On the growth of homogeneously nucleated water droplets in nitrogen: an experimental study

Abstract

A pulse-expansion wave tube method to determine homogeneous nucleation rates of water droplets has been improved. In particular, by accounting for background scattering, the experimental light scattering can be fitted extremely well with the Mie scattering theory. This results in an accurate determination of the droplet growth curve, which is well defined owing to the sharp monodispersity of the droplet cloud generated by the nucleation pulse method. With this method, water condensation is effectively decoupled in birth (nucleation) and growth of droplets. Droplet growth curves yield information on the diffusion coefficient, which only depends on pressure and temperature and on the supersaturation of the individual experiments. Here, we propose to use this information in the interpretation of nucleation rate data. Experimental results are given for homogeneous nucleation rates of supercooled water droplets at nucleation temperature 240 K and pressure 1.0 MPa and for growth of supercooled water droplets at temperature 247 K and pressure 1.1 MPa. The supersaturation was varied between 10 and 14, resulting in nucleation rates varying between 10$^{14}$ m$^{-3}$ s$^{-1}$ and 10$^{17}$ m$^{-3}$ s$^{-1}$. For the diffusion coefficient, a value of 1.51 $\pm$ 0.03 mm$^2$ s$^{-1}$ was found (247 K, 1.1 MPa) in agreement with previously reported results. It is discussed how the information from droplet growth data can be used to assess the quality of the individual water nucleation experiments.

Appendices

Appendix 1: Freezing of droplets that grow by vapor diffusion

In a nucleation experiment, a large number $N_0$ of monodisperse droplets are generated. They grow by condensing the available vapor. At a time $t, N$ droplets are still liquid, or more precisely, no nucleation event occurred in them until time $t$. With respect to the purity of the system, we can assume that only homogeneous ice nucleation can occur. As noted by Sigurbjörnsson and Signorell (2008), for nanometer-sized droplets present experimental techniques do not allow to distinguish between ice nucleation on the surface of droplets and inside them. The droplets observed in the present experiments are in the micron range, for which it is generally accepted that most nucleation events occur inside the droplets. Consequently, the number of droplets in which a nucleation event will occur during an (infinitesimally short) time interval ${\mathrm d} t$ is proportional to their volume:

$$\begin{aligned} N \, J \, V {\mathrm d} t\,. \end{aligned}$$

(17)

Here, $J$ is the homogeneous ice nucleation rate, i.e., the number of ice nuclei formed per unit of time (s) in a volume unit (m$^3$), and $V$ is the droplet’s volume. In other words, the number of liquid droplets $N$ decreases during the time interval $\hbox {d}t$ by:

$$\begin{aligned} \hbox{d} N= - N \, J \, V \hbox{d} t \,. \end{aligned}$$

(18)

By integrating Eq. 18, we obtain for the fraction of liquid droplets

$$\begin{aligned} P(t)=\frac{N}{N_0}=\exp \left( -\int \limits _0^t J \, V(t) \, {\mathrm d} t'\right) \,. \end{aligned}$$

(19)

$P(t)$ can also be interpreted as a probability that a given droplet remains liquid at time $t$.

As already discussed, the growth of droplets in the reported experiments is controlled by diffusion of the vapor toward the phase interface. By integrating Eq. 6 without considering the vapor depletion, one finds that the square of the droplet radius $r$ increases proportionally to time $t$,

$$\begin{aligned} r^2=\fancyscript{G} t \,. \end{aligned}$$

(20)

The growth rate $\fancyscript{G}$ is given as

$$\begin{aligned} \fancyscript{G}=2 \frac{\rho _{\mathrm{{g}}}^*}{\rho _{\mathrm{{l}}}^*} D(y-y_{\text {eq}}),\end{aligned}$$

(21)

and its magnitude is about $10^{-11}\,\mathrm {m^2\,s^{-1}}$. Since the temperature and pressure are constant during the growth, the nucleation rate $J$ can be assumed constant. As a consequence of the growth law given by Eq. 20, the volume increases in time as

$$\begin{aligned} V(t)=\frac{4}{3}\pi \, r^3=\frac{4}{3}\pi \, \fancyscript{G}^{3/2} t^{3/2} \,. \end{aligned}$$

(22)

Performing the integration in Eq. 19, we obtain

$$\begin{aligned} P(t)=\exp (-a t^{5/2})\,, \end{aligned}$$

(23)

where we defined the parameter $a$ as

$$\begin{aligned} a=\frac{8\pi }{15} \fancyscript{G}^{3/2} J\,. \end{aligned}$$

(24)

To represent the time in a dimensionless form, we propose a scaling time

$$\begin{aligned} t_{1/e}=a^{-2/5} \end{aligned}$$

(25)

and define the dimensionless time $\tau$ as

$$\begin{aligned} \tau = \frac{t}{t_{1/e}}\,. \end{aligned}$$

(26)

