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 \(C_{s f}\)

We first note down the given parameters: - Electrical wire diameter (D) = \(1\times 10^{-3} \mathrm{~m}\) - Heat dissipation rate (q) = \(4100 \mathrm{~W/m}\) - Surface temperature (T) = \(128^{\circ} \mathrm{C}\) - Experimental constant (n) = 1

We'll be using the Rohsenow's expression for nucleate boiling heat transfer, which is given by: \(h_{N B}=C_{s f}\left(\frac{q}{L}\right)^{n}\) Where: - \(h_{N B}\): Nucleate boiling heat transfer coefficient - \(C_{s f}\): Experimental constant we need to determine - q: Heat dissipation rate - L: Length of the electrical wire - n: Experimental constant of fluid We know the value of 'q' and 'n', but we have to find the length (L) of the electrical wire. Since the diameter (D) is given as 1 mm, we will assume the length of the wire is equal to its diameter.

Assuming the length is equal to the diameter of the electrical wire, L = D = 1 mm = \(1\times 10^{-3} \mathrm{~m}\).

We can now calculate the value of \(h_{N B}\) using the given values for 'q', 'n', and 'L'. But we still need 'C_{s f}'. To do that, we will first calculate the value of \(h_{N B}\) . Using the Rohsenow's expression: \(h_{N B}=C_{s f}\left(\frac{q}{L}\right)^{n}\) We have the following values: - q = \(4100 \mathrm{~W/m}\) - L = \(1\times 10^{-3} \mathrm{~m}\) - n = 1 Plugging in the values: \(h_{N B}=C_{s f} \left(\frac{4100}{1\times 10^{-3}}\right)^{1}\) We now use the following information to find the value of \(h_{N B}\): Heat dissipation rate per unit area (q') = \(q/(\pi D L)\) Plugging in the values for q, D and L: \(q'=\frac{4100}{(\pi\cdot 1\times 10^{-3}\cdot 1\times 10^{-3})} = 1.304 \times 10^6 \mathrm{~W/m^{2}}\) Using the heat transfer formula: \(q'=h_{N B}(T_{s}-T_{\infty})\) We know the values of T_s and q'. T_s = \(128^{\circ} \mathrm{C}\) = 401.15 K (converted to kelvin) The saturation temperature of water, \(T_{\infty}\), can be found using a steam table. For atmospheric pressure, the saturation temperature is \(100^{\circ} \mathrm{C}\) = 373.15 K (converted to kelvin). Plugging these values into the heat transfer formula, we can find \(h_{N B}\). \(1.304 \times 10^6 = h_{N B} (401.15 - 373.15)\) Solving for \(h_{N B}\), we get: \(h_{N B} = 4.47 \times 10^4 \mathrm{~W/m^2K}\)

Now that we have found the value of \(h_{N B}\), we can use the Rohsenow's expression to calculate the value of \(C_{s f}\). \(h_{N B}=C_{s f}\left(\frac{q}{L}\right)^{n}\) Plugging in the known values for \(h_{N B}\), q, and L: \(4.47 \times 10^4 = C_{s f} \left(\frac{4100}{1\times 10^{-3}}\right)^{1}\) Solving for \(C_{s f}\), we get: \(C_{s f} = 0.0109\)

Therefore, we have found the nucleate boiling heat transfer coefficient and the experimental constant \(C_{s f}\) to be the following: Nucleate boiling heat transfer coefficient (\(h_{N B}\)) = \(4.47\times 10^4 \mathrm{~W/m^2K}\) Experimental constant (\(C_{s f}\)) = 0.0109