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Consider a \(15-\mathrm{cm} \times 20\)-cm printed circuit board \((\mathrm{PCB})\) that has electronic components on one side. The board is placed in a room at \(20^{\circ} \mathrm{C}\). The heat loss from the back surface of the board is negligible. If the circuit board is dissipating \(8 \mathrm{~W}\) of power in steady operation, determine the average temperature of the hot surface of the board, assuming the board is \((a)\) vertical, \((b)\) horizontal with hot surface facing up, and (c) horizontal with hot surface facing down. Take the emissivity of the surface of the board to be \(0.8\) and assume the surrounding surfaces to be at the same temperature as the air in the room. Evaluate air properties at a film temperature of \(32.5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

Short Answer

Expert verified
Question: Calculate the hot surface temperature of the printed circuit board for different orientations: vertical, horizontal with hot surface facing up, and horizontal with hot surface facing down. Assume a film temperature of 32.5°C for air properties calculations. Note: The student should perform Steps 4 to 6 of the provided solution and obtain the hot surface temperature for each orientation. Also, the reflection on the assumption of the film temperature should be discussed in the answer.

Step by step solution

01

Calculate the area of the PCB and the heat loss rate per unit area

First, we determine the area of the PCB and the heat loss rate per unit area: Area: \(A = width \times height = 15 \times 20 = 300\,\mathrm{cm^2}\) Heat loss rate per unit area: \(q'' = \dfrac{8\,\mathrm{W}}{A} = \dfrac{8\,\mathrm{W}}{300\,\mathrm{cm^2}} \cdot \dfrac{10^4\,\mathrm{cm^2}}{1\,\mathrm{m^2}} = \dfrac{8}{0.03}\,\mathrm{W/m^2} = 266.67\,\mathrm{W/m^2}\)
02

Obtain convection correlations for different orientations

We will use the following correlations for natural convection heat transfer over a flat surface: a) Vertical orientation (correlation for vertical plates): \(Nu_{L} = 0.825 + \dfrac{0.387Ra_{L}^{1/6}}{[1+(0.492/Pr)^{9/16}]^{8/27}}\) b) Horizontal orientation with hot surface facing up (correlation for horizontal plates facing upward): \(Nu_{L} = 0.54Ra_{L}^{1/4}\) c) Horizontal orientation with hot surface facing down (correlation for horizontal plates facing downward): \(Nu_{L} = 0.27Ra_{L}^{1/4}\) Here, \(Nu_{L}\) is the Nusselt number based on length, \(Ra_{L}\) is the Rayleigh number based on length, and \(Pr\) is the Prandtl number.
03

Calculate the air properties at the film temperature

We will calculate the air properties at the film temperature of \(32.5^{\circ} \mathrm{C}\) and \(1\,\mathrm{atm}\) pressure. We can use any thermodynamic property table or online resource to obtain the properties: \(T_f = 32.5^{\circ} \mathrm{C}\) Properties at film temperature: - Kinematic viscosity: \(\nu = 1.61 \times 10^{-5}\,\mathrm{m^2/s}\) - Thermal conductivity: \(k = 0.026\,\mathrm{W/(m\cdot K)}\) - Prandtl number: \(Pr = 0.7\) - Coefficient of thermal expansion: \(\beta = 1/\left(273.15 + T_f\right) = 1/305.65\,\mathrm{K^{-1}} = 3.27 \times 10^{-3}\,\mathrm{K^{-1}}\)
04

Calculate the Rayleigh number for each orientation and find the Nusselt number

Using the length-based Grashof number (\(Gr_{L}\)), we can calculate the Rayleigh number. \(Gr_{L} = \dfrac{g\beta\Delta T L^3}{\nu^2}\) \(Ra_{L} = Gr_L \cdot Pr\) For each orientation (a, b, and c), calculate \(Gr_{L}\) and \(Ra_{L}\), and use the appropriate \(Nu_{L}\) correlations (from Step 2) to find their values.
05

Calculate the average convective heat transfer coefficients

Calculate the average convective heat transfer coefficient (\(h\)) for each orientation: \(h = \dfrac{k}{L}\cdot Nu_{L}\)
06

