32 - Boiling

Example 1: Critical Heat Flux and Critical Temperature Difference For water boiling at 1 atm at 100^oC determine the value of excess temperature in the film nucleate boiling region for a heat flux equal to the critical heat flux for this condition. (AU-Apr-2017)

Solution: Heat transfer correlations and data from: Heat and Mass Transfer Data Book (7th ed.) – C. P. Kothandaraman & S. Subramanyan From page no. 143:
Critical heat flux (\(q_c\)) is given by \[q_c = 0.18 \lambda\, \rho_v\left[\frac{\sigma g (\rho_l-\rho_v)}{\rho_v^2} \right]^{0.25} \tag*{(1)}\] where, the properties are evaluated at (\(T_w+T_{\text{sat}}\))/2.

Given: \(T_{\text{sat}}=100^{\circ}C\). Since, we don’t know \(T_w\), we shall assume that \(T_w=150^{\circ}C\), and proceed with. Hence, we need to get the fluid properties at \((150+100)/2=125^{\circ}C\).

From the data book, (page no. 147) \[\begin{aligned} \sigma &= \text{surface tension of water with its vapor} = \frac{0.0589+0.0487}{2}\\ &=0.0538 \text{ N/m} \end{aligned}\] From Steam Tables, at 125 \[\begin{aligned} \rho_v &= \text{density of water vapor (i.e., steam)} = 1.299 \text{ kg/m$^3$} \\ \rho_l &= \text{density of liquid water} = 939 \text{ kg/m$^3$} \\ \lambda &= \text{latent heat of vaporization of water} = 2188000 \text{ J/kg} \end{aligned}\] Substituting these in Eqn.(1), we get \[\begin{aligned} q_c &= 0.18\times2188000\times1.299\times\left[\frac{0.0538\times9.812\times(939-1.299)}{1.299^2} \right]^{0.25} \\ &= {\bf 2117266} \text{ W/m$^2$} \end{aligned}\]

From the data book, (page no: 143), for the nucleate pool boiling, \[q = \mu_l\lambda\left[g\frac{(\rho_l-\rho_v)}{\sigma} \right]^{0.5}\left[\frac{C_{Pl}}{C_{sf}\lambda\; \text{Pr}}\right]^3(\Delta T)^3 \tag*{(2)}\] From page no. 22, of data book, we get \[\nu_l = \text{kinematic viscosity of liquid water} = 0.23\times10^{-6} \text{ m$^2$/s}\] From which, \[\mu_l = \nu_l \rho_l =0.23\times10^{-6} \times 939 = 2.2\times10^{-4} \text{ kg/m.s}\] Data for liquid water from data book: (page no. 22): \[C_{Pl} = 4266.5 \text{ J/kg.K} \qquad \text{and} \qquad \text{Pr}_l = 1.3435\] From page no. 144 of data book, \[C_{sf} = 0.013\]

Substituting the data in Eqn.(2), we get \[\begin{aligned} q &= 2.2\times10^{-4}\times2188000\times\left[9.812\times\frac{(939-1.299)}{0.0538}\right]^{0.5}\times \\ &\ \qquad\qquad \qquad \qquad \left[\frac{4266.5}{0.013\times2188000\times1.3435} \right]^3\times(\Delta T)^3 \\ &= 481.36\times413.5\times1.392\times10^{-3}\times(\Delta T)^3 = 277.1(\Delta T)^3 \end{aligned}\] Equating \(q\) to \(q_c\), we get \[2117300 = 277.1(\Delta T_c)^3 \qquad \Longrightarrow \quad \Delta T_c= 19.7^{\circ}C\] Therefore, the excess temperature = \(\Delta T=T_w-T_{\text{sat}}={\bf 19.7^{\circ}C}\)