Question

A 1 -mm-diameter long electrical wire submerged in water at atmospheric pressure is dissipating \(4100 \mathrm{~W} / \mathrm{m}\) of heat, and the surface temperature reaches \(128^{\circ} \mathrm{C}\). If the experimental constant that depends on the fluid is \(n=1\), determine the nucleate boiling heat transfer coefficient and the value of the experimental constant $C_{\text {sf. }}$.

Short answer

Answer: The nucleate boiling heat transfer coefficient (h) is 65286.4 W/m²K, and the experimental constant Csf is 3264.32 W/m²K².

Step-by-step solution

Calculate the wire's surface area per unit length

The surface area per unit length (As) for a cylindrical wire can be calculated using the formula: \(A_{s} = \pi d\) Where d is the diameter of the wire, given as 1mm or 0.001m: \(A_{s} = \pi * 0.001 \mathrm{m} = 0.00314 \mathrm{m^2/m}\)

Determine the heat dissipation per unit surface area of the wire

Now, we will find the heat dissipation per unit surface area (q) by dividing the given heat dissipation per unit length (q'') by the wire's surface area per unit length (As): \(q = \frac{q''}{A_s} = \frac{4100 \mathrm{W/m}}{0.00314 \mathrm{m^2/m}} = 1305728 \mathrm{W/m^2}\)

Calculate the saturation temperature of water at atmospheric pressure

At atmospheric pressure (1 atm or 101325 Pa), the saturation temperature of water is approximately: \(T_{sat} = 100^{\circ} \mathrm{C}\)

Find the nucleate boiling heat transfer coefficient (h)

With the formula for the nucleate boiling heat transfer coefficient (h), the experimental constant n as 1, and the temperature difference between the wire's surface and the saturation temperature, we have: \(h_{NB} = C_{sf}(T_s - T_{sat})^n\) We need to rearrange the formula to solve for Csf: \(C_{sf} = \frac{h_{NB}}{(T_s - T_{sat})^n}\) First, let's find the value of h: \(h = \frac{q}{(T_s - T_{sat})} = \frac{1305728 \mathrm{W/m^2}}{(128^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C})} = 65286.4 \mathrm{W/m^2K}\)

Determine the experimental constant Csf

Now we can find the value of Csf using the rearranged formula and the calculated h: \(C_{sf} = \frac{65286.4 \mathrm{W/m^2K}}{(128^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C})^1} = 3264.32 \mathrm{W/m^2K^2}\) Therefore, the nucleate boiling heat transfer coefficient (h) is \(65286.4 \mathrm{W/m^2K}\), and the experimental constant Csf is \(3264.32 \mathrm{W/m^2K^2}\).