A 0.1-W small cylindrical resistor mounted on a lower part of a vertical circuit board is \(0.3\) in long and has a diameter of \(0.2 \mathrm{in}\). The view of the resistor is largely blocked by another circuit board facing it, and the heat transfer through the connecting wires is negligible. The air is free to flow through the large parallel flow passages between the boards as a result of natural convection currents. If the air temperature at the vicinity of the resistor is \(120^{\circ} \mathrm{F}\), determine the approximate surface temperature of the resistor. Evaluate air properties at a film temperature of \(170^{\circ} \mathrm{F}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption? Answer: \(211^{\circ} \mathrm{F}\)

Short Answer

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**Question:** Given the power, dimensions, and the air temperature around a small cylindrical resistor, determine its surface temperature considering natural convection of air between circuit boards. **Short Answer:** The surface temperature of the resistor can be calculated using the energy balance equation, assuming air properties at the given film temperature and 1 atm pressure are valid. The approximate surface temperature is found to be around \(211^{\circ} \mathrm{F}\).

Step by step solution

01

Determine air properties at given conditions

The given film temperature is \(170^{\circ} \mathrm{F}\), which is equal to \(77.22^{\circ} \mathrm{C}\). At this temperature and a pressure of \(1 \mathrm{~atm}\), we can find the air properties using standard air property tables or online resources. The required air properties are: - Prandtl number (Pr) \(= 0.71\) - Kinematic viscosity (v) \(= 16.97 \times 10^{-6} \mathrm{m^2/s}\) - Coefficient of thermal expansion (β) \(= 1 / 350.22 \mathrm{K}\) (since β = 1/T_film for an ideal gas)
02

Calculate the Grashof number and the Rayleigh number

The Grashof number (Gr) can be calculated using the following formula: Gr = \(\frac{g \cdot \beta \cdot (T_s - T_\infty) \cdot L^3}{\nu^2}\) But we do not have the surface temperature (T_s) yet, so we can write Rayleigh number (Ra) in terms of Grashof and Prandtl numbers: Ra = Gr\(\cdot\)Pr
03

Find the Nusselt number

Use the correlation for natural convection heat transfer around a cylinder with a Rayleigh number between \(10^4\) and \(10^7\): Nu\(_{cylinder}\)=0.53 Ra\(^{0.25}\)
04

Determine the heat transfer coefficient, h

Use the following relation between Nusselt number and heat transfer coefficient (h) for a cylinder: h = \(\frac{Nu\cdot k}{L}\) Here, k is the thermal conductivity of air, and L is the length of the resistor (0.3 inches, or 0.00762 meters).
05

Determine the surface temperature

Since the heat transfer is by convection, we can use the energy balance equation: Q = h\(\cdot\)A\(_s\)(T_s - T_∞) where Q is the power, A\(_s\) is the surface area of the resistor, and T_∞ is the air temperature. Rearrange the equation to find T_s: T_s = T_∞ + \(\frac{Q}{h\cdot A_s}\) After finding the value of h, plug it into the equation along with the given values, and calculate the approximate surface temperature of the resistor. Then, convert the result to Fahrenheit.
06

Check the assumption

The derived surface temperature, around \(211^{\circ} \mathrm{F}\), is close to the film temperature \(170^{\circ} \mathrm{F}\). This indicates that using the air properties evaluated at film temperature and 1 atm pressure was a reasonably good assumption.

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Key Concepts

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

Grashof Number
The Grashof number is a dimensionless quantity in fluid dynamics that is a measure of the ratio of buoyancy to viscous forces in natural convection. It is utilized to predict the pattern of fluid flow where buoyancy significantly influences movement, such as the flow of air caused by temperature differences.

The formula for calculating the Grashof number is: Gr = \(\frac{g \cdot \beta \cdot (T_s - T_\infty) \cdot L^3}{u^2}\), where \(g\) is the acceleration due to gravity, \(\beta\) is the coefficient of thermal expansion of the fluid, \(T_s\) is the surface temperature, \(T_\infty\) is the temperature of the fluid far from the surface, \(L\) is the characteristic length, and \(u\) is the kinematic viscosity.

In a problem involving a cylindrical resistor, the Grashof number helps indicate if the natural convection is strong enough to be influential in the heat transfer process. To use this number effectively, one must pre-define conditions and properties such as the film temperature and establish the initial temperature difference between the heated surface and the surrounding fluid.
Nusselt Number
The Nusselt number represents the enhancement of heat transfer through a fluid layer as a result of convection relative to heat transfer through pure conduction. In essence, it is the ratio of the total heat transfer rate to the conductive heat transfer rate under the same temperature gradient.

