In order to calculate the terminal velocity and collision efficiency, the equations of motion need to be solved. Droplet motion depends on the following three forces: gravity, the flow drag force, and the electrostatic force due to droplet charge and the external electric field. The equations of motion for a pair of droplets are as follows: 4adv1dt=g-C6πr1μm1v1-u2+Fe1m1 4bdv2dt=g-C6πr2μm2v2-u1+Fe2m2, where subscripts 1 and 2 denote droplet 1 and droplet 2, respectively, g is the gravitational acceleration, v is the velocity of the droplet relative to the flow if there are no other droplets present, u is the flow velocity field induced by the droplet, μ is the dynamic viscosity of air, C is the drag coefficient, which is a function of Reynolds number, r is the droplet radius, and m is the droplet mass, with m=4πr3ρ/3, and Fe is the electrostatic force. We set the air temperature at T=283 K and pressure p=900 hPa in this study for the calculation of air dynamic viscosity.

The flow drag force is described by the second term on the right hand side of Eq. (4), which assumes a simple hydrodynamic interaction of the two droplets. That is, each droplet moves in the flow field induced by the other one moving alone, and it is called the superposition method in cloud physics. This method has been successfully used in many studies for the calculation of collision efficiencies (Pruppacher and Klett, 1997). The superposition method can also ensure that the stream function satisfies the no-slip boundary condition (i.e., Wang et al., 2005). To calculate the flow drag force, the induced flow field u is required. The method for obtaining the induced flow field u is discussed below.

Considering a rigid sphere moving in a viscous fluid with a velocity U relative to the flow, the stream function depends on the Reynolds number of this spherical particle, NRe=2rvρμ, where ρ is the density of the air, and μ is the dynamic viscosity of the air. It is known that when the Reynolds number is small, the flow is considered as being a Stokes flow, and the stream function can be expressed as follows: 5ψs=U14R̃-3R̃4sin⁡2θ0, where R̃=R/r is the normalized distance (R is the distance from the sphere center, and r is the droplet radius), θ0 is the angle between the droplet velocity and vector R pointing from the sphere center. U is the value of the droplet velocity relative to the flow, i.e., U1=|v1-u2| for droplet 1 and U2=|v2-u1| for droplet 2. However, this stream function for the Stokes flow does not apply to the system with a large Reynolds number. Hamielec and Johnson (1962, 1963) gave the stream function ψh induced by a moving rigid sphere, which can be used for flows with large Reynolds numbers as follows: 6ψh=UA1R̃+A3R̃2+A3R̃3+A4R̃4sin⁡2θ0-UB1R̃+B3R̃2+B3R̃3+B4R̃4sin⁡2θ0cosθ0, where A1,…,B4 are functions only of the Reynolds number NRe for each droplet. The method is valid for NRe<5000. But the solution deviates from the Stokes flow solution when NRe→0 for small droplets. Cloud droplets with radii ranging from 2 to 1024 µm typically have a Reynolds number ranging from 10-4 to 103. Therefore, it is necessary to construct a stream function that applies to a wide range of NRe. This work adopts a stream function that is a linear combination of ψh and a Stokes stream function ψs (Pinsky and Khain, 2000) as follows: 7ψ=NReψh+NRe-1ψsNRe+NRe-1, which converges to a Stokes flow when NRe→0. Then the induced flow field u is derived as follows: 8u=-1R̃2sin⁡θ0∂ψ∂θ0e^R+1R̃sin⁡θ0∂ψ∂R̃e^θ=uRe^R+uθe^θ, where e^R and e^θ are the unit vectors in the polar coordinate (R, θ0). It can also be expressed in Cartesian coordinates (x, z) as follows: 9u=(uRcos⁡φ-uθsin⁡φ)e^z+(uRsin⁡φ+uθcos⁡φ)e^x, where the direction of e^z is vertically down, the same as gravitation. φ is the angle between e^z and the droplet velocity v.

