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Chapter 9: Problem 6

The heat transfer rate due to free convection from a vertical surface, \(1 \mathrm{~m}\) high and \(0.6 \mathrm{~m}\) wide, to quiescent air that is \(20 \mathrm{~K}\) colder than the surface is known. What is the ratio of the heat transfer rate for that situation to the rate corresponding to a vertical surface, \(0.6 \mathrm{~m}\) high and \(1 \mathrm{~m}\) wide, when the quiescent air is \(20 \mathrm{~K}\) warmer than the surface? Neglect heat transfer by radiation and any influence of temperature on the relevant thermophysical properties of air.

Short Answer

Expert verified
The ratio of the heat transfer rate for the given situations is -0.6. This indicates that the heat flow direction in the second case is opposite to that in the first case.

Step by step solution

01

Use the Nusselt Number formula for vertical surfaces

The dimensionless Nusselt number (Nu) represents the ratio of convective heat transfer to conductive heat transfer and is calculated using the following formula for free convection on vertical surfaces: \(Nu = f(Ra_pr)\), where \(Ra_pr\) is the Rayleigh Product (combined Grashof and Prandtl numbers) and \(f\) is a function that depends on the specific problem.
02

Recall the formula for heat transfer rate

The heat transfer rate (Q) is related to the Nusselt number by the following formula: \(Q = hA\Delta T\), where \(h = k\frac{Nu}{L}\) is the convective heat transfer coefficient, \(k\) is the thermal conductivity of the fluid, \(A\) is the surface area, and \(\Delta T\) is the temperature difference.
03

Apply the given information to the formula

For the two surfaces given in the exercise, we can apply their dimensions and temperature differences to find the following heat transfer rates: - For the 1 m high and 0.6 m wide surface: \(Q_1 = h_1 A_1 \Delta T_1\) - For the 0.6 m high and 1 m wide surface: \(Q_2 = h_2 A_2 \Delta T_2\) Notice that the temperature difference in the second case is negative since the air is warmer than the surface.
04

Find the relationship between heat transfer coefficients

Since the Nusselt number is the same for both situations, we can express the heat transfer coefficients as: \(h_1 = k\frac{Nu}{L_1}\) and \(h_2 = k\frac{Nu}{L_2}\), where \(L_1\) is the height of surface 1 and \(L_2\) is the height of surface 2. From this, we can find their ratio: \(\frac{h_1}{h_2} = \frac{L_2}{L_1}\)
05

Calculate the ratio of heat transfer rates

Now, we will find the ratio of the heat transfer rates: \(\frac{Q_1}{Q_2} = \frac{h_1 A_1 \Delta T_1}{h_2 A_2 \Delta T_2}\) Using the relationship of heat transfer coefficients and knowing the surface dimensions and temperature differences, we can plug in the values: \(\frac{Q_1}{Q_2} = \frac{L_2 A_1 \Delta T_1}{L_1 A_2 \Delta T_2} = \frac{0.6 \times 1 \times 0.6 \times -20}{1 \times 0.6 \times 1 \times 20}\)
06

Simplify and find the answer

Simplify the expression and find the ratio: \(\frac{Q_1}{Q_2} = \frac{-0.6}{1} = -0.6\) Therefore, the ratio of the heat transfer rate for the two situations is -0.6. The negative sign indicates that the direction of the heat flow in the second case is opposite to that in the first case.

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Key Concepts

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

Free Convection
Free convection, also known as natural convection, is a mechanism of heat transfer that occurs without the assistance of an external force, like a fan or pump. The movement of fluid in free convection is driven by buoyancy forces that result from density differences in the fluid. These differences are often caused by temperature gradients, leading to the warmer and less dense fluid rising while the cooler and denser fluid descends.

In the context of the exercise, free convection occurs between a heated vertical surface and the surrounding quiescent air, causing the air to move naturally due to the temperature difference. The rate at which heat is transferred from the surface to the air by free convection can be influenced by several factors including the size and orientation of the surface, the properties of the fluid, and the temperature difference between the surface and the fluid.
Nusselt Number
The Nusselt number (Nu) is a dimensionless quantity used in heat transfer calculations to characterize the convective heat transfer relative to conductive heat transfer across a fluid boundary. It is named after Wilhelm Nusselt, a pioneer in the study of heat transfer.

Mathematically, the Nusselt number is expressed as the ratio of convective to conductive heat transfer at a boundary in a fluid. It is represented as:
\( Nu = \frac{hL}{k} \)
where \(h\) is the convective heat transfer coefficient, \(L\) is the characteristic length (such as the height of a vertical surface), and \(k\) is the thermal conductivity of the fluid. A higher Nusselt number indicates a more effective convective heat transfer. In the solved exercise, the Nusselt number plays a crucial role in quantifying how effectively heat is transferred from the vertical surfaces to the surrounding air by convection.
Convective Heat Transfer
Convective heat transfer occurs when heat is transported by the motion of a fluid, which can be a liquid or a gas. This mode of heat transfer can be either free, driven by buoyancy forces (as mentioned in free convection), or forced, driven by external means such as a fan or pump.

