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Chapter 1: Problem 68

An electric current of 5 A passing through a resistor has a measured voltage of \(6 \mathrm{~V}\) across the resistor. The resistor is cylindrical with a diameter of \(2.5 \mathrm{~cm}\) and length of \(15 \mathrm{~cm}\). The resistor has a uniform temperature of \(90^{\circ} \mathrm{C}\), and the room air temperature is \(20^{\circ} \mathrm{C}\). Assuming that heat transfer by radiation is negligible, determine the heat transfer coefficient by convection.

Short Answer

Expert verified
Answer: The heat transfer coefficient by convection is approximately 0.00416 W/(m²·K).

Step by step solution

01

Determine the power generated in the resistor

We can find the power generated in the resistor using Ohm's Law, which states that the power generated (P) is equal to the product of the current (I) and voltage (V) across the resistor. In this case, I = 5 A and V = 6 V. So, the power generated P = IV.
02

Calculate the rate of heat transfer by convection (Q)

We know that the rate of heat transfer is equal to power generated, so Q = P.
03

Determine the area of the cylinder (A)

A cylinder's surface area, A, can be calculated using the formula A = 2πrh + 2πr^2, where r is the radius and h is the height of the cylinder. In this case, the diameter is given as 2.5 cm, which means the radius r = 1.25 cm, and the height h = 15 cm.
04

Calculate the temperature difference between the resistor and room air(ΔT)

The temperature difference (ΔT) is given by the difference between the temperature of the resistor and the room air temperature. In this case, ΔT = (90°C - 20°C).
05

Determine the heat transfer coefficient by convection (h)

The heat transfer coefficient by convection (h) can be found using the following formula: Q = hAΔT. We have calculated the values of Q, A, and ΔT already. Solve for h by rearranging the formula to h = Q / (AΔT).
06

Plug in the values and calculate h

Now, plug in the values for Q, A, and ΔT into the formula for h and solve for the heat transfer coefficient by convection. Now let's plug in the values and calculate the results. Step 1: P = (5 A)(6 V) = 30 W Step 2: Q = P = 30 W Step 3: A = 2π(1.25 cm)(15 cm) + 2π(1.25 cm)^2 ≈ 103.67 cm² Step 4: ΔT = 90°C - 20°C = 70°C Step 5: h = Q / (AΔT) = 30 W / (103.67 cm²)(70°C) * 10000 = 0.00416 W/(m²·K) The heat transfer coefficient by convection is approximately 0.00416 W/(m²·K).

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Most popular questions from this chapter

Solve this system of two equations with two unknowns using appropriate software: $$ \begin{aligned} &x^{3}-y^{2}=10.5 \\ &3 x y+y=4.6 \end{aligned} $$

A cylindrical fuel rod \(2 \mathrm{~cm}\) in diameter is encased in a concentric tube and cooled by water. The fuel generates heat uniformly at a rate of $150 \mathrm{MW} / \mathrm{m}^{3}$. The convection heat transfer coefficient on the fuel rod is \(5000 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\), and the average temperature of the cooling water, sufficiently far from the fuel rod, is \(70^{\circ} \mathrm{C}\). Determine the surface temperature of the fuel rod, and discuss whether the value of the given convection heat transfer coefficient on the fuel rod is reasonable.

On a still, clear night, the sky appears to be a blackbody with an equivalent temperature of \(250 \mathrm{~K}\). What is the air temperature when a strawberry field cools to \(0^{\circ} \mathrm{C}\) and freezes if the heat transfer coefficient between the plants and air is $6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ because of a light breeze and the plants have an emissivity of \(0.9\) ? (a) \(14^{\circ} \mathrm{C}\) (b) \(7^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-3^{\circ} \mathrm{C}\)

A series of experiments were conducted by passing \(40^{\circ} \mathrm{C}\) air over a long 25 -mm-diameter cylinder with an embedded electrical heater. The objective of these experiments was to determine the power per unit length required \((W / L)\) to maintain the surface temperature of the cylinder at \(300^{\circ} \mathrm{C}\) for different air velocities \((V)\). The results of these experiments are given in the following table: $$ \begin{array}{lccccc} \hline V(\mathrm{~m} / \mathrm{s}) & 1 & 2 & 4 & 8 & 12 \\ W / L(\mathrm{~W} / \mathrm{m}) & 450 & 658 & 983 & 1507 & 1963 \\ \hline \end{array} $$ (a) Assuming a uniform temperature over the cylinder, negligible radiation between the cylinder surface and surroundings, and steady-state conditions, determine the convection heat transfer coefficient \((h)\) for each velocity \((V)\). Plot the results in terms of $h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\( vs. \)V(\mathrm{~m} / \mathrm{s})$. Provide a computer- generated graph for the display of your results, and tabulate the data used for the graph. (b) Assume that the heat transfer coefficient and velocity can be expressed in the form \(h=C V^{n}\). Determine the values of the constants \(C\) and \(n\) from the results of part (a) by plotting \(h\) vs. \(V\) on log-log coordinates and choosing a \(C\) value that assures a match at \(V=1 \mathrm{~m} / \mathrm{s}\) and then varying \(n\) to get the best fit.

Air enters a 12-m-long, \(7-\mathrm{cm}\)-diameter pipe at $50^{\circ} \mathrm{C}\( at a rate of \)0.06 \mathrm{~kg} / \mathrm{s}$. The air is cooled at an average rate of \(400 \mathrm{~W}\) per square meter surface area of the pipe. The air temperature at the exit of the pipe is (a) \(4.3^{\circ} \mathrm{C}\) (b) \(17.5^{\circ} \mathrm{C}\) (c) \(32.5^{\circ} \mathrm{C}\) (d) \(43.4^{\circ} \mathrm{C}\) (e) \(45.8^{\circ} \mathrm{C}\)
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