The scaling time has the physical meaning of the time at which the fraction of liquid droplets decreases to $1/e \, \doteq 0.3679$. Equation 23 can be rewritten in terms of the dimensionless time as

$$\begin{aligned} P(\tau )\,=\,\exp (-\tau ^{5/2})\,. \end{aligned}$$

(27)

Equation 27 can be easily inverted to give the dimensionless time at which the fraction of liquid droplets decreases to a given level:

$$\begin{aligned} \tau =(-\ln P)^{2/5}\,. \end{aligned}$$

(28)

For the purpose of the present experiments, we assume that the light scattering signal will be hardly affected if the fraction of frozen droplets is smaller that about 5 %, i.e., limiting the experimental time to $\tau \le \tau _{0.95}\,\doteq\,0.3048$.

The probability density $f$ of nucleation event at dimensionless time $\tau$ can be obtained as

$$\begin{aligned} f(\tau )=\frac{\mathrm d [1-P(\tau )]}{\mathrm d \tau } =\frac{5}{2}\tau ^{3/2} \exp (-\tau ^{5/2})\,. \end{aligned}$$

(29)

The number of ice nucleation events during a time interval ${\mathrm d} t$ is $N_0 f(\tau ) {\mathrm d} t/ t_{1/e}$. Its maximum value is reached at $\tau _{\text {max}}=(3/5)^{2/5} \, \doteq \, 0.8152$. The fraction of liquid droplets $P$ and the probability distribution function $f$ are shown in Fig. 26. The mean dimensionless lifetime of a liquid droplet can be computed as

$$\begin{aligned} \tau _{\text {m}}=\int \limits _0^\infty \tau \, f(\tau ) \mathrm \, d \tau =\frac{2}{5}\Gamma (2/5) \, \doteq \, 0.8873\,. \end{aligned}$$

(30)

Fig. 26

Probability $P(\tau )$ (solid line) that a droplet remains liquid at dimensionless time $\tau$ given by Eq. 27. Probability density $f(\tau )$ (dashed line) of an ice nucleation event at dimensionless time $\tau$ given by Eq. 29

Murray et al. (2010) measured ice nucleation rates in the range of temperatures from 234.9 to 236.7 K and provided a parametrization of the nucleation rate in a broad range of temperatures. Two variants of the parametrization are provided, differing in the exponent in their formula for the temperature dependency of the interfacial tension between liquid water and cubic ice. We computed the ice nucleation rate $J=5.9\times 10^{-18}\,\mathrm {m^{-3}\,s^{-1}}$ at 247 K using exponent 0.3, which best matches data by Murray et al. (2010) and other ice nucleation data at small supercoolings. For our growth rate $\fancyscript{G}=10^{-11}\,\mathrm {m^2\,s^{-1}}$, we found a mean lifetime of the liquid droplet $t_{\text {m}}=2.2\times 10^{13}$ s, corresponding to 70,000 years! Hence, freezing by homogeneous ice nucleation can be completely excluded at 247 K.

Ice nucleation, however, comes into play at temperatures not much lower. For a growth temperature of 237 K, roughly corresponding to a nucleation pulse temperature of 230 K, we computed the ice nucleation rate $1.3\times 10^{12}\,\mathrm {m^{-3}\,s^{-1}}$ and a mean liquid droplet lifetime of $41$ s. For a growth temperature of 227 K, corresponding to a nucleation temperature of 220 K, the nucleation rate increases to $7.1\times 10^{22}\,\mathrm {m^{-3}\,s^{-1}}$ and the mean lifetime of a liquid droplet shrinks to 2.1 ms. For the latter temperature, we adopted an exponent of $0.97$ for the interfacial tension formula given by Murray et al. (2010), since the nucleation rates computed using this value fit the data at deep supercoolings by Huang and Bartell (1995) and the recent data by Manka et al. (2012) well. It has been noted by Němec (2013) that the Laplace pressure in small droplets significantly increases the formation work of ice nuclei formed inside the droplets and, thus, decreases the predicted nucleation rate. In this way, the data at high supercoolings could be represented with an exponent of 0.35 in the interfacial tension formula that fits the data at small supercoolings as well. In the present work, the pressure inside the droplets is increased by the surrounding gas pressure (roughly 1.0 MPa). The pressure effect decreases the ice nucleation rate and, consequently, increases the lifetime of liquid droplets. Furthermore, vapor depletion, neglected in the growth law (Eq. 20), decreases and, ultimately, stops the growth and increases the lifetime of liquid droplets. These facts reconfirm that homogeneous ice nucleation is practically impossible for the experiments presented here.