Both the Stokes and Hamielec stream functions satisfy the no-slip boundary condition, i.e., the fluid velocity on the surface of the droplet is equal to the velocity of the droplet. The Hamielec stream function is no-slip because those functions A1,…,B4 in Eq. (6) satisfy A1+2A2+3A3+4A4=1 and B1+2B2+3B3+4B4=0, as long as the droplet is considered as a rigid sphere (Hamielec, 1963). These relations ensure that uθ=-Usin⁡θ0 at the surface of the droplet, which means the no-slip boundary condition. (Note that uθ is the tangential component of the velocity of the fluid, and Usin⁡θ0 is the tangential velocity of the droplet surface.)

According to an empirical equation of Beard (1976), the drag coefficient C in Eq. (4) is a function of NRe as follows: 10C=1+exp⁡a0+a1X+a2X2, where X=ln⁡ (NRe) and fitting constants a0,a1,a2 are from Table 1 of Beard (1976). The drag coefficient increases with the Reynolds number. For example, the terminal velocity of a droplet of 2 µm in radius is 4.92 × 10-4 cm s-1, with NRe=1.23 × 10-4 and C=1.00001; the terminal velocity of a droplet of 32 µm in radius is 11.8 cm s-1, with NRe=0.47 and C=1.07; the terminal velocity of a droplet of 1024 µm in radius is 715 cm s-1, with NRe=915 and C=18.0.

For droplets with r<10 µm, the assumption of no-slip boundary condition is no longer valid because droplet sizes are comparable with the mean free path of air molecules. Air cannot be considered as a continuous medium. The flow slips on the droplet surface. To take this effect into consideration, the drag coefficient should be multiplied by another coefficient (Lamb and Verlinde, 2011, p. 386) as follows: 11C′=C⋅1+1.26λr-1, where λ is the free path of air molecules, and r is the droplet radius.

The electric force is described by the third term on the right side of Eq. (4). The electric force includes the interactive force between the two charged droplets, and it is also an external electric force if there is an external electric field. For two point particles, we apply Coulomb's law as follows: 12Fe=-14πε0q1q2R2e^R, where Fe is the interactive force between point charges q1 and q2, and R is the distance between the two point charges. However, this inverse-square law does not apply to uneven charge distribution such as the case of charged cloud droplets.

The interaction between charged conductors is a complex mathematical problem in physics. Davis (1964) demonstrated an appropriate computational method for an electric force between two spherical conductors in a uniform external field. The electric force depends on droplet radius (r1,r2), charge (q1,q2), center distance R, electric field E0, and the angle θ between the electric field and the line connecting the centers of two droplets (note that θ=θ0+φ). The resultant electric force acting on droplet 2 is expressed as follows: 13Fe2=E0q2cos⁡θe^R+E0q2sin⁡θeθ^+{r22E02F1cos⁡2θ+F2sin⁡2θ+E0cos⁡θ(F3q1+F4q2)+1r22F5q12+F6q1q2+F7q22}eR^+{r22E02F8sin⁡2θ+E0sinθF9q1+F10q2}eθ^, where eR^ is the radial unit vector, eθ^ is the tangential unit vector, E0 is the eternal electric field, and parameters F1 to F10 are a series of complicated functions of geometric parameters (r1,r2,R; Davis, 1964).

The electric force directly from the external field is shown as two terms in Eq. (13) and can be simply written as E0q2 if the two terms are combined. The second half of Eq. (13) represents the interactive force from droplet 1 in the radial direction and tangential direction, respectively. Note that the third term represents the interactive force from droplet 1 if there is no external electric field. Except for this term, all the other terms in Eq. (13) are the interactive forces from droplet 1 due the induction from the external field.

Similarly, the resultant electric force Fe1 acting on droplet 1 includes both the force directly from the external field and the interactive force from droplet 2. The sum of the electric forces on the two droplets, Fe1+Fe2, must equal to the external electric force acting on the system, which can be expressed as E0q1+q2, because the two droplets can be considered as a system. Then, the electric force acting on droplet 1 can be derived immediately as follows: 14Fe1=E0q1+q2-Fe2. Figure 2 is a schematic diagram showing the forces acting on each droplet in a pair. Also shown in Fig. 2 is the velocity of each droplet relative to the flow, if there are no other droplets present (v), and the flow velocity induced by the other droplet (u). Droplet velocity relative to the flow is v-u. The electric field E0 is in the downward direction, which is the same as gravity. Droplet 1 has a positive charge and droplet 2 has a negative charge in this example. The forces acting on each droplet include gravity, flow drag force, and the electrostatic force, as seen on the right side of Eq. (4). For droplet 1, the electric force directly from the external field is in the downward direction and is shown as E0q1 in the figure. The interactive electric force from droplet 2, shown as Finter in the figure, has a radial component and a tangential component, so that it is in a direction that does not necessarily align with the line connecting the two droplets. Because of the interactive electric force from droplet 2, the velocity v of droplet 1 is not in the vertical direction. The electrostatic force between charged droplets tends to make the droplets attract each other. This force is particularly strong when droplets are close to each other, thus enhancing collisions. The flow drag force on droplet 1 is in the opposite direction with v-u.