The heat transfer rate \(Q\) during convection is described by the formula:
\( Q = hA\Delta T \)
where \(h\) is the convective heat transfer coefficient, \(A\) is the surface area over which heat transfer takes place, and \(\Delta T\) is the temperature difference between the surface and the fluid. The convective heat transfer coefficient \(h\) is dependent on fluid properties and flow characteristics, and it directly affects the efficiency of heat transfer. In our exercise, \(h\) is critical in determining the rate of heat transfer for both of the vertical surfaces and thus affects both the calculation and the final ratio of the heat transfer rates.
Rayleigh Product
The Rayleigh Product, often simply referred to as Rayleigh number (Ra), is an essential dimensionless number in the study of free convection heat transfer. It is the product of the Grashof number (Gr) and the Prandtl number (Pr), which represent the buoyancy and thermal diffusivity effects, respectively:
\( Ra = Gr \cdot Pr \)
The Grashof number quantifies the relative significance of buoyancy forces compared to viscous forces within the fluid, and the Prandtl number represents the ratio of momentum diffusivity to thermal diffusivity. High Rayleigh numbers indicate that the buoyancy forces are significant in driving fluid movement, thereby affecting convective heat transfer.

In the given exercise, the term \(Ra_pr\) is likely a representation of the Rayleigh Product, which factors into the Nusselt number relationship \(Nu = f(Ra_pr)\). The understanding and application of the Rayleigh Product help determine how the changes in surface dimensions and temperature difference influence the convective heat transfer rate.

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Most popular questions from this chapter

To reduce heat losses, a horizontal rectangular duct that is \(W=0.80 \mathrm{~m}\) wide and \(H=0.3 \mathrm{~m}\) high is encased in a metal radiation shield. The duct wall and shield are separated by an air gap of thickness \(t=0.06 \mathrm{~m}\). For a duct wall temperature of \(T_{d}=40^{\circ} \mathrm{C}\) and a shield temperature of \(T_{\mathrm{sh}}=20^{\circ} \mathrm{C}\), determine the convection heat loss per unit length from the duct.

The front door of a dishwasher of width \(580 \mathrm{~mm}\) has a vertical air vent that is \(500 \mathrm{~mm}\) in height with a \(20-\mathrm{mm}\) spacing between the inner tub operating at \(52^{\circ} \mathrm{C}\) and an outer plate that is thermally insulated. (a) Determine the heat loss from the tub surface when the ambient air is \(27^{\circ} \mathrm{C}\). (b) A change in the design of the door provides the opportunity to increase or decrease the \(20-\mathrm{mm}\) spacing by \(10 \mathrm{~mm}\). What recommendations would you offer with regard to how the change in spacing will alter heat losses?

As is evident from the property data of Tables A.3 and A.4, the thermal conductivity of glass at room temperature is more than 50 times larger than that of air. It is therefore desirable to use windows of double-pane construction, for which the two panes of glass enclose an air space. If heat transfer across the air space is by conduction, the corresponding thermal resistance may be increased by increasing the thickness \(L\) of the space. However, there are limits to the efficacy of such a measure, since convection currents are induced if \(L\) exceeds a critical value, beyond which the thermal resistance decreases. Consider atmospheric air enclosed by vertical panes at temperatures of \(T_{1}=22^{\circ} \mathrm{C}\) and \(T_{2}=-20^{\circ} \mathrm{C}\). If the critical Rayleigh number for the onset of convection is \(R a_{L} \approx 2000\), what is the maximum allowable spacing for conduction across the air? How is this spacing affected by the temperatures of the panes? How is it affected by the pressure of the air, as, for example, by partial evacuation of the space?

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.

Air at \(3 \mathrm{~atm}\) and \(100^{\circ} \mathrm{C}\) is discharged from a compressor into a vertical receiver of \(2.5-\mathrm{m}\) height and \(0.75-\mathrm{m}\) diameter. Assume that the receiver wall has negligible thermal resistance, is at a uniform temperature, and that heat transfer at its inner and outer surfaces is by free convection from a vertical plate. Neglect radiation exchange and any losses from the top. (a) Estimate the receiver wall temperature and the heat transfer to the ambient air at \(25^{\circ} \mathrm{C}\). To facilitate use of the free convection correlations with appropriate film temperatures, assume that the receiver wall temperature is \(60^{\circ} \mathrm{C}\). (b) Were the assumed film temperatures of part (a) reasonable? If not, use an iteration procedure to find consistent values. (c) Now consider two features of the receiver neglected in the previous analysis: (i) radiation exchange from the exterior surface of emissivity \(0.85\) to large surroundings, also at \(25^{\circ} \mathrm{C}\); and (ii) the thermal resistance of a 20 -mm-thick wall with a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Represent the system by a thermal circuit and estimate the wall temperatures and the heat transfer rate.
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