Question

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) \(0.76 \mathrm{~kW} / \mathrm{m}^{2}, \mathrm{~K}\) (e) \(256 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short answer

Answer: (a) 108 kW/m²·K

Step-by-step solution

Calculate the surface area of the wire

We first need the surface area of the wire. The wire is a cylinder with a length of 40 cm and a diameter of 0.4 cm. We can use the following formula for the surface area of a cylinder: \(A = 2\pi rL\), where \(r\) is the radius and \(L\) is the length. Convert the given measurements to meters by dividing them by 100: \(L = 0.4\mathrm{~m}\) and \(d = 0.004\mathrm{~m}\). Then find the radius: \(r = 0.002\mathrm{~m}\). Now calculate the surface area: \(A = 2\pi (0.002\mathrm{~m})(0.4\mathrm{~m}) \approx 0.00503\mathrm{~m}^2\).

Determine the temperature difference

We need to find the temperature difference \(\Delta T\) between the wire and the surrounding water. Since the wire is submerged in boiling water at \(1\mathrm{~atm}\) pressure, we know the water's temperature is \(100^{\circ}\mathrm{C}\). The wire has a surface temperature of \(114^{\circ}\mathrm{C}\). So, the temperature difference is: \(\Delta T = 114 - 100 = 14\mathrm{~K}\).

Calculate the heat transfer coefficient

Now that we have the surface area and temperature difference, we can find the heat transfer coefficient (h) using the formula \(Q = hA \Delta T\). The power consumption \(Q\) is given as \(7.6\mathrm{~kW}\). To use the same units as the other quantities, convert \(Q\) to Watts: \(Q = 7600\mathrm{~W}\). Rearrange the formula to solve for h: \(h = \frac{Q}{A\Delta T}\). Finally, compute the heat transfer coefficient: \(h = \frac{7600\mathrm{~W}}{(0.00503\mathrm{~m}^2)(14\mathrm{~K})} \approx 108480 \mathrm{~W/m}^2\mathrm{~K}\). When rounded and expressed in kilowatts: \(h \approx 108\mathrm{~kW/m}^2\mathrm{~K}\). The correct answer is (a) \(108\mathrm{~kW/m^2\cdot K}\).