If there is no external electric field but only a charge effect, Eq. (13) is reduced to the following: 15Fe2=1r22F5q12+F6q1q2+F7q22eR^. To illustrate it, a comparison between the electrostatic forces derived by the inverse-square law and conductor model without electric field (i.e., Eq. 15) is shown in Fig. 3, where the electric force between droplets with opposite-sign charges (dashed lines) and with same-sign charges (solid lines) varies with distance. When R≫r1,r2, we have F5,F7→0,F6→r22/R2, and it is also shown that two models are basically identical in remote distance. But when the spheres approach closely, the conductor interaction (blue lines) changes to strong attraction, because of electrostatic induction. The interaction is always attraction at a small distance, regardless of the sign of charges. If there is only an inverse-square law without electrostatic induction, it is obvious that the same-sign charges must decrease collision efficiency. However, after taking electrostatic induction into account, the effects of same-sign and opposite-sign charges need to be reconsidered.

The equations of motion (Eq. 4), along with the other equations in this section, are used to calculate the terminal velocities of charged droplets. Note that the terminal velocity refers to the steady-state velocity of a droplet relative to the flow when there are no other droplets present, as mentioned earlier. Therefore, by setting the induced flow u to be 0, Eq. (4) can be integrated to obtain the terminal velocity of the droplets with electric charge and field.

Equation (4), along with other equations, is also integrated to obtain the trajectories for the two droplets in any possible droplet pair (r1,q1 and r2,q2) for different strengths of the downward electric field (0, 200, and 400 V cm-1). The second-order Runge–Kutta method is used for the integration. The initial settings of droplet positions and velocities and the flow velocities are required. To save computational power, the initial vertical distance is set to be 30 (r1+r2), as an approximation of infinity. The initial flow velocity field u1 and u2 are set to be zero. Initial velocities of the two droplets are set to be the terminal velocities V1 and V2. We vary the initial horizontal distance using the bisection method until we find a threshold distance rc, which is the maximum horizontal distance at which the two droplets can collide. The threshold distance is found with a precision of 0.1 %. The collision cross section Sc=πrc2 and collision efficiency E are than calculated.

After computing the collision efficiencies E for a droplet pair with (r1,q1) and (r2,q2), the collection kernel K(r1,q1,r2,q2) is then derived. With the collection kernel K(r1,q1,r2,q2), the effect of the electric charges and fields on the droplet collision is determined by solving the SCE.

To solve the stochastic collection equation (Eq. 3) numerically, droplet radius and charge are both divided into discrete bins that are logarithmically equidistant. Droplet radius, ranging from 2 to 1024 µm, is divided into 37 bins, with the radius increasing by a factor of 21/4 from one bin to the next. Droplets with radii larger than 1024 µm are assumed to precipitate out and are not included in the size distribution.

In each radius bin, droplets may have different amounts and different signs of charges. For the bin of radius r, droplet charge ranges from -32 r2 to +32 r2 (in units of elementary charge and r in µm). This means that smaller droplets have a smaller range of charge. The setting here is based on the observations that the charge amount is proportional to the square of droplet radius, as discussed in the Introduction. The upper limit charge bin of 32 r2 is close to the thunderstorm condition of 42 r2. The charge range is then divided into 15 bins, with the center bin having zero charge, seven bins to the right having positive charges, and seven bins to the left having negative charges. For the positive charge bins, the one next to the center bin has a charge of +0.5 r2. The charge amount is increased by a factor of 2 from this bin to the next, until the upper limit of 32 r2 is reached. The setting for the negative charge bins is symmetric to the positive charge bins. For the size bins and charge bins described above, a large matrix of kernel Kr1,q1,r2,q2 is computed in advance as a lookup table for solving the SCE.