Appendix 2: Nucleation data

The nucleation data are given in Table 2. The quantities $p_{\mathrm{{0}}}$ and $T_{\mathrm{{0}}}$ denote the initial pressure and temperature in the high-pressure section prior to the experiment. The pressure $p$ and temperature $T$ during the nucleation pulse are listed in the 4th and 5th column. Furthermore, the water vapor fractions for both methods (MPD and PEWT) $y$ are tabulated. The pulse duration $\Delta t$, the droplet number density $n_{\mathrm{{d}}}$, and the experimental nucleation rate $J_{\mathrm{{expt}}}$ are listed next. The corrected supersaturations $S'$ for both methods (MPD and PEWT) are given in the last two columns.

Table 2 Water nucleation data in nitrogen at 240 K and 1.0 MPa

Appendix 3: Correction of the supersaturation

In the PEWT setup, it is impossible in practice to obtain exactly the same nucleation temperature repeatedly. For each experiment, we therefore corrected the individual nucleation temperature $T$ with the average temperature $\bar{T}$. For this purpose, we assume that the Classical Nucleation Theory (Sect. 2.1) predicts the correct changes in the nucleation rate for small changes in supersaturation and temperature. Then, it is possible to estimate the supersaturation $S'$ that is obtained when the temperature is made equal to the mean temperature $\bar{T}$, while keeping the nucleation rate $J$ constant. A change in temperature leads to a different value for the surface tension $\sigma$ and liquid density $\rho _{\mathrm{{l}}}$. This is taken into account in the following corrected supersaturation:

$$\begin{aligned} S' = S + \frac{S \, \ln S}{2} \Big ( \frac{3}{\sigma } \frac{\partial \sigma }{\partial T} - \frac{2}{\rho _{\mathrm{{l}}}}\frac{\partial \rho _{\mathrm{{l}}}}{\partial T} - \frac{3}{T} \Big ) (\bar{T} - T). \end{aligned}$$

(31)

Below, we discuss the derivation of Eq. 31. In first order, the correction to the supersaturation is given by

$$\begin{aligned} S' = S + \Big ( \frac{\partial S}{\partial T} \Big ) _{J} (\bar{T}-T). \end{aligned}$$

(32)

The derivative $( \partial S / \partial T)_{J}$ is calculated by means of the Classical Nucleation Theory, which defines the nucleation rate as

$$\begin{aligned} J = J_{\mathrm{{0}}} \exp (-A \Psi ), \end{aligned}$$

(33)

with

$$\begin{aligned} A = \frac{16 \pi }{3} \frac{M^2}{N_{\mathrm{{A}}}^2 k^3} \end{aligned}$$

(34)

and $\Psi$ a function of the temperature given by

$$\begin{aligned} \Psi = \frac{\sigma ^3}{T^3 (\ln S)^2 \rho _{\mathrm{{l}}}^2}. \end{aligned}$$

(35)

Here, we have rewritten Eqs. 1 and 2. The Euler chain relation relates the derivative $( \partial S / \partial T)_{\mathrm{{\Psi }}}$ to the following ratio:

$$\begin{aligned} \Big ( \frac{\partial S}{\partial T} \Big ) _\Psi = \frac{ -\Big ( \frac{\partial \Psi }{\partial T} \Big ) _S}{\Big ( \frac{\partial \Psi }{\partial S} \Big ) _T}. \end{aligned}$$

(36)

Simple algebra gives the following results for the derivatives:

$$\begin{aligned} \Big ( \frac{\partial \Psi }{\partial T} \Big ) _S = \frac{3 \, \sigma ^3}{T^3 (\ln S)^2 \rho_{\mathrm{{l}}} ^2} \Big ( \frac{1}{\sigma } \frac{\partial \sigma }{\partial T} - \frac{1}{T} - \frac{2}{3}\frac{1}{\rho _{\mathrm{1}}}\frac{\partial \rho _{\mathrm{1}}}{\partial T} \Big ),\end{aligned}$$

(37)

$$\begin{aligned} \Big ( \frac{\partial \Psi }{\partial S} \Big ) _T = \frac{-2 \, \sigma ^3}{S \; \ln (S) \; T^3 \rho_{\mathrm{{l}}} ^2}. \end{aligned}$$

(38)

If Eqs. 37 and 38 are combined, the result is

$$\begin{aligned} \Big ( \frac{\partial S}{\partial T} \Big ) _{J} = \Big ( \frac{\partial S}{\partial T} \Big ) _\Psi = \frac{S \; \ln S}{2} \Big ( \frac{3}{\sigma } \frac{\partial \sigma }{\partial T} - \frac{2}{\rho _{\mathrm{{l}}}}\frac{\partial \rho _{\mathrm{{l}}}}{\partial T} - \frac{3}{T} \Big ). \end{aligned}$$

(39)

Equation 31 is retrieved when substituting Eq. 39 in Eq. 32.