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Chapter 3: Problem 21

A cylindrical resistor element on a circuit board dissipates $0.15 \mathrm{~W}\( of power in an environment at \)40^{\circ} \mathrm{C}$. The resistor is \(1.2 \mathrm{~cm}\) long, and has a diameter of \(0.3 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a 24 -h period; (b) the heat flux on the surface of the resistor, in $\mathrm{W} / \mathrm{m}^{2}$; and (c) the surface temperature of the resistor for a combined convection and radiation heat transfer coefficient of $9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$.

Short Answer

Expert verified
Answer: During a 24-hour period, the resistor dissipates approximately 12,960 J of heat. The heat flux on the surface of the resistor is approximately 1326.376 W/m², and the surface temperature of the resistor is approximately 43.153°C.

Step by step solution

01

Calculate the heat dissipated during a 24-hour period

To find the heat dissipated (Q) during a 24-hour period, we need to multiply the power dissipated (P) by the period of time (t): Q = Pt where P = 0.15 W (power dissipation) t = 24 hours × 3600 s/hour = 86400 s Q = 0.15 W × 86400 s ≈ 12960 J So, the resistor dissipates approximately 12,960 J of heat during a 24-hour period.
02

Calculate the surface area of the resistor

The surface area (A) of a cylindrical resistor can be calculated using the formula: A = πdL where d = 0.3 cm = 0.003 m (diameter) L = 1.2 cm = 0.012 m (length) A = π × 0.003 m × 0.012 m ≈ 0.0001131 m²
03

Calculate the heat flux on the surface of the resistor

We can now find the heat flux (q') on the surface of the resistor by using the formula: q' = q / A where q = 0.15 W (heat transfer rate) A ≈ 0.0001131 m² (surface area) q' = 0.15 W / 0.0001131 m² ≈ 1326.376 W/m² The heat flux on the surface of the resistor is approximately 1326.376 W/m².
04

Calculate the surface temperature of the resistor

To find the surface temperature (T_s) of the resistor, we will use the combined heat transfer coefficient (h) and the following heat transfer equation: q = hA(T_s - T_inf) where q = 0.15 W (heat transfer rate) h = 9 W/m²·K (combined convection and radiation heat transfer coefficient) A ≈ 0.0001131 m² (surface area) T_inf = 40°C (environment temperature) Solving for T_s: T_s = (q / (hA)) + T_inf T_s = (0.15 W / (9 W/m²·K × 0.0001131 m²)) + 40°C ≈ 43.153°C The surface temperature of the resistor is approximately 43.153°C.

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Most popular questions from this chapter

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5\)-cm-thick aluminum \((k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a 10 -cm-thick layer of insulation \((k=0.02\) $\mathrm{W} / \mathrm{m} \cdot \mathrm{K})$. The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The rate at which the liquid oxygen gains heat is (a) \(141 \mathrm{~W}\) (b) \(176 \mathrm{~W}\) (c) \(181 \mathrm{~W}\) (d) \(201 \mathrm{~W}\) (e) \(221 \mathrm{~W}\)

Consider two walls, \(A\) and \(B\), with the same surface areas and the same temperature drops across their thicknesses. The ratio of thermal conductivities is \(k_{A} / k_{B}=4\) and the ratio of the wall thicknesses is \(L_{A} / L_{B}=2\). The ratio of heat transfer rates through the walls \(\dot{Q}_{A} / \dot{Q}_{B}\) is (a) \(0.5\) (b) 1 (c) 2 (d) 4 (e) 8

Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. Which of the following conditions is accurate for the efficiency and effectiveness of the first finned surface relative to the second one? (a) Higher efficiency and higher effectiveness (b) Higher efficiency but lower effectiveness (c) Lower efficiency but higher effectiveness (d) Lower efficiency and lower effectiveness (e) Equal efficiency and equal effectiveness

Hot water at an average temperature of \(80^{\circ} \mathrm{C}\) and an average velocity of \(1.5 \mathrm{~m} / \mathrm{s}\) is flowing through a \(25-\mathrm{m}\) section of a pipe that has an outer diameter of $5 \mathrm{~cm}\(. The pipe extends \)2 \mathrm{~m}$ in the ambient air above the ground, dips into the ground $(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( ) vertically for \)3 \mathrm{~m}$, and continues horizontally at this depth for \(20 \mathrm{~m}\) more before it enters the next building.

Explain how the fins enhance heat transfer from a surface. Also, explain how the addition of fins may actually decrease heat transfer from a surface.
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