Droplet size and charge after collision–coalescence usually do not fall into any existing bins. A simple method is to linearly redistribute the droplets to the two neighboring bins (Khain et al., 2004). We first redistribute the droplets to the size bins. The ratio of redistribution is simultaneously based on total mass conservation and droplet number conservation. For example, to redistribute droplets with mass m (mi<m<mi+1) and number Δn, a proportion of Δni=mi+1-mmi+1-miΔn is added to the ith bin, and Δni+1=m-mimi+1-miΔn is added to the (i+1)th bin. These droplets are then redistributed to the charge bins within each size bin, satisfying the total charge conservation and droplet number conservation. For example, to redistribute droplets with charge q (qi,j<q<qi,j+1) within the ith size bin, a proportion of Δni,j=qi,j+1-qqi,j+1-qi,jΔni is added to the bin of (i,j), and a proportion of Δni,j+1=q-qi,jqi,j+1-qi,jΔni is added to the bin of (i,j+1).

As shown in Fig. 4, the collision–coalescence between bin (r1,q1) and bin (r2,q2), shown with black dots, generates droplets shown with the red dot. These newly generated droplets are then redistributed into two size bins and are further redistributed into two charge bins within each of the size bins, as shown with the blue dots. Note that the numbers close to each of the blue dots in Fig. 4 are the percentages of droplets that are redistributed into that bin. In fact, this method only reaches first-order accuracy. Although Bott (1998) compared several methods for redistributing droplets with higher-order correction, the two-parameter distribution is too complicated to do the higher-order correction in this study.

The initial droplet size distribution used in this study is derived based on an exponential function in Bott (1998) as follows: 16nm=Lm¯2exp⁡-mm¯, where nm is the distribution of droplet number concentration over droplet mass, L is the liquid water content, and m¯ is the mean mass of droplets. This function is used to derive nln⁡(r), which is the distribution of droplet number concentration over droplet radius. With the definitions of nm and nln⁡(r), and m=4πr3ρ/3, where ρ is droplet density, we can derive nln⁡(r) as follows: 17nln⁡(r)=dNdln⁡(r)=rdNdr=rdNdm4πρr2=4πρr3nm. By substituting Eq. (16) into Eq. (17) and assuming that m¯=4πr¯3ρ/3, where r¯ is the mean radius, we have the following: 18n(ln⁡(r))=L9r34πr¯6exp⁡-r3r¯3. Equation (18) is used as the initial droplet size distribution for the calculations of collision–coalescence in this study. It has two parameters, L and r¯, and can be considered as a gamma distribution. Using parameters L and r¯ in the initial size distribution has the advantage of representing the aerosol effect. The parameter L can be set as a constant. Using a different mean radius can represent different aerosol conditions and a different number concentration of cloud droplets.

A total of 12 cases with different initial conditions are considered for studying the evolution of droplet distribution. The mean droplet radius r¯ is set with three different sizes, namely 15, 9, and 6.5 µm, where r¯=15 µm case represents clean conditions and 6.5 µm represents polluted conditions. The liquid water content in our study is set at L=1 g m-3, which is a typical value in warm clouds, according to observations (Warner, 1955; Miles et al., 2000). With the fixed liquid water content, a smaller mean radius corresponds to a larger number concentration. As shown in Table 1, r¯=15, 9, and 6.5 µm give an initial droplet number concentration of 71, 325, and 851 cm-3, respectively.

