Chapter 1 - Problems - Page 32: 1.43

Consider the following assumptions: 1. Density of the air, $\rho_{\text {air }}=1 \mathrm{~kg} / \mathrm{m}^{3}$. 2. Density of the water, $\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^{3}$. 3. Radius of the rain droplet can be determined based on the capillarity. Calculate the percentage by which air water mixture density is greater than that of just still air. $$ \begin{aligned} \frac{\text { Air-water density }}{\text { Air density }} &=\frac{\rho_{\text {air }}\left(K_{\text {air }}-K_{\text {droplets }}\right)+\rho_{\text {droplets }} K_{\text {dropless }}}{\rho_{\text {air }} K_{\text {air }}} \\ &=\frac{\rho_{\text {air }}\left(\frac{4}{3} \pi R^{3}-N \frac{4}{3} \pi r^{3}\right)+\rho_{\text {droplees }}\left(N \frac{4}{3} \pi r^{3}\right)}{\rho_{\text {air }}\left(\frac{4}{3} \pi R^{3}\right)} \end{aligned} $$ Here, $K$ is the volume, $R$ is the arbitrary radius of the still air sphere, $r$ is the radius of the rain droplet, and $N$ is the number of droplets in the air water mixture sphere. Simplify the above equation to get the ratio in only two parameters that is the number of rain droplets and the radius of the arbitrary volume. $$ \begin{aligned} \frac{\text { Air-water density }}{\text { Air density }} &=\frac{\rho_{\text {air }}\left(\frac{4}{3} \pi R^{3}-N \frac{4}{3} \pi r^{3}\right)+\rho_{\text {droplets }}\left(N \frac{4}{3} \pi r^{3}\right)}{\rho_{\text {air }}\left(\frac{4}{3} \pi R^{3}\right)} \\ &=\frac{\rho_{\text {air }}\left(R^{3}-r^{3} N\right)+\rho_{\text {dropless }} N r^{3}}{\rho_{\text {air }} R^{3}} \\ &=1-\frac{r^{3} N}{R^{3}}+\frac{\rho_{\text {dropless }} N r^{3}}{\rho_{\text {air }} R^{3}} \end{aligned} $$ Here, the only inputs to be substituted are the number of rain droplets per arbitrary volume, and radius of the arbitrary volume, and $r$ can be estimated using capillarity of water.

Consider the following assumptions: 1. Density of the air, $\rho_{\text {air }}=1 \mathrm{~kg} / \mathrm{m}^{3}$. 2. Density of the water, $\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^{3}$. 3. Radius of the rain droplet can be determined based on the capillarity. Calculate the percentage by which air water mixture density is greater than that of just still air. $$ \begin{aligned} \frac{\text { Air-water density }}{\text { Air density }} &=\frac{\rho_{\text {air }}\left(K_{\text {air }}-K_{\text {droplets }}\right)+\rho_{\text {droplets }} K_{\text {dropless }}}{\rho_{\text {air }} K_{\text {air }}} \\ &=\frac{\rho_{\text {air }}\left(\frac{4}{3} \pi R^{3}-N \frac{4}{3} \pi r^{3}\right)+\rho_{\text {droplees }}\left(N \frac{4}{3} \pi r^{3}\right)}{\rho_{\text {air }}\left(\frac{4}{3} \pi R^{3}\right)} \end{aligned} $$ Here, $K$ is the volume, $R$ is the arbitrary radius of the still air sphere, $r$ is the radius of the rain droplet, and $N$ is the number of droplets in the air water mixture sphere. Simplify the above equation to get the ratio in only two parameters that is the number of rain droplets and the radius of the arbitrary volume. $$ \begin{aligned} \frac{\text { Air-water density }}{\text { Air density }} &=\frac{\rho_{\text {air }}\left(\frac{4}{3} \pi R^{3}-N \frac{4}{3} \pi r^{3}\right)+\rho_{\text {droplets }}\left(N \frac{4}{3} \pi r^{3}\right)}{\rho_{\text {air }}\left(\frac{4}{3} \pi R^{3}\right)} \\ &=\frac{\rho_{\text {air }}\left(R^{3}-r^{3} N\right)+\rho_{\text {dropless }} N r^{3}}{\rho_{\text {air }} R^{3}} \\ &=1-\frac{r^{3} N}{R^{3}}+\frac{\rho_{\text {dropless }} N r^{3}}{\rho_{\text {air }} R^{3}} \end{aligned} $$ Here, the only inputs to be substituted are the number of rain droplets per arbitrary volume, and radius of the arbitrary volume, and $r$ can be estimated using capillarity of water.