91Ó°ÊÓ

The given data can be listed below as,

• The elevation of point 1 is, h₁= 10.0m .
• The elevation of point 2 and 3 is.h₂= h₃= 2.0m .
• The cross-sectional area at point 2 is, A₂= 0.0480m² .
• The cross-sectional area at point 3 is. A₃= 0.0160m².

Part (a)

The speed of efflux is expressed as,

$v=\sqrt{2gh}$

Here h is the distance of the hole below the surface of the fluid, and g is the gravitational acceleration.

The elevation at point 3 is,

$h=h_1−h_3$

The velocity at point 3 is expressed as,

$v_3=\sqrt{2g\left(h_1−h_3\right)}$

Here, v₃ is the velocity of water at point 3.

The discharge rate at point 3 is expressed as,

$\stackrel{Ë™}{Q}=A_3{v}_3$

Substitute the value of v₃ in the above equation.

$\begin{array}{l}\stackrel{˙}{Q}=A_3{v}_3\\ \stackrel{˙}{Q}=A_3×\sqrt{2g\left(h_1−h_3\right)}\end{array}$

Substitute $0.0160{\text{″¾}}^{\text{2}}{\text{forA}}_3\text{,}9.81\text{″¾}/{\text{s}}^{\text{2}}\text{forg}$, 10.0 m for h₁ and 2.0 m for h₃ in the above equation.

$\begin{array}{c}\stackrel{Ë™}{Q}=0.0160{\text{″¾}}^{\text{2}}×\sqrt{2×9.81\text{″¾}/{\text{s}}^{\text{2}}×\left(10.0\text{″¾-}2.0\text{″¾}\right)}\\ =0.0160{\text{″¾}}^{\text{2}}×\sqrt{2×9.81\text{″¾}/{\text{s}}^{\text{2}}×8.0\text{″¾}}\\ =0.0160{\text{″¾}}^{\text{2}}×12.52\text{″¾}/\text{s}\\ =0.20{\text{″¾}}^{\text{3}}/\text{s}\end{array}$

Hence the discharge rate is 0.20 m³/s.

Part (b)

At the point 3 the pressure is atmospheric and point 2 the pressure is gauge pressure. The elevation for both points is same then, it is expressed as,

$\begin{array}{c}{p}_2+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2+\mathrm{ÒÏ}g{h}_2=p_3+\frac{1}{2}\mathrm{ÒÏ}{v}_3^2+\mathrm{ÒÏ}g{h}_3\\ p_2=\frac{1}{2}\mathrm{ÒÏ}\left(v_3^2−v_2^2\right)\\ =\frac{1}{2}\mathrm{ÒÏ}{v}_3^2\left(1−\frac{v_2^2}{v_3^2}\right)\end{array}$

Here, v₂ is the velocity of water at point 2.

For the continuity equation at point 2 and 3 is expressed as,

$\begin{array}{c}{A}_2{v}_2=A_3{v}_3\\ \frac{v_2}{v_3}=\frac{A_3}{A_2}\end{array}$

Here A₂ and A₃ are the area at point 2 and 3.

This above equation put in gauge pressure equation.

$p_2=\frac{1}{2}\mathrm{ÒÏ}{v}_3^2\left(1−{\left(\frac{A_3}{A_2}\right)}^2\right)$

Substitute $v_3=\sqrt{2g\left(h_1−h_3\right)}$ in the above equation.

${\text{p}}_2=\frac{1}{2}×\mathrm{ÒÏ}×2g\left(h_1−h_3\right)×\left(1−{\left(\frac{A_3}{A_2}\right)}^2\right)$

Substitute 0.0160 m² for A₃ , 0.0480m² for A₂ , 9.81 m/s²for g, 1000kg/m³for $\mathrm{ÒÏ}$ , 10.0 m for h₁and 2.0 for h₃in the above equation.

$\begin{array}{c}{p}_2=\frac{1}{2}×\mathrm{ÒÏ}×2g\left(h_1−h_3\right)×\left(1−{\left(\frac{0.0160{\text{″¾}}^{\text{2}}}{0.0480{\text{″¾}}^{\text{2}}}\right)}^2\right)\\ =\frac{8}{9}×\mathrm{ÒÏ}×g\left(h_1−h_3\right)\\ =\frac{8}{9}×1000\text{ k²µ}/{\text{m}}^{\text{3}}×9.81\text{″¾}/{\text{s}}^{\text{2}}×\left(10.0\text{″¾}−2.0\text{″¾}\right)\\ =6.97×{10}^4\text{ P²¹}\end{array}$

Hence, the gauge pressure at point 2 is 6.97 x 10⁴ Pa .