For each r¯, comparisons are made among four different electric conditions in which (a) droplets are uncharged, (b) droplets are charged but have no external electric field, (c) droplets are charged and also have an external downward electric field of 200 V cm-1, and (d) droplets are charged and also have an external downward electric field of 400 V cm-1. For the uncharged cloud, the initial distribution is shown in Fig. 5a, where all droplets are put in the bins with no charge. For the charged clouds, an initial charge distribution shown in Fig. 5b is made as follows. To simulate an early stage of the warm cloud precipitation, we need to distribute the droplets in each size bin to different charge bins so that these droplets have different charges. Since there is little data on this, we assume a Gaussian distribution as follows: 19Nq=N02πσexp⁡-q22σ2, where N0 is the number concentration in the size bin, and σ is the standard deviation of the Gaussian distribution in that size bin. Nq represents the number concentration of droplets with charge q. This distribution satisfies electric neutrality q¯=0. For different size bins, the droplet number concentration N0 is different. We purposely set the standard deviation σ to be different for different size bins. For a larger size, the charge amount is larger, based on |q|¯=1.31 r2 (q in units of elementary charge and r in µm) as stated in the Introduction. Therefore, we set a larger standard deviation σ for the larger size bins. With this setting of droplet charge, the total amount of charge in each case is shown in Table 1. The r¯=15, 9, and 6.5 µm cases have an initial charge concentration of 9438, 15 638, and 21 634e cm-3, respectively, for both positive charge and negative charge.

The initial electric charges and electric field strength are set according to the conditions in warm clouds or in the early stage of thunderstorms. In fact, in some extreme thunderstorm cases, both the electric charge and field could be 1 order of magnitude larger (Takahashi, 1973) than the values used in this study. Furthermore, in natural clouds, the electric charge on a droplet leaks away gradually. In this study, the charge leakage is assumed to be a process of exponential decay (Pruppacher and Klett, 1997), and the relaxation time is set to τ=120 min. Thus, all the bins lose Δtτ of electric charge in each time step of Δt=1 s.

Here we present collision efficiencies for typical droplet pairs to illustrate the electrostatic effects. During the evolution of droplet size distribution, the radius and charge amount of colliding droplets have large variability. In addition, the charge sign of the colliding droplets may be the same or the opposite. Therefore, only some examples are shown.

The collision efficiencies for droplet pairs with no electric charge and field are presented in Fig. 6 as a reference. Collector droplets with radii larger than 30 µm are shown here to represent the precipitating droplets. The calculated collision efficiencies from this study are also compared with the measurements from previous studies. It is seen that results from this study are generally consistent with the measurements. Collision efficiencies increase as r2 changes from 2 to 14 µm and also increase as r1 changes from 30 to 305 µm. For two droplets that are both large enough, the collision efficiency could be close to one.

Figure 7 shows the collision efficiencies for droplet pairs with electric charge and field. Basically, droplet pairs that have no charge, same-sign charges, and opposite-sign charges are selected here and under the 0 and 400 V m-1 electric fields. Results for the collector droplet with a radius of 30 µm (Fig. 7a) and 40 µm (Fig. 7b) are shown. When comparing Fig. 7a and b, it can be seen that electrostatic effects are less significant for a larger collector. The electrostatic effects are even weaker for a collector radius larger than 40 µm (figures not shown). Therefore, we use the 30 µm collector as an example to explain the electrostatic effects on collision efficiencies below.

For the collector droplet with a radius of 30 µm (Fig. 7a), a noticeable, and sometimes significant, electrostatic effect can be seen. Compared to the droplet pair with no charge (line 1), the positively charged pair under no electric field (line 2) has a slightly smaller collision efficiency due to the repulsive force. As can be seen in Fig. 3, when the charged droplets move together, they first experience a repulsive force and then an attractive force at a small distance. The net effect is that the droplets have a smaller collision efficiency. The results for negatively charged pair under no electric field are identical to line 2 and are therefore not shown. When a downward electric field of 400 V m-1 is added, the positively charged pair (line 3) has a collision efficiency very close to the pair with no charge. This implies that the enhancement of collision efficiency by the electric field offsets the repulsive force effect. For a negatively charged pair in a downward electric field (line 4), the collision efficiency with a small r2 is significantly enhanced. This could be easily explained by electrostatic induction, i.e., the strong downward electric field induces a positive charge on the lower part of the collector droplet (even though it is negatively charged overall), so the negatively charged collected droplet below experiences an attractive force. In other words, we can approximately consider the collected droplet as a negative monopole (since it is very small) and consider the collector as a negative monopole plus a downward dipole that is induced by the electric field. When the two droplets are relativity far, the monopole–monopole interaction is dominant so that the force is repulsive. But when the two droplets come close, the monopole–dipole interaction becomes dominant in certain circumstances, so the force changes to attractive.

