Water flows steadily from an open tank as in Fig. P12.81. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 0.0480 m²; at point 3 it is 0.0160 m². The area of the tank is very large compared with the cross-sectional area of the pipe. Assuming that Bernoulli’s equation applies, compute (a) the discharge rate in cubic meters per second and (b) the gauge pressure at point 2.

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,

$\dot{Q}=A_3{v}_3$

Substitute the value of v₃ in the above equation.

$\begin{array}{l}\dot{Q}=A_3{v}_3\\ \dot{Q}=A_3\times \sqrt{2g\left(h_1-h_3\right)}\end{array}$

Substitute $0.0160{\text{\hspace{0.17em}m}}^{\text{2}}{\text{forA}}_3\text{,}9.81\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\text{forg}$, 10.0 m for h₁ and 2.0 m for h₃ in the above equation.

$\begin{array}{c}\dot{Q}=0.0160{\text{\hspace{0.17em}m}}^{\text{2}}\times \sqrt{2\times 9.81\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times \left(10.0\text{\hspace{0.17em}m-}2.0\text{\hspace{0.17em}m}\right)}\\ =0.0160{\text{\hspace{0.17em}m}}^{\text{2}}\times \sqrt{2\times 9.81\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times 8.0\text{\hspace{0.17em}m}}\\ =0.0160{\text{\hspace{0.17em}m}}^{\text{2}}\times 12.52\text{\hspace{0.17em}m}/\text{s}\\ =0.20{\text{\hspace{0.17em}m}}^{\text{3}}/\text{s}\end{array}$

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,

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

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.

$p_2=\frac{1}{2}\rho v_3^2\left(1-{\left(\frac{A_3}{A_2}\right)}^2\right)$

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

${\text{p}}_2=\frac{1}{2}\times \rho \times 2g\left(h_1-h_3\right)\times \left(1-{\left(\frac{A_3}{A_2}\right)}^2\right)$

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

$\begin{array}{c}{p}_2=\frac{1}{2}\times \rho \times 2g\left(h_1-h_3\right)\times \left(1-{\left(\frac{0.0160{\text{\hspace{0.17em}m}}^{\text{2}}}{0.0480{\text{\hspace{0.17em}m}}^{\text{2}}}\right)}^2\right)\\ =\frac{8}{9}\times \rho \times g\left(h_1-h_3\right)\\ =\frac{8}{9}\times 1000\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\times 9.81\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times \left(10.0\text{\hspace{0.17em}m}-2.0\text{\hspace{0.17em}m}\right)\\ =6.97\times {10}^4\text{\hspace{0.17em}Pa}\end{array}$

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