Правка разделов «Statement», «Answer»
en/4.5.17.md
+2 −3
| @@ -1,7 +1,6 @@ | |||
| ### Statement | |||
| − | $4.5.17.$ [Insert the problem statement] | ||
| − | |||
| + | $4.5.17.$ Determine the maximum and minimum pressure inside a spherical liquid droplet immersed in another liquid. The distance from the center of the droplet to the free surface of the liquid is $h$, the radius of the droplet is $R$, the density of both liquids is $\rho$, and the surface tension at the liquid-liquid interface is $\sigma$. | ||
| ### Solution | |||
| The pressure inside the droplet is the sum of the Laplace pressure and the pressure of the surrounding liquid. Let $x$ be the distance from the liquid surface to the point inside the droplet,then the pressure inside the droplet: | |||
| \begin{equation} | |||
| P(x)=\rho gx+\frac{2\sigma}{R} | |||
| \end{equation} | |||
| Now we see that the resulting function is linear,so we get that the pressure reaches the minimum value when $x=h-R$ and maximum value when $x=h+R$,so: | |||
| \begin{equation} | |||
| P_{max}=\rho g(h+R)+\frac{2\sigma}{R} | |||
| \end{equation} | |||
| \begin{equation} | |||
| P_{min}=\rho g(h-R)+\frac{2\sigma}{R} | |||
| \end{equation} | |||
| @@ -26,4 +25,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $P_{max}=\rho g(h+R)+\frac{2\sigma}{R}$.$P_{min}=\rho g(h-R)+\frac{2\sigma}{R}$ | ||
| @@ -1,7 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $4.5.17.$ [Insert the problem statement] | $4.5.17.$ Determine the maximum and minimum pressure inside a spherical liquid droplet immersed in another liquid. The distance from the center of the droplet to the free surface of the liquid is $h$, the radius of the droplet is $R$, the density of both liquids is $\rho$, and the surface tension at the liquid-liquid interface is $\sigma$. | ||
| ### Solution | ### Solution | ||
| The pressure inside the droplet is the sum of the Laplace pressure and the pressure of the surrounding liquid. Let $x$ be the distance from the liquid surface to the point inside the droplet,then the pressure inside the droplet: | The pressure inside the droplet is the sum of the Laplace pressure and the pressure of the surrounding liquid. Let $x$ be the distance from the liquid surface to the point inside the droplet,then the pressure inside the droplet: | ||
| \begin{equation} | \begin{equation} | ||
| P(x)=\rho gx+\frac{2\sigma}{R} | P(x)=\rho gx+\frac{2\sigma}{R} | ||
| \end{equation} | \end{equation} | ||
| Now we see that the resulting function is linear,so we get that the pressure reaches the minimum value when $x=h-R$ and maximum value when $x=h+R$,so: | Now we see that the resulting function is linear,so we get that the pressure reaches the minimum value when $x=h-R$ and maximum value when $x=h+R$,so: | ||
| \begin{equation} | \begin{equation} | ||
| P_{max}=\rho g(h+R)+\frac{2\sigma}{R} | P_{max}=\rho g(h+R)+\frac{2\sigma}{R} | ||
| \end{equation} | \end{equation} | ||
| \begin{equation} | \begin{equation} | ||
| P_{min}=\rho g(h-R)+\frac{2\sigma}{R} | P_{min}=\rho g(h-R)+\frac{2\sigma}{R} | ||
| \end{equation} | \end{equation} | ||
| @@ -26,4 +25,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $P_{max}=\rho g(h+R)+\frac{2\sigma}{R}$.$P_{min}=\rho g(h-R)+\frac{2\sigma}{R}$ | ||