As for a pair with opposite-sign charges, line 5 in Fig. 7a shows that the collision efficiency is enhanced by the electrostatic effect even when there is no electric field. The collision efficiency is nearly 1 order of magnitude higher, with r2<5 µm. Line 6 in Fig. 7a shows that, with an electric field of 400 V cm-1, the electrostatic effect for the pairs with opposite-sign charges is even stronger. There is also an interesting feature in Fig. 7a; as the collector and collected droplets have similar sizes, the collision efficiency is high for the pairs with opposite-sign charges. This is quite different from the other four lines, where collision efficiencies are very low for droplet pairs with similar sizes.

Figure 8 shows the collision efficiencies for droplet pairs with charge and field and with smaller collectors. The collector droplet has a radius of 10 µm (Fig. 8a) and 20 µm (Fig. 8b) here. Collision efficiencies for these smaller collectors are much smaller than one when there is no charge (line 1 in Fig. 8a and b), which is already well known in the cloud physics community. However, the electrostatic effects are so strong that the collision efficiencies could be significantly changed for these collectors. For the collector droplet with a radius of 10 µm (Fig. 8a), the positively charged pair has a very small collision efficiency that is out of the scale in the figure, due to the dominating effect of the repulsive force, as discussed above. For the positively charged pair under a downward electric field, the collision efficiencies have a similar order of magnitude to the pair with no charge. For the negatively charged pair under the downward electric field, and for the pairs with opposite-sign charges, the electrostatic effects are very strong. The negatively charged pair even exhibits collision efficiency increases of as much as 2 orders of magnitude. Similarly, for the collector droplet with a radius of 20 µm (Fig. 8b), the electrostatic effect can lead to 1 order of magnitude increase in collision efficiencies.

It is evident that droplet charge and field can significantly affect the collision efficiency, especially for smaller collectors. This means that the electrostatic effects depend on the radius of collector droplets, and it mainly affects small droplets. The section below provides a detailed description on how these electrostatic effects can influence droplet size distributions.

This section shows the electrostatic effects on the evolution of different droplet size distributions. As discussed in Sect. 4, this study uses three initial size distributions, where r¯=15, 9, and 6.5 µm, respectively. For each initial size distribution, comparisons are made among four different electric conditions, namely uncharged droplets, charged droplets without electric field, charged droplets with a 200 V cm-1 electric field, and charged droplets with a 400 V cm-1 electric field. Note that charged droplets here refers to the initial charge distribution shown in Fig. 5. We also compare the results of the uncharged clouds with r¯=15, 9, and 6.5 µm, which represent the aerosol effects, and then investigate whether the electrostatic effects can mitigate the aerosol effects during the collision–coalescence process.

Figure 9 shows the evolution of the droplet size distribution with initial r¯=15 µm, which has an initial droplet number concentration of 71 cm-3. The four rows show different times (t=7.5, 15, 22.5, and 30 min) during the simulated evolution. The left column shows the size distribution of droplet mass concentration M(ln⁡r), and the right column shows the size distribution of droplet number concentration n(ln⁡r). They are related as M(ln⁡r) =4πr3ρ/3⋅n(ln⁡r). A second mode in size distribution gradually forms as droplets undergo the collision–coalescence process from t=7.5 to 30 min. Although the second mode can clearly be seen in the plots of n(ln⁡r), we show M(ln⁡r) here so that the second mode can be seen as a peak. In each panel, the dotted line denotes the initial size distribution (t=0 min) for reference. It is seen that droplet size distributions under four electric conditions have a similar behavior for initial r¯=15 µm in that they all evolve to a double-peak form, regardless of the electric charge or field. At 30 min, the four cases all have a modal radius of about 200 µm (Fig. 9d). The electrostatic effect is not notable for large droplets in the r¯=15 µm cases because the initial radius is large enough to start gravitational collision–coalescence quickly.

