Question

A 1-mm-diameter nickel wire with electrical resistance of \(0.129 \mathrm{\Omega} / \mathrm{m}\) is submerged horizontally in water at atmospheric pressure. Determine the electrical current at which the wire would be in danger of burnout in nucleate boiling.

Short answer

In this problem, we are given a nickel wire with a diameter of 1 mm and an electrical resistance per length of 0.129 Ω/m. Determine the electrical current at which the wire would be in danger of burnout in nucleate boiling. After analyzing the given information, it has been determined that the electrical current at which the wire would be in danger of burnout in nucleate boiling is approximately 6.98 A.

Step-by-step solution

Determine the wire's cross-sectional area

Given the diameter of the wire (\(1\,\mathrm{mm}\)), first we need to determine the cross-sectional area of the wire. The formula for the cross-sectional area of a cylinder is \(A=\pi r^2\). Assuming the wire is circular in cross-section, we can solve for the area. Given, diameter \(d = 1\,\mathrm{mm}\), so the radius is \(r = 0.5\,\mathrm{mm}\). Converting to meters, \(r = 0.0005\,\mathrm{m}\). $$A = \pi \times (0.0005\,\mathrm{m})^2 \approx 7.854 \times 10^{-7}\,\mathrm{m^2}$$

Determine the wire's electric resistance

We are given the electrical resistance per length, \(R_l = 0.129\,\mathrm{\Omega / m}\). To find the total resistance, \(R\), we need to know the length of the wire, \(L\). Since this is not given, we will assume a unit length, \(L = 1\,\mathrm{m}\). So, the total resistance is: $$R = R_l \times L = 0.129\,\mathrm{\Omega / m} \times 1\,\mathrm{m} = 0.129\,\mathrm{\Omega}$$

Find the electric power using Joule's law

Joule's law for heating relates the electric power, \(P\), to the current, \(I\), and resistance, \(R\): $$P = I^2R$$

Determine the heat transfer rate

In order to cause nucleate boiling, the heat transfer rate \(\dot{Q}\) must equal the heat required to raise the temperature of the water to the boiling point. Therefore, \(\dot{Q} = P\). Using the result from step 3, we have: $$\dot{Q} = I^2 \times 0.129\,\mathrm{\Omega}$$

Estimate the boiling point and associated heat transfer rate

Since we don't have a specific temperature but we know it's atmospheric pressure, we can estimate the nucleate boiling point at \(100^\circ\mathrm{C}\). Based on typical values for heat transfer coefficients in nucleate boiling (\(h_{NB}\)), let's estimate it as \(10^5\,\mathrm{W/m^2K}\). The boiling heat transfer rate is: $$\dot{Q}_{NB} = h_{NB}A(T_{NB} - T_w)$$ where \(T_{NB}\) is the temperature at the nucleate boiling point, and \(T_w\) is the wall temperature of the wire. For our case, let's assume \(T_w = 20^\circ\mathrm{C}\) (room temperature). Thus, the boiling heat transfer rate is: $$\dot{Q}_{NB} = 10^5\,\mathrm{\dfrac{W}{m^2K}} \times 7.854 \times 10^{-7}\,\mathrm{m^2} \times (100 - 20)^\circ\mathrm{C} \approx 6.283 \,\mathrm{W}$$

Solve for the current at nucleate boiling

Using the result from step 4, we can set the heat transfer rates equal to each other and solve for the current: $$I^2 \times 0.129\,\mathrm{\Omega} = 6.283\,\mathrm{W}$$ $$I^2 = \dfrac{6.283\,\mathrm{W}}{0.129\,\mathrm{\Omega}}$$ $$I^2 \approx 48.7$$ $$I \approx \sqrt{48.7} \approx 6.98\,\mathrm{A}$$ Therefore, the electrical current at which the wire would be in danger of burnout in nucleate boiling is approximately \(6.98\,\mathrm{A}\).

Key concepts

Joule's Law

Joule's law describes the relationship between electric current, resistance, and the heat generated in a conductor. It is a fundamental principle in understanding how electrical energy is converted into thermal energy. According to Joule's law, the heat produced (\(P\text{ in watts}\)) in a resistor is directly proportional to the square of the current (\(I\text{ in amperes}\)) flowing through it and its resistance (\(R\text{ in ohms}\)). Mathematically, this is written as \[P = I^2R\]. In the context of our exercise, Joule's law allows us to calculate the electrical power generated by the current in the nickel wire, which is critical in determining if the wire is at risk of burnout when submerged in water and subjected to nucleate boiling.

In simple terms, if the resistance of the wire or the current passing through it is too high, excessive heat will be generated. This excess heat could cause the water to boil at the surface of the wire and potentially lead to burnout, a situation where the wire is damaged due to overheating.

Electrical Resistance

Electrical resistance (\(R\text{ in ohms}\)) is a measure of the opposition to the flow of electric current through a conductor. It can be thought of as the electrical 'friction' that electrons encounter as they move through a material. The resistance of a wire depends on its material, length, and cross-sectional area. For a uniform wire of length (\(L\text{ in meters}\)) and cross-sectional area (\(A\text{ in square meters}\)), the resistance is given by \[R = R_l \times L\] where \(R_l\) is the resistance per unit length. In our example, we're given the resistance per meter for nickel, and by multiplying it by the length of the wire, we can find the total resistance the current faces.

Understanding how resistance affects the current and heat generation is key to preventing wire burnout. The resistance of materials is a fixed property but will vary slightly with temperature changes. With increasing resistance, for the same amount of current, more heat will be generated by the wire.

Boiling Point at Atmospheric Pressure

The boiling point of a liquid is the temperature at which its vapor pressure equals the surrounding atmospheric pressure. At this point, the liquid starts to turn into a gas at a rapid rate, otherwise known as boiling. For water at sea level (\text{atmospheric pressure of 101.325 kPa}), the boiling point is typically around \(100^\text{o}C\).

In our exercise, we assume the wire is in contact with water at atmospheric pressure. Therefore, when estimating the conditions for nucleate boiling—the onset of vigorous bubble formation on the wire's surface—we use the standard boiling point as our reference temperature. This assumption simplifies the calculations and is consistent with the typical behavior of water under such conditions. When the wire's temperature rises to the boiling point of the water, nucleate boiling begins, and heat is rapidly transferred from the wire to the water.

Heat Transfer Coefficient

The heat transfer coefficient (\(h_{NB}\text{ in W/m^2K}\)) is a measure of a material's ability to conduct thermal energy. It quantifies how effectively heat is transferred between a solid surface and a fluid (or between two fluids). In the example of nucleate boiling, the heat transfer coefficient is essential for calculating the heat transfer rate from the hot wire to the boiling water.

The formula for heat transfer in the context of boiling over a surface is expressed as \[\dot{Q}_{NB} = h_{NB}A(T_{NB} - T_w)\], where \(\dot{Q}_{NB}\) is the heat transfer rate, \(A\) is the surface area, \(T_{NB}\) is the temperature at the nucleate boiling point, and \(T_w\) is the temperature of the wire's surface. High values of \(h_{NB}\) represent more efficient heat transfer, which means the wire can dissipate more heat into the water before reaching dangerous temperatures. The numeric value chosen for \(h_{NB}\) in our problem (\(10^5 W/m^2K\)) represents a typical value for water in nucleate boiling. By calculating the heat transfer rate, we can determine the wire's capability to withstand a certain electrical current without overheating.