Question

water is flowing through a pipe with a velocity of 0.212m/s and 0.477m/s at section 1-1 (bottom end) and section 2-2(upper end) respectively as shown in figure-03. The pressure at section 1-1 is 350KN/m². Determine pressure at section 2-2, if the difference in elevation between section 1-1 and section 2-2 is 83.5ft.​

Answer 1

Velocity of water at lower end, $\sf{v_1=0.212\ m\,s^{-1}.}$

Velocity of water at upper end, $\sf{v_2=0.477\ m\,s^{-1}.}$

Pressure at lower end, $\sf{P_1=350\ kN\,m^{-2}=3.5\times10^5\ N\,m^{-2}.}$

Height difference between the ends, $\sf{\Delta h=83.5\ ft=25.45\ m.}$

By Bernoulli's Theorem,

$\sf{\longrightarrow P_1+\dfrac{1}{2}\,\rho(v_1)^2=P_2+\dfrac{1}{2}\,\rho(v_2)^2+\rho g\Delta h}$

where $\sf{\rho=10^3\ kg\,m^{-3}}$ is density of water and $\sf{g=10\ m\,s^{-2}}$ is acceleration due to gravity.

Then, pressure at upper end,

$\sf{\longrightarrow P_2=P_1+\dfrac{1}{2}\,\rho\left[(v_1)^2-(v_2)^2\right]-\rho g\Delta h}$

$\sf{\longrightarrow P_2=3.5\times10^5+\dfrac{1}{2}\times10^3\left[0.212^2-0.477^2\right]-10^3\times10\times25.45}$

$\sf{\longrightarrow\underline{\underline{P_2=95.4\ kN\,m^{-2}}}}$

Answer 2

GiVeN ;-

• Water is flowing through a pipe with a velocity of 0.212m/s at bottom .

• Water is flowing through a pipe with a velocity of 0.477m/s at upper .

• The pressure at bottom portion is 350kN/m² .

• Height difference between the two ends is 83.5 ft. .

To FiNd ;-

• The pressure at upper portion .

SoLuTiOn ;-

We know that,

➤ By Bernoulli's Theorem,

$\\ \huge\star$ $\bf\green{\triangle{P}\:=\:\dfrac{\rho\:(v^2_l\:-\:v^2_u)}{2}\:-\:\rho\:g\:\triangle{h}\:}$

$\bf{\red{\longrightarrow}}\:P_u\:-\:P_l\:=\:\dfrac{\rho\:(v^2_l\:-\:v^2_u)}{2}\:-\:\rho\:g\:\triangle{h}\:$

Where,

• $\bf{P_l}$ = 350 kN/m² = 35 × 10⁴ N/m²

• $\bf{v_l}$ = 0.212 m/s

• $\bf{v_u}$ = 0.477 m/s

• $\bf{\rho}$ = 10³ kg/m³ [density of water ]

• $\bf{g}$ = 10 m/s²

• $\bf{\triangle{h}}$ = 83.5 ft. = 25.45 m

$\bf{\red{\longrightarrow}}\:P_u\:=\:P_l\:+\:\dfrac{\rho\:(v^2_l\:-\:v^2_u)}{2}\:-\:\rho\:g\:\triangle{h}\:$

$\bf{\red{\longrightarrow}}\:P_u\:=\:35\times{10^4}\:+\:\dfrac{10^3\:\Big((0.212)^2\:-\:(0.477)^2\Big)}{2}\:-\:\Big(10^3\times{10}\times{25.45}\Big)\:$

$\bf{\red{\longrightarrow}}\:P_u\:=\:95.4\:kN.m^{-2}\:$

∴ The pressure at upper portion is '95.4 kN/m²' .