A 40-cm-long, 0.4-cm-diameter electric resistance wire submerged in water is used to determine the convection heat transfer coefficient in water during boiling at \(1 \mathrm{~atm}\) pressure. The surface temperature of the wire is measured to be \(114^{\circ} \mathrm{C}\) when a wattmeter indicates the electric power consumption to be \(7.6 \mathrm{~kW}\). The heat transfer coefficient is (a) \(108 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(13.3 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(68.1 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(0.76 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(256 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Step by step solution

01

Calculate the surface area of the wire

To find the surface area of the wire, we will use the formula for the surface area of a cylinder: \(A = 2 \pi r L\), where \(r\) is the radius of the wire, and \(L\) is its length. The diameter is given as \(0.4 \mathrm{~cm}\), so the radius \(r = 0.2 \mathrm{~cm} = 0.002 \mathrm{~m}\). Thus, the surface area of the wire is $$ A = 2 \pi (0.002 \text{ m}) (0.40 \text{ m}) = 0.004\pi\text{ m}^2. $$

02

Convert the temperature to the absolute scale

We have to convert the given temperature from the Celsius scale to Kelvin, as the difference between temperature values is important in heat transfer calculations. Therefore, \(T_s = 114^{\circ} \mathrm{C}+ 273.15 \mathrm{K} = 387.15 \mathrm{K}\) and \(T_{\infty} = 100^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 373.15 \mathrm{K}\).

03

Calculate the heat transfer coefficient

We are given the electric power consumption \(Q = 7.6\text{ kW}\). Now we can use the heat transfer by convection equation \(Q = hA(T_s - T_{\infty})\) and solve for \(h\). $$ h = \frac{Q}{A(T_s - T_{\infty})} = \frac{7.6\text{ kW}}{0.004\pi\text{ m}^2 (387.15 \mathrm{K} - 373.15 \mathrm{K})} $$ Evaluating this expression yields, $$ h \approx 13.29 \frac {\mathrm{kW}}{\mathrm{m}^{2} \cdot \mathrm{K}} $$ Considering the given options, we can round the answer to \(13.3 \frac{\mathrm{kW}}{\mathrm{m}^{2} \cdot \mathrm{K}}\). So, the correct answer is (b) \(13.3 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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