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

Short answer

Question: A copper wire of length 40 cm and diameter 0.4 cm is immersed vertically in boiling water at 1 atm. The surface temperature of the wire is 114°C, and its electric power consumption is 7.6 kW. Find the convection heat transfer coefficient. a) 12.7 kW/m²·K b) 13.3 kW/m²·K c) 14.1 kW/m²·K d) 14.7 kW/m²·K Answer: b) 13.3 kW/m²·K

Step-by-step solution

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 $$ A = 2 \pi (0.002 \text{ m}) (0.40 \text{ m}) = 0.004\pi\text{ m}^2. $$

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 \(T_{\infty} = 100^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 373.15 \mathrm{K}\).

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) \(13.3 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Key concepts

Convection Heat Transfer

Convection heat transfer is a mechanism where heat is transferred from a solid surface to a fluid (or vice versa). You can imagine it like the transfer of warmth from a heated wire to the surrounding water in this exercise. The process involves the motion of fluid, enhancing the heat transfer as compared to conduction alone.
The heat transfer coefficient, often notated as \( h \), plays a crucial role here. It quantifies the efficiency of heat transfer through convection. In our case, this coefficient is determined using the known values of electric power consumption and the temperature difference across the surfaces.

• The higher the heat transfer coefficient, the better the system is at transferring heat.
• Factors such as fluid velocity, temperature difference, and surface area influence the effectiveness of convection heat transfer.

Boiling Heat Transfer

Boiling heat transfer is a special type of convection where the liquid changes to vapor. This transformation enhances heat absorption significantly. Imagine boiling water, where heat from a wire causes water to boil and turn into steam, effectively transferring the heat away.
In the context of this exercise, the wire submerged in water reaches a temperature above the boiling point of water at 1 atm, causing the water around it to boil. This adds complexity to heat transfer as there are more factors at play.

• The phase change from water to steam involves high energy exchange.
• Boiling significantly increases the heat transfer rate because of high latent heat.

The boiling process is affected by parameters such as pressure and fluid properties apart from the temperature difference.

Surface Area Calculation

Calculating the surface area is an essential step in determining the heat transfer rate. We use the formula for the surface area of a cylinder, since the electric resistance wire can be modeled as a cylinder.
In this exercise, the formula used is:\[A = 2 \pi r L\]where:

• \( r \) is the radius of the wire, and
• \( L \) is the length of the wire.

The surface area determines how much heat can be transferred by convection from the wire to the water. A smaller surface area would mean less interaction with the surrounding medium, while a larger area provides more opportunity for heat transfer.

Temperature Conversion

Temperature conversion is vital in thermal calculations because it ensures accurate results. In the provided exercise, temperature values are converted from Celsius to Kelvin for compatibility in heat transfer formulas.
Kelvin is the preferred temperature scale for these calculations because it is an absolute scale. Unlike Celsius, it starts from absolute zero, making temperature differences meaningful in physical equations.

• To convert from Celsius to Kelvin, add 273.15 to the Celsius value.
• Using Kelvin avoids potential pitfalls related to negative temperatures, ensuring all calculations are based on absolute values.

This conversion is particularly critical when dealing with temperature differences in equations, as seen in this exercise where the temperatures of the wire and boiling water are adjusted accordingly.