Determine the hot surface temperature

Now that we have the convective heat transfer coefficient, we express the total heat transfer from the PCB (through convection and radiation) as: \(q_{conv} + q_{rad} = q\) Use the following formulas: \(q_{conv} = h A \Delta T = hA(T_s - T_{\infty})\) \(q_{rad} = \sigma\varepsilon A \left(T_s^4 - T_{\infty}^4\right)\) where \(\sigma = 5.67 \times 10^{-8}\,\mathrm{W/(m^2\cdot K^4)}\) is the Stefan-Boltzmann constant, \(\varepsilon = 0.8\) is the emissivity, and \(T_{\infty} = 20^{\circ} \mathrm{C}\). Combine the two equations and solve for the hot surface temperature \(T_s\) for each orientation.
07

Reflect on the assumption of film temperature for air properties

After finding the hot surface temperature, evaluate if the assumption of the film temperature as \(32.5^{\circ}\,\mathrm{C}\) for calculating air properties is reasonable. If the calculated values of \(T_s\) are close to this temperature, then the assumption is justified. Otherwise, air properties might need to be recalculated at a new film temperature, and the calculations repeated.

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Key Concepts

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

Convection Heat Transfer
Convection heat transfer plays an essential role in dissipating heat from surfaces like a printed circuit board (PCB). This process occurs when heat is transferred through a fluid, in this case, air, near a solid surface like a PCB. The movement of air helps carry heat away from the surface, reducing its temperature. There are two types of convection: natural convection and forced convection. In natural convection, fluid motion is driven by buoyancy forces due to density differences created by temperature variations.
To calculate convection heat transfer, we often use the Nusselt number ( Nu extsubscript{L} ), which provides a measure of the convective heat transfer relative to conductive heat transfer across the fluid's boundary layer. The higher the Nusselt number, the more effective the convection heat exchange. For different orientations of the PCB, various empirical correlations are used to calculate Nu , such as the one proposed for vertical plates or horizontal surfaces facing up or down.
Radiation Heat Transfer
Radiation heat transfer is another crucial mechanism in determining the temperature of the circuit board. It occurs when heat is emitted as electromagnetic waves (infrared radiation). Unlike convection, radiation does not depend on a medium and can occur in a vacuum. Every object emits radiation based on its temperature.
The amount of radiation emitted by a surface is affected by its emissivity ( ε d), a dimensionless factor ranging from 0 to 1, which represents a material's effectiveness in emitting energy as thermal radiation. A higher emissivity means more heat is radiated away from the surface, hence reducing the PCB's temperature. In our exercise, it's given as 0.8, indicating it is a good emitter of thermal radiation. The Stefan-Boltzmann Law helps quantify the radiation heat transfer, shown in the formula q_{rad} = σ ε A(T_s^4 - T_{ ∞}^{4}) .
Thermodynamic Properties
Understanding the thermodynamic properties of air is vital for accurately calculating heat transfer rates. These properties include kinematic viscosity ( ν i), thermal conductivity ( k i), and the coefficient of thermal expansion ( β i).
The film temperature method is used to simplify computations by approximating thermal properties at a temperature average between the surface and the ambient air temperature. In the given exercise, a film temperature of 32.5°C is used. Properties such as thermal conductivity directly affect the heat transfer rate. Higher thermal conductivity in a fluid means more efficient heat removal from the PCB.
Moreover, the kinematic viscosity and thermal expansion coefficient influence convection heat transfer rates by affecting the dynamics of air movement near the board's surface.
Prandtl Number
The Prandtl number ( Pr i) is a dimensionless number that reflects the ratio of momentum diffusivity (viscosity) to thermal diffusivity. It is a critical factor in heat transfer analysis because it signifies how efficiently heat diffuses in relation to momentum within a fluid. The number helps decide the thickness of the thermal and velocity boundary layers.
For air at standard conditions, the Prandtl number is around 0.7, indicating that momentum and heat diffuse at similar rates. In our exercise, this value allows us to use specific correlations to calculate the Nusselt number and analyze how convection varies based on the PCB's orientation. A lower Prandtl number suggests a thicker thermal boundary layer, affecting how heat transfer is conducted through convection.
Nusselt Number
As mentioned earlier, the Nusselt number ( Nu i) quantifies the enhancement of heat transfer through convection over conduction. This dimensionless number is crucial because it helps determine the convective heat transfer coefficient ( h i), which is pivotal in calculating the heat dissipated from a PCB.
Various empirical correlations relate the Nusselt number to the Rayleigh number and Prandtl number, depending on the configuration and orientation of the heat-transferring surface. A higher Nusselt number means a more effective convection, leading to lower surface temperatures.
Calculating Nu i involves complex mathematical relations depending on the geometry and orientation of the PCB. In our exercise, the Nusselt number is computed using established formulas for different mounting positions, giving insights into the thermal management of electronic components.

