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Chapter 9: Problem 85

Consider a 3-m-high rectangular enclosure consisting of two surfaces separated by a \(0.1-\mathrm{m}\) air gap at \(1 \mathrm{~atm}\). If the surface temperatures across the air gap are \(30^{\circ} \mathrm{C}\) and \(-10^{\circ} \mathrm{C}\), determine the ratio of the heat transfer rate for the horizontal orientation (with hotter surface at the bottom) to that for vertical orientation.

Short Answer

Expert verified
Answer: The ratio of heat transfer rates between the horizontal orientation and the vertical orientation can be obtained by following the steps mentioned in the solution. It requires the calculation of Grashof, Nusselt, and heat transfer coefficients for both orientations, and eventually, the ratio of heat transfer rates can be determined using these results.

Step by step solution

01

Define given parameters and constants

First, let's list all the given parameters and some requiredconstantsused in this problem. 1. Height of the enclosure: \(h = 3\,\text{m}\) 2. Air gap between surfaces: \(L = 0.1\,\text{m}\) 3. Temperatures of surfaces: \(T_1 = 30^{\circ}\text{C}\) and \(T_2=-10^{\circ}\text{C}\) 4. Atmospheric pressure: 1 atm 5. Gravitational acceleration: \(g = 9.81\,\text{m/s}^2\) 6. Thermal expansion coefficient for air: \(\beta = 1/T_{mean}\), where \(T_{mean}\) is the mean temperature between surfaces \(T_0 = \frac{T_1 + T_2}{2} = 10^{\circ}\text{C}\).
02

Calculate the Grashof number for both orientations

Next, we need to calculate the Grashof number, which gives an indication of the significance of natural convection. We need to find the Grashof number for both orientations (vertical and horizontal). For vertical orientation: \(Gr_v = \frac{g\beta\Delta T (h^3)}{\nu^2}\) For horizontal orientation (hot surface at the bottom): \(Gr_h = \frac{g\beta\Delta T (L^3)}{\nu^2}\) Where \(\Delta T = T_1 - T_2 = 40^{\circ}\,\text{C}\) and \(\nu\) is the kinematic viscosity of air which can be looked up from the table for the mean temperature \(T_0\).
03

Calculate the Nusselt number for both orientations

Based on the Grashof number, we can now calculate the Nusselt number for both orientations. The Nusselt number is a measure of the effectiveness of convection compared to conduction. For vertical orientation: \(Nu_v = 0.59(Gr_v\cdot Pr)^{1/4}\) For horizontal orientation: \(Nu_h = 0.15(Gr_h\cdot Pr)^{1/3}\) Here, \(Pr\) is the Prandtl number for air which can be looked up from the table for the mean temperature \(T_0\).
04

Calculate heat transfer coefficients and heat transfer rates

Now, we can calculate the heat transfer coefficients and heat transfer rates for both orientations using the Nusselt number and the thermal conductivity of air, \(k\), which can be looked up from the table for the mean temperature \(T_0\). For vertical orientation: \(h_v = \frac{Nu_v \cdot k}{h}\) and \(q_v = h_v\cdot A\cdot\Delta T\) For horizontal orientation: \(h_h = \frac{Nu_h \cdot k}{L}\) and \(q_h = h_h\cdot A\cdot\Delta T\) Here, \(A\) is the area of the enclosure, which can be calculated as \(A = h\cdot L = 3\,\text{m}\times 0.1\,\text{m}=0.3\,\text{m}^2\).
05

Calculate the ratio of the heat transfer rates

Finally, we will calculate the ratio of the heat transfer rates between the horizontal orientation and the vertical orientation: \(\frac{q_h}{q_v} = \frac{h_h}{h_v} = \frac{\frac{Nu_h \cdot k}{L}}{\frac{Nu_v \cdot k}{h}}\) Using the results from Steps 2, 3, and 4, calculate the ratio of heat transfer rates. This ratio represents the heat transfer in the horizontal orientation compared to the vertical orientation.

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Most popular questions from this chapter

In an ordinary double-pane window, about half of the heat transfer is by radiation. Describe a practical way of reducing the radiation component of heat transfer.

A circular grill of diameter \(0.25 \mathrm{~m}\) has an emissivity of \(0.8\). If the surface temperature is maintained at \(150^{\circ} \mathrm{C}\), determine the required electrical power when the room air and surroundings are at \(30^{\circ} \mathrm{C}\).

A 4-m-diameter spherical tank contains iced water at \(0^{\circ} \mathrm{C}\). The tank is thin-shelled and thus its outer surface temperature may be assumed to be same as the temperature of the iced water inside. Now the tank is placed in a large lake at \(20^{\circ} \mathrm{C}\). The rate at which the ice melts is (a) \(0.42 \mathrm{~kg} / \mathrm{s}\) (b) \(0.58 \mathrm{~kg} / \mathrm{s}\) (c) \(0.70 \mathrm{~kg} / \mathrm{s}\) (d) \(0.83 \mathrm{~kg} / \mathrm{s}\) (e) \(0.98 \mathrm{~kg} / \mathrm{s}\) (For lake water, use \(k=0.580 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=9.45, v=\) \(0.1307 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \beta=0.138 \times 10^{-3} \mathrm{~K}^{-1}\) )

A vertical 4-ft-high and 6-ft-wide double-pane window consists of two sheets of glass separated by a 1-in air gap at atmospheric pressure. If the glass surface temperatures across the air gap are measured to be \(65^{\circ} \mathrm{F}\) and \(40^{\circ} \mathrm{F}\), determine the rate of heat transfer through the window by \((a)\) natural convection and (b) radiation. Also, determine the \(R\)-value of insulation of this window such that multiplying the inverse of the \(R\)-value by the surface area and the temperature difference gives the total rate of heat transfer through the window. The effective emissivity for use in radiation calculations between two large parallel glass plates can be taken to be \(0.82\).

What is natural convection? How does it differ from forced convection? What force causes natural convection currents?
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