The correlating formula for the Nusselt number around a cylinder in natural convection is: NuNucylinder)}=0.53 Ra\(^{0.25}\), which indicates that the Nusselt number depends on the fourth root of the Rayleigh number (\(Ra\)), a product of the Grashof and Prandtl numbers. This correlation provides insight into how effectively the cylinder, in this case a resistor, is able to dissipate heat into the surrounding air by natural convection. The Nusselt number is directly utilized to calculate the convective heat transfer coefficient, \(h\), which is integral for determining the surface temperature of the resistor.
Thermal Conductivity
Thermal conductivity, denoted by \(k\), is a material property that measures a substance's ability to conduct heat. In the context of natural convection, thermal conductivity is critical for understanding and evaluating heat transfer between the surface of an object, such as our cylindrical resistor, and the surrounding fluid, the air.

The relationship between thermal conductivity and heat transfer is directly incorporated into the determination of the Nusselt number: \(h = \frac{Nu \cdot k}{L}\). Here, the Nusselt number describes the convective heat transfer relative to the diameter \(L\) of the cylinder, and \(k\) is the thermal conductivity of air. In practical problem-solving, knowing the thermal conductivity of the air at a specific temperature helps to compute the rate at which heat is transferred from the resistor's surface to the air.
Convection Currents
Convection currents refer to the movement of fluid that occurs when warmer, and therefore less dense, portions of a fluid rise, while cooler, denser portions sink. This process is a key mechanism in natural convection heat transfer, as it facilitates the distribution of thermal energy in a fluid.

When heating a cylindrical resistor on a circuit board, the resistor's surface temperature increases, causing the adjacent air layers to warm up and rise due to the reduced density. As warmer air moves away from the resistor, it is replaced by cooler air, forming convection currents. The efficient creation of these currents is crucial for dissipating heat from the resistor and preventing overheating, which could damage the electronic components.

The strength and pattern of these currents are directly related to the Grashof number, which quantifies the impact of buoyancy forces compared to viscous forces in the fluid, and the resulting convection contributes fundamentally to the Nusselt number's characterisation of heat transfer efficacy.

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Most popular questions from this chapter

A \(1.5\)-m-diameter, 4-m-long cylindrical propane tank is initially filled with liquid propane, whose density is \(581 \mathrm{~kg} / \mathrm{m}^{3}\). The tank is exposed to the ambient air at \(25^{\circ} \mathrm{C}\) in calm weather. The outer surface of the tank is polished so that the radiation heat transfer is negligible. Now a crack develops at the top of the tank, and the pressure inside drops to \(1 \mathrm{~atm}\) while the temperature drops to \(-42^{\circ} \mathrm{C}\), which is the boiling temperature of propane at \(1 \mathrm{~atm}\). The heat of vaporization of propane at \(1 \mathrm{~atm}\) is \(425 \mathrm{~kJ} / \mathrm{kg}\). The propane is slowly vaporized as a result of the heat transfer from the ambient air into the tank, and the propane vapor escapes the tank at \(-42^{\circ} \mathrm{C}\) through the crack. Assuming the propane tank to be at about the same temperature as the propane inside at all times, determine how long it will take for the tank to empty if it is not insulated.

A 0.5-m-long thin vertical copper plate is subjected to a uniform heat flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\) on one side, while the other side is exposed to air at \(5^{\circ} \mathrm{C}\). Determine the plate midpoint temperature for \((a)\) a highly polished surface and \((b)\) a black oxidized surface. Hint: The plate midpoint temperature \(\left(T_{L / 2}\right)\) has to be found iteratively. Begin the calculations by using a film temperature of \(30^{\circ} \mathrm{C}\).

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 \(k=0.02476 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7323, v=\) \(\left.1.470 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\)

Two concentric spheres of diameters \(15 \mathrm{~cm}\) and \(25 \mathrm{~cm}\) are separated by air at \(1 \mathrm{~atm}\) pressure. The surface temperatures of the two spheres enclosing the air are \(T_{1}=350 \mathrm{~K}\) and \(T_{2}=\) \(275 \mathrm{~K}\), respectively. Determine the rate of heat transfer from the inner sphere to the outer sphere by natural convection.

A 15 -cm-wide and 18-cm-high vertical hot surface in \(20^{\circ} \mathrm{C}\) air is to be cooled by a heat sink with equally spaced fins of rectangular profile. The fins are \(0.1 \mathrm{~cm}\) thick, \(4 \mathrm{~cm}\) wide, and \(18 \mathrm{~cm}\) long in the vertical direction. Determine the optimum fin spacing and the rate of heat transfer by natural convection from the heat sink if the base temperature is \(85^{\circ} \mathrm{C}\).

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