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 \(C_{s f}\)

Short answer

Question: Determine the nucleate boiling heat transfer coefficient and the value of the experimental constant \(C_{s f}\) for an electrical wire submerged in water with a given diameter of 1 mm, heat dissipation rate of 4100 W/m, surface temperature of 128°C, and an experimental constant of the fluid of 1. Answer: The nucleate boiling heat transfer coefficient (\(h_{N B}\)) is \(4.47\times 10^4 \mathrm{~W/m^2K}\) and the experimental constant (\(C_{s f}\)) is 0.0109.

Step-by-step solution

Understand the given information

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

Rohsenow's expression for nucleate boiling heat transfer

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.

Calculate the length of the electrical wire

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

Calculate the nucleate boiling heat transfer coefficient

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: \(h_{N B} = 4.47 \times 10^4 \mathrm{~W/m^2K}\)

Calculate the value of experimental constant \(C_{s f}\)

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: \(C_{s f} = 0.0109\)

Final results

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

Key concepts

Rohsenow's correlation

Rohsenow's correlation provides a way to predict heat transfer during nucleate boiling.
It describes the relationship between the heat transfer coefficient in nucleate boiling and the physical properties of the fluid and surface in contact.

The correlation is formulated as:\[h_{NB} = C_{sf} \left( \frac{q}{L} \right)^{n}\]where:

• \(h_{NB}\) is the nucleate boiling heat transfer coefficient.
• \(C_{sf}\) is an empirical constant related to the fluid and surface characteristics.
• \(q\) denotes the heat flux.
• \(L\) indicates the characteristic length, often the diameter of the electric wire in this context.
• \(n\) is the empirical constant associated with the fluid.

This equation helps calculate how efficiently a surface can transfer heat during boiling by relying heavily on the values of \(C_{sf}\) and \(n\), both of which must be determined experimentally under specified conditions.
Understanding this correlation is essential for thermal systems where phase changes affect thermal performance.

Heat transfer coefficient

The heat transfer coefficient, denoted as \(h\), is a crucial parameter in determining how effectively heat is transferred between a solid surface and a fluid.

In the context of nucleate boiling, it indicates the amount of heat per unit area per unit time that can be transferred through boiling.
This measurement is significant, as it allows engineers to predict the thermal performance of heating systems and make necessary design adjustments.
The formula to calculate it in nucleate boiling scenarios reads as:\[q' = h_{NB} \times (T_s - T_{\infty})\]where:

• \(q'\) is the heat flux per unit area.
• \(h_{NB}\) is the nucleate boiling heat transfer coefficient derived from Rohsenow's correlation.
• \(T_s\) is the surface temperature.
• \(T_{\infty}\) is the saturation temperature of the fluid.

This coefficient is pivotal in determining the design and safe operation criteria of any equipment involving boiling due to its direct impact on heat transfer efficiency and surface integrity.

Boiling heat transfer

Boiling heat transfer is a form of convective heat transfer that occurs during the phase change from liquid to vapor.
This process significantly enhances heat transfer efficiency because of the large amount of energy absorbed during vaporization.

There are various regimes of boiling, with nucleate boiling being the most efficient, where bubbles form at distinct points on the heated surface and subsequently detach to rise in the liquid.
This regime happens at surface temperatures higher than the saturation temperature but lower than the critical heat flux point, which, if exceeded, will lead to surface deterioration and inefficient heat transfer.
Key factors influencing boiling heat transfer include:

• Surface material and condition: Influences the nucleation sites and how bubbles form.
• Fluid properties: Viscosity, thermal conductivity, and latent heat of vaporization play significant roles.
• Pressure: Affects the saturation temperature and thus the boiling dynamics.

Understanding boiling heat transfer is essential for engineering applications like nuclear reactors, heat exchangers, and any system where efficient thermal management is critical.