ? If a constant vehicle speed is maintained, such that \(u_{\infty}=30 \mathrm{~m} / \mathrm{s}\), how will \(\dot{q}\) and \(T_{s, o}\) vary with \(T_{\infty, i}\) for \(5 \leq T_{\infty, i} \leq 20^{\circ} \mathrm{C}\) and \(T_{\infty, o}=-25,-10\), and \(5^{\circ} \mathrm{C}\) ?
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