91Ó°ÊÓ

1. The speed of water flows into the atmosphere, $v_o=15m/s$.

2. The diameter of the inlet ,$d_i=5.0cm$.

3. The diameter of the outlet,$d_o=3.0cm$.

4. The pipe is horizontal.

5. The pipe outlet is open to the atmosphere.

Find the volume rate of flow per sec and the speed of water flowing into the pipe by using the continuity equation. Then, find the gauge pressure of water by using Bernoulli’s principle. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

Formulae are as follows:

$pâˆ\P Ä+\frac{1}{2}p{g}^{2}h+\mathrm{co}ns\tan t$

$A_v=cons\tan t=R_v=Volumerateofflowpersec$

Where, p is pressure, v is velocity, h is height, g is the acceleration due to gravity, h is height, A is the area, R is the rate of flow, and $\mathrm{ÒÏ}$is density.

The volume rate of flow of water can be calculated by using the continuity equation.

So,

$$\begin{gathered} R_v=Volumerateofflowpersec=A_v \\ {A}_v=\frac{{\mathrm{Ï€»å}}_{\mathrm{o}}^2}{4}{v}_o \\ =\frac{\mathrm{Ï€}{\left(3.0×{10}^{-2}\mathrm{m}\right)}^2×15\mathrm{m}/\mathrm{s}}{4} \\ =1.1×{10}^{-2}{m}^3/s \end{gathered}$$

Hence, the volume of water flowing into the atmosphere in 10 min is given by,

$$\begin{gathered} V=R_v×10×60 \\ =1.1×{10}^{-2}{m}^3/s×10×60 \\ =6.36 \\ ≈6.4m^3 \end{gathered}$$

Hence, the volume of water that flows into the atmosphere during 10 min period is localid="1657694542906" $6.4m^3$.

The flow of water obeys the continuity equation.

So,

$A_i{v}_i=A_o{v}_o$

Hence,

$$\begin{gathered} v_i=\frac{A_o{v}_o}{A_i} \\ =\frac{{\left(3.0×{10}^{-2}{m}^2\right)}^2×15m/s}{{\left(5.0×{10}^{-2}{m}^2\right)}^2} \\ =5.4m/s \end{gathered}$$

Hence, the speed of water flowing into the pipe is $5.4m/s$.

Now,

Gauge pressure = actual pressure – atmospheric pressure

Gauge pressure $\left(∆p\right)$can be calculated using Bernoulli’s principle.

$p_{i}v+\frac{1}{2}\mathrm{ÒÏ}{g}_i^{2}h+p_1=p_{o}v+\frac{1}{2}\mathrm{ÒÏ}{g}_o^{2}h+p_2$

Here,

$h_1=h_{2}and{p}_i=p_o$

Hence,

role="math" localid="1657694208416" $$\begin{gathered} p_{o}v=p_{i}v+\frac{1}{2}\left(g_i^2-g_o^2\right) \\ =1.01×{10}^{5}Pa+\frac{1}{2}\left(1000kg/m^3\right)\left({\left(15m/s\right)}^2-{\left(5.4m/s\right)}^2\right) \\ =1.99×{10}^{5}Pa \\ =1.97atm \end{gathered}$$

Using given information and simplifying, we get

$$\begin{gathered} ∆p=p_i-p_o \\ =\left(1.97-1.00\right)atm \\ =0.97atm \\ =9.8×{10}^{4}Pa \end{gathered}$$

Hence, the gauge pressure of water is $9.8×{10}^{4}Pa$.