The evolution of droplet total number concentration and total positive charge concentration (also equal to the total negative charge concentration) is shown in Fig. 10. It is evident that droplet total number concentration decreases from 71 to less than 5 cm-3 in 30 min and is nearly not affected by the four different electric conditions. Both the positive charge and negative charge concentration decrease from 9384 to about 1000e cm-3 as droplets with opposite-sign charges go through collision–coalescence and charge neutrality occurs.

Figure 11 shows the evolution of the droplet size distribution with initial r¯=9 µm. For the uncharged cloud, it takes 60 min to have the second peak grow to about 200 µm. Therefore, the four panels of Fig. 11 show the simulated evolution for t=15, 30, 45, and 60 min. The charges and the electric fields have a more significant effect in the r¯=9 µm case than in the r¯=15 µm case. It is seen that, at 15 and 30 min, the clouds with different electric conditions evidently differ from each other, but the second mode is not obvious. At 45 min, the electrostatic effects on the second peak are evident. The charged cloud (red line) evolves more quickly than the uncharged cloud, as can been from the lower first peak and the growing second peak. Moreover, the downward electric fields further boost the collision–coalescence process of charged droplets (green and purple lines). At 60 min, the modal radius of the second peak is about 200 µm for the uncharged cloud, 300 µm for the charged cloud without an electric field, 500 µm for the charged cloud with a field of 200 V cm-1, and 700 µm for the charged cloud with a field of 400 V cm-1, respectively.

As for the evolution of the droplet total number concentration and charge concentration, Fig. 12 shows that they are distinctly affected by the four different electric conditions. The charged cloud with a field of 400 V cm-1 has a very low droplet number concentration and charge concentration at 60 min. The electrostatic effects play an important role in converting smaller droplets to larger droplets. The 2D distribution of droplet mass concentration for r¯=9 µm at 60 min is shown in Fig. 13. Figure 13a shows the uncharged situation. Figure 13b, c, and d show the situations with charges and with electric fields of 0, 200, 400 V cm-1, respectively. After 60 min of evolution, the distribution of mass over the charge bins is still symmetric. It is also shown that both mass and charges are transported from smaller droplets to larger droplets during collision–coalescence. Note that the integration of this 2D distribution along the charge bins gives a 1D distribution over the droplet size at 60 min, as shown in Fig. 11d.

Figure 14 shows the evolution of the droplet size distribution with initial r¯=6.5 µm. For the uncharged cloud, it takes 120 min to have the second peak grow to about 200 µm. Therefore, the four panels of Fig. 14 show the simulated evolution for t=30, 60, 90 and 120 min. The enhancement by the electric field on the collision–coalescence process is much more obvious than r¯=9 µm. After 90 min of evolution, the uncharged cloud (blue line) and charged cloud without a field (red line) are almost the same as the initial distribution. This is because the droplets are too small to initiate gravitational collision. At 120 min, a second peak has formed for the situations with no charge and with charge but no field. In contrast, under the external electric field of 200 and 400 V cm-1 (green and purple lines), the cloud droplets grow much more quickly than in the no-field situations. Some droplets have even evolved to larger than 1024 µm, which are supposed to precipitate out from the clouds. The evolution of droplet total number concentration and charge concentration is shown in Fig. 15, which indicates that droplet total number concentrations and charge concentration are strongly affected by the electrostatic effects. These results show that, the electric field would remarkably trigger the collision–coalescence process for the small droplets.

As for the initial mean droplet radius r¯<6 µm (figure not shown), similar to Fig. 14, the droplet size distribution of uncharged and charged clouds without electric field would have nearly no difference, while the effect of electric fields is much stronger. This means that the charge effect is relatively small compared to the electric fields when the initial droplet radius of the cloud is small enough.

Now we compare the electrostatic effects shown above with the aerosol effects. Let us take the cases with r¯=15 µm and r¯=9 µm as examples. When there are no electrostatic effects, the case with r¯=15 µm can develop a significant second peak in the size distribution in less than 30 min, while it takes about 60 min for the r¯=9 µm case to develop a similar second peak, as can be seen in Figs. 9 and 11. This can be regarded as an aerosol effect. When considering the electrostatic effects, it only takes about 45 min for the r¯=9 µm case to develop a similar second peak, as can be seen in Fig. 9. Therefore, the aerosol-induced precipitation suppression effect is mitigated by the electrostatic effects.