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Most popular questions from this chapter

A 4-m-long section of a 5-cm-diameter horizontal pipe in which a refrigerant flows passes through a room at \(20^{\circ} \mathrm{C}\). The pipe is not well insulated and the outer surface temperature of the pipe is observed to be \(-10^{\circ} \mathrm{C}\). The emissivity of the pipe surface is \(0.85\), and the surrounding surfaces are at \(15^{\circ} \mathrm{C}\). The fraction of heat transferred to the pipe by radiation is \(\begin{array}{lllll}\text { (a) } 0.24 & \text { (b) } 0.30 & \text { (c) } 0.37 & \text { (d) } 0.48 & \text { (e) } 0.58\end{array}\) (For air, use \(k=0.02401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.735, v=\) \(1.382 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) )

During a visit to a plastic sheeting plant, it was observed that a 60 -m-long section of a 2 -in nominal \((6.03\)-cm-outerdiameter) steam pipe extended from one end of the plant to the other with no insulation on it. The temperature measurements at several locations revealed that the average temperature of the exposed surfaces of the steam pipe was \(170^{\circ} \mathrm{C}\), while the temperature of the surrounding air was \(20^{\circ} \mathrm{C}\). The outer surface of the pipe appeared to be oxidized, and its emissivity can be taken to be \(0.7\). Taking the temperature of the surrounding surfaces to be \(20^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the steam pipe. Steam is generated in a gas furnace that has an efficiency of 78 percent, and the plant pays \(\$ 1.10\) per therm ( 1 therm \(=\) \(105,500 \mathrm{~kJ}\) ) of natural gas. The plant operates \(24 \mathrm{~h}\) a day 365 days a year, and thus \(8760 \mathrm{~h}\) a year. Determine the annual cost of the heat losses from the steam pipe for this facility.

Consider a hot boiled egg in a spacecraft that is filled with air at atmospheric pressure and temperature at all times. Will the egg cool faster or slower when the spacecraft is in space instead of on the ground? Explain.

A hot object suspended by a string is to be cooled by natural convection in fluids whose volume changes differently with temperature at constant pressure. In which fluid will the rate of cooling be lowest? With increasing temperature, a fluid whose volume (a) increases a lot (b) increases slightly (c) does not change (d) decreases slightly (e) decreases a lot.

Consider an industrial furnace that resembles a 13 -ft-long horizontal cylindrical enclosure \(8 \mathrm{ft}\) in diameter whose end surfaces are well insulated. The furnace burns natural gas at a rate of 48 therms/h. The combustion efficiency of the furnace is 82 percent (i.e., 18 percent of the chemical energy of the fuel is lost through the flue gases as a result of incomplete combustion and the flue gases leaving the furnace at high temperature). If the heat loss from the outer surfaces of the furnace by natural convection and radiation is not to exceed 1 percent of the heat generated inside, determine the highest allowable surface temperature of the furnace. Assume the air and wall surface temperature of the room to be \(75^{\circ} \mathrm{F}\), and take the emissivity of the outer surface of the furnace to be \(0.85\). If the cost of natural gas is \(\$ 1.15 /\) therm and the furnace operates \(2800 \mathrm{~h}\) per year, determine the annual cost of this heat loss to the plant. Evaluate properties of air at a film temperature of \(107.5^{\circ} \mathrm{F}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

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