Question

Figure 14-56 shows a siphon,which is a device for removing liquid from a container. Tube ABCmust initially be filled, but once this has been done, liquid will flow through the tube until the liquid surface in the container is level with the tube opening at A.The liquid has density $\mathbf{1000}\mathbf{}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$and negligible viscosity. The distances shown are, $\mathit{h}\mathbf{1}\mathbf{=}\mathbf{25}\mathbf{}\mathit{c}\mathit{m}$d=12cm, and.$\mathit{h}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{40}\mathbf{}\mathit{c}\mathit{m}$ (a) With what speed does the liquid emerge from the tube at C? (b) If the atmospheric pressure is $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}$Pa, what is the pressure in the liquid at the topmost point B?(c)Theoretically, what is the greatest possible heightthat a siphon can lift water?

Short answer

1. The speed of the liquid at point C is $3.2\text{m/s}$.
2. The pressure in the liquid at the topmost point B is$9.3\times {10}^{4}Pa$.
3. The greatest possible height h₁ that a siphon can lift water is$10.3\text{m}$ .

Step-by-step solution

Listing the given quantities

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, $h_2=40cm=0.4m$.

Understanding the Bernoulli’s Principle

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:

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

Here, p is pressure, ρ 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.

(a) Calculation of speed of liquid at point c

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

Applying Bernoulli’s equation to points C and D yields

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

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

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

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

(b) Calculation of the pressure in the liquid at the topmost point B

Applying Bernoulli’s equation to points B and C yields

$P_B+\frac{1}{2}\rho v_B^2+\rho g{h}_B=P_C+\frac{1}{2}\rho v_C^2+\rho g{h}_C$

So,

$P_B=P_C+\frac{1}{2}\rho v_C^2+\rho g{h}_C-\frac{1}{2}\rho v_B^2-\rho g{h}_B$

By the equation of continuity, we can write as

$v_B=v_C$

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

$P_B=P_{air}+\rho g(h_C-h_B)$

From the given figure, we can write

$P_B=P_{air}-\rho g(h_1+h_2+d)$

Substituting the given values in the above equation we get,

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

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

(c) Calculation of greatest possible height that a siphon can lift water

From part b, we can write that$P_B⩾0.$Then,

$P_{air}-\rho g(h_1+h_2+d)⩾0$

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

$P_{air}-\rho g\left(h_2+d\right)⩾0$

$\frac{P_{air}}{\rho g}-\left(h_2+d\right)⩾0$

But,

$\frac{P_{air}}{\rho g}-\left(h_2+d\right)⩽\frac{P_{air}}{\rho g}$

Substituting the given values in the above equation, we get

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

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