Question

A 50 -cm-long, 0.2 -cm-diameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally. The surface temperature of the wire is measured to be \(130^{\circ} \mathrm{C}\) when a wattmeter indicates the electric power consumption to be 4.1 \(\mathrm{kW}\). Then the heat transfer coefficient is \((a) 43,500 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) \((b) 137 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) \((c) 68,330 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) \((d) 10,038 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\) \((e) 37,540 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C}\)

Short answer

(a) 43,500 $\frac{W}{m^2 \cdot ^{\circ} C}$

Step-by-step solution

Find the surface area of the wire

The surface area of a cylinder can be found using the formula: $$A = 2\pi rL + 2\pi r^2$$ Where \(r\) is the radius and \(L\) is the length of the cylinder. In our case, the diameter is \(0.2\: cm\), so the radius is: $$r = \frac{0.2\: cm}{2} = 0.1\: cm$$ Now, we will convert the radius and length from centimeters to meters, and then use the formula to find the surface area: $$r_m = 0.1\: cm * \frac{1\: m}{100\: cm} = 0.001\: m$$ $$L_m = 50\: cm * \frac{1\: m}{100\: cm} = 0.5\: m$$ $$A = 2\pi (0.001\: m)(0.5\: m) + 2\pi (0.001\: m)^2 = 0.00314\: m^2$$

Calculate the heat transfer rate

We're given the power consumption as \(4.1\: kW\). To use it in the heat transfer equation, we need to convert it to watts: $$Q = 4.1\: kW * \frac{1000\: W}{1\: kW} = 4100\: W$$

Calculate the temperature difference

We are given the surface temperature of the wire, \(130^{\circ} C\). We know that the water is at its boiling point at \(1\) atm, which is \(100^{\circ} C\). So the temperature difference is: $$\Delta T = T_s - T_\infty = 130^{\circ} C - 100^{\circ} C = 30^{\circ} C$$

Find the heat transfer coefficient

Now, using the heat transfer equation, we can calculate the heat transfer coefficient: $$Q = hA(T_s - T_\infty)$$ Solving for \(h\), we get: $$h = \frac{Q}{A\Delta T} = \frac{4100\: W}{(0.00314\: m^2)(30^{\circ} C)} = 43500\: \frac{W}{m^2 \cdot ^{\circ} C}$$ So, the correct answer is (a) \(43,500\: \frac{W}{m^2 \cdot ^{\circ} C}\).