Question

To understand the burnout phenomenon, boiling experiments are conducted in water at atmospheric pressure using an electrically heated 30 -cm-long, 4-mm- diameter nickel-plated horizontal wire. Determine \((a)\) the critical heat flux and \((b)\) the increase in the temperature of the wire as the operating point jumps from the nucleate boiling to the film boiling regime at the critical heat flux. Take the emissivity of the wire to be 0.3.

Short answer

The critical heat flux is 92829.37 W/m², and the increase in the temperature of the wire as the operating point jumps from nucleate boiling to film boiling regime at the critical heat flux is 202.48 K.

Step-by-step solution

Determine the critical heat flux (q_crit)

To calculate the critical heat flux, we will use the Kutateladze correlation for a horizontal cylinder (\(Dh\) for diameter and \(h_fg\) for latent heat of vaporization): $$ q_{crit} = 0.131 \sqrt{g Dh^5 \rho (\rho_h - \rho_c) h_{fg} / \mu_c} $$ At atmospheric pressure, we have the following properties for water and vapor: - Density of water, \(\rho_h = 958 kg/m^3\) - Density of vapor, \(\rho_c = 0.6 kg/m^3\) - Latent heat of vaporization, \(h_{fg} = 2.26 \times 10^6 J/kg\) - Dynamic viscosity of vapor, \(\mu_c = 1.256 \times 10^{-5} kg/m.s\) Plug in these values and the given diameter \(Dh = 0.004 m\) into the equation: $$ q_{crit} = 0.131 \sqrt{9.81 \cdot (0.004)^5 \cdot 958 \cdot (958 - 0.6) \cdot 2.26 \times 10^6 / 1.256 \times 10^{-5}} $$ Calculate the critical heat flux, \(q_{crit}\): $$ q_{crit} = 92829.37 W/m^2 $$

Calculate the heat transfer area (A_s)

Calculate the surface area \(A_s\) of the wire using its diameter and length (\(L\)): $$ A_s = \pi Dh L $$ Plug in the given diameter \(Dh = 0.004m\) and length \(L = 0.3m\): $$ A_s = \pi (0.004)(0.3) $$ Calculate the heat transfer area, \(A_s\): $$ A_s = 0.0037699 m^2 $$

Calculate the total heat transfer rate (Q_crit)

To calculate the total heat transfer rate at the critical heat flux, multiply the critical heat flux by the heat transfer area: $$ Q_{crit} = q_{crit} \cdot A_s $$ Plug in the values of \(q_{crit}\) and \(A_s\): $$ Q_{crit} = 92829.37 \cdot 0.0037699 $$ Calculate the total heat transfer rate, \(Q_{crit}\): $$ Q_{crit} = 349.84 W $$

Calculate the increase in the temperature of wire (ΔT)

As the operating point jumps from the nucleate boiling to the film boiling regime at the critical heat flux, we need to find the increase in the temperature of the wire. To do that, we have to use the following equation for the heat transfer rate in the film boiling regime: $$ Q_{crit} = \sigma \epsilon A_s (T_w^4 - T_{sat}^4) $$ Where, - \(\sigma = 5.67 \times 10^{-8} W/m^2.K^4\) is the Stefan-Boltzmann constant - \(\epsilon = 0.3\) is the emissivity of the wire - \(T_w\) is the temperature of the wire - \(T_{sat} = 373.15 K\) is the saturation temperature of water at atmospheric pressure Solve the equation for \(T_w\): $$ T_w^4 = Q_{crit} / (\sigma \epsilon A_s) + T_{sat}^4 $$ Calculate the temperature of the wire, \(T_w\): $$ T_w = \sqrt[4]{349.84 / (5.67 \times 10^{-8} \cdot 0.3 \cdot 0.0037699) + 373.15^4} = 575.63 K $$ Now, we can find the increase in the temperature of the wire, \(\Delta T\): $$ \Delta T = T_w - T_{sat} = 575.63 - 373.15 = 202.48 K $$ The critical heat flux is \(q_{crit} = 92829.37 W/m^2\), and the increase in the temperature of the wire as the operating point jumps from the nucleate boiling to the film boiling regime at the critical heat flux is \(\Delta T = 202.48 K\).