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Chapter 3: Problem 22

Consider a power transistor that dissipates \(0.2 \mathrm{~W}\) of power in an environment at \(30^{\circ} \mathrm{C}\). The transistor is \(0.4 \mathrm{~cm}\) long and has a diameter of \(0.5 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a \(24-h\) period, in kWh; \((b)\) the heat flux on the surface of the transistor, in \(\mathrm{W} / \mathrm{m}^{2}\); and \((c)\) the surface temperature of the resistor for a combined convection and radiation heat transfer coefficient of \(18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Short Answer

Expert verified
Question: Calculate the surface temperature of a power transistor dissipating 0.2 W of power in an environment with a combined convection and radiation heat transfer coefficient of 18 W/m²·K. The transistor has a cylindrical shape with a length of 0.4 cm and a diameter of 0.5 cm. Answer: The surface temperature of the power transistor is approximately 30.27°C.

Step by step solution

01

Calculate the amount of heat dissipated during a 24-hour period

We are given that the power dissipation of the transistor is 0.2 W. To find the amount of heat dissipated during a 24-hour period, we can use the formula: Heat Dissipated = Power × Time We need to convert the time from hours to seconds: Time = 24 hours × 3600 seconds/hour = 86400 seconds Now, calculate the heat dissipated: Heat Dissipated = 0.2 W × 86400 seconds = 17280 Joules Now, we need to convert Joules to kWh. We know that 1 Joule = 2.778 × 10^{-7} kWh. Therefore: Heat Dissipated = 17280 Joules × 2.778 × 10^{-7} kWh/Joule ≈ 0.0048 kWh
02

Calculate the surface area of the transistor

The transistor has a cylindrical shape, with a length of 0.4 cm and a diameter of 0.5 cm. To find the surface area, we can use the formula for the surface area of a cylinder: Surface Area = 2πr(h + r) First, convert the length (h) and diameter to meters: Length (h) = 0.4 cm × 0.01 m/cm = 0.004 m Diameter = 0.5 cm × 0.01 m/cm = 0.005 m Now calculate the radius (r): Radius (r) = Diameter / 2 = 0.005 m / 2 = 0.0025 m Finally, find the surface area: Surface Area = 2π(0.0025 m)(0.004 m + 0.0025 m) ≈ 4.07 × 10^{-5} m²
03

Calculate the heat flux on the surface of the transistor

Given the power dissipation (0.2 W) and the surface area, we can now calculate the heat flux on the surface. The formula for heat flux is: Heat Flux = Power / Surface Area Substitute the values we found: Heat Flux = 0.2 W / (4.07 × 10^{-5} m²) ≈ 4915 W/m²
04

Calculate the surface temperature of the resistor

Now, we will find the surface temperature of the resistor using the given combined convection and radiation heat transfer coefficient (18 W/m²·K). We can use the formula: Heat Transfer = h × A × (T_surface - T_environment) Rearrange the equation to find T_surface: T_surface = Heat Transfer / (h × A) + T_environment The heat transfer is equal to the power dissipation: Heat Transfer = 0.2 W Substitute the known values: T_surface = 0.2 W / (18 W/m²·K × 4.07 × 10^{-5} m²) + 30°C ≈ 30.27°C The surface temperature of the resistor is approximately 30.27°C.

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Most popular questions from this chapter

We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

Steam at \(450^{\circ} \mathrm{F}\) is flowing through a steel pipe \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) whose inner and outer diameters are \(3.5\) in and \(4.0\) in, respectively, in an environment at \(55^{\circ} \mathrm{F}\). The pipe is insulated with 2 -in-thick fiberglass insulation \((k=\) \(\left.0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and \(5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations?

Hot water at an average temperature of \(85^{\circ} \mathrm{C}\) passes through a row of eight parallel pipes that are \(4 \mathrm{~m}\) long and have an outer diameter of \(3 \mathrm{~cm}\), located vertically in the middle of a concrete wall \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(4 \mathrm{~m}\) high, \(8 \mathrm{~m}\) long, and \(15 \mathrm{~cm}\) thick. If the surfaces of the concrete walls are exposed to a medium at \(32^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the hot water and the surface temperature of the wall.

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-thick plastic cover whose thermal conductivity is \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Steam in a heating system flows through tubes whose outer diameter is \(5 \mathrm{~cm}\) and whose walls are maintained at a temperature of \(180^{\circ} \mathrm{C}\). Circular aluminum alloy 2024-T6 fins \((k=186 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of outer diameter \(6 \mathrm{~cm}\) and constant thickness \(1 \mathrm{~mm}\) are attached to the tube. The space between the fins is \(3 \mathrm{~mm}\), and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.
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