91Ó°ÊÓ

The height $h_1$as shown in the figure is,$h_1=25cm=0.25m,$

The height das shown in the figure is,$d=12cm=0.12m,$

The height $h_2$as shown in the figure is, $h_2=40cm=0.4m$.

Applying the principle of conservation of mechanical energy to the flow of an ideal fluid leads to Bernoulli’s equation along any tube of flow:

$p+\frac{1}{2}\mathrm{ÒÏ}{v}^2+\mathrm{ÒÏ}gy=\text{Constant}$

Here, p is pressure, ÒÏ is density, v is velocity, g is the acceleration due to gravity and y is height.

We can find the speed of the liquid at point C by applying Bernoulli’s equation to points B and C. Then we can find the pressure in the liquid at the topmost point B.By applying Bernoulli’s equation to points B and C, we can find the greatest possible height h₁ that a siphon can lift water by using the result obtained for part b.

Let’s consider point D on the surface of the liquid in the container.

Applying Bernoulli’s equation to points C and D yields

$$\begin{gathered} P_D+\frac{1}{2}\mathrm{ÒÏ}{v}_D^2+\mathrm{ÒÏ}g{h}_D=P_C+\frac{1}{2}\mathrm{ÒÏ}{v}_C^2+\mathrm{ÒÏ}g{h}_C \\ \frac{1}{2}\mathrm{ÒÏ}{v}_C^2=P_D+\frac{1}{2}\mathrm{ÒÏ}{v}_D^2+\mathrm{ÒÏ}g{h}_D-P_C-\mathrm{ÒÏ}g{h}_C \\ {v}_C^2=\frac{2\left(P_D-P_C\right)}{\mathrm{ÒÏ}}+v_D^2+2g\left(h_D-h_C\right) \\ {v}_C=\sqrt{\frac{2\left(P_D-P_C\right)}{\mathrm{ÒÏ}}+v_D^2+2g\left(h_D-h_C\right)} \end{gathered}$$

Since $P_D=P_C=P_{air}$and $v_D~0,$the above equation becomes

$v_C=\sqrt{2g\left(h_D-h_C\right)}$

From the given figure, we can write

$v_C=\sqrt{2g\left(d+h_2\right)}$

Substituting the given values in the above equation, we get

$\begin{array}{rcl}{v}_C& =& \sqrt{2\left(9.8{\text{m/s}}^2\right)\left(0.4\text{m}+0.12\text{m}\right)}\\ & =& 3.19\text{m/s}\\ & ≈& 3.2\text{m/s}\\ & & \end{array}$

Therefore, the speed of the liquid at point C is$3.2\text{m/s}$ .

Applying Bernoulli’s equation to points B and C yields

$P_B+\frac{1}{2}\mathrm{ÒÏ}{v}_B^2+\mathrm{ÒÏ}g{h}_B=P_C+\frac{1}{2}\mathrm{ÒÏ}{v}_C^2+\mathrm{ÒÏ}g{h}_C$

So,

$P_B=P_C+\frac{1}{2}\mathrm{ÒÏ}{v}_C^2+\mathrm{ÒÏ}g{h}_C-\frac{1}{2}\mathrm{ÒÏ}{v}_B^2-\mathrm{ÒÏ}g{h}_B$

By the equation of continuity, we can write as

$v_B=v_C$

Since, $P_C=P_{air}$the above equation becomes,

$P_B=P_{air}+\mathrm{ÒÏ}g(h_C-h_B)$

From the given figure, we can write

$P_B=P_{air}-\mathrm{ÒÏ}g(h_1+h_2+d)$

Substituting the given values in the above equation we get,

$\begin{array}{rcl}{P}_B& =& 1.01×{10}^5-\left(1000{\text{kg/m}}^3\right)\left(9.8{\text{m/s}}^2\right)\left(0.25\text{m}+0.4\text{m}+0.12\text{m}\right)\\ & =& 93454\text{Pa}\\ & ≈& 9.3×{10}^{4}Pa\\ & & \end{array}$

Therefore, the pressure in the liquid at the topmost point B is$9.3×{10}^{4}Pa.$

From part b, we can write that$P_Bâ\copyright ¾0.$Then,

$P_{air}-\mathrm{ÒÏ}g(h_1+h_2+d)â\copyright ¾0$

Since,$h_1<h_{1,max}$the above inequality becomes

$P_{air}-\mathrm{ÒÏ}g\left(h_2+d\right)â\copyright ¾0$

$\frac{P_{air}}{\mathrm{ÒÏ}g}-\left(h_2+d\right)â\copyright ¾0$

But,

$\frac{P_{air}}{\mathrm{ÒÏ}g}-\left(h_2+d\right)â\copyright ½\frac{P_{air}}{\mathrm{ÒÏ}g}$

Substituting the given values in the above equation, we get

$\begin{array}{rcl}\frac{P_{air}}{\mathrm{ÒÏ}g}& =& \frac{1.01×{10}^5\text{Pa}}{\left(1000{\text{kg/m}}^3\right)\left(9.8{\text{m/s}}^2\right)}\\ & =& 10.3\text{m}\\ & & \end{array}$

Therefore, the greatest possible height h₁ that a siphon can lift water is$10.